Thank you for answering this one...
If s is the product of the integers from 100 to 200 inclusive, and if t is the product of the integers from 100 to 201 inclusive, what is 1/s + 1/t in terms of t?
A. (201)^2/t
B. {(202)(201)}/t
C. 201/t
D. 202/t
E. {(202)(201)}/t^2
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If s is the product of the integers from 100 to 200 inclusiv
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shashank.ism
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S = 100...200imane81 wrote:Thank you for answering this one...
If s is the product of the integers from 100 to 200 inclusive, and if t is the product of the integers from 100 to 201 inclusive, what is 1/s + 1/t in terms of t?
A. (201)^2/t
B. {(202)(201)}/t
C. 201/t
D. 202/t
E. {(202)(201)}/t^2
t = 100...201
t= s.201
1/s+1/t = 1/(t/201)+1/t = 201/t +1/t = [spoiler]202/t[/spoiler]
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Scott@TargetTestPrep
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We are given that s is the product of the integers from 100 to 200 inclusive and that t is the product of the integers from 100 to 201 inclusive. Thus, we know the following:imane81 wrote:Thank you for answering this one...
If s is the product of the integers from 100 to 200 inclusive, and if t is the product of the integers from 100 to 201 inclusive, what is 1/s + 1/t in terms of t?
A. (201)^2/t
B. {(202)(201)}/t
C. 201/t
D. 202/t
E. {(202)(201)}/t^2
s x 201 = t
s = t/201
We must determine the value of 1/s + 1/t in terms of t. Since we know that s = t/201, we can substitute t/201 for s in the expression 1/s + 1/t:
1/(t/201) + 1/t
201/t + 1/t
202/t
Answer: D
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Matt@VeritasPrep
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s = 100 * 101 * ... * 199 * 200
t = 100 * 101 * ... * 199 * 200 * 201
t = s * 201
t/201 = s
so:
1/t + 1/s is the same as
1/t + 1/(t/201) is the same as
1/t + 201/t is the same as
202/t
t = 100 * 101 * ... * 199 * 200 * 201
t = s * 201
t/201 = s
so:
1/t + 1/s is the same as
1/t + 1/(t/201) is the same as
1/t + 201/t is the same as
202/t
