TG 10th Pair of Linear Equations

10th Maths Pair of Linear Equations In Two Variables Exercise 4.3 Solutions

zaheerpashamd

Exercise 4.3 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on solving pairs of equations by reducing them to linear equations and solving word problems. Mathematical expressions are rendered using MathJax.

1. Solve each of the following pairs of equations by reducing them to a pair of linear equations.

(i) \( \frac{5}{x-1} + \frac{1}{y-2} = 2 \), \( \frac{6}{x-1} – \frac{3}{y-2} = 1 \)

Substitute \( u = \frac{1}{x-1} \), \( v = \frac{1}{y-2} \).

Rewrite equations: \( 5u + v = 2 \), \( 6u – 3v = 1 \).

Multiply the first by 3: \( 15u + 3v = 6 \).

Add to the second: \( (15u + 3v) + (6u – 3v) = 6 + 1 \implies 21u = 7 \implies u = \frac{1}{3} \).

Substitute \( u = \frac{1}{3} \) into \( 5u + v = 2 \): \( 5 \left(\frac{1}{3}\right) + v = 2 \implies \frac{5}{3} + v = 2 \implies v = 2 – \frac{5}{3} = \frac{1}{3} \).

Solve for \( x \), \( y \): \( u = \frac{1}{x-1} = \frac{1}{3} \implies x – 1 = 3 \implies x = 4 \).

\( v = \frac{1}{y-2} = \frac{1}{3} \implies y – 2 = 3 \implies y = 5 \).

Check: \( \frac{5}{4-1} + \frac{1}{5-2} = \frac{5}{3} + \frac{1}{3} = 2 \), \( \frac{6}{4-1} – \frac{3}{5-2} = \frac{6}{3} – \frac{3}{3} = 1 \), both true.

Solution: \( (x, y) = (4, 5) \)

(ii) \( \frac{x+y}{xy} = 2 \), \( \frac{x-y}{xy} = 6 \)

Simplify: First equation: \( \frac{x+y}{xy} = \frac{1}{y} + \frac{1}{x} = 2 \).

Second equation: \( \frac{x-y}{xy} = \frac{1}{y} – \frac{1}{x} = 6 \).

Let \( u = \frac{1}{x} \), \( v = \frac{1}{y} \). Then: \( u + v = 2 \), \( -u + v = 6 \).

Add the equations: \( (u + v) + (-u + v) = 2 + 6 \implies 2v = 8 \implies v = 4 \).

Substitute \( v = 4 \) into \( u + v = 2 \): \( u + 4 = 2 \implies u = -2 \).

Solve: \( u = \frac{1}{x} = -2 \implies x = -\frac{1}{2} \), \( v = \frac{1}{y} = 4 \implies y = \frac{1}{4} \).

Check: \( \frac{-\frac{1}{2} + \frac{1}{4}}{-\frac{1}{2} \cdot \frac{1}{4}} = \frac{-\frac{1}{4}}{-\frac{1}{8}} = 2 \), \( \frac{-\frac{1}{2} – \frac{1}{4}}{-\frac{1}{8}} = \frac{-\frac{3}{4}}{-\frac{1}{8}} = 6 \), both true.

Solution: \( (x, y) = \left(-\frac{1}{2}, \frac{1}{4}\right) \)

(iii) \( \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \), \( \frac{4}{\sqrt{x}} – \frac{9}{\sqrt{y}} = -1 \)

Substitute \( u = \frac{1}{\sqrt{x}} \), \( v = \frac{1}{\sqrt{y}} \). Then \( u^2 = \frac{1}{x} \), \( v^2 = \frac{1}{y} \).

Rewrite: \( 2u + 3v = 2 \), \( 4u – 9v = -1 \).

Multiply the first by 3: \( 6u + 9v = 6 \).

Add to the second: \( (6u + 9v) + (4u – 9v) = 6 – 1 \implies 10u = 5 \implies u = \frac{1}{2} \).

Substitute \( u = \frac{1}{2} \) into \( 2u + 3v = 2 \): \( 2 \left(\frac{1}{2}\right) + 3v = 2 \implies 1 + 3v = 2 \implies 3v = 1 \implies v = \frac{1}{3} \).

Solve: \( u = \frac{1}{\sqrt{x}} = \frac{1}{2} \implies \sqrt{x} = 2 \implies x = 4 \).

\( v = \frac{1}{\sqrt{y}} = \frac{1}{3} \implies \sqrt{y} = 3 \implies y = 9 \).

Check: \( \frac{2}{\sqrt{4}} + \frac{3}{\sqrt{9}} = 1 + 1 = 2 \), \( \frac{4}{\sqrt{4}} – \frac{9}{\sqrt{9}} = 2 – 3 = -1 \), both true.

Solution: \( (x, y) = (4, 9) \)

(iv) \( 6x + 3y = 6xy \), \( 2x + 4y = 5xy \)

Divide the first by \( xy \): \( \frac{6x}{xy} + \frac{3y}{xy} = \frac{6xy}{xy} \implies \frac{6}{y} + \frac{3}{x} = 6 \).

Divide the second by \( xy \): \( \frac{2}{y} + \frac{4}{x} = 5 \).

Let \( u = \frac{1}{x} \), \( v = \frac{1}{y} \). Then: \( 3u + 6v = 6 \implies u + 2v = 2 \), \( 4u + 2v = 5 \).

Subtract: \( (4u + 2v) – (u + 2v) = 5 – 2 \implies 3u = 3 \implies u = 1 \).

Substitute \( u = 1 \) into \( u + 2v = 2 \): \( 1 + 2v = 2 \implies 2v = 1 \implies v = \frac{1}{2} \).

Solve: \( u = \frac{1}{x} = 1 \implies x = 1 \), \( v = \frac{1}{y} = \frac{1}{2} \implies y = 2 \).

Check: \( 6(1) + 3(2) = 6 \cdot 1 \cdot 2 \implies 12 = 12 \), \( 2(1) + 4(2) = 5 \cdot 1 \cdot 2 \implies 10 = 10 \), both true.

Solution: \( (x, y) = (1, 2) \)

(v) \( \frac{5}{x+y} – \frac{2}{x-y} = -1 \), \( \frac{15}{x+y} + \frac{7}{x-y} = 10 \)

Substitute \( u = \frac{1}{x+y} \), \( v = \frac{1}{x-y} \).

Rewrite: \( 5u – 2v = -1 \), \( 15u + 7v = 10 \).

Multiply the first by 7 and the second by 2: \( 35u – 14v = -7 \), \( 30u + 14v = 20 \).

Add: \( 65u = 13 \implies u = \frac{1}{5} \).

Substitute \( u = \frac{1}{5} \) into \( 5u – 2v = -1 \): \( 5 \left(\frac{1}{5}\right) – 2v = -1 \implies 1 – 2v = -1 \implies 2v = 2 \implies v = 1 \).

Solve: \( u = \frac{1}{x+y} = \frac{1}{5} \implies x + y = 5 \).

\( v = \frac{1}{x-y} = 1 \implies x – y = 1 \).

Add the resulting equations: \( 2x = 6 \implies x = 3 \), \( y = 5 – 3 = 2 \).

Check: \( \frac{5}{3+2} – \frac{2}{3-2} = 1 – 2 = -1 \), \( \frac{15}{5} + \frac{7}{1} = 3 + 7 = 10 \), both true.

Solution: \( (x, y) = (3, 2) \)

(vi) \( \frac{2}{x} + \frac{3}{y} = 13 \), \( \frac{5}{x} – \frac{4}{y} = -2 \)

Substitute \( u = \frac{1}{x} \), \( v = \frac{1}{y} \).

Rewrite: \( 2u + 3v = 13 \), \( 5u – 4v = -2 \).

Multiply the first by 4 and the second by 3: \( 8u + 12v = 52 \), \( 15u – 12v = -6 \).

Add: \( 23u = 46 \implies u = 2 \).

Substitute \( u = 2 \) into \( 2u + 3v = 13 \): \( 2(2) + 3v = 13 \implies 4 + 3v = 13 \implies 3v = 9 \implies v = 3 \).

Solve: \( u = \frac{1}{x} = 2 \implies x = \frac{1}{2} \), \( v = \frac{1}{y} = 3 \implies y = \frac{1}{3} \).

Check: \( \frac{2}{\frac{1}{2}} + \frac{3}{\frac{1}{3}} = 4 + 9 = 13 \), \( \frac{5}{\frac{1}{2}} – \frac{4}{\frac{1}{3}} = 10 – 12 = -2 \), both true.

Solution: \( (x, y) = \left(\frac{1}{2}, \frac{1}{3}\right) \)

(vii) \( \frac{10}{x+y} + \frac{2}{x-y} = 4 \), \( \frac{15}{x+y} – \frac{5}{x-y} = -2 \)

Substitute \( u = \frac{1}{x+y} \), \( v = \frac{1}{x-y} \).

Rewrite: \( 10u + 2v = 4 \implies 5u + v = 2 \), \( 15u – 5v = -2 \).

Multiply the first by 5: \( 25u + 5v = 10 \).

Add to the second: \( (25u + 5v) + (15u – 5v) = 10 – 2 \implies 40u = 8 \implies u = \frac{1}{5} \).

Substitute \( u = \frac{1}{5} \) into \( 5u + v = 2 \): \( 5 \left(\frac{1}{5}\right) + v = 2 \implies 1 + v = 2 \implies v = 1 \).

Solve: \( u = \frac{1}{x+y} = \frac{1}{5} \implies x + y = 5 \), \( v = \frac{1}{x-y} = 1 \implies x – y = 1 \).

Solve the linear system: \( x = 3 \), \( y = 2 \) (same as in (v), confirming consistency).

Solution: \( (x, y) = (3, 2) \)

(viii) \( \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \), \( \frac{1}{2(3x+y)} – \frac{1}{2(3x-y)} = \frac{-1}{8} \)

Substitute \( u = 3x + y \), \( v = 3x – y \). Then \( u + v = 6x \), \( u – v = 2y \).

First equation: \( \frac{1}{u} + \frac{1}{v} = \frac{3}{4} \implies \frac{u+v}{uv} = \frac{3}{4} \implies 4(u + v) = 3uv \).

Second equation: \( \frac{1}{2u} – \frac{1}{2v} = \frac{-1}{8} \implies \frac{v – u}{2uv} = \frac{-1}{8} \implies v – u = -\frac{uv}{4} \).

Let \( p = \frac{1}{u} \), \( q = \frac{1}{v} \). Then: \( p + q = \frac{3}{4} \), \( \frac{q – p}{2} = \frac{-1}{8} \implies q – p = \frac{-1}{4} \).

Solve: Add the equations: \( 2q = \frac{3}{4} – \frac{1}{4} = \frac{1}{2} \implies q = \frac{1}{4} \). Then \( p = \frac{3}{4} – \frac{1}{4} = \frac{1}{2} \).

Solve: \( p = \frac{1}{u} = \frac{1}{2} \implies u = 2 \), \( q = \frac{1}{v} = \frac{1}{4} \implies v = 4 \).

Then: \( u + v = 6x \implies 6x = 2 + 4 = 6 \implies x = 1 \).

\( u – v = 2y \implies 2 – 4 = 2y \implies -2 = 2y \implies y = -1 \).

Check: \( 3x + y = 3(1) – 1 = 2 \), \( 3x – y = 3 + 1 = 4 \). First: \( \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \), second: \( \frac{1}{4} – \frac{1}{8} = \frac{1}{8} \), so \( \frac{1}{8} = -\left(-\frac{1}{8}\right) \), true.

Solution: \( (x, y) = (1, -1) \)

2. Formulate the following problems as a pair of equations and then find their solutions.

(i) A 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 the stream and that of the boat in still water.

Let the speed of the boat in still water be \( x \) km/h, speed of the stream be \( y \) km/h.

Upstream speed: \( x – y \), downstream speed: \( x + y \).

First condition: \( \frac{30}{x-y} + \frac{44}{x+y} = 10 \).

Second condition: \( \frac{40}{x-y} + \frac{55}{x+y} = 13 \).

Substitute \( u = \frac{1}{x-y} \), \( v = \frac{1}{x+y} \).

Rewrite: \( 30u + 44v = 10 \), \( 40u + 55v = 13 \).

Multiply the first by 5 and the second by 4: \( 150u + 220v = 50 \), \( 160u + 220v = 52 \).

Subtract: \( 160u – 150u = 52 – 50 \implies 10u = 2 \implies u = \frac{1}{5} \).

Substitute \( u = \frac{1}{5} \) into \( 30u + 44v = 10 \): \( 30 \left(\frac{1}{5}\right) + 44v = 10 \implies 6 + 44v = 10 \implies 44v = 4 \implies v = \frac{1}{11} \).

Solve: \( u = \frac{1}{x-y} = \frac{1}{5} \implies x – y = 5 \), \( v = \frac{1}{x+y} = \frac{1}{11} \implies x + y = 11 \).

Add: \( 2x = 16 \implies x = 8 \), \( y = 11 – 8 = 3 \).

Check: First: \( \frac{30}{5} + \frac{44}{11} = 6 + 4 = 10 \), second: \( \frac{40}{5} + \frac{55}{11} = 8 + 5 = 13 \), both true.

Speed of boat: 8 km/h, Speed of stream: 3 km/h

(ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels 200 km by train and rest by car. Find the speed of the train and the car.

Let the speed of the train be \( x \) km/h, speed of the car be \( y \) km/h.

First condition: 120 km by train, 480 km by car, time = 8 hours: \( \frac{120}{x} + \frac{480}{y} = 8 \).

Second condition: 200 km by train, 400 km by car, time = 8 hours 20 minutes = \( 8 + \frac{20}{60} = \frac{25}{3} \) hours: \( \frac{200}{x} + \frac{400}{y} = \frac{25}{3} \).

Substitute \( u = \frac{1}{x} \), \( v = \frac{1}{y} \).

Rewrite: \( 120u + 480v = 8 \implies 15u + 60v = 1 \), \( 200u + 400v = \frac{25}{3} \implies 24u + 48v = 1 \).

Multiply the first by 4 and subtract the second: \( (60u + 240v) – (24u + 48v) = 4 – 1 \implies 36u + 192v = 3 \).

Simplify: \( 3u + 16v = \frac{1}{4} \). Multiply the second original by 3: \( 72u + 144v = 3 \implies u + 2v = \frac{1}{24} \).

Solve: \( 3u + 16v = \frac{1}{4} \), \( u + 2v = \frac{1}{24} \). Multiply the second by 3: \( 3u + 6v = \frac{1}{8} \).

Subtract: \( (3u + 16v) – (3u + 6v) = \frac{1}{4} – \frac{1}{8} \implies 10v = \frac{1}{8} \implies v = \frac{1}{80} \).

Substitute \( v = \frac{1}{80} \) into \( u + 2v = \frac{1}{24} \): \( u + 2 \left(\frac{1}{80}\right) = \frac{1}{24} \implies u + \frac{1}{40} = \frac{1}{24} \implies u = \frac{1}{24} – \frac{1}{40} = \frac{5-3}{120} = \frac{1}{60} \).

Solve: \( u = \frac{1}{x} = \frac{1}{60} \implies x = 60 \), \( v = \frac{1}{y} = \frac{1}{80} \implies y = 80 \).

Check: First: \( \frac{120}{60} + \frac{480}{80} = 2 + 6 = 8 \), second: \( \frac{200}{60} + \frac{400}{80} = \frac{10}{3} + 5 = \frac{25}{3} \), both true.

Speed of train: 60 km/h, Speed of car: 80 km/h

(iii) 2 women and 5 men can together finish an embroidery work in 4 days while 3 women and 6 men can finish it in 3 days. Find the time to be taken by 1 woman alone and 1 man alone to finish the work.

Let 1 woman finish the work in \( w \) days, 1 man in \( m \) days.

Work rate of 1 woman = \( \frac{1}{w} \), 1 man = \( \frac{1}{m} \).

First condition: 2 women and 5 men in 4 days: \( 4 \left( \frac{2}{w} + \frac{5}{m} \right) = 1 \implies \frac{2}{w} + \frac{5}{m} = \frac{1}{4} \).

Second condition: 3 women and 6 men in 3 days: \( 3 \left( \frac{3}{w} + \frac{6}{m} \right) = 1 \implies \frac{3}{w} + \frac{6}{m} = \frac{1}{3} \).

Substitute \( u = \frac{1}{w} \), \( v = \frac{1}{m} \).

Rewrite: \( 2u + 5v = \frac{1}{4} \), \( 3u + 6v = \frac{1}{3} \).

Multiply the first by 3 and the second by 2: \( 6u + 15v = \frac{3}{4} \), \( 6u + 12v = \frac{2}{3} \).

Subtract: \( 15v – 12v = \frac{3}{4} – \frac{2}{3} \implies 3v = \frac{9-8}{12} = \frac{1}{12} \implies v = \frac{1}{36} \).

Substitute \( v = \frac{1}{36} \) into \( 2u + 5v = \frac{1}{4} \): \( 2u + 5 \left(\frac{1}{36}\right) = \frac{1}{4} \implies 2u + \frac{5}{36} = \frac{1}{4} \implies 2u = \frac{1}{4} – \frac{5}{36} = \frac{9-5}{36} = \frac{1}{9} \implies u = \frac{1}{18} \).

Solve: \( u = \frac{1}{w} = \frac{1}{18} \implies w = 18 \), \( v = \frac{1}{m} = \frac{1}{36} \implies m = 36 \).

Check: First: \( \frac{2}{18} + \frac{5}{36} = \frac{1}{9} + \frac{5}{36} = \frac{4+5}{36} = \frac{1}{4} \), second: \( \frac{3}{18} + \frac{6}{36} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \), both true.

1 woman alone: 18 days, 1 man alone: 36 days

10th Maths Pair of Linear Equations In Two Variables Exercise 4.2 Solutions

zaheerpashamd

Exercise 4.2 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on forming and solving pairs of linear equations in two variables. Mathematical expressions are rendered using MathJax.

1. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹2000 per month, find their monthly incomes.

Let the incomes of the two persons be \( 9x \) and \( 7x \), and their expenditures be \( 4y \) and \( 3y \).

Savings = Income – Expenditure. Given savings are ₹2000 for both:

First person: \( 9x – 4y = 2000 \).

Second person: \( 7x – 3y = 2000 \).

Solve the system: Multiply the first equation by 3 and the second by 4 to eliminate \( y \).

\( 27x – 12y = 6000 \), \( 28x – 12y = 8000 \).

Subtract: \( (28x – 12y) – (27x – 12y) = 8000 – 6000 \implies x = 2000 \).

Substitute \( x = 2000 \) into \( 9x – 4y = 2000 \): \( 9(2000) – 4y = 2000 \implies 18000 – 4y = 2000 \implies 4y = 16000 \implies y = 4000 \).

Incomes: First person: \( 9x = 9 \cdot 2000 = 18000 \), Second person: \( 7x = 7 \cdot 2000 = 14000 \).

Check: Expenditures: \( 4y = 4 \cdot 4000 = 16000 \), \( 3y = 3 \cdot 4000 = 12000 \). Savings: \( 18000 – 16000 = 2000 \), \( 14000 – 12000 = 2000 \), both correct.

Monthly incomes: ₹18000 and ₹14000

2. The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

Let the tens digit be \( x \), units digit be \( y \). The number is \( 10x + y \), reversed number is \( 10y + x \).

First condition: \( (10x + y) + (10y + x) = 66 \implies 11x + 11y = 66 \implies x + y = 6 \).

Second condition: The digits differ by 2, so \( x – y = 2 \) or \( y – x = 2 \).

Case 1: \( x – y = 2 \). Solve with \( x + y = 6 \): Add the equations: \( 2x = 8 \implies x = 4 \), \( y = 6 – 4 = 2 \). Number: \( 10 \cdot 4 + 2 = 42 \).

Case 2: \( y – x = 2 \). Solve with \( x + y = 6 \): Subtract: \( (y – x) – (x + y) = 2 – 6 \implies -2x = -4 \implies x = 2 \), \( y = 6 – 2 = 4 \). Number: \( 10 \cdot 2 + 4 = 24 \).

Check: \( 42 + 24 = 66 \), \( |4 – 2| = 2 \); \( 24 + 42 = 66 \), \( |4 – 2| = 2 \). Both satisfy.

There are 2 such numbers: 42 and 24.

Numbers: 42 and 24, Total: 2 numbers

3. The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.

Supplementary angles sum to 180°. Let the smaller angle be \( x \), larger be \( y \).

\( x + y = 180 \).

Larger exceeds smaller by 18°: \( y = x + 18 \).

Substitute: \( x + (x + 18) = 180 \implies 2x + 18 = 180 \implies 2x = 162 \implies x = 81 \).

Then, \( y = 81 + 18 = 99 \).

Check: \( 81 + 99 = 180 \), \( 99 – 81 = 18 \), both true.

Angles: 81° and 99°

4. The taxi charges in Hyderabad are fixed, along with the charge for the distance covered. Up to the first 3 km you will be charged a certain minimum amount. From there onwards you have to pay additionally for every kilometer travelled. For the first 10 km, the charge paid is ₹166. For a journey of 15 km the charge paid is ₹256.

(i) What are the fixed charges and charge per km?

Let the fixed charge for the first 3 km be \( x \) ₹, and the charge per km after that be \( y \) ₹/km.

For 10 km: First 3 km at \( x \), next 7 km at \( y \): \( x + 7y = 166 \).

For 15 km: First 3 km at \( x \), next 12 km at \( y \): \( x + 12y = 256 \).

Subtract the first from the second: \( (x + 12y) – (x + 7y) = 256 – 166 \implies 5y = 90 \implies y = 18 \).

Substitute \( y = 18 \) into \( x + 7y = 166 \): \( x + 7(18) = 166 \implies x + 126 = 166 \implies x = 40 \).

Check: For 15 km: \( 40 + 12(18) = 40 + 216 = 256 \), matches.

Fixed charges: ₹40, Charge per km: ₹18

(ii) How much does a person have to pay for travelling a distance of 25 km?

For 25 km: First 3 km at ₹40, next \( 25 – 3 = 22 \) km at ₹18/km.

Total cost: \( 40 + 22 \cdot 18 = 40 + 396 = 436 \).

Cost for 25 km: ₹436

5. A fraction will be equal to \( \frac{4}{5} \) if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction will be equal to \( \frac{1}{2} \). What is the fraction?

Let the fraction be \( \frac{x}{y} \).

First condition: \( \frac{x+1}{y+1} = \frac{4}{5} \implies 5(x + 1) = 4(y + 1) \implies 5x + 5 = 4y + 4 \implies 5x – 4y = -1 \).

Second condition: \( \frac{x-5}{y-5} = \frac{1}{2} \implies 2(x – 5) = (y – 5) \implies 2x – 10 = y – 5 \implies 2x – y = 5 \).

Solve: Multiply the second by 4: \( 8x – 4y = 20 \).

Subtract the first: \( (8x – 4y) – (5x – 4y) = 20 – (-1) \implies 3x = 21 \implies x = 7 \).

Substitute \( x = 7 \) into \( 2x – y = 5 \): \( 2(7) – y = 5 \implies 14 – y = 5 \implies y = 9 \).

Fraction: \( \frac{7}{9} \). Check: \( \frac{7+1}{9+1} = \frac{8}{10} = \frac{4}{5} \), \( \frac{7-5}{9-5} = \frac{2}{4} = \frac{1}{2} \), both true.

Fraction: \( \frac{7}{9} \)

6. Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time at different speeds. If the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Let the speed of the car from A be \( x \) km/h, from B be \( y \) km/h.

Same direction (relative speed \( x – y \), assuming \( x > y \)): Distance = 100 km, time = 5 hours: \( 5(x – y) = 100 \implies x – y = 20 \).

Towards each other (relative speed \( x + y \)): Distance = 100 km, time = 1 hour: \( 1(x + y) = 100 \implies x + y = 100 \).

Solve: Add the equations: \( (x – y) + (x + y) = 20 + 100 \implies 2x = 120 \implies x = 60 \).

Substitute \( x = 60 \) into \( x + y = 100 \): \( 60 + y = 100 \implies y = 40 \).

Check: Same direction: \( 60 – 40 = 20 \), \( 5 \cdot 20 = 100 \). Towards each other: \( 60 + 40 = 100 \), matches.

Speeds: 60 km/h and 40 km/h

7. Two angles are complementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the measure of each angle.

Complementary angles sum to 90°. Let the smaller angle be \( x \), larger be \( y \).

\( x + y = 90 \).

Larger is 3° less than twice the smaller: \( y = 2x – 3 \).

Substitute: \( x + (2x – 3) = 90 \implies 3x – 3 = 90 \implies 3x = 93 \implies x = 31 \).

Then, \( y = 90 – 31 = 59 \).

Check: \( y = 2 \cdot 31 – 3 = 62 – 3 = 59 \), matches.

Angles: 31° and 59°

8. A dictionary has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages more than the first part. How many pages are in each part of the book?

Let the first part have \( x \) pages, second part have \( y \) pages.

Total pages: \( x + y = 1382 \).

Second part has 64 more pages: \( y = x + 64 \).

Substitute: \( x + (x + 64) = 1382 \implies 2x + 64 = 1382 \implies 2x = 1318 \implies x = 659 \).

Then, \( y = 659 + 64 = 723 \).

Check: \( 659 + 723 = 1382 \), \( 723 – 659 = 64 \), both true.

First part: 659 pages, Second part: 723 pages

9. A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68% solution?

Let \( x \) ml of 50% solution and \( y \) ml of 80% solution be used.

Total volume: \( x + y = 100 \).

Acid contribution: \( 0.5x + 0.8y = 0.68 \cdot 100 = 68 \).

Solve: From the first, \( y = 100 – x \). Substitute into the second: \( 0.5x + 0.8(100 – x) = 68 \).

\( 0.5x + 80 – 0.8x = 68 \implies -0.3x + 80 = 68 \implies -0.3x = -12 \implies x = 40 \).

Then, \( y = 100 – 40 = 60 \).

Check: Acid: \( 0.5 \cdot 40 + 0.8 \cdot 60 = 20 + 48 = 68 \), matches 68% of 100 ml.

50% solution: 40 ml, 80% solution: 60 ml

10. You have ₹12,000/- saved amount, and wants to invest it in two schemes yielding 10% and 15% interest. How much amount should be invested in each scheme so that you should get overall 12% interest?

Let \( x \) be invested at 10%, \( y \) at 15%.

Total amount: \( x + y = 12000 \).

Total interest at 12%: \( 0.1x + 0.15y = 0.12 \cdot 12000 = 1440 \).

Solve: From the first, \( y = 12000 – x \). Substitute: \( 0.1x + 0.15(12000 – x) = 1440 \).

\( 0.1x + 1800 – 0.15x = 1440 \implies -0.05x + 1800 = 1440 \implies -0.05x = -360 \implies x = 7200 \).

Then, \( y = 12000 – 7200 = 4800 \).

Check: Interest: \( 0.1 \cdot 7200 + 0.15 \cdot 4800 = 720 + 720 = 1440 \), matches 12% of 12000.

10% scheme: ₹7200, 15% scheme: ₹4800

10th Maths Pair of Linear Equations In Two Variables Exercise 4.1 Solutions

zaheerpashamd

Exercise 4.1 Solutions – Class X Mathematics

These solutions are based on the Telangana State Class X Mathematics textbook, focusing on pairs of linear equations in two variables. Mathematical expressions are rendered using MathJax.

1. By comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), state whether the lines represented by the following pairs of linear equations intersect at a point, are parallel, or are coincident.

(a) \( 5x – 4y + 8 = 0 \), \( 7x – 6y – 9 = 0 \)

Rewrite in standard form \( a_1 x + b_1 y + c_1 = 0 \), \( a_2 x + b_2 y + c_2 = 0 \):

Equation 1: \( 5x – 4y + 8 = 0 \), so \( a_1 = 5 \), \( b_1 = -4 \), \( c_1 = 8 \).

Equation 2: \( 7x – 6y – 9 = 0 \), so \( a_2 = 7 \), \( b_2 = -6 \), \( c_2 = -9 \).

Compute ratios: \( \frac{a_1}{a_2} = \frac{5}{7} \), \( \frac{b_1}{b_2} = \frac{-4}{-6} = \frac{2}{3} \), \( \frac{c_1}{c_2} = \frac{8}{-9} \).

Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at a point (unique solution).

Conclusion: The lines intersect at a point.

(b) \( 9x + 3y + 12 = 0 \), \( 18x + 6y + 24 = 0 \)

Equation 1: \( 9x + 3y + 12 = 0 \), so \( a_1 = 9 \), \( b_1 = 3 \), \( c_1 = 12 \).

Equation 2: \( 18x + 6y + 24 = 0 \), so \( a_2 = 18 \), \( b_2 = 6 \), \( c_2 = 24 \).

Compute ratios: \( \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2} \).

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident (infinite solutions).

Conclusion: The lines are coincident.

(c) \( 6x – 3y + 10 = 0 \), \( 2x – y + 9 = 0 \)

Equation 1: \( 6x – 3y + 10 = 0 \), so \( a_1 = 6 \), \( b_1 = -3 \), \( c_1 = 10 \).

Equation 2: \( 2x – y + 9 = 0 \), so \( a_2 = 2 \), \( b_2 = -1 \), \( c_2 = 9 \).

Compute ratios: \( \frac{a_1}{a_2} = \frac{6}{2} = 3 \), \( \frac{b_1}{b_2} = \frac{-3}{-1} = 3 \), \( \frac{c_1}{c_2} = \frac{10}{9} \).

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel (no solution).

Conclusion: The lines are parallel.

2. Check whether the following equations are consistent or inconsistent. Solve them graphically.

For graphical solution, plot each pair of equations as lines on a graph. Consistency is determined by whether they intersect (consistent, unique solution), are coincident (consistent, infinite solutions), or are parallel (inconsistent, no solution). Here, I’ll solve algebraically to determine consistency, then describe the graphical approach.

(a) \( 3x + 2y = 5 \), \( 2x – 3y = 7 \)

Rewrite: \( 3x + 2y – 5 = 0 \), \( 2x – 3y – 7 = 0 \).

Check ratios: \( a_1 = 3 \), \( b_1 = 2 \), \( c_1 = -5 \); \( a_2 = 2 \), \( b_2 = -3 \), \( c_2 = -7 \).

\( \frac{a_1}{a_2} = \frac{3}{2} \), \( \frac{b_1}{b_2} = \frac{2}{-3} \), \( \frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7} \).

Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the equations are consistent (intersect at a point).

Solve algebraically: Multiply first by 3 and second by 2 to make \( y \)-coefficients opposites.

\( 9x + 6y = 15 \), \( 4x – 6y = 14 \).

Add: \( 13x = 29 \implies x = \frac{29}{13} \).

Substitute into \( 3x + 2y = 5 \): \( 3 \left(\frac{29}{13}\right) + 2y = 5 \implies \frac{87}{13} + 2y = 5 \implies 2y = 5 – \frac{87}{13} = \frac{65 – 87}{13} = -\frac{22}{13} \implies y = -\frac{11}{13} \).

Solution: \( (x, y) = \left(\frac{29}{13}, -\frac{11}{13}\right) \).

Graphically: For \( 3x + 2y = 5 \), points are \( (0, 2.5) \), \( (1, 1) \). For \( 2x – 3y = 7 \), points are \( (0, -\frac{7}{3}) \), \( (1, -1.67) \). The lines intersect at \( \left(\frac{29}{13}, -\frac{11}{13}\right) \), confirming consistency.

Conclusion: Consistent, Solution: \( \left(\frac{29}{13}, -\frac{11}{13}\right) \)

(b) \( 2x – 3y = 8 \), \( 4x – 6y = 9 \)

Rewrite: \( 2x – 3y – 8 = 0 \), \( 4x – 6y – 9 = 0 \).

Check ratios: \( a_1 = 2 \), \( b_1 = -3 \), \( c_1 = -8 \); \( a_2 = 4 \), \( b_2 = -6 \), \( c_2 = -9 \).

\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9} \).

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the equations are inconsistent (parallel lines).

Graphically: For \( 2x – 3y = 8 \), points are \( (0, -\frac{8}{3}) \), \( (1, -2) \). For \( 4x – 6y = 9 \), points are \( (0, -\frac{3}{2}) \), \( (1, -\frac{5}{6}) \). The lines are parallel, confirming inconsistency.

Conclusion: Inconsistent

(c) \( \frac{3}{2}x – \frac{5}{3}y = 7 \), \( 9x – 10y = 12 \)

Clear fractions in the first equation: \( 3x – \frac{5}{3}y = 7 \implies 9x – 5y = 21 \).

Second equation: \( 9x – 10y = 12 \).

Check ratios: \( a_1 = 9 \), \( b_1 = -5 \), \( c_1 = -21 \); \( a_2 = 9 \), \( b_2 = -10 \), \( c_2 = -12 \).

\( \frac{a_1}{a_2} = \frac{9}{9} = 1 \), \( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-21}{-12} = \frac{7}{4} \).

Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the equations are consistent.

Solve: Subtract the second from the first: \( (9x – 5y) – (9x – 10y) = 21 – 12 \implies 5y = 9 \implies y = \frac{9}{5} \).

Substitute into \( 9x – 10y = 12 \): \( 9x – 10 \left(\frac{9}{5}\right) = 12 \implies 9x – 18 = 12 \implies 9x = 30 \implies x = \frac{10}{3} \).

Graphically: For \( 9x – 5y = 21 \), points are \( (0, -\frac{21}{5}) \), \( (1, -\frac{12}{5}) \). For \( 9x – 10y = 12 \), points are \( (0, -\frac{6}{5}) \), \( (1, -\frac{3}{10}) \). Intersect at \( \left(\frac{10}{3}, \frac{9}{5}\right) \).

Conclusion: Consistent, Solution: \( \left(\frac{10}{3}, \frac{9}{5}\right) \)

(d) \( 5x – 3y = 11 \), \( -10x + 6y = -22 \)

Check ratios: \( a_1 = 5 \), \( b_1 = -3 \), \( c_1 = -11 \); \( a_2 = -10 \), \( b_2 = 6 \), \( c_2 = 22 \).

\( \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2} \).

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the equations are coincident (consistent, infinite solutions).

Graphically: For \( 5x – 3y = 11 \), points are \( (0, -\frac{11}{3}) \), \( (1, -2) \). The second equation is a multiple (\( -2 \times \text{first} \)), so the lines coincide.

Conclusion: Consistent (coincident)

(e) \( \frac{4}{3}x + 2y = 8 \), \( 2x + 3y = 12 \)

Clear fraction: \( \frac{4}{3}x + 2y = 8 \implies 4x + 6y = 24 \).

Second equation: \( 2x + 3y = 12 \).

Check ratios: \( a_1 = 4 \), \( b_1 = 6 \), \( c_1 = -24 \); \( a_2 = 2 \), \( b_2 = 3 \), \( c_2 = -12 \).

\( \frac{a_1}{a_2} = \frac{4}{2} = 2 \), \( \frac{b_1}{b_2} = \frac{6}{3} = 2 \), \( \frac{c_1}{c_2} = \frac{-24}{-12} = 2 \). Coincident, but solve to confirm.

Solve: Multiply second by 2: \( 4x + 6y = 24 \), which is the first equation. Thus, coincident.

Graphically: For \( 2x + 3y = 12 \), points are \( (0, 4) \), \( (6, 0) \). The first equation plots the same line, confirming infinite solutions.

Conclusion: Consistent (coincident)

(f) \( x + y = 5 \), \( 2x + 2y = 10 \)

Check ratios: \( a_1 = 1 \), \( b_1 = 1 \), \( c_1 = -5 \); \( a_2 = 2 \), \( b_2 = 2 \), \( c_2 = -10 \).

\( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2} \). Coincident.

Graphically: For \( x + y = 5 \), points are \( (0, 5) \), \( (5, 0) \). The second equation is the same line, confirming infinite solutions.

Conclusion: Consistent (coincident)

(g) \( x – y = 8 \), \( 3x – 3y = 16 \)

Check ratios: \( a_1 = 1 \), \( b_1 = -1 \), \( c_1 = -8 \); \( a_2 = 3 \), \( b_2 = -3 \), \( c_2 = -16 \).

\( \frac{a_1}{a_2} = \frac{1}{3} \), \( \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3} \), \( \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \). Parallel, inconsistent.

Graphically: For \( x – y = 8 \), points are \( (0, -8) \), \( (8, 0) \). For \( 3x – 3y = 16 \), points are \( (0, -\frac{16}{3}) \), \( (\frac{16}{3}, 0) \). Parallel lines.

Conclusion: Inconsistent

(h) \( 2x + y – 6 = 0 \), \( 4x + 2y – 4 = 0 \)

Check ratios: \( a_1 = 2 \), \( b_1 = 1 \), \( c_1 = -6 \); \( a_2 = 4 \), \( b_2 = 2 \), \( c_2 = -4 \).

\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-6}{-4} = \frac{3}{2} \). Parallel, inconsistent.

Graphically: For \( 2x + y = 6 \), points are \( (0, 6) \), \( (3, 0) \). For \( 4x + 2y = 4 \), points are \( (0, 2) \), \( (1, 0) \). Parallel lines.

Conclusion: Inconsistent

(i) \( 2x – 2y – 2 = 0 \), \( 4x – 4y – 5 = 0 \)

Check ratios: \( a_1 = 2 \), \( b_1 = -2 \), \( c_1 = -2 \); \( a_2 = 4 \), \( b_2 = -4 \), \( c_2 = -5 \).

\( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5} \). Parallel, inconsistent.

Graphically: For \( x – y = 1 \), points are \( (0, -1) \), \( (1, 0) \). For \( 4x – 4y = 5 \), points are \( (0, -\frac{5}{4}) \), \( (1, -\frac{1}{4}) \). Parallel lines.

Conclusion: Inconsistent

3. Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered, “the number of skirts are two less than twice the number of pants purchased and the number of skirts is four less than four times the number of pants purchased.” Help her friend to find how many pants and skirts Neha bought.

Let \( x \) be the number of pants, \( y \) be the number of skirts.

First statement: \( y = 2x – 2 \).

Second statement: \( y = 4x – 4 \).

Equate the two expressions for \( y \): \( 2x – 2 = 4x – 4 \).

Solve: \( 2x – 4x = -4 + 2 \implies -2x = -2 \implies x = 1 \).

Substitute \( x = 1 \) into \( y = 2x – 2 \): \( y = 2(1) – 2 = 0 \).

So, Neha bought 1 pant and 0 skirts. Check the second equation: \( y = 4(1) – 4 = 0 \), which matches.

Number of pants: 1, Number of skirts: 0

4. 10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.

Let \( x \) be the number of boys, \( y \) be the number of girls.

Equation 1: \( x + y = 10 \) (total students).

Equation 2: \( y = x + 4 \) (girls are 4 more than boys).

Substitute \( y = x + 4 \) into the first equation: \( x + (x + 4) = 10 \).

Solve: \( 2x + 4 = 10 \implies 2x = 6 \implies x = 3 \).

Then, \( y = x + 4 = 3 + 4 = 7 \).

Check: \( 3 + 7 = 10 \), and \( 7 = 3 + 4 \), both true.

Number of boys: 3, Number of girls: 7

5. 5 pencils and 7 pens together cost ₹50 whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.

Let \( x \) be the cost of one pencil, \( y \) be the cost of one pen (in ₹).

Equation 1: \( 5x + 7y = 50 \).

Equation 2: \( 7x + 5y = 46 \).

Add the equations: \( (5x + 7y) + (7x + 5y) = 50 + 46 \implies 12x + 12y = 96 \implies x + y = 8 \).

Subtract the second from the first: \( (5x + 7y) – (7x + 5y) = 50 – 46 \implies -2x + 2y = 4 \implies -x + y = 2 \).

Solve the system: \( x + y = 8 \), \( -x + y = 2 \).

Add: \( 2y = 10 \implies y = 5 \).

Substitute \( y = 5 \) into \( x + y = 8 \): \( x + 5 = 8 \implies x = 3 \).

Check: \( 5(3) + 7(5) = 15 + 35 = 50 \), \( 7(3) + 5(5) = 21 + 25 = 46 \), both true.

Cost of one pencil: ₹3, Cost of one pen: ₹5

6. Half the perimeter of a rectangular garden is 36 m. If the length is 4 m more than its width, find the dimensions of the garden.

Let the width be \( x \) m, length be \( y \) m.

Half the perimeter: \( x + y = 36 \).

Length is 4 m more than width: \( y = x + 4 \).

Substitute \( y = x + 4 \) into \( x + y = 36 \): \( x + (x + 4) = 36 \implies 2x + 4 = 36 \implies 2x = 32 \implies x = 16 \).

Then, \( y = x + 4 = 16 + 4 = 20 \).

Check: Perimeter = \( 2(16 + 20) = 72 \), half = 36, and \( 20 = 16 + 4 \), both true.

Dimensions: Length = 20 m, Width = 16 m

7. We have a linear equation \( 2x + 3y – 8 = 0 \). Write another linear equation in two variables \( x \) and \( y \) such that the geometrical representation of the pair so formed is intersecting lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.

Given equation: \( 2x + 3y – 8 = 0 \).

For intersecting lines, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Choose \( a_2 = 3 \), \( b_2 = 2 \), \( c_2 = -7 \): \( 3x + 2y – 7 = 0 \).

Check: \( \frac{2}{3} \neq \frac{3}{2} \), so they intersect.

For parallel lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Multiply by 2 and change the constant: \( 4x + 6y – 10 = 0 \). Check: \( \frac{2}{4} = \frac{3}{6} \neq \frac{-8}{-10} \).

For coincident lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). Multiply by 3: \( 6x + 9y – 24 = 0 \). Check: \( \frac{2}{6} = \frac{3}{9} = \frac{-8}{-24} = \frac{1}{3} \).

Intersecting: \( 3x + 2y – 7 = 0 \), Parallel: \( 4x + 6y – 10 = 0 \), Coincident: \( 6x + 9y – 24 = 0 \)

8. The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area increases by 50 sq units. Find the length and breadth of the rectangle.

Let length be \( x \) units, breadth be \( y \) units. Area = \( xy \).

First condition: \( (x – 5)(y + 2) = xy – 80 \).

Expand: \( xy + 2x – 5y – 10 = xy – 80 \implies 2x – 5y = -70 \).

Second condition: \( (x + 10)(y – 5) = xy + 50 \).

Expand: \( xy – 5x + 10y – 50 = xy + 50 \implies -5x + 10y = 100 \implies -x + 2y = 20 \).

Solve the system: \( 2x – 5y = -70 \), \( -x + 2y = 20 \).

Multiply the second by 2: \( -2x + 4y = 40 \).

Add to the first: \( (2x – 5y) + (-2x + 4y) = -70 + 40 \implies -y = -30 \implies y = 30 \).

Substitute \( y = 30 \) into \( -x + 2y = 20 \): \( -x + 2(30) = 20 \implies -x + 60 = 20 \implies -x = -40 \implies x = 40 \).

Check: First: \( (40 – 5)(30 + 2) = 35 \cdot 32 = 1120 \), \( 40 \cdot 30 – 80 = 1200 – 80 = 1120 \), matches. Second: \( (40 + 10)(30 – 5) = 50 \cdot 25 = 1250 \), \( 1200 + 50 = 1250 \), matches.

Length: 40 units, Breadth: 30 units