Quant by Arun Sharma

MBA CAT 2024
5486

Can anyone correct me if i am wrong.

1.Arun Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the beginning?
a. 11
b. 10
c. 9
d. 12

The ans i got was 20... but there is no such option only and the ans given in the book is 10

5487

log (2x-2) -log (11.66-x) = 1 + log3

Find x.
(LOD-1, 25;logs;429)

Kindly provide the explanation as well. I know this must be easy but my math is pathetic. Thanks guys.

5488

find x, if 0.01^x= 2
(lod 1, 20; logs;429; ans acc to author: -2log2)
Please explain the solution. I've managed to solve the rest of LOD 1 except for these two.

5489

yes, I am also getting B=20. But this can only be possible when A starts out with 60 coins, B with 20 coins and C with 0 coins.
In this case, for the first iteration, we will get A=20, B=60, C=0
For the second iteration we will get A=60,B=20, C=0.
which is the initial condition.

5490
@neerajagayatri said:

Can anyone correct me if i am wrong.

1.Arun Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the beginning? a. 11 b. 10 c. 9 d. 12

The ans i got was 20... but there is no such option only and the ans given in the book is 10

yes, I am also getting B=20. But this can only be possible when A starts out with 60 coins, B with 20 coins and C with 0 coins.
In this case, for the first iteration, we will get A=20, B=60, C=0
For the second iteration we will get A=60,B=20, C=0.
which is the initial condition.
Sorry, new to this format, forgot to quote the post, hence posting again.
5491
@sputnik said:log (2x-2) -log (11.66-x) = 1 + log3
Find x.
(LOD-1, 25;logs;429)
Kindly provide the explanation as well. I know this must be easy but my math is pathetic. Thanks guys.

(2x-2)/(11.66-x) = 30...Now solve for x

5492


@YouMadFellow said:
There are N different books ... B1 B2.. Bn
and there are P books of each kind.. i.e. p books of B1, p books of B2 etc...
Now, you want to select one or more books from this whole library, so you can select B1 in (P+1) ways i.e take 0 B1 books or take 1 B1 books or take 2 B1 books or... or take P .. B1 books
similarly all other books can also be taken in (P+1) ways
So, (P+1)^N
But one of this would be the case when you select 0 B1 books, 0 B2 books, ... 0 Bn books , and that has to be subtracted
So, (P+1)^N - 1
Why can't we do it like this:-
There are N different books. P books of each kind.
Total books=n*p
Now, for each book, we have a choice of picking up or not.
therefore total cases, 2^(N*P)-1
Kindly point out if there is anything wrong in the approach.
5493

doesnt your approach forgets to consider that P books of each are identical

so if u have to selct one book

ur cases consider selcting 1st copy of the book from p books different from the case where one selects 2nd book form thr p copies. hence repetition

point me if i am wrong

@[468589:Sauron20]

@YouMadFellow said:
Why can't we do it like this:-
There are N different books. P books of each kind.
Total books=n*p
Now, for each book, we have a choice of picking up or not.
therefore total cases, 2^(N*P)-1
Kindly point out if there is anything wrong in the approach.
5494
@Sauron20 said:
@YouMadFellow said:
Why can't we do it like this:- There are N different books. P books of each kind. Total books=n*p Now, for each book, we have a choice of picking up or not. therefore total cases, 2^(N*P)-1 Kindly point out if there is anything wrong in the approach.
The flaw here is the approach of having a choice of picking up of not picking up works when all the things are different from each other.

Say there are 3 kinds of books - B1 , B2 , B3
And each of them have 4 books of each. So, the total number of books is 12.

In the library, the books will look like this.
B1B1B1B1 B2B2B2B2 B3B3B3B3
Now, when you say 2^12 -1, you are double counting few cases, for eg:-

You don't take any from first four books , you don't take the next four too, but then two take the 9th book and the last book. - This is one case and you will have B3 B3 as the selection

Next, you don't take any from first four books, you don't take the next four too, but you take the 10th book and the 11th book - This is another case , when you will have B3 B3 as the selection, and this is the same as above and hence you counted cases like this twice.

So, instead of considering every book's choice, you do it for books of different kind, in this case- B1, B2 , B3 where each one of them have 5 ways to pick up ( pick 1, pick2, .. pick nothing).. hence answer for this would be 5^3 - 1 = 124
5495
@techsurge said:

doesnt your approach forgets to consider that P books of each are identical

so if u have to selct one book

ur cases consider selcting 1st copy of the book from p books different from the case where one selects 2nd book form thr p copies. hence repetition

point me if i am wrong

@Sauron20

Ok..got it, in case it was mentioned to select one or more copies of the book, then this approach could have been used.
Thanks


5496

@[468589:Sauron20]

no i think this approach was right if it was given the p copies are differnet.... like volume 1,2 .....p

@Sauron20 said:Ok..got it, in case it was mentioned to select one or more copies of the book, then this approach could have been used.
Thanks
5497

@[124022:pr_chetana]:- Basically R.S Aggarwal is only for build the basic concepts... Arun Sharma is the best one..

5498

@[75523:abhijit111]:-its 'C'.

5499


@sputnik said:log (2x-2) -log (11.66-x) = 1 + log3
Find x.
(LOD-1, 25;logs;429)
Kindly provide the explanation as well. I know this must be easy but my math is pathetic. Thanks guys.

(2x-2)/(11.66-x) = 30...Now solve for x


2x-2=30(11.66-x)

2x-2=349.8-30x

x=351.8/32



acc to arun sharma answer is 11. also it gives 350/32 as one of the options. my answer is close to both. how do i pick the right choice without taking too much time. also what would be the correct strategy for exams.

5500
@rkshtsurana said:
confusion ll occur in 6, 8 , 9
so 9 numbers are made from dem..
total 2 digit numbers = 90
90-9 = 81
Answer is 71


but i am getting different answer
the numbers which causes confusion..
will be 16890

as it's a two digit number....
total numbers will be 4*4 = 16

as two numbers 69 & 96 will be same..
as from up side down both numbers will be same
so
16-2 =14

so
total numbers will be
90-14 = 74

but the overall answer is 71
puys

plz help...
do tell where i am wrong
5501
@ajayarora116 said:
Answer is 71
but i am getting different answer the numbers which causes confusion.. will be 16890as it's a two digit number.... total numbers will be 4*4 = 16as two numbers 69 & 96 will be same.. as from up side down both numbers will be same
so
16-2 =14
so total numbers will be 90-14 = 74but the overall answer is 71 puys
plz help...
do tell where i am wrong
how 1 can cause confusion..see 1 as displayed not 1 as we write like dis " | "
and 0 ll also not cause confusion bcz if wriiten upside down,,0 ll come on to ten's place which ll nt be a two digit number
5502
@Fuhgetaboutit said: Two Sides of a square lie on the lines x+y=1 and x+y=2 What is its area ?
1/2
5503

guys another one from logs: lod2,2;430:

If log 3=.4771, find log(.81)^2*log(27/10)^2/3 / log 9.

5504
@Fuhgetaboutit said: If P and Q are two points on the line 3x+4y = -15 such that OP = OQ = 9 units. The area of the Triangle POQ will be
18rt2
5505
@Fuhgetaboutit said: How ?
1/2 BH ...find H by 15/sqrt(3^2+4^2)= 3....lemgth of perpendicular from origin to a line
now find B by pythagorus.....u will get 12sqrt2
rest u knw......