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655-705 (Hard)|   Algebra|      
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Bunuel
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If p is the product of the reciprocals of integers from 150 to 250, in 21 Apr 2022, 08:15
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If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of \((p^{−1} + q^{−1})\) in terms of p?

(A) p/(251)^2

(B) 251*252*p

(C) 252p

(D) 252/p

(E) 251*252*p^2


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D
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If p is the product of the reciprocals of integers from 150 to 250, in 21 Apr 2022, 08:36
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we can write q = p*(1/251)
1/p+1/q ; 1/p + 251/p
252/p
option D


Bunuel
If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of \((p^{−1} + q^{−1})\) in terms of p?

(A) p/(251)^2

(B) 251*252*p

(C) 252p

(D) 252/p

(E) 251*252*p^2


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If p is the product of the reciprocals of integers from 150 to 250, in 09 May 2022, 20:42
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[quote="Archit3110"]we can write q = p*(1/251)
1/p+1/q ; 1/p + 251/p
252/p
option D


[quote="Bunuel"]If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of \((p^{−1} + q^{−1})\) in terms of p?

(A) p/(251)^2

(B) 251*252*p

(C) 252p

(D) 252/p

(E) 251*252*p^2


Hi Archit3110

Can you please explain your answer. How did you get the q = p*(1/251). I think I did not get the question quite well.
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If p is the product of the reciprocals of integers from 150 to 250, in 09 May 2022, 21:31
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Well p is product of reciprocals from 150 to 250 and

q is product of reciprocals from 150 to 251
So we can say that


q= p*1/251
Anvesh99

Hope this helps..

Anvesh99
Archit3110
we can write q = p*(1/251)
1/p+1/q ; 1/p + 251/p
252/p
option D


Bunuel
If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of \((p^{−1} + q^{−1})\) in terms of p?

(A) p/(251)^2

(B) 251*252*p

(C) 252p

(D) 252/p

(E) 251*252*p^2


Hi Archit3110

Can you please explain your answer. How did you get the q = p*(1/251). I think I did not get the question quite well.
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If p is the product of the reciprocals of integers from 150 to 250, in 24 Aug 2022, 11:14
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Anvesh99
Archit3110
we can write q = p*(1/251)
1/p+1/q ; 1/p + 251/p
252/p
option D


Bunuel
If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of \((p^{−1} + q^{−1})\) in terms of p?

(A) p/(251)^2

(B) 251*252*p

(C) 252p

(D) 252/p

(E) 251*252*p^2


Hi Archit3110

Can you please explain your answer. How did you get the q = p*(1/251). I think I did not get the question quite well.
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p = (1/150)*(1/151)*(1/152).......................*(1/249)*(1/250).
q= (1/150)*(1/151)*(1/152).......................*(1/249)*(1/250)*(1/251)

Replacing (1/150)...........(1/250) with p
q = p * (1/251)

Asked - 1/p + 1/q
-> (1/p)+ (1/(p * (1/251)))
-> (1/p)+(251/p)
-> 252/p -> Option D
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If p is the product of the reciprocals of integers from 150 to 250, in 24 Aug 2022, 22:13
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Bunuel
If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of \((p^{−1} + q^{−1})\) in terms of p?

(A) p/(251)^2

(B) 251*252*p

(C) 252p

(D) 252/p

(E) 251*252*p^2

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Solution:

We are given \(p=\frac{1}{150}\times \frac{1}{151}\times \frac{1}{152}\times .......\times \frac{1}{249}\times \frac{1}{250}\)

And \(q=\frac{1}{150}\times \frac{1}{151}\times \frac{1}{152}\times .......\times \frac{1}{249}\times \frac{1}{250}\times \frac{1}{251}\)

Or we can write \(q=p\times \frac{1}{251}\)

\(⇒q=\frac{p}{251}\)......\((i)\)

We need the value \((p^{−1} + q^{−1})\)

\(=\frac{1}{p}+\frac{1}{q}\)

\(=\frac{1}{p}+\frac{251}{p}\)

\(=\frac{252}{p}\)

Hence the right answer is Option D
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