Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:39 am

If \(X\) is the hundredths digit in the decimal \(0.1X\) and if \(Y\) is the thousandths digit in the decimal \(0.02Y,\) where \(X\) and \(Y\) are nonzero digits, which of the following is closest to the greatest possible value of \(\dfrac{0.1X}{0.02Y}?\)

A. 4
B. 5
C. 6
D. 9
E. 10

Answer: D

Source: Official Guide

To MAXIMIZE the value of 0.1X/0.02Y, we must MAXIMIZE the numerator (0.1X) and MINIMIZE the denominator (0.02Y)

So, the numerator is maximized when X = 9. So, the numerator is 0.19
The denominator is minimized when Y = 1. So, the denominator is 0.021

So, we must determine the value of 0.19/0.021

IMPORTANT: We need not calculate the value of 0.19/0.021
Instead, just recognize that 0.19/0.02 = 9.5, which is halfway between 9 and 10
Since 0.021 is a bit bigger than 0.02, we know that 0.19/0.021 is a bit LESS THAN 9.5
So, 0.19/0.021 must be closest to 9

Answer: D

Gmat_mission wrote: ↑

Fri Jan 08, 2021 4:39 am

If \(X\) is the hundredths digit in the decimal \(0.1X\) and if \(Y\) is the thousandths digit in the decimal \(0.02Y,\) where \(X\) and \(Y\) are nonzero digits, which of the following is closest to the greatest possible value of \(\dfrac{0.1X}{0.02Y}?\)

A. 4
B. 5
C. 6
D. 9
E. 10

Answer: D

Source: Official Guide

Solution:

We are given two decimals:

0.1X and 0.02Y

To make 0.1X the greatest, we can let X = 9 and we have:

0.19

To make 0.02Y the smallest, we can make Y = 1 (since Y = 0 is not allowed) and we have:

0.021

Thus:

0.19/0.021 = 190/21 = 9 1/21

So, the greatest possible value is about 9.

Answer: D