If a, b, c, and d are positive, is ac + bd bc + ad?

Question

If a, b, c, and d are positive, is ac + bd > bc + ad?

(1) b > a
(2) c > d

Please explain reasoning and answer choice. Please note, i am not looking to test different numbers. I can see that as an approach. but, rather would like to see an algebraic way of solving the problem. Also, what is the estimate difficulty level of this question and similar question types?

Any help is appreciated.

Thanks

Bunuel · Answer

If a, b, c, and d are positive, is ac + bd > bc + ad?

Is  ac + bd > bc + ad ? --> is  ac + bd - bc - ad>0 ? --> is  a(c-d) - b(c-d)>0 ? --> is  (a-b)(c-d)>0 ?

(1) b > a -->  a-b<0 . Not sufficient.
(2) c > d -->  c-d>0 . Not sufficient.

(1)+(2)  (a-b)(c-d)=negative*positive=negative . Sufficient.

Answer: C.

I'd say it's 600 level question.

audiogal101 · Answer

Thanks so much for the quick response. that was indeed easier than i thought!!

audiogal101 · Answer

Bunuel,

I was trying the question with a similar approach and here is what I have. Please review my reasoning and tell me what you think.

If a, b, c, and d are positive, is ac + bd > bc + ad?

1) b>a
2) c>d

The prompt " is ac + bd > bc + ad?" can be expressed as "is c(a-b) > d(a-b)?

Statement 1) says b>a, and therefore we know (a-b) is negative. 
So, if knowing (a-b) is negative, we simplify the prompt further, we get "is c<d"? 
Since we don't know values of c and d, the statement is insufficient.

Again question is , "is c(a-b) > d(a-b)?"
Statement 2) says c>d
which means that (a-b) should be positive to maintain the inequality. And we know, that any equality when multiplied by a positive number, the inequality is preserved. So shouldn't answer be B?? That is statement 2 is sufficient.

I must be having some error in my reasoning, cause I perfectly agree with Bunuel's approach and solution. How am I arriving at something different. Even when using testing cases, I get C as answer. I would like to understand where my reasoning is falling apart!!

Bunuel · Answer

For (2) we have that c>d. 
IF a-b is positive, then c(a-b) > d(a-b) and the answer to the question is YES.
IF a-b is negative, then c(a-b) < d(a-b) and the answer to the question is NO.

Hope it's clear.

irda · Answer

Bunuel, a quick question. I did the same as the poster above, and I think we can approach the question in his/her manner too. I have tried to explain, but obviously I can be wrong. If a, b, c, and d are positive, is ac + bd > bc + ad? 1) b>a 2) c>d The prompt " is ac + bd > bc + ad?" can be expressed as "is c(a-b) > d(a-b)? The question basically asks if c > b for this we need to have (a-b) > 0 , or if we find that ( a-b) < 0 then c(a-b) > d(a-b) is no --->( a,b, c, d all positive; thus, none is zero and negative.. we could not prove this way if they were) Statement 1) says b>a, and therefore we know (a-b) is negative. So, if knowing (a-b) is negative, we simplify the prompt further, we get "is c d(a-b)?" Statement 2) says c>d which means that (a-b) should be positive to maintain the inequality. Since we do not know any relation between a & b we can't be sure. For eg: if a>b , then a-b>0 and we can divide both sides of c(a-b) > d(a-b) by (a-b) However, if a< b then (a-b) <0 and the inequality changes to -c > -b , hence c < b But taking both together we know that b > a; hence (a-b) <0 and the inequality changes to -c > -b , hence c < b From this we can say c < d

mvictor · Answer

If a, b, c, and d are positive, is ac + bd > bc + ad?

we can rewrite it as:
ac-ad>bc-bd?
factor a and b
a(c-d)>b(c-d)?

since all are positive, the question is:
is a>b if c-d is >0?
or is a<b if c-d is <0?

so as we can see..we need 2 pieces of information...

(1) b > a
gives only first piece..
suppose c-d=-1.
a*-1 > b*-1

suppose c-d=1
a<b

2 outcomes...so insufficient.

(2) c > d
clearly insufficient. we only know that c-d is positive, but what about the relation between a and b?

1+2 -> sufficient.

re-did the question after 6 months; done in the same way - took 1 minute...

bumpbot · Answer

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If a, b, c, and d are positive, is ac + bd > bc + ad?

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