Quant Basics for Zeros in Quant - CAT 2012

MBA CAT 2024
490
freakonomist Says
Thanks for the reply,what would have been your approach if it was 3 people in 10 chairs with no adjacent sitting..


Again there is no option but to count the cases individually.
Select 3 chairs out of 10 such that no two are consecutive. This is done by subtracting cases when 2 consecutive chairs are selected from total no. of ways of selecting 3 chairs out of 10, which is 10 C 3 = 120.

To count ways where atleast 2 are consecutive =
Nos. in brackets are serial nos. of chairs selected, and no. in front is the no. of ways of selecting those 2 chairs.
(1,2) = 8 ways, (1,2,3); (1,2,4) etc.
(2,3) = 7
(3,4) = 6
(4,5) = 5
(5,6) = 4
(6,7) = 3
(7, = 2
(8,9) = 1

Total = 36

So, required ways = 120-36 = 84
And total ways of seating 3 people = 84X3! = 504
491
That will not work....the last assumption you made is itself wrong "any number ending with unit digit 5 divided by 4 will have remainder 1" - for example consider 15 divided by 4, leaves remainder 3! Units digit alone is not sufficient - this is an easy trap to fall into (speaking from experience here 😃 )

~scrabbler

thanks buddy ,that was really a silly mistake 😃
492
_x_x_x_x_x_

out of 6 dash we need to select 5....so 6C5*5!



Junta please note that the solution quoted above is true, and mine was wrong.

I had missed out on 4 additional cases:
1,3,5,7,10
1,3,5,8,10
1,3,6,8,10
1,4,6,8,10
So total no. of ways = 6, and no. of seating arrangements = 6X5! = 720.

The above quoted text is obviously a better way to go about this problem.

Thanks to MadaboutIIMs for pointing out my bad 😃
493

Somebody please solve this, Thanks.

How many even integers n, where 100

494
For Qa please participate in this thread also to get your fundas clear...

http://www.pagalguy.com/discussions/how-to-crack-section-i-of-the-cat-25082497

ps- Number system week
495
Somebody please solve this, Thanks.

How many even integers n, where 100

is it 39

total even numbers 51
numbers divisible by 7 -7
numbers divisible by 9- 6
and one number is common in both so total even numbers will be
51-(7+6-1)
39
496

what is the remainder when 18^323/325?

497
sameer verma Says
what is the remainder when 18^323/325?



=>18^322*18
=>(324)^161*18 (taking 18^2=324)
=>(325-1)^161*18
=>(-1)^161*18
=>-1*18=-18

Remanider = 325-18=307

What is the answer?
498

-----------------------

499

--------------------------

500
Again there is no option but to count the cases individually.
Select 3 chairs out of 10 such that no two are consecutive. This is done by subtracting cases when 2 consecutive chairs are selected from total no. of ways of selecting 3 chairs out of 10, which is 10 C 3 = 120.

To count ways where atleast 2 are consecutive =
Nos. in brackets are serial nos. of chairs selected, and no. in front is the no. of ways of selecting those 2 chairs.
(1,2) = 8 ways, (1,2,3); (1,2,4) etc.
(2,3) = 7
(3,4) = 6
(4,5) = 5
(5,6) = 4
(6,7) = 3
(7, 8 ) = 2
(8,9) = 1

Total = 36

So, required ways = 120-36 = 84
And total ways of seating 3 people = 84X3! = 504


Hi crab365,

I am afraid this won't work. When you have counted the 36 cases, you have missed cases where the 2nd and 3rd chairs are consecutive. This adds another 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 cases such as (1, 3, 4), (1, 4, 5) (2, 4, 5)....till (7, 9, 10)

So we must do 120 - 36 - 28 = 56 cases x 3! = 336 ways.

I also have another way of trying this, but will describe that in a separate post as this one is already confused...that also gives 336 as an answer (and will work for the earlier problem too)

~scrabbler
501
freakonomist Says
Thanks for the reply,what would have been your approach if it was 3 people in 10 chairs with no adjacent sitting..


An alternate approach (seems logical, though I can't quite prove it correct):

Let me represent a person by a "1" and a chair by a "0". So first I will put in the sequence "1 0 1 0 1" which satisfies the condition that the people should be separated. Now the remaining 5 "0"s can be placed in any of the positions - before the first person, between any of the people, or beyond the last; in other words we can distribute the remaining 5 identical "0"s into 4 groups which can be done in (5+4-1)C(4-1) or 8C3 or 56 ways. In each case the 3 people can be arranged in 3! ways ie 56 x 6 = 336 ways.

If I try it with the earlier question i.e. the 5 people 10 chairs case, I start with "1 0 1 0 1 0 1 0 1" and say that the remaining chair can be put into any of the 6 groups formed by the 5 people in (1+6-1)C(6-1) or 6 ways (this could also have been counted manually, in this case!) and hence we get 6 x 5! = 720 ways. Of course, for this question, MadaboutIIMs method is far superior as it is much quicker and more elegant and easier to understand! :)

~scrabbler
502
3) Let us assume that each gram costs 1 Rs .
So The price of 1 kg = 1000.
He says that the price is Increased by 20%.
Now the new price = 1200
But the Actual weight is 800gm. Its cost = 800
Profit = 1200 - 800 = 400
Profit % = 400/800 * 100 = 50%

4) Let quantity of milk be 100 Liters.
Let the cost of each liter be 1 Rs. So the total cost = 100.
He wants 25% gain which means 100 + 25% of 100 = 125
Now he must add 25 liters of water to the milk.
Percentage of water in the mixture is 25/125 * 100 = 20%

5) Profit % = S.P - C.P / C.P * 100
22 = (S.P - 200 / 200) * 100
44 = S.P - 200
S.P = 244


I had done "5" with "S.P. = (100 + profit%) C.P 100 "
and my answer is 266.
Plz correct me ??
503
good work puys!!!!!!!!!

here are some fundas/concepts which could be useful..

PS: I found them in old QUANT threads in PG... hope it is useful..


Ton of thanks....
504
Hi;
We get 2 Equations. as per the question solving which we get the req. ans.
(s-c)/c = 20/100 ............................(1)
(s-100)-(c-100)=(c-100)(104/100).....(2)

Solve simultaneous eq. to get the ans..


'Q2 a person sells his table at a profit of 25/2% and the chair at a loss of 25/3% but on the whole he gains rs 25 . on the other hand if he sells the table at a loss of 25/3% and the chair at a profit of 25/2% then he neither gain nor loss
find the cost price .'

Let CP be x,y respectively::again 2 simultaneous equations:-
x(1+25/200) + y(1-25/300) - (x+y) =25......(1)
x(1-25/300) + y(1+25/200) = (x+y)............(2)



Regards,
Never Back Down



Having Doubt...
Plz explain
In this
"
(s-c)/c = 20/100 ............................(1)
(s-100)-(c-100)=(c-100)(104/100).....(2)
"
Why its not 120/100 in eq (1) ???

And
How these two eq made ??
"
x(1+25/200) + y(1-25/300) - (x+y) =25......(1)
x(1-25/300) + y(1+25/200) = (x+y)............(2)
"
TIA
505
SI / CI :

1) what annual installment will discharge a debt of Rs 1815 due in 3 years at 10% simple interest ?

a - 500
b - 520
c - 550
d - 580



2) A person invested Rs 20,000 in a bank which is offering 10% per annum simple interest. After two years, he withdrew the money from the bank and deposited the total amount in another bank which gives an interest rate of r% p.a compounded annually. After 2 years, he received an amount of Rs 2460 more than what he had invested in that bank. Find r ?

a - 5
b - 10
c - 15
d - 12


3) A person borrowed a certain sum at the rate of 20% p.a compound interest. He paid a total of Rs 4000 in two annual equal installments. what is the interest paid ?

a - 955
b - 944
c - 933
d - 922


4) What annual installment will discharge a debt Rs 717.60 due in 4 years at 20% p.a simple interest, if installments are paid at the end of each year?

a - 128
b - 134
c - 138
d - 144

Kindly do post the explanations. thanks puys

506
SI / CI :

1) what annual installment will discharge a debt of Rs 1815 due in 3 years at 10% simple interest ?

a - 500
b - 520
c - 550
d - 580


4) What annual installment will discharge a debt Rs 717.60 due in 4 years at 20% p.a simple interest, if installments are paid at the end of each year?

a - 128
b - 134
c - 138
d - 144

Kindly do post the explanations. thanks puys



soln....

1. there will be three installments.....

on 1st installment he will get 2 years interest
on 2nd installment he will get 1 year interest
on 3rd installment he will get no interest as it would be final installment

let each installment be x
interest on 1st installment for the vendor = 0.2x=> amount =1.2 x
interest on 2nd installment for the vendor = 0.1x=> amount =1.1 x
3rd installment = x

so total money the vendor will make from 3 installments = 3.3 x =1815
x=550

4. there will be four installments.....

on 1st installment vendor will get 3 years interest
on 2nd installment vendor will get 2 year interest
on 3rd installment vendor will get 1 year interest
on 4th installment vendor will get no interest as it would be final installment

let each installment be x
interest on 1st installment for the vendor = 0.6x=> amount =1.6 x
interest on 2nd installment for the vendor = 0.4x=> amount =1.4 x
interest on 2nd installment for the vendor = 0.2x=> amount =1.2 x
4th installment = x

so total money the vendor will make from 3 installments = 5.2 x =717.6
x=138


3) A person borrowed a certain sum at the rate of 20% p.a compound interest. He paid a total of Rs 4000 in two annual equal installments. what is the interest paid ?

a - 955
b - 944
c - 933
d - 922

could anyone tell me what's wrong my approach
let loan be P and compounded annually for 2 years on 20 % means

A= P(1.2)^2 =1.44 P
Now let installment be equal to x
x=4000
interest on 1st installment =0.2x => amount =1.2x= 4800
no interest on 2nd installment so amount =4000
total =8800

8800 =1.44 P
P = 6111.11
interest =8000-6111.11 = 1888.888

where am i wrong
507

A train encounters 38 intermediate stations between two cities. In how many ways can the train stop at exactly 3 intermediate stations such that no two of them are consecutive?

1) 6545
2) 5984
3) 7140
4) 8436

508

The answer says its 36C3.But that is in the case of no consecutive stations as

3 stations can be considered as 1
38-3=35

35+1(3 taken together) = 36

36C3

How is it the answer in case of no consecutive stations??

509

No matter hw hard I study I get only 50-60%ile in mocks... Pls help wat to do?