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kevincan
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If x/16 = 1,199.625 28 May 2026, 20:07
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D
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55% (hard)

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35% (02:17) correct 65% (03:05) wrong based on 17 sessions

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If \(\frac{x}{16}=1,199.625\) and \(\frac{y}{25}=29.12,\) what is the sum of the remainders when \(x\) and \(y\) are divided by \(24\) and \(15\), respectively?

A. 18
B. 20
C. 24
D. 26
E. 28
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D
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ankushsambare
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If x/16 = 1,199.625 29 May 2026, 20:01
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x / 16 = 1199.625
So, x = 1199.625 × 16 = 19194
Now divide 19194 by 24:
24 × 799 = 19176
Remainder = 19194 − 19176 = 18
So remainder of x ÷ 24 = 18
y / 25 = 29.12
So, y = 29.12 × 25 = 728
Now divide 728 by 15:
15 × 48 = 720
Remainder = 728 − 720 = 8
So remainder of y ÷ 15 = 8
Sum of remainders = 18 + 8 = 26
Answer: D. 26
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If x/16 = 1,199.625 29 May 2026, 22:04
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This question requires alot of calculation, is there a faster way of doing this?
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If x/16 = 1,199.625 29 May 2026, 23:03
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Yes, there's a cleaner way to do each part. The trick is working from a nearby round number rather than brute-forcing the division.

For x: x = 16 * 1199.625 = 19194. Now you need x mod 24. Instead of dividing 19194 by 24 from scratch, notice that 19200 = 800 * 24 exactly (since 800 * 24 = 19200). So 19194 = 19200 - 6, and the remainder is 24 - 6 = 18. Done in about 5 seconds.

For y: y = 25 * 29.12 = 728. You need y mod 15. Notice 720 = 48 * 15 exactly. So 728 = 720 + 8, remainder = 8.

Sum = 18 + 8 = 26. Answer D.

The general move here: once you have your big number, ask "what's the nearest multiple of the divisor that I can compute quickly?" Usually a round number like 800 or 48 jumps out faster than dividing directly. Remainders on the GMAT almost always come out cleanly this way — the question writers set the numbers so that the nearest multiple isn't hard to find.
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