for LETS CRACK CAT 2004 junta ----QUESTIONNAIRE

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hey guys can any1 provide solution to this question Q. a sum of Rs10 is lent to be returned in 11 months instalments of Re1 each,interest being simple.the rate of interest is?
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can u plz tell me the remainder wen:
a) 3^1001 is divided by 1001?
b) 18!/ 247?


Thanks to the sheer depth of the forum, not only threads, but even QA problems with the same numbers are starting to repeat themselves! Go here....

http://www.pagalguy.com/forum/showthread.php?t=7663&highlight;=3%5E1001
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can u plz tell me the remainder wen:
a) 3^1001 is divided by 1001?
b) 18!/ 247?


hey!check out in numbersystem thread...solutions are already there..anyway..iam givin outline ..

a) 1001=7*11*13

so find rem of 3^1001 divided by 7,11&13 individually ...then in turn find the rem when divided by LCM of divisors..say i take an eg..
rem when 41 divided by 6 ?(ans is 5)
6=2*3
41 divided by 2 rem =1 such nos are...(2n+1 ..i.e 1,3,5..)
41 divided by 3 rem =2 such nos are...(3n+2 ..i.e 2,5..)
first common no is 5.. next nos are in form 6n+5..
so when divided by 6 the rem is 5..

b) 247= 13*19

so 18! divided by 13*19 is 18!(without 13) divided by 19
i.e 1*2*...12*14*15...*18 divided by 19...remainders are..
1*2*3*4*5*6*7*8*9*-9*-8*-7*-5*-4*-3*-2*-1
i.e (5!)^2 *6*7^2*8^2*9^2 i.e 6^2*6*-8*7*5 i.e -2*6*-8*7*5
i.e 96*35 i.e 1*-3 =-3 i.e 16 so when
1*2*...12*14*15...*18 divided by 19 the remainder is 16 so 18! divided by 247 the rem is 16*13 =208

check out previous post for detailed solns
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can u plz tell me the remainder wen:
a) 3^1001 is divided by 1001?
b) 18!/ 247?

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Hi Meti,
i am not sure man...
this question has a formula....
i do not remember that...
have to go through it.....

but if u apply that formula the answer should be near 9.8%...
i will confirm and let u know


Hey there

The formula is

S.I. = n*a +((r*a/100y)(n*(n-1)/2))

n = number of instalments.
a= amount paid in each instalment.
r=rate of interest.
y=1 if paid yearly, 12 if paid monthly,2 if paid by yearly.


By this we get 10.09%.

Thanks.
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For the maximum answer, why why both a and b have to be 1?
What if I consider a to be equal to 9 and d to be equal to 0?



dunno abt 'a' ..... bt if v tk d = 0 then the reverse of this number won't b a four digit no..... so the condition given in the ques wud b violated.........which can't obviously b done........

infact i feel thusu typed it wrongly saying dat 'a' hs to b one.........he hs in fact taken 'a' to b equal to 9.....which is obvious..........
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The answer must be of the form (1000a+100b+10c+d)-(1000d+100c+10b+a)
or, (999a+90b)-(90c+999d)
Here, a and d must both be equal to one for the maximum answer
So, the answer is of the form 9_ _ 1.
Now, to get the max. value of 90b-90c, b must be 9 and c = 0..
So, the answer must be 9901.


For the maximum answer, why why both a and b have to be 1?
What if I consider a to be equal to 9 and d to be equal to 0?
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hi all,

try this prob...
this is an easy one...just time urself

what is the reaminder when 8^88 is divided by 7???



Hi there,
The remainder is 1.

8 = 1 mod 7

8 = 1 mod 7

8^2 = 1 mod 7

.
.
.

8^88 = 1 mod 7


Hence 1.
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hii ppl

this is anya sah

am doin computer engg (3rd yr) from mumbai.

will be taking CAT in 2005.

just wanted to introduce myself.

ciao

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hey guys
can any1 provide solution to this question
Q. a sum of Rs10 is lent to be returned in 11 months instalments of Re1 each,interest being simple.the rate of interest is?


I have an ingeniuos way to tackle such problem...may be super6 also solves it in the same way but let me explain it for the newcomers ( this has a selfish motive, MY REVISION )

Denote lender by L and borrower by B

Imagine there was no transaction. Nobody gave anything to anyone. After 11 months, lender would have 10 rupees intact and in addition, he would earn an interest on Rs 10 for 11 months

Now, consider the deal to take place. After 11 months, lender would have got Rs 11 ( Re 1 installment for 11 months ). In addition, he earned the following interest :

On Re 1 for 10 months ( this is the money he got as first installment )
On Re 1 for 9 months ( 2nd installment amount )
On Re 1 for 8 months
On Re 1 for 7 months
---------
-------
---
On Re 1 for 1 month.

Lender should not incur any losses because of this transaction, so, both the cases should equate.

=> 10 + SI on Rs 10 for 11 months = 11 + SI on Re 1 for ( 10 + 9 + .... + 2 + 1 ) months
=> 10 + SI on Re 1 for 110 months = 11 + SI on Re 1 for 55 months
=> SI on Re 1 for 55 months = 1
=> 1*r*55/(12*100) = 1
=> r = 21 9/11 % p.a.

This method looks BIG but is very systematic and would never confuse me in such problems. It's analogous to ENERGY CONSERVATION 😛
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hi abhijit ur solution of 8^88 % 7 is th eright process..
thats the way to do it..
bye

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