Official Quant thread for CAT 2013

MBA CAT 2024
21326
@hippophobia said:
all the terms are in form :-11+11+1+21+1+2+3+4.....................+1551+(155*156/2)12091plz correct me if wrong
Nice....faster method :)

regards
scrabbler
21327
@iLoveTorres said:
bro how did you get that the height to point "Y" is half that of height to point "D"?
Look at AYE and ADG. They are similar with sides in ratio 1 : 2 hence ht too will be in 1 : 2. Check the height from base AE or AG...

regards
scrabbler
21328
Two motorcyclists, Ajay and Vijay, start simultaneously from a point S on an oval track and drive around the track in the same direction, with speeds of 29 km/hr and 19 km/hr resp. Every time Ajay overtakes Vijay (anywhere on the track), both of them decrease their respective speeds by 1 km/hr. If the length of the track is 1 km, how many times do they meet at the starting point before Vijay comes to rest?
21329
@krum a+b=197 ayega
Soln:
@jain4444 :is ans 197??
S=1-(97/2^100)
So, a=97 which is odd and +ve and b=100;
21330
In how many ways can 4 boys and 20 girls stand in a line so that no 3 of the boys are next to each other?
21331

Question:

Vikram asked Mayank to count the number of mangoes on a mango tree near their college.
Mayank said, €œThe number obtained by reversing the digits in the number of mangoes that
are one third of half the total number of mangoes on the tree is 10 more than two third of
half the mangoes on the tree. Also the number of mangoes is not less than the square of the
lowest two digit number and not more than double of the highest two digit number €?. Find
the sum of digits in the number of mangoes on the tree


Give the approach please... Don't know the answer !
21332
The expression 1/2^2 + 2/2^3 + 3/2^4 +…+ 99/2^100 can be expressed as 1− (a/2^b), where a is an odd positive integer, and b is a non-negative integer. What is the value of a+b ?
Let S=1/2^2 + 2/2^3 + 3/2^4 +…+ 99/2^100........(1)
S/2= 1/2^3+2/2^4+........................+99/2^101.......(2)
1-2, we get,
S/2=1/2^3(1 + 2/2^1 + 3/2^2+ ...................+ 99/2^98)
or 4S= P= 1 + 2/2^1 + 3/2^2+ ...................+ 99/2^98 .................(3)
P/2= 1/2^1 + 2/2^2+.......................+98/2^98 +99/2^99.........(4)
(3)-(4),
we get
P/2=1+ 1/2{ [1-(1/2)^98]/(1/2) } + 99/2^99
or,S=1- 1/2^99 + 99/2^100 [as P=4S assumed]
THUS S=1-97/2^100 hence a+b=197

21333
@jain4444 said:
In how many ways can 4 boys and 20 girls stand in a line so that no 3 of the boys are next to each other?
Options? Kuch to bada sa aata hai, can't calculate value! Something like 24! - 21!*(4! + 20*3!)

regards
scrabbler
21334
@tani90 said:
Question:Vikram asked Mayank to count the number of mangoes on a mango tree near their college.Mayank said, €œThe number obtained by reversing the digits in the number of mangoes thatare one third of half the total number of mangoes on the tree is 10 more than two third ofhalf the mangoes on the tree. Also the number of mangoes is not less than the square of thelowest two digit number and not more than double of the highest two digit number €?. Findthe sum of digits in the number of mangoes on the treeGive the approach please... Don't know the answer !
12 I get...

Let the number be x. We are told x between 100 and 198. Also 1/6th of x, when reversed is 10 more than 1/3rd of x. So x is divisible by 6.

Let x=6y. So y between 100/6 and 198/6 => between 17 and 33 inclusive. Also, y reversed = 2y+10. And y is 2 digit with secnond digit more (else reverse can't be greater). A little trial and error yields y = 26, so x = 156, so sum of digits = 12.

regards
scrabbler
21335

1.In how many ways can a person send invitation cards to 6 of his friends if he has 4 servants to distribute the cards?



2.In how many ways can 5 prizes be distributed to 8 students if each student can get any no. of prizes?


Plz share approach
21336
@scrabbler Yeah thanks ! I was taking it as a two digit but later on realized the >100 condition ! the answer is one of the options so fine
21337
1.In how many ways can a person send invitation cards to 6 of his friends if he has 4 servants to distribute the cards?

2.In how many ways can 5 prizes be distributed to 8 students if each student can get any no. of prizes?

Plz share approach
21338
@maroof10 said:
1.In how many ways can a person send invitation cards to 6 of his friends if he has 4 servants to distribute the cards?

Assuming the invitations are distinct - each inv can go to any of the 4 servants so 4 * 4 * 4...= 4^6 = 4096.
@maroof10 said:
2.In how many ways can 5 prizes be distributed to 8 students if each student can get any no. of prizes?

If prizes distinct then 8^5 = 32768 as above. If identical then (5+7)C7 = 792.

regards
scrabbler
21339
@scrabbler said:
Options? Kuch to bada sa aata hai, can't calculate value! Something like 24! - 21!*(4! + 20*3!)regardsscrabbler
please explain 21!*20*3!
21340
@maroof10 said:
1.In how many ways can a person send invitation cards to 6 of his friends if he has 4 servants to distribute the cards?2.In how many ways can 5 prizes be distributed to 8 students if each student can get any no. of prizes?Plz share approach
1> 4^6
2> 8^5
21341
@hippophobia said:
please explain 21!*20*3!
Is for exactly 3 boys together. 20! ways for girls. Around them 21 spaces so grp of 3 boys can go in 21 ways and single boy in 20 ways. And the 3 can be arranged internally in 3!. So 20! * 21 * 20 *3! = 21! *20*3!

regards
scrabbler
21342
@scrabbler thanks

21343
@amresh_maverick right....thanks
21344
@jain4444 said:
Two motorcyclists, Ajay and Vijay, start simultaneously from a point S on an oval track and drive around the track in the same direction, with speeds of 29 km/hr and 19 km/hr resp. Every time Ajay overtakes Vijay (anywhere on the track), both of them decrease their respective speeds by 1 km/hr. If the length of the track is 1 km, how many times do they meet at the starting point before Vijay comes to rest?
@scrabbler how to solve this
21345
0,1,1,2,0,5,2,2,4,2,2,4,3,7,2,?