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Page No 3.101:
Question 1:
Point A and B are 70 km. a part on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if the travel towards each other, the meet in one hour. Find the speed of the two cars.
Answer:
We have to find the speed of car
Let and be two cars starting from pointsand respectively. Let the speed of car be x km/hr and that of carbe y km/hr.
Case I: When two cars move in the same directions:
Suppose two cars meet at point, Then,
Distance travelled by car
Distance travelled by car
It is given that two cars meet in 7 hours.
Therefore, Distance travelled by car X in hours = km
Distance traveled by car y in 7 hours = km
Clearly
Dividing both sides by common factor 7 we get,
Case II : When two cars move in opposite direction
Suppose two cars meet at point. Then,
Distance travelled by car,
Distance travelled by car.
In this case, two cars meet in 1 hour
Therefore Distance travelled by car X in1 hour km
Distance travelled by car Y in 1 hour km
From the above clearly,
...(ii)
By solving equation (i) and (ii), we get
Substituting in equation (ii) we get
Hence, the speed of car starting from point A is
The speed of car starting from point B is.
Page No 3.101:
Question 2:
A sailor goes 8 km downstream in 40 minutes and returns in 1 hours. Determine the speed of the sailor in still water and the speed of the current.
Answer:
Let the speed of the sailor in still water be x km/hr and the speed of the current be y km/hr
Speed upstream
Speed downstream
Now, Time taken to cover 8km down stream =
Time taken to cover 8km upstream=
But, time taken to cover 8 km downstream in 40 minutes or that is
Dividing both sides by common factor 2 we get
Time taken to cover 8km upstream in1hour
...(ii)
By solving these equation and we get
Substitute in equationwe get
Hence, the speed of sailor is
The speed of current is
Page No 3.101:
Question 3:
The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water.
Answer:
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr
Speed upstream =km/hr
Speed down stream =km/hr
Now,
Time taken to cover 30 km upstream = hrs
Time taken to cover 44 km down stream = hrs
But total time of journey is 10 hours
Time taken to cover 40 km upstream=
Time taken to cover 55 km down stream =
In this case total time of journey is given to be 13 hours
Therefore, ...(ii)
Putting = u and in equation and we get
Solving these equations by cross multiplication we get
and
Now,
By solving equation and we get ,
Substituting in equation we get ,
Hence, speed of the boat in still water is
Speed of the stream is
Page No 3.102:
Question 4:
A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in $6\frac{1}{2}\mathrm{hrs}$. Find the speed of the boat in still water and also speed of the stream.
Answer:
We have to find the speed of the boat in still water and speed of the stream
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr then
Speed upstream
Sped down stream
Now, Time taken to cover km down stream =
Time taken to cover km upstream =
But, total time of journey is 6 hours
Time taken to cover km upstream =
Time taken to cover km down stream =
In this case total time of journey is given to or
...(ii)
By and in equation (i) and (ii) we get
Solving these equations by cross multiplication we get
and
and
Now,
and
By solving equation and we get,
By substituting in equation we get
Hence, the speed of the stream is
The speed of boat is
Page No 3.102:
Question 5:
A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.
Answer:
Let the actual speed of the train be x Km/hr and the actual time taken by y hours. Then,
Distance covered=
If the speed is increased by, then time of journey is reduced by 1 hour i.e., when speed is , time of journey is
Distance covered= km
...(ii)
When the speed is reduced by, then the time of journey is increased by i.e., when speed is, time of journey is
Distance covered =
Thus we obtain the following equations
By using elimination method, we have
Putting the value in equation (iii) we get
Putting the value of x and y in equations (i) we get
Distance covered =
=
Hence, the distance is,
The speed of walking is .
Page No 3.102:
Question 6:
A person rowing at the rate of 5 km/h in still water , takes thrice as much time in going 40 km upstream as in going 40 km downstream . Find the speed of the stream .
Answer:
Speed of the boat in still water = 5 km/h
Let the speed of stream = x km/h
∴ Speed of boat upstream = (5 − x) km/h
Speed of boat downstream = (5 + x) km/h
Time taken to row 40 km upstream = $\frac{40}{5-x}$
Time taken to row 40 km downstream = $\frac{40}{5+x}$
According to the given condition,
$\frac{40}{5-x}=3\left(\frac{{\displaystyle 40}}{{\displaystyle 5+x}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{5-x}=\frac{3}{5+x}\phantom{\rule{0ex}{0ex}}\Rightarrow 5+x=15-3x\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=10\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{10}{4}=2.5\mathrm{km}/\mathrm{h}$
Therefore, the speed of the stream is 2.5 km/h.
Page No 3.102:
Question 7:
Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours if he travels 160 km. by train and the rest by car. He takes 12 minutes more if the travels 240 km by train and the rest by car. Find the speed of the train and car respectively.
Answer:
Let the speed of the train be x km/hour that of the car be y km/hr, we have the following cases
Case I: When Ramesh travels 760 Km by train and the rest by car
Time taken by Ramesh to travel 160 Km by train =
Time taken by Ramesh to travel (760-160) =600 Km by car =
Total time taken by Ramesh to cover 760Km = +
It is given that total time taken in 8 hours
Case II: When Ramesh travels 240Km by train and the rest by car
Time taken by Ramesh to travel 240 Km by train =
Time taken by Ramesh to travel (760-240) =520Km by car =
In this case total time of the journey is 8 hours 12 minutes
...(ii)
Putting and, , the equations and reduces to
Multiplying equation (iii) by 6 and (iv)by 20 the above system of equation becomes
Subtracting equation from we get
Putting in equation, we get
Now
and
Hence, the speed of the train is ,
The speed of the car is .
Page No 3.102:
Question 8:
A man travels 600 km partly by train and partly by car. If the covers 400 km by train and the rest by car, it takes him 6 hours 30 minutes. But, if the travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.
Answer:
Let the speed of the train be x km/hr that of the car be y km/hr, we have the following cases:
Case I: When a man travels 600Km by train and the rest by car
Time taken by a man to travel 400 Km by train =
Time taken by a man to travel (600-400) =200Km by car =hrs
Total time taken by a man to cover 600Km =
It is given that total time taken in 8 hours
Case II: When a man travels 200Km by train and the rest by car
Time taken by a man to travel 200 Km by train =
Time taken by a man to travel (600-200) = 400 Km by car
In this case, total time of the journey in 6 hours 30 minutes + 30 minutes that is 7 hours,
...(ii)
Putting and, , the equations and reduces to
Multiplying equation (iii) by 6 the above system of equation becomes
Substituting equation and, we get
Putting in equation, we get
Now
and
Hence, the speed of the train is,
The speed of the car is.
Page No 3.102:
Question 9:
Places A and B are 80 km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find the speeds of the cars.
Answer:
Let x and y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of the car Y be y km/hr.
Case I: When two cars move in the same directions:
Suppose two cars meet at point Q, then,
Distance travelled by car X = AQ
Distance travelled by car Y = BQ
It is given that two cars meet in 8 hours.
Distance travelled by car X in 8 hours =km
AQ=
Distance travelled by car Y in 8 hours =km
BQ =
Clearly AQ-BQ = AB
Both sides divided by 8, we get
Case II: When two cars move in opposite direction
Suppose two cars meet at point P, then,
Distance travelled by X car X=AP
Distance travelled by Y car Y=BP
In this case, two cars meet in 1 hour 20 minutes, we can write it as 1 hour or
hours that is hours.
Therefore,
Distance travelled by car y in hours = km
Distance travelled by car y in hours = km
...(ii)
By solving (i) and (ii) we get,
By substituting in equation (ii), we get
Hence, speed of car X is_{ } , speed of car Y is.
Page No 3.102:
Question 10:
A boat goes 12 km upstream and 40 km downstream in 8 hours. I can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
Answer:
We have to find the speed of the boat in still water and speed of the stream
Let the speed of the boat in still water be km/hr and the speed of the stream be km/hr then
Speed upstream
Sped down stream
Now, Time taken to cover km upstream =
Time taken to cover km down stream =
But, total time of journey is 8 hours
Time taken to cover km upstream =
Time taken to cover km down stream =
In this case total time of journey is given to
...(ii)
By and in equation (i) and (ii) we get
Solving these equations by cross multiplication we get
and
and
Now,
and
By solving equation and we get,
By substituting in equation we get
Hence, the speed of boat in still water is ,
The speed of the stream is.
Page No 3.102:
Question 11:
Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Answer:
Let the speed of the train be x km/hour that of the bus be y km/hr, we have the following cases
Case I: When Roohi travels 300 Km by train and the rest by bus
Time taken by Roohi to travel 60 Km by train =
Time taken by Roohi to travel (300-60) =240 Km by bus =
Total time taken by Roohi to cover 300Km= +
It is given that total time taken in 4 hours
Case II: When Roohi travels 100 km by train and the rest by bus
Time taken by Roohi to travel 100 Km by train =
Time taken by Roohi to travel (300-100) =200Km by bus =
In this case total time of the journey is 4 hours 10 minutes
...(i)
Putting and, , the equations and reduces to
Subtracting equation (iv) from (iii)we get
Putting in equation (iii), we get
Now
and
Hence, the speed of the train is,
The speed of the bus is.
Page No 3.102:
Question 12:
Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Answer:
Let the speed of rowing in still water be x km/hr and the speed of the current be y km/hr
Speed upstream
Speed downstream
Now,
Time taken to cover km down stream =
Time taken to cover km upstream =
But, time taken to cover km downstream in
Time taken to cover km upstream in 2 hours
...(i)
By solving these equation (i) and (ii) we get
Substitute in equation (i)we get
Hence, the speed of rowing in still water is,
The speed of current is .4 km/ hr
Page No 3.102:
Question 13:
A motor boat can travel 30 km upstream and 28 km downstream in 7 hours . It can travel 21 km upstream and return in 5 hours . Find the speed of the boat in still water and the speed of the upstream .
Answer:
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
Speed of boat upstream = x − y
Speed of boat downstream = x + y
It is given that, the boat travels 30 km upstream and 28 km downstream in 7 hours.
$\therefore \frac{30}{x-y}+\frac{28}{x+y}=7$
Also, the boat travels 21 km upstream and return in 5 hours.
$\therefore \frac{21}{x-y}+\frac{21}{x+y}=5$
Let $\frac{1}{x-y}=u\mathrm{and}\frac{1}{x+y}=v$.
So, the equation becomes
$30u+28v=7.....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}21u+21v=5.....\left(\mathrm{ii}\right)$
Multiplying (i) by 21 and (ii) by 30, we get
$630u+588v=147...\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}630u+630v=150...\left(\mathrm{iv}\right)$
Solving (iii) and (iv), we get
$v=\frac{1}{14}$ and $u=\frac{1}{6}$
But, $\frac{1}{x-y}=u\mathrm{and}\frac{1}{x+y}=v$
So,
$\frac{1}{x-y}=\frac{1}{6}\mathrm{and}\frac{1}{x+y}=\frac{1}{14}\phantom{\rule{0ex}{0ex}}\Rightarrow x-y=6\mathrm{and}x+y=14$
Solving these two equations, we get
x = 10 and y = 4
So, the speed of boat in still water = 10 km/h and speed of stream = 4 km/h.
Page No 3.102:
Question 14:
Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi he takes 6 minutes longer. Find the speed of the train and that of the taxi.
Answer:
Let the speed of the train be x km/hour that of the taxi be y km/hr, we have the following cases
Case I: When Abdul travels 300 Km by train and the 200 Km by taxi
Time taken by Abdul to travel 300 Km by train =
Time taken by Abdul to travel 200 Km by taxi =
Total time taken by Abdul to cover 500 Km =
It is given that total time taken in 5 hours 30 minutes
Case II: When Abdul travels 260 Km by train and the 240 km by taxi
Time taken by Abdul to travel 260 Km by train =
Time taken by Abdul to travel 240 Km by taxi =
In this case total time of the journey is 5 hours 36 minutes
Putting and, , the equations and reduces to
Multiplying equation by 6 the above system of equation becomes
Subtracting equation from we get
Putting in equation, we get
Now
and
Hence, the speed of the train is ,
The speed of the taxi is .
Page No 3.102:
Question 15:
A train covered a certain distance at a uniform speed. If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/hr; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Answer:
Let the actual speed of the train be and the actual time taken by y hours. Then,
Distance covered=
If the speed is increased by, then time of journey is reduced by 2 hours
when speed is , time of journey is
Distance covered=
When the speed is reduced by, then the time of journey is increased by when speed is, time of journey is
Distance covered=
Thus, we obtain the following system of equations:
By using cross multiplication, we have
Putting the values of x and y in equation (i), we obtain
Distance=
Hence, the length of the journey is .
Page No 3.102:
Question 16:
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of two cars.
Answer:
Let x and y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of the car Y be y km/hr.
Case I: When two cars move in the same directions:
Suppose two cars meet at point Q, then,
Distance travelled by car X=AQ
Distance travelled by car Y=BQ
It is given that two cars meet in 5 hours.
Distance travelled by car X in 5 hours =km AQ=
Distance travelled by car Y in 5 hours =km BQ=
Clearly AQ-BQ = AB
Both sides divided by 5, we get
Case II: When two cars move in opposite direction
Suppose two cars meet at point P, then,
Distance travelled by X car X=AP
Distance travelled by Y car Y=BP
In this case, two cars meet in 1 hour
Therefore,
Distance travelled by car y in1 hours = km
Distance travelled by car y in 1 hours = km
By solving (i) and (ii) we get,
By substituting in equation (ii), we get
Hence, speed of car X is , speed of car Y is.
Page No 3.103:
Question 17:
While covering a distance of 30 km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit. Find their speeds of walking.
Answer:
Let the speed of Ajeet and Amit be x Km/hr respectively. Then,
Time taken by Ajeet to cover
Time taken by Amit to cover
By the given conditions, we have
If Ajeet doubles his speed, then speed of Ajeet is
Time taken by Ajeet to cover
Time taken by Amit to cover
According to the given condition, we have
...(ii)
Putting and, in equation (i) and (ii), we get
Adding equations (iii) and (iv), we get
Putting in equation (iii), we get
Now,
and
Hence, the speed of Ajeet is
The speed of Amit is
Page No 3.103:
Question 18:
A takes 3 hours more than B to walk a distance of 30 km. But, if A doubles his pace (speed) he is ahead of B by $1\frac{1}{2}$ hours. Find the speeds of A and B.
Answer:
Let the speed of A and B be x Km/hr and y Km/hr respectively. Then,
Time taken by A to cover ,
And, Time taken by B to cover .
By the given conditions, we have
If A doubles his pace, then speed of A is
Time taken by A to cover ,
Time taken by B to cover .
According to the given condition, we have
Putting and, in equation (i) and (ii), we get
Adding equations (iii) and (iv), we get,
Putting in equation (iii), we get
Now,
and,
Hence, the A’s speed is,
The B’s speed is.
Page No 3.111:
Question 1:
If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.
Answer:
Let the length and breadth of the rectangle be and units respectively
Then, area of rectangle square units
If length is increased and breadth reduced each by units, then the area is reduced by square units
$\left(x+2\right)\left(y-2\right)=xy-28\phantom{\rule{0ex}{0ex}}\Rightarrow xy-2x+2y-4=xy-28\phantom{\rule{0ex}{0ex}}\Rightarrow -2x+2y-4+28=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2x+2y+24=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-2y-24=0$
Therefore,
Then the length is reduced by unit and breadth is increased by units then the area is increased by square units
$\left(x-1\right)\left(y+2\right)=xy+33\phantom{\rule{0ex}{0ex}}\Rightarrow xy+2x-y-2=xy+33\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-y-2-33=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-y-35=0$
Therefore, $2x-y-35=0.....\left(ii\right)$
Thus we get the following system of linear equation
$2x-2y-24=0\phantom{\rule{0ex}{0ex}}2x-y-35=0$
By using cross multiplication, we have
and
The length of rectangle is units.
The breadth of rectangle is units.
Area of rectangle =lengthbreadth,
square units
Hence, the area of rectangle is square units
Page No 3.111:
Question 2:
The area of a rectangle remains the same if the length is increased by 7 meters and the breadth is decreased by 3 meters. The area remains unaffected if the length is decreased by 7 meters and breadth in increased by 5 meters. Find the dimensions of the rectangle.
Answer:
Let the length and breadth of the rectangle be and units respectively
Then, area of rectangle =square units
If length is increased by meters and breadth is decreased by meters when the area of a rectangle remains the same
Therefore,
If the length is decreased by meters and breadth is increased by meters when the area remains unaffected, then
Thus we get the following system of linear equation
By using cross-multiplication, we have
and
Hence, the length of rectangle ismeters
The breadth of rectangle is meters
Page No 3.111:
Question 3:
In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 meters, the area of the rectangle is reduced by 67 square meters. If length is reduced by 1 meter and breadth is increased by 4 meters, the area is increased by 89 Sq. meters. Find the dimensions of the rectangle.
Answer:
Let the length and breadth of the rectangle be and units respectively
Then, area of rectangle = square units
If the length is increased by meters and breath is reduced each by square meters the area is reduced by square units
Therefore,
Then the length is reduced by meter and breadth is increased by meter then the area is increased by square units
Therefore,
Thus, we get the following system of linear equation
By using cross multiplication we have
and
Hence, the length of rectangle ismeter,
The breath of rectangle is meter.
Page No 3.111:
Question 4:
The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs 1250, find their incomes.
Answer:
Let the income of be Rs and the income of be Rs.further let the expenditure of be and the expenditure of be respectively then,
Saving of =
Saving of =
Solving equation and by cross- multiplication, we have
The monthly income of =
The monthly income of
Hence the monthly income of is Rs
The monthly income of is Rs
Page No 3.111:
Question 5:
A and B each has some money. If A gives Rs 30 to B, then B will have twice the money left with A. But, if B gives Rs 10 to A, then A will have thrice as much as is left with B. How much money does each have?
Answer:
Let the money with A be Rs x and the money with B be Rs y.
If A gives Rs 30 to B, Then B will have twice the money left with A, According to the condition we have,
If B gives Rs 10 to A, then A will have thrice as much as is left with B,
By multiplying equation with 2 we get,
By subtracting from we get,
By substituting in equation we get
Hence the money with A be and the money with B be
Page No 3.111:
Question 6:
ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y − 5)°, ∠C = (−4x)° and ∠D = (7x + 5)°. Find the four angles.
Answer:
We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral angles and and angles and pairs of opposite angles
Therefore
and
By substituting and we get
Divide both sides of equation by 4 we get
By substituting and we get
By multiplying equation by 3 we get
By subtracting equation from we get
By substituting in equation we get
The angles of a cyclic quadrilateral are
Hence the angles of quadrilateral are
Page No 3.111:
Question 7:
2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it?
Answer:
A man can alone finish the work in days and one boy alone can finish it in days then
One mans one days work =
One boys one days work=
2men one day work=
7boys one day work=
Since 2 men and 7 boys can finish the work in 4 days
Again 4 men and 4 boys can finish the work in 3 days
Putting and in equation and we get
By using cross multiplication we have
Now,
and
Hence, one man alone can finish the work in and one boy alone can finish the work in _{.}
Page No 3.111:
Question 8:
In a ∆ABC, ∠A = x°, ∠B = (3x − 2)°, ∠C = y°. Also, ∠C − ∠B = 9°. Find the three angles.
Answer:
Let , and
$\angle C-\angle B={9}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C={9}^{\xb0}+\angle B\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C={9}^{\xb0}+3{x}^{\xb0}-{2}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C={7}^{\xb0}+3{x}^{\xb0}$
Substitute in above equation we get ,
${y}^{\xb0}={7}^{\xb0}+3{x}^{\xb0}$
$\angle A+\angle B+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{\xb0}+(3{x}^{\xb0}-{2}^{\xb0})+({7}^{\xb0}+3{x}^{\xb0})={180}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow 7{x}^{\xb0}+{5}^{\xb0}={180}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow 7{x}^{\xb0}={180}^{\xb0}-{5}^{\xb0}={175}^{\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{\circ}=\frac{{175}^{\circ}}{{7}^{\circ}}={25}^{\circ}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle A={x}^{\circ}={25}^{\circ}\phantom{\rule{0ex}{0ex}}\angle B=(3x-2{)}^{\circ}=3({25}^{\circ})-{2}^{\circ}={73}^{\circ}\phantom{\rule{0ex}{0ex}}\angle C=({7}^{\xb0}+3{x}^{\xb0})={7}^{\circ}+3(25{)}^{\circ}={82}^{\circ}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle A={25}^{\circ},\angle B={73}^{\circ},\angle C={82}^{\circ}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{answer}.$
Page No 3.111:
Question 9:
In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x − 5)°. Find the four angles.
Answer:
We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral, angles and and angles and pairs of opposite angles
Therefore and
Taking
By substituting and we get
Taking
By substituting and we get
By multiplying equation by we get
By subtracting equation from we get
By substituting in equation we get
The angles of a cyclic quadrilateral are
Hence, the angles of cyclic quadrilateral ABCD are .
Page No 3.111:
Question 10:
Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awanded for each correct answer and 2 marks been deducted for each incorrect answer, the Yash would have scored 50 marks. How many question were there in the test?
Answer:
Let take right answer will beand wrong answer will be .
Hence total number of questions will be
If yash scored marks in atleast getting marks for each right answer and losing mark for each wrong answer then
If 4 marks awarded for each right answer and 2 marks deduced for each wrong answer the he scored 50 marks
By multiplying equation by 2 we get
By subtracting from we get
Putting in equation we have
Total number question will be
Hence, the total number of question is .
Page No 3.111:
Question 11:
In a ∆ABC, ∠A = x°, ∠B = 3x° and ∠C = y°. If 3y − 5x = 30, prove that the triangle is right angled.
Answer:
We have to prove that the triangle is right
Given and
Sum of three angles in triangle are
By solving with we get,
Multiplying equation by 3 we get
Subtracting equation from
Substituting in equation we get
Angles and are
A right angled triangle is a triangle in which one side should has a right angle that is in it.
Hence, The triangle ABC is right angled
Page No 3.111:
Question 12:
The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and the journey of 20 km, the charge paid is Rs 145. What will a person have to pay for travelling a distance of 30 km?
Answer:
Let the fixed charges of car be per km and the running charges be km/hr
According to the given condition we have
Putting in equation we get
Therefore, Total charges for travelling distance of km
= Rs
Hence, A person have to pay for travelling a distance of km.
Page No 3.112:
Question 13:
A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes foods for 20 days, he has to pay Rs 1000 as hostel charges whereas a students B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.
Answer:
Let the fixed charges of hostel be and the cost of food charges be per day
According to the given condition we have,
Subtracting equation from equation we get
Putting in equation we get
Hence, the fixed charges of hostel is _{.}
The cost of food per day is _{.}
Page No 3.112:
Question 14:
Half the perimeter of a garden, whose length is 4 more than its width is 36 m. Find the dimension of the garden.
Answer:
Let perimeter of rectangular garden will be .if half the perimeter of a garden will be
When the length is four more than its width then
Substituting in equation we get
Putting in equation we get
Hence, the dimensions of rectangular garden are and
Page No 3.112:
Question 15:
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer:
We know that the sum of supplementary angles will be.
Let the longer supplementary angles will be.
Then,
If larger of supplementary angles exceeds the smaller by degree, According to the given condition. We have,
Substitute in equation, we get,
Put equation, we get,
Hence, the larger supplementary angle is ,
The smaller supplementary angle is.
Page No 3.112:
Question 16:
2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery, and that taken by 1 man alone.
Answer:
1 women alone can finish the work in days and 1 man alone can finish it in days .then
One woman one day work=
One man one days work =
2 women’s one days work=
5 man’s one days work =
Since 2 women and 5 men can finish the work in 4 days
3 women and 6 men can finish the work in 3 days
Putting and in equation and we get
By using cross multiplication we have
Now ,
Hence, the time taken by 1 woman alone to finish the embroidery is,
The time taken by 1 man alone to finish the embroidery is .
Page No 3.112:
Question 17:
Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes Rs 50 and Rs 100 she received.
Answer:
Let be the notes of and notes will be
If Meena ask for and notes only, then the equation will be,
Divide both sides by then we get,
If Meena got 25 notes in all then the equation will be,
By subtracting the equation from we get,
Substituting in equation, we get
Therefore and
Hence, Meena has notes of and notes of
Page No 3.112:
Question 18:
There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.
Answer:
Let us take the A examination room will be x and the B examination room will be y
If 10 candidates are sent from A to B, the number of students in each room is same. According to the above condition equation will be
If 20 candidates are sent from B to A, the number of students in A is double the number of students in B, then equation will be,
By subtracting the equationfromwe get,
Substituting in equation, we get
Hence candidates are in A examination Room,
candidates are in B examination Room.
Page No 3.112:
Question 19:
A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs 216 and one full and one half reserved first class tickets cost Rs 327. What is the basic first class full fare and what is the reservation charge?
Answer:
Let take first class full of fare is Rs and reservation charge is Rs per ticket
Then half of the ticket as on full ticket =
According to the given condition we have
Multiplying equation by 2 we have
Subtracting from we get
Putting in equation we get
Hence, the basic first class full fare is
The reservation charge is .
Page No 3.112:
Question 20:
A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that'. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the the stake of money each of the cock-owners have.
Answer:
Let the strike money of first cock-owner be and of second cock-owner be respectively. Then we have,
For second cock-owner according to given condition we have,
By subtracting from, we have,
Putting in equation we get,
Hence the stake of money first cock-owner is and of second cock-owner is respectively.
Page No 3.112:
Question 21:
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of student in the class.
Answer:
Let the number of students be and the number of row be .then,
Number of students in each row
Where three students is extra in each row, there are one row less that is when each row has students the number of rows is
Total number of students =no. of rowsno. of students in each row
If three students are less in each row then there are rows more that is when each row has
Therefore, total number of students=Number of rowsNumber of students in each row
Putting in and equation we get
Adding and equation we get
Putting in equation we get
Hence, the number of students in the class is.
Page No 3.112:
Question 22:
One says. "give me hundred, friend! I shall then become twice as rich as you" The other replies, "If you give me ten, I shall be six times as rich as you". Tell me what is the amount of their respective capital?
Answer:
21. Let the money with first person be and the money with the second person be. Then,
If first person gives to second person then the second person will become twice as rich as first person, According to the given condition, we have,
if second person gives to first person then the first person will becomes six times as rich as second person, According to given condition, we have,
Multiplying equation by we get,
By subtracting from, we get
Putting in equation, we get,
Hence, first person’s capital will be ,
Second person’s capital will be .
Page No 3.112:
Question 23:
A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of ₹1008. If she had sold the saree at 10% profit and the sweater at 8% discount , she would have got ₹ 1028 . Find the cost price of the saree and the list price (price before discount) of the sweater.
Answer:
Let the CP of saree be ₹x and the list price of sweater be ₹y.
Case I: When saree is sold at 8% profit and sweater at 10% discount.
SP = CP + Profit
⇒ SP of saree = x + $\frac{8}{100}x$ = $\frac{108}{100}x$
SP of sweater = List price − Discount
⇒ SP of sweater = $y-\frac{10}{100}y$ = $\frac{90}{100}y$
Total sum received by the shopkeeper = ₹1008
⇒ $\frac{108}{100}x$ + $\frac{90}{100}y$ = 1008
⇒ 108x + 90y = 100800
⇒ 6x + 5y = 5600 .....(i)
Case II: When saree is sold at 10% profit and sweater at 8% discount.
⇒ SP of saree = x + $\frac{10}{100}x$ = $\frac{110}{100}x$
SP of sweater = List price − Discount
⇒ SP of sweater = $y-\frac{8}{100}y$ = $\frac{92}{100}y$
Total sum received by the shopkeeper = ₹1028
⇒ $\frac{110}{100}x$ + $\frac{92}{100}y$ = 1028
⇒ 110x + 92y = 102800 .....(ii)
Multiplying (i) by 110 and (ii) by 6, we get
660x + 550y = 616000 .....(iii)
660x + 552y = 616800 .....(iv)
Subtracting (iii) from (iv), we get
2y = 800
⇒ y = 400
Putting y = 400 in (i), we get
6x + 2000 = 5600
⇒ 6x = 5600 − 2000 = 3600
⇒ x = 600
Hence, CP of saree = ₹600 and list price of sweater = ₹400.
Page No 3.112:
Question 24:
In a competitive examination , one mark is awarded for each correct answer while 1/2 mark is deducted for everey wrong answer . Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly.
Answer:
Let the number of correct answers be x and the number of wrong answers be y.
Total questions Jayanthi answered = 120
So, x + y = 120 .....(i)
Now, marks obtained for answering correctly = 1 × x = x
Marks deducted for answering incorrectly = $\frac{1}{2}\times y$ = $\frac{y}{2}$
Total marks obtained = $x-\frac{y}{2}=90$ .....(ii)
Subtracting (ii) from (i), we get
$y+\frac{y}{2}=30\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3y}{2}=30\phantom{\rule{0ex}{0ex}}\Rightarrow y=20$
Putting y = 20 in (i), we get
x + 20 = 120
⇒ x = 100
Thus, Jayanti answered 100 questions correctly.
Page No 3.113:
Question 25:
A shopkeeper gives books on rent for reading . She takes a fixed charge for the first two days, and an additional charge for each day thereafter . Latika paid ₹22 for a book kept for 6 days, while Anand paid ₹16 for the book kept for four days . Find the fixed charges and charge for each extraday.
Answer:
Let the fixed charge for first two days be ₹x and the additional charge for each day extra be ₹y.
It is given that Latika kept the book for 6 days and paid ₹22.
So,
Fixed charge for the first 2 days + Additional charge for 4 days = ₹22
∴ x + 4y = 22 .....(i)
Anand kept the book for 4 days and paid ₹16. So,
Fixed charge for the first 2 days + Additional charge for 2 days = ₹16
∴ x + 2y = 16 .....(ii)
Subtracting (ii) from (i), we get
2y = 6
⇒ y = 3
Putting y = 3 in (i), we get
x + 12 = 22
⇒ x = 10
Thus, the fixed charge is ₹10 and the additional charge for each extraday is ₹3.
Page No 3.113:
Question 1:
The value of k for which the system of equations has a unique solution, is
kx − y = 2
6x − 2y = 3
(a) =3
(b) ≠3
(c) ≠0
(d) =0
Answer:
The given system of equations are
for unique solution
Here
By cross multiply we get
Hence, the correct choice is.
Page No 3.113:
Question 2:
The value of k for which the system, of equations has infinite number of solutions, is
2x + 3y = 5
4x + ky = 10
(a) 1
(b) 3
(c) 6
(d) 0
Answer:
The given system of equations are
For the equations to have infinite number of solutions,
Here,
Therefore
By cross multiplication of we get,
And
Therefore the value of k is 6
Hence, the correct choice is .
Page No 3.113:
Question 3:
The value of k for which the system of equations x + 2y − 3 = 0 and 5x + ky + 7 = 0 has no solution, is
(a) 10
(b) 6
(c) 3
(d) 1
Answer:
The given system of equations are
For the equations to have no solutions,
If we take
Therefore the value of k is10.
Hence, correct choice is.
Page No 3.114:
Question 4:
The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has non-zero solution, is
(a) 0
(b) 2
(c) 6
(d) 8
Answer:
The given system of equations are,
Here,
By cross multiply we get
Therefore the value of k is 6,
Hence, the correct choice is.
Page No 3.114:
Question 5:
If the system of equations has infinitely many solutions, then
2x + 3y = 7
(a + b)x + (2a − b)y = 21
(a) a = 1, b = 5
(b) a = 5, b = 1
(c) a = −1, b = 5
(d) a = 5, b = −1
Answer:
The given systems of equations are
For the equations to have infinite number of solutions,
Here ,
Let us take
By cross multiplication we get,
Now take
By cross multiplication we get,
Substitute in the above equation
Substitute in equation we get,
Therefore and.
Hence, the correct choice is.
Page No 3.114:
Question 6:
If the system of equations is inconsistent, then k =
$3x+y=1\phantom{\rule{0ex}{0ex}}\left(2k-1\right)x+\left(k-1\right)y=2k+1$
(a) 1
(b) 0
(c) −1
(d) 2
Answer:
The given system of equations is inconsistent,
If the system of equations is in consistent, we have
Therefore, the value of k is2.
Hence, the correct choice is .
Page No 3.114:
Question 7:
If am ≠ bl, then the system of equations
$ax+by=c\phantom{\rule{0ex}{0ex}}lx+my=n$
(a) has a unique solution
(b) has no solution
(c) has infinitely many solutions
(d) may or may not have a solution
Answer:
Given the system of equations has
We know that intersecting lines have unique solution
Here
Therefore intersecting lines, have unique solution
Hence, the correct choice is
Page No 3.114:
Question 8:
If the system of equations has infinitely many solutions, then
$2x+3y=7\phantom{\rule{0ex}{0ex}}2ax+\left(a+b\right)y=28$
(a) a = 2b
(b) b = 2a
(c) a + 2b = 0
(d) 2a + b = 0
Answer:
Given the system of equations are
For the equations to have infinite number of solutions,
By cross multiplication we have
Divide both sides by 2. we get
Hence, the correct choice is .
Page No 3.114:
Question 9:
The value of k for which the system of equations has no solution is
$x+2y=5\phantom{\rule{0ex}{0ex}}3x+ky+15=0$
(a) 6
(b) −6
(c) 3/2
(d) None of these
Answer:
The given system of equation is
If then the equation have no solution.
By cross multiply we get
Hence, the correct choice is.
Page No 3.114:
Question 10:
If 2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b represent coincident lines, then a and b satisfy the equation
(a) a + 5b = 0
(b) 5a + b = 0
(c) a − 5b = 0
(d) 5a − b = 0
Answer:
The given system of equations are
For coincident lines , infinite number of solution
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{b2}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{\left(a+b\right)}=\frac{-3}{-\left(a+b-3\right)}=\frac{7}{4a+b}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(a+b-3\right)=3\left(a+b\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+2b-6=3a+3b\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+2b-3a-3b=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow -a-b=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow a+b=-6---\left(i\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3\left(4a+b\right)=7\left(a+b-3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 12a+3b=7a+7b-21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 5a-4b=-21---\left(ii\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{multiply}\mathrm{equation}\left(i\right)\mathrm{by}5,\mathrm{we}\mathrm{get}5a+5b=-30---\left(iii\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{subtract}\left(ii\right)\mathrm{from}\left(iii\right),\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5a+5b\right)-\left(5a-4b\right)=-30+21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 5a+5b-5a+4b=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 9b=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow b=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{substitute}b=-1\mathrm{in}\mathrm{equation}\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}a+\left(-1\right)=-6\phantom{\rule{0ex}{0ex}}\Rightarrow a=-6+1=-5$
Option A.:
Option B:
$5a+b=0\phantom{\rule{0ex}{0ex}}5\left(-5\right)+\left(-1\right)=-25-1=-26\ne 0$
Option.C:
a - b = 0
-5 - (-1) = -4 $\ne $0
None of the option satisfies the values.
Page No 3.114:
Question 11:
If a pair of linear equations in two variables is consistent, then the lines represented by two equations are
(a) intersecting
(b) parallel
(c) always coincident
(d) intersecting or coincident
Answer:
If a pair of linear equations in two variables is consistent, then its solution exists.
∴The lines represented by the equations are either intersecting or coincident.
Hence, correct choice is.
Page No 3.114:
Question 12:
The area of the triangle formed by the line $\frac{x}{a}+\frac{y}{b}=1$ with the coordinate axes is
(a) ab
(b) 2ab
(c) $\frac{1}{2}ab$
(d) $\frac{1}{4}ab$
Answer:
Given the area of the triangle formed by the line
If in the equation either A and B approaches infinity, The line become parallel to either x axis or y axis respectively,
Therefore
Area of triangle
Hence, the correct choice is .
Page No 3.114:
Question 13:
The area of the triangle formed by the lines y = x, x = 6 and y = 0 is
(a) 36 sq. units
(b) 18 sq. units
(c) 9 sq. units
(d) 72 sq. units
Answer:
Given and
We have plotting points as when
Therefore, area of
Area of triangle is square units
Hence, the correct choice is _{.}
Page No 3.114:
Question 14:
If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =
(a) 1
(b) ½
(c) 3
(d) 6
Answer:
The given system of equations
For the equations to have infinite number of solutions
If we take
And
Therefore, the value of k is 6.
Hence, the correct choice is .
Page No 3.115:
Question 15:
If the system of equations kx − 5y = 2, 6x + 2y = 7 has no solution, then k =
(a) −10
(b) −5
(c) −6
(d) −15
Answer:
The given systems of equations are
If
Here
Hence, the correct choice is .
Page No 3.115:
Question 16:
The area of the triangle formed by the lines x = 3, y = 4 and x = y is
(a) ½ sq. unit
(b) 1 sq. unit
(c) 2 sq. unit
(d) None of these
Answer:
Given and
We have plotting points as when
Therefore, area of
Area of triangle is square units
Hence, the correct choice is
Page No 3.115:
Question 17:
The area of the triangle formed by the lines 2x + 3y = 12, x − y − 1 = 0 and x = 0 (as shown in Fig. 3.23), is
(a) 7 sq. units
(b) 7.5 sq. units
(c) 6.5 sq. units
(d) 6 sq. units
Answer:
Given and
If We have plotting points as
Therefore, area of
Area of triangle is square units
Hence, the correct choice is
Page No 3.115:
Question 18:
The sum of the digits of a two digit number is 9 . If 27 is added to it , the digits of the number get reversed . th number is
(a) 25 (b) 72 (c) 63 (d) 36
Answer:
Let the digits at the tens and the ones place be x and y, respectively. So, the two digit number is 10x + y.
Now,
x + y = 9 .....(i)
Also,
10x + y + 27 = 10y + x
⇒ 9x − 9y = −27
⇒ x − y = −3 .....(ii)
Adding (i) and (ii), we get
2x = 6
⇒ x = 3
Putting x = 3 in (i), we get
3 + y = 9
⇒ y = 6
Thus, the required number is 10 × 3 + 6 = 36.
Hence, the correct answer is option (d).
Page No 3.115:
Question 19:
If $x=a,y=b$ is the solution of the systems of equations $x-y=2$ and $x+y=4$ , then the values of $a$ and $b$ are, respectively
(a) 3 and 1 (b) 3 and 5 (c) 5 and 3 (d) $-$ 1 and $-$3
Answer:
The given equations are
$x-y=2.....\left(1\right)\phantom{\rule{0ex}{0ex}}x+y=4.....\left(2\right)$
Adding (1) and (2), we get
2x = 6
⇒ x = 3
Putting x = 3 in (1), we get
3 + y = 4
⇒ y = 1
So, x = a = 3 and y = b = 1.
Thus, the values of a and b are 3 and 1, respectively.
Hence, the correct answer is option (a).
Page No 3.115:
Question 20:
For what value k , do the equations $3x-y+8=0$ and $6x-ky+16=0$ reperesent coincident lines ?
(a) $\frac{1}{2}$ (b) $-\frac{1}{2}$ (c) 2 (d) $-2$
Answer:
The given system of equations is
$3x-y+8=0\phantom{\rule{0ex}{0ex}}6x-ky+16=0\phantom{\rule{0ex}{0ex}}$
We know that the lines
${a}_{1}x+{b}_{1}y+{c}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}x+{b}_{2}y+{c}_{2}=0$
are coincident iff
$\frac{{a}_{1}}{{a}_{2}}=\frac{{\displaystyle {b}_{1}}}{{\displaystyle {b}_{2}}}=\frac{{\displaystyle {c}_{1}}}{{\displaystyle {c}_{2}}}$
$\therefore \frac{3}{6}=\frac{{\displaystyle -1}}{{\displaystyle -k}}=\frac{8}{16}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{1}{k}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow k=2$
Thus, the value of k = 2.
Hence, the correct answer is option (c).
Page No 3.116:
Question 21:
Aruna has only ₹1 and ₹2 coins with her . If the total number of coins that she has is 50 and the amount of money with her is ₹75 , then the number of ₹1 and ₹2 coins are , respectively
(a) 35 and 15 (b) 35 and 20 (c) 15 and 35 (d) 25 and 25
Answer:
Let the number of ₹1 coins be x and that of ₹2 coins be y.
Now,
Total number of coins = 50
So, x + y = 50 .....(i)
Also,
₹1 × x + ₹2 × y = ₹75
∴ x + 2y = 75 .....(ii)
Subtracting (i) from (ii), we get
y = 25
Putting y = 25 in (i), we get
x + 25 = 50
⇒ x = 25
So, the number of ₹1 coins and ₹2 coins are 25 and 25, respectively.
Hence, the correct answer is option (d).
Disclaimer: The answer given in the book does not match with the one obtained.
Page No 3.116:
Question 1:
The pair of equations y = 0 and y = –7 has _________ solution.
Answer:
The equation y = 0 is the x-axis and y = –7 is the line parallel to x-axis.
Therefore, both the lines are parallel lines.
Hence, the pair of equations y = 0 and y = –7 has no solution.
Page No 3.116:
Question 2:
The pair of equations x = a and y = b has solution ________.
Answer:
The equation x = a is the line parallel to y-axis and y = b is the line parallel to x-axis.
Therefore, both the lines intersect each other.
The point of intersection is x = a and y = b i.e. (a, b).
Hence, the pair of equations x = a and y = b has solution (a, b).
Page No 3.116:
Question 3:
If the pair of equations 3x + 2ky = 2 and 2x + 5y + 1 = 0 has no solution, then k = _________.
Answer:
If the system of equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0\mathrm{and}{a}_{2}x+{b}_{2}y+{c}_{2}=0$ has no solution, then $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$.
Since, the system of equations 3x + 2ky = 2 and 2x + 5y + 1 = 0 has no solution
Therefore,
$\frac{3}{2}=\frac{2k}{5}\ne \frac{-2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{2}=\frac{2k}{5}\mathrm{and}\frac{2k}{5}\ne \frac{-2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow 15=4k\mathrm{and}2k\ne -10\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{15}{4}\mathrm{and}k\ne -5$
Hence, k = $\overline{)\frac{15}{4}}.$
Page No 3.116:
Question 4:
If a pair of linear equations is consistent, then the lines representing them are either _______ or ________.
Answer:
If a pair of linear equations is consistent, then two case arises:
Case 1: When the lines are coincident, they have infinitely many solutions.
Case 2: When the lines are intersecting, they have a unique solution.
Hence, If a pair of linear equations is consistent, then the lines representing them are either coincident or intersecting.
Page No 3.116:
Question 5:
There is (are) ________ value(s) of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 have infinitely many solutions.
Answer:
If the system of equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0\mathrm{and}{a}_{2}x+{b}_{2}y+{c}_{2}=0$ has infinitely many solutions, then $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.
Since, the system of equations cx – y = 2 and 6x – 2y = 3 has infinitely many solutions
Therefore,
$\frac{c}{6}=\frac{-1}{-2}=\frac{-2}{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{c}{6}=\frac{1}{2}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\mathrm{But},\frac{1}{2}\ne \frac{2}{3}$
Therefore, for no value of c the pair of equations cx – y = 2 and 6x – 2y = 3 have infinitely many solutions.
Hence, there is (are) no value(s) of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 have infinitely many solutions.
Page No 3.116:
Question 6:
If one equation of a pair of dependent linear equations is 5x – 7y + 2 = 0, then the second equation is given by _______.
Answer:
To get the dependent linear equation, multiply the given equation by any non-zero integer.
If we multiply the given equation by any non-zero integer a, we get
5ax – 7ay + 2a = 0
Hence, the second equation is given by 5ax – 7ay + 2a = 0, where a is any non-zero integer.
Page No 3.116:
Question 7:
If a pair of linear equations is consistent with a unique solution, then the lines representing them are _______.
Answer:
If a pair of linear equations is consistent, then two case arises:
Case 1: When the lines are coincident, they have infinitely many solutions.
Case 2: When the lines are intersecting, they have a unique solution.
Hence, if a pair of linear equations is consistent with a unique solution, then the lines representing them are intersecting.
Page No 3.116:
Question 8:
The pair of equations λx + 3y = 7, 2x + 6y = 14 will have infinitely many solutions for λ = ________.
Answer:
If the system of equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0\mathrm{and}{a}_{2}x+{b}_{2}y+{c}_{2}=0$ has infinitely many solutions, then $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$.
Since, the system of equations λx + 3y = 7, 2x + 6y = 14 has infinitely many solutions
Therefore,
$\frac{\lambda}{2}=\frac{3}{6}=\frac{-7}{-14}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\lambda}{2}=\frac{1}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\lambda}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =1$
Hence, the pair of equations λx + 3y = 7, 2x + 6y = 14 will have infinitely many solutions for λ = 1 .
Page No 3.116:
Question 9:
If the pair of equations 2x + 3y – 5 = 0 and px – 6y – 8 = 0 has a unique solution for all real values of p except _______.
Answer:
If the system of equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0\mathrm{and}{a}_{2}x+{b}_{2}y+{c}_{2}=0$ has a unique solution, then $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$.
Since, the system of equations 2x + 3y – 5 = 0 and px – 6y – 8 = 0 has a unique solution
Therefore,
$\frac{2}{p}\ne \frac{3}{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{p}\ne \frac{-1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\ne -p\phantom{\rule{0ex}{0ex}}\Rightarrow p\ne -4$
Hence, if the pair of equations 2x + 3y – 5 = 0 and px – 6y – 8 = 0 has a unique solution for all real values of p except –4.
Page No 3.116:
Question 10:
If the lines represented by the equations 3x – y – 5 = 0 and 6x – 2y – p = 0 are parallel, then p is equal to ________.
Answer:
If the system of equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0\mathrm{and}{a}_{2}x+{b}_{2}y+{c}_{2}=0$ has no solution, i.e., they are parallel, then $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$.
Since, the system of equations 3x – y – 5 = 0 and 6x – 2y – p = 0 are parallel
Therefore,
$\frac{3}{6}=\frac{-1}{-2}\ne \frac{-5}{-p}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{1}{2}\ne \frac{5}{p}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\ne \frac{5}{p}\phantom{\rule{0ex}{0ex}}\Rightarrow p\ne 10$
Hence, if the lines represented by the equations 3x – y – 5 = 0 and 6x – 2y – p = 0 are parallel, then p is equal to all real values except 10.
Page No 3.116:
Question 11:
If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4, then a = _______ and b = _______.
Answer:
Since, x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4
Therefore, x = a, y = b satisfies the equations x – y = 2 and x + y = 4.
$a-b=2...\left(1\right)\phantom{\rule{0ex}{0ex}}a+b=4...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}a=3\mathrm{and}b=1$
Hence, a = 3 and b = 1.
Page No 3.116:
Question 12:
If the pair of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines, then ab = ________.
Answer:
If the system of equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0\mathrm{and}{a}_{2}x+{b}_{2}y+{c}_{2}=0$ has no solution, i.e., they are parallel, then $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$.
Since, the system of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines
Therefore,
$\frac{a}{3}=\frac{2}{b}\ne \frac{-7}{-16}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{3}=\frac{2}{b}\ne \frac{7}{16}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{3}=\frac{2}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow ab=6$
Hence, If the pair of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines, then ab = 6.
Page No 3.116:
Question 13:
The line 4x + 3y – 12 = 0 cuts the coordinate axes at A and B. The area of ΔOAB is _________.
Answer:
The given equation is 4x + 3y – 12 = 0 ...(i)
Solving equation (i), we get
$4x+3y-12=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3y=12-4x\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{12-4x}{3}$
When x = 0, y = 4
When x = 3, y = 0
x | 0 | 3 |
y | 4 | 0 |
On plotting these points on a graph, we get
We can see that OA = 3 units and OB = 4 units
Area of ΔOAB = $\frac{1}{2}\times OA\times OB$
= $\frac{1}{2}\times 3\times 4$
= 6
Hence, the area of ΔOAB is 6 square units.
Page No 3.116:
Question 14:
If 2^{x+y} = 2^{x–y} = $\sqrt{8}$, then x = _______ and y = _______.
Answer:
Given: 2^{x+y} = 2^{x–y} = $\sqrt{8}$
Now,
${2}^{x+y}=\sqrt{8}\phantom{\rule{0ex}{0ex}}\Rightarrow {2}^{x+y}=\sqrt{{2}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {2}^{x+y}={2}^{\frac{3}{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow x+y=\frac{3}{2}...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{2}^{x-y}=\sqrt{8}\phantom{\rule{0ex}{0ex}}\Rightarrow {2}^{x-y}=\sqrt{{2}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {2}^{x-y}={2}^{\frac{3}{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow x-y=\frac{3}{2}...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}x=\frac{3}{2}\mathrm{and}y=0$
Hence, x = $\overline{)\frac{3}{2}}$ and y = 0.
Page No 3.116:
Question 15:
The system of equations ax + 3y = 1, –12x + ay = 2 has ________ for all real values of a.
Answer:
The given system of equations are:
ax + 3y = 1 ...(i)
–12x + ay = 2 ...(ii)
Now,
$\frac{{a}_{1}}{{a}_{2}}=\frac{a}{-12}\phantom{\rule{0ex}{0ex}}\frac{{b}_{1}}{{b}_{2}}=\frac{3}{a}\phantom{\rule{0ex}{0ex}}\frac{{c}_{1}}{{c}_{2}}=\frac{-1}{-2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{If}\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{-12}=\frac{3}{a}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=-36\phantom{\rule{0ex}{0ex}}\mathrm{Which}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{for}\mathrm{any}\mathrm{value}\mathrm{of}a.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}\mathrm{for}\mathrm{all}\mathrm{real}\mathrm{values}\mathrm{of}a.$
Hence, the system of equations ax + 3y = 1, –12x + ay = 2 has a unique solution for all real values of a.
Page No 3.117:
Question 1:
Write the value of k for which the system of equations x + y − 4 = 0 and 2x + ky − 3 = 0 has no solution.
Answer:
The given system of equations is
.
For the equations to have no solutions
By cross multiplication we get,
Hence, the value of k is when system equations has no solution.
Page No 3.117:
Question 2:
Write the value of k for which the system of equations has infinitely many solutions.
$2x-y=5\phantom{\rule{0ex}{0ex}}6x+ky=15$
Answer:
The given systems of equations are
For the equations to have infinite number of solutions,
By cross Multiplication we get,
Hence the value of k is when equations has infinitely many solutions.
Page No 3.117:
Question 3:
Write the value of k for which the system of equations 3x − 2y = 0 and kx + 5y = 0 has infinitely may solutions.
Answer:
The given equations are
For the equations to have infinite number of solutions,
Therefore,
By cross multiplication we have
Hence, the value of k for the system of equation and is.
Page No 3.117:
Question 4:
Write the value of k for which the system of equations x + ky = 0, 2x − y = 0 has unique solution.
Answer:
The given equations are
$x+ky=0\phantom{\rule{0ex}{0ex}}2x-y=0\phantom{\rule{0ex}{0ex}}{a}_{1}=1,{a}_{2}=2,{b}_{1}=k,{b}_{2}=-1\phantom{\rule{0ex}{0ex}}\frac{{a}_{1}}{{a}_{2}}=\frac{1}{2},\phantom{\rule{0ex}{0ex}}\frac{{b}_{1}}{{b}_{2}}=\frac{k}{-1}$
For unique solution
For all real values of k, except the equations have unique solutions.
Page No 3.117:
Question 5:
Write the set of values of a and b for which the following system of equations has infinitely many solutions.
$2x+3y=7\phantom{\rule{0ex}{0ex}}2ax+\left(a+b\right)y=28$
Answer:
The given equations are
For the equations to have infinite number of solutions,
Therefore
Let us take
By dividing both the sides by 7 we get,
By multiplying equations by 2 we get
Substituting from we get
Subtracting in equation we have
Hence, the value of when system of equations has infinity many solutions.
Page No 3.117:
Question 6:
For what value of k, the following pair of linear equation has infinitely many solutions?
$10x+5y-\left(k-5\right)=0\phantom{\rule{0ex}{0ex}}20x+10y-k=0$
Answer:
The given equations are
For the equations to have infinite number of solutions
Let us take
Hence, the value of when the pair of linear equations has infinitely many solutions.
Page No 3.117:
Question 7:
Write the number of solution of the following pair of linear equations:
x + 2y − 8 = 0
2x + 4y = 16
Answer:
The given equations are
Every solution of the second equation is also a solution of the first equation.
Hence, there are, the system equation is consistent.
Page No 3.117:
Question 8:
Write the number of solutions of the following pair of linear equations:
x + 3y − 4 = 0
2x + 6y = 7
Answer:
The given linear pair of equations are
If then
Hence, the number of solutions of the pair of linear equation is.
Therefore, the equations have no solution.
Page No 3.12:
Question 1:
Akhila went to a fair in her village. She wanted to enjoy rides in the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it.) The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.
Answer:
Let no. of ride is and no. of Hoopla is .He paid Rs 20 for ride and for Hoopla.
The cost of ride is Rs and cost of Hoopla is Rs.then
The number of Hoopla is the half number of ride, then
Hence algebraic equations are and
Now, we draw the graph for algebraic equations.
Page No 3.12:
Question 2:
Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be". Is not this interesting? Represent this situation algebraically and graphically.
Answer:
Let age of Aftab is years and age of his daughter is years. Years ago his age wastimes older as her daughter was. Then
Three years from now, he will be three times older as his daughter will be, then
Hence the algebraic representation are and
Page No 3.12:
Question 3:
The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically.
Answer:
The given equation are and.
In order to represent the above pair of linear equation graphically, we need
Two points on the line representing each equation. That is, we find two solutions
of each equation as given below:
We have,
Putting we get
Putting we get
Thus, two solution of equation are
We have
Putting we get
Putting we get
Thus, two solution of equation are
8 | ||
6 |
Now we plot the pointand and draw a line passing through
These two points to get the graph o the line represented by equation
We also plot the points and and draw a line passing through
These two points to get the graph O the line represented by equation
We observe that the line parallel and they do not intersect anywhere.
Page No 3.12:
Question 4:
Gloria is walking along the path joining (−2, 3) and (2, −2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.
Answer:
Gloria is walking the path joining and
Suresh is walking the path joining and
0 | 4 | |
5 | 0 |
The graphical representations are
Page No 3.12:
Question 5:
On comparing the ratios $\frac{{a}_{1}}{{a}_{2}},\frac{{b}_{1}}{{b}_{2}}\mathrm{and}\frac{{c}_{1}}{{c}_{2}}$, and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide :
(i) 5x − 4y + 8 = 0
7x + 6y − 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x − 3y + 10 = 0
2x − y + 9 = 0
Answer:
(i) Given equation are: 5x + 4y + 8 = 0
7x + 6y − 9 = 0
We have And
Thus the pair of linear equation is intersecting.
(ii) Given equation are: 9x + 3y + 12 = 0
18x + 6y + 24 = 0
We have
Thus the pair of linear is coincident lines.
(iii) Given equation are:
We have
Thus the pair of line is parallel lines.
Page No 3.12:
Question 6:
Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Answer:
(i) Given the linear equation are:
We know that intersecting condition:
Where
Hence the equation of other line is
(ii) We know that parallel line condition is:
Where
Hence the equation is
(iii) We know that coincident line condition is:
Where
Hence the equation is
Page No 3.12:
Question 7:
The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2kg of grapes is Rs. 300 Represent th situation algebraically and geometrically.
Answer:
Let the cost of 1 kg of apples be Rs x.
And, cost of 1 kg of grapes = Rs y
According to the question, the algebraic representation is
For ,
The solution table is
x | 50 | 60 | 70 |
y | 60 | 40 | 20 |
For 4x + 2y = 300,
The solution table is
x | 70 | 80 | 75 |
y | 10 | –10 | 0 |
The graphical representation is as follows.
Page No 3.29:
Question 1:
Solve the following systems of equations graphically:
x + y = 3
2x + 5y = 12
Answer:
The given equations are:
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 3 | |
3 | 0 |
Draw the graph by plotting the two points from table.
Graph of the equation:
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 6 | |
12/5 | 0 |
Draw the graph by plotting the two points from the table.
The two lines intersect at point P.
Hence, and is the solution.
Page No 3.29:
Question 2:
Solve the following systems of equations graphically:
x − 2y = 5
2x + 3y = 10
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 5 | |
0 |
Draw the graph by plotting the two points from table.
Graph the equation
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | 5 | |
10/3 | 0 |
Draw the graph by plotting the two points from table.
The two lines intersects at point B
Hence is the solution
Page No 3.29:
Question 3:
Solve the following systems of equations graphically:
3x + y + 1 = 0
2x − 3y + 8 = 0
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
0 | –1/3 | |
–1 | 0 |
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we ge
Putting in equationwe get
Use the following table to draw the graph.
0 | –4 | |
8/3 | 0 |
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence is the solution.
Page No 3.29:
Question 4:
Solve the following systems of equations graphically:
2x + y − 3 = 0
2x − 3y − 7 = 0
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
0 | 7/2 | |
–7/3 | 0 |
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence is the solution.
Page No 3.29:
Question 5:
Solve the following systems of equations graphically:
x – y + 1 = 0
3x + 2y – 12 = 0
Answer:
The given equations are:
x – y + 1 = 0 ...(i)
3x + 2y – 12 = 0 ...(ii)
Solving equation (i), we get
$x-y+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow y=x+1$
When x = 0, y = 1
When x = 1, y = 2
x | 0 | 1 |
y | 1 | 2 |
Solving equation (ii), we get
$3x+2y-12=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2y=12-3x\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{12-3x}{2}$
When x = 0, y = 6
When x = 4, y = 0
x | 0 | 4 |
y | 6 | 0 |
On plotting these points on a graph, we get
Hence, point A(2, 3) is the point of intersection.
Page No 3.29:
Question 6:
Solve the following systems of equations graphically:
x − 2y = 6
3x − 6y = 0
Answer:
The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Plotting the two points equation (i) can be drawn.
Graph of the equation….
Putting in equation, we get:
Putting x=2 in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
We see that the two lines are parallel, so they won’t intersect
Hence there is no solution
Page No 3.29:
Question 7:
Solve the following systems of equations graphically:
x + y = 4
2x − 3y = 3
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence is the solution
Page No 3.29:
Question 8:
Solve the following systems of equations graphically:
2x + 3y = 4
x − y + 3 = 0
Answer:
The given equations are:
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
The two lines intersect at points P.
Hence, and is the solution.
Page No 3.29:
Question 9:
Solve the following systems of equations graphically:
2x − 3y + 13 = 0
3x − 2y + 12 = 0
Answer:
The given equations are:
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equationwe get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at points P.
Hence, and is the solution.
Page No 3.29:
Question 10:
Solve the following systems of equations graphically:
2x + 3y + 5 = 0
3x − 2y − 12 = 0
Answer:
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equationwe get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersects at points P
Hence, and is the solution.
Page No 3.29:
Question 11:
Show graphically that each one of the following systems of equations has infinitely many solutions:
2x + 3y = 6
4x + 6y = 12
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Question 12:
Show graphically that each one of the following systems of equations has infinitely many solutions:
x − 2y = 5
3x − 6y = 15
Answer:
The given equations are
Putting in equation we get:
Putting in equationswe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Graph of the equation….
Putting in equations we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Question 13:
Show graphically that each one of the following systems of equations has infinitely many solutions:
3x + y = 8
6x + 2y = 16
Answer:
The given equations are
Putting in equation we get:
Putting in equationswe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equations we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Question 14:
Show graphically that each one of the following systems of equations has infinitely many solutions:
x − 2y + 11 = 0
3x − 6y + 33 = 0
Answer:
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Thus the graph of the two equations are coincide
Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
Page No 3.29:
Question 15:
Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :
3x − 5y = 20
6x − 10y = −40
Answer:
The given equations are
Putting in equation we get:
Putting in equationwe get
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here we see that the two lines are parallel
Hence the given system of equations has no solution.
Page No 3.29:
Question 16:
Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :
x − 2y = 6
3x − 6y = 0
Answer:
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the two points _{.}
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here the two lines are parallel and so there is no point in common
Hence the given system of equations has no solution.
Page No 3.29:
Question 17:
Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :
2y − x = 9
6y − 3x = 21
Answer:
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here two lines are parallel and so don’t have common points
Hence the given system of equations has no solution.
Page No 3.29:
Question 18:
Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :
3x − 4y − 1 = 0
$2x-\frac{8}{3}y+5=0$
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the two points_{.}
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Here, the two lines are parallel.
Hence the given system of equations is inconsistent.
Page No 3.29:
Question 19:
Determine graphically the vertices of the triangle, the equations of whose sides are given below :
(i) 2y − x = 8, 5y − x = 14 and y − 2x = 1
(ii) y = x, y = 0 and 3x + 3y = 10
Answer:
(i) Draw the 3 lines as given by equations
By taking x=1 = 1 cm on x−axis
And y =1=1cm on y−axis
Clearly from graph points of intersection three lines are
(−4,2) , (1,3), (2,5)
(ii) Draw the 3 lines as given by equations
By taking x=1 = 1 cm on x−axis
And y =1=1cm on y−axis
From graph point of intersection are (0,0) (10/3,0) (5/3,5/3)
Page No 3.29:
Question 20:
Determine, graphically whether the system of equations x − 2y = 2, 4x − 2y = 5 is consistent or in-consistent.
Answer:
The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
0 | 5/4 | |
–5/2 | 0 |
Draw the graph by plotting the two points from table.
It has unique solution.
Hence the system of equations is consistent
Page No 3.29:
Question 21:
Determine, by drawing graphs, whether the following system of linear equations has a unique solution or not :
(i) 2x − 3y = 6, x + y = 1
(ii) 2y = 4x − 6, 2x = y + 3
Answer:
(i) The given equations are
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of the equation….
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at point .
Hence the equations have unique solution.
(ii) The equations of graphs is
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the two points .
Graph of the equation
Putting in equation we get.
Putting in equation we get.
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines are coincident.
Hence the equations have infinitely much solution.
Hence the system is consistent
Page No 3.29:
Question 22:
Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.
(i) 2x − 5y + 4 = 0,
2x + y − 8 = 0
(ii) 3x + 2y = 12,
5x − 2y = 4
(iii) 2x + y − 11 = 0,
x − y − 1 = 0
(iv) x + 2y − 7 = 0,
2x − y − 4 = 0
(v) 3x + y − 5 = 0,
2x − y − 5 = 0
(vi) 2x − y − 5 = 0,
x − y − 3 = 0
Answer:
(i) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
−2 | 8 | |
0 | _{4} |
_{The two points satisfying (ii) can be listed in a table as,}
4 | 2 | |
0 | 4 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 3, y = 2.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(ii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
4 | 6 | |
0 | –3 |
_{The two points satisfying (ii) can be listed in a table as,}
3 | 2 | |
5.5 | 3 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = 3.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(iii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
3 | 1 | |
5 | 9 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 5 | |
0 | 4 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 4, y = 3.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(iv) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
5 | 7 | |
1 | 0 |
_{The two points satisfying (ii) can be listed in a table as,}
2 | 1 | |
0 | –2 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 3, y = 2.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.(v) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
1 | 3 | |
2 | –4 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 4 | |
–3 | 3 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = −1.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
(vi) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
1 | 3 | |
–3 | 1 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 5 | |
–2 | 2 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = −1.
Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.
Page No 3.30:
Question 23:
Solve the following system of linear equation graphically and shade the region between the two lines and x-axis:
(i) 2x + 3y = 12,
x − y = 1
(ii) 3x + 2y − 4 = 0,
2x − 3y − 7 = 0
(iii) 3x + 2y − 11 = 0
2x − 3y + 10 = 0
Answer:
The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at. The region enclosed by the lines represented by the given equations and x−axis are shown in the above figure
Hence, and is the solution.
(ii) The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
0 | ||
2 | 0 |
The graph of (i) can be obtained by plotting the two points.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
− |
Draw the graph by plotting the two points from table.
The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.
Hence, and is the solution.
(iii) The given equations are:
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.
Hence, and is the solution.
Page No 3.30:
Question 24:
Draw the graphs of the following equations on the same graph paper.
2x + 3y = 12,
x − y = 1
Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis.
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Draw the graph by plotting the two points from table.
The intersection point is P(3, 2)
Three points of the triangle are.
Hence the value of and
Page No 3.30:
Question 25:
Draw the graphs of x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines and x-axis.
Answer:
The given equations are
Putting in equation we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Now, Required area = Area of shaded region
Required area = Area of PBD
Required area =
Required area =
Required area =
Hence the area =
Page No 3.30:
Question 26:
Solve graphically the system of linear equations:
4x − 3y + 4 = 0
4x + 3y − 20 = 0
Find the area bounded by these lines and x-axis.
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the points (0, 4/3), (−1, 0).
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Hence is the solution of the given equations.
Now,
Required area = Area of PBD
Required area =
Required area =
Required area =
Hence, the area =
Page No 3.30:
Question 27:
Solve the following system of linear equations graphically :
3x + y − 11 = 0, x − y − 1 = 0.
Shade the region bounded by these lines and y-axis. Also, find the area of the region bounded by the these lines and y-axis.
Answer:
The given equations are
Putting in equation we get:
Putting in equation we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Hence is the solution of the given equations
The area enclosed by the lines represented by the given equations and the y−axis is shaded region in the figure
Now, Required area = Area of shaded region
Required area = Area of PAC
Required area =
Required area =
Required area =
Hence the required area is sq. unit
Page No 3.30:
Question 28:
Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system.
(i) 2x + y = 6
x − 2y = −2
(ii) 2x − y = 2
4x − y = 8
(iii) x + 2y = 5
2x − 3y = −4
(iv) 2x + 3y = 8
x − 2y = −3
Answer:
(i)
The given equations are
_{The two points satisfying (i) can be listed in a table as,}
4 | _{0} | |
_{−2} | _{6} |
_{The two points satisfying (ii) can be listed in a table as,}
4 | 6 | |
_{3} | _{4} |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
(ii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | 2 | |
2 | 2 |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 3 | |
–4 | 4 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 3, y = 4.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
(iii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | _{−1} | |
_{2.5} | _{3} |
_{The two points satisfying (ii) can be listed in a table as,}
4 | –5 | |
4 | –2 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 1, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
Solution is missing
(iv) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
–2 | 7 | |
_{4} | –2 |
_{The two points satisfying (ii) can be listed in a table as,}
–1 | 3 | |
1 | 3 |
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 1, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
Page No 3.30:
Question 29:
Draw the graphs of the following equations:
2x − 3y + 6 = 0
2x + 3y − 18 = 0
y − 2 = 0
Find the vertices of the triangle so obtained. Also, find the area of the triangle.
Answer:
The given equations are
_{The two points satisfying (i) can be listed in a table as,}
−3 | 6 | |
_{0} | _{6} |
The two points satisfying (ii) can be listed in a table as,
0 | 9 | |
6 | _{0} |
The two points satisfying (iii) can be listed in a table as,
−1 | 8 | |
_{2} | _{2} |
Now, graph of equations (i), (ii) and (iii) can be drawn as,
It is seen that the coordinates of the vertices of the obtained triangle are
_{∴Area of ΔABC = }
Page No 3.30:
Question 30:
Solve the following system of equations graphically.
2x − 3y + 6 = 0
2x + 3y − 18 = 0
Also, find the area of the region bounded by these two lines and y-axis.
Answer:
The given equations are:
Putting in equation (i) we get:
Putting in equation (i) we get:
Use the following table to draw the graph
Draw the graph by plotting the two points from table.
Putting in equationwe get:
Putting in equationwe get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at.
Hence is the solution of the given equations.
The area enclosed by the lines represented by the given equations and the y−axis
Now,
Required area = Area of PCA
Required area =
Required area =
Required area =
Hence the required area is
Page No 3.30:
Question 31:
Solve the following system of linear equations graphically.
4x − 5y − 20 = 0
3x + 5y − 15 = 0
Determine the vertices of the triangle formed by the lines representing the above equation and the y-axis.
Answer:
The given equations are:
Putting in equation we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
x |
0 |
5 |
y |
0 |
Draw the graph by plotting the two points from table
Putting in equation (ii) we get:
Putting in equation (ii) we get:
Use the following table to draw the graph.
x |
0 |
5 |
y |
3 |
0 |
Draw the graph by plotting the two points from table.
The three vertices of the triangle are.
Hence the solution of the equation is and
Page No 3.31:
Question 32:
Draw the graphs of the equations 5x − y = 5 and 3x − y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and y-axis. Calculate the area of the triangle so formed.
Answer:
The given equations are:
Putting in equation (i) we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table
Putting in equation we get:
Putting in equationwe get:
Use the following table to draw the graph.
–3 |
Draw the graph by plotting the two points from table.
Hence the vertices of the required triangle are.
Now,
Required area = Area of PCA
Required area =
Required area =
Hence the required area is
Page No 3.31:
Question 33:
Form the pair of linear equations in the following problems, and find their solution graphically:
(i) 10 students of class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and a pen.
(iii) Champa went to a 'sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased." Help her friends to find how many pants and skirts Champa bought.
Answer:
(i) Let the number of girls be x and the number of boys be y.
According to the question, the algebraic representation is
x + y = 10
x − y = 4
For x + y = 10,
x = 10 − y
x | 5 | 4 | 6 |
y | 5 | 6 | 4 |
For x − y = 4,
x = 4 + y
x | 5 | 4 | 3 |
y | 1 | 0 | −1 |
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines intersect each other at point (7, 3).
Therefore, the number of girls and boys in the class are 7 and 3 respectively.
(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.
According to the question, the algebraic representation is
5x + 7y = 50
7x + 5y = 46
For 5x + 7y = 50,
x | 3 | 10 | − 4 |
y | 5 | 0 | 10 |
7x + 5y = 46
x | 8 | 3 | − 2 |
y | − 2 | 5 | 12 |
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines intersect each other at point (3, 5).
Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.
(iii) Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :
y = 2x − 2 … (i)
y = 4x − 4 … (ii)
The graphs of the equations (i) and (ii) can be drawn by finding two solutions for each of the equations. They are given in the following table.
x | 2 | 0 |
y = 2x − 2 | 2 | −2 |
Hence, the graphic representation is as follows.
The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.
Page No 3.31:
Question 34:
Solve the following system of equations graphically:
Shade the region between the lines and the y-axis
(i) 3x − 4y = 7
5x + 2y = 3
(ii) 4x − y = 4
3x + 2y = 14
Answer:
The given equations are:
Putting in equation (i) we get:
Putting in equation (i) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Putting in equation (ii) we get:
Putting in equation (ii) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at points of y−axis.
Hence, and is the Solution.
(ii) The equations are:
Putting in equation (1) we get:
Putting in equation (1) we get:
Use the following table to draw the graph:
Draw the graph by plotting the two points from table.
Putting in equation (2) we get:
Putting in equation (2) we get:
Use the following table to draw the graph.
x | ||
Draw the graph by plotting the two points from table.
Two lines intersect at points of y−axis.
Hence and is the solution.
Page No 3.31:
Question 35:
Represent the following pair of equations graphically and write the coordinates of points where the lines intersects y-axis.
x + 3y = 6
2x − 3y = 12
Answer:
The given equations are
Putting in equation (i) we get:
Putting in equationwe get:
Use the following table to draw the graph.
x | ||
The graph of (i) can be obtained by plotting the two points.
Putting in equation (ii) we get:
Putting in equation (ii) we get:
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
Graph of lines represented by the equations meet y−axis at respectively.
Page No 3.31:
Question 36:
Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is
(i) intersecting lines
(ii) Parallel lines
(iii) coincident lines
Answer:
(i) For intersecting lines,
Equation of another intersecting line to the given line is−
Since, condition for intersecting lines and unique solution is−
(ii) For parallel lines,
Equation of another parallel line to the given line is−
Since, condition for parallel lines and no solution is−
(iii) For co−incident lines,
Equation of another coincident line to the given line is−
Since, condition for coincident lines and infinite solution is−
Page No 3.31:
Question 37:
Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are :
(i) y = x, y = 2x and y + x = 6
(ii) y = x, 3y = x, x + y = 8
Answer:
(i) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | _{1} | |
0 | _{1} |
_{The two points satisfying (ii) can be listed in a table as,}
1 | 3 | |
2 | 6 |
_{The two points satisfying (iii) can be listed in a table as,}
0 | 6 | |
6 | 6 |
Now, graph of equations (i), (ii) and (iii) can be drawn as,
It is seen that the coordinates of the vertices of the obtained triangle are
(ii) The given equations are
_{The two points satisfying (i) can be listed in a table as,}
0 | 2 | |
0 | 2 |
_{The two points satisfying (ii) can be listed in a table as,}
3 | –3 | |
1 | –1 |
_{The two points satisfying (iii) can be listed in a table as,}
3 | 5 | |
5 | 3 |
Now, graph of equations (i), (ii) and (iii) can be drawn as,
It is seen that the coordinates of the obtained triangle are
Page No 3.31:
Question 38:
Graphically , solve the following pair of equations:
$2x+y=6\phantom{\rule{0ex}{0ex}}2x-y+2=0$
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.
Answer:
The given linear equations are:
$2x+y=6.....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}2x-y+2=0.....\left(\mathrm{ii}\right)$
For (i), we have
x | 0 | 3 |
y | 6 | 0 |
For (ii), we have
x | 0 | −1 |
y | 2 | 0 |
Thus, we plot the graph for these two equations and mark the point where these two lines intersect.
From the graph we see that the two lines intersect at point E(1, 4).
Now, the area of triangle CEB is
${A}_{1}=\frac{1}{2}\times 4\times 4=8\mathrm{square}\mathrm{unit}$
The area of triangle AED is
${A}_{2}=\frac{1}{2}\times 4\times 1=2\mathrm{square}\mathrm{unit}$
So, the ratio of the areas of the two triangles will be
$\frac{{A}_{1}}{{A}_{2}}=\frac{8}{2}=\frac{4}{1}$
Thus, the required ratio is 4 : 1.
Page No 3.31:
Question 39:
Determine, graphically, the vertices of the triangle formed by the lines $y=x,3y=x,x+y=8.$
Answer:
The given lines are:
$y=x.....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}3y=x.....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}x+y=8.....\left(\mathrm{iii}\right)$
From (i), we have
x | 1 | 2 |
y | 1 | 2 |
For (ii), we have
x | 0 | 3 |
y | 0 | 1 |
For (iii), we have
x | 0 | 8 |
y | 8 | 0 |
Thus, we plot the graph for these three equations and mark the point where these two lines intersect.
From the graph we find that the vertices of the triangle thus formed are H(4, 4), I(6, 2) and D(0, 0).
Page No 3.31:
Question 40:
Draw the graph of the equations x = 3 , x = 5 and 2x − y − 4 = 0 . Also, find the area of the quadrilateral formed by the lines and the x-axis.
Answer:
The given equations are
x = 3 .....(i)
x = 5 .....(ii)
2x − y − 4 = 0 .....(iii)
For (iii), we have
x | 0 | 2 |
y | −4 | 0 |
We plot the lines on the graph as follows:
The vertices of the quadrilateral thus formed are A(5, 0), B(5, 6), C(3, 2) and D(3, 0).
∴ Area of the quadrilateral DABC
= Area of a trapezium
= $\frac{1}{2}h\left(a+b\right)$
$=\frac{1}{2}\times 2\times \left(2+6\right)\phantom{\rule{0ex}{0ex}}=8\mathrm{square}\mathrm{units}$
Thus, the area of the quadrilateral formed by the given lines and the x-axis is 8 square units.
Page No 3.31:
Question 41:
Draw the graph of the lines x = $-$2 and y = 3 . Write the vertices of the figure formed by these lines , the x-axis and the y-axis . Also , find the area of the figure.
Answer:
The given lines are
$x=-2.....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}y=3.....\left(\mathrm{ii}\right)$
The graph thus obtained will be as follows:
The figure thus obtained is a rectangle ABCD. The vertices of the rectangle are A(−2, 3), B(0, 3), C(0, 0) and D(−2, 0).
Length of the rectangle = AD = BC = 3 units
Breadth of the rectangle = CD = AB = 2 units
∴ Area of rectangle ABCD
$=3\times 2\phantom{\rule{0ex}{0ex}}=6\mathrm{square}\mathrm{units}$
Page No 3.32:
Question 42:
Draw the graphs of the pair of linear equations $x-y+2=0$ and $4x-y-4=0$ . Calculate the area of the triangle formed by the lines so drawn and the x-axis .
Answer:
The given linear equations are
x − y + 2 = 0 .....(i)
4x − y − 4 = 0 .....(ii)
For (i), we have
x | 0 | −2 |
y | 2 | 0 |
For (ii), we have
x | 0 | 1 |
y | −4 | 0 |
The graph of the lines represented by the given equations is shown below:
The triangle thus obtained has the vertices A(2, 4), B(−2, 0) and D(1, 0).
∴ Area of triangle ADB
$=\frac{1}{2}\times 3\times 4\phantom{\rule{0ex}{0ex}}=6\mathrm{square}\mathrm{units}$
Thus, the area of the triangle formed by the given lines and the x-axis is 6 square units.
Page No 3.44:
Question 1:
Solve the following systems of equations:
$11x+15y+23=0\phantom{\rule{0ex}{0ex}}7x-2y-20=0$
Answer:
The given equations are:
Multiply equation by 2 and equation by 15, and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.44:
Question 2:
Solve the following systems of equations:
3x − 7y + 10 = 0
y − 2x − 3 = 0
Answer:
The given equations are:
…
…
Multiply equation by 2 and equation by, and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.44:
Question 3:
Solve the following systems of equations:
0.4x + 0.3y = 1.7
0.7x − 0.2y = 0.8
Answer:
The given equations are:
…
…
Multiply equation by 2 and equation by, and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.44:
Question 4:
Solve the following systems of equations:
$\frac{x}{2}+y=0.8\phantom{\rule{0ex}{0ex}}\frac{7}{x+{\displaystyle \frac{y}{2}}}=10$
Answer:
The given equations are:
…
Subtract (ii) from (i) we get
Put the value of in equation we get
Hence the value of and .
Page No 3.44:
Question 5:
Solve the following systems of equations:
7(y + 3) − 2(x + 2) = 14
4(y − 2) + 3(x − 3) = 2
Answer:
The given equations are:
…
…
Multiply equation by and equation by and add both equations we get
Put the value of in equationwe get
Hence the value of and
Page No 3.44:
Question 6:
Solve the following systems of equations:
$\frac{x}{7}+\frac{y}{3}=5\phantom{\rule{0ex}{0ex}}\frac{x}{2}-\frac{y}{9}=6$
Answer:
The given equations are:
…
…
Multiply equation by and add both equations we get
Put the value of in equation we get
Hence the value of and .
Page No 3.44:
Question 7:
Solve the following systems of equations:
$\frac{x}{3}+\frac{y}{4}=11\phantom{\rule{0ex}{0ex}}\frac{5x}{6}-\frac{y}{3}=-7$
Answer:
The given equations are:
…
…
Multiply equation by and equation by and add both equations we get
Put the value of in equation we get
Hence the value of and .
Page No 3.44:
Question 8:
Solve the following systems of equations:
$\frac{4}{x}+3y=8\phantom{\rule{0ex}{0ex}}\frac{6}{x}-4y=-5$
Answer:
The given equations are:
…
…
Multiply equation by and equation by and add both equations we get
Put the value of in equation we get
Hence the value of and .
Page No 3.44:
Question 9:
Solve the following systems of equations:
$x+\frac{y}{2}=4\phantom{\rule{0ex}{0ex}}\frac{x}{3}+2y=5$
Answer:
The given equations are:
…
…
Multiply equation by and subtract equations, we get
Put the value of in equation, we get
Hence the value of x and y are and
Page No 3.44:
Question 10:
Solve the following systems of equations:
$x+2y=\frac{3}{2}\phantom{\rule{0ex}{0ex}}2x+y=\frac{3}{2}$
Answer:
The given equations are:
…
…
Multiply equation by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Question 11:
Solve the following systems of equations:
$\sqrt{2}x-\sqrt{3}y=0\phantom{\rule{0ex}{0ex}}\sqrt{3}x-\sqrt{8}y=0$
Answer:
The given equations are:
…
…
Multiply equation by and equation by and subtract equation (ii) from (i), we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Question 12:
Solve the following systems of equations:
$3x-\frac{y+7}{11}+2=10\phantom{\rule{0ex}{0ex}}2y+\frac{x+11}{7}=10$
Answer:
The given equations are:
$3x-\frac{y+7}{11}+2=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 3x-\frac{\left(y+7\right)}{11}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{33x-y-7}{11}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 33x-y-7=88\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 33x-y=95........\left(1\right)$
$2y+\frac{x+11}{7}=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{14y+x+11}{7}=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y+x+11=70\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y+x=59\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow x+14y=59.........\left(2\right)$
Multiply equation (1) by , we get
$462x-14y=1330......\left(3\right)$
adding (2) and (3), we get$\left(x+14y\right)+\left(462x-14y\right)=59+1330\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 463x=1389\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow x=3$
$\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3+14y=59\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y=59-3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 14y=56\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow y=4\phantom{\rule{0ex}{0ex}}$
Hence the value of x and y are and
Page No 3.44:
Question 13:
Solve the following systems of equations:
$2x-\frac{3}{y}=9\phantom{\rule{0ex}{0ex}}3x+\frac{7}{y}=2,y\ne 0$
Answer:
The given equations are:
…
…
Multiply equation by and by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.44:
Question 14:
Solve the following systems of equations:
0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
Answer:
The given equations are:
…
…
Multiply equation by and by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 15:
Solve the following systems of equations:
$\frac{1}{7x}+\frac{1}{6y}=3\phantom{\rule{0ex}{0ex}}\frac{1}{2x}-\frac{1}{3y}=5$
Answer:
The given equations are:
…
…
Multiply equation by and add both equations we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 16:
Solve the following systems of equations:
$\frac{1}{2x}+\frac{1}{3y}=2\phantom{\rule{0ex}{0ex}}\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$
Answer:
The given equations are:
…
…
Multiply equation by and by and subtract equation (ii) from (i) we get
Put the value of in equation, we get
$\frac{1}{2\times {\displaystyle \frac{1}{2}}}+\frac{1}{3y}=2\phantom{\rule{0ex}{0ex}}\frac{1}{3y}=2-1\phantom{\rule{0ex}{0ex}}\frac{1}{3y}=1\phantom{\rule{0ex}{0ex}}y=\frac{1}{3}\phantom{\rule{0ex}{0ex}}$
Hence the value of and $y=\frac{1}{3}$
Page No 3.45:
Question 17:
Solve the following systems of equations:
$\frac{15}{u}+\frac{2}{\nu}=17\phantom{\rule{0ex}{0ex}}\frac{1}{u}+\frac{1}{\nu}=\frac{36}{5}$
Answer:
The given equations are:
…
…
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Question 18:
Solve the following systems of equations:
$\frac{3}{x}-\frac{1}{y}=-9\phantom{\rule{0ex}{0ex}}\frac{2}{x}+\frac{3}{y}=5$
Answer:
The given equations are:
…
…
Multiply equation by and add both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 19:
Solve the following systems of equations:
$\frac{2}{x}+\frac{5}{y}=1\phantom{\rule{0ex}{0ex}}\frac{60}{x}+\frac{40}{y}=19,x\ne 0,y\ne 0$
Answer:
The given equations are:
Multiply equation by and subtract (ii) from equation (i), we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Question 20:
Solve the following systems of equations:
$\frac{1}{5x}+\frac{1}{6y}=12\phantom{\rule{0ex}{0ex}}\frac{1}{3x}-\frac{3}{7y}=8,x\ne 0,y\ne 0$
Answer:
The given equations are:
…
…
Multiply equation by and equation by, add both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 21:
Solve the following systems of equations:
$\frac{4}{x}+3y=14\phantom{\rule{0ex}{0ex}}\frac{3}{x}-4y=23$
Answer:
The given equations are:
…
…
Multiply equation by and equation by, add both equations, we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Question 22:
Solve the following systems of equations:
$\frac{4}{x}+5y=7\phantom{\rule{0ex}{0ex}}\frac{3}{x}+4y=5$
Answer:
The given equations are:
…
…
Multiply equation by and equation by and subtract (ii) from (i) we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 23:
Solve the following systems of equations:
$\frac{2}{x}+\frac{3}{y}=13\phantom{\rule{0ex}{0ex}}\frac{5}{x}-\frac{4}{y}=-2$
Answer:
The given equations are:
…
…
Multiply equation by and equation by 3 and add both equations we get
Put the value of in equation, we get
Hence the value of and .
Page No 3.45:
Question 24:
Solve the following systems of equations:
$\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\phantom{\rule{0ex}{0ex}}\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$
Answer:
The given equations are:
…
…
Multiply equation by and add both equations we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 25:
Solve the following systems of equations:
$\frac{x+y}{xy}=2\phantom{\rule{0ex}{0ex}}\frac{x-y}{xy}=6$
Answer:
The given equations are:
Adding both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 26:
Solve the following systems of equations:
$\frac{2}{x}+\frac{3}{y}=\frac{9}{xy}\phantom{\rule{0ex}{0ex}}\frac{4}{x}+\frac{9}{y}=\frac{21}{xy},x\ne 0,y\ne 0$
Answer:
The given equations are:
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 27:
Solve the following systems of equations:
$\frac{6}{x+y}=\frac{7}{x-y}+3\phantom{\rule{0ex}{0ex}}\frac{1}{2\left(x+y\right)}=\frac{1}{3\left(x-y\right)\text{'}}$
where x + y ≠ 0 and x − y ≠ 0
Answer:
The given equations are:
Let and then equations are
…
…
Multiply equation by and subtract (ii) from (i), we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in second equation, we get
Hence the value of and.
Page No 3.45:
Question 28:
Solve the following systems of equations:
$\frac{xy}{x+y}=\frac{6}{5}\phantom{\rule{0ex}{0ex}}\frac{xy}{y-x}=6,$
where x + y ≠ 0, y − x ≠ 0
Answer:
The given equations are:
Add both equations, we get
Put the value of in equation, we get
Hence the value of and
Page No 3.45:
Question 29:
Solve the following systems of equations:
$\frac{22}{x+y}+\frac{15}{x-y}=5\phantom{\rule{0ex}{0ex}}\frac{55}{x+y}+\frac{45}{x-y}=14$
Answer:
The given equations are:
Let and then equations are
…
…
Multiply equation by and subtracting (ii) from (i), we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in second equation, we get
Hence the value of and .
Page No 3.45:
Question 30:
Solve the following systems of equations:
$\frac{5}{x+y}-\frac{2}{x-y}=-1\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}+\frac{7}{x-y}=10$
Answer:
The given equations are:
Let and then equations are
…
…
Multiply equation by and equation by and add both equations, we get
Put the value of in equation, we get
Then
Add both equations, we get
Put the value of in first equation, we get
Hence the value of and .
Page No 3.45:
Question 31:
Solve the following systems of equations:
$\frac{3}{x+y}+\frac{2}{x-y}=2\phantom{\rule{0ex}{0ex}}\frac{9}{x+y}-\frac{4}{x-y}=1$
Answer:
The given equations are:
Let and then equations are
…