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I'm starting a new thread as the discussions will happen on this new thread for next 50 questions and the sectional test :-) Happy solving :smiley:

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Q- in an examination, the maximum marks for each of the three papers is 50 each. the maximum marks for the fourth paper is 100. find the number of ways with which a student can score 60% in aggregate.

a) 330850 b)233551 c) 110551 d) 220800 e) none of these

hi........:grin:

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Ya I'm aware of that formula for GP..thanks :)actually I was confused coz I assumed that there are 6 terms in GP including 4 GM....Any luck in solving Quant q 98 that I had posted earlier...

MaskedMenace Saysyeah...for a gp common ratio is term2/term1 ....nd it remain same through out d series...tht is wht i hav done....if not clear just revert

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this is regd following q :

Four Geometrical means are inserted between 1/8 and 128.find the 3rd Geometrical mean.

hey just one query....regd the way u calculated common ratio..

I thnk it shuld be...calculated using nth term of GP formula..In that putting first term as 1/8 and nth term as 128 and n =6we get common ratio as : 4....In your Method what I found dubious was that equaiton that you have formed....can you elaborate it a bit....coz assumin " x" is common ratio I thnk that equation can't be formed....

yeah...for a gp common ratio is term2/term1 ....nd it remain same through out d series...tht is wht i hav done....if not clear just revert

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this is regd following q :

Four Geometrical means are inserted between 1/8 and 128.find the 3rd Geometrical mean.

hey just one query....regd the way u calculated common ratio..

I thnk it shuld be...calculated using nth term of GP formula..In that putting first term as 1/8 and nth term as 128 and n =6we get common ratio as : 4....In your Method what I found dubious was that equaiton that you have formed....can you elaborate it a bit....coz assumin " x" is common ratio I thnk that equation can't be formed....

Just find the common ratio..n keep multiplying it ...

this is how it can b done...

1/8, _ , _ , _ , _ ,128

Now,128/x = 8x == >x =4 (common ratio)

so, 1/8,1/2,2,8,32,128 ....3rd gm is 8..

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hello all,

quant Question no 98 :**------------------------------------------------------****Quantitative Question # 098****------------------------------------------------------ **

**Consider two distinct positive integers x and y having integer arithmetic, geometric and harmonic means. The minimum value of x - y is **

And my logic for solving this is described in my post above...but as per that I'm not able to get an answer...

Need help .....

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Hello all,

I'm Deepa..

just had this query regd the Quant Q 98 :It says AM,GM,HM all should be integral..so that means the no's x & y should be an even perfect square ...x & y can be both positive or both negative....coz otherwise we can't get an intergral square root...

now goin by all this....lets consider even perfect squares like 4,16,36,64....and x&y; can be (2,2) ; (4,4) ; (-2,-2)....like wise and if we go by this method we can't get an answer....:...can sum1 sugg a way to solve it......

hoping for quick replies ;)..

i cldnt able to find q.no 98...please post tht ques...

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Four Geometrical means are inserted between 1/8 and 128.find the 3rd Geometrical mean.

I kno its a simple question but just cant find the answer!

Just find the common ratio..n keep multiplying it ...

this is how it can b done...

1/8, _ , _ , _ , _ ,128

Now,128/x = 8x == >x =4 (common ratio)

so, 1/8,1/2,2,8,32,128 ....3rd gm is 8..

Commenting on this post has been disabled.

I'm Deepa..

just had this query regd the Quant Q 98 :It says AM,GM,HM all should be integral..so that means the no's x & y should be an even perfect square ...x & y can be both positive or both negative....coz otherwise we can't get an intergral square root...

now goin by all this....lets consider even perfect squares like 4,16,36,64....and x&y; can be (2,2) ; (4,4) ; (-2,-2)....like wise and if we go by this method we can't get an answer....:...can sum1 sugg a way to solve it......

hoping for quick replies ;)..

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Bhaiyon Aur Behno,

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