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s, t, and u are integers, and 10≤ s < t < u ≤ 20 [#permalink]
20 Dec 2017, 15:34
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s, t, and u are integers, and 10≤ s < t < u ≤ 20
Quantity A 
Quantity B 
\(s + \frac{t}{u}\) 
11 
A. Quantity A is greater B. Quantity B is greater. C. The two quantities are equal D. The relationship cannot be determined from the information given. kudo for the right solution and explanation
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Re: s, t, and u are integers, and 10≤ s < t < u ≤ 20 [#permalink]
20 Dec 2017, 20:38
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we can deduce deduce the max value of A and min value of A for Max value: S=18 t=19 u=20 in this case A>B for Min Value: S=10 t=11 u=20 In that case A=10+(11/20)<11 So in this case B>A
Hence Ans D



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Re: s, t, and u are integers, and 10≤ s < t < u ≤ 20 [#permalink]
21 Dec 2017, 16:32
Added the OA. It is D. Regards
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Re: s, t, and u are integers, and 10≤ s < t < u ≤ 20 [#permalink]
17 Feb 2018, 01:25
Such questions don't need any sort of calculation. What it need is just thinking. Here's how: Minimum s can be 10. And the ratio of t to u is always less than 1. Thus the result will be between 10 and 11 i.e less than 11. On other hand, s can be greater than 11, and thus Quantity A will become greater in that case because the ratio of t to u cannot be negative. So, information is not sufficient to answer the question. Choice D correct.
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Re: s, t, and u are integers, and 10≤ s < t < u ≤ 20 [#permalink]
17 Feb 2018, 16:00
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GREMasterBlaster wrote: Such questions don't need any sort of calculation. What it need is just thinking. Here's how: As minimum s can be 10. And the ration of t and u is always less than 1. Thus the result will be between 10 and 11 i.e less than 11. On other hand, s can be greater than 11, and thus Quantity A will become greater in that case because the ratio of t and u cannot be negative.
So, information is not sufficient to answer the question. Choice D correct. I got this one wrong because I didn't look closely enough and see that it was either greater or equals, and only noticed the greater bit



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Re: s, t, and u are integers, and 10≤ s < t < u ≤ 20 [#permalink]
04 Jun 2018, 19:16
Carcass wrote: s, t, and u are integers, and 10≤ s < t < u ≤ 20
Quantity A 
Quantity B 
\(s + \frac{t}{u}\) 
11 
A. Quantity A is greater B. Quantity B is greater. C. The two quantities are equal D. The relationship cannot be determined from the information given. kudo for the right solution and explanationFor me, I calculated this by making s, t, & u equal by applying their Maximum and Minimum values of 20 and 10 respectively. Is this reasoning appropriate?



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Re: s, t, and u are integers, and 10≤ s < t < u ≤ 20 [#permalink]
05 Jun 2018, 17:19
Answer: D We assess the biggest and lowest possible amounts for A to be able to compare it with 11.
If we want to maximize A: u should be the lowest and t and s should be the biggest possible. But there might be a tradeoff between u and (s and t). Minimum value for u can be 12 while t is 11 and s is 10. Then A=S+T/U= 10+11/12 which is less than B, but this equation can have a bigger value, for example when s=11 and t=12 and u is whatever between 13 to 20, the value of A is bigger than 11. (The maximum value for A is when s is maximum, because u/t is always something less than 2 and doesn’t have so much effect. And it is when s has the minimum value. For u=20, t=19 and s=18 we have the maximum value of A which is 18+19/20.)
If we want to minimize A: u should be maximum value and t&s should be minimum. But again there might be a tradeoff between u and (s and t) Maximum value that u can have is 20 (as it can be equal to the upper bound(20) and the minimum value for s can be 10 (as it can be equal to the lower bound(10) and t can’t be equal to 10, the Lowes possible value for t can be 11. Thus we have: A = S+T/U = 10 + 11/20 that is less than 11 (B)
Answer is D. As A is between 10+11/20 and 18+19/20
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Re: s, t, and u are integers, and 10≤ s < t < u ≤ 20
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