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sandeepk123
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What is the value of a?

Given \(a^2 + b^2\) = 116

1) a and b are integers
\(4^2 + 10^2 = 116\)
a could take values of either 4 or 10 (Insufficient)

2) a < b
Case 1: a = 4, b = 10
Case 2: a = -4, b = 10
Here a could be +4 or -4 because \(-4^2 = 4^2 =16\) (Insufficient)

On combining the information from both the statements,
a could take be -4 or 4. (Insufficient - Option E)
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Solution



We are given:

    • \(a^2+b^2 = 116\)
    • \(10^2+4^2 = 116\), the possible values of \(a\) and \(b\) are:
      o \(a= 10, b=4\)
      o \(a= 10, b= -4\)
      o \(a= -10, b=4\)
      o \(a= 10, b=-4\)
      o \(a= 4, b= 10\)
      o \(a= 4, b= -10\)
      o \(a=-4, b=10\)
      o \(a= -4, b= -10\)

Statement -1: “\(a\) and \(b\) are integers

In all the above cases, \(a\) and \(b\) are integer.

We cannot arrive at a unique value of \(a\).
Hence, Statement 1 is not sufficient to answer the question.

Statement -2:a < b

The value of \(a\) is less than \(b\) for various cases.
    • \(a= -10, b=4\)
    • \(a= 4, b= 10\)
    • \(a=-4, b=10\)

We cannot arrive at a unique value of \(a\).
Hence, Statement 2 is not sufficient to answer the question.

Combining both the statements together:

After combining the statements, \(a\)= \(4 or -4 or -10\).
Hence, we can not answer the question even after combining both the statements.

Answer: E
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\(a^2 + b^2\) = 116

Statement 1
a and b are integers
The integers should be less than 11 as 11^2 = 121, and 116< 121
Squares could be 0,1,4,9,16,25,36,49,64,81,100
a^2 could be 16 and b^2 could be 100 or vice versa
a, b could be (4,10), (-4,10), (-4,-10), (4,-10), (10,4), (10,-4), (-10,4), (-10,-4)
Insufficient

Statment 2
a<b
But we don't know if a and b are integers
Insufficient

Together we have a, b could be (4,10), (-4,10), (-10,4), (-10,-4)
Insufficient

Answer E
sandeepk123
What is the value of a?

Given \(a^2 + b^2\) = 116

1) a and b are integers
2) a < b
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