CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

srikar2097

Quote:

Originally Posted by himanshuarora

hiw much time did u take 2 solve d ques????
plz reply

I think it took me abt. 9-10min. Why ask? Are you bothered about speed right now?

nitin360

my take for QQAD 40

let they meet at R in first case...
in one hour A will go 54 and B will go 36 km so remining distance is (PQ-90)
after accident speed of A is 36 and b is also 36 so remaining distance cover by both will be same (PQ-90)/2

so PR =54+(PQ-90)/2

again in second case
in 3/2 hour total distance by both will be (54+36)/2*3=135

remaining will be (PQ-135)
similarly PR'=81+(PQ-135)/2
again PR'-PR=PQ/30
so PQ comes out to be 135 km
in case of B and C
in 3/2 hr distance traveled by B is 54 kn so remining distance is 81 km

since both car is having same speed so they will travel same distance in same time
so distance traveled by C is 81/2=41.5
so option C is right

ghar

from wat i hv concluded d answer cums out 2 b 48 kms..i.e. pt no. 2...
plz lemme kno d exact ans so dat i cn verify my mistake...thanx

Aarav

Very good response on a weekend day -> keep the spirits up

Here is the official solution to 040

Solution:

Two cars meet at mid-point of PQ if they were at same speed. If the starting point of A shifts by l1 towards Q and that of
B by l2 towards Q, where l2 < l1), the meeting point shifts by (l1 - l2)/2 towards Q.
=> meeting point shifts by (27-18 )/2 km towards Q => PQ = 30*(4.5) km = 135 km

B covers 3/2*36 = 54 km from Q when C starts.
Cars B and C are (135 - 54)km apart when C starts => PR = 81km. If C and B meet => they meet in the mid-way as they have
same speed.

=> Choice (3) is the right answer

pnvishal

QQAD 40

first we will find the distance betwn P and Q.

Let the distance between car A and B be d at the time of break down in car A.

speed of car A after break down = 2/3*54 = 36kmph

this is equal to the speed of car B, hence they will meet at half of this distance d i.e d/2.

if the beak down is delayed by 30 min. car A will travel 54*0.5 = 27 km and car B will travel 36*.5= 18 km. so distance betwn cars when the delayed break down occured = d -45.

theywill meet at (d-45)/2.

difference between two meeting point (d-45)/2 +27 -d/2 = 9/2

now 9/2 = PQ*1/30 => PQ = 135km

in 90 min car B will travel 54 Km. as car C is mooving with equal speed of car B they will meet at the half of the remaining distance 81km.

hence they will meet at 40.5 km from P. Hence option c.

rajeev_hts

Did somebody get QQAD 41? If yes plz post it here. Arav I did not get QQAD 41 yet.

Aarav

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Quantitative Question # 041
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N people vote for one of 27 candidates. Each candidate's vote % is atleast one less than his/her number of votes. What is the smallest possible value of N?

(a) 108 (b) 127 (c) 134 (d) 162 (e) none of these

rajeev_hts

Let 1st candidate's votes =x1
2nd's = x2 and so on...

so N= x1+x2+x3+.....+x27

since (x1*100)/N >= x1-1
or x1 <= N/(N-100)
such as x2 <= N/(N-100)

adding all Xs. (N*27) /(N-100) <= N
N(N-100) >=27N
or N^2 -127N >= 0
N(N-127)>=0
N>=127

So least value of N=127 (b)

Ailurophobia

Why didn't I get Q41 in my mail yet?

nsitaditya

Even i have not received the mail yet