**Fractions** Class- 7th RS Aggarwal Exe-2B Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-2 Fractions for ICSE Class-7 **Foundation RS Aggarwal Mathematics** of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-2B to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-7 Mathematics.

## Fractions Class- 7th RS Aggarwal Exe-2B Goyal Brothers ICSE Math Solution

Board | ICSE |

Publications | Goyal brothers Prakashan |

Subject | Maths |

Class | 7th |

Chapter-2 | Fractions |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of Exe-2B |

Academic Session | 2023 – 2024 |

**Exercise – 2B**

Fractions Class- 7th RS Aggarwal Exe-2B Goyal Brothers ICSE Math Solution

**Que: 1. Find the reciprocal of :**

**(i) 13/17**

**Solution: **13/17

= 1(4/13)

**(ii) 3/313**

**Solution:** 3/313

= 313/3

= 104(1/3)

**(iii) 217**

**Solution: **217

= 1/217

**(iv) 1/1024**

**Solution:** 1/1024

= 1024

**(v) 3(1/3)**

**Solution: **3(1/3)

= 10/3

= 3/10

**(vi) 11(1/11)**

**Solution: **11(1/11)

= 122/11

= 11/122

**(vii) 1(1/2)**

**Solution: **1(1/2)

= 3/2

= 2/3

**(viii) 125(1/8)**

**Solution: **125(1/8)

= 1008/8

= 8/1008

**2. Divide :**

##### (**i) (1/2) ÷ (1/3)**

**Solution: **(1/2) ÷ (1/3)

= (1/2) × (3/1)

= 3/2

= 1(1/2)

**(ii) (3/14) ÷ (11/42)**

**Solution: **(3/14) ÷ (11/42)

= (3/14) × (42/11)

= (3/1) × (3/11)

= 9/11

**(iii) 1 ÷ 6(2/3)**

**Solution: **1 ÷ 6(2/3)

= 1 ÷ (20/3)

= 1 × (3/20)

= 3/20

**(iv) 13(1/3) ÷ 10**

**Solution: **13(1/3) ÷ 10

= (40/3) ÷ 10

= (40/3) × 10

= 4/3 = 1(1/3)

**(v) 12(1/12) ÷ (5/36)**

**Solution: **12(1/12) ÷ (5/36)

= (145/12) ÷ (5/36)

= (145/12) × (36/5)

= 29 × 3

= 87

**(vi) (13/46) ÷ 2(1/11)**

**Solution: **(13/46) ÷ 2(1/11)

= (13/46) ÷ (23/11)

= (13/46) × (11/23)

= 143/1058

**(vii) (20/31) ÷ 54**

**Solution:** (20/31) ÷ 54

= (20/31) × (1/54)

= (10/31) × (1/7)

= 10/837

**(viii) 9(4/5) ÷ 3(23/25)**

**Solution: **9(4/5) ÷ 3(23/25)

= (49/5) ÷ (98/25)

= (1/1) × (5/2)

= 5/2 = 2(1/2)

**(ix) 32(1/2) ÷ 8(3/4)**

**Solution: **32(1/2) ÷ 8(3/4)

= (65/2) ÷ (35/4)

= (65/2) × (4/35)

= 26/7 = 3(5/7)

**(x) 8(1/21) ÷ 1(6/7)**

**Solution: **8(1/21) ÷ 1(6/7)

= (169/21) ÷ (13/7)

= (169/21) × (7/13)

= (13/3) × (1/1)

= 13/3 = 4(1/3)

**(xi) 6(3/16) ÷ 3(1/7)**

**Solution: **6(3/16) ÷ 3(1/7)

= (99/16) ÷ (22/7)

= (99/16) × (7/22)

= 63/32 = 1(31/32)

**(xii) 3(11/15) ÷ 19(3/5)**

**Solution: **3(11/15) ÷ 19(3/5)

= (56/15) ÷ (98/5)

= (56/15) ×(5/98)

= 28/147 = 4/21

**3. By what fraction should 9/35 be multiplied to get 7/15 ?**

**Solution: **First No. = 9/35

Second No. = (7/5) ÷ (9/35)

= (7/5) × (35/9)

= (7 × 7)/(3 × 9)

= 49/27 = 1(22/27)

**4. If the cost of 6(1/4) m of cloth is ₹1421(7/8), Then find the cost of 1 meter of cloth.**

**Solution: **6(1/4) m of cloth cost = ₹1421(7/8)

1 m of cloth cost = ₹1421(7/8) ÷ 6(1/4)

= ₹(11375/8) ÷ (25/4)

= ₹(11375/8) × (4/25)

= ₹455/2 = ₹227(1/2)

**5. How many pieces of length 3(5/7) m each can be cut from a wall paper 52 m long?**

**Solution: **Total wall paper = 52m

length of pieces = 3(5/7) m = 26/7m

Number of pieces = 26/7 ÷ 52

= (52/1) × (7/26)

= 14m

**6. A log of wood 6(2/7) m in length, was cut into 11 pieces. What is the length of each pieces?**

**Solution: **L of wood = 6(2/7) m = 44/7 m

Cutting pieces = 11 pieces

L of each pieces = 44/7 ÷ 11

= (44/7) × (1/11)

= 4/7 m

**7. A car covers 641(1/4) km in 43(1/8) liters of fuel. How much distance can this car cover in 1 liter of fuel?**

**Solution: **Car covered distance in 43(1/8) = 44/8 liters fuel = 641(1/4) km = 2565/4

Distance covered in 1 liter fuel = ((2565/4) ÷ 43(1/8))km

= 2565/4 ÷ 345/8

= 342/23

= 14(20/23)km

**8. The area of a rectangular plot of land is 817(4/5) sq. m. If its breadth is 21(3/4) m, find its length.**

**Solution: **Plot area 817(4/5) sq. m.

Plot breadth = 21(3/4) m

Length of plot = (817(4/5) ÷ 21(3/4)

= (4089/5) ÷ (87/4)

= (4089/5) × (87/4)

= (47 × 4)/(5 × 1)

= 188/5m

= 37(3/5)m

**9. The area of a sheet of paper is 623(7/10) sq. cm. If its length is 29(7/10) cm, find its width.**

**Solution: **Area of sheet of paper = 623(7/10) sq. = 6237/10 sq. cm.

Length of sheet of paper = 29(7/10) cm = 297/10 m.

width = ((6237/10) ÷ (297/10))

= (6237/10) × (10/297)

= (21 × 1)/(1 × 1)

= 21 m.

**10. A bundle of 500 sheets of paper has a net weight of 2(3/10) kg. Find the weight of 1 sheet of paper.**

**Solution: **Weight of 500 sheet of paper = 2(3/10) = 23/10 kg

Weight of 1 sheet of paper = ((23/10) ÷ 500)

= (23/10) ÷ 500

= (23/10) × (1/500)

= 23/5000

**11. A car travels 283(1/2) km in 4(2/3) hours. How far does this car go in 1 hour, travelling at the same speed?**

**Solution: **Distance covered by car in 4(2/3) hours = 283(1/2) = 567/2km

Distance covered by car in 1 hours = ((567/2) ÷ (14/3))km

= (567/2) × (3/14)

= (81 × 1)/(1 × 2)

= 81/2

= 40(1/2)

**12. The product of two fractions is 8(7/25). If one of them is 3(1/15), find the other.**

**Solution: **Two fractions = 8(7/25) = 207/25

First fractions = 3(1/15) = 46/15

Second number = ((207/25) ÷ (46/15))

= (207/25) × (15/46)

= 621/230

= 27/10

= 2(7/10)

**— : end of Fractions Class- 7th RS Aggarwal Exe-2B Goyal Brothers ICSE Math Solution :–**

**Return to- ICSE Class -7 RS Aggarwal Goyal Brothers Math Solutions**

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