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

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

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​

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)

= 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 :–

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