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Jeoren
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 15 Nov 2016, 13:47
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest?

(1) r = s =1

(2) r - s is the least of the numbers
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 16 Nov 2016, 03:08
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S1 - r=s=1, therefore r+s=2; r-s=0; r x s=1; r/s = 1
hence greatest is r+s.

S2 - r-s is the least.
When r=s=1, r+s is greatest.
Suppose, when r=3,s=2
r+s=5; r-s=1; r x s=6; r/s = 3/2 which gives r x s as the greatest.
Since we get two different values, so B not sufficient.

Hence answer is A.
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jyotsnamahajan
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 19 Apr 2019, 09:52
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Is there a better way to solve the second statement? Using numbers as examples takes time and leaves rool for making error
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 13 Aug 2019, 01:13
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Is there a better way to solve the second statement? Using numbers as examples takes time and leaves rool for making error
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For Statement 1, solving for r+s, r-s, rs and r/s clearly indicates r+s as the largest number.

For Statement 2, lets assume the following cases.

Case 1: r is +ve, and s is +ve
=> r+s and r*s is definitely bigger than r-s
=> but its hard to discern which one will be the largest.
=>Please note, you can only take cases where r-s is < than r/s because r/s may or may not be bigger than r-s
eg: if r=4 and s=1, r-s<r/s; but if r=6 and s=2, r-s>r/s

Case 2: r is +ve and s is -ve
=> r-s will always be greater than r+s
=> since this case does not satisfy the condition in Statement 2 (r-s is the least number), we can disregard this case.

Case 3: r is -ve and s is -ve
=> Here too r+s will always be greater than r-s
=> since this case does not satisfy the condition in Statement 2 (r-s is the least number), we can disregard this case.

Case 4: r is -ve and s is +ve
=> analysing the r+s, r-s, rs and r/s no particular pattern is discernible.
=> so we can choose numbers to find cases where r-s is the least but there are different answers for the max

All in all, after looking at the cases, Case 2 and Case 3 can be disregarded and Case 1 and Case 4 does not give you a constant maximum.
During the test, If you start with Case 1, given that case 1 does not give you a number which will always be maximum, you can strike Statement 2 as insufficient
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 20 Oct 2020, 19:00
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest?

(1) r = s =1

We know the values of r and s, so sufficient.

(2) r - s is the least of the numbers

just because r-s is the least, doesn't indicate to us whether r or s is negative or positive either, so r*s and r/s could have varying results. insufficient.
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 15 Aug 2023, 04:26
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KarishmaB


Hi Karishma,

At the first glance of such question how can I solve the above mentioned question through reasoning or number line rather taking different values.

Thanks,
Himanshu
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 21 Aug 2023, 10:50
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Chipping in with my 2 cents.

Statement 1: Straight forward so no sense of adding aditional information.

Statement 2: Since it is stated that "r - s is the least of the numbers" we can set up the following equation --> r + s > r - s. From here we need to consider that S > 0.
Now we can test cases (but with the restriction that S is positive. No other solutuon mentions this restriction).
    R = 10, S = 1 --> r+s is greatest
    R = 10, S = 3 --> r*s is greatest
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Of the numbers r + s, r - s, r*s, and r/s, which is greatest? 06 Jan 2025, 09:31
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