Page 48 - Maths Class 06

EX AM PLE 3. The HCF of two num bers is 23 and their LCM is 1449. If one of the num bers is 161, find the
other num ber.

SOLUTION : From Prop erty 7 stated above, we have
One number ´ the other number = HCF ´ LCM
23 ´1449
\ Required Number = = 207
161

Exercise 3.5

1. Find the LCM of:
(a) 12 and 18 (b) 45 and 75 (c) 24 and 80
(d) 24 and 100 (e) 20, 25 and 30 (f) 40, 48 and 45

2. Find the LCM of:
(a) 28, 36, 45, 60 (b) 144, 180, 384
3. Find the HCF and LCM of:
(a) 186, 403 (b) 490, 1155

4. Can two numbers have 16 as their HCF and 204 as their LCM? Give reasons in support of your
answer.

5. The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the
other.

6. Is the prod uct of three numbers always equal to the prod uct of their HCF and LCM? Give reasons
in support of your answer.

Applications of HCF and LCM

In our day-to-day life, the concept of HCF and LCM is widely used. This can be illustrated by various
examples given below.
EXAMPLE 1. Three ropes of different lengths 4 m 78 cm, 2 m 48 cm and 3 m 28 cm are to be cut into
equal lengths. What is the greatest possible length of each piece?

SOLUTION : The three ropes are to be cut in such a way that they 248 478 1
are equal in length so its length must be exact divisor –248
of all three ropes. Moreover, the greatest possible 230 248 1
length is required. Thus, greatest possible length is –230
HCF of 478, 248 and 328. 18 230 12
–216
We know,
14 18 1
4 m 78 cm = 4 m + 78 cm –14
= 400 cm + 78 cm = 478 cm 4 14 3

2 m 48 cm = 248 cm –12
2 328 164 2 4 2
3 m 28 cm = 328 cm
–328 –4
\ Greatest possible length of each rope is 2 cm. 00 0

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