If x/y = 5.32, where x and y are integers, what is the sum of all the

Question

If x/y = 5.32, where x and y are integers, what is the sum of all the values of Y that yield a remainder less than 30, when X is divided by Y ?

A) 25
B) 50
C) 75
D) 100
E) 150

e-GMAT

chetan2u · Answer

The remainder will be 0.32*Y, and as 0.32*100>30, y<100..

Now 0.32=\frac{32}{100}=\frac{2^5}{100} ..
For each 2, we require a 5 to remove one decimal. Therefore, two 5s are required to remove two decimals..
Thus y should be a multiple of 25..
Possible values of y are 25, 50 and 75. SUM = 25+50+75=150

E

gvij2017 · Answer

Please explain how Y can be 75?

I did this way. 
32/100, 16/50, 8/25
So for reminder less than 30, y can be 50 and 25 
Therefore the sum is 75.

Please tell me where I made mistake.

Rachit92 · Answer

gvij2017  :

8Y/25 should be an integer less than 30

When y=25, remainder is 8 (<30)
When y = 50, remainder is 16(<30)
When y = 75, remainder is 24 (<30)
When y = 100, remainder is 32 (>30)

Therefore, the sum of possible values of y = 25+50+75 = 150

Magdalena1609 · Answer

Struggling to understand why this does not work too.
Could someone clarify, please?

Thank you

kashika101 · Answer

X gets its remainder from the 0.32Y term ( X= 5Y+ 0.32Y ).
0.32Y = 8Y/25
Now, 8Y/25 should be a number <30 
therefore, Y = 25,50,75 and sum of Y = 150.

GMATGuruNY · Answer

When one positive integer is divided by another, we typically represent what's left over either as a REMAINDER or as a DECIMAL.
There is a relationship between the two representations:

\frac{remainder}{divisor} = decimal

When 5 is divided by 2:
Remainder representation:  \frac{5}{2} =  2 R1

Decimal representations:  \frac{5}{2} = 2.5

\frac{remainder}{divisor} = \frac{1}{2}

decimal = 0.5

Since the two values are equal:
 \frac{remainder}{divisor} = decimal

In most cases, it is helpful to represent the decimal AS A FRACTION IN ITS MOST REDUCED FORM.

In the problem above:
remainder = R
divisor = Y
decimal  = 0.32 = \frac{32}{100} = \frac{8}{25}

Plugging these values into  \frac{remainder}{divisor} = decimal , we get:
 \frac{R}{Y }= \frac{8}{25}

Since R must be less than 30, the following options are possible for  \frac{R}{Y} :
 \frac{R}{Y} = \frac{8}{25} = \frac{16}{50} = \frac{24}{75}

Sum of the options for Y = 25 + 50 + 75 = 150

E

rsrighosh · Answer

Best way to find the solution:-

\frac{ X}{Y} = 5.32

\frac{X}{Y} = 5 + 0.32 = 5 + \frac{32}{100} = 5 + \frac{8}{25}  (Simplify)

Now lets start counting

\frac{8*1}{25*1} OR \frac{8*2}{25*2} OR \frac{8*3}{25* 3} OR \frac{8*4}{25*4}  .. and so on

==>  \frac{8}{25}  OR   \frac{16}{50}  OR  \frac{24}{75}  OR   \frac{32}{100}   .. and so on

Hence, sum of possible divisors:-  25 + 50 + 75 = 150

Ahmed9955 · Answer

X/Y = 5.32 => X/Y = 5+32/100 We need to find Sum of values of y = y1+y2+y3...... for r<30 Comparing above eq with -> Dividend/Divisor = quotient + remainder/Divisor => r/y = 32/100 =8/25 => 25r =8y for equations to hold true y must be multiple of 25 => y=25a (a is variable) Now, r = (8* 25a )/25 for a=1=> y = 25 ; r = 8 a=2 => y= 50 ; r =16 a=3 => y= 75 ; r=24 a=4 => y=100 ; r=32 because r<30 For above values sum of y= 150 IMO- E

Akshaynandurkar · Answer

Answer is zero if you consider the negative values of X and Y . Therefore question should mention that Y is always positive or X and Y are both positive

Kinshook · Answer

Rachit92

If X/Y = 5.32,where x and y are integers, what is the sum of all the values of Y that yield a remainder less than 30, when X is divided by Y ?

X/Y = 532/100 = 266/50 = 133/25

X/Y = Quotient + Remainder/Y = 133/25 = 5 + 8/25

8/25 = 16/50 = 24/75

Sum of all values of Y that yield a remainder less than 30, when X is divided by Y = 25 + 50 + 75 = 150

IMO E

voicesinyourwall · Answer

sorry, what???? this one came up in my forum quiz and I'm completely baffled. where are you getting 8 and 25 from out of this question? seems to be completely nonsensical other than to the person who wrote the question.

If x/y = 5.32, where x and y are integers, what is the sum of all the

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