Rd Sharma 2021 for Class 10 Maths Chapter 3 - Pair Of Linear Equations In Two Variables

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.

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

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

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

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 .

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, 
$$\begin{gathered} \frac{40}{5-x}=3\left(\frac{{\displaystyle 40}}{{\displaystyle 5+x}}\right) \\ \Rightarrow \frac{1}{5-x}=\frac{3}{5+x} \\ \Rightarrow 5+x=15-3x \\ \Rightarrow 4x=10 \\ \Rightarrow x=\frac{10}{4}=2.5 \mathrm{km}/\mathrm{h} \end{gathered}$$
Therefore, the speed of the stream is 2.5 km/h.

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 .

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.

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.

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.

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.

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

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 
$$\begin{gathered} 30u+28v=7 .....\left(\mathrm{i}\right) \\ 21u+21v=5 .....\left(\mathrm{ii}\right) \end{gathered}$$
Multiplying (i) by 21 and (ii) by 30, we get
$$\begin{gathered} 630u+588v=147 ...\left(\mathrm{iii}\right) \\ 630u+630v=150 ...\left(\mathrm{iv}\right) \end{gathered}$$
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,
 $$\begin{gathered} \frac{1}{x-y}=\frac{1}{6} \mathrm{and} \frac{1}{x+y}=\frac{1}{14} \\ \Rightarrow x-y=6 \mathrm{and} x+y=14 \end{gathered}$$
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.

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 .

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 .

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.

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

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.

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

$$\begin{gathered} \left(x + 2\right) \left(y - 2\right) = xy - 28 \\ \Rightarrow xy - 2x + 2y - 4 = xy - 28 \\ \Rightarrow -2x + 2y - 4 + 28 = 0 \\ \Rightarrow -2x + 2y + 24 = 0 \\ \Rightarrow 2x - 2y - 24 = 0 \end{gathered}$$

Therefore,

Then the length is reduced by unit and breadth is increased by units then the area is increased by square units

$$\begin{gathered} \left(x - 1\right) \left(y + 2\right) = xy + 33 \\ \Rightarrow xy + 2x - y - 2 = xy + 33 \\ \Rightarrow 2x - y -2 -33 = 0 \\ \Rightarrow 2x - y -35 = 0 \end{gathered}$$
 

Therefore, $2x - y - 35 = 0 .....\left(ii\right)$

Thus we get the following system of linear equation

$$\begin{gathered} 2x - 2y - 24 = 0 \\ 2x - y - 35 = 0 \end{gathered}$$

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

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

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.

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

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

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

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 _.

Answer:

Let , and

$$\begin{gathered} \angle C-\angle B=9^{°} \\ \Rightarrow \angle C=9^{°}+\angle B \\ \Rightarrow \angle C=9^{°}+3x^{°}-2^{°} \\ \Rightarrow \angle C=7^{°}+3x^{°} \end{gathered}$$

Substitute in above equation we get ,

$y^{°}=7^{°}+3x^{°}$

$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow x^{°}+ (3x^{°}-2^{°})+(7^{°}+3x^{°}) = {180}^{°} \\ \Rightarrow 7x^{°}+5^{°}={180}^{°} \\ \Rightarrow 7x^{°}={180}^{°}-5^{°}={175}^{°} \\ \Rightarrow x^{\circ }=\frac{{175}^{\circ }}{7^{\circ }}={25}^{\circ } \\ \angle A = x^{\circ } = {25}^{\circ } \\ \angle B = (3x-2{)}^{\circ } = 3({25}^{\circ })-2^{\circ }={73}^{\circ } \\ \angle C = (7^{°}+3x^{°}) = 7^{\circ }+3(25{)}^{\circ }={82}^{\circ } \\ \angle A={25}^{\circ }, \angle B={73}^{\circ }, \angle C={82}^{\circ } \\ \mathrm{Hence}, \mathrm{the} \mathrm{answer}. \end{gathered}$$

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 .

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 .

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

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.

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 _.

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

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.

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 .

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

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.

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 .

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.

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.

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 .

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.

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
$$\begin{gathered} y+\frac{y}{2}=30 \\ \Rightarrow \frac{3y}{2}=30 \\ \Rightarrow y=20 \end{gathered}$$
Putting y = 20 in (i), we get
x + 20 = 120
⇒ x = 100
Thus, Jayanti answered 100 questions correctly.

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.

Answer:

The given system of equations are

for unique solution

Here

By cross multiply we get

Hence, the correct choice is.

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 .

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.

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.

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.

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 .

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

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 .

Answer:

The given system of equation is

If then the equation have no solution.

By cross multiply we get

Hence, the correct choice is.

Answer:

The given system of equations are

For coincident lines , infinite number of solution

Option A.:

Option B:

$$\begin{gathered} 5a+b=0 \\ 5\left(-5\right)+\left(-1\right) = -25-1 = -26\ne 0 \end{gathered}$$

Option.C:

a - b = 0

-5 - (-1) = -4 $\ne$0

None of the option satisfies the values.

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.

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 .

Answer:

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is _.

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 .

Answer:

The given systems of equations are

If

Here

Hence, the correct choice is .

Answer:

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

Answer:

Given and

If We have plotting points as

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

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).

Answer:

The given equations are 
$$\begin{gathered} x-y=2 .....\left(1\right) \\ x+y=4 .....\left(2\right) \end{gathered}$$
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).

Answer:

The given system of equations is
$$\begin{gathered} 3x-y+8=0 \\ 6x-ky+16=0 \end{gathered}$$
We know that the lines 
$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$
are coincident iff
$\frac{a_1}{a_2}=\frac{{\displaystyle b_1}}{{\displaystyle b_2}}=\frac{{\displaystyle c_1}}{{\displaystyle c_2}}$
$$\begin{gathered} \therefore \frac{3}{6}=\frac{{\displaystyle -1}}{{\displaystyle -k}}=\frac{8}{16} \\ \Rightarrow \frac{1}{2}=\frac{1}{k}=\frac{1}{2} \\ \Rightarrow k=2 \end{gathered}$$
Thus, the value of k = 2.

Hence, the correct answer is option (c).

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.

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.

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).

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,
$$\begin{gathered} \frac{3}{2}=\frac{2k}{5}\ne \frac{-2}{1} \\ \Rightarrow \frac{3}{2}=\frac{2k}{5} \mathrm{and} \frac{2k}{5}\ne \frac{-2}{1} \\ \Rightarrow 15=4k \mathrm{and} 2k\ne -10 \\ \Rightarrow k=\frac{15}{4} \mathrm{and} k\ne -5 \end{gathered}$$

Hence, k = $\underline{\frac{15}{4}}.$


 

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.

 

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,
$$\begin{gathered} \frac{c}{6}=\frac{-1}{-2}=\frac{-2}{-3} \\ \Rightarrow \frac{c}{6}=\frac{1}{2}=\frac{2}{3} \\ \mathrm{But}, \frac{1}{2}\ne \frac{2}{3} \end{gathered}$$

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.

 

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.

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.

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,
$$\begin{gathered} \frac{\lambda }{2}=\frac{3}{6}=\frac{-7}{-14} \\ \Rightarrow \frac{\lambda }{2}=\frac{1}{2}=\frac{1}{2} \\ \Rightarrow \frac{\lambda }{2}=\frac{1}{2} \\ \Rightarrow \lambda =1 \end{gathered}$$

Hence, the pair of equations λx + 3y = 7, 2x + 6y = 14 will have infinitely many solutions for λ =  1 .

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,
$$\begin{gathered} \frac{2}{p}\ne \frac{3}{-6} \\ \Rightarrow \frac{2}{p}\ne \frac{-1}{2} \\ \Rightarrow 4\ne -p \\ \Rightarrow p\ne -4 \end{gathered}$$

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.

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,
$$\begin{gathered} \frac{3}{6}=\frac{-1}{-2}\ne \frac{-5}{-p} \\ \Rightarrow \frac{1}{2}=\frac{1}{2} \ne \frac{5}{p} \\ \Rightarrow \frac{1}{2} \ne \frac{5}{p} \\ \Rightarrow p\ne 10 \end{gathered}$$

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.
 

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.

$$\begin{gathered} a-b=2 ...\left(1\right) \\ a+b=4 ...\left(2\right) \\ \mathrm{Solving} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ a=3 \mathrm{and} b=1 \end{gathered}$$

Hence, a = 3 and b = 1.

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,
$$\begin{gathered} \frac{a}{3}=\frac{2}{b}\ne \frac{-7}{-16} \\ \Rightarrow \frac{a}{3}=\frac{2}{b}\ne \frac{7}{16} \\ \Rightarrow \frac{a}{3}=\frac{2}{b} \\ \Rightarrow ab=6 \end{gathered}$$

Hence, If the pair of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines, then ab = 6.

Answer:

The given equation is 4x + 3y – 12 = 0    ...(i)

Solving equation (i), we get

$$\begin{gathered} 4x+3y-12=0 \\ \Rightarrow 3y=12-4x \\ \Rightarrow y=\frac{12-4x}{3} \end{gathered}$$

When x = 0, y = 4
When x = 3, y = 0
 

Answer:

Given: 2^x+y = 2^x–y = $\sqrt{8}$

Now,
$$\begin{gathered} 2^{x+y}=\sqrt{8} \\ \Rightarrow 2^{x+y}=\sqrt{2^3} \\ \Rightarrow 2^{x+y}=2^{\frac{3}{2}} \\ \Rightarrow x+y=\frac{3}{2} ...\left(1\right) \\ {2}^{x-y}=\sqrt{8} \\ \Rightarrow 2^{x-y}=\sqrt{2^3} \\ \Rightarrow 2^{x-y}=2^{\frac{3}{2}} \\ \Rightarrow x-y=\frac{3}{2} ...\left(2\right) \\ \mathrm{Solving} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ x=\frac{3}{2} \mathrm{and} y=0 \end{gathered}$$


Hence, x = $\underline{\frac{3}{2}}$ and y = 0.

Answer:

The given system of equations are:
ax + 3y = 1     ...(i)
–12x + ay = 2     ...(ii)

Now, 
$$\begin{gathered} \frac{a_1}{a_2}=\frac{a}{-12} \\ \frac{b_1}{b_2}=\frac{3}{a} \\ \frac{c_1}{c_2}=\frac{-1}{-2}=\frac{1}{2} \\ \mathrm{If} \frac{a_1}{a_2}=\frac{b_1}{b_2} \\ \Rightarrow \frac{a}{-12}=\frac{3}{a} \\ \Rightarrow a^2=-36 \\ \mathrm{Which} \mathrm{is} \mathrm{not} \mathrm{possible} \mathrm{for} \mathrm{any} \mathrm{value} \mathrm{of} a. \\ \mathrm{Thus}, \frac{a_1}{a_2}\ne \frac{b_1}{b_2} \mathrm{for} \mathrm{all} \mathrm{real} \mathrm{values} \mathrm{of} a. \end{gathered}$$

Hence, the system of equations ax + 3y = 1, –12x + ay = 2 has a unique solution for all real values of a.



 

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.

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.

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.

Answer:

The given equations are

$$\begin{gathered} x+ky=0 \\ 2x-y=0 \\ {a}_1=1, a_2=2, b_1=k, b_2=-1 \\ \frac{a_1}{a_2}=\frac{1}{2}, \\ \frac{b_1}{b_2}=\frac{k}{-1} \end{gathered}$$
For unique solution

For all real values of k, except the equations have unique solutions.

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.

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.

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.

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.

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.

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

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.

Answer:

Gloria is walking the path joining and

Suresh is walking the path joining and

	0	4
	5	0

The graphical representations are

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.

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

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

For 4x + 2y = 300,

The solution table is

The graphical representation is as follows.

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.

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

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.

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.

Answer:

The given equations are:

x – y + 1 = 0    ...(i)
3x + 2y – 12 = 0     ...(ii)

Solving equation (i), we get

$$\begin{gathered} x-y+1=0 \\ \Rightarrow y=x+1 \end{gathered}$$

When x = 0, y = 1
When x = 1, y = 2
 

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

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

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.