Official Quant thread for CAT 2013

MBA CAT 2024
8141
@bhatkushal said:
Q)At 7:00 AM,Ravish started from A towards B and Amit started from B towards A.At 9:00 AM they crossed each other and continued towards respective destination.if time taken by Ravish is 3 hours more than time taken by Amit. Find ratio of Ravish speed to Amit's speed?Could someone explain it using V1/V2=sqrt(t1/t2)?
D/(r+a) = 2
D/r - D/a = 3

=>2(r+a)/r - 2(r+a)/a = 3
=>2+2a/r-2-2r/a=3
=>a/r-r/a=3/2=2-1/2

1/2
8142
@chillfactor said:
I think bold underlined word should be numbersabc, a can be 1, 6, 7, 8, 9b can be 0, 1, 6, 7, 8, 9c can be 0, 1, 6, 7, 8, 9So, 5*6*6 = 180 such numbersA man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3
3/8???

8143
@gautam22 said:
There are 6 points in a rectangle with the sides, 3 and 4. Prove that the distance between at least two of these points is smaller than the square root of 5.
I'm not sure, but if we consider one point P at the center and then draw a circle of radius ห†ลก5 with center as P, then we can easily prove that we can put at max 4 points on the boundary of the circle.
8144
@gautam22 said:
Prove that for any natural N, 1000^N - 1 cannot be a divisor of 1978^N - 1
1000^N - 1 = (999 + 1)^N - 1
Using binomial, we can say that exponent of 3 in 1000^N - 1 is {3 + (exponent of 3 in N)}

Similarly, 1987^N - 1 = (1986 + 1)^N - 1
Using binomial, we can say that exponent of 3 in 1987^N - 1 is {1 + (exponent of 3 in N)}

So, exponent of 3 is always higher in 1000^N - 1 in comparison to 1987^N - 1. Hence 1000^N - 1 will never divide 1987^N - 1
8145
@mailtoankit said:
3^4highest power 4PS: cat mein aaya tha..
thank god mene bhi yehi mara tha..ek toh sahi hoga.
8146
@chillfactor
ok

8147
Find the length of the side of a square
A. Total cost of flooring the square is Rs. 1,000 at the rate of Rs. 40 per square meter.
B. A rectangle whose perimeter is 20 m is equal to the perimeter of the square.
C. A square of area equal to 50 sq. m can be made on the diagonal of the given square.
a) Any of them b) Any two of them c) Either B or C only d) Either A or B only
e) All three of them
8148
@chillfactor said:
Use Allegations75000 ............... 64000........... 700006000 ..................5000So, ratio is 6:564000............... 73000.............690004000................. 5000So, ratio is 4:5Ratio for T : B : R will be 24:20:25Average = (24*75000 + 20*64000 + 25*73000)/(24 + 20 + 25) = 71086.95Find all natural numbers (a, b) such that a/b + 21b/25a is a natural number
a/b+21b/25a=k

Multiply by 25ab

25a^2+21b^2=25kab
25a^2+25b^2-25kab=4b^2

k=2

(5a-5b)^2=4b^2
25(a-b)^2=4*b^2
(a-b)^2/b^2=4/25

(a-b)/b=2/5
5a=7b
b=5,a=7
b=10,a=14
b=5k,a=7k

8149
The lengths of the hour hand and the minute hand of a clock are 3.5 cm and 5.25 cm respectively. If the hour hand covers an area of 7.7 cm2, then find the approximate area (in cm2) covered by the minute hand during the same time period.

a17
b158
c260
d208
8150
@mbajamesbond said:
The lengths of the hour hand and the minute hand of a clock are 3.5 cm and 5.25 cm respectively. If the hour hand covers an area of 7.7 cm2, then find the approximate area (in cm2) covered by the minute hand during the same time period.a17 b158 c260 d208
208
8151
There are 13 equidistant bus stops on a straight road. A bus running at 60 km/hr is some distance away from the 1st stop from where it will move towards the 13th stop. Two cars start running from the 6th stop in opposite directions with the same speed. If the bus meets one of the cars at the 1st stop and the other at the 13th stop, then find the speed of the cars.

a10 km/hr
b20 km/hr
c30 km/hr
dCannot be determined
8152
@gnehagarg said:
a/b+21b/25a=kMultiply by 25ab25a^2+21b^2=25kab25a^2+25b^2-25kab=4b^2k=2(5a-5b)^2=4b^225(a-b)^2=4*b^2(a-b)^2/b^2=4/25(a-b)/b=2/55a=7bb=5,a=7b=10,a=14b=5k,a=7k
You missed the case (a - b)/b = -2/5
From here you will get a/b = 3/5
So, (3k, 5k)
8153
@mbajamesbond said:
There are 13 equidistant bus stops on a straight road. A bus running at 60 km/hr is some distance away from the 1st stop from where it will move towards the 13th stop. Two cars start running from the 6th stop in opposite directions with the same speed. If the bus meets one of the cars at the 1st stop and the other at the 13th stop, then find the speed of the cars.a10 km/hr b20 km/hr c30 km/hr dCannot be determined
10?
8154
@mbajamesbond said:
There are 13 equidistant bus stops on a straight road. A bus running at 60 km/hr is some distance away from the 1st stop from where it will move towards the 13th stop. Two cars start running from the 6th stop in opposite directions with the same speed. If the bus meets one of the cars at the 1st stop and the other at the 13th stop, then find the speed of the cars.a10 km/hr b20 km/hr c30 km/hr dCannot be determined
So when the bus meets the car at Stop 1, the car 2 is at 11th stop.
And the bus meets the car 2 at 13th stop.

So the Bus travels 12 stops at 60 kmph and the car 2 travels 2 stops in the same time.

d - distance between the stops

12d/Sb = 2d/Sc
12d/60 = 2d/Sc

Sc = 10 Kmph
8155

Q)K and T are moving along two straight roads that intersect perpendicularly at X. When K is at X.T is 500 m from X.After 2 mins both K and T are at equidistant from X,again after 8 mins both are equidistant from X.

What is ratio of K and T's speed?

Bhailog explaination needed....
8156
@mbajamesbond bhai 10 ayega na?
8157
@bhatkushal bhai option hai kya ??
8158
@bhatkushal said:
Q)K and T are moving along two straight roads that intersect perpendicularly at X. When K is at X.T is 500 m from X.After 2 mins both K and T are at equidistant from X,again after 8 mins both are equidistant from X.What is ratio of K and T's speed?Bhailog explaination needed....
a)4:5 b)2:3 c)5:8 d) 1:2
8159
@bhatkushal said:
Q)K and T are moving along two straight roads that intersect perpendicularly at X. When K is at X.T is 500 m from X.After 2 mins both K and T are at equidistant from X,again after 8 mins both are equidistant from X.What is ratio of K and T's speed?Bhailog explaination needed....
is it 4:3 ?
8160
@bhatkushal said:
Q)K and T are moving along two straight roads that intersect perpendicularly at X. When K is at X.T is 500 m from X.After 2 mins both K and T are at equidistant from X,again after 8 mins both are equidistant from X.What is ratio of K and T's speed?Bhailog explaination needed....
Let the speed of T is t m/min and that of K is v m/min

in two minutes, T will cover 2t and K will cover 2v
so, 2v=500-2t=> v+t=250

in next 8 minutes, T will cover 8t and K will cover 8v
so, 8v+2v=8t-(500-2t)=> -v+t=50

=> v=100 m/min, t=150 m/min