Official Quant thread for CAT 2013

MBA CAT 2024
15623
@ravi.theja said:
101 prime ==> any digit repeated (p-1) multiple times ==> rem =1 ..so 1??
??????????
Here a particular digit is not repeated. A number i.e. 1234 is repeated.
15624

find the hypotenuse of the triangle if base=36015cm & perpendicular=48020cm ?
any less calculation intensive method ?

15625
The first and the last term of an arithmetic progression, having at least three terms, are 5 and 25 respectively. If all the terms of this arithmetic progression are integers, find the number of different values that the common difference of the arithmetic progression can take.
15626
@bs0409

Shld be 5..

Given, 5 + (n-1)d = 25, or, (n-1)d = 20

When d=1, n=21...When d=2, n=11, When d=4, n=6...When d=5, n=5..When d=10, n=3..

Thus a total of 5 unique values..
15627
@TootaHuaDil said:
12341234..........upto 400 digits.......find the remaindr when it is divided by 101
@gs4890 said:
79 ?
@pyashraj said:
@TootaHuaDilIs it 79??
@techgeek2050 said:
1 ?
@ravi.theja said:
101 prime ==> any digit repeated (p-1) multiple times ==> rem =1 ..so 1??
@mailtoankit said:
79?
@catconquerer said:
@TootaHuaDil is it 0?rem[1234/101]=22
Answer is 79
15628
@somnathbhatta

I was able to reduce the calculation to this level..Hope it helps..

36015 = 7^4*5*3...48020= 7^4*2^2*5

Thus, hyp^2 = 7^8*5^2*3^2 + 7^8*2^4*5^2 = 7^8*5^2*(2^4 + 3^2) = 7^8*5^2*5^2

=>hypo = 7^4*5^2 = 2401*25 = 60025 cm..
15629
@bs0409 said:
The actual question is10000!=C*(100!)^kwhere C is a constant. Find the maximum value of k?
Sorry my bad...Did a mistake while posting.
15630

@pyashraj said:
@somnathbhattaI was able to reduce the calculation to this level..Hope it helps..36015 = 7^4*5*3...48020= 7^4*2^2*5Thus, hyp^2 = 7^8*5^2*3^2 + 7^8*2^4*5^2 = 7^8*5^2*(2^4 + 3^2) = 7^8*5^2*5^2=>hypo = 7^4*5^2 = 2401*25 = 60025 cm..
did this itself . thought i'm missing some trick.
mock test question.
15631
@bs0409 said:
The first and the last term of an arithmetic progression, having at least three terms, are 5 and 25 respectively. If all the terms of this arithmetic progression are integers, find the number of different values that the common difference of the arithmetic progression can take.
5 values?
15632
@bs0409 said:
The first and the last term of an arithmetic progression, having at least three terms, are 5 and 25 respectively. If all the terms of this arithmetic progression are integers, find the number of different values that the common difference of the arithmetic progression can take.
5 values?

5+(n-1)d = 25
(n-1)d = 20

n=3,d=10
n=5,d=5
n=6,d=4
n=11,d=2
n=21,d=1
15633
@somnathbhatta said:
find the hypotenuse of the triangle if base=36015cm & perpendicular=48020cm ? any less calculation intensive method ?
factorise the values
36015=3.5.7^4
48020=2^2.5.7^4
Remove the common part i.e. 5.7^4
u r left with 3 and 2^2 i.e 3,4
Now hypotunes for 3,4 will be 5
Now multiply 5 with the common part i.e. 5.7^4
Hence 5.5.7^4 = 60025
15634

@somnathbhatta said:
find the hypotenuse of the triangle if base=36015cm & perpendicular=48020cm ? any less calculation intensive method ?
dont think there is any shortcut...probably u can find it using the choices...

ans will be close to 60000
15635
@bs0409 said:
The first and the last term of an arithmetic progression, having at least three terms, are 5 and 25 respectively. If all the terms of this arithmetic progression are integers, find the number of different values that the common difference of the arithmetic progression can take.
5 ?
15636

@sos2god said:
dont think there is any shortcut...probably u can find it using the choices...ans will be close to 60000
prob 3 was 3 out of 4 choices were close to 60000
15637

The difference between simple and compound interest on a sum of money at 5% per annum is Rs 25. What is the sum?


(a) Rs 5,000 (b) Rs 10,000 (c) Rs 4,000 (d) Data insufficient.

My ans came out to be option (d) but in Arun Sharma book ans is option (b).
15638
@saurav.kgp said:
The difference between simple and compound interest on a sum of money at 5% per annum is Rs 25. What is the sum?(a) Rs 5,000 (b) Rs 10,000 (c) Rs 4,000 (d) Data insufficient.My ans came out to be option (d) but in Arun Sharma book ans is option (b).
The interest is calculated for how many years ?
15639
@saurav.kgp

Bhai i just crss chkd with my Arun Sharma..Option D hi toh diya hua hai in LOD I answer script.. :P
15640
@bs0409 said:
The first and the last term of an arithmetic progression, having at least three terms, are 5 and 25 respectively. If all the terms of this arithmetic progression are integers, find the number of different values that the common difference of the arithmetic progression can take.
1,2,4,5,10 ... 5 values
15641
@pyashraj said:
@bs0409Shld be 5..Given, 5 + (n-1)d = 25, or, (n-1)d = 20When d=1, n=21...When d=2, n=11, When d=4, n=6...When d=5, n=5..When d=10, n=3..Thus a total of 5 unique values..
@karan20 said:
5 ?
@mailtoankit said:
5 values?5+(n-1)d = 25(n-1)d = 20n=3,d=10n=5,d=5n=6,d=4n=11,d=2n=21,d=1
OA=5..............

A starts from a point P on a circular track ,at 6:00 am and runs around the track in clockwise direction .At 6:15 am ,B starts from the same point P and runs around the track in counter clockwise direction and meets A for the first time at 7:09 am.If B started at 6 am then he would have met A for the first time at 7 am.If both A & B start from P at 7:00 am and run in the opposite direction,then find the time at which they would meet at the starting point for the first time?
15642
@bs0409

Shld be at 12:00 Noon..

Let the speed of A be x m/min n that of B be y m/min..

Thus, Length of the Track = (x+y)*60 m..

Now, By 6:15 am A wud have covered = 15x m..Thus, Length of the remaining track = 60(x+y) - 15x m=> 15(3x + 4y) m..

Now, 15(3x + 4y)/(x+y) = 54, or, 2y = 3x..Thus, the ratio of their speed is 2:3..

Now, If they simultaneously started at 7 am..Time taken by X to complete one lap = 150 min..

Again, Time Taken by B to complete one Lap = 100 min..

Thus, Time taken for them to meet at P for the 1st tym = LCM(150, 100) = 300 min..

Thus, after 5 hrs from 7 am, i.e at 12:00 Noon will they meet 4 the 1st tym at P..