NCERT Solutions for Class 10 Maths Chapter 3 Ex 3.4 | Updated 2026-27

Elimination Method:

Step 1: Write the two equations:

\[ x + y = 5 \quad \text{…(1)} \]
\[ 2x – 3y = 4 \quad \text{…(2)} \]

Step 2: Multiply equation (1) by 3 to make the coefficient of \( y \) equal in magnitude:

\[ 3x + 3y = 15 \quad \text{…(3)} \]

Step 3: Add equation (2) and equation (3) to eliminate \( y \):

\[ (2x – 3y) + (3x + 3y) = 4 + 15 \]
\[ 5x = 19 \]
\[ x = \frac{19}{5} \]

Step 4: Substitute \( x = \frac{19}{5} \) into equation (1):

\[ \frac{19}{5} + y = 5 \]
\[ y = 5 – \frac{19}{5} = \frac{25 – 19}{5} = \frac{6}{5} \]

Substitution Method:

Step 1: From equation (1): \( x = 5 – y \)

Step 2: Substitute into equation (2):

\[ 2(5 – y) – 3y = 4 \]
\[ 10 – 2y – 3y = 4 \]
\[ 10 – 5y = 4 \]
\[ 5y = 6 \]
\[ y = \frac{6}{5} \]

Step 3: \( x = 5 – \frac{6}{5} = \frac{19}{5} \)

Verification: \( \frac{19}{5} + \frac{6}{5} = \frac{25}{5} = 5 \) ✓ and \( 2 \times \frac{19}{5} – 3 \times \frac{6}{5} = \frac{38-18}{5} = \frac{20}{5} = 4 \) ✓

Elimination Method:

Step 1: Write the equations:

\[ 3x + 4y = 10 \quad \text{…(1)} \]
\[ 2x – 2y = 2 \quad \text{…(2)} \]

Step 2: Multiply equation (2) by 2:

\[ 4x – 4y = 4 \quad \text{…(3)} \]

Step 3: Add equations (1) and (3) to eliminate \( y \):

\[ (3x + 4y) + (4x – 4y) = 10 + 4 \]
\[ 7x = 14 \]
\[ x = 2 \]

Step 4: Substitute \( x = 2 \) into equation (2):

\[ 2(2) – 2y = 2 \]
\[ 4 – 2y = 2 \]
\[ 2y = 2 \]
\[ y = 1 \]

Substitution Method:

Step 1: From equation (2): \( 2x – 2y = 2 \Rightarrow x – y = 1 \Rightarrow x = 1 + y \)

Step 2: Substitute into equation (1):

\[ 3(1 + y) + 4y = 10 \]
\[ 3 + 3y + 4y = 10 \]
\[ 7y = 7 \]
\[ y = 1 \]

Step 3: \( x = 1 + 1 = 2 \)

Verification: \( 3(2) + 4(1) = 6 + 4 = 10 \) ✓ and \( 2(2) – 2(1) = 4 – 2 = 2 \) ✓

Step 1: Rewrite in standard form:

\[ 3x – 5y = 4 \quad \text{…(1)} \]
\[ 9x – 2y = 7 \quad \text{…(2)} \]

Elimination Method:

Step 2: Multiply equation (1) by 3:

\[ 9x – 15y = 12 \quad \text{…(3)} \]

Step 3: Subtract equation (2) from equation (3):

\[ (9x – 15y) – (9x – 2y) = 12 – 7 \]
\[ -13y = 5 \]
\[ y = -\frac{5}{13} \]

Step 4: Substitute into equation (1):

\[ 3x – 5\left(-\frac{5}{13}\right) = 4 \]
\[ 3x + \frac{25}{13} = 4 \]
\[ 3x = 4 – \frac{25}{13} = \frac{52 – 25}{13} = \frac{27}{13} \]
\[ x = \frac{9}{13} \]

Substitution Method:

Step 1: From equation (1): \( x = \frac{4 + 5y}{3} \)

Step 2: Substitute into equation (2):

\[ 9 \times \frac{4 + 5y}{3} – 2y = 7 \]
\[ 3(4 + 5y) – 2y = 7 \]
\[ 12 + 15y – 2y = 7 \]
\[ 13y = -5 \]
\[ y = -\frac{5}{13} \]

Step 3: \( x = \frac{4 + 5(-\frac{5}{13})}{3} = \frac{4 – \frac{25}{13}}{3} = \frac{\frac{27}{13}}{3} = \frac{9}{13} \)

Step 1: Clear fractions. Multiply equation (1) by 6:

\[ 3x + 4y = -6 \quad \text{…(A)} \]

Multiply equation (2) by 3:

\[ 3x – y = 9 \quad \text{…(B)} \]

Elimination Method:

Step 2: Subtract equation (B) from equation (A):

\[ (3x + 4y) – (3x – y) = -6 – 9 \]
\[ 5y = -15 \]
\[ y = -3 \]

Step 3: Substitute \( y = -3 \) into equation (B):

\[ 3x – (-3) = 9 \]
\[ 3x + 3 = 9 \]
\[ 3x = 6 \]
\[ x = 2 \]

Substitution Method:

Step 1: From equation (B): \( 3x = 9 + y \Rightarrow x = \frac{9 + y}{3} \)

Step 2: Substitute into equation (A):

\[ 3 \times \frac{9 + y}{3} + 4y = -6 \]
\[ 9 + y + 4y = -6 \]
\[ 5y = -15 \]
\[ y = -3 \]

Step 3: \( x = \frac{9 + (-3)}{3} = \frac{6}{3} = 2 \)

Verification: \( \frac{2}{2} + \frac{2(-3)}{3} = 1 – 2 = -1 \) ✓ and \( 2 – \frac{-3}{3} = 2 + 1 = 3 \) ✓

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \), if we only add 1 to the denominator. What is the fraction?

Step 1: Define variables. Let the fraction be \( \frac{x}{y} \), where \( x \) = numerator and \( y \) = denominator.

Step 2: Form equations.

Condition 1: Add 1 to numerator, subtract 1 from denominator → fraction = 1:

\[ \frac{x + 1}{y – 1} = 1 \]
\[ x + 1 = y – 1 \]
\[ x – y = -2 \quad \text{…(1)} \]

Condition 2: Add 1 to denominator → fraction = \( \frac{1}{2} \):

\[ \frac{x}{y + 1} = \frac{1}{2} \]
\[ 2x = y + 1 \]
\[ 2x – y = 1 \quad \text{…(2)} \]

Step 3: Elimination. Subtract equation (1) from equation (2):

\[ (2x – y) – (x – y) = 1 – (-2) \]
\[ x = 3 \]

Step 4: Substitute \( x = 3 \) into equation (1):

\[ 3 – y = -2 \]
\[ y = 5 \]

Verification: \( \frac{3+1}{5-1} = \frac{4}{4} = 1 \) ✓ and \( \frac{3}{5+1} = \frac{3}{6} = \frac{1}{2} \) ✓

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?

Step 1: Define variables. Let Nuri’s present age = \( x \) years and Sonu’s present age = \( y \) years.

Step 2: Form equations.

Five years ago: \( x – 5 = 3(y – 5) \)

\[ x – 5 = 3y – 15 \]
\[ x – 3y = -10 \quad \text{…(1)} \]

Ten years later: \( x + 10 = 2(y + 10) \)

\[ x + 10 = 2y + 20 \]
\[ x – 2y = 10 \quad \text{…(2)} \]

\[ (x – 2y) – (x – 3y) = 10 – (-10) \]
\[ y = 20 \]

Step 4: Substitute \( y = 20 \) into equation (2):

\[ x – 2(20) = 10 \]
\[ x = 50 \]

Verification: 5 years ago: Nuri = 45, Sonu = 15. \( 45 = 3 \times 15 \) ✓. 10 years later: Nuri = 60, Sonu = 30. \( 60 = 2 \times 30 \) ✓

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.

Step 1: Define variables. Let the tens digit = \( x \) and units digit = \( y \). The original number = \( 10x + y \). Reversed number = \( 10y + x \).

Step 2: Form equations.

Condition 1: Sum of digits = 9:

\[ x + y = 9 \quad \text{…(1)} \]

Condition 2: Nine times the number = twice the reversed number:

\[ 9(10x + y) = 2(10y + x) \]
\[ 90x + 9y = 20y + 2x \]
\[ 88x – 11y = 0 \]
\[ 8x – y = 0 \quad \text{…(2)} \]

Step 3: Elimination. Add equations (1) and (2):

\[ (x + y) + (8x – y) = 9 + 0 \]
\[ 9x = 9 \]
\[ x = 1 \]

Step 4: Substitute \( x = 1 \) into equation (1):

\[ 1 + y = 9 \]
\[ y = 8 \]

Verification: Number = 18. \( 1 + 8 = 9 \) ✓. \( 9 \times 18 = 162 \) and \( 2 \times 81 = 162 \) ✓

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.

Step 1: Define variables. Let the number of ₹50 notes = \( x \) and number of ₹100 notes = \( y \).

Step 2: Form equations.

Total notes = 25:

\[ x + y = 25 \quad \text{…(1)} \]

Total amount = ₹2000:

\[ 50x + 100y = 2000 \]
\[ x + 2y = 40 \quad \text{…(2)} \]

\[ (x + 2y) – (x + y) = 40 – 25 \]
\[ y = 15 \]

Step 4: Substitute \( y = 15 \) into equation (1):

\[ x + 15 = 25 \]
\[ x = 10 \]

Verification: \( 10 \times 50 + 15 \times 100 = 500 + 1500 = 2000 \) ✓ and \( 10 + 15 = 25 \) ✓

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.

Step 1: Define variables. Let the fixed charge for the first 3 days = ₹\( x \) and the charge for each extra day = ₹\( y \).

Step 2: Form equations.

Saritha kept the book for 7 days → extra days = \( 7 – 3 = 4 \) days:

\[ x + 4y = 27 \quad \text{…(1)} \]

Susy kept the book for 5 days → extra days = \( 5 – 3 = 2 \) days:

\[ x + 2y = 21 \quad \text{…(2)} \]

Step 3: Elimination. Subtract equation (2) from equation (1):

\[ (x + 4y) – (x + 2y) = 27 – 21 \]
\[ 2y = 6 \]
\[ y = 3 \]

Step 4: Substitute \( y = 3 \) into equation (2):

\[ x + 2(3) = 21 \]
\[ x + 6 = 21 \]
\[ x = 15 \]

Verification: Saritha: \( 15 + 4 \times 3 = 15 + 12 = 27 \) ✓. Susy: \( 15 + 2 \times 3 = 15 + 6 = 21 \) ✓