Of the numbers r + s, r - s, r*s, and r/s, which is greatest?

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

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.

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

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