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01 May 2018, 01:11
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
89% (01:12) correct
11%
(01:17)
wrong
based on 52
sessions
History
Date
Time
Result
Not Attempted Yet
Babia has seven more marbles than three times the number of marbles that Joe has. If Babia has B marbles and Joe has J marbles, which of the following is a possible value for B?
I. 10
II. 18
III. 49
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
I. 10
II. 18
III. 49
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
01 May 2018, 01:41
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Bunuel
Babie = b
Joe = j
b = 7 + 3j
i.e. b - 7 = 3j = multiple of 3
I. 10 i.e. 10 - 7 = 3 which is a multiple of 3 hence a possible value of b
II. 18 i.e. 18 - 7 = 11 which is NOT a multiple of 3 hence NOT a possible value of b
III. 49 i.e. 49 - 7 = 42 which is a multiple of 3 hence a possible value of b
Answer: Option D
01 May 2018, 02:22
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Bunuel
Suppose the number of marbles for Babita is B and for Joe is J (Please note, the number of marbles must be an integer)
Now as per question, B = 3J + 7
now B can take up values like 7 (when J =0)
B = 10 (J =1)
B = 13 (J=2) and so on
We don't actually need to find the whole series, we can check the options provided
B =10 (we already got that at J =1)
B = 18 or 18 = 3J+7 or J = \(\frac{11}{3}\), but the number of marbles must be an integer...so B = 18 not possible
Lastly, B = 49, or 49 = 3J +7, or J = 14
So from the given values of B, only 10 and 49 is possible
Hence option D
