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

[implex]

Originally Posted by Ailurophobia

Using the concept of relative velocity between A&B:
[81+(PQ-81-54)*36/(36+36)] - [54+(PQ-54-36)*36/(36+36)]=PQ/30
=>4.5=PQ/30
=>PQ=135

Now,
Since C & B are travelling at the same speed, they will meet at exactly half the distance remaining from P after B has travelled for 1.5 hrs, which is
= (135 - 1.5*36)/2
= 81/2
= 40.5Km....Ans.

can you explain the part in bold
why not PQ=600??

[v-factor]

Let PQ be x Km

Case 1: A slows down after 60 min.

A's distance in 60 min = 54 Km and B's = 36 Km
They meet after further t1 min
A's and B's distance is 36t1

90+72t1 = x --->(1)

Case 2:A slows down after 90 min
Similarly,
A's total distance = 81+36t2
B's total distance = 54+36t2

135+72t2 = x---->(2)

From (1) and (2) we get
t2 - t1 = 5/8 hrs.

Also the difference A's distance in 2 cases is x/30
So
(81+36t2) - (54+36t1) = x/30

36+ x/30 = 36(t2 - t1)
36 + x/30 = 36 * 5/8
x= 135 Km

Let C and B meet at y Km from P after time T
So,
y = 36T
135 - y = 36(T+1.5)
2Y = 81
Y = 40.5 Km
Answer (C)

[shivam_01]

Originally Posted by srikar2097

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Quantitative Question # 040
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Two cars A and B started from P and Q respectively towards each other at the same time. Car A was travelling at a speed of 54km/h but due to some problem reduced its speed by 1/3rd after travelling for 60 minutes. Car B was travelling at a speed of 36km/h. Had the technical problem in car A had arisen 30 minutes later, they would have met at a distance which is (1/30*PQ) more than towards Q than where they met earlier(PQ > 120km). Anothet car C starts from P, 90 minutes after car B started at Q, and car C travels towards Q with a speed of 36km/h, at what distance from P will cars B and C meet?

(a) 63 km (b) 48 km (c) 40.5 km (d) none of the foregoing

Case 1: When car A breaks down after 1 hour.
PQ is the distance
54 + 36t + 36t + 36 = PQ

72t = PQ - 90 ---(1)

Case 2 : When car A break after 1.5 hour
Let t' be the time taken

A's total distance = 81+36t'
B's total distance = 54+36t'
135+72t'=PQ
72t' = PQ-135 ---(2)

t-t'=5/8

distance traveled by A in t hrs - distance travelled by A in t' hrs = 1/30 PQ

(81+36t') - (54+36t') = x/30

=>PQ=135

When C starts B has already covered 54 km. there fore distance from A (135-54)/2 = 40.5

thus option C

[adarshiitm]

My solution for QQAD #40

suppose the distance between P and Q is x
the relative speed of the cars A and B before the problem= 54+36 = 90 km/h
the relative speed of the cars after the problem= 54*2/3+36 = 72 km/h

case 1: when the problem in car A arises after 60 minutes
total time taken before the cars meet(in hr) = 1 + (x-90)/72 = (x-18 )/72
the distance between Q and meeting point M1 = (x-18 )/72*speed of B
= [(x-18 )/72]*36 = x/2-9

case 2: when the problem in car A arises after 90 minutes
time taken(in hr) = 3/2 + (x-90*3/2)/72 = (x-27)/72
the distance between Q and meeting point M2 = (x-27)/72*36 = x/2 - 27/2

given that QM1 - QM2 = 1/30*x
27/2 - 9 = x/30
=> x = 135
distance between C and B when C starts = 135 - 3/2*36 = 81
relative speed of C and B = 36+36 = 72
time taken by C to meet B = 81/72
distance from P = 81/72*36 = 40.5 kms
so option (c)

[Daniel Ocean]

Total speed before technical problem = 90 kph
Total speed after technical problem = 72 kph
So, distance traveled by B:

case I : [(PQ-90)/72 + 1] = D1
case II: [(PQ-135)/72 + 3/2] = D2

given that D2-D1= PQ/30
solving, PQ = 135 km

consider B:
travels @ 36kph all the time.
distance covered in first 90 min = 54km
distance remaining = 81km
now B & C travel at same speed. so they meet exactly in the middle of 81 km = 40.5 km from P

so ans is (C) 40.5

ps: why's is the answer always C????

[slam]

Originally Posted by srikar2097

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Quantitative Question # 040
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Two cars A and B started from P and Q respectively towards each other at the same time. Car A was travelling at a speed of 54km/h but due to some problem reduced its speed by 1/3rd after travelling for 60 minutes. Car B was travelling at a speed of 36km/h. Had the technical problem in car A had arisen 30 minutes later, they would have met at a distance which is (1/30*PQ) more than towards Q than where they met earlier(PQ > 120km). Anothet car C starts from P, 90 minutes after car B started at Q, and car C travels towards Q with a speed of 36km/h, at what distance from P will cars B and C meet?

(a) 63 km (b) 48 km (c) 40.5 km (d) none of the foregoing

Hello people, back after a day's(and a night's break). I solved this as follows:
distance from P at which A and B meet = D
distance PQ = x
So, using the info given,
D - 54 = 36 ( x - 90 )/72
D1 - 81 = 36 ( x - 135 )/72

Also, D1 = D + x/30
Substitute this in the second equation and solve for x, to get x = 135 km
For the case of B and C, B would've covered 54k by the time C starts
So, distance from P at which they meet = 36 * ( 135 - 54 )/72 = 40.5 km.
That's it.

[Topgear]

Case 1 -
(P) l---------l-----l--------------l (Q)
<--54---><-x->
<--------------D------------->

Distance b = D-54-x
if speed wasn't reduced by 1/3rd.. A wood travel = 54+3x/2

Aslo ratio of speeds = ratio of distances

54/36 = [54+3x/2] % [D-54-x]

Case 2-

(P) l----------l-------l----------l(Q)
<---81---><--y-->
<-------------D------------>

Distance b= D-81-y
Distance a with original speed = 81+3y/2

Here also

54/36 = [81+3y/2] % [D-81-y]

-------------------------------------------------------

Also

Distance A case 1 - Distance A case 2 = D/30

.....solving we get d=135..

here we need not calculate x and y.. Only PQ is needed to answer the remaining Question..

[m2shines]

M is pt when dey meet first case and M' pt when they meet in 2nd case.
so MM' = PQ/30
speed of car A = 54 and speed of car B = 36
reduced speed of car A = 36
extra distance travelled by CAR A due to extra 30 mins= (54-36)*30/60 = 9

so in 2nd case the pt of meet will 9/2 km away from first pt since both cars were travelling with 36 in first case.

so PQ/30 = 4.5 PQ = 135KmsThus the pt of meet of CAR C and CAR b will be mid pt of distance = (135 - 36*1.5)/2 = 81/2 = 40.5

[Topgear]

Similarly by taking time as a variable instead of distances..

Because we are already given speeds.. if we aren't given speeds then we cud have done with the above approach.

Distance A2 - A1 = d/30

=> 81+36t2 - 54+36t1 = d/30 => 27-36(t1-t2) = d/30 ----------1st eqn

Also.. 54+36t1+36t1+36 = 81+36t2+54+36t2 = d

=> 90+72t1 = 135+72t2 = d

=> 72(t1-t2) = 45 => 36(t1-t2) = 22.5--------- 2nd eqn..

From 1st..

D = 135..

Hence solving second part of question..

B and C have same speed so they need to meet at midpoint after 1.5 hrs..

D' = (135-36*1.5)/2
D' = 40.5 Ans..

[vishrock]

My first post here....


In 1st case :-

Let t1 be the total time by A & B:-
Then,
54 + 36(t1-1) + 36t1 = X --- (1) Dist. of A + Dist. of B

2nd Case :-

Let t2 be the time taken by A & B :-
Then,
54(3/2) + 36(t2-3/2) + 36t2 = X ---- (2)

From 1 & 2
==>
72t1 + 18 = 72t2 + 27
t1 - t2 = 1/8 ----- (3)

Now using the case where diff of dstance covered is X/30 :-

(1) - (2) = X/30

==>
t1 = 4.5 ---- (4)

which gives X =135. ---- (5)

Finally, since dist of C + dist of B = 135

==> 36(t3 - 3/2) + 36t3 = 135
gives t3 = 81/72 --- (6)

hence, point of C&B meeting is (t3-3/2)*36(speed of c) = 40.5

Ans --- (c).


I think this is the longest possible solution for this question