Official Quant thread for CAT 2013

MBA CAT 2024
28188
@Subhashdec2 Bhai i dont think you'll find a thread on pg for this....
Torrent par you'll get...try it...:)
28189
@ScareCrow28 yaar wo shayd all possible values of n ka sum pooch ra h..
and n=11 is just one of those values of n
28190
@rrcboss said:
@ScareCrow28 yaar wo shayd all possible values of n ka sum pooch ra h..and n=11 is just one of those values of n
Only n=11 is possible :)
28191
@Asfakul said:
thank you sir . but should we not consider 1,2,3,4 and 4,2,3,1 as one ??
@ScareCrow28 said:
Sir, it contains repeated solutions since 1234 is same aa 1243, please correct me if i am wrong sir
No, here ABCD is mentioned, so 1, 2, 3, 4 and 2, 3, 4, 1 will be different.
28192
@chillfactor said:
No, here ABCD is mentioned, so 1, 2, 3, 4 and 2, 3, 4, 1 will be different.
Okay Sir! :)
Sir if A, B, C and D had not been mentioned, then what could have been done?
28193

GOT IT 😃

28194
@chillfactor said:
ABCD is a quadrilateral with integral sides and perimeter of 36 cm . How many different quadrilaterals ABCD can be formed?Sides can not be more than 17 (as if a side is 18, then sum of remaining three sides will also be 18 which is not possible). Suppose sides are 17 - a, 17 - b, 17 - c, 17 - dSo, (17 - a) + (17 - b) + (17 - c) + (17 - d) = 36=> a + b + c + d = 32So, C(35, 3) = 9139 such quadrilaterals
Sir, ek doubt hai. Wouldn't this 35C3 give us also cases where a or b or c or d go greater than 17, i.e. the side becomes negative in length?

Also, I think the question is not really solvable as a quadrilateral, unlike a triangle, has multiple degrees of freedom - so given 3 sides we get a unique triangle, but given 4 sides we do not get a unique quadrilateral (for example if all 4 sides in the above question are 9, we will get a square and several rhombi.)

regards
scrabbler
28195
@albiesriram said:
1*2*11*-2*-1 = 44

so sum = 11
28196
A & B - having speed in ratio 3 : 5.
Both run on a circular track wid circumference = 1.5 km
Distance travelled by A when A passes B for the 7th time. ?
28197
@The_Loser said:
A & B - having speed in ratio 3 : 5.Both run on a circular track wid circumference = 1.5 kmDistance travelled by A when A passes B for the 7th time. ?
same direction or opp direction?
28198
@scrabbler said:
Sir, ek doubt hai. Wouldn't this 35C3 give us also cases where a or b or c or d go greater than 17, i.e. the side becomes negative in length?Also, I think the question is not really solvable as a quadrilateral, unlike a triangle, has multiple degrees of freedom - so given 3 sides we get a unique triangle, but given 4 sides we do not get a unique quadrilateral (for example if all 4 sides in the above question are 9, we will get a square and several rhombi.)regardsscrabbler
Oops forgot to consider those cases.

a + b + c + d = 32
So, C(35, 3) = 9139 cases

When a is 17 or more, say a = 17 + a'
a' + b + c + d = 15
C(18, 3) = 816 ways

=> Total quadrilaterals = 9139 - 4*816 = 5875
28199
@ScareCrow28 said:
Okay Sir! Sir if A, B, C and D had not been mentioned, then what could have been done?
Then it will be complicated, it will involve circular arrangements also.
28200
@pratskool said:
same direction or opp direction?
not mentioned as such.
I guess should be in same direction
28202
@subhakimi said:
A beats B by 20 meters, B beats C by 21 meters and A beats C by 34 meters in a race. Find the length of the race.Try NOT to use equations.
60m?? did it without using eq... when A reaches, distance b/w B and C is 14m .. then B beats C by 21m finally.... 7m more distance, hence B has already covered twice of what it would cover after A reaches.. hence 20 + 20*2 = 60m
28203
@subhakimi said:
A beats B by 20 meters, B beats C by 21 meters and A beats C by 34 meters in a race. Find the length of the race.Try NOT to use equations.
60 m ?

A....B.....C
d....d - 20....d - 34
.......d..........d - 21

d - 20/d - 34 = d/d - 21....yaha se waise options se bhi kar sakte hai..

d = 60
28205
@subhakimi said:
A beats B by 20 meters, B beats C by 21 meters and A beats C by 34 meters in a race. Find the length of the race.Try NOT to use equations.
When A finishes the race, he is 20 ahead of B and 34 ahead of C i.e. B is 14 ahead of C.

When B finishes the race, he is 21 ahead of C, which is 14 *3/2.

In the second case, therefore, B must have traveled 3/2 times as far as in the first case. But in the first case he was 20 metres from the finish and hence he must have earlier traveled 40 and in the second case 60.

So length of track is 60.

regards
scrabbler
28206
@The_Loser said:
A & B - having speed in ratio 3 : 5.Both run on a circular track wid circumference = 1.5 kmDistance travelled by A when A passes B for the 7th time. ?
if they are in opp direction.. starting from same point, den A would have covered (1.5*5*7)/8 .....
28208
@subhakimi said:
1.5 years ago the ratio of ages of father to that of his son was 1:4 and 15 years hence it will be 1:2. What is the present age of son ?Again try by without equation. (I know it's easy)

Father to son is 1 : 4 😲 Progressive world this is but...!!!

Older 40:10, newer 60 : 30, current 45:15 so I go with 15 years.

@subhakimi said:
2.Akbar says to Birbal "I am four times as old as you were when I was as old as you are". Birbal replies "10 years back I was 9 yours younger to you". Find the age of Birbal.

15 here too I guess, 24 is 4 times 6.

regards
scrabbler
28210
@subhakimi said:
3. How many zeros are there at the end of 626! - 625! ?
625 * 625! so....4 + 125 + 25 + 5 + 1 = 160?

regards
scrabbler
28212
@subhakimi said:
1.5 years ago the ratio of ages of father to that of his son was 4:1 and 15 years hence it will be 2:1. What is the present age of son ?Again try by without equation. (I know it's easy)2.Akbar says to Birbal "I am four times as old as you were when I was as old as you are". Birbal replies "10 years back I was 9 yours younger to you". Find the age of Birbal.3. How many zeros are there at the end of 626! - 625! ?
3rd one:

625/5 + 625/25 + 625/125 + 625/625 + 4 = 160