Table of Contents
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
