A seemingly easy question that's very hard

[Stockmoose16]

(49 + 49)/ (49+49^2)


The OA: 1/25

But I don't understand, shouldn't it be:

98/147

Please explain.

[parallel_chase]

lets solve this separately
49+49 =98

49+49^2 = 49 (1+49) = 49*50

98/(49*50) = 2/50 =1/25

Hope this helps.

[vishubn]

(49 + 49)/ (49+49^2)
-->49(1+1)/49(1+49)

now we end up with 2/50 => 1/25

the whole idea is to factorize

vishu

[Stockmoose16]

(49 + 49)/ (49+49^2)
-->49(1+1)/49(1+49)

now we end up with 2/50 => 1/25

the whole idea is to factorize

vishu

Why wouldn't it be:

49+49 / 49 + (49*49)

Since 49^2 = 49*49, you'd have to multiply that out first (by PEMDAS)... why is this wrong?

[Stockmoose16]

(49 + 49)/ (49+49^2)
-->49(1+1)/49(1+49)

now we end up with 2/50 => 1/25

the whole idea is to factorize

vishu

In the denominator, why would you factorize (49+49^2) into 49 (1+49)? Wouldn't it be 49 (1+1^2)

That would mean the equation would be:

49 (1+1) /49 (1+1^2)

that would then be: 49 *2 / 49 *2 = 1

Why is this wrong? Ugh, this is so frustrating.

[parallel_chase]

Dude, keep it cool. Dont worry you'll get it.

In the denominator, why would you factorize (49+49^2) into 49 (1+49)? Wouldn't it be 49 (1+1^2)

first of all what is (49+49^2) = 49 + 49*49

Now you take 49 common 49 (1+49) = 49*50

Let me know if you still have any doubts.

[vishubn]

why wouldi do that 49(1+1^2) is again 49 ( 1+1) .. when we multiply it back we get (49+49) and not (49+49^2)
coz eg (3+3^2) is equal to 3 (1+3)=12 and not 3 (1+1^2)= 3(2) =6


i hope it helps

Vishu

[Stuart@KaplanGMAT]

Why wouldn't it be:

49+49 / 49 + (49*49)

Since 49^2 = 49*49, you'd have to multiply that out first (by PEMDAS)... why is this wrong?

That's fine... but you still do the same thing next:

2*49/(1*49 + 49*49)

Then factoring out a 49 in the denominator:

2*49/(1+49)*49

2*49/50*49

We can "cancel out" a 49 on the top and bottom to get:

2/50 = 1/25

Here's another way you can think of it... what if the original had been:

(x + x)/(x + x^2)

we'd then have:

2x/x(1+x)

or

2/(1+x)

Now sub in x=49 and you'll see that the exact same mathelogic applies!