Quant questions for CAT
December 2nd, 2009 Posted in Quantitative Ability
Email This Post
Quant questions for CAT
Direction for next 2 Quant questions for CAT:
1. The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?
a) 20
b) 25
c) 40
d) 45
e) 75
2. An express train traveled at an average speed of 100 kilometers per hour, stopping for 3 minutes after every 75 kilometers. A local train traveled at an average speed of 50 kilometers, stopping for 1 minute after every 25 kilometers. If the trains began traveling at the same time, how many kilometers did the local train travel in the time it took the express train to travel 600 kilometers?
a. 300
b. 305
c. 307.5
d. 1200
e. 1236
Related posts:
Tags: CAT 2009 Math Questions, CAT 2009 Quant Questions, CAT Quantitative Ability Test for MBA, CAT Sample Quantitative Ability Questions, CAT Sample Quantitative Questions, math CAT questions., math questions, Quantitative Ability, Quantitative Ability Questions, Quantitative Ability Sample Questions for CAT, Quantitative Questions and Answers, Quants question, Sample Quantitative Ability
December 2nd, 2009 at 11:44 pm
1) answer = 70 not 75
2) option(c)
December 2nd, 2009 at 11:47 pm
Sln 1)
To make one maximum.. make the rest minimum. i.e. 45
(45*4 + t) = 50*5 = 250
t = 250-180 = 70
December 2nd, 2009 at 11:55 pm
1.range=max-min
so ans=25(b)
2.(c)
December 3rd, 2009 at 12:23 am
1. b
2. c
December 3rd, 2009 at 1:56 am
1. B
2. C
December 3rd, 2009 at 2:19 am
1. B
(x1+x2+x3+x4+x5)= 50*5 = 250
x5-x1
= 250-(2*45) – (x2+x3+x4)
= 160 – (x2+x3+x4)
To maximize (x2+x3+x4) should be min, min x is 45
thus x2+x3+x4=3*45=135
x5-x1=160-135=25
HITESH you have calculated the value of D. Range is asked.
2. C 307.5 km
December 3rd, 2009 at 2:21 am
2 ka answer kaise aaya
December 3rd, 2009 at 4:29 am
1 )b=25 as done by sumedha
2) c= 307.5 km
time taken 2 travel 600kms by xprees= 6 hrs + 21 min of stopage
now in 6 hrs local vl travel 300 kms with speed of 50 km|hr but it vl take stop for 12 mins as 1 min for each 25 kms…
hence 300 kms in 6 hr 12 mins
in reamining 9 min distance =7.5 kms thus total =307.5 kms
December 3rd, 2009 at 5:08 am
My take is same as you guys..
But why so much of calculations for 1st one..?? when we get the maximim temeprature to be 70..The range is Max-Min = 25.
December 3rd, 2009 at 6:16 am
@ thanks Sumedha… u r right.. i thought maximum value is asked..
December 3rd, 2009 at 6:19 am
1.b
2.c
December 3rd, 2009 at 11:09 pm
December 7th, 2009 at 12:18 am
Social comments and analytics for this post…
This post was mentioned on Twitter by farhan786aslam: Published a new post: Quant questions for CAT http://tinyurl.com/yz5tvbt...
December 8th, 2009 at 6:10 pm
1.b
1.c