CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

sudeepdeb

Quote:

Originally Posted by implex

answer is 108

thanks implex.

n=1,10,19 doesn't satisfy.

kunalvakil

Consider a string of n 7s, 7777....77, into which + signs are inserted to produce an arithmetic expression. For example, 7 + 777 + 7 = 791 could be obtained from five 7s in this way. For how many values of n is it possible to insert + signs so that the resulting expression has value 7000?(1) 105 (2) 106 (3) 108 (4) 109 (5) 111I feel answer is (4) 109.Solution. 777x+77y+7z=7000 or, 111x+11y+z=1000.and second inequation from options can be, 3x+2y+z>100 ,but near to hundred. Now, if you go on putting values of x,at x=5 we get,y=40 and z=5 and with these values 3x+2y+z becomes exactly 100.Now keeping the value of x=5 and next lower value of y,i.e. y=39 we get z=16.These values exactly satisfy 111x+11y+z=1000 and with these values we get 3x+2y+z=109 i.e. (4) option

sudeepdeb

kunal,
read the question again carefully. it says "how many values of n", u did not have to find a value of n >100.

But the size of the set of values which n can satisfy.

for eg: 777*9 + 7*1 = 7000
here n = 28.

abhisaba

Dear MODS

The unsubscribe link on the newsletter doesn't seem to be working. Kindly rectify it.

rajkamal.roka

I am getting 109: the approach is

7000=777X9 + 7

now by some trial n errors;
7000=777X5 + 777X4 + 7
=777X5 + [ 77X10 + 7 ] 4 times + 7
=15 7's + 84 7's + 1 (seven) = 100 7's

now we need some more seven's but not many, so we take on of the 77 and write it as 7X10 + 7; hence we take out two seven's and add 11 more- so effectively its 9;
hence the answer is 100 -2 + 11 = 109 7's
hope its correct

am sure thrs some better approach.

kharesaurabh84

i think it shud b 111

see,
a+11b+111c=1000
=>9b+108c=(1000-N)/9
=>b+12c=111+(1-N)/9
=>b+12c=111-k [N=9k+1]
=>k can have any value from 1 to 111 to have non negative values of b and c.

Aarav

Quote:

Originally Posted by abhisaba

Dear MODS

The unsubscribe link on the newsletter doesn't seem to be working. Kindly rectify it.

Thanks for pointing this out - we will resolve this after the weekend. But just a thought - why did you choose to click unsubscribe link?

implex

Quote:

Originally Posted by Aarav

Thanks for pointing this out - we will resolve this after the weekend. But just a thought - why did you choose to click unsubscribe link?

I had the same question, but I dared not ask, but couldn't hold my curiosity now that you have fired the gun!

OperationBLACKI · 11:31 PM

Quote:

Originally Posted by Aarav

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Quantitative Question # 002
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Consider a string of n 7s, 7777....77, into which + signs are inserted to produce an arithmetic expression. For example, 7 + 777 + 7 = 791 could be obtained from five 7s in this way. For how many values of n is it possible to insert + signs so that the resulting expression has value 7000?

(1) 105 (2) 106 (3) 108 (4) 109 (5) 111

Finally got the time to chip in my answer . Was late from office so couldnt post it earlier . I might probably be the last one to answer today . Here's my approach .

From the problem statement we can frame an equation like 7A+77B+777C =7000

Which is nothing but A+11B+111C =1000

We have to find value of N for different value of A , B and C

I have found out a few values or rather ranges of N to find out how many 'n' can be possible . N = No of digits in the series 7777777777....N and 'n' = No of such different series possible .

B -----C------ A------- N

90----- 0------- 10----- 180+10----- 190
0------- 0-------- 1000--1000-------- 1000
0------- 9-------- 1------- 28---------- 28

To get the value of N all we need to do is find the number of terms in the AP

28=1000+(n-1)*-9

Which gives n = 109

I haven't gone through the other solutions here but am sure someone might have come up with a better solution . Alas was happy that I got the answer to todays question .

@ Aarav : This was an amazing question that really required one to think out of the box . Wonderful question . We need more of this kinds in the QQAD . Rather than terming it as a difficult question I would call it a smart question . Even though it took me time to figure out the approach I am happy with the knowledge I gained .

Aarav

Quote:

Originally Posted by OperationBLACKI

To get the value of N all we need to do is find the number of terms in the AP

28=1000+(n-1)*-9

Which gives n = 109

I haven't gone through the other solutions here but am sure someone might have come up with a better solution . Alas was happy that I got the answer to todays question .

@ Aarav : This was an amazing question that really required one to think out of the box . Wonderful question . We need more of this kinds in the QQAD . Rather than terming it as a difficult question I would call it a smart question . Even though it took me time to figure out the approach I am happy with the knowledge I gained .

Actually - I'm rather delighted to see most of you get the answer almost correctly. This was a problem that required 100% concentration. You might be amused to know that n = 37 doesn't hold true and hence we have in all 108 solutions.

Two smart questions follow in the weekend - again I will be more interested in the approach rather than the answer. Next week will be mixture of some swing, googly and bouncer besides some straight ones