If x/y = 2.12, where x and y are positive integers, then what could be

Question

If  \frac{x}{y} = 2.12 , where x and y are positive integers, then what could be the remainder when  x  is divided by  y ?

I. 6
II. 9
III. 12

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

chetan2u · Answer

\frac{x}{y} = 2.12 
 x=2.12y=2y+.12y ....x=quotient*2+remainder
remainder=0.12y

I. 6 = 0.12*50...Yes, when y=50
II. 9= 0.12*75...Yes, when y=75
III. 12=0.12*100..Yes, when y=100

All possible.

E

GMATinsight · Answer

\frac{x}{y} = 2.12 = \frac{212}{100} = \frac{53}{25}

53 when divided by 25 leaves remainder 3

i.e Remainder must be a multiple of 3

i.e. 6, 9, 12 all are possible remainder

Answer: Option E

athenaaa · Answer

\frac{x}{y}= 2.12xy=2.12
x=2.12y=2y+.12y
= 2y+ \frac{12}{100y}

=>\frac{12}{100} = \frac{6}{25}

As 6 is a multiple of 3, all reminders should be divisible by 3.

AnkithaSrinivas · Answer

12/100 = 3/25

so any number below 25 is remainder
is this approach right?

GMATinsight · Answer

AnkithaSrinivas

The denominator is the divisor's reflection and numerator is the reflection of remainder

i.e. Divisor must be a multiple of 25 and
The remainder must be a multiple of 3

sujoykrdatta · Answer

x/y = 2.12

The quotient is 2. Let the remainder is r. We have:

x/y = 2 + r/y = 2 + 0.12

Thus: r/y = 0.12 = 12/100 = 3/25

This, possible cases: 
y = 25, r = 3
y = 50, r = 6
y = 75, r = 9
y = 100, r = 12
y = 125, r = 15, and so on

Thus, 6, 9 and 12 are possible remainders

Answer E

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bumpbot · Answer

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If x/y = 2.12, where x and y are positive integers, then what could be

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If x/y = 2.12, where x and y are positive integers, then what could be

Prep Toolkit

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