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23 Dec 2022, 07:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (02:00) correct
68%
(01:57)
wrong
based on 113
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12 Days of Christmas GMAT Competition with Lots of Fun
If x and y are non-zero integers, is \(\frac{x}{30} + \frac{y}{50} = 10\)?
(1) \(x + y = 500\)
(2) \(y – x = 20\)
If x and y are non-zero integers, is \(\frac{x}{30} + \frac{y}{50} = 10\)?
(1) \(x + y = 500\)
(2) \(y – x = 20\)
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24 Dec 2022, 04:02
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Bunuel
GMATWhiz Official Explanation:
Step 1: Analyse Question Stem
- x and y are non-zero integers
- We are given x/30+y/50=10
- ⇒ (5x+3y)/150=10
⇒ 5x+3y=1500
⇒ 3y=1500-5x
⇒ y=(1500-5x)/3
⇒ y=500-5x/3 - Since y is an integer, therefore 5x/3 has to be an integer for which we can assume x=3k where k is non-zero integer
- Plugging x = 3k, we get y=500-(5×3k)/3 ⇒ y=500-5k
- So, for x/30+y/50=10 to exist, the value of x and y has to be of the form 3k and 500 – 5k respectively
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: x + y = 500
- If x=3k and y=500-5k is true:
- o We know x + y = 3k + 500 – 5k = 500 – 2k
o 500 – 2k is never equal to 500 because k is non-zero - So, x=3k and y=500-5k is not true
- Thus statement 1 alone is sufficient and we can eliminate options B, C and E
Statement 2: y – x = 20
- If x=3k and y=500-5k is true:
- o We know y – x = 500 – 5k – 3k = 500 – 8k
- 500 – 8k = 20 when k = 60
- And 500 – 8k ≠ 20 for other values of k.
- 500 – 8k = 20 when k = 60
Hence the right answer is Option A.
General Discussion
UtkarshAnand
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23 Dec 2022, 08:29
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Bunuel
Solution:
Given, x and y are non-zero integers
\(\frac{x}{30} + \frac{y}{50} = 10\)?
or 5x + 3y = 1500?
Statement (1) \(x + y = 500\)
Multiplying the equation by 3 we have, 3x + 3y = 1500. This when compared to 5x + 3y = 1500 has a difference of 2x unaccounted for.
As x and y are non-zero integers, clearly 2x will have some value and hence, either 5x + 3y>1500 or 5x + 3y<1500. But 5x + 3y cannot equal 1500 (as x cannot be equal to 0).
Sufficient.
Statement (2) \(y – x = 20\)
As the difference is given, we cannot manipulate the statement to arrive at something similar to question statement, 5x + 3y = 1500. Clearly, insufficient.
ANSWER A.
