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25 Dec 2020, 23:03
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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:
40% (03:02) correct
60%
(03:05)
wrong
based on 156
sessions
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Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is
a. 33
b. 27
c. 30
d. 36
e. 28
a. 33
b. 27
c. 30
d. 36
e. 28
Updated on: 02 Jul 2024, 04:18
Originally posted by BrushMyQuant on 25 Dec 2020, 23:25.
Last edited by BrushMyQuant on 02 Jul 2024, 04:18, edited 1 time in total.
Last edited by BrushMyQuant on 02 Jul 2024, 04:18, edited 1 time in total.
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Aron
Let Aron buys p pencils and s sharpeners
Let cost of one pencil is x so cost of one sharpener is x+2
Total Money spent by Aron = p*x + s*(x+2)
Aditya
Aditya bought twice as many pencils and 10 less sharpeners
So, Pencils bought by Aditya = 2p
Sharpeners bought by Aditya = s-10
Total Money spent by Aditya= 2p*x + (s-10)*(x+2)
Both have spent the same amount of money
=> p*x + s*(x+2) = 2p*x + (s-10)*(x+2)
=> (s-10)*(x+2) - s*(x+2) + 2p*x - p*x = 0
=> -10x - 20 + px = 0
=> px - 10x - 20 = 0
=> p = \(\frac{(10x + 20)}{x}\)
=> p = 10 + \(\frac{20}{x}\)
Now p has to be integer, \(\frac{20}{x}\) has to be integer => x is a factor of 20
We need to find minimum number of pencils bought by both of them => Min value of p+2p = 3p
=> We need to find minimum value of p
=> p = 10 + \(\frac{20}{x}\)
=> For p to be minimum x has to be maximum => 20 (As x is a factor of 20 and maximum value of factor of 20 is 20)
=> p = 10 + 1 = 11
3p = 3*11 = 33
So, Answer will be A
Hope it helps!
Let Aron buys p pencils and s sharpeners
Let cost of one pencil is x so cost of one sharpener is x+2
Total Money spent by Aron = p*x + s*(x+2)
Aditya
Aditya bought twice as many pencils and 10 less sharpeners
So, Pencils bought by Aditya = 2p
Sharpeners bought by Aditya = s-10
Total Money spent by Aditya= 2p*x + (s-10)*(x+2)
Both have spent the same amount of money
=> p*x + s*(x+2) = 2p*x + (s-10)*(x+2)
=> (s-10)*(x+2) - s*(x+2) + 2p*x - p*x = 0
=> -10x - 20 + px = 0
=> px - 10x - 20 = 0
=> p = \(\frac{(10x + 20)}{x}\)
=> p = 10 + \(\frac{20}{x}\)
Now p has to be integer, \(\frac{20}{x}\) has to be integer => x is a factor of 20
We need to find minimum number of pencils bought by both of them => Min value of p+2p = 3p
=> We need to find minimum value of p
=> p = 10 + \(\frac{20}{x}\)
=> For p to be minimum x has to be maximum => 20 (As x is a factor of 20 and maximum value of factor of 20 is 20)
=> p = 10 + 1 = 11
3p = 3*11 = 33
So, Answer will be A
Hope it helps!
General Discussion
02 Oct 2025, 06:17
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There's always an easy way out
a(x) + b(x+2) = 2a(x) + (b-10)(x+2)
On simplifying, you'll get
x(a-10) = 20
x & (a-10) should be integer and we want the value of "a" as minimum
Clearly it'll be 11
Hence, a+2a = 3a = 33
GMAT-Club-Forum-vtu54gbx.png [ 2.87 KiB | Viewed 686 times ]
a(x) + b(x+2) = 2a(x) + (b-10)(x+2)
On simplifying, you'll get
x(a-10) = 20
x & (a-10) should be integer and we want the value of "a" as minimum
Clearly it'll be 11
Hence, a+2a = 3a = 33
ShashankRawat007
Attachment:
GMAT-Club-Forum-vtu54gbx.png [ 2.87 KiB | Viewed 686 times ]
