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10 Jul 2020, 04:59
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C
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Difficulty:
15%
(low)
Question Stats:
82% (01:19) correct
18%
(01:24)
wrong
based on 88
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The difference between two positive consecutive integers, when each is squared, equals 29. What is the smaller number?
A. 12
B. 13
C. 14
D. 15
E. 16
A. 12
B. 13
C. 14
D. 15
E. 16
10 Jul 2020, 05:03
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(n+1)^2 - n^2 = 29
A^2 - B^2 = (A+B)(A-B)
So,
(n+1)^2 - n^2 = 29
=> (n+1+n)(n+1-n) = 29
=> (2n+1)(1) = 29
=> 2n + 1 = 29
=> 2n = 28
=> n = 14
Answer: C
A^2 - B^2 = (A+B)(A-B)
So,
(n+1)^2 - n^2 = 29
=> (n+1+n)(n+1-n) = 29
=> (2n+1)(1) = 29
=> 2n + 1 = 29
=> 2n = 28
=> n = 14
Answer: C
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10 Jul 2020, 05:03
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Bunuel
Let x and x+1 be the consecutive numbers
as per the question stem, (x+1)^2 - x^2 = 29
=> 2x+1 = 29
=>x=14 & x+1 = 15
Smallest number = 14
OA: C
