Official Quant thread for CAT 2013

MBA CAT 2024
24788

Still no. Guys please work on the hint that i provided earlier. I shall provide the full solution for this by night.

Team BV - Vineet

24789
@bodhi_vriksha said:
Guys, this problem is yet to be solved correctly...Team BV - Vineet
My take will be 4.

Let take the roots of 1st eq a and b and 2nd eq b and c

So
a+b = 2
ab = -4(m^2+1)

Similarly
b+c = 4
bc = -2m(m^2+1)

Now c-a =2
and
c/a = m/2

If we solve further:
a(m-2) = 4

So a or (m-2) can take either 1, 2, 4,-1,-2,-4(if negative both should be negative)
So we can make out that m can have 4 values : 0,1,3,4
24790
@Pradeep_Rss said:
A Baba went to a Jungle with a certain number of toffees. He distributed the toffees equally among the Monkeys; he was left with 7 toffees. Next day he carried the same number of toffees and found that there was twice the number of Monkeys present on the first day. On dividing them equally among the Monkeys, he was left with 47 toffees. Find the number of Monkeys present on the first day
40 monkeys initially ... total 127 toffees..... rem is 7
next day
80 monkeys..... rem is 47 toffees......
so 40..
24791
@gautam22 said:
seen u mentioned 1 on facebook but i guess its 2 for 3 and -1..............10 , -4, -6(in case of 3 ) and for -1 as well there are three values
I have not yet provided solution or hint to this on fb :)
Team BV - Vineet
24792
@Pradeep_Rss said:
A Baba went to a Jungle with a certain number of toffees. He distributed the toffees equally among the Monkeys; he was left with 7 toffees. Next day he carried the same number of toffees and found that there was twice the number of Monkeys present on the first day. On dividing them equally among the Monkeys, he was left with 47 toffees. Find the number of Monkeys present on the first day
M*x + 7 = n
2M*y + 47 = n

M*x + 7 = 2M*y + 47

M*x - 2M*y = 40

M*(x - 2y) = 40

from here M could be 40 only

thanks to @Dexian @Calvin4ever @ChirpiBird
24793
@jain4444 said:
M*x + 7 = n
2M*y + 47 = n

M*x + 7 = 2M*y + 47

M*x - 2M*y = 40

M*(x - 2y) = 40

from here M could be any factor of 40

say M = 40
x - 2y = 1
take , x = 5 and y = 2

or

M = 20
x = 22 , y = 10
total toffees = 247

so , CBD
but sir in the second case that u have mentioned where M=20... in that case we will never have 47 toffees remaining na...
24794
@Dexian said:
but sir in the second case that u have mentioned where M=20... in that case we will never have 47 toffees remaining na...
okay take x = 100 and y = 49

so , we have total toffees = 2*20*49 + 47 = 2007

now we have 47 toffees left right
24795
@jain4444 said:
okay take x = 100 and y = 49

so , we have total toffees = 2*20*49 + 47 = 2007

now we have 47 toffees left right
haan.... i was taking the least possible value for X and Y....
24796
@jain4444 said:
M*x + 7 = n 2M*y + 47 = n M*x + 7 = 2M*y + 47 M*x - 2M*y = 40 M*(x - 2y) = 40 from here M could be any factor of 40 say M = 40 x - 2y = 1 take , x = 5 and y = 2 or M = 20 x = 22 , y = 10 total toffees = 247 so , CBD
2M > 47 ; Hence M> 23 . M cannot be 20. Correct me if I am wrong !!
24797
@jain4444 said:
okay take x = 100 and y = 49 so , we have total toffees = 2*20*49 + 47 = 2007now we have 47 toffees left right
If there are only 40 monkeys the second time, how wud 47 be remaining ?? Wouldn't he distribute one more to each of them??
24798
@Pradeep_Rss said:
A Baba went to a Jungle with a certain number of toffees. He distributed the toffees equally among the Monkeys; he was left with 7 toffees. Next day he carried the same number of toffees and found that there was twice the number of Monkeys present on the first day. On dividing them equally among the Monkeys, he was left with 47 toffees. Find the number of Monkeys present on the first day
40??

jugaadu method. [Image]
see. when number of monkeys is doubled , number of items they receive would be half.
say u need to distribute 13 things among 5 people equally.. each gets 2. ... [3 remaining]
but if u need to distribute 13 things among 10 ...each gets 1...[3 remaining]

here we know M(x-2y)=40
2y=x/2
M*x/2=40
M*X=80

X has to be 2 or multiple of 2. because when monkeys are doubled this X will be halved.

taking first case, x=2
M=40

taking second case x=4. M=20.
but m>23 ... since 2M > 47. else, distribute kar sakte hain.

So i m guessing it should be 40 only.

24799
@ChirpiBird said:
40??jugaadu method. see. when number of monkeys is doubled , number of items they receive would be half.say u need to distribute 13 things among 5 people.. each gets 2. ... [3 remaining]but if u need to distribute 13 things among 10 ...each gets 1...[3 remaining]here we know M(x-2y)=402y=x/2 [halving]M*x/2=40M*X=80X has to be 2 or multiple of 2. because when monkeys are doubled this X will be halved.taking first case, x=2M=40taking second case x=4. M=20. but m>23 ... since 2M > 47. else, distribute kar sakte hain.So i m guessing it should be 40 only.
let say total toffees = 2007

M = 20

on 1st visit he distribute 100 toffees each and on 2nd visit he distribute 49 toffees each

20*100 + 7 = 2007
20*49 + 47 = 2007

whats wrong in it ?
24800
@jain4444 said:
let say total toffees = 2007 M = 20 on 1st visit he distribute 100 toffees each and on 2nd visit he distribute 49 toffees each 20*100 + 7 = 2007 20*49 + 47 = 2007 whats wrong in it ?
2007/40...
wo 50 toffees each monkey dega na.. 49 kyun dega when he can distribute..

40*50 = 2000...remainder bcomes 7 not 47. :)


24801
@ChirpiBird said:
2007/40... wo 50 toffees each monkey dega na.. 49 kyun dega when he can distribute..40*50 = 2000...remainder bcomes 7 not 47
hmm did mistake 😛

chalo ye try karo

A jar is filled with red, white, and blue tokens that are same but their color. The chance of randomly selecting a red token, replacing it, and then randomly selecting a white token is the same as the chance of randomly selecting a blue token. If the number of tokens of every color is a multiple of 3, what is the smallest possible total number of tokens in the jar?

via - kamal/vineet sir

24802
@jain4444 said:
let say total toffees = 2007 M = 20 on 1st visit he distribute 100 toffees each and on 2nd visit he distribute 49 toffees each 20*100 + 7 = 2007 20*49 + 47 = 2007 whats wrong in it ?
Monkey double hongey na dusre case mai?
40*50+7=2007
24803
If a bunch of positive integers adds up to 20, what is the greatest possible product of these integers?
a. 1024 b. 625 c. 1458 d. 2187
24804
@shattereddream said:
If a bunch of positive integers adds up to 20, what is the greatest possible product of these integers?a. 1024 b. 625 c. 1458 d. 2187
1024?
24805
@shattereddream said:
If a bunch of positive integers adds up to 20, what is the greatest possible product of these integers?a. 1024 b. 625 c. 1458 d. 2187
3 + 3 + 3 + 3 + 3 + 3 + 2 = 20

product = 3^6 * 2 = 1458

In how many ways can 7^11 be expressed as a product of 3 numbers?
24806
@shattereddream said:
If a bunch of positive integers adds up to 20, what is the greatest possible product of these integers?a. 1024 b. 625 c. 1458 d. 2187
C ..
729*2
(3^6*2^1)
24807
@jain4444 said:
hmm did mistake chalo ye try karo A jar is filled with red, white, and blue tokens that are same but their color. The chance of randomly selecting a red token, replacing it, and then randomly selecting a white token is the same as the chance of randomly selecting a blue token. If the number of tokens of every color is a multiple of 3, what is the smallest possible total number of tokens in the jar?via - kamal/vineet sir
24 ??