Question of The Day14-12-2020

What will be the largest 4-digit number, which when divided by 12, 15, 20, 25 respectively leaves remainder 7 in each case.

Correct Answer : a ) 9907

Explanation :

We know that,

The number, which when divided by x, y, z respectively and leaves r remainder in each case is given as:

Number = LCM of (x, y, z) × K + r, where K = 1, 2, 3, …. and so on

Thus, required number = (LCM of 12, 15, 20, and 25) × K + 7

LCM of 12, 15, 20, and 25 will be 300

So, the required number = 300K + 7 …. (1)

Largest four-digit number is 9999

⇒K=Remainder of$$9999 \over 300$$=99

So, Largest number divisible by 300 will be = 9999 – 99 = 9900

In equation (1) putting k = 99

Required number = 300 × 99 + 7 = 9900 + 7 = 9907

Hence, (a) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA etc.

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