6th Maths : Term 2 Unit 1 : Numbers : HCF and LCM : Exercise 1.2 : Text Book Back Exercises Questions with Answers, Solution

__Exercise
1.2__

** **

**1. Fill in the blanks**

(i) The HCF of 45 and 75 is __15__**.**

(ii) The HCF of two successive even numbers is __2__**.**

(iii) If the LCM of 3 and 9 is 9, then their HCF
is __3__**.**

(iv) The LCM of 26, 39 and 52 is __156__**.**

(v) The least number that should be added to 57
so that the sum is exactly divisible by 2,3,
4 and 5 is __3__**.**

** **

**2. Say
True or False**

(i) The numbers 57 and 69 are co-primes.**[False]**

*(Common factor of 57 and 69
is 3)*

(ii) The HCF of 17 and 18 is 1.**[True]**

(iii) The LCM of two successive numbers is the product
of the numbers.**[True]**

(iv) The LCM of two co-primes is the sum of the
numbers.**[False]**

*(Product of two co−primes)*

(v) The HCF of two numbers is always a factor of
their LCM .**[True]**

** **

**3.
Find the HCF of each set of numbers using prime factorisation method.**

**(i)
18,24 (ii) 51,85 (iii) 61,76 (iv) 84,120 (v) 27,45,81 (vi) 45,55,95**

**(i) 18, 24**

Product of common factors = 2 × 3 = 6

H.C.F. of 18 and 24 is 6

**(ii) 51, 85**

Common factors = 17

H.C.F. of 15 and 85 is 17

**(iii) 61, 76**

Common factors is 1

H.C.F. of 61 and 76 is 1

**(iv) 84, 120**

Product of common factors = 2 × 2 × 3 = 12

H.C.F. of 84 and 120 is 12

**(v) 27, 45, 81**

Product of common factors = 3 × 3 = 9

H.C.F. of 27, 45 and 81
is 9

**(vi) 45, 55, 95**

Common factor = 5

H.C.F. of 45, 55 and 95 = 5

** **

**4.
Find the LCM of each set of numbers using prime factorisation method.**

**(i)
6,9 (ii) 8,12 (iii) 10,15 (iv) 14,42 (v) 30,40,60 (vi) 15,25,75**

**(i) 6, 9**

L.C.M. = Common factor × Product of uncommon factors

= 3 × (2 × 3) = 18

L.C.M. of 6 and 9 is 18

**(ii) 8, 12**

L.C.M. = Product of common factors × Product of uncommon factors

= (2 × 2) × (2 × 3) = 24

L.C.M. of 8 and 12 = 24

**(iii) 10, 15**

L.C.M. = Common factor × Product of uncommon factors

= 5 × (2 × 3) = 30

L.C.M. of 10 and 15 = 30

**(iv) 14, 42**

L.C.M. = Product of common factor × uncommon factor

= (2 × 7) × 3 = 42

L.C.M. of 14 and 42 = 42

**(v) 30, 40, 60**

L.C.M. = Product of common factors of all the three numbers × common
factors of two numbers × Uncommon factor

= (2 × 5) × (2 × 3) × 2 = 120

L.C.M. of 30, 40 and 60 = 120

**(vi) 15, 25, 75**

L.C.M. = Product of common factors of all the three numbers × product
of common factors of two numbers

= 5 × (5 × 3) = 75

L.C.M. of 15, 25 and 75 = 75

** **

**5.
Find the HCF and the LCM of the numbers 154, 198 and 286.**

H.C.F. = Product of common factors of all the three numbers

= 2 × 11 = 22

L.C.M. = Product of common factors of all the three numbers × Product
of uncommon factors = (2 × 11) × 7 × 3 × 3 × 13 = 18018

**H.C.F. of 154,198 and 286 is 22**

**L.C.M. of 154,198 and 286 is 18018**

** **

**6.
What is the greatest possible volume of a vessel that can be used to measure exactly
the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?**

Solve the problem by using H.C.F.

H.C.F.of 80, 100 and 120 is 2 × 2 × 5 = 20

**The greatest possible volume of a vessel that can be used to
measure exactly the volume of milk in cans = 20 litres**

** **

**7.
The traffic lights at three different road functions change after every 40 seconds,
60 seconds and 72 seconds respectively. If they changed simultaneously together
at 8 a.m at the unctions, at what time will they simultaneously change together
again?**

This sum is related to L.C.M. sum. So calculate the L.C.M.

L.C.M. of 40, 60 and 72 = 2 × 2 × 2 × 5 × 3 × 3

= 360 seconds

= 6 × 60 seconds = 6 minutes

All the three traffic lights change together again after 360
seconds

(6 minutes)

**At 8.06 a.m the three traffic lights change together again.**

** **

**8.
The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible?**

The L.C.M of two numbers =210

H.C.F = 14

Possible pairs = 2

They are 210, 14

70, 42

** **

**9.
The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers
is 36, then find the other number.**

H.C.F of two numbers = 12

L.C.M. = 6 times their H.C.F

= 6 × 12 = 72

One of the numbers = 36

Product of two numbers = Product of H.C.F. × L.C.M.

36 × the other number = 12 × 72

The other number = [12 × 72] / 36 = 24

The other number = 24.

** **

__Objective
Type Questions__

** **

**10.
Which of the following pairs is co-prime?**

**a)
51, 63**

**b)
52, 91 **

**c)
71, 81 **

**d)
81, 99**

*Answer:***c) 71, 81**

** **

**11.
The greatest 4 digit number which is exactly divisible by 8, 9 and 12 is**

**a)
9999 **

**b)
9996**

**c)
9696**

**d)
9936**

*Answer:***d) 9936**

** **

**12.
The HCF of two numbers is 2 and their LCM is 154. If the difference between the
numbers is 8, then the sum is**

**a)
26**

**b)
36**

**c)
46 **

**d)
56**

*Answer:***b) 36**

** **

**13.
Which of the following cannot be the HCF of two numbers whose LCM is 120**

**a)
60**

**b)
40 **

**c)
80 **

**d)
30**

*Answer:***c) 80**

** **

__ANSWERS: __

__Exercise 1.2 __

**1. i) 15 ii) 2 iii)
3 iv) 156 v) 3**

**2. i) False ii)
True iii) True iv) False v) True**

**3. i) 6 ii) 17 iii)
1 iv) 12 v) 9 vi) 5 **

**4. i) 18 ii) 24
iii) 30 iv) 42 v) 120 vi) 75 **

**5. HCF → 22; LCM →
18018**

**6. HCF = 20 litres**

**7. After 360
seconds (6 min), at 8.06 a.m**

**8. 2 pairs possible**

**9. 24**

**Objective Type
Questions**

**10. c) 71, 81 11.
d) 9936**

**12. b) 36 **

**13. c) 80**

** **

Tags : Questions with Answers, Solution | HCF and LCM | Term 2 Chapter 1 | 6th Maths , 6th Maths : Term 2 Unit 1 : Numbers

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6th Maths : Term 2 Unit 1 : Numbers : Exercise 1.2 | Questions with Answers, Solution | HCF and LCM | Term 2 Chapter 1 | 6th Maths

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