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

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 3F.

### RS Aggarwal Solutions Class 10 Chapter 3

- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
- RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

**Very-Short and Short-Answer Questions**

**Question 1.**

**Solution:**

**Question 2.**

**Solution:**

**Question 3.**

**Solution:**

**Question 4.**

**Solution:**

**Question 5.**

**Solution:**

**Question 6.**

**Solution:**

**Question 7.**

**Solution:**

Let first, number = x

and second number = y

x – y = 5

**Question 8.**

**Solution:**

Let cost of one pen = ₹ x

and cost of one pencil = ₹ y

According to the conditions,

5x + 8y = 120 …(i)

8x + 5y = 153 …(ii)

Adding, we get

13x + 13y = 273

x + y = 21 …(iii) (Dividing by 13)

and subtracting (i) from (ii),

3x – 3y = 33

⇒ x – y = 11 …….(iv) (Dividing by 3)

Again adding (iii) and (iv),

2x = 32 ⇒ x = 16

Subtracting,

2y = 10 ⇒ y = 5

Cost of 1 pen = ₹ 16

and cost of 1 pencil = ₹ 5

**Question 9.**

**Solution:**

Let first number = x

and second number = y

According to the conditions,

and x + y = 80

⇒ x + 15 = 80

x = 80 – 15 = 65

Numbers are : 65, 15

**Question 10.**

**Solution:**

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

and ten’s digit = y

Number = x + 10y

By reversing its digits One’s digit = y

and ten’s digit = x

Then number = y + 10x

According to the conditions,

x + y = 10 …(i)

x + 10y – 18 = y + 10x

x+ 10y – y – 10x = 18

⇒ -9x + 9y = 18

⇒ x – y = -2 (Dividing by -9) …..(ii)

Adding (i) and (ii),

2x = 8 ⇒ x = 4

and by subtracting,

2y = 12 ⇒ y = 6

Number = x + 10y = 4 + 10 x 6 = 4 + 60 = 64

**Question 11.**

**Solution:**

Let number of stamps of 20p = x

and stamps of 25 p = y

According to the conditions,

x + y = 47 …..(i)

20x + 25y = 1000

4x + 5y = 200 …(ii)

From (i), x = 47 – y

Substituting the value of x in (ii),

4 (47 – y) + 5y = 200

188 – 4y + 5y = 200

⇒ y = 200 – 188 = 12

and x + y = 47

⇒ x + 12 = 47

⇒ x = 47 – 12 = 35

Hence, number of stamps of 20 p = 35

and number of stamps of 25 p = 12

**Question 12.**

**Solution:**

Let number of hens = x

and number of cows = y

According to the conditions,

x + y = 48 …..(i)

x x 2 + y x 4 = 140

⇒ 2x + 4y = 140

⇒ x + 2y = 70 ……(ii)

Subtracting (i) from (ii),

y = 22

and x + y = 48

⇒ x + 22 = 48

⇒ x = 48 – 22 = 26

Number of hens = 26

and number of cows = 22

**Question 13.**

**Solution:**

**Question 14.**

**Solution:**

**Question 15.**

**Solution:**

12x + 17y = 53 …(i)

17x + 12y = 63 …(ii)

Adding, 29x + 29y = 116

Dividing by 29,

x + y = 4 …(iii)

Subtracting,

-5x + 5y = -10

⇒ x – y = 2 …(iv) (Dividing by -5)

Adding (iii) and (iv)

2x = 6 ⇒ x = 3

Subtracting,

2y = 2 ⇒ y = 1

x = 3, y = 1

x + y = 3 + 1 = 4

**Question 16.**

**Solution:**

**Question 17.**

**Solution:**

kx – y = 2

6x – 2y = 3

**Question 18.**

**Solution:**

**Question 19.**

**Solution:**

**Question 20.**

**Solution:**

**Question 21.**

**Solution:**

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

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