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30 May 2022, 09:41
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (01:36) correct
39%
(01:48)
wrong
based on 166
sessions
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List A contains \(x_1, \ x_2, \ ...., \ x_{10}\); list B contains \(y_1, y_2, ..., y_{10}\), where numbers in B are the remainders when the numbers in A is divided by 2, respectively. What is sum of numbers in B?
(1) Sum of numbers in A is 100
(2) There are 6 even numbers in B
(1) Sum of numbers in A is 100
(2) There are 6 even numbers in B
30 May 2022, 10:29
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Question stem states that 10 numbers in B are remainders when the Elements in A are divided by 2
There can be only two remainders 0 or 1
So to get the sum we need to have some idea about elements in List B
Only statement 2 gives the idea. 6 numbers are EVEN . 0 is an EVEN number. Implies 6 elements are 0 and 4 are 1 so sum is 4
Answer b
There can be only two remainders 0 or 1
So to get the sum we need to have some idea about elements in List B
Only statement 2 gives the idea. 6 numbers are EVEN . 0 is an EVEN number. Implies 6 elements are 0 and 4 are 1 so sum is 4
Answer b
22 Sep 2023, 23:11
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Bunuel
Solution:
Pre Analysis:
- \(y_1, y_2, ..., y_{10}\) are remainders when \(x_1, \ x_2, \ ...., \ x_{10}\) are divided by 2 respectively
- So, the values of \(y_1, y_2, ..., y_{10}\) will be either 0 and 1 when \(x_1, \ x_2, \ ...., \ x_{10}\) are even and odd respectively
- Therefore to get the answer, we need to know how many of \(x_1, \ x_2, \ ...., \ x_{10}\) are even and odd
Statement 1: Sum of numbers in A is 100
- According to this statement, \(x_1 +\ x_2 + \ .... + \ x_{10}=100\)
- Not gonna help us get the answer
- Thus, statement 1 alone is not sufficient and we can eliminate options A and D
Statement 2: There are 6 even numbers in B
- According to this statement, 6 out of \(x_1, \ x_2, \ ...., \ x_{10}\) are even while the rest 4 are odd
- So, sum of numbers in \(B =6\times 0+4\times 1=4\)
- Thus, statement 2 alone is sufficient
Hence the right answer is Option B
