NCERT Solutions for Class 10 Maths Ex 3.3

# NCERT Solutions for Class 10 Maths Ex 3.3

In this post, you will find the NCERT solutions for class 10 maths ex 3.3. These solutions are based on the latest syllabus of NCERT Maths class 10.

## NCERT Solutions for Class 10 Maths Ex 3.3

1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4                   (ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3 5– 4 = 0 and 9= 2+ 7       (iv)

Solution:  (i) x + y = 5 … (1)

2x – 3y = 4 … (2)

Elimination Method:

Multiplying equation (1) by 3, we get

3+ 3= 15 … (3)

2 3= 4 … (2)

Adding equations (2) and (3), we get

5= 19

19/5

Putting the value of x in equation (1), we get

19/5 + y = 5

= 5  19/5 = 6/5

Therefore, = 19/5 and = 6/5.

Substitution Method:

= 5 … (1)

2 3= 4 … (2)

From equation (1), we get

= 5  y

Putting this value of x in equation (2), we get

2(5  y 3= 4

10  2 3= 4

5= 6

= 6/5

Putting the value of y in equation (1), we get

= 5  6/5 = 19/5

Therefore, = 19/5 and = 6/5.

(ii) 3x + 4y = 10 … (1)

2x – 2y = 2 … (2)

Elimination Method:

Multiplying equation (2) by 2, we get

4 4= 4 … (3)

3+ 4= 10 … (1)

Adding equations (3) and (1), we get

7= 14

= 2

Putting the value of x in equation (1), we get

3(2) + 4= 10

4= 10 – 6 = 4

= 1

Therefore, = 2 and = 1.

Substitution Method:

3+ 4= 10 … (1)

2 2= 2 … (2)

From equation (2), we get

2= 2 + 2y

= 1 + y … (3)

Putting this value of x in equation (1), we get

3(1 + y) + 4= 10

3 + 3+ 4= 10

7= 7

= 1

Putting the value of y in equation (3), we get

= 1 + 1 = 2

= 2

Therefore, = 2 and = 1.

(iii) 3 5– 4 = 0 … (1)

9= 2+ 7… (2)

Elimination Method:

Multiplying equation (1) by 3, we get

9 15– 12 = 0 … (3)

Equation (2) can be written as

9 2– 7 = 0 … (2)

Subtracting equation (2) from equation (3), we get

13– 5 = 0

13= 5

−5/13

Putting the value of y in equation (1), we get

3– 5(−5/13)  4 = 0

3= 4 – 25/13 = (52 – 25)/13 = 27/13

= 27/39 = 9/13

Therefore, = 9/13 and −5/13.

Substitution Method:

3 5– 4 = 0 … (1)

9= 2+ 7… (2)

Equation (2) can be written as

9– 2= 7… (2)

From equation (1), we have

3= 4 + 5y

= (4 + 5y)/3

Putting this value of x in equation (2), we get

9(4 + 5y)/3  2= 7

12 + 15 2= 7

135

5/13

Putting the value of y in equation (1), we get

3– 5(5/13) = 4

3= 4  25/13 = (52 – 25)/13 = 27/13

= 27/39 = 9/13

Therefore, = 9/13 and 5/13.

(iv)

Elimination Method:

Multiplying equation (2) by 2, we get

Adding equations (3) and (1), we get

5x/2 = 5

= 2

Putting the value of x in equation (2), we get

y/3 = 3

3

Therefore, = 2 and 3.

Substitution Method:

From equation (2), we have

Putting this value of x in equation (1), we get

5+ 9 = 6

515

3

Putting the value of y in equation (1), we get

= 2

Therefore, = 2 and 3.

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw  2000. She asked the cashier to give her  50 and  100 notes only. Meena got 25 notes in all. Find how many notes of  50 and  100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid  27 for a book kept for seven days, while Susy paid  21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:  (i) Let the numerator of the required fraction be and the denominator be y.

According to the given conditions, we have

+ 1 = – 1 and 2+ 1

– … (1)

and 2– = 1… (2)

Subtracting equation (1) from equation (2), we get

x = 3

Putting the value of x in equation (1), we get

3 – y = − 2

5

= 5

Therefore, the required fraction is 3/5.

(ii) Let the present age of Nuri be x years and the present age of Sonu be y years.

5 years ago, the age of Nuri = (x – 5) years

5 years ago, the age of Sonu = (y – 5) years

According to the first condition, we have

( 5) = 3( 5)

– 5 = 3– 15

310 …… (1)

10 years later, the age of Nuri = (+ 10) years

10 years later, the age of Sonu = (+ 10) years

According to the second condition, we have

(+ 10) = 2(+ 10)

+ 10 = 2+ 20

2= 10 …… (2)

Subtracting equation (1) from equation (2), we get

= 10  (10) = 20 years

Putting the value of y in equation (1), we get

– 3(20) = 10

– 60 = 10

= 50 years

Therefore, the present age of Nuri is 50 years and the present age of Sonu is 20 years.

(iii) Let the digit at the tens place be and the digit at the ones place be y.

Number = 10x + y

Number obtained by reversing the order of the digits = 10y + x

According to the first condition, we have

= 9 …… (1)

According to the second condition, we have

9 × Number = 2 × Number obtained by reversing the order of the digits

9(10y) = 2(10x)

90+ 9= 20+ 2x

88= 11y

8y

8– = 0 …… (2)

Adding equations (1) and (2), we get

9= 9

= 1

Putting the value of x in equation (1), we get

1 + = 9

= 9 – 1 = 8

Therefore, the number = 10= 10(1) + 8 = 10 + 8 = 18

(iv) Let the number of  100 notes be and the number of  50 notes be y.

According to the given conditions, we have

= 25 …… (1)

and 100+ 50= 2000

2= 40 …… (2)

Subtracting equation (2) from equation (1), we get

15

= 15

Putting the value of x in equation (1), we get

15 + = 25

= 25 – 15 = 10

Therefore, the number of  100 notes is 15 and the number of  50 notes is 10.

(v) Let the fixed charge for 3 days be  and the additional charge for each day thereafter be  y.

According to given condition, we have

+ 4= 27  …… (1)

+ 2= 21  …… (2)

Subtracting equation (2) from equation (1), we get

2= 6

= 3

Putting the value of y in equation (1), we get

+ 4(3) = 27

= 27 – 12 = 15

Therefore, the fixed charge for 3 days is  15 and the additional charge for each day thereafter is  3.