This document provides solutions to 15 worksheets on integers from the mathematics textbook for Class VII. Each worksheet contains multiple-choice and short answer questions testing concepts like addition, subtraction, multiplication and division of integers, properties of integers, and solving word problems involving integers. The solutions provide step-by-step workings and explanations for arriving at the answers to help students understand the concepts.

1. Solution to Me N Mine Mathematics VII Oct 2011 By Sunil
Solutions to
Mathematics
PULLOUT WORKSHEETS
FOR CLASS VII
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2. CONTENTS
1. Integers
Worksheets (1 to 7) ..................................................................................................... 5
2. Fractions
Worksheets (8 to 14) ................................................................................................. 14
3. Decimals
Worksheets (15 to 21) ............................................................................................... 27
4. Data Handling
Worksheets (22 to 28) ............................................................................................... 37
5. Simple Equations
Worksheets (29 to 36) ............................................................................................... 52
6. Lines and Angles
Worksheets (37 to 43) ............................................................................................... 67
7. Triangles
Worksheets (44 to 50) ............................................................................................... 78
8. Congruence
Worksheets (51 to 56) ............................................................................................... 89
9. Comparing Quantities
Worksheets (57 to 69) ............................................................................................... 97
–2–

3. 10. Rational Numbers
Worksheets (70 to 76) ............................................................................................. 116
11. Symmetry and Practical Geometry
Worksheets (77 to 83) ............................................................................................. 128
12. Perimeter and Area
Worksheets (84 to 91) ............................................................................................. 140
13. Algebraic Expressions
Worksheets (92 to 98) ............................................................................................. 155
14. Exponents and Powers
Worksheets (99 to 105) ........................................................................................... 163
15. Visualising Solid Shapes
Worksheets (106 to 111) ......................................................................................... 172
PRACTICE PAPERS (1 to 5) ................................................................................... 179
–3–

5. Chapter
1 INTEGERS
WORKSHEET–1 5. (C)When two negative integers are
added, we always get a negative integer,
1. (A) (– 21) + (– 29) = – 21 – 29
e.g.,
= – (21 + 29)
(– 7) + (– 13) = – 7 – 13 = – (7 + 13)
= – (50) = – 50.
= – 20
2. (C)Let us take all the options one by
= a negative integer.
one.
6. (A) On a number line when we add a
175 175
(A)175 ÷ (–175) = =– positive integer, we always move to the
–175 175
right.
= – 1.
7. (B) Let the additive inverse of – 6 is a ,
(B) (– 16) × 10 = – (16 × 10) = – (160)
then
= – 160.
– 6 + a = 0 ∴ a = 6.
– 70 70 8. (D) 7 + 3 = 10 ≠ – 10.
(C)(–70) ÷ (– 10) = = = 7.
– 10 10
9. (A) Let us take option (A).
3. (B) Clearly, second term – 3 × 1 = – (3 × 1) = – (3) = – 3
= First term – 3 1 × (– 3) = – (1 × 3) = – (3) = – 3
= 10 – 3 = 7 Hence, – 3 × 1 = – 3 = 1 × (– 3) is correct.
Also, third term = Second term – 3 10. (C) Let the blank space be filled by a,
=7–3=4 then
Similarly, fourth term = Third term – 3 a × (– 9) = – 72 ⇒ – 9a = – 72
=4–3 = 1
– 72 72
and fifth term = Fourth term – 3 ⇒ a= = ⇒ a = 8.
–9 9
= 1 – 3 = – 2.
11. (B) If in a fraction, 0 is at the place of
4. (B) (– 3) + 7 – (19) = – 3 + 7 – 19 denominator, then the fraction is not
= 7 – 3 – 19 defined.
= 7 – (3 + 19) = 7 – 22 a
∴ a÷ 0 = is not defined.
= – 15 0
15 – 8 + (– 9) = 15 – 8 – 9 a
12. (B) a ÷ 48 = – 1 or =–1
= 15 – (8 + 9) 48
= 15 – 17 = – 2 or a = – 1 × 48 or a = – 48.
Clearly, – 15 is less than – 2 13. (A) (– 41) ÷ [(– 40) + (– 1)]
so, (– 3) + 7 – (19) is less than 15 – 8 + (– 9) – 41
= – 41 ÷ [– 40 – 1) = = 1.
∴ (– 3) + 7 – (19) < 15 – 8 + (– 9). – 41
I N T E G E R S 5

6. 14. (D) The additive identity of every 3. In this case, the negative integer must
integer is 0. be less than – 10. Suppose this is – 16.
15. (B) As the additive identity of every Now,
integer is zero, the additive identity of – 16 + Positive integer = – 10
– 23 is 0. ∴ Positive integer = – 10 – (– 16)
16. (C) Let us take option (C). = – 10 + 16
LHS = (– 12) + 2 + 10 = – 12 + (2 + 10) = + 6.
= – 12 + 12 = 0 Hence, the required pair is – 16 and 6.
RHS = 12 + (– 2) + (– 10) = 12 – 2 – 10 4. (i) 400 + (– 31) + (– 71)
= 12 – (2 + 10) = 12 – 12 = 0 = 400 – 31 – 71
Clearly, LHS = RHS. = 400 – (31 + 71)
= 400 – 102 = 298.
17. (D) – 212 + 99 – 87 = 99 – 87 – 212
(ii) 937 + (– 37) + 100 + (– 200) + 300
= 99 – (87 + 212)
= 937 – 37 + 100 – 200 + 300
= 99 – 299
= 937 + 100 + 300 – 37 – 200
= – 200.
= (937 + 100 + 300) – (37 + 200)
18. (D) Let us take option (D).
= 1337 – 237 = 1100.
 – 16  5. (i) First integer = – 27
[(– 16) ÷ 4] ÷ (– 2) =  ÷ (– 2)
 4   Second integer = – 54
= [– 4] ÷ (– 2) Second integer – First integer
= – 54 – (– 27)
–4
= = 2. = – 54 + 27 = – 27.
–2
which is greater than zero. (ii) First integer = 12
Hence, [(– 16) ÷ 4] ÷ (– 2) < 0 is incorrect. Second integer = – 7
Second integer – First integer
19. (D) Since the multiplicative identity of
any integer is 1, therefore, the multipli- = – 7 – (12)
cative identity of 7 is 1. = – 7 – 12 = – 19.
6. (i) (– 14) × (– 11) × 10
20. (C) We know that addition is commuta-
tive for integers, so a + b = b + a is true Since the number of negative integers
for any integers a and b. in the product is even (here 2), therefore,
their product must be positive.
WORKSHEET–2 ∴ (– 14) × (– 11) × 10 = 14 × 11 × 10
= 154 × 10
1. All integers between – 2 and 2 are – 1, 0
(∵ 14 × 11 = 154)
and 1.
= 1540.
2. The successor of – 380 = – 380 + 1
(ii) (– 4) × (– 5) × (– 2) × (– 1)
= – 379 Since the number of negative integers
The predecessor of – 380 = – 380 – 1 is even (here 4), so their product must
= – 381. be positive.
6 M A T H E M A T I C S – VII

7. ∴ (– 4) × (– 5) × (– 2) × (– 1)
40
=4×5 × 2×1 10. (i) 40 ÷ – 1 = = – 40.
–1
= 4 × 5 × 2 (∵ 2 × 1 = 2)
= 4 × 10 (∵ 5 × 2 = 10) – 37
= 40.
(ii) – 37 ÷ (– 1) = 1 = 37.
7. (– 2 – 5) × (– 6) = (– 7) × (– 6) –1
= 7 × 6 = 42
[∵ (– a) × (– b) = a × b] WORKSHEET – 3
(– 2) – 5 × (– 6) = – 2 – [5 × (– 6)]
35 35
= – 2 – [– 5 × 6] 1. (i) 35 ÷ (– 5) = =– = – 7.
–5 5
[∵ a × (– b) = – a × b]
= – 2 – (– 30) (ii) 0 × (– 2) = 0.
= – 2 + 30 (iii) – 275 + x = 1 ⇒ x = 1 + 275 = 276.
[∵ a – (– b) = a + b] (iv) (– 59) + 1 = – 59 + 1 = – 58.
= 28 2. – 8 on the number line = 8 steps towards
Clearly, 42 > 28 the left of 0.
Therefore, (– 2 – 5) × (– 6) is greater. + 3 on the number line = 3 steps towards
the right of 0.
– 20
8. (i) (– 20) ÷ (– 10) =
– 10
20  –a a 
= = 2.  ∵ = 
10  –b b 
∴ – 8 + 3 = 8 steps towards the left of
– 15 0 and then 3 steps towards
(ii) (– 15) ÷ (– 3) =
–3 the right
15  –a a  = – 5.
= = 5.  ∵ = 
3  –b b  3. The sign of the product depends only
9. (i) 20 × 12 + 20 × (– 4) = 20 × (12 – 4)
on the number of negative numbers.
LHS = 20 × 12 + 20 × (– 4)
= 20 × 12 – 20 × 4 (i) There is even number of negative
[∵ a × (– b) = – a × b] integers, so the product must be
= 20 × (12 – 4) positive.
[∵ a × b – a × c = a × (b – c)] (ii) There is odd number of negative
= RHS. Hence proved. integers, so the product must be
(ii) 14 × 10 + 14 × (– 20) = 14 × (10 – 20) negative.
LHS = 14 × 10 + 14 × (– 20) 4. There are seven days in a week.
= 14 × 10 – 14 × 20
Temperature after the 1st day
[∵ a × (– b) = – a × b]
= 14 × (10 – 20) = 42 °C – 2 °C = 40 °C
[∵ a × b – a × c = a × (b – c] Temperature after the 2nd day
= RHS. Hence proved. = 40 °C – 2 °C = 38 °C
I N T E G E R S 7

8. Temperature after the 3rd day which is RHS.
= 38 °C – 2 °C = 36 °C Let us take right hand side.
Temperature after the 4th day RHS = (– 16) × (18 + 2)
= 36 °C – 2 °C = 34 °C
= (– 16) × 18 + (– 16) × 2
Temperature after the 5th day
[Distributivity]
= 34 °C – 2 °C = 32 °C
Temperature after the 6th day = 18 × (– 16) + 2 × (– 16)
= 32 °C – 2 °C = 30 °C [Commutativity]
Temperature after the 7th day which is LHS. Hence proved.
= 30 °C – 2 °C = 28 °C 8. a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Thus, the temperature after the whole
Let us take LHS of this inequality.
week is 28 °C.
LHS = a ÷ (b + c)
5. (i) 120 – (– 80) = 120 + 80 [ä– (– a) = a]
= 200. Substituting a = 15, b = – 3 and c = 1,
(ii) 0 – (– 50) = 0 + 50 = 50. we get
6. (i) [124 × (– 2)] × (– 5) LHS = 15 ÷ (– 3 + 1) = 15 ÷ (– 2)
= 124 × [(– 2) × (– 5)]
15 15
(Associativity) = =–
–2 2
= 124 × [ 2 × 5]
On the same way,
[∵ (– a) × (– b) = a × b]
RHS = (a ÷ b) + (a ÷ c)
= 124 × 10 = 1240.
= [15 ÷ (– 3)] + (15 ÷ 1)
(ii) [(– 1) × {217 × (– 20)}] × 5
= [{(– 1) × 217} × (– 20)] × 5  15  15
=  +   = – 5 + 15 = 10.
 
(Associativity)  – 3
  1 
= {(– 217) × (– 20)} × 5 Clearly, LHS ≠ RHS
= (– 217) × {(– 20) × 5)} i.e., a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c).
(Associativity)
= (– 217) × (– 20 × 5) 9. a ÷ b = – 4
= (– 217) × (– 100) a
or = –4 or a = –4×b
= 217 × 100 = 21700. b
[∵ (– a) × (– b) = a × b] If b = 1, then a = – 4 × 1 = – 4
7. 18 × (– 16) + 2 × (– 16) = (– 16) × (18 + 2) If b = 2, then a = – 4 × 2 = – 8
Let us take left hand side.
If b = 3, then a = – 4 × 3 = – 12
LHS = 18 × (– 16) + 2 × (– 16)
= (18 + 2) × (– 16) Thus, three pairs of integers (a, b) are
[∵ a × c + b × c = (a + b) × c] (– 4, 1), (– 8, 2) and (– 12, 3).
= (– 16) × (18 + 2) 1 9 3
10. × (– 9) = – =– .
[Commutativity] 12 12 4
8 M A T H E M A T I C S – VII

9. WORKSHEET – 4 4. (i) – 4 × 16 × 25 × 3
= 16 × (– 4) × 25 × 3
–2 2
1. (i) × 25 × (– 1) = × 25 × 1 (Commutativity)
5 5
= 16 × [(– 4) × 25 × 3]
[∵ – a × b × (– c) = a × b × c]
= 16 × [{(– 4) × 25} × 3]
2 2
= × (25 × 1) = × 25 = 16 × [(– 100) × 3]
5 5
= [16 × (– 100)] × 3
[∵ a × 1 = a]
(Associativity)
25
=2× = 2 × 5 = 10. = – 1600 × 3 = – 4800.
5
(ii) 4 + (– 8) + 6 + (– 2)
3 3
(ii) × (– 4) × (– 1) = × 4 × 1 = [4 + (– 8) + 6] + (– 2)
2 2
= [{4 + (– 8)} + 6] + (– 2)
3 4 = [(– 4) + 6] + (– 2)
=×4 = 3×
2 2
= (– 4) + [6 + (– 2)]
= 3 × 2 = 6.
(Associativity)
2. Let three negative integers be – 2, – 3
= – 4 + 4 = 0.
and – 4.
5. Let the other number be a.
Their product = (– 2) × (– 3) × (– 4)
∴ 60 × a = – 180
= (– 2) × [(– 3) × (– 4)]
Dividing both sides by 60, we get
= (– 2) × [3 × 4]
– 180 180
[∵ (– a) × (– b) = a × b] a= =–
60 60
= (– 2) × 12
or a = – 3.
[∵ (– a) × b = – (a × b)]
6. Let the number be b.
= – (24) = – 24.
b
= Negative integer. According to the question, = 14
3
Hence, the product of three negative Multiplying both sides by 3, we get
integer and is a negative integer.
b = 3 × 14 or b = 42
3. (i) – 800000 ÷ (– 200) 7. (i) 34 × (– 1) = – (34 × 1)
– 800000 800000 ∵ – a = a 
[∵ a × (– b) = – (a × b)]
= =  –b b  = – 34. [∵ a × 1 = a]
– 200 200  
8000 (ii) (– 12) × (– 1) = 12 × 1
= = 4000. [∵ (– a) × (– b) = a × b]
2
343 343 = 12. [∵ a × 1 = a]
(ii) 343 ÷ (– 49) = =
– 49 49 8. (i) – 66 – (– 22) = – 66 + 22 = – 44.
∵ a = – a  (ii) 100 – (– 42 + 39)
 –b
 b = 100 – (– 42) – (+ 39)
49 = 100 + 42 – 39
=– = – 7. = 142 – 39 = 103.
7
I N T E G E R S 9

10. – 55 55 = – (20 × 80) = – 1600
9. (i) (– 55) ÷ 11 = =– = – 5.
11 11 RHS = (20 × 5) × (– 16) = 100 × (– 16)
– 77 77 = – (100 × 16) = – 1600.
(ii) =– = – 11.
7 7 So, 20 × [5 × (– 16)] = (20 × 5) × (– 16).
10. 1 hour = 60 minutes
(ii) 18 × [100 + (– 5)] = 18 × 100 + 18 × (– 5)
2 hours = 2 × 60 minutes
Here 18 × [100 + (– 5)]
= 120 minutes.
= 18 × [100 – 5]
∵ In 1 minute the elevator covers a
= 18 × 95 = 1710
depth of 6 metres
And 18 × 100 + 18 × (– 5)
∴ In 2 hours the elevator will cover a
depth of 6 × 120 metres = (18 × 100) – (18 × 5)
i.e., 720 metres. = 1800 – 90 = 1710.
Thus, the elevator will be 720 metres So, 18 × [100 + (– 5)] = 18 × 100 + 18 × (– 5).
below the initial position. 5. Let your home be at O. You was at A.
Now, you are at B.
– 88 88
11. (i) = = 11. AO = 8 km, OB = 4 km
–8 8
You travelled from A to B via O.
– 25 25
(ii) =– = – 5.
5 5
WORKSHEET – 5
1. (i) [4 × (– 112)] × 5
= [– (4 × 112)] × 5
[∵ a × (– b) = – (a × b)]
= [– (448)] × 5 = (– 448) × 5
= – (448 × 5) = – 2240.
(ii) 19 + (– 13 + 3) = 19 + (– 10)
= 19 – 10 = 9. ∴ Required distance travelled by you
2. (i) 25 × 7 × 4 × 3 = 25 × 4 × 7 × 3 = AO + OB = 8 km + 4 km.
= (25 × 4) × (7 × 3) = 12 km.
= 100 × 21 = 2100. 6. Let the position of bird be at A and the
(ii) (– 15) + 24 + 5 + (– 4) position of fish at B. AB is a vertical
straight line.
= (– 15) + 5 + 24 + (– 4)
= (– 15 + 5) + (24 – 4)
= – 10 + 20 = 10.
3. (i) Additive inverse of 15 = – 15.
(ii) Additive inverse of – 23 = 23.
(iii) Additive inverse of 0 = 0.
4. (i) 20 × [5 × (– 16)] = (20 × 5) × (– 16)
LHS = 20 × [5 × (– 16)] = 20 × [– 80]
10 M A T H E M A T I C S – VII

11. Now, required distance WORKSHEET – 6
= AB
1. (i) [13 × 19] × (– 3) = 13 × [19 × (– 3)]
= 6000 m + 1600 m = 7600 m.
(Associativity of multiplication)
a
7. a ÷ b = – 3 or = – 3 Thus, the blank space is filled with 19.
b
or a = – 3b (ii) (– 10) × 9 × (– 10) × 1
[Multiplying both sides by b] = – 10 × 9 × [(– 10) × 1]
If b = 1, then a = – 3(1) = – 3 = – 10 × 9 × (– 10)
∴ (a, b) = (– 3, 1). [∵ (– a) × 1 = – a]
8. 423 × (– 63) – [63 × (– 423)] = – 10 × [9 × (– 10)]
= 423 × (– 63) – [(– 423) × 63] = – 10 × [– (9 × 10)]
(Commutativity) = – 10 × (– 90)
= – (423 × 63) – [– (423 × 63)] Thus, the blank space is filled with – 90.
= – (423 × 63) + (423 × 63) 2. (i) The additive inverse of – 13 = 13.
[∵ – a – (– a) = – a + a] (ii) The additive inverse of 22 = – 22.
= 0. 3. 30125 × 99 – (– 30125)
9. Let one of the two integers be 4. Then = 30125 × 99 + 30125
according to the question, [∵ – (– a) = a]
4 + another integer = – 20 = 30125 × 99 + 30125 × 1
∴ Another integer = – 20 – 4 = – 24 [∵ a = a × 1]
= 30125 × (99 + 1)
Hence, the required pair is – 24, 4.
= 30125 × 100 = 3012500.
10. (i) 80 × [5 × (– 36)] = (80 × 5) × (– 36) 4. (i) (– 5) + (– 3) + 2
Let us take left hand side (LHS). = – 5 – 3 + 2 = – (5 + 3) + 2
LHS = 80 × [5 × (– 36)] = – 8 + 2 = – 6.
= (80 × 5) × (– 36) (ii) (– 613) + (– 111) + (– 500)
[Associativity for multiplication] = – 613 – 111 – 500
= RHS Hence proved. = – (613 + 111 + 500)
(ii) – 4 × 16 × 25 × 3 = {(– 4) × 25} × (16 × 3) = – (1224) = – 1224.
Let us take left hand side (LHS). 5. The difference of – 19 and – 43
= – 19 – (– 43) = – 19 + 43 = 24
LHS = – 4 × 16 × 25 × 3
Now, required value = – 63 + 24
= – 4 × (16 × 25) × 3
= – 39.
= – 4 × (25 × 16) × 3
6. Ascending order is
[Commutativity of multiplication]
– 33, – 10, – 7, – 5, – 3, 0, 4, 6, 11, 19.
= – 4 × 25 × 16 × 3
7. The product of (– 5) × (6) × (– 7) × (– 20)
= {(– 4) × 25} × (16 × 3)
has an odd number of negative integers,
= RHS. Hence proved. so its value must be negative.
I N T E G E R S 11

12. ∴ (– 5) × (6) × (– 7) × (– 20) ∴ (– 4) × (– 5) × (– 2) × (– 1)
= – 5 × 6 × 7 × 20 = 4 × 5 × 2 × 1 = 40.
= – (5 × 20) × (6 × 7) (ii) 2 × (– 5) × (– 7) × 4
= – 100 × 42 = – 4200. Here the number of negative integers
8. – 4, – 3, – 2, – 1 and 0. is even.
9. Let the one negative integer be –10. ∴ 2 × (– 5) × (– 7) × 4
Then, – 10 – (Other negative integer) = 2×5×7×4
= (2 × 5) × (7 × 4)
= 18
= 10 × 28 = 280.
∴ Other negative integer
(iii) (– 4) × (– 11) × 10
= – 10 – 18 = – 28
Here the number of negative integers is
Hence the integers are – 10 and – 28. even.
10. To find balance finally, we add the ∴ (– 4) × (– 11) × 10 = 4 × 11 × 10 = 440.
deposits and subtract the withdrawls. 4. Difference of 0 and –10 = 0 – (– 10) = 10
∴ So, Anita's balance Sum of 0 and –10 = 0 + (– 10) = – 10
= ` 3148 + ` 1500 – ` 2100 Thus, the required pair is (0, – 10).
+ ` 2000 – ` 1550
5. (i) Rise in the temperature
= ` (3148 + 1500 + 2000)
= 6 °C – (– 3 °C)
– ` (2100 + 1550)
= 6 °C + 3 °C
= ` 6648 – ` 3650 = ` 2, 998.
= 9°C.
WORKSHEET – 7 (ii) The temperature at the end of the
afternoon = 5°C – 7°C
1. (i) 738 + (– 99) + 100 – (– 400)
= – 2°C.
= 738 – 99 + 100 + 400
= (738 + 100 + 400) – 99 7009
6. (i) 7009 ÷ (– 7009) =
= 1238 – 99 = 1139. – 7009
(ii) 76 × (– 18) + 76 × 18 7009   a  a 
= –  ∵ – b = –  b  
= 76 × (– 18 + 18)  7009    
= 76 × 0 = – 1.
= 0. [∵ a × 0 = 0] (ii) (– 808) × [110 + (– 33)]
2. (i) (– 100 + 7) – 63 = – 100 + 7 – 63 = – 808 × [110 – 33]
= 7 – (100 + 63) = – 808 × 77 = – 62216.
= 7 – 163 = – 156. 7. The temperature of water will be 20 °C
(ii) – 666 – (– 222) = – 666 + 222 after a change of 20 °C – 80° C = – 60 °C
[∵ – a – (– b) = – a + b] ∵ Time taken in the change of – 4°C
= – 444. = 10 minutes
3. (i) (– 4) × (– 5) × (– 2) × (– 1) ∴ Time taken in the change of – 1°C
Here the number of negative integers is 10
= minutes
even. 4
12 M A T H E M A T I C S – VII

13. ∴ Time taken in the change of – 60°C
2000 litres
10 = = 500.
= × 60 minutes = 150 minutes. 4 litres
4
10. (i) 336 × (– 2) × (– 5)
8. We know that the product of a positive
integer with the negative integer is = 336 × [(– 2) × (– 5)]
negative. So, the required number will = 336 × (2 × 5)
be positive. As twice of the required [∵ (– a) × (– b) = a × b]
number is 150, the number will be the = 336 × 10
half of 150.
= 3360.
150 (ii) 114 × 0 × (– 2) = 114 × [0 × (– 2)]
So, the required number = = 75.
2 = 114 × 0
9. Required time in hours
[∵ 0 × Any integer = 0]
Capacity of the tank = 0.
=
Quantity of water reduced per hour ❏❏
I N T E G E R S 13

14. Chapter
2 FRACTIONS
WORKSHEET–8 1 17 1× 17 17
7. (D) × = = .
3 8 3×8 24
4
1. (A) In , numerator < denominator.
5 4 5 4 5
8. (A) of = ×
4 5 21 5 21
∴ is the proper fraction.
5 4×5 4
= = .
7 4 × 8 – 7 32 – 7 25 5 × 21 21
2. (B) 4 – = = =
8 8 8 8 1 1 10 + 1 5+1
9. (C) 2 ÷1 = ÷
1 5 5 5 5
= 3 .
8
11
1 2 4+1 9+2 5 11 11 6 5
3. (C) 2 + 3 = + = + = ÷ =
2 3 2 3 2 3 5 5 6
5
5 × 3 11× 2
= +
2×3 3×2 11× 5 11 5
= = = 1 .
[∵ LCM of 2 and 3 = 2 × 3 = 6] 5×6 6 6
10. (B) ∵ Reciprocal of a non-zero whole
15 22 37 1
= + = = 6 . 1
6 6 6 6 number =
Whole number
4
4. (A) 3 × = a 1
7 ∴ Reciprocal of a = .
a
4 3×4
⇒ a=3× ⇒ a=
7 7 9 1 7
11. (A) Reciprocal of = = .
⇒ a=
12
7
5
⇒ a=1 .
7
7 9
7 ()
9
3 3 4 4 1 4 ×1 1
5. (D) of 16 = × 16 12. (A) ÷4= × = = .
4 4 5 5 4 5×4 5
3 × 16 48 1 1
= = = 12. 13. (B) is a reciprocal of = 2.
4 4 2
3 5
()
1
2
1 1 1 3 14. (D) ∵ × =1
6. (B) + + = 5 3
4 4 4 4 3 5
Therefore, and are reciprocals of
1 3 5 3
⇒ 3× = .
4 4 each other.
14 M A T H E M A T I C S – VII

15. 1 1 12 × 2 + 31 × 1
15. (C) Total weight = 2 kg + 3 kg =
2 5 14
4+1 15 + 1 [∵ LCM of 7 and 14 = 14]
= kg + kg
2 5 24 + 31 55 13
= = = 3 .
14 14 14
5 16
= kg + kg
2 5 9 3
2. (i) –
8 4
5×5 16 × 2
= kg + kg 2 4, 8
2×5 5×2
[∵ LCM of 2 and 5 = 2 × 5 = 10] 2 2, 4
2 1, 2
25 32
= kg + kg 1, 1
10 10
∴ LCM of 4 and 8 = 2 × 2 × 2 = 8
57 7
= kg = 5 kg. 9 3 9×1– 3× 2 9–6 3
10 10 ∴ – = = = .
8 4 8 8 8
16. (B) The distance covered by scooter in
1 1
1 litre of petrol = 40 km (ii) –
2 4
The scooter will cover the distance in
3 2 2, 4
3 litres of petrol 2 1, 2
4
3 1, 1
= 40 × 3 km
4 ∴ LCM of 2 and 4 = 2 × 2 = 4
15 40 × 15
= 40 × km = km 1 1 1× 2 –1×1 2–1 1
4 4 ∴ – = = = .
2 4 4 4 4
= 10 × 15 km = 150 km.
6 7+6 13
WORKSHEET – 9 3. (i) 1 × 21 = × 21 = × 21
7 7 7
1 1 2×3 +1 1× 2 + 1 13 × 21
1. (i) 2 +1 = + = = 13 × 3 = 39.
3 2 3 2 7
6+1 2+1 7 3 3 2 4 4+3 2 4
= + = + (ii) 1 × × = × ×
3 2 3 2 4 3 28 4 3 28
[∵ LCM of 2 and 3 = 2 × 3 = 6] 7 2 4
= × ×
1 1 7×2 3×3 4 3 28
∴2 +1 = +
3 2 3×2 2×3
7 2 1
14 9 23 5 = × ×
= + = = 3 . 4 3 7
6 6 6 6
5 3 7+5 28 + 3 7× 2×1
(ii) 1 + 2 = + =
7 14 7 14 4×3×7
12 31 2 1 1
= + = = = .
7 14 4×3 2×3 6
F R A C T I O N S 15

16. 1 1 81 1
4. (i) of 81 = × 81 = = 9. 6. (i) Reciprocal of 4 = .
9 9 9 4
1 12 1 12 2 1 5
(ii) Reciprocal of = = .
()
(ii) of = × 5 2
4 15 4 15 2
5
1 × 12 3 1
= = = .
4 × 15 15 5 1 1 1 2 1× 2 2
7. (i) ÷ = × = = .
5 2 5 1 5 ×1 5
7 7  7 × 64
(iii) of ` 64 = `  × 64  = ` 4 5 18 × 5
8 8  8 (ii) 18 ÷ = 18 × =
5 4 4
= ` (7 × 8) = ` 56.
9×5 45 1
= = = 22 .
5. (i)
4 1 5
× +  2 2 2
15 4 6 5
8. Length of each part = m÷2
2 4, 6 12
2 2, 3 5 1
3 1, 3 = m×
12 2
1, 1
5 ×1
= m
∴ LCM of 4 and 6 = 2 × 2 × 3 = 12 12 × 2
∴
4 1 5
× +  =
15  4 6 
4
15
×
12 (
1× 3 + 5 × 2
) =
5
24
m.
4  3 + 10  4 13 WORKSHEET – 10
= ×  = 15 × 12
15  12 
–2 –7 –2 – 7 –9 1
4 × 13 13 13 1. (i) + = = =– .
= = = . 18 18 18 18 2
15 × 12 15 × 3 45
4 3 1 11 –2 11 – 2 9
(ii)  2 + 1  × 1
 
(ii)
25
+
25
=
25
=
25
.
 5 10  2
– 7 18 – 7 × 5 + 18 × 1
10 + 4 10 + 3  2 + 1 2. (i) + =
=
 +  × 4 20 20
 5 10  2
– 35 + 18 – 17
14 13 3 = = .
= +  ×
  20 20
 5 10  2
–5 4 –5 × 2 + 4 ×1 – 10 + 4
14 × 2 + 13 × 1  3
=
  ×
(ii)
7
+
14
=
14
=
14
 10  2
28 + 13  3 41 3 –6 –3
=
 × = × =
14
=
7
.
 10  2 10 2
–7  1 1
=
41× 3
=
123 3
= 6 . 3. (i) 
  + 6 + 4
10 × 2 20 20  8 
16 M A T H E M A T I C S – VII

17. 2 4, 6, 8 5. The given fractions are:
2 2, 3, 4 –6 –1 3 7
, , 1 , 2
2 1, 3, 2 11 22 11 33
3 1, 3, 1 – 6 – 1 11 + 3 66 + 7
1, 1, 1 or , , ,
11 22 11 33
So, LCM of 4, 6 and 8 = 2 × 2 × 2 × 3 – 6 – 1 14 73
= 24 or , , ,
11 22 11 33
–7  1 1
Now,   + + 4 2 11, 22, 33
 8  6
3 11, 11, 33
–7 × 3 + 1× 4 + 1× 6
= 11 11, 11, 11
24
1, 1, 1
– 21 + 4 + 6 – 11
= = .
24 24 LCM of 11, 22 and 33 = 2 × 3 × 11 = 66
1 –3 5 –6 –6×6 – 36
(ii) + + = =
3 4 8 11 11 × 6 66
2 3, 4, 8
–1 –1× 3 –3
2 3, 2, 4 = =
22 22 × 3 66
2 3, 1, 2
3 3, 1, 1 14 14 × 6 84
= =
1, 1, 1 11 11× 6 66
So, LCM of 3, 4 and 8 = 2 × 2 × 2 × 3 73 73 × 2 146
= =
= 24 33 33 × 2 66
1 –3 5 ∵ – 36 < – 3 < 84 < 146
Now, + +
3 4 8 – 36 –3 84 146
∴ < < <
1× 8 – 3 × 6 + 5 × 3 66 66 66 66
=
24 –6 –1 3 7
i.e, < < 1 < 2 .
8 – 18 + 15 5 11 22 11 33
= = .
24 24
6. Perimeter of rectangle
9 4
4. (i) – = 2 × (length + breadth)
10 5
LCM of 5 and 10 = 10  1 3
= 2 ×  14 + 10  m
 2 4
9 4 9×1 – 4 × 2 9–8 1
∴ – = = = .
29 43   58 + 43  m
= 2× 
10 5 10 10 10
 +  m = 2×  
7 2  2 4   4 
(ii) – –
18 9 101 101 1
= 2× m= m = 50 m.
LCM of 18 and 9 = 18 4 2 2
–7 2 –7 – 4 –11 1 20 × 1
∴ – = = . 7. (i) 20 × = = 4.
18 9 18 18 5 5
F R A C T I O N S 17

18. 11 11× 6 11 1 (ii) No.
(ii) ×6 = = = 5 .
12 12 2 2 11
Example:- is an improper fraction
5
3 3 3 × 76
8. (i) of 76 = × 76 =
4 4 4 11 5
Reciprocal of is .
= 3 × 19 = 57. 5 11
4 4 4 × 70 5
(ii) of 70 = × 70 = Clearly, is not an improper
5 5 5 11
= 4 × 14 = 56. fraction.
8 4 8 15 8 15
7 1 7 3 +1 7 4 3. (i) ÷ = × = ×
9. (i) × 1 = × = × 9 15 9 4 4 9
3 3 3 3 3 3
1 2 5 2×5 10
7×4 28 = × = =
= = = 3 . 1 3 1× 3 3
3×3 9 9
3 8 3×8 3 1 1
(ii) × = = = . =3 .
8 9 8×9 9 3 3
1 1 3×4+1 1×6 +1
WORKSHEET – 11 (ii) 3 ÷1 = ÷
4 6 4 6
2 10 2 10 1 2 13 7 13 6
1. (i) × = × = ×2= = ÷ = ×
5 18 18 5 9 9 4 6 4 7
Here 2 < 9
13 6 13 3
i.e., numerator < denominator = × = ×
7 4 7 2
2
So, is less than 1. 13 × 3 39 11
9 = = =2 .
7×2 14 14
7 6 7 12 7 14
(ii) ÷ = × = ×2= 4 4 4 21
3 12 3 6 3 3 4. (i) of 21 = × 21 = ×
7 7 7 1
Here 14 > 3
4 21 4 4×3
i.e., numerator > denominator = × = ×3=
1 7 1 1
14 12
So, is greater than 1. = = 12.
3 1
2. (i) No.
7 14 5 14 5
(ii) 14 ÷ = × = ×
4 5 1 7 7 1
Example:- is a proper fraction
7 2 2×5
5 10
= × = =
4 7 3 1 1 1× 1 1
Reciprocal of is or 1
7 4 4 = 10.
3 1 2
Clearly, 1 is not a proper fraction. (iii) 14 ÷ 12
4 5 25
18 M A T H E M A T I C S – VII

19. 14 × 5 + 1 12 × 25 + 2 71 302 2 2 × 14 28 70
= ÷ = ÷ and = = (∵ = 14)
5 25 5 25 5 5 × 14 70 5
71 25 71 25 28 5
= × = × ∵ 28 > 5 ∴ >
5 302 302 5 70 70
71 5 71 × 5 355 53 3 2
= × = = =1 . So, of is greater.
302 1 302 302 302 5 3
2 2 4 1 8 1 8 4
5. of 2 hours = × 2 = hours (ii) of = × =
3 3 3 2 9 2 9 9
∵ 1 hour = 60 minutes 4 10 4 10 8
of = × =
= 60 × 60 seconds 5 11 5 11 11
(∵ 1 minute = 60 seconds) LCM of 9 and 11 = 9 × 11 = 99
4 4 4 4 11 44
∴ hours = × 60 × 60 seconds Now, = × =
3 3 9 9 11 99
= 4 × 20 × 60 seconds 99
(∵ = 11)
= 4800 seconds. 9
8 8 9 72
1 and = × =
6. ∵ In 1 hour Akshit reads = part 11 11 9 99
3
99
1 1 1 (∵ = 9)
∴ In 2 hour he will read = ×2 11
8 3 8 ∵ 72 > 44
1 17 17 72 44
= × = part. ∴ >
3 8 24 99 99
17
So, Akshit read part of the book. 4 10
24 So, of is greater.
5 11
3 1 5 9. First we have to find LCM of 3, 9 and
7. (i) Reciprocal of = = .
5
()
3
5
3 12
2 3, 9, 12
12 1 11 2 3, 9, 6
(ii) Reciprocal of = = .
11
( )
12
11
12 3 3, 9, 3
3 1, 3, 1
3 1 1, 1, 1
1 3 1
8. (i) of = × =
7 6 7 6 14
∴ LCM = 2 × 2 × 3 × 3 = 36.
3 2 3 2 2
of = × = 1 1 12 12 36
5 3 5 3 5 Now, = × = (∵ = 12)
LCM of 14 and 5 = 14 × 5 = 70 3 3 12 36 3
1 1× 5 5 70 4 4 4 16 36
Now, = = (∵ = 5) = × = (∵ = 4)
14 14 × 5 70 14 9 9 4 36 9
F R A C T I O N S 19

20. 5 5 3 15 36 1
and = × = (∵ = 3) 2. ∵ Cost of litre = ` 20
12 12 3 36 12 4
∵ 16 > 15 > 12 ` 20
∴ Cost of 1 litre = = ` (20 × 4)
16 15 12 1
∴ > > 4
36 36 36
4 5 1 = ` 80
So, , , is the descending order.
9 12 3 1 1
∴ Cost of 5 litres = ` 80 × 5
10. Length of main strip = 6 cm 2 2
3 11
Length of smaller strip = cm = ` 80 ×
2 2
Number of strips = ` 440.
3 27
Length of main strip 3. (i) × =
= 5 40
Length of smaller strip
Let =a
6 2 3 27
= =6× =2×2
()
Then × a=
3 3 5 40
2
27 5 27 × 5 9
= 4 strips. ⇒ a= × = =
40 3 40 × 3 8
WORKSHEET – 12 1 1
⇒ a= 1 ∴ =1 .
8 8
1 1 1 4 12
1. (i) 15 ×6 ×1 (ii) + =
2 5 10 5 10
15 × 2 + 1 6 × 5 + 1 1× 10 + 1 Let =b
= × ×
2 5 10 4 12
Then +b=
30 + 1 30 + 1 10 + 1 5 10
= × ×
2 5 10 12 4 12 × 1 – 4 × 2
⇒ b= – =
31 31 11 10 5 10
= × ×
2 5 10 12 – 8 4 2
⇒ b= = =
31 × 31 × 11 10571 71 10 10 5
= = = 105 .
2 × 5 × 10 100 100 2
∴ = .
5
2 3 2 3
(ii) of of 26 = × × 26 1
3 4 3 4 4. Weight of apples = 3 kg
2
2 × 3 × 26 2 × 26 3×2 +1
= = = kg
3× 4 4 2
52 7
= = 13. = kg
4 2
20 M A T H E M A T I C S – VII

21. 3 cm
3 12
Weight of oranges = 6 kg 4
4 E D
1 1 cm
6×4 + 3 4 cm 4
8 8
= kg
4 A C
27 3 cm
= kg 5 8
1 cm
4 4
B
2
LCM of 2 and 4 = 4
Total weight of fruits 3 12 × 4 + 3 51
DE = 12 cm = = cm.
4 4 4
= weight of apples + weight of oranges
33
7 27 EA = CD = cm
= kg + kg 8
2 4 Now, perimeter of figure ABCDE
14 + 27 41 1 = AB + BC + CD + DE + EA
= kg = kg = 10 kg.
4 4 4
 23 + 17 + 33 + 51 + 33  cm
= 
1 2×3 +1 7  4 2 8 4 8 
5. 2 = =
3 3 3 LCM of 2, 4 and 2 2, 4, 8
8 = 2×2×2 = 8 2 1, 2, 4
1 1× 6 + 1 7
1 = = 2 1, 1, 2
6 6 6
1, 1, 1
11 3 × 12 + 11 47 2 3, 6, 12
3 = =
12 12 12 2 3, 3, 6 23 × 2 + 17 × 4 + 33 × 1 + 51 × 2 + 33 × 1
3 3, 3, 3 = cm
5 1× 6 + 5 11 8
1 = = 1, 1, 1
6 6 6 46 + 68 + 33 + 102 + 33
∴ LCM of 3, 6 and 12 = 2 × 2 × 3 = 12 = cm
8
1 1 11 5 282 2 1
Now, 2 +1 + 3 – 1 = cm = 35 cm i.e., 35 cm.
3 6 12 6 8 4 2
7 7 47 11 7. 2 dozen = 2 × 12 = 24
= + + –
3 6 12 6
1 1
7 × 4 + 7 × 2 + 47 × 1 – 11 × 2 of the oranges = × 24 = 8 oranges
= 3 3
12 1 1
of the total oranges = × 24
28 + 14 + 47 – 22 67 7 4 4
= = = 5 .
12 12 12 = 6 oranges
3 5× 4 + 3 23 Number of sold oranges = 8 + 6 = 14
6. AB = 5 cm = cm = cm
4 4 4 Number of left of the oranges = Total
1 8×2 +1 17 number of oranges – Number of sold
BC = 8 cm = cm = cm
2 2 2 oranges
= 24 – 14
1 4×8 +1 33
CD = 4 cm = cm = cm = 10 oranges.
8 8 8
F R A C T I O N S 21

22. 1 5 × 6 + 1 30 + 1 31 6 14 15
8. Here, 5 = = = ∴ , , are in ascending order
6 6 6 6 48 48 48
1 9 31 9 1 7 5
Now, 5 ÷ = ÷ ∴ , , are in ascending order.
6 2 6 2 8 24 16
31 2 31 × 2
= × = WORKSHEET – 13
6 9 6×9
31 31 3 3
= = 1. (i) of 32 kg = × 32 kg
3×9 27 4 4
4
=1 . 3 × 32
27 = kg = 3 × 8 kg
4
9. Initially, Shyam has money = ` 240.
= 24 kg.
2 2
Money spent by shyam part. 2
8 (ii) of 1 week = of 7 days
7 7
Money left with shyam (∵ 1 week = 7 days)
2 6 3 2
=1– = = part. = × 7 days
8 8 4 7
Money left with Shyam = 2 days.
3 4 4
= of ` 240 (iii) of ` 120 = × ` 120
4 5 5
3 × 240 4 × 120
=` = ` 180. =` = ` 4 × 24
4 5
2 8, 16, 24 = ` 96.
10.
2 4, 8, 12
3 2 × 4 + 3 11
2 2, 4, 6 2. (i) 2 = = ;
4 4 4
2 1, 2, 3
3 1, 1, 3 1 1× 3 + 1 4
1 = =
1, 1, 1 3 3 3
∴ LCM of 8, 16 and 24 3 1 3 11 4 3
= 2 × 2 × 2 × 2 × 3 = 48 Now,  2 – 1  × =  –  ×
   
 4 3 4  4 3 4
1 1× 6 6
( 11 × 312 4 × 4 ) × 3
∴ = = –
8 8×6 48 =
4
5 5×3 15
= =
16 16 × 3 48  33 – 16  3 17 × 3 51
=  × = =
7 7×2 14  12  4 12 × 4 48
and = =
24 24 × 2 48 17 1
= =1 .
∵ 6, 14, 15 are in ascending order 16 16
22 M A T H E M A T I C S – VII

23. 1 3 × 4 + 1 13 1
(ii) 3 = = Breadth of rectangle = b = 4 cm
4 4 4 2
1 2×3 +1 7 4×2+1
2 = = = cm
3 3 3 2
1 1× 4 + 1 5 9
1 = = = cm
4 4 4 2
1 1 1 1 Area of rectangle = l × b
Now, 3 × 2 –1 ×
4 3 4 5 39 9 39 × 9
= cm × cm = cm2
13 7 5 1 4 2 4×2
= × – ×
4 3 4 5 351 7
= cm2 = 43 cm2.
 13 7  5 1 8 8
=  ×  – × 
 
 4 3 4 5 1 1× 2 + 1 3
4. 1 = =
13 × 7 5×1 91 1 2 2 2
= – = –
4×3 4×5 12 4
1 8×4 +1 33
91 × 1 – 1 × 3 91 – 3 and =
8 =
= = 4 4 4
12 12 ∵ Weight of 1 watermelon
88 22 1 1 3
= = = 7 . = 1 kg = kg
12 3 3 2 2
3. (i) Length of rectangle = l = 4 cm
1
1 ∴ Weight of 8 watermelons
Breadth of rectangle = b = 1 cm 4
2
3 1
1× 2 + 1 = × 8 kg
= cm 2 4
2
3 3 33 3 × 33
= cm = × kg = kg
2 2 4 2×4
Area of rectangle = l × b 99 3
= kg = 12 kg.
3 8 8
= 4 cm × cm
2 3 3 3 × 30
4×3 5. of 30 km = × 30 km = km
= cm2 5 5 5
2 90
= 2 × 3 cm2 = km = 18 km.
5
= 6 cm2. 2 2 2 × 40
of 40 km = × 40 km = km
3 8 8 8
(ii) Length of rectangle = l = 9 cm
4 80
=km = 10 km
9× 4 + 3 8
= cm Since, 18 km is greater than 10 km
4
39 ∴ Difference = (18 – 10) km
= cm
4 = 8 km.
F R A C T I O N S 23

24. 1 2×3 +1 6+1 7 4 35 9 2 7 9
6. (i) ∵ 2 = = = Now, × × = × ×
3 3 3 3 18 20 14 9 4 14
1 7 3 8×3 2×7×9 1
∴ 8÷ 2 = 8÷ = 8× = = = .
3 3 7 7 9 × 4 × 14 4
24 3 2 9× 3 + 2 29
= =3 . (ii) 9 = = ;
7 7 3 3 3
1 8×4 +1 32 + 1 33 1 1 × 29 + 1 30 6 2
(ii) 8 = = = 1 = = ; = .
4 4 4 4 29 29 29 15 5
5 1× 6 + 5 6+5 11 1 7×5+1 36
1 = = = 7 = =
6 6 6 6 5 5 5
1 5 33 11 2 1 6 1
Now, 8 ÷1 = ÷ Now, 9 ×1 × ×7
4 6 4 6 3 29 15 5
33 6 33 × 6 29 30 2 36
= × = = × × ×
4 11 4 × 11 3 29 5 5
3×3 9 1 29 × 30 × 2 × 36
= = = 4 . =
2 2 2 3 × 29 × 5 × 5
29 30 36
4 4×5 + 4 = × × ×2
7. Side = 4 cm = cm 29 5×5 3
5 5
6
20 + 4 24 =1× × 12 × 2
= cm = cm 5
5 5
6 × 12 × 2 144 4
Perimeter = 4 × Side = = = 28 .
5 5 5
24 4 × 24
= 4× cm = cm
5 5 WORKSHEET – 14
96 1
= cm = 19 cm. 5 15 5 14 5 14
5 5 1. (i) ÷ = × = ×
7 14 7 15 15 7
8. Let the man initially had ` a.
1 2
= ×2 = .
2 2 2a 3 3
Expenditure = of a = ×a=
5 5 5
2 9 6×5 + 2 9 32 9
(ii) 6 ÷ = ÷ = ÷
2a 3a 5 7 5 7 5 7
So, money left with him = a – =
5 5
32 7 32 × 7 224
= × = =
3a 120 × 5 5 9 5×9 45
∴ = 120 or a = = 200.
5 3 44
= 4 .
Thus, the man initially had ` 200. 45
4 2 35 7 7 8
9. (i) = ; = 2. (i) 8 × = × 7 = 4 × 7 = 28.
18 9 20 4 2 2
24 M A T H E M A T I C S – VII

25. 5 40 27 × 2 + 1
(ii) 40 × = × 5 = 5 × 5 = 25. = m
8 8 2
6 2 × 13 + 6 32 55
3. (i) 2 = = ; = m
13 13 13 2
1 1 × 26 + 1 27 3 2× 4 + 3
1 = = Length of 1 piece = 2 m= m
26 26 26 4 4
6 1 32 27 11
Now, 2 ÷1 = ÷ = m
13 26 13 26 4
32 26 Number of pieces
= ×
13 27
Length of ribbon
64 10 =
= = 2 . Length of 1 piece
27 27
55
3 4×7 + 3 31 55 4
(ii) 4 = = ; = 2 = ×
7 7 7 11 2 11
1 1× 7 + 1 8 4
1 = =
7 7 7 55 4
= ×
3 1 31 8 31 7 11 2
Now, 4 ÷ 1 = ÷ = ×
7 7 7 7 7 8 = 5 × 2 = 10 pieces.
31 7 31 2 2×3 + 2 8
= × = ×1 6. 2 = =
8 7 8 3 3 3
31 7
= = 3 . 2
8 8 Marks got by Bulbul = 2 of mark got
3
4. 1 hour = 60 minutes
= 60 × 60 seconds 8
by Kanika = × 75
3
5 5
of 3 hours = × 3 hours 75
8 8 =8× = 8 × 25
3
5
= × 3 × 60 × 60 seconds = 200 marks.
8
60 7. First find LCM of 8, 9, 16 and 36
= (5 × 3 × 60) × seconds
8 2 8, 9, 16, 36
15 2 4, 9, 8, 18
= 900 × seconds
2 2 2, 9, 4, 9
900 2 1, 9, 2, 9
= × 15 seconds
2 3 1, 9, 1, 9
= 450 × 15 seconds 3 1, 3, 1, 3
= 6750 seconds. 1, 1, 1, 1
1 ∴ LCM of 8, 9, 16 and 36
5. Length of ribbon = 27 m
2 = 2 × 2 × 2 × 2 × 3 × 3 = 144.
F R A C T I O N S 25