official quant thread for cat08

[implex]

Originally Posted by harneet9

Good question -
australia scored a total of x runs in 50 overs . india tied the score in 20% less overs . If India's average run rate had been 33.33% higher the scores would have been tied 10 overs earlier . Find how many runs were scored by austalia ?
a) 250
b) 240
c) 200
d) 175
e) none of above

(x/40)(4/3)=(x/30)
its an identity
so any x will do for us

none of the options are correct.
none of the above is not a correct option
it should be all of the above !

[raghav507]

Originally Posted by implex

(x/40)(4/3)=(x/30)
its an identity
so any x will do for us

none of the options are correct.
none of the above is not a correct option
it should be all of the above !

yes, its an identity because it is satisfying any of the given values...

[srikar2097]

New Problem:

Find the smallest +ve cube that ends with digits 2-0-0-7 (that's two thousand and seven)?

(a) 7543 (b) 4549 (c) 7343 (d) 8343 (e) None

[raghav507]

Originally Posted by srikar2097

New Problem:

Find the smallest +ve cube that ends with digits 2-0-0-7 (that's two thousand and seven)?

(a) 7543 (b) 4549 (c) 7343 (d) 8343 (e) None

hi srikar,
does the ques mean (number)^3=XX...XX2007

if this is the case then i am getting 7543 as the answer...
just want to see the approach you all are following...coz i have done it by actual multiplication...

[srikar2097]

Originally Posted by raghav507

hi srikar,
does the ques mean (number)^3=XX...XX2007

if this is the case then i am getting 7543 as the answer...
just want to see the approach you all are following...coz i have done it by actual multiplication...

Yup, your answer is correct. By multiplication! I'm assuming while multiplying you have considered digits till the thousandth place or else you will be lacking paper in exam

I have done it differently though...

[implex]

Originally Posted by srikar2097

New Problem:

Find the smallest +ve cube that ends with digits 2-0-0-7 (that's two thousand and seven)?

(a) 7543 (b) 4549 (c) 7343 (d) 8343 (e) None

just find out remainders on 10,100 1000

we will be done !
1849.43=49.43=2007!!
we r done with the first option itself

[raghav507]

Originally Posted by srikar2097

Yup, your answer is correct. By multiplication! I'm assuming while multiplying you have considered digits till the thousandth place or else you will be lacking paper in exam

I have done it differently though...

hi,
you are right...its a long calculation...
but i did it using the first five digits (from units place) to arrive at the answer...
of course it needed the right choice of the options to carry on with...here it was the first one itself...so i was a bit lucky...
anyways how did u do it...can you post your approach...?

[ankaj]

Originally Posted by srikar2097

Yup, your answer is correct. By multiplication! I'm assuming while multiplying you have considered digits till the thousandth place or else you will be lacking paper in exam

I have done it differently though...

We can neglect the option B as it does not fit in the cyclicity for the last digit.....then we just need to check for the tens and hundreds place...

but how to check the cyclicity for hundreds place????

[srikar2097]

Originally Posted by implex

just find out remainders on 10,100 1000

we will be done !
1849.43=49.43=2007!!
we r done with the first option itself

@Ankaj: I don't think cyclicity is the way to go here. At least I don't see a way. You could give it a try if you know the cyclicity for cubes... Implex seems to have a nice approach...

@Raghav: My sol. is the longest possible. I could have done it through options but I wanted to try out the long method. Cannot type all of it now. Probably you guys could hold on for a couple of hours? I'll post it then...

I didn't quite get your approach. If I take N = 7543, and div. by 10 I get rem(3), by 100 get rem(43) and b 1000 rem(543). then what. Could you elaborate?

[shivam_01]

Originally Posted by implex

just find out remainders on 10,100 1000

we will be done !
1849.43=49.43=2007!!
we r done with the first option itself

49*43 = 2107 how come u got the conclusion >>>>> we need to multiply atleast three digits