Quantative Ability Questions
November 2nd, 2009 Posted in Daily Question & Gyan
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1. A test has 50 questions. A student scores 1 mark for a correct answer, -1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than
a. 6
b.12
c. 3
d. 9
2. The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?
a. 1st
b. 9th
c. 12th
d. None of the above.

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Tags: CAT 2009 Math Questions, CAT 2009 Quant Questions, CAT Quantitative Ability Test for MBA, CAT Sample Quantitative Ability Questions, CAT Sample Quantitative Questions, Quantitative Ability, Quantitative Ability Questions, Quantitative Ability Sample Questions for CAT, Quantitative Questions and Answers, Sample Quantitative Ability
November 3rd, 2009 at 12:17 am
1-c
2-c
November 3rd, 2009 at 12:26 am
1.c
2.c
November 3rd, 2009 at 12:27 am
Ans 1 c
Ans 2 d
very easy qus
November 3rd, 2009 at 12:27 am
answer of 2nd question is 13th term
November 3rd, 2009 at 12:54 am
answer for both the question is c
November 3rd, 2009 at 3:20 am
Answer is
1.c
2.c
November 3rd, 2009 at 9:11 am
ans1) c
ans2) c
November 3rd, 2009 at 1:50 pm
1. c
2. c
November 3rd, 2009 at 11:25 pm
1. Ans. c
Let the number of correct answer be ‘x’ number of wrong answers be ‘y’ and the number of questions not attempted be ‘z’
Thus, x + y + z = 50 …………….(i)
And x – y/3 z-6 = 32
The second equations we get,
7x – y = 242/7 + y
Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can have is 3.
2. Ans. c.
If we consider the third term to be ‘x’
The 15th term will be (x + 12d)
6th term will be (x + 3d)
11th term will be (x + 8d) and
13th term will be (x + 10d)
Thus, as per the given condition, 2x + 12d = 3x + 21d.
Or x + 9d = 0
x + 9d will be the 12th term.
November 4th, 2009 at 5:38 am
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November 8th, 2009 at 11:44 pm
2. solution
If first term is ‘a’ then
3rd term = a+2d
6th term = a+5d
11th term = a+10d
13th term = a+12d
15th term = a+14d
Given,
(a+2d)+(a+14d) = (a+5d)+(a+10d)+(a+12d)
a+11d = 0 ,ie the 12th term