QOTD | Quantitative Aptitude | Number system

Question of The Day13-04-2020

The least possible number which when successively been divided by 5, 3, 4, 7 gives the remainder 2, 2, 1, 2 respectively will be

Answer

Correct Answer : c ) 567

Explanation :

We know that

Dividend (D) = Divisor (d) * Quotient (Q) + Remainder (R)

Here, Remainder should always be less than divisor.

Let us assume the last dividend which will be divided by 7 is x

Thus, our required number or dividend will be

((((x * 7) + 2 ) * 4 + 1) * 3 + 2 ) * 5 + 2)

So, for the number to be minimum x = 1

So, the least required number will be

=((((1 * 7) + 2 ) * 4 + 1) * 3 + 2 ) * 5 + 2)

= ((37 * 3 + 2) * 5 +2 )

= (113×5+2)

= 567

Hence, (c) 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 etc.

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