dave13 wrote:

Bunuel wrote:

What is the value of b + c ?

(1) ab + cd + ac + bd = 6 --> \((ab+bd)+(cd+ac)=6\) --> \(b(a+d)+c(a+d)=6\) --> \((a+d)(b+c)=6\) --> if \(a+d=1\), then \(b+c=6\) but if \(a+d=6\), then \(b+c=1\). Not sufficient.

(2) a + d = 4. Nothing about \(b\) and \(c\). Not sufficient.

(1)+(2) From (1) we have that \((a+d)(b+c)=6\) and from (2) we have that \(a + d = 4\), thus \(4(b+c)=6\) --> \(b+c=1.5\). Sufficient.

Answer: C.

pushpitkc hello

can you help me please

sos

cant understand the logic of Bunuel`s solution...

how from this \(b(a+d)+c(a+d)=6\) he gets this \((a+d)(b+c)=6\)

Alsi \(a+d=1\) how can a+d equal to 1

why a = 0.5 and d = 0.5

?

and he says if \(b+c=6\) ... but first statement says that whole expression is 6 ----> ab + cd + ac + bd = 6

and from combing two statements how can \(b+c=1.5\).

why decimal and not integers

and why 1.5

Hey

dave13To begin with, \(b(a+d)+c(a+d)\) can also be written as \((b+c)(a+d)\)

because if you take (a+d) as common, you will get the expression (b+c)

Whenever you have trouble with such expression - put some random values for a,b,c, and d

If a=1,b=2,c=3, and d=4

The right-hand side \(b(a+d)+c(a+d)\) is \(2(1+4) + 3(1+4) = 2*5 + 3*5 = 25\)

Similarly,the left hand side \((b+c)(a+d)\) is \((2+3)(1+4) = 5*5 = 25\) and we can prove the same

Coming to the second part of your question, in Bunuel's solution it has been given

If \(a+d=1\) then \(b+c=6\) since in statement 1 it is written that \((a+d)(b+c) = 6\)

The reason it is not sufficient is there is a second possibility when \(a+d=6\) and \(b+c=1\)

So, (b+c) can have two values - 1 or 6

Hope this helps clear your confusion!

_________________

Stay hungry, Stay foolish

2017-2018 MBA Deadlines

Class of 2020: Rotman Thread | Schulich Thread

Class of 2019: Sauder Thread