RS Aggarwal Solutions Class 10 Ex 3J

Question 1: If twice the son’s age in years is added to the mother’s age, the sum is 70. But, if twice the mother’s age is added to the son’s age, the sum is 95. Find the ages of mother and son.

Solution:

Let the present ages of the mother and her son be x and y respectively.

According to the given question:

x + 2y = 70 …….. (1)

And, 2x + y = 95 …….. (2)

Multiplying (1) by 1 and (2) by 2, we get

x + 2y = 70 …….. (3)

4x + 2y = 190 …….. (4)

Subtracting (3) from (4), we get

3x = 120 \( \Rightarrow \) y = 40

Putting x = 40 in (1), we get

40 + 2y = 70

2y = 30

y = 15

x = 40, y = 15.

Hence, the ages of the mother and the son are 40 years and 15 years respectively.

Question 2: The monthly incomes of A and B are in the ratio 5:4 and their monthly expenditures are in the ratio 7:5. If each saves Rs.3000 per month, find the monthly income of each.

Solution:

Let the monthly income of A and B be Rs. 5x and Rs. 4x respectively and let their expenditure be Rs. 7y and Rs. 5y respectively.

Then,

5x – 7y = 3000 ……..(1)

4x – 5y = 3000 ……..(2)

Multiplying (1) by 5 and (2) by 7, we get

25x – 35y = 15000 ……..(3)

28x – 35y = 21000 ……..(4)

Subtracting (3) from (4), we get

3x = 6000

x = 2000

Putting x = 2000 in (1), we  get

(5)(2000) – 7y = 3000

-7y = 3000 – 10000

\( y =  \frac{-7000}{-7} = 1000 \)

x = 2000, y = 1000

Income of A = 5x = Rs. 10000

Income of B = 4x = Rs. 8000

Question 3: A man sold a chair and a table together for Rs.760, thereby making a profit of 25% on a chair and 10% on the table. By selling them together for Rs.767.50, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.

Solution:

Let Rs. x and Rs. y be the CP of a chair and table respectively.

If profit is 25%, then SP of chair = \( \frac{100 + 25}{100} \times x = Rs. \frac{125x}{100} \)

If profit is 10%, then SP of chair = \( \frac{100 + 10}{100} \times x = Rs. \frac{110y}{100} \)

SP of a chair and table = Rs. 760

Therefore,

\( \frac{125x}{100} + \frac{110y}{100} = 760 \)

\( \Rightarrow \frac{25x}{20} + \frac{22y}{20} = 760 \)

\( \Rightarrow 25x + 22y = 15200 \) ……..(1)

Further, if profit is 10%, then the SP of a chair = \( \frac{100 + 10}{100} \times x = Rs. \frac{110}{100} \)

if profit is 25%, then the SP of a chair = \( \frac{100 + 25}{100} \times y = Rs. \frac{125}{100} \)

SP of a chair and table = Rs. 767.50

\( \frac{110x}{100} + \frac{125y}{100} = 767.50 \)

\( \Rightarrow \frac{22x}{20} + \frac{25y}{20} = 767.50 \)

\( \Rightarrow 22x + 25y = 15350 \) ……..(2)

Adding (1) and (2), we get

47(x + y) = 30550

x + y = \( \frac{30550}{47} = 650 \) ……..(3)

Subtracting (2) from (1), we get

3(x – y) = 15200 – 15350

3(x – y) = -150

x – y = -50 ……..(4)

Adding (3) and (4), we get

2x = 640 – 50

2x = 600

x = 300

Subtracting (4) from (3), we get

2y = 650 + 50

2y = 700

y = 350

Hence, CP of a chair is Rs 300 and CP of the table is Rs. 350

Question 4: A man invested an amount at 12% per annum and another amount at 10% per annum simple interest. Thus, he received Rs.1145 as annual interest. Had he interchanged the amounts invested, he would have received Rs.90 less as interest. What amounts did he invest at different rates?

Solution:

Let the amounts invested at 12% and 10% be Rs. x and Rs. y respectively.

Then,

First case:

SI on Rs. x at 12% p.a. For 1 year = \( \frac{x \times 12 \times 1}{100} = \frac{3x}{25} \)

SI on Rs. y at 10% p.a. For 1 year = \( \frac{y \times 10 \times 1}{100} = \frac{y}{10} \)

Total SI = Rs. 1145

\( \Rightarrow \frac{3x}{25} + \frac{y}{10} = 1145 \)

\( \Rightarrow \frac{6x + 5y}{50} = 1145 \)

\( \Rightarrow 6x + 5y = 57250 \) ……..(1)

Second case:

SI on Rs. x at 10% p.a. For 1 year = \( \frac{x \times 10 \times 1}{100} = \frac{x}{10} \)

SI on Rs. y at 12% p.a. For 1 year = \( \frac{y \times 12 \times 1}{100} = \frac{3y}{25} \)

Total SI = Rs. (1145 – 90) = 1055

\( \Rightarrow \frac{x}{10} + \frac{3y}{25} = 1055 \)

\( \Rightarrow \frac{5x + 6y}{50} = 1055 \)

\( \Rightarrow 5x + 6y = 52750 \) ……..(2)

Multiplying (1) by 6 and (2) by 5, we get

36x + 30y = 343500 ……..(3)

25x + 30y = 263750 ……..(4)

Subtracting (4) from (3), we get

11x = 79750

\( \frac{79750}{11} = 7250 \)<

Putting x = 7250 in (1), we get

(6)(7250) + 5y = 57250

5y = 13750

y = 2750

x = 7250, y = 2750

Hence, amount invested at 12% = Rs. 7250

And amount invested at 10% = Rs. 2750

Question 5: There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.

Solution:

Let the number of student in class room A and B be x and y respectively.

When 10 students are transferred from A to B:

x – 10 = y + 10

x – y = 20 – (-1)

When 20 students are transferred from B to A:

2(y – 20) = x + 20

2y – 40 = x + 20

-x + 2y = 60 – 2

Adding (1) and (2), we get

y = 80

Putting y = 80 in (1), we get

x – 80 = 20

x = 100

Hence, number of students of A and B are 100 and 80 respectively.

Question 6: Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.

Solution:

Let P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.

Case 1:

When the cars P and Q move in the same direction.

Distance covered by the car P in 7 hours = 7x km

Distance covered by the car Q in 7 hours = 7y km

AM = 7x km and BM = 7y km

AM – BM = AB

7x – 7y = 70

7(x – y) = 70

x – y = 10 ……..(1)

Case 2:

When the cars P and Q move in opposite directions.

Distance covered by P in 1 hour = x km

Distance covered by Q in 1 hour = y km

In this case, let the cars meet at a point N.

AN = x km and BN = y km

AN + BN = AB

x + y = 70 …….. (2)

Adding (1) and (2), we get

2x = 80

x = 40

Putting x = 40 in (1), we get

40 – y = 10

y = (40 – 10) = 30

x = 40, y = 30

Hence, the speeds of these cars are 40 km/hr and 30 km/hr respectively.

Question 7: A train covered a certain distance at a uniform speed. If the train had been 5 kph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kph, it would have taken 3 hours more than the scheduled time. Find the length of the journey.

Solution:

Let the original speed be x km/h and time taken be y hours.

Then, length of journey = xy km

Case 1:

Speed = (x + 5) km/h and time taken = (y – 3)hour

Distance covered = (x + 5)(y – 3)km

(x + 5)(y – 3) = xy

xy + 5y – 3y – 15 = xy

5y – 3y = 15 …….. (1)

Case 2:

Speed (x – 4) km/hr and time taken = (y + 3) hours

Distance covered = (x – 4)(y + 3) km

(x – 4)(y + 3) = xy

xy – 4y + 3x – 12 = xy

3x – 4y = 12 …….. (2)

Multiplying (1) by 4 and (2) by 5, we get

20y × 12x = 60 …….. (3)

-20y + 15x = 60 …….. (4)

Adding (3) and (4), we get

3x = 120

Or x = 40

Putting x = 40 in (1), we get

5y – 3(40) = 15

5y = 135 = 27

Hence, length of the journey is (40 * 27) = 1080 km.


Practise This Question

Variations that affect germ cells are called Somatic variation.