Mathematics Part i Solutions Solutions for Class 9 Maths Chapter 5 Linear Equations In Two Variables are provided here with simple step-by-step explanations. These solutions for Linear Equations In Two Variables are extremely popular among class 9 students for Maths Linear Equations In Two Variables Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 9 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 86:
Question 1:
By using variables x and y form any five linear equations in two variables.
Answer:
Linear equations in two variables x and y will be
Page No 86:
Question 2:
Write five solutions of the equation x + y = 7.
Answer:
The given equation is x + y = 7.
The possible solution are
x | 0 | 7 | 1 | 6 | 4 |
y | 7 | 0 | 6 | 1 | 3 |
Page No 86:
Question 3:
(i) x + y = 4 ; 2x 5y = 1
(ii) 2x + y = 5; 3x y = 5
(iv) 2yx =0; 10x + 15y = 105
(v) 2x + 3y + 4 = 0; x 5y = 11
(vi) 2x 7y = 7; 3x + y = 22
Answer:
2x 5y = 1 .....(II)
From (I) we have
Putting this value of x in (II)
Putting the value of y in (I) we have
Thus,
(ii) 2x + y = 5 .....(I)
3x y = 5 .....(II)
Adding (I) and (II)
Putting this value in (I) we have
x 3y = 8 .....(II)
Multiply (II) with 3
.....(III)
(I) (III)
10x + 15y = 105 .....(II)
Taking out 5 common from (II)
.....(III)
From (I) and (III)
(v) 2x + 3y + 4 = 0 .....(I)
x 5y = 11 .....(II)
From (I) we have
2x + 3y = 4 .....(III)
Multiply (II) with 2
.....(IV)
From (III) and (IV) we have
2x + 3y = 4
(III) (IV)
x = 11 + 5y
x = 1
(vi) 2x 7y = 7 .....(I)
3x + y = 22 .....(II)
Multiply (II) with 7
21x + 7y = 154 .....(III)
(I) + (III)
23x = 161
x = 7, y = 1
Page No 90:
Question 1:
In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.
Answer:
Disclaimer: There is an error in the given question. Instead of "Number of 5 rupee notes are less by 10 than number of 10 rupee notes"
the sentence should have been "Number of 5 rupee notes are greater by 10 than number of 10 rupee notes".
Let the number of Rs 5 notes be x and the number of Rs 10 notes be y.
Number of 5 rupee notes are greater by 10 than number of 10 rupee notes
So,
.....(II)
From (I) and (II) we have
Thus, x = 10 + 20 = 30
Number of Rs 5 notes = 30, number of Rs 10 notes = 20
Page No 90:
Question 2:
The denominator of a fraction is 1 more than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 1 : 2. Find the fraction.
Answer:
Disclaimer: There is an error in the given question. With the given statements, the two linear equations formed will be the same.
Page No 90:
Question 3:
The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today's ages.
Answer:
Let the age of Priyanka be x years and that of Deepika be y years.
The sum of ages of Priyanka and Deepika is 34 years.
.....(I)
Priyanka is elder to Deepika by 6 years.
.....(II)
Adding (I) and (II) we have
Putting the value of x in (I) we have
y = 14
Thus, Priyanka's age = 20 years and Deepika's age = 14 years.
Page No 90:
Question 4:
The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is140. Then find the number of lions and peacocks in the zoo.
Answer:
Let the number of lions be x and the number of peacocks be y.
The total number of lions and peacocks in a certain zoo is 50.
So, .....(I)
The total number of their legs is140.
Lion has 4 legs while peacock has 2.
Subtracting (II) from (I) we have
Thus, the number of lions = 20 and the number of peacocks = 30.
Page No 90:
Question 5:
Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment.
Answer:
Let the fixed monthly income be Rs x.
Annual increament be Rs y.
After 4 years, his monthly salary was Rs. 4500
Monthly salary + annual increament of 4 years = 4500
.....(I)
After 10 years his monthly salary became 5400 rupees
Monthly salary + annual increament of 10 years = 5400
.....(II)
(II) (I)
6y = 900
y = 150
x = 4500 150 4 = 3900
Thus, the monthly salary = Rs 3900
Annual increament = Rs 150
Page No 90:
Question 6:
The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the price of 2 chairs and 2 tables.
Answer:
Let the price of 1 chair be Rs x and the price of 1 table be Rs y.
The price of 3 chairs and 2 tables is 4500 rupees
Price of 5 chairs and 3 tables is 7000 rupees
Multiply (I) with 3 and (II) with 2
Price of 2 chairs and 2 tables will be
Thus, the price of 2 chairs and 2 tables will be Rs 4000
Page No 91:
Question 7:
The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Answer:
Let the number at the ten's place be x and that at the one's place be y.
Number obtained = 10x + y
The sum of the digits in a two-digits number is 9.
.....(I)
The number obtained by interchanging the digits exceeds the original number by 27.
Adding (I) and (II)
x = 3
Thus, the number is 36.
Page No 91:
Question 8:
Answer:
A = B + C .....(I)
We know by angle sum property,
A + B + C = 180
From (I)
B + C +B + C = 180
And A = B + C = .
Page No 91:
Question 9:
Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to of the larger part. Then find the length of the larger part.
Answer:
Let the smaller part be x cm and larger part be y cm.
.....(I)
Twice the length of the smaller part is equal to of the larger part.
Putting the value of y in (I) we have
And thus,
Thus, the larger part = 420 cm and smaller part is 140 cm.
Page No 91:
Question 10:
In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong ?
Answer:
Let the number of correct answers be x and the number of incorrect answers be y.
Total answers = Number of correct + number of incorrect = 60
.....(I)
Total marks obtained = 90
.....(II)
Adding (I) and (II) we have
Number of questions he got wrong is 10.
Page No 91:
Question 1:
(i) If 3x + 5y = 9 and 5x + 3y= 7 then What is the value of x + y ?
(ii) 'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.' What is the mathematical form of the statement ?
(iii) Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's age ?
Answer:
(i) 3x + 5y = 9 .....(I)
5x + 3y= 7 .....(II)
Adding (I) and (II) we have
Hence, the correct answer is option (A).
(ii) Let the length be x and the breadth be y.
Perimeter of the rectangle = 2(l + b) = 26
2(l + b) = 26
Hence, the correct answer is option (C).
(iii) Let Ajay's age be x years and Vijay's age be y years.
Ajay is younger than Vijay by 5 years.
Sum of their ages is 25 years.
Adding (I) and (II) we have
Thus, Ajay's age is 10 years.
Hence, the correct answer is option (C).
Page No 91:
Question 2:
(i) 2x + y = 5 ; 3x y = 5
(ii) x 2 y = 1 ; 2x y = 7
(iv) 2x + y = 2 ; 3x y = 7
(vi) x 2y = 2 ; x + 2y = 10
Answer:
3x y = 5 .....(II)
(I) + (II)
Putting the value of x in (I)
(ii) x 2y = 1 .....(I)
2x y = 7 .....(II)
Multiply (I) with 2
.....(III)
Subtracting (III) from (II)
Putting the value of y in (I) we get
2x 3y = 7 .....(II)
Multiply (I) with 3
.....(III)
Adding (II) and (III)
Putting the value of x in (I)
(iv) 2x + y = 2 .....(I)
3x y = 7 .....(II)
Adding (I) with (II)
Putting the value of x in (I)
3x + 2y = 11 .....(II)
Multiplying (I) with 2 we get
.....(III)
Adding (II) with (III)
Putting the value of x in (I) we get
(vi) x 2y = 2 .....(I)
x + 2y = 10 .....(II)
Adding (I) with (II)
Putting the value of x in (I) we get
Page No 91:
Question 3:
(i) 3x 4 y = 7; 5x + 2y = 3
(ii) 5x + 7 y = 17 ; 3x 2y = 4
(iii) x 2y = 10; 3x 5y = 12
(iv) 4x + y = 34 ; x + 4y = 16
Answer:
5x + 2y = 3 .....(II)
Multiply (II) with 2
10x + 4y = 6 .....(III)
Adding (I) and (III)
Putting the value of x in (I) we get
Thus, x = 1 and .
(ii) 5x + 7 y = 17 .....(I)
3x 2y = 4 .....(II)
Multiply (I) with 3 and (II) with 5
15x + 21 y = 51 .....(III)
15x 10y = 20 .....(IV)
Subtracting (IV) from (III) we get
Putting this value of y in (I) we get
Thus, .
(iii) x 2y = 10 .....(I)
3x 5y = 12 .....(II)
Multiply (I) with 3
.....(III)
Subtracting (II) from (III) we get
Putting the value of y in (I) we get
Thus,
(iv) 4x + y = 34 .....(I)
x + 4y = 16 .....(II)
Adding (I) and (II) we get
Subtracting (II) from (I) we get
(III) + (IV)
Putting the value of x in (I) we get
Thus,
Page No 91:
Question 4:
Solve the following simultaneous equations.
(i)
(ii)
(iii)
Answer:
(i) .....(I)
.....(II)
Multiplying (I) with LCM of 3 and 4 which is 12 we get
.....(III)
Multiplying (II) with LCM of 2 and 4 which is 4 we get
.....(IV)
Multiplying (IV) with 2
.....(V)
Subtracting (V) from (III)
Putting this value of y in (IV) we get
(ii)
Multiplying (III) with 4 we get
.....(V)
Subtracting (IV) from (V)
Putting the value of y in (IV) we get
(iii)
Let
So, the equations obtained are
Page No 92:
Question 5:
A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Answer:
Let the two digit number be 10x + y.
A two digit number is 3 more than 4 times the sum of its digits.
If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits.
(I) (II)
Thus, the number obtained is 35.
Page No 92:
Question 6:
The total cost of 6 books and 7 pens is 79 rupees and the total cost of 7 books and 5 pens is 77 rupeess. Find the cost of 1 book and 2 pens.
Answer:
Disclaimer: There is an error in the given question. Instead of 7 books and 5 pens there should be 7 books and 6 pens.
Let the cost of 1 book be x and the cost of 1 pen be y.
The total cost of 6 books and 7 pens is 79 rupees.
Total cost of 7 books and 6 pens is 77 rupeess.
Adding (I) and (II)
Subtracting (II) from (I)
Adding (III) and (IV)
y = 7
Cost of 1 book and 2 pens is 5 + 14 = Rs 19.
Page No 92:
Question 7:
Answer:
Let the incomes of the two persons be x and y.
Savings of the two persons = Rs 200
Expenses = Income Savings
Multiply (I) with 3 and (II) with 7.
Putting the value y in (I) we get
x = 1800
Thus, the income of the two persons is Rs 1800 and Rs 1400.
Page No 92:
Question 8:
If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Answer:
Let the length be l and the breadth be b.
Area of rectangle = lb
Length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units.
If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units.
Multiply (I) with 2 and (II) with 3
Putting this value in (I) we get
Thus, length = 344 units and breadth = 205 units.
Page No 92:
Question 9:
Answer:
Let the speed of the faster car be x km/h and that of the slower car be y km/h.
Total distance between them is 70 km.
When both cars travel in the same direction, they meet in 7 hours.
Total speed = x − y
When both cars move towards each other,
Total speed = x + y
Adding (I) and (II)
Putting the value of x in (I)
Thus, speed of the cars are 40 km/h and 30 km/h.
Page No 92:
Question 10:
The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Answer:
Let the two digit number be 10x + y.
Number obtained on interchanging the digits = 10y + x.
The sum of a two digit number and the number obtained by interchanging its digits is 99.
10x + y + 10y + x = 99
With the given information, only one equation can be formed. So, the number can be
18, 81, 54, 45, 27, 72, 36, 63.
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