Robert Bosch LCM & HCF Quiz-4

LCM of 4, 6, 8, 14
= 168 seconds
= 2 minutes 48 seconds
They ring again at 12 + 2 min. 48 sec.
= 12 hrs. 2 min. 48 sec.

LCM of 20, 30 and 40 minutes = 120 minutes
Hence, the bells will toll together again after 2 hours i.e. at 1 p.m.

The difference between the divisor and the corresponding remainder is equal.
LCM of 3, 5, 7 and 9 = 315
Largest 4-digit number = 9999
\therefore Number divisible by 315
= 9999 – 234 = 9765
Required number
= 9765 – 2 = 9763

Required time = LCM of 6, 7, 8, 9 and 12 seconds
= 504 seconds

LCM of 3, 4, 5, 6, 7, 8
= 840
Since, the remainder 760 is more than half of the divisor 840.
\therefore The nearest number
= 10000 + (840 – 760) = 10080

The smallest number divisible by 16, 20 and 24
= LCM of 16, 20 and 24
\therefore LCM = 2 \times 2 \times 2 \times 2 \times 5 \times 3
= 2^{2} \times 2^{2} \times 5 \times 3
\therefore Required complete square number = 2^{2} \times 2^{2} \times 5^{2} \times 3^{2} = 3600

LCM of 25, 50 and 75 = 150
On dividing 43582 by 150, remainder = 82
\therefore Required number
= 43582 + (150 – 82) = 43650

Required number = (LCM of 24, 32, 36 and 54) – 5
LCM = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 4
= 864
\therefore Required number = 864 – 5
= 859

\therefore LCM of 20,28,32 and 35 is = 2 \times 2 \times 5 \times 7 \times 8
= 1120
\therefore Required number
= 5834 – 1120 = 4714

The LCM of 6, 12 and 18
= 36 = 6^{2}