official quant thread for cat08

[vishrock]

Originally Posted by bindaas_life

puys ... soln to the Q3 is c) 6 13/15 ...
how about other 2 probs ..

(1 - p + q - pq)(1+r) = 1 - p + q - pq + r - rp + rq - pqr ---1
(1 - q + p - pq)(1+r) = 1 - q + p - pq + r - rq + rp - pqr ---2
(1 - r + p - rp)(1+q) = 1 - r + p - rp + q - rq + pq - pqr ---3


Adding all,

3 + p + q + r - pq - rp - rq - 3pqr

3 + (0) - (-1) - 3(1) = 1

(1+p)(1+q)(1+r) = 1 + p + q + pq(1+r) = 1 + p + q + pq + r + rp + rq + pqr = 1 + (-1) + 0 + 1 = 1

=> ANs = 1

[ankaj]

Originally Posted by masoom

an easy one :

two cars start simultaneoulsy from 2 points p and Q opposite to each other...after crossing each other on the way they take 16 hours and 25 hours respectively... if the speed of car starting from Q is 40 km/hr find the distance PQ...

1800KM is the answer

[ankaj]

Originally Posted by naamakul

Guys please help me with this

Prob:-In a class there are 150 students. at least 38% of them play hockey,40% play basketball and at least 32% play football.if 47 students do not play then at most how many students play exactly one game?
1.72 2.43 3.56 4.28

Students who play Hockey = 57, basketball=60 football=48

let the students who play both hockey and Basketball=a [blue]
let the students who play both hockey and football=b [Yellow]
let the students who play both Football and Basketball= [Green]
All three= D

there fore
57= Hockey only+A+B+D
60= Bsaketball only+A+C+D
48= Football only+B+D+C

Also,
Hockey only+Bsaketball only+Football only+B+D+C+A= 150-47=103
Hockey only+Bsaketball only+Football only=X

165= X+ 2 (A+B+C+D) & 103= X+ (A+B+C+D)
X=206-165=41....
not sure where i missed....

[bindaas_life]

@vishrock .. even i followed the same approach ... but the ans is c) 2 ... it seems ...
and the method is as follows :
(1-p)/(1+p) + (1-q)/(1+q) + (1-r)/(1+r) = 2*(1/(1+p) + 1/(1+q) + 1/(1+r)) -3.
If P(x) = x3 -x -1 has roots p, q, r then P(x-1) has roots as p+1, q+1, r+1
P(x-1) = x3 - 3x2 +2x -1
In a cubic x3 + px2 + qx + r = 0, product of the roots is -r, sum taken 2 at a time is q, and the sum of the roots is -p.
Thus 1/(1+p) + 1/(1+q) + 1/(1+r) = (sum taken 2 at a time)/(product) = 2/1 = 2.
Hence, choice (c) is the right answer.

please let me know if you get it ..

Originally Posted by vishrock

(1 - p + q - pq)(1+r) = 1 - p + q - pq + r - rp + rq - pqr ---1
(1 - q + p - pq)(1+r) = 1 - q + p - pq + r - rq + rp - pqr ---2
(1 - r + p - rp)(1+q) = 1 - r + p - rp + q - rq + pq - pqr ---3


Adding all,

3 + p + q + r - pq - rp - rq - 3pqr

3 + (0) - (-1) - 3(1) = 1

(1+p)(1+q)(1+r) = 1 + p + q + pq(1+r) = 1 + p + q + pq + r + rp + rq + pqr = 1 + (-1) + 0 + 1 = 1

=> ANs = 1

[naga25french]

Originally Posted by ankaj

Students who play Hockey = 57, basketball=60 football=48

let the students who play both hockey and Basketball=a [blue]
let the students who play both hockey and football=b [Yellow]
let the students who play both Football and Basketball= [Green]
All three= D

there fore
57= Hockey only+A+B+D
60= Bsaketball only+A+C+D
48= Football only+B+D+C

Also,
Hockey only+Bsaketball only+Football only+B+D+C+A= 150-47=103
Hockey only+Bsaketball only+Football only=X

165= X+ 2 (A+B+C+D) & 103= X+ (A+B+C+D)
X=206-165=41....
not sure where i missed....

check the bold part.. 2(A+B+C) +3D...

ANSWER IS 43..

[ankaj]

Originally Posted by bindaas_life

Q3 : A takes 10 days,B takes 20 days and C takes 25 days to complete a job.What is the minimum time taken to finish the job if not more than two of them work simultaneously on any single day ? [No two consecutive days have the same pair of people working.]
a) 3 13/29 days b) 3 13/15 days c) 6 13/15 days d) none of these.

when a& b work together one day work is 3/20=15/100
b&c one day work 9/100
c& a = 14/100
consider, the pair AB and CA working alternatively...the work done in first 6 days is 87/100
amount of work left 13/100 ....looks close to C...but how do we map it

[bindaas_life]

@ankaj ... here is my approach :

let the total work is 100 units. Then , work done in units in one day by
A => 100/10 => 10
B => 100/20 => 5
C => 100/25 => 4

therefore, for min time, we can choose the pairs (A+B) and (A+C).
Now (A+B) => 10+5 => 15 units
(A+C) => 10+4 => 14 units . there total is 29 units in 2 days.

For 6 days work done is => 6 X 29 => 87 units . Remaining work (100-87) => 13 units will be taken up by (A+B) pair and hence ans is 6 13/15 days......
i hope it helps...

Originally Posted by ankaj

when a& b work together one day work is 3/20=15/100
b&c one day work 9/100
c& a = 14/100
consider, the pair AB and CA working alternatively...the work done in first 6 days is 87/100
amount of work left 13/100 ....looks close to C...but how do we map it

[bindaas_life]

One more Q puys ....

Q : The function Add[x,y] is defined as increase in the volume of the solution by x%, by adding water only for y number of times. If we operate this function on a 3:2 solution of milk and water with the function being Add[25,2], what is the final ratio of milk to water in the solution ??

a) 67:55 b) 48:77 c) 25:9 d) 25:16

[naiquevin]

Question:
Tarun is standing 2 steps to the left of a red mark and three steps to the right of a blue mark. He tosses a coin. If a head comes up, he moves one step to the right; otherwise he moves one step to the left. He keeps doing this until he reaches one of the two marks, and then he stops. At which mark does he stop if he tosses the coin 21 times in all?


This not a quant problem by the way, taken from a DS problem...
Although I have the ans , the solution is vaguely explained........does anyone know the concept?

[naga25french]

Originally Posted by bindaas_life

One more Q puys ....

Q : The function Add[x,y] is defined as increase in the volume of the solution by x%, by adding water only for y number of times. If we operate this function on a 3:2 solution of milk and water with the function being Add[25,2], what is the final ratio of milk to water in the solution ??

a) 67:55 b) 48:77 c) 25:9 d) 25:16

is it b) 48 : 77 ?