I keep getting B but the answer is A. Where am I going wrong?

Is the best way to assess this problem by prime factoring?

(2^4 * 7)^2 - (7 * 5)^2 vs (7 * 11)^2

Simplified…

(2^8 * 7^2) - (7^2 *5^2) vs 7^2 * 11^2

Thanks for any help!

I keep getting B but the answer is A. Where am I going wrong?

Is the best way to assess this problem by prime factoring?

(2^4 * 7)^2 - (7 * 5)^2 vs (7 * 11)^2

Simplified…

(2^8 * 7^2) - (7^2 *5^2) vs 7^2 * 11^2

Thanks for any help!

(2^8 * 7^2) - (7^2 *5^2) vs 7^2 * 11^2

Up to this, you are correct.

Now take common on both sides, i.e,

(7^2)*[(2^8) - (5^2)] and (7^2)*(11^2)

Now just get rid of what they share in common, i.e, (7^2)

So, the equation becomes,

(2^8) - (5^2) and (11^2)

=> 256 - 25 and => 121

=> 231 and => 121

Therefore, A>B

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You can use various strategies along with prime factorization to solve this question. Here, is my approach :

As we can clearly see from the question that the two identities that we require to solve are :

(a-b)^2 = (a+b)(a-b)\\
(a-b)^2 =a^2-b^2-2(a\times b)

Let’s us first simplify the quantities using these equations:

**Quantity A:**

(112)^2-(35^2) = (112+35)(112-35) = (147)\times(77)\\
\text{Now, let us approximate that 147 is closer to 150 and 77 is closer to 80}\\
= 15 \times 8 \times 100 = 12,000 \\

**Quantity B :**

(112-35)^2 = (112)^2 + 35^2 - 2(112\times35)\\
\text{let us approximate that 112 is close to 110 and}\\
\text{35 ^2 should be the somewhere in between 30^2 = 900 and 40^2 = 1600}\\
\text{so, let us choose the value to be somewhere around 35^2=1200}\\
(112-35)^2 = (110)^2 + 1200 - 2 (35\times110)\\
\text{11^2 = 121 thus, } (110)^2 = 11 \times 11 \times 100=12,100\\
= 12,100 + 1600 - 2 (35 \times110)\\
= 13,700-2 (35 \times110)\\
\text{Now, if we do 2 x (30 x 100) = 2x3000 = 6000} \\
\text{ and here we got a slightly bigger quantity than the values we picked then also , this estimations take our quantity B to}\\
= 13,700- 6000 = 7,700

Hence, Quantity A > Quantity B.

Now, you can also first prime factorize then use approximation to do this too.

Also, once you get a hang of it , approximating values to make our calculations easier is a very hefty tool in GRE

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