Quantitative Ability Questions
Quantitative Ability Questions
Direction for next 2 Quantitative Ability Questions:
1. A business school club, Friends of Foam, is throwing a party at a local bar. Of the business school students at the bar, 40% are first year students and 60% are second year students. Of the first year students, 40% are drinking beer, 40% are drinking mixed drinks, and 20% are drinking both. Of the second year students, 30% are drinking beer, 30% are drinking mixed drinks, and 20% are drinking both. A business school student is chosen at random. If the student is drinking beer, what is the probability that he or she is also drinking mixed drinks?
a) 2/5
b) 4/7
c) 10/17
d) 7/24
e) 7/10
2. If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there?
a) 42,500
b) 37,500
c) 45,000
d) 40,000
e) 72,000
Related posts:
- Quantitative Ability Questions and Answers
- Quantitative Ability Questions
- Quantitative Questions For CAT
December 1st, 2009 at 11:31 pm
c
d
December 1st, 2009 at 11:44 pm
1.e (7/10)
2.d.40,000
December 1st, 2009 at 11:54 pm
2. D
Can any one paste the solution for Q1?
December 1st, 2009 at 11:56 pm
expalantion to question 1.
For 1st year the distribution is ,16,16 & 8 ..The probablity of having both the drinks is 8/40 =1/5 …..(1)
For 2nd year student the distribution is 18,18,12 .
In second year, some students don’t take anything .So the number of students who take either of the drink is 18+18-12 =24. The probablity of students who take both the drinks from second year is 12/24 = 1/2…………………….(2)
The required probablity is (1) +(2) = 1/5 + 1/2 =7/10
second one is simple…8*10*10*10*5 = 40,000
December 1st, 2009 at 11:57 pm
Hitesh, can you expalin your approach to question 1..I’m NOT sure on my part..
December 2nd, 2009 at 12:33 am
Not Clear Govind..
December 2nd, 2009 at 12:58 am
Ans 1. c
Lets say there are 100 students:
CASE 1:
40% in 1st year => 40 students in 1st year
1. 40% of 1st year students drink beer = 40% * 40 = 16
2. 20% of 1st year students drink both = 20% * 40 = 8
CASE 2:
60% in 2nd year => 60 students in 2nd year
1. 30% of 2nd year students drink beer = 30% * 60 = 18
2. 20% of 2nd year students drink both = 20% * 60 = 12
Total students drinking beer = 16 + 18 = 34
Total students drinking both = 12 + 8 = 20
Thus, required probability = 20/34 = 10/17
———————————————————
Ans 2. d
No. of ways of filling unit’s digit = 5 (i.e. 1,3,5,7,9)
No. of ways of filling ten’s hundrends and thousand’s digit = 10 * 10 * 10 (any digit from 0 to 10)
No. of ways of filling ten thousand’th digit = 8 (i.e. 1-9 excluding 1 and 5)
Thus total = (5*10*10*10*8) = 40,000
———————————————————-
December 2nd, 2009 at 1:07 am
sumedha….Are you sure of the way you have calculated the probablity…?? We have to select one student out of the total lot…I’m confused..May be i’m wrong..!!
December 2nd, 2009 at 3:55 am
Yes Govind I am 100% sure
Okay I’ll give u simpler explanation
Venn diagram types, the problem is I can’t draw it.
Okay here, I think u are getting confused with Mix,
consider mix as VODKA and beer as BEER
(Take total students as 100)
1st YR 2nd YR
Total 60 40
students
BEER 16 18
VODKA 16 18
BOTH 8 12
Only BEER 16-8=8 18-12=6
Only VODKA 16-8=8 18-12=6
So venn diagram
8 8 8 6 12 6
| | | | | |
BEER BOTH VODKA BEER BOTH VODKA
ONLY ONLY ONLY ONLY
Now reqd probabliity
= No. of students drinking both / No. of students drinking beer
= (8+12)/((8+8)+(6+12))
= 20/34
= 10/17
Incase u still don’t get it, mail me ur Id, I’ll post a venn diagram…
All the best for CAT. I am equally scared!
December 2nd, 2009 at 3:59 am
Oh okay hold on… I tried to put gaps but looks like it did not accept the white spaces and it looks shabby. I hope u understand now.
(Take total students as 100)
1st YR
Total 60
students
BEER 16
VODKA 16
BOTH 8
Only BEER 16-8=8
Only VODKA 16-8=8
———————–
2nd YR
Total 40
students
BEER 18
VODKA 18
BOTH 12
Only BEER 18-12=6
Only VODKA 18-12=6
———————–
Now make venn diagram of what I have written for both 1st and 2nd year students
————————–
Now reqd probabliity
= No. of students drinking both / No. of students drinking beer
= (8+12)/((8+8)+(6+12))
= 20/34
= 10/17
December 2nd, 2009 at 5:01 am
Thanks…sumedha..May be i got confused with that “ïf” “and also” thing..
Yaar i have NOT prepared well..My D day on 5th…i haven’t written any MOCKL too..:-(
December 2nd, 2009 at 5:36 am
Ok.. let me explain the simplest way for solving qus 1
Suppose we have 100 students total
40 in 1st year and
60 in 2nd year
in first year 40% means 16 drinks beer..
And in 2nd year 30% means 18 drinks bear..
Total who likes beer = 16+18 = 34.. ok
Now who drinks both are ..
in first yr 20% means 8
And in 2nd yr 20% means 12
SO total who likes both 8+12= 20
Hence ; out of 34 beer takers ; 20 likes both..
P = 20/34
= 10/17… simple.. ( no need to consider prob of selecting first year or second year. since we have already selected one who likes beer..)
December 2nd, 2009 at 6:02 am
Thanks Hitest, referring my first post, that’s exactly what I have explained but had to go by the lengthy approach, sometimes that gives a better understanding!
Anyhow, can someone tell me how to go about DI…
As in any clues from the recent CAT papers!
December 2nd, 2009 at 6:52 am
Hi,
Can any body share soft copy of CAT materials/mock test papers with answes…
It will be very helpful to me.
Email:-haldar.uttam@gmail.com
December 2nd, 2009 at 10:37 am
Uttam, just go to topcoaching.com or minglebox, you’ll get plenty of material to study from.
December 2nd, 2009 at 11:12 pm
December 7th, 2009 at 8:38 pm
1.c
2.d
I’m agree with Sumedha’s description
December 7th, 2009 at 10:49 pm