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Source: — Data Sufficiency |

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by clock60 » Wed May 11, 2011 11:31 am
hi
to me it is neither D not E but rather A
we need product of two -ve numbers that are as far as possible from 0, as their product will be +ve and thus greatest
it is p*q
in d, q*s=(-)*(+)=(-)
also in E it is r*s=(-)*(+)=(-) they can`t be greatest as their product is -ve, but we have also +ve products in consideration

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by Whitney Garner » Wed May 11, 2011 12:41 pm
clock60 wrote:hi
to me it is neither D not E but rather A
we need product of two -ve numbers that are as far as possible from 0, as their product will be +ve and thus greatest
it is p*q
in d, q*s=(-)*(+)=(-)
also in E it is r*s=(-)*(+)=(-) they can`t be greatest as their product is -ve, but we have also +ve products in consideration
Absolutely clock60!

We can systematically eliminate possibilities by starting with positive vs. negative products.

- Because s is the only positive number, anything we multiply it by (other than itself) will have to be negative. Therefore a (+)(-) = (-). There are multiple negative choices, so whenever we multiply them together we will get a (+). Therefore, any product with s cannot be the greatest! Eliminate D and E.

- Now we are only dealing with positive products in A, B and C (product of a neg. times a neg.). That means that we need to find the 2 largest negative numbers and multiply them. (large negative)*(large negative) = large positive. The 2 largest negative numbers are p and q so their product must be the largest.

The correct answer is A.
Whitney Garner
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