Quantitative ability questions
Email This Post
Quantitative ability questions
1. What is the remainder when 265 + 275 + 285 + 295 is divisible by 110?
a) 0
b) 2
c) 55
d) 4
e) 7
2. Some men and women participate in a tennis tournament. In the first round every man has to play every other man. In the second round every woman has to play every other woman. Then, in the next round every man has to play every woman. How many matches are possible in this round if there were 120 and 91 matches in the 1st and 2nd rounds respectively?
a) 224
b) 200
c) 190
d) 211
e) None of these
3. 1 man, 3 women and 4 boys can complete a work in 96 hrs, 2 men and 8 boys can complete it in 80 hrs, 2 men & 3 women can do it in 120hr. How many hours would it take for 5 Men & 12 boys to complete it?
a) 39 1/11 hrs
b) 42 7/11 hrs
c) 43 8/11 days
d) 44hrs
e) 43 7/11 hrs.
Related posts:
- Quantitative Ability Questions and Answers
- Quantitative Ability Questions
- Quantitative Ability Questions
December 4th, 2009 at 1:39 am
1. A
2. E
3. E
December 4th, 2009 at 4:05 am
1.26^5 + 27^5 + 28^5 + 29^5 is divisible by 55 and 2 individually..
2.n*(n-1)/2 = 120 and
n*(n-1)/2 = 91 gives n = 16 & 14 respectively.
Therefore, the number of mathes in third round =16*14=324.
Not sure though!!!!
3.Let one women = x women, one boy = y boy.
(3x + 4y + 1) …… 96 days
(8y +2) ………….80 days
(3x + 2)………….120 days..
(3x + 4y + 1)*96 = (8y +2)*80 = (3x + 2)*120
=>>> x= 2/3 and y= 1/2
Now,
(5+12y )…………..?????
Now 4 Man===== 120 days
11 Man ====480/11
December 5th, 2009 at 6:39 am
1. A
110=2*5*11
26^5 + 27^5 + 28^5 + 29^5 is divisible by 2,5 and 11
So ans=A
2. A
2.n*(n-1)/2 = 120 and
n*(n-1)/2 = 91 gives n = 16 & 14 respectively.
Therefore, the number of mathes in third round =16*14=224
So ans = A
3. E
December 6th, 2009 at 7:16 am
A
E
E
December 6th, 2009 at 10:57 pm