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Answer = B

\(\frac{(58^2 + 58^2)}{28^2} = \frac{2 * 58^2}{28^2}\)

\(= \frac{2 * (60-2)^2}{(30-2)^2}\)

\(= \frac{2 (60^2 - 2*60*2 + 4)}{30^2 - 2*30*2 + 4}\)

\(= \frac{2 * 60^2}{30^2}\)

\(= 2 * 2^2 = 8\)
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By approximation, 58/28 = ~2 (lower side)
2^2 + 2^2 = 8
Answer B
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There was a typo. Question reads: (58^2 + 58^2)/29^2 =

Edited.
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Wow. With that corrected type this becomes really easy, as we don't need to make any approximations:

\(\frac{2*(58^2*58^2)}{29^2}=\frac{2*2*2*29^2}{29^2}=8\)
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= {(29 x 2)^2 + (29x2)^2}/29^2
= {(29^2) X (2^2)/(29^2)} +{(29^2) X (2^2)/(29^2)}
= 2^2+2^2
= 8
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Bunuel

Tough and Tricky questions: Arithmetic.



(58^2 + 58^2)/29^2 =

A. 4
B. 8
C. 29
D. 58
E. 116

Kudos for a correct solution.

(58^2 + 58^2)/29^2 = 2 (58^2)/29^2 = 2 * (2^2 * 29^2) / 29^2 = 2 * (2^2) = 2 * 4 = 8

Ans 8
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Ans is 8
my approach was:
(58^2 + 58^2)/29^2 =58(58+58)/29*29
=58*116/29*29=2*4=8
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(58^2 + 58^2)/29^2
--> 2 * (58^2) / 29^2
--> 2 * ((29 * 2)^2) / 29^2
--> (2^3) * (29^2) / 29^2
--> 2^3 = 8
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\(29^2(2^2+2^2)/29^2\)
simplify by \(29^2\)

we get \(2^2+2^2=8\)
answ. B
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the given (58^2+58^2)/29^2= 2*58^2/29^2 == 2*29^2*2^2/29^2 = 2*2*2= 8
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Bunuel

Tough and Tricky questions: Arithmetic.



(58^2 + 58^2)/29^2 =

A. 4
B. 8
C. 29
D. 58
E. 116

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OFFICIAL SOLUTION:

(B) There are many ways to solve this problem. The easiest is probably to factor the numerator and then cancel terms.
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29^2 * (2^2 + 2^2)/29^2=
4+4=8
Option B

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Bunuel

Tough and Tricky questions: Arithmetic.



(58^2 + 58^2)/29^2 =

A. 4
B. 8
C. 29
D. 58
E. 116

Kudos for a correct solution.

29^2(4+4)/29^2= 8

IMO B

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58^2 can be taken as 'a'

equation becomes

a^2+a^2/(a/2)^2

2a^2/a^2/4

8a^2/a^2=8
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(58ˆ2 + 58ˆ2) / (29ˆ2)
= [2 * (60 - 2)ˆ2] / [(30 - 1)ˆ2]
= 2 * [{(30 - 1) * 2} ˆ2] / [(30 - 1)ˆ2]
=2*4 * [ (30-1)ˆ2 ] / [ (30-1)ˆ2]
=8
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