Class 6 Maths | Chapter 3 Solutions | Exercise 3.6 and 3.7

EXERCISE 3.6

1. Find the HCF of the following numbers:

a) 18, 48
Ans. a) Factors of 18 = 2 x 3 x3

Factors of 48 = 2 x 2 x2 x 2 x 3
HCF (18, 48) = 2 X 3 = 6

b) 30, 42

Ans. Factors of 30 = 2 x 3 x5
Factors of 42 = 2 x 3 x 7
HCF (30, 42) = 2 X3 =6

c) 18, 60

Ans. Factors of 18 = 2 x 3 x 3
Factors of 60 = 2 x 2 x 3 x 5
HCF (18, 60) = 2 X 3 =6

d) 27, 63

Ans. Factors of 27 = 3 x 3 x 3
Factors of 63 = 3 x 3 x7
HCF (27, 63) =3 X 3 = 9

e) 36, 84

Ans. Factors of 36 = 2 x 2 x 3x 3
Factors of 84 = 2 x 2 x3 x 7
HCF (36, 84) = 2 X 2 X3 = 12

f) 34, 102

Ans. Factors of 34 = 2 x 17
Factors of 102 = 2 x 3 x 17
HCF 934, 102) = 2 x 17 = 34

g) 70, 105, 175

Ans. Factors of 70 = 2 x 5 x7
Factors of 105 = 3 x 5 x7
Factors of 175 = 5 x 5 x7
HCF (70, 105, 175) = 5 X 7 = 35

h) 91, 112, 49

Ans. Factors of 91 = 7 x 13
Factors of 112 = 2 x 2 x 2 x2x7
Factors of 49 = 7 x7
HCF (91, 112, 49) = 7

i) 18, 54, 81

Ans. Factors of 18 = 2 x 3 x 3
Factors of 54 = 2 x 3 x 3 x3
Factors of 81 = 3 x 3 x 3 x3
HCF (18, 54, 81) = 3 X 3 =9

j) 12, 45, 75

Ans. Factors of 12 = 2 x2 x 3
Factors of 45 = 3 x 3 x5
Factors of 75 = 3 x 5 x 5
HCF (12, 45, 75) = 3

2. What is the HCF of two consecutive

a) numbers?
b) even numbers?
c) odd numbers?

Ans.

a) One
b) Two
c) One

3. HCF of co –prime numbers 4 and 15 was found as follows by factorization: 4 = 2 x 2 and 15 = 3 x 5 since there is no common prime factor, so HCF of 4 and 15 is 0.Is the answer correct? If not, what is the correct HCF?

Ans. No, the correct HCF is one.

Exercise 3.7

1. Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.

Ans. To find the maximum value of weight, we have to find the HCF of 75 and 69.

Factors of 75 = 3 x 5 x 5
Factors of 69 = 3 x 23
HCF (75, 69) = 3
Therefore the required weight = 3 kg

2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Ans. To find the minimum distance, we have to find the LCM of 63, 70 and 77.

LCM = 2 X 3 X 3 X 5 X 7 X 11 = 6930 cm
Therefore the minimum distance = 6930 cm

3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Ans. To find the longest tape, we have to find the HCF of 825, 675 and 450.

Factors of 825 = 3 x 5 x 5 x 11
Factors of 675 = 3 x3 x 3 x 5 x 5
Factors of 450 = 2 x 3 x 3 x 5 x 5
HCF = 3 X 5 X 5 = 75CM

Therefore the longest tape is 75 cm.

4. Determine the smallest 3- digit number which is exactly divisible by 6, 8 and 12.

Ans. We have to find the LCM of 6, 8 and 12.
LCM = 2 x 2 x 2 x 3 = 24
Now find the multiples of 24.
It can be seen that 24 x 4 = 96 and 24 x 5 = 120.
Hence the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Ans. We have to find the LCM of 8, 10 and 12.
LCM = 2 x 2 x 2 x3 x 5 = 120
Now find the multiples of 120.
It can be seen that 120 x 8 = 960 and 120 x 9 = 1080.
Hence the greatest 3-digit number which is exactly divisible by 8, 10 and 12 is 960.

6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m.at what time will they change simultaneously again?

Ans. Here we have to find the LCM of 48, 72 and 108.
LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432
They will change together after every 432 seconds, i.e. 7 minute 12 seconds

7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Ans. To find the maximum capacity of a container, we have to find the HCF of 403, 434 and 465.
Factors of 403 = 13 x 31
Factors of 434 = 2 x 7 x 31
Factors of 465 = 3 x 5 x 31
HCF = 31
So the maximum capacity of the container = 31 litre

8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Ans. To find the least number, we have to find the LCM of 6, 15 and 18.
LCM = 2 x 3 x 3 x5 = 90
So required number = 90 + 5 = 95

9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Ans. We have to find the LCM of 18, 24 and 32
LCM = 2 x 2 x 2 x 2 x2 x 3 x 3 = 288
Now find the multiples of 288.
It can be seen that 288 x 3 = 864 and 288 x 4 = 1152.
Therefore the smallest 4-digit number which is divisible by 18, 24 and 32 is 1152.

10. Find the LCM of the following numbers:

a) 9 and 4
b) 12 and 5
c) 6 and 5
d) 15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Ans.

a) LCM of 9 and 4 = 2 x 2 x 3 x3 = 36
b) LCM of 12 and 5 = 2 x 2 x 3 x 5 = 60
c) LCM of 6 and 5 = 2 x 3 x 5 = 30
d) LCM of 15 and 4 = 2 x 2 x 3 x 5 = 60

Here in each case LCM is a multiple of 3.

Yes, in each case LCM = the product of two numbers

11. Find the LCM of the following numbers in which one number is the factor of the other.

a) 5, 20
b) 6, 18
c) 12, 48
d) 9, 45

What do you observe in the results obtained?

a) LCM of 5 and 20 = 2 x 2 x 5 = 20
b) LCM of 6 and 18 = 2 x 3 x 3 = 18
c) LCM of 12 and 48 = 2 x 2 x 2 x 2 x 3 = 48
d) LCM of 9 and 45 = 3 x 3 x 5 = 45

The LCM of the given numbers in each case is the larger of the two numbers.

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