LCM and HCF
 The smallest number, which when increased by 5 is divisible by each of 24,32, 36 and 564, is

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Required number = (LCM of 24, 32, 36 and 54) – 5
Now,LCM of 24, 32, 36 and 54
LCM = 2 × 2 × 2 × 3 × 3 × 3 × 4 = 864Correct Option: B
Required number = (LCM of 24, 32, 36 and 54) – 5
Now,LCM of 24, 32, 36 and 54
LCM = 2 × 2 × 2 × 3 × 3 × 3 × 4 = 864
∴ Required number = 864 – 5 = 859
 The number nearest to 43582 divisible by each of 25, 50 and 75 is :

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LCM of 25, 50 and 75 = 150
On dividing 43582 by 150, remainder = 82
∴ Required number = 43582 + (150 – 82)Correct Option: B
LCM of 25, 50 and 75 = 150
On dividing 43582 by 150, remainder = 82
∴ Required number = 43582 + (150 – 82) = 43650
 The least number, which is a perfect square and is divisible by each of the numbers 16, 20 and 24, is

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The smallest number divisible by 16, 20 and 24
= LCM of 16, 20 and 24
∴ LCM = 2 × 2 × 2 × 2 × 5 × 3
LCM = 2^{2} × 2^{2} × 5 × 3Correct Option: B
The smallest number divisible by 16, 20 and 24
= LCM of 16, 20 and 24
∴ LCM = 2 × 2 × 2 × 2 × 5 × 3
LCM = 2^{2} × 2^{2} × 5 × 3
∴ Required complete square number = 2^{2} × 2^{2} × 5^{2} × 3^{2} = 3600
 The largest 4digit number exactly divisible by each of 12, 15, 18 and 27 is

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As we know that Greatest n digit number which when divided by three numbers p,q,r leaves no remainder will be Required Number = (n – digit greatest number) – R , R is the remainder obtained on dividing greatest n digit number by L.C.M of p.q,r.
The largest number of 4digits is 9999. L.C.M. of divisors
LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540Correct Option: B
As we know that Greatest n digit number which when divided by three numbers p,q,r leaves no remainder will be Required Number = (n – digit greatest number) – R , R is the remainder obtained on dividing greatest n digit number by L.C.M of p.q,r.
The largest number of 4digits is 9999. L.C.M. of divisors
LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540
Divide 9999 by 540, now we get 279 as remainder.
9999 – 279 = 9720
Hence, 9720 is the largest 4digit number exactly divisible by each of 12, 15, 18 and 27.
 From a point on a circular track 5 km long A, B and C started running in the same direction at the same time with speed of 2^{1}/_{2} km per hour, 3 km per hour and 2 km per hour respectively. Then on the starting point all three will meet again after

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A makes one complete round of the circular track in
5 = 2 hours, 5 2 B in 5 hours and C in 5 hours. 3 2
That is after 2 hours A is at the starting point,B after 5 hours and C after 5 hours. 3 2 Hence the required time = LCM of 2, 5 and 5 hours 3 2 Required time = LCM of 2, 5, 5 HCF of 1 , 3 , 2
Correct Option: C
A makes one complete round of the circular track in
5 = 2 hours, 5 2 B in 5 hours and C in 5 hours. 3 2
That is after 2 hours A is at the starting point,B after 5 hours and C after 5 hours. 3 2 Hence the required time = LCM of 2, 5 and 5 hours 3 2 Required time = LCM of 2, 5, 5 HCF of 1 , 3 , 2 Required time = 10 = 10 hours. 1