LCM and HCF
 The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and same number of pencils, is :

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As we know that ,
Maximum number of students = The greatest common divisor = HCF of 1001 and 910Correct Option: A
As we know that ,
Maximum number of students = The greatest common divisor = HCF of 1001 and 910 = 91
 Three electronic devices make a beep after every 48 seconds, 72 seconds and 108 seconds respectively. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest is

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Required time = LCM of 48, 72 and 108 seconds
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds = 7 minutes 12 secondCorrect Option: A
Required time = LCM of 48, 72 and 108 seconds
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds = 7 minutes 12 second
∴ Required time = 10 a.m. + 7 minutes 12 second = 10 : 07 : 12 hours
 The LCM fo two prime numbers p and q, (p > q) is 161. The value of (3q – p) :

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LCM of p and q = 161
∴ pq = 23 × 7
∴ p = 23; q = 7
∴ 3q – p = 3 × 7 – 23Correct Option: A
LCM of p and q = 161
∴ pq = 23 × 7
∴ p = 23; q = 7
∴ 3q – p = 3 × 7 – 23
The value of (3q – p) = 21 – 23 = – 2
 The LCM of four consecutive numbers is 60. The sum of the first two numbers is equal to the fourth number. What is the sum of four numbers?

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We find LCM of 60 ,
∴ 60 = 2 × 2 × 3 × 5
i.e., Numbers = 2, 3, 4 and 5Correct Option: B
We find LCM of 60 ,
∴ 60 = 2 × 2 × 3 × 5
i.e., Numbers = 2, 3, 4 and 5
∴ Required sum = 2 + 3 + 4 + 5 = 14
 Three bells ring at intervals of 36 seconds, 40 seconds and 48 seconds respectively. They start ringing together at a particular time. They will ring together after every

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Required answer = LCM of 36, 40 and 48 seconds
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720Correct Option: B
Required answer = LCM of 36, 40 and 48 seconds
= 720 secondsRequired answer = 720 minutes = 12 minutes 60
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720