## RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E. You must go through **NCERT Solutions for Class 10 Maths** to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

### RS Aggarwal Solutions Class 10 Chapter 3

**Question 1.**

**5 chairs and 4 tables together cost ₹ 5600, while 4 chairs and 3 tables together cost ₹ 4340. Find the cost of a chair and that of a table.**

**Solution:
**Let cost of one chair = ₹ x

and cost of one table = ₹ y

According to the conditions,

5x + 4y = ₹ 5600 …(i)

4x + 3y = ₹ 4340 …(ii)

x = -560

and from (i)

5 x 560 + 4y = 5600

2800 + 4y = 5600

⇒ 4y = 5600 – 2800

⇒ 4y = 2800

⇒ y = 700

Cost of one chair = ₹ 560

and cost of one table = ₹ 700

**Question 2.**

**23 spoons and 17 forks together cost ₹ 1770, while 17 spoons and 23 forks together cost ₹ 1830. Find the cost of a spoon and that of a fork.**

**Solution:**

Let the cost of one spoon = ₹ x and cost of one fork = ₹ y

According to the conditions,

23x + 17y = 1770 …(i)

17x + 23y = 1830 …(ii)

Adding, we get

40x + 40y = 3600

Dividing by 40,

x + y = 90 …(iii)

and subtracting,

6x – 6y = -60

Dividing by 6,

x – y = -10 …(iv)

Adding (iii) and (iv)

2x = 80 ⇒ x = 40

and subtracting,

2y = 100 ⇒ y = 50

Cost of one spoon = ₹ 40

and cost of one fork = ₹ 50

**Question 3.**

**A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all totalling ₹ 19.50, how many coins of each kind does she have?**

**Solution:**

Let number of 25-paisa coins = x

and number 50-paisa coins = y

Total number of coins = 50

and total amount = ₹ 19.50 = 1950 paisa

x + y = 50 …(i)

25x + 50y = 1950

⇒ x + 2y = 78 …(ii)

Subtracting (i) from (ii), y = 28

x = 50 – y = 50 – 28 = 22

Number of 25-paisa coins = 22

and 50-paisa coins = 28

**Question 4.**

**The sum of two numbers is 137 and their difference is 43. Find the numbers.**

**Solution:**

Sum of two numbers = 137

and difference = 43

Let first number = x

and second number = y

x + y = 137 …..(i)

x – y = 43 ……(ii)

Adding, we get

2x = 180 ⇒ x = 90

and subtracting,

2y = 94

y = 47

First number = 90

and second number = 47

**Question 5.**

**Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.**

**Solution:**

Let first number = x

and second number = y

According to the conditions,

2x + 3y = 92 …(i)

4x – 7y = 2 …(ii)

Multiply (i) by 2 and (ii) by 1

4x + 6y = 184 …..(iii)

4x – 7y = 2 …….(iv)

Subtracting (iii) from (iv),

13y = 182

y = 14

From (i), 2x + 3y = 92

2x + 3 x 14 = 92

⇒ 2x + 42 = 92

⇒ 2x = 92 – 42 = 50

⇒ x = 25

First number = 25

Second number = 14

**Question 6.**

**Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.**

**Solution:**

Let first number = x

and second number = y

According to the conditions,

3x + y=142 …(i)

4x – y = 138 …(ii)

Adding, we get

7x = 280

⇒ x = 40

and from (i)

3 x 40 + y = 142

⇒ 120 + y = 142

⇒ y = 142 – 120 = 22

First number = 40,

second number = 22

**Question 7.**

**If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.**

**Solution:**

Let first greater number = x

and second smaller number = y

According to the conditions,

2x – 45 = y …(i)

2y – 21 = x …(ii)

Substituting the value of y in (ii),

2 (2x – 45) – 21 = x

⇒ 4x – 90 – 21 = x

⇒ 4x – x = 111

⇒ 3x = 111

⇒ x = 37

From (i),

y = 2 x 37 – 45 = 74 – 45 = 29

The numbers are 37, 29

**Question 8.**

**If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.**

**Solution:**

Let larger number = x

and smaller number = y

According to the conditions,

3x = 4 x y + 8 ⇒ 3x = 4y + 8 …….(i)

5y = x x 3 + 5 ⇒ 5y = 3x + 5 …(ii)

Substitute the value of 3x in (ii),

5y = 4y + 8 + 5

⇒ 5y – 4y = 13

⇒ y = 13

and 3x = 4 x 13 + 8 = 60

⇒ x = 20

Larger number = 20

and smaller number = 13

**Question 9.**

**If 2 is added to each of two given numbers, their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.**

**Solution:**

Let first number = x and

second number = y

According to the conditions,

⇒ 11x – 44 = 5(2x + 2) – 20

⇒ 11x – 44 = 10x + 10 – 20

⇒ 11x – 10x = 10 – 20 + 44

⇒ x = 34

and y = 2 x 34 + 2 = 68 + 2 = 70

Numbers are 34 and 70

**Question 10.**

**The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.**

**Solution:**

Let first number = x

and second number (smaller) = y

According to the conditions,

x – y = 14

and x² – y² = 448

⇒ (x + y) (x – y) = 448

⇒ (x + y) x 14 = 448

⇒ x + y = 32 ……(i)

and x – y = 14 ……(ii)

Adding (i) and (ii),

2x = 46 ⇒ x = 23

and subtracting (i) and (ii),

2y = 18 ⇒ y = 9

Numbers are 23, 9

**Question 11.**

**The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number. [CBSE 2006]**

**Solution:**

Let ones digit of a two digit number = x

and tens digit = y

Number = x + 10y

By interchanging the digits,

Ones digit = y

and tens digit = x

Number = y + 10x

According to the conditions,

x + y = 12 ………. (i)

y + 10x = x + 10y + 18

⇒ y + 10x – x – 10y = 18

⇒ 9x – 9y = 18

⇒ x – y = 2 …(ii) (Dividing by 9)

Adding (i) and (ii),

2x = 14 ⇒ x = 7

and subtracting,

2y = 10 ⇒ y = 5

Number = 7 + 10 x 5 = 7 + 50 = 57

**Question 12.**

**A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.**

**Solution:**

Let one’s digit of a two digit number = x

and ten’s digit = y

Then number = x + 10y

After reversing the digits,

Ones digit = y

and ten’s digit = x

and number = y + 10x

According to the conditions,

x + 10y – 27 = y + 10x

⇒ y + 10x – x – 10y = -27

⇒ 9x – 9y = -27

⇒ x – y = -3 …(i)

and 7 (x + y) = x + 10y

7x + 7y = x+ 10y

⇒ 7x – x = 10y – 7y

⇒ 6x = 3y

⇒ 2x = y …(ii)

Substituting the value of y in (i)

x – 2x = -3

⇒ -x = -3

⇒ x = 3

y = 2x = 2 x 3 = 6

Number = x + 10y = 3 + 10 x 6 = 3 + 60 = 63

**Question 13.**

**The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number. [CBSE 2004]**

**Solution:**

Let one’s digit of a two digit number = x

and ten’s digit = y

Then number = x + 10y

After interchanging the digits,

One’s digit = y

and ten’s digit = x

Then number = y + 10x

According to the conditions,

y + 10x = x + 10y + 9

⇒ y + 10x – x – 10y = 9

⇒ 9x – 9y = 9

⇒ x – y = 1 …(i)

and x + y= 15 …(ii)

Adding, we get

2x = 16

x = 8

and subtracting,

2y = 14

⇒ y = 7

Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

**Question 14.**

**A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.**

**Solution:**

Let one’s digit of the two digit number = x

and ten’s digit = y

Then number = x + 10y

By reversing the digits,

One’s digit = y

and ten’s digit = x

Then number = y + 10x

Now, according to the conditions,

x + 10y + 18 = y + 10x

⇒ 18 = y + 10x – x – 10y

⇒ 9x – 9y = 18

⇒ x – y = 2 …(i)

and 4(x + y) + 3 = x + 10y

4x + 4y + 3 = x + 10y

⇒ 4x + 4y – x – 10y = -3

3x – 6y = -3

⇒ x – 2y = -1 ……..(ii)

Subtracting,

y = 3

and x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5

Number = x + 10y = 5 + 10 x 3 = 5 + 30 = 35

**Question 15.**

**A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number. [CBSE 1999C]**

**Solution:**

Let ones digit of a two digit number = x

and tens digit = y

Then number = x + 10y

By reversing the digits,

One’s digit = y

and ten’s digit = x

and number = y + 10x

According to the conditions,

x + 10y – 9 = y + 10x

⇒ x + 10y – y – 10x = 9

⇒ -9x + 9y = 9

⇒x – y = -1 …(i) (Dividing by -9)

**Question 16.**

**A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number. [CBSE 2006]**

**Solution:**

Let the one’s digit of a two digit number = x

and ten’s digit = y

Then number = x + 10y

By interchanging the digits,

One’s digit = y

and ten’s digit = x

Then number = y + 10x

According to the conditions,

x + 10y + 18 = y + 10x

⇒ 18 = y + 10x – x – 10y

⇒ 9x – 9y = 18

⇒ x – y = 2 …(i)

and xy = 35 …(ii)

Now, (x + y)² = (x – y)² + 4xy = (2)² + 4 x 35 = 4 + 140 = 144 = (12)²

⇒ (x + y) = 12 …(iii)

Subtracting (i) from (iii), we get

**Question 17.**

**A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number. [CBSE 2006C]**

**Solution:**

Let one’s digit of a two digit number = x

and ten’s digit = y

Then number = x + 10y

After interchanging the digits One’s digit = y

Ten’s digit = x

Then number = y + 10x

According to the conditions,

x + 10y – 63 = y + 10x

**Question 18.**

**The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.**

**Solution:**

Let one’s digit of a two digit number = x

and ten’s digit = y

Number = x + 10y

By reversing the digits,

One’s digit = y

and ten’s digit = x

Number = y + 10x

According to the conditions,

x + 10y + y + 10x = 121

⇒ 11x + 11y = 121

⇒ x + y = 11 …(i)

x – y = 3 …(ii)

Adding, we get

2x = 14 ⇒ x = 7

Subtracting,

2y = 8 ⇒ y = 4

Number = 7 + 10 x 4 = 7 + 40 = 47

or 4 + 10 x 7 = 4 + 70 = 74

**Question 19.**

**The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes . Find the fraction. [CBSE 2003]**

**Solution:**

Let numerator of a fraction = x

and denominator = y

Then fraction =

According to the conditions,

x + y = 8 …(i)

**Question 20.**

**If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fraction.**

**Solution:**

Let numerator of a fraction = x

and denominator = y

Then fraction =

According to the conditions,

=

=

⇒ 2x + 4 = y …(i)

3x = y – 1 …(ii)

⇒ 3x = 2x + 4 – 1

⇒ 3x = 2x + 3

⇒ 3x – 2x = 3

⇒ x = 3

and y = 2x + 4 = 2 x 3 + 4 = 6 + 4 = 10

Fraction =

**Question 21.**

**The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes . Find the fraction.**

**Solution:**

Let numerator of a fraction = x

and denominator = y

Then fraction =

According to the conditions,

y – x = 11

y = 11 + x …(i)

**Question 22.**

**Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.**

**Solution:**

Let numerator of a fraction = x

and denominator = y

Then fraction =

According to the conditions,

**Question 23.**

**The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction. [CBSE 2010]**

**Solution:**

Let numerator of a fraction = x

and denominator = y

Then fraction =

According to the conditions,

x + y = 4 + 2x

⇒ y = 4 + x …(i)

Fraction = =

**Question 24.**

**The sum of two numbers is 16 and sum of their reciprocal is . Find the numbers. [CBSE 2005]**

**Solution:**

Let first number = x

and second number = y

According to the conditions,

x + y = 16

**Question 25.**

**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 in classroom A, the number of students = x

and in classroom B = y

According to the conditions,

x – 10 = y + 10

⇒ x – y = 10 + 10 = 20

⇒ x – y = 20 …(i)

and x + 20 = 2 (y – 20)

⇒ x + 20 = 2y – 40

⇒ x – 2y = -(40 + 20) = -60

x – 2y = -60 …(ii)

Subtracting, y = 80

and x – y = 20

⇒ x – 80 = 20

⇒ x = 20 + 80 = 100

Number of students in classroom A = 100 and in B = 80

**Question 26.**

**Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80 km, he pays ₹ 1330, and travelling 90 km, he pays ₹ 1490. Find the fixed charges and rate per km.**

**Solution:**

Let fixed charges = ₹ x

and other charges = ₹ y per km

According to the conditions,

For 80 km,

x + 80y = ₹ 1330 …(i)

and x + 90y = ₹ 1490 …(ii)

Subtracting (i) from (ii),

10y = 160 ⇒ y = 16

and from (i)

x + 80 x 16 = 1330

⇒ x + 1280 = 1330

⇒ x = 1330 – 1280 = 50

Fixed charges = ₹ 50

and rate per km = ₹ 16

**Question 27.**

**A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay ₹ 4500, whereas a student B who takes food for 30 days, has to pay ₹ 5200. Find the fixed charges per month and the cost of the food per day.**

**Solution:**

Let fixed charges of the hostel = ₹ x

and other charges per day = ₹ y

According to the conditions,

x + 25y = 4500 ……..(i)

x + 30y = 5200 ……(ii)

Subtracting (i) from (ii),

5y = 700

y = 140

and from (i),

x + 25 x 140 = 4500

⇒ x + 3500 = 4500

⇒ x = 4500 – 3500 = 1000

Fixed charges = ₹ 1000

and per day charges = ₹ 140

**Question 28.**

**A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received ₹ 1350 as annual interest. Had he interchanged the amounts invested, he would have received ₹ 45 less as interest. What amounts did he invest at different rates?**

**Solution:**

Let first investment = ₹ x

and second investment = ₹ y

Rate of interest = 10% p.a. for first kind and 8% per second

Interest is for the first investment = ₹ 1350

and for the second = ₹ 1350 – ₹45 = ₹ 1305

According to the conditions,

**Question 29.**

**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 ₹ 9000 per month, find the monthly income of each.**

**Solution:**

Ratio in the income of A and B = 5 : 4

Let A’s income = ₹ 5x and

B’s income = ₹ 4x

and ratio in their expenditures = 7 : 5

Let A’s expenditure = 7y

and B’s expenditure = 5y

According to the conditions,

5x – 7y = 9000 …(i)

and 4x – 5y = 9000 …(ii)

Multiply (i) by 5 and (ii) by 7,

25x – 35y = 45000

28x – 37y = 63000

Subtracting, we get

3x = 18000

⇒ x = 6000

A’s income = 5x = 5 x 6000 = ₹ 30000

and B’s income = 4x = 4 x 6000 = ₹ 24000

**Question 30.**

**A man sold a chair and a table together for ₹ 1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for ₹ 1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.**

**Solution:**

Let cost of one chair = ₹ x

and cost of one table = ₹ y

In first case,

Profit on chair = 25%

and on table = 10%

and selling price = ₹ 1520

According to the conditions,

**Question 31.**

**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. [CBSE 2007C]**

**Solution:**

Distance between two stations A and B = 70 km

Let speed of first car (starting from A) = x km/hr

and speed of second car = y km/hr

According to the conditions,

7x – 7y = 70

⇒ x – y = 10 …(i)

and x + y = 70 …(ii)

Adding (i) and (ii),

2x = 80 ⇒ x = 40

Subtracting (i) and (ii),

2y = 60 ⇒ y = 30

Speed of car A = 40 km/hr

and speed of car B = 30 km/hr

**Question 32.**

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

**Solution:**

Let uniform speed of the train = x km/hr

and time taken = y hours

Distance = x x y = xy km

Case I:

Speed = (x + 5) km/hr

and Time = (y – 3) hours

Distance = (x + 5) (y – 3)

(x + 5) (y – 3) = xy

⇒ xy – 3x + 5y – 15 = xy

-3x + 5y = 15 …(i)

Case II:

Speed = (x – 4) km/hr

and Time = (y + 3) hours

Distance = (x – 4) (y + 3)

(x – 4) (y + 3) = xy

⇒ xy + 3x – 4y – 12 = xy

3x – 4y = 12 …(ii)

Adding (i) and (ii),

y = 27

and from (i),

-3x + 5 x 27 = 15

⇒ -3x + 135 = 15

⇒ -3x = 15 – 135 = -120

⇒ x = 40

Speed of the train = 40 km/hr

and distance = 27 x 40 = 1080 km

**Question 33.**

**Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi. [CBSE 2006C]**

**Solution:**

Let the speed of the train = x km/hr

and speed of taxi = y km/hr

According to the conditions,

**Question 34.**

**Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car. [CBSE 2009C]**

**Solution:**

Distance between stations A and B = 160 km

Let the speed of the car starts from A = x km/hr

and speed of car starts from B = y km/hr

8x – 8y = 160

⇒ x – y = 20 …(i)

and 2x + 2y = 160

⇒ x + y = 80 …(ii)

Adding (i) and (ii)

2x = 100 ⇒ x = 50

and subtracting,

2y = 60 ⇒ y = 30

Speed of car starting from A = 50 km/hr

and from B = 30 km/hr

**Question 35.**

**A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.**

**Solution:**

Distance = 8 km

Let speed of sailor in still water = x km/hr

and speed of water = y km/hr

According to the conditions,

**Question 36.**

**A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.**

**Solution:**

Let speed of a boat = x km/hr

and speed of stream = y km/hr

According to the conditions,

**Question 37.**

**2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.**

**Solution:**

Let a man can do a work in x days

His 1 day’s work =

and a boy can do a work in y days

His 1 day’s work =

According to the conditions,

**Question 38.**

**The length of a room exceeds its breadth by 3 metres. If the length is increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.**

**Solution:**

Let length of a room = x m

and breadth = y m

and area = xy m²

According to the conditions,

x = y + 3 …(i)

(x + 3) (y – 2) = xy

xy – 2x + 3y – 6 = xy

-2x + 3y = 6 …(ii)

-2 (y + 3) + 3y = 6 [From (i)]

-2y – 6 + 3y = 6

⇒ y = 6 + 6 = 12

x = y + 3 = 12 + 3 = 15 …(ii)

Length of room = 15 m

and breadth = 12 m

**Question 39.**

**The area of a rectangle gets reduced by 8 m², when its length is reduced by 5 m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m². Find the length and the breadth of the rectangle.**

**Solution:**

Let length of a rectangle = x m

and breadth = y m

Then area = x x y = xy m²

According to the conditions,

(x – 5) (y + 3) = xy – 8

⇒ xy + 3x – 5y – 15 = xy – 8

⇒ 3x – 5y = -8 + 15 = 7 …..(i)

and (x + 3) (y + 2) = xy + 74

⇒ xy + 2x + 3y + 6 = xy + 74

⇒ 2x + 3y = 74 – 6 = 68 …(ii)

Multiply (i) by 3 and (ii) by 5

**Question 40.**

**The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 in. If the length is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.**

**Solution:**

Let length of a rectangle = x m

and breadth = y m

Then area = xy m²

According to the conditions,

(x + 3) (y – 4) = xy – 67

⇒ xy – 4x + 3y – 12 = xy – 67

⇒ -4x + 3y = -67 + 12 = -55

⇒ 4x – 3y = 55 …(i)

and (x – 1) (y + 4) = xy + 89

⇒ xy + 4x – y – 4 = xy + 89

⇒ 4x – y = 89 + 4 = 93 ….(ii)

⇒ y = 4x – 93

Substituting the value of y in (i),

4x – 3(4x – 93) = 55

⇒ 4x – 12x + 279 = 55

⇒ -8x = 55 – 279 = -224

⇒ x = 28

and y = 4x – 93 = 4 x 28 – 93 = 112 – 93 = 19

Length of rectangle = 28 m

and breadth = 19 m

**Question 41.**

**A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs ₹ 4150 while one full and one half reserved first class tickets cost ₹ 6255. What is the basic first class full fare and what is the reservation charge? [HOTS]**

**Solution:**

Let reservation charges = ₹ x

and cost of full ticket from Mumbai to Delhi

According to the conditions,

x + y = 4150 …(i)

2x + y = 6255

⇒ 4x + 3y = 12510 …(ii)

From (i), x = 4150 – y

Substituting the value of x in (ii),

4 (4150 – y) + 3y = 12510

⇒ 16600 – 4y + 3y = 12510

-y = 12510 – 16600

-y = -4090

⇒ y = 4090

and x = 4150 – y = 4150 – 4090 = 60

Reservation charges = ₹ 60

and cost of 1 ticket = ₹ 4090

**Question 42.**

**Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.**

**Solution:**

Let present age of a man = x years

and age of a son = y years

5 year’s hence,

Man’s age = x + 5 years

and son’s age = y + 5 years

x + 5 = 3 (y + 5) = 3y + 15

⇒ x – 3y = 15 – 5 = 10

x = 10 + 3y …(i)

and 5 years ago,

Man’s age = x – 5 years

and son’s age = y – 5 years

x – 5 = 7 (y – 5) = 7y – 35

x = 7y – 35 + 5 = 7y – 30 …(ii)

From (i) and (ii),

10 + 3y = 7y – 30

⇒ 7y – 3y = 10 + 30

⇒ 4y = 40

⇒ y = 10

and x = 10 + 3y = 10 + 3 x 10 = 10 + 30 = 40

Present age of a man = 40 years

and of son’s age = 10 years

**Question 43.**

**Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find their present ages. [CBSE 2008]**

**Solution:**

Let present age of a man = x years

and age of his son = y years

2 years ago,

Man’s age = x – 2 years

Son’s age = y – 2 years

x – 2 = 5 (y – 2)

⇒ x – 2 = 5y – 10

x = 5y – 10 + 2 = 5y – 8 …(i)

2 years later,

Man’s age = x + 2 years

and son’s age = y + 2 years

x + 2 = 3(y + 2) + 8

x + 2 = 3y + 6 + 8

⇒ x = 3y + 6 + 8 – 2 = 3y + 12 …(ii)

From (i) and (ii),

5y – 8 = 3y + 12

⇒ 5y – 3y = 12 + 8

⇒ 2y = 20

⇒ y = 10

and x = 5y – 8 = 5 x 10 – 8 = 50 – 8 = 42

Present age of man = 42 years

and age of son = 10 years

**Question 44.**

**If twice the son’s age in years is added to the father’s age, the sum is 70. But, if twice the father’s age is added to the son’s age, the sum is 95. Find the ages of father and son.**

**Solution:**

Let age of father = x years

and age of his son = y years

According to the conditions,

2y + x = 10 …(i)

2x + y = 95 …(ii)

From (i),

x = 70 – 2y

Substituting the value of x in (ii),

2 (70 – 2y) + y = 95

⇒ 140 – 4y + y = 95

⇒ -3y = 95 – 140 = -45

⇒ -3y = -45

⇒ y = 15

and x = 70 – 2y = 70 – 2 x 15 = 70 – 30 = 40

Age of father = 40 years

and age of his son = 15 years

**Question 45.**

**The present age of a woman is 3 years more than three times the age of her daughter. Three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.**

**Solution:**

Let present age of a woman = x years

and age of her daughter = y years

According to the conditions,

x = 3y + 3 …(i)

3 years hence,

Age of woman = x + 3 years

and age of her daughter = y + 3 years

x + 3 = 2 (y + 3) + 10

⇒ x + 3 = 2y + 6 + 10

⇒x = 2y + 16 – 3 = 2y + 13 …(ii)

From (i),

3y + 3 = 2y + 13

⇒ 3y – 2y = 13 – 3

⇒ y = 10

and x = 3y + 3 = 3 x 10 + 3 = 30 + 3 = 33

Present age of woman = 33 years

and age of her daughter = 10 years

**Question 46.**

**On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery seller gains ₹ 7. If he sells the tea set at 5% gain and the lemon set at 10% gain, he gains ₹ 13. Find the actual price of each of the tea set and the lemon set. [HOTS]**

**Solution:**

Let cost price of tea set = ₹ x

and of lemon set = ₹ y

According to the conditions,

**Question 47.**

**A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Mona paid ₹ 21 for a book kept for 7 days, while Tanvy paid ₹ 21 for the book she kept for 5 days. Find the fixed charge and the charge for each extra day.**

**Solution:**

Let fixed charges = ₹ x (for first three days)

and then additional charges for each day = ₹ y

According to the conditions,

Mona paid ₹ 27 for 7 dyas

x + (7 – 3) x y = 27

⇒ x + 4y = 27

and Tanvy paid ₹ 21 for 5 days

x + (5 – 3) y = 21

⇒ x + 2y = 21 …(ii)

Subtracting,

2y = 6 ⇒ y = 3

But x + 2y = 21

⇒ x + 2 x 3 = 21

⇒ x + 6 = 21

⇒ x = 21 – 6 = 15

Fixed charges = ₹ 15

and additional charges per day = ₹ 3

**Question 48.**

**A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of a 40% acid solution? [HOTS]**

**Solution:**

Let x litres of 50% solution be mixed with y litres of 25% solution, then

x + y = 10 …(i)

Subtracting (i) from (ii),

x = 6

and x + y = 10

⇒ 6 + y = 10

⇒ y = 10 – 6 = 4

50% solution = 6 litres

and 25% solution = 4 litres

**Question 49.**

**A jeweller has bars of 18-carat gold and 12- carat gold. How much of each must be melted together to obtain a bar of 16-carat gold, weighing 120 g? (Given : Pure gold is 24-carat). [HOTS]**

**Solution:**

Let x g of 18 carat be mixed with y g of 12 carat gold to get 120 g of 16 carat gold, then

x + y = 120 …(i)

Now, gold % in 18-carat gold = x 100 = 75%

⇒ 3x + 2y = 320 …(ii)

From (i),

x = 120 – y

Substituting the value of x in (ii),

3 (120 – y) + 2y = 320

⇒ 360 – 3y + 2y = 320

⇒ -y = 320 – 360

⇒ -y = -40

⇒ y = 40

and 40 + x = 120

⇒ x = 120 – 40 = 80

Hence, 18 carat gold = 80 g

and 12-carat gold = 40 g

**Question 50.**

**90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of acids to be mixed to form the mixture. [HOTS]**

**Solution:**

Let x litres of 90% pure solution be mixed withy litres of 97% pure solution to get 21 litres of 95% pure solution. Then,

x + y = 21 …(i)

⇒ 90x + 97y = 1995

From (i), x = 21 – y

Substituting the value of x in (ii),

90 (21 – y) + 97y = 1995

⇒ 1890 – 90y + 97y = 1995

⇒ 7y = 1995 – 1890 = 105

⇒ y =15

and x = 21 – y = 21 – 15 = 6

90% pure solution = 6 litres

and 97% pure solution = 15 litres

**Question 51.**

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

**Solution:**

Let larger supplementary angle = x°

and smaller angle = y°

According to the conditions,

x + y = 180° …(i)

x = y + 18° …(ii)

From (i),

y + 18° + y = 180°

⇒ 2y = 180° – 18° = 162°

⇒ 2y = 162°

⇒ y = 81°

and x= 180°- 81° = 99°

Hence, angles are 99° and 81°

**Question 52.**

**In a ∆ABC, ∠A = x°, ∠B = (3x – 2)°, ∠C = y° and ∠C – ∠B = 9°. Find the three angles.**

**Solution:**

In ∆ABC,

∠A = x, ∠B = (3x – 2)°, ∠C = y°, ∠C – ∠B = 9°

**Question 53.**

**In a cyclic quadrilateral ABCD, it is given that ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)° and ∠D = (4x – 5)°. Find the four angles.**

**Solution:**

In a cyclic quadrilateral ABCD,

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E are helpful to complete your math homework.

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