Official Quant thread for CAT 2013

MBA CAT 2024
10981
@nole said:
without stoppage,a train travels at an average speed of 75km/h,with stoppages it covers the same distance at an average speed of 60 km/h.how many minutes per hour does the train stop.I didn't understand what the question meant.How to solve this ?
If the distance to be travelled is D, then difference in the time taken to cover the distance is D/60-D/75=D/300 hrs.
Average stoppage per hr= (D/300)/(D/75)=1/4 Hrs= 15 min.

PS: this is my interpretation of the question...let me know if the answer is rite.
10982
@VINAYAK108 said:
If the distance to be travelled is D, then difference in the time taken to cover the distance is D/60-D/75=D/300 hrs.Average stoppage per hr= (D/300)/(D/75)=1/4 Hrs= 15 min.PS: this is my interpretation of the question...let me know if the answer is rite.
Oops..it's shud be (D/300)/(D/60)=1/5 hrs = 12 min...
10983
coefficient of x^18 in the expansion (x+x^2+x^3+....+x^6)^4 ?
10984
Without stoppage, a train travels at an average speed of 75km/h. With stoppages it covers the same distance at an average speed of 60 km/h. How many minutes per hour does the train stop?

cc: @nole

This seems like a pretty straightforward question but can confuse because of the way it is asked.

The best way to understand it, in my opinion, is to assume a distance.

Let us say that the distance that the train covers is 300 km. (I am taking 300 km because it is divisible by both 75 & 60)

Case 1: Time taken without stoppages = 300 / 75 = 4 hours.
Case 2: Time taken with stoppages = 300 / 60 = 5 hours.

In the second case, the train was moving for 240 minutes and static for 60 minutes. So, the ratio of stopped time to motion time is 60:240 or 1:4.
=> The train stops for 1/5th of the time travelled.
=> The train stops for 12 minutes per hour.

10985
@tani90 said:
@pari19O1 Half of this is correct, but half is wrong... See, you said EC= 8- root 12, now EC = the tangent till the point O strikes a perpendicular on BC, if you take that point as D, then EC=CD....But, you can't surely say that BD would be equal to CD or not.... The remaining value might differ....Is that fine ?
AD is the angle bisector of an isoceles triangle....
so BD and CD have to be equal...
:\

isn't it??
10986

Which among the following is the largest 2^1/2 , 3^1/3 , 4^1/4 , 6^1/6 and 12^1/12 ?

10987
@kingsleyx said:
Which among the following is the largest 2^1/2 , 3^1/3 , 4^1/4 , 6^1/6 and 12^1/12 ?
Raise each term to the power of 12 (LCM of 2,3,4,6,12)

>> 3^1/3
10988
@kingsleyx 3^1/3
10989
@shanu18 said:
@Enigma001 3^1/3
Tagging the wrong guy dude/dudette :P
10990
@Enigma001 han bhai galti se ho gya ...
10991
@kingsleyx said:
coefficient of x^18 in the expansion (x+x^2+x^3+....+x^6)^4 ?
Coefficient of x^18 in (x+x^2+x^3+....+x^6)^4
= Coefficient of x^14 in (1 + x+x^2+....+x^5)^4
= Coefficient of x^14 in (x^6 - 1)^4 * ( x - 1)^(-4)
= Coefficient of x^14 in (1 - 4C1*(x^6) + 4C2*(x^12) + .. higher terms) * (x-1)^(-4)

1/(1-x) = 1 + x + x^2 + .....
1/(1-x)^2 = 1 + 2x + 3x^2 + ....
1/(1-x)^4 = 1 + 4x + 10x^2 + .... T(n) = n*(n+1)*(n+2)/6

So, Coefficient of x^14 = 1*T(15) - 4C1*T(9) + 4C2*T(3)
= (15*16*17/6) - 4*(9*10*11/6) + 6*(3*4*5/6)
= 80 ?

There could be calculation mistakes 😐 ..
10992
6. ABCD is a square. P is the midpoint of AB. The line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R. If AB = 2 then PR = ______ ?
10993
If x > 0, then minimum value of (đ?‘Ľ+1/đ?‘Ľ)^ 6 ˆ' (đ?‘Ľ^6+1/đ?‘Ľ^6) ˆ'2 / (đ?‘Ľ+1/đ?‘Ľ)^ 3+ (đ?‘Ľ^3+1/đ?‘Ľ^3) is......
a. 3
b. 2
c. 6
d. CBD
10994
@tiwarishiva said:
6. ABCD is a square. P is the midpoint of AB. The line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R. If AB = 2 then PR = ______ ?
sol:-
=>Triangle ABR & APD will be similar.
so Applyng Similarity condition,

=>(AP/BR)=(AD/AB).....So BR=1
and in triangle PBR
=> PR=sqrt(1^2+1^2)

Therefore PR=sqrt2
10995
@tiwarishiva said:
If x > 0, then minimum value of (đ?‘Ľ+1/đ?‘Ľ)^ 6 ˆ' (đ?‘Ľ^6+1/đ?‘Ľ^6) ˆ'2 / (đ?‘Ľ+1/đ?‘Ľ)^ 3+ (đ?‘Ľ^3+1/đ?‘Ľ^3) is......a. 3b. 2c. 6d. CBD
Can you write down the question instead of copy pasting, the boxes mean what 😐 :splat: ?
10996
@tiwarishiva said:
6. ABCD is a square. P is the midpoint of AB. The line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R. If AB = 2 then PR = ______ ?
root2
10997
@tiwarishiva said:
If x > 0, then minimum value of (đ?‘Ľ+1/đ?‘Ľ)^ 6 ˆ' (đ?‘Ľ^6+1/đ?‘Ľ^6) ˆ'2 / (đ?‘Ľ+1/đ?‘Ľ)^ 3+ (đ?‘Ľ^3+1/đ?‘Ľ^3) is......a. 3b. 2c. 6d. CBD
6?
10998
@deedeedudu said:
6?
sahi hai boss.....
approach plz
10999
@tiwarishiva said:
sahi hai boss.....approach plz
Make denominator max by x=1/x=1
11000
@tiwarishiva said:
6. ABCD is a square. P is the midpoint of AB. The line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R. If AB = 2 then PR = ______ ?
Put A to origin
A(0,0) B(2,0) C(2,2) D(0,2) P(1,0) let R be (2,y)
Slope AR * Slope DP =-1
y/x * 2/-1 =-1
y=x/2
y=2/2=1
PR=(1,0)------(2,1)
=Sqrt(1+1)=Sqrt(2)