Algebra!

[ankit.kohli87]

whAT is the value of (a!+b!)(c!+d!)?
1.b!d!=4(a!d!)
2.60(b!c!)=(b!d!)

Pls explain the concept.

[messi10]

Hi,

Interesting one...

1) b!d!=4(a!d!)
=> b! = 4a!
=> b!/a! = 4

The only way that the above is possible if b = 4 and a = 3.

4 x 3 x 2 x 1
------------ = 4
3 x 2 x 1

But we do not know anything about c and d so statement 1 is insufficient

I am slightly doubtful about statement 2 because of the presence of 60 but will try and use the same principle

2) 60(b!c!)=(b!d!)
=>60c! = d!
=> d!/c! = 60

d = 60 and c = 59.

Again, statement 2 is insufficient as we do not know anything about a and b.

1 and 2 together, we have all the values of the variables so the answer is C

Please post the OA

Thanks

Sunil

[karan13]

wow what a question.. where did you find this one...

i have tried to solve this ....

whAT is the value of (a!+b!)(c!+d!)?
1.b!d!=4(a!d!)
2.60(b!c!)=(b!d!)

for writing faster i will not write the factorial sign but please assume its there.

from the question we can get ac+ad+bc+bd
1. now bd=4ad not enough so insufficient
2. again insuffcient.

now when we combine them assume bd=x so ad=x/4 ,bc=x/60
now divinde 1 and 2 we get bd/60bc=4ad/bd we get bd=240ac therfore ac=x/240.

now if substitute all values above we get x+x/4+x/60+x/240 but as we do not have a value of x for me the answer is E.

any suggestions... let me know the original ans... but if i had to guess on the gmat i wud go with a C... it has to bring out a solution and i think sunils solution is correct....

[jnelson0612]

Please do post the original source and answer. If this question does not have an original source posted within seven days I will be forced to delete this thread per forum guidelines.