Fractions Class 6 RS Aggarwal Exe-4D Addition and Subtraction Goyal Brothers Prakashan ICSE Foundation Maths Solutions. We provide step by step Solutions of ICSE Class-6 **Foundation RS Aggarwal Mathematics** of Goyal Brothers Prakashan. Visit official Website **CISCE** for detail information about ICSE Board Class-6 Mathematics.

## Fractions Class 6 RS Aggarwal Exe-4D Addition and Subtraction Goyal Brothers Prakashan ICSE Foundation Maths Solutions

Board | ICSE |

Publications | Goyal brothers Prakashan |

Subject | Maths |

Class | 6th |

Ch-4 | Fraction |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of Exe-4D |

Academic Session | 2023 – 2024 |

### Addition and Subtraction of Fractions

Fractions Class 6 RS Aggarwal Exe-4D Goyal Brothers Prakashan ICSE Foundation Maths Solutions.

**Page- 73,74**

**Exercise- 4D**

**Find the sum :**

**Que-1: (i) 2/7 + 3/7 (ii) 5/8 + 1/8 (iii) 7/9 + 4/9 (iv) 1(5/6) + 1/6**

**Solution- **(i) 4/7

(ii) 6/8 = 3/4

(iii) 11/9 = 1*2/9

(iv) 11/6 + 1/6 = 12/6 = 2

**Que-2: (i) 5/14 + 3/14 + 1/14 (ii) 7/12 + 5/12 + 11/12 (iii) 1(7/8) + 3/8 + 1(5/8)**

**Solution- **(i) 9/14

(ii) 23/12 = 1*11/12

(iii) 15/8 + 3/8 + 13/8 = 31/8 = 3*7/8

**Que-3: (i) 4/9 + 5/6 (ii) 5/12 + 9/16 (iii) 7/12 + 13/18**

**Solution- **(i) the least common multiple of 9 and 6 is 18.

(4×2)/(9×2) + (5×3)/(6×3)

= 8/18 + 15/18

= 23/18 = 1*5/18

(ii) L.C.M of 12 and 16 is 48

(5×4)/(12×4) + (9×3)/(16×3)

= 20/48 + 27/48 = 47/48.

(iii) L.C.M of 12 and 18 is 36.

(7×3)/(12×3) + (13×2)/(18×2)

= 21/36 + 26/36

= 47/36 = 1*11/36.

**Que-4: (i) 3/10 + 8/15 + 7/20 (ii) 7/8 + 9/16 + 17/24 (iii) 5/6 + 8/9 + 11/18 + 13/27**

**Solution- **(i) The least common denominator (LCD) for 10, 15, and 20 is 60.

So, rewrite each fraction with the denominator of 60:

3/10 = (3×6)/(10×6) = 18/60

8/15 = (8×4)/(15×4) = 32/60

7/20 = (7×3)/(20×3) = 21/60

Now, add the fractions:

18/60 + 32/60 + 21/60

=(18+32+21)/60

=71/60 = 1(11/60)

So, 3/10+8/15+7/20 = 71/60 = 1(11/60).

(ii) LCM (8, 16, 24) = 48

= (7×6)/(8×6) + (9×3)/(16×3) + (17×2)/(24×2)

= (42 + 27 + 34)/48

= 103/48 = 2(7/48).

(iii) LCM of 6,9,18 and 27 is 54.

= (5×9)/(6×9) + (8×6)/(9×6) + (11×3)/(18×3) + (12×2)/(27×2)

= 45/54+48/54+33/54+26/54

= 152/54 = 76/27 = 2(22/27)

**Que-5: (i) 4(1/6) + 2(5/8) + 3(7/12) (ii) 1(1/2) + 2(2/3) + 3(3/4) + 4(4/5) (iii) 3(1/3) + 2(2/9) + 4(1/2) + 1(13/18)**

**Solution- **(i) 25/6 + 21/8 + 43/12

LCM of 6,8 and 12 is 24

= (25×4)/(6×4) + (21×3)/(8×3) + (43×2)/(12×2)

= 100/24 + 63/24 + 86/24

= 249/24 = 83/8 =10(3/8).

(ii) 3/2 + 8/3 + 15/4 + 24/5

LCM of 2,3,4 and 5 is 60.

= (3×30)/(2×30) + (8×20)/(3×20) + (15×15)/(4×15) + (24×12)/(5×12)

= 90/60 + 160/60 + 225/60 + 288/60

= 763/60 = 12(43/60)

(iii) 10/3 + 20/9 + 9/2 + 31/18

LCM of 3,9,2 and 18 is 18

= (10×6)/(3×6) + (20×2)/(9×2) + (9×9)/(2×9) + (31×1)/(18×1)

= 60/18 + 40/18 + 81/18 + 31/18

= 212/18 = 106/9 = 11(7/9).

**Find the difference :**

**Que-6: (i) 9/11 – 5/11 (ii) 8(5/7) – 6/7 (iii) 3(3/8) – 1(7/8) (iv) 7/4 – 3/4**

**Solution- **(i) 4/11

(ii) 61/7 – 6/7 = 55/7 = 7(6/7)

(iii) 27/8 – 15/8 = 12/8 = 3/2 = 1(1/2)

(iv) 4/4 = 1

**Que-7: (i) 7/15 – 9/20 (ii) 4(1/7) – 2(1/14) (iii) 1(7/8) – 5/12 (iv) 5 – 2(3/4) **

**Solution- **(i) LCM of 15 and 20 is 60.

= (7×4)/(15×4) – (9×3)/(20×3)

= 28/60 – 27/60

= 1/60

(ii) 29/7 – 29/14

LCM of 7 and 14 is 14.

= (29×2)/(7×2) – (29×1)/(14×1)

= 58/14 – 29/14

= 29/14 = 2(1/14).

(iii) 15/8 – 5/12

LCM of 8 and 12 is 24.

= (15×3)/(8×3) – (5×2)/(12×2)

= 45/24 – 10/24

= 35/24 = 1(11/24)

(iv) 5/1 – 11/4

LCM of 1 and 4 is 4.

= (5×4)/(1×4) – (11×1)/(4×1)

= 20/4 – 11/4

= 9/4 = 2(1/4).

**Que-8: 5(1/8) – 4(1/12) (ii) 6(1/6) – 3(7/10) (iii) 12 – 6(5/8)**

**Solution- **(i) 41/8 – 49/12

LCM of 8 and 12 is 24.

= (41×3)/(8×3) – (49×2)/(12×2)

= 123/24 – 98/24

= 25/24 = 1(1/24)

(ii) 37/6 – 37/10

LCM of 6 and 10 is 30.

= (37×5)/(6×5) – (37×3)/(10×3)

= 185/30 – 111/30

= 74/30 = 37/15 = 2(7/15).

(iii) 12/1 – 53/8

LCM of 1 and 8 is 8.

= (12×8)/(1×8) – (53×1)/(8×1)

= 96/8 – 53/8

= 43/8 = 5(3/8).

**Simplify :**

**Que-9: 4/9 – 5/12 + 1/4**

**Solution- **The least common multiple (LCM) of 9, 12, and 4 is 36.

= (4×4)/(9×4) – (5×3)/(12×3) + (1×9)/(4×9)

= 16/36 – 15/36 + 9/36

= (16 – 15 + 9)/36

= 10/36 = 5/18 Ans,

**Que-10: 6(2/3) + 4(1/6) – 2(2/9)**

**Solution- **20/3 + 25/6 – 20/9

LCM of 3,6 and 9 is 18

= (20×6)/(3×6) + (25×3)/(6×3) – (20×2)/(9×2)

= 120/18 + 75/18 – 40/18

= (120 + 75 – 40)/18

= 155/18 = 8(11/18) Ans.

**Que-11: 9 – 4(1/2) – 2(1/5)**

**Solution- **9/1 – 9/2 – 11/5

LCM of 1,2 and 5 is 10

= (9×10)/(1×10) – (9×5)/(2×5) – (11×2)/(5×2)

= 90/10 – 45/10 – 22/10

= (90 – 45 – 22)10

= 23/10 = 2(3/10) Ans.

**Que-12: 10(3/4) – 5(1/8) – 4(5/12)**

**Solution- **43/4 – 41/8 – 53/12

LCM of 4,8 and 12 is 24.

= (43×6)/(4×6) – (41×3)/(8×3) – (53×2)/(12×2)

= 258/24 – 123/24 – 106/24

= (258 – 123 – 106)/24

= 29/24 = 1(5/24) Ans.

**Que-13: 6(4/5) – 3(4/15) + 4(3/10)**

**Solution- **34/5 – 49/15 – 43/10

LCM of 5,10 and 15 is 30.

= (34×6)/(5×6) – (49×2)/(15×2) + (43×3)/(10×3)

= 204/36 – 98/30 + 129/30

= (204 – 98 + 129)/30

= 235/30 = 47/6 = 7(5/6) Ans.

**Que-14: 4(5/12) + 3(11/18) – 2(7/24)**

**Solution- **53/12 + 65/18 – 55/24

LCM of 12,18 and 24 is 72.

= (53×6)/(12×6) + (65×4)/(18×4) – (55×3)/(24×3)

= 318/72 + 260/72 – 165/72

= (318 + 260 – 165)/72

= 413/72 = 5(53/72) Ans.

**Que-15: Subtract the sum of 9(3/4) and 3(5/6) from 15(7/12)**

**Solution- **First add the numbers

= 9(3/4) +3(5/6)

= 39/4 + 23/6

= 117/12 + 46/12

= 163/12

Now, subtract the number from a given number.

= 15 (7/12) – 163/12

= 187/12 – 163/12

= 24/12 = 2 Ans.

**Que-16: Subtract the sum of 2(5/12) and 3(3/4) from the sum of 7(1/3) and 5(1/6).**

**Solution- **The sum of 2(5/12) and 3(3/4) is

= 2(5/12) + 3(3/4) = 29/12 + 15/4

= (29 + 45)/12

= 74/12 ——(1)

The sum of 7(1/3) and 5(1/6) is

= 7(1/3) + 5(1/6) = 22/3 + 31/6

= (44 + 31)/6

= 75/6 ——(2)

Now subtracting (1) from (2)

= 75/6 – 74/12

= 150/12 – 74/12

= (150 – 74)/12

= 76/12 = 19/3 = 6(1/3) Ans.

**Que-17: What should be added to 9(4/7) to get 16?**

**Solution- **Let the number be y

y + 9(4/7) = 16

y = 16 – 9(4/7)

y = 16 – 67/7

y = (112 – 67)/7

y = 45/7 = 6(3/7) Ans.

**Que-18: What must be subtracted from 9(1/6) to get 6(1/9)?**

**Solution- **Let the subtracted no. be y

55/6 – y = 55/9

55/6 – 55/9 = y

y = (55×9)/(6×9) – (55×6)/(9×6)

y = 495/54 – 330/56

y = 165/56 = 55/18 = 3(1/8) Ans.

**Que-19: Of 17/20 and 21/25, which is greater than and by how much?**

**Solution- **17/20 and 21/25

Cross multiplying

17×25>20×21

425>420

Therefore, 17/20 is greater.

(17/20)-(21/25)

={(17×5)-(21×4)}/100

=(85-84)/100

=1/100

17/20 is greater by 1/100.

**Que-20: The sum of two fractions is 14(5/12). If one of them is 7(2/3), find the other.**

**Solution-** m = 14(5/12) = 173

other = y

then, y + 7(2/3) = 14(5/12)

y = 14(5/12) – 7(2/3)

y = 173/12 – 23/3

y = (173 – 92)/12

y = 81/12 = 27/4 = 6(3/4) Ans.

**Que-21: From a piece of wire 12(3/4) m long, a small piece of length 3(5/6) m has been cut off. What is the length of remaining piece?**

**Solution- **small piece = 3(5/6) = 23/6

remaining piece be y

total piece = 12(3/4) = 51/4

y + 23/6 = 51/4

y = 51/4 – 23/6

y = (153 – 46)/12

y = 107/12 = 8(11/12)m Ans.

**Que-22: Three boxes weigh 9(1/2) kg, 14(1/5) kg, and 18(3/4) kg respectively. A porter carriers all the three boxes. What is the total weight carried by the porter?**

**Solution- **Total weight carried by the porter = sum of all boxes

= 9(1/2) + 14(1/5) + 18(3/4)

= 19/2 + 71/5 + 75/4

L.C.M of 2, 5 and 4 is 20.

= (19×10)/(2×10) + (71×4)/(5×4) + (75×5)/(4×5)

= 190/20 + 284/20 + 375/20

= 849/20 = 42(6/9) kg Ans.

**Que-23: On one day, a labourer earned Rs125. Out of this money, he spent Rs68(1/2) on food, Rs20(3/4) on tea and Rs16(2/5) on other eatables. How much does he save on that day?**

**Solution- **Based on the given problem, the money saved by the labourer can be given as:

Money saved by the labourer = Total earned money – (Money spent on food + Money spent on tea + Money spent on other eatables)

Substitute the known values in the above equation.

Money saved by the labourer = 125 – [68(1/2) + 20(3/4) + 16(2/5)]

Money saved by the labourer = 125 -[137/2 + 83/4 + 82/5]

Money saved by the labourer = 125 -[1370 + 415 + 328]/20

On further solving,

Money saved by the labourer = 125 -[2113/20]

Money saved by the labourer = (2500 – 2113)/20

Money saved by the labourer = 387/20

Money saved by the labourer = 19(7/20)Rs Ans.

**— : end of Fractions Class 6 RS Aggarwal Exe-4D Addition and Subtraction Goyal Brothers Publication :–**

**Return to :- ICSE Class -6 RS Aggarwal Goyal Brothers Math Solutions**

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