Quant by Arun Sharma

MBA CAT 2024
2438
Which is the largest 4 digit number that can be added to 7249 in order to make the derived number divisible by each of 12,14,21,33 and 54?

Please explain how to solve this one.



Hi,

take lcm of 12,14,21,33 and 54 .it will come 8316.

now subtract 7249 from 8316 then u get answer.

8316-7249 = 1067.

Answer 1067.
2439
What is the maximum value of 'n' such that the expression
42*57*92*91*52*62*63*64*65*66*67 is divisible by 42^n ??

Please give me funda for the same...



Take factors of each term 42,57,92,91,52,62,63,64,65,66 and 67.

now factors of 42 = 7 * 2 * 3

Total factors of 7 is minimum .so value of n should be equal to factor of 7.

42 = 7 * 6
91 = 13 * 7
63 =9 *7

Only three terms contain factor of 7 thus ,n =3.
2440
dinesh1787 Says
It is a problem on circles Im unable to insert the figure So I mentioned the Problem no and chapter else I would have posted the problem

Simple use PT^2=PA x PB ...
2441
Hi,

take lcm of 12,14,21,33 and 54 .it will come 8316.

now subtract 7249 from 8316 then u get answer.

8316-7249 = 1067.

Answer 1067.


are u sure 1067 is the largest no?
2442
urvashi.kapale Says
are u sure 1067 is the largest no?



Yes Next bigger no is 8316 * 2 = 16632 i.e is five digit no.

if you have any thoughts share with me.
2443
urvashi.kapale Says
are u sure 1067 is the largest no?

Since the ques asks for the largest 4 digit no.
add 8316 to 1067
i.e. 8316 + 1067 = 9383
so 9383 is the largest 4 digit no. that can be added to 7249.
2444

9383 is the answer.

The solution given is:

7249+n=8316x2 ... (8316 being the LCM)
=>n=9383

But I did not get the funda behind it. Why multiply by 2?

2445

If a man saves 1000 each year & invests at end of year at 5% compound interest how much amount will be at end of 15 years?
a.21478b.21578c.22578d.22478e.22178
ans=b
pls help me solve this

2446

dis book has too many mistakes and qustion quality is also not gud..wat u say puys...

2447
If a man saves 1000 each year & invests at end of year at 5% compound interest how much amount will be at end of 15 years?
a.21478b.21578c.22578d.22478e.22178
ans=b
pls help me solve this

Now 1000/- investment/year can be seen as 15 separate investments over period of 15 yrs.
Since he invests in the end of the year he will get interest for 14yrs on his first investment and none on the 15th investment.
Amount after 15 yrs will be sum of the returns

= 1000*^14 + 1000*^13 +....... +1000*^1 +1000
=a + a.r+a.r^2 +... + a.r^14
This is a GP with a= 1000 & r= =105/100
sum of the GP of 15 terms is
= a * (r^n -1)/(r-1)
=1000 * (^15 -1)/( -1)
= 21578.56 Actually i cudn't do the calc. on paper had to use calci.
Please help if anyone has better idea of calculating ^15

ans : b. 21578
2448
9383 is the answer.

The solution given is:

7249+n=8316x2 ... (8316 being the LCM)
=>n=9383

But I did not get the funda behind it. Why multiply by 2?

Actually the funda is in the question itself.
It says what is largest 4 digit number that can be added to 7249.

so 7249 + n should be a multiple of 8316.
i.e. 7249 + n = 8316 * k where k = 1,2,3,4....
Now for k =1 , n= 1067
k= 2, n= 9383
k= 3, n= 17699

So, 9383 is the largest 4 digit no.
If the ques was for the smallest then 1067 would have been the answer.
2449
Dear PUYs i hav just started CAT preps but with just 3 months left i don't know whether there is any merit in doing this.............
I have started quants with Arun Sharma but i m even stuck up in LOD I.
Please help me in following ques:
Q1: If a,b,c,d are in continued proportion then (a-d)/(b-c) =>x.
What is the value of x.
(a) 2 (b) 3 (c) 1 (d) 0
ans b

Q2: If a,b,c,d are proportional, then (a-b)*(a-c)/a = ?
(a)a+c+d (b)a+d-b-c (c) a+b+c+d (d) none of these
ans b

Please help me as I am getting desperate..............
:

Please help........................:banghead:
2450
A sum of RS 36.90 is made up of 90 coins that are either 20p coins or 50p coins.Find out how many 20 paise coins are there in the total amount?..(a)47 (b)43 (c)27 (d)63 (e)70........pls do show the entire process
2451

A man purchased a cow and a calf for RS 1300.He sold the calf at a profit of 20% and the cow st a profit of 25%.In this way,his total profit was 300/13%.Find the cost price of the cow.(a)RS1100 (b)RS 600 (c)RS 500 (d)RS 400 (e)RS 800...........KINDLY SHOW THE ENTIRE PROCESS.

2452
rahul005 Says
A sum of RS 36.90 is made up of 90 coins that are either 20p coins or 50p coins.Find out how many 20 paise coins are there in the total amount?..(a)47 (b)43 (c)27 (d)63 (e)70........pls do show the entire process


Total amount is 3690 paise

Let there be x coins of 20 paise and y coins of 50 paise

x+y=90

20x+50y=3690

Solving, you get

x=27
y=63

Option (C)
rahul005 Says
A man purchased a cow and a calf for RS 1300.He sold the calf at a profit of 20% and the cow st a profit of 25%.In this way,his total profit was 300/13%.Find the cost price of the cow.(a)RS1100 (b)RS 600 (c)RS 500 (d)RS 400 (e)RS 800...........KINDLY SHOW THE ENTIRE PROCESS.


Let cost price of cow be x

So, that of calf is (1300-x)

Selling price of cow= 1.25x

Selling price of calf= 1.2(1300-x)

Total selling price was 1300+1300*300/1300= 1600

So,

1.25x-1.2x+1560=1600

0.05x=40

x=800

Option (E)

Please post your questions in a single post next time onwards.
2453

Two candles of the same length are lighted at 12 noon.The first is consumed in 6hrs and the second in 4 hrs.Assuming that each candle burns at a constant rate ,in how many hours after being lighted,was the first candle twice the length of the second?ANS. 3p.m.................can u pls show how the ans is coming?

2454
rahul005 Says
Two candles of the same length are lighted at 12 noon.The first is consumed in 6hrs and the second in 4 hrs.Assuming that each candle burns at a constant rate ,in how many hours after being lighted,was the first candle twice the length of the second?ANS. 3p.m.................can u pls show how the ans is coming?


The answer is 3 PM. This is because after the three hours, the first candle would have burned 1/2 of itself which means there is 1/2 remaining.

The second candle would have burned 3/4 of itself till then i.e 1/4 parts of the candle are remaining.

As you can see for yourself, 1/2 is double of 1/4. Hence, the answer is 3 PM.

Roshan Jha
IIMA Aspirant
2455

Hi Shashank,
Please have a look at problems posted by me (pg 246) also.

Thanks in advance.......

2456
A person purchased a cupboard and a cot for RS 18,000.He sold the cupboard at a profit of 20% and the cot at a profit of 30%.If his total profit was 25.833%,find the cost price of the cupboard(ans. is RS 7500).............can u solve dis with the process of alligation??if yes,then pls show.
2457
rahul005 Says
A man purchased a cow and a calf for RS 1300.He sold the calf at a profit of 20% and the cow st a profit of 25%.In this way,his total profit was 300/13%.Find the cost price of the cow.(a)RS1100 (b)RS 600 (c)RS 500 (d)RS 400 (e)RS 800...........KINDLY SHOW THE ENTIRE PROCESS.


Hi,Let assume cost of cow =x

cost of calf =y

then from first condition x + y =1300---------1
Total SP = C.P + 300/13% of 1300 =1600

Now SP = 1.20* x + 1.25* y = 1600----------2

Now solve both equation and find out the answer.