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JEE Mathematics 2025 Question with Solution

If the mean and the variance of 6,4,8,8,b,12,10,136, 4, 8, 8, b, 12, 10, 13 are 99 and 9.259.25 respectively, then a+b+aba + b + ab is equal to:

  • A

    105105

  • B

    103103

  • C

    100100

  • D

    106106

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The observations are {6,4,a,8,b,12,10,13}\{6, 4, a, 8, b, 12, 10, 13\}, mean is 99, and variance is 9.259.25.

Find: The value of a+b+aba + b + ab.

Use the mean formula:

xˉ=∑xin\bar{x} = \frac{\sum x_i}{n}

Here, n=8n = 8 and

∑xi=6+4+a+8+b+12+10+13=53+a+b\sum x_i = 6 + 4 + a + 8 + b + 12 + 10 + 13 = 53 + a + b

So,

9=53+a+b89 = \frac{53 + a + b}{8} 72=53+a+b72 = 53 + a + b a+b=19a + b = 19

Now use the variance formula:

σ2=∑xi2n−(xˉ)2\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2

Also,

∑xi2=62+42+a2+82+b2+122+102+132=529+a2+b2\sum x_i^2 = 6^2 + 4^2 + a^2 + 8^2 + b^2 + 12^2 + 10^2 + 13^2 = 529 + a^2 + b^2

Hence,

9.25=529+a2+b28−819.25 = \frac{529 + a^2 + b^2}{8} - 81 90.25=529+a2+b2890.25 = \frac{529 + a^2 + b^2}{8} 722=529+a2+b2722 = 529 + a^2 + b^2 a2+b2=193a^2 + b^2 = 193

Now use

(a+b)2=a2+b2+2ab(a+b)^2 = a^2 + b^2 + 2ab

Substituting,

192=193+2ab19^2 = 193 + 2ab 361=193+2ab361 = 193 + 2ab 2ab=1682ab = 168 ab=84ab = 84

Therefore,

a+b+ab=19+84=103a + b + ab = 19 + 84 = 103

So, the correct option is B.

Using mean and variance equations

Given: Mean of the eight observations is 99 and variance is 9.259.25.

Find: a+b+aba + b + ab.

From the mean condition:

6+4+a+8+b+12+10+138=9\frac{6 + 4 + a + 8 + b + 12 + 10 + 13}{8} = 9 53+a+b8=9\frac{53 + a + b}{8} = 9 a+b=19a + b = 19

From the variance condition:

62+42+a2+82+b2+122+102+1328−92=9.25\frac{6^2 + 4^2 + a^2 + 8^2 + b^2 + 12^2 + 10^2 + 13^2}{8} - 9^2 = 9.25 529+a2+b28−81=9.25\frac{529 + a^2 + b^2}{8} - 81 = 9.25 529+a2+b28=90.25\frac{529 + a^2 + b^2}{8} = 90.25 a2+b2=193a^2 + b^2 = 193

Now,

(a+b)2=a2+b2+2ab(a+b)^2 = a^2 + b^2 + 2ab 192=193+2ab19^2 = 193 + 2ab 2ab=1682ab = 168 ab=84ab = 84

Finally,

a+b+ab=19+84=103a + b + ab = 19 + 84 = 103

Therefore, the required value is 103103.

Common mistakes

  • Using the listed numbers incorrectly from the question statement. The solution works with observations {6,4,a,8,b,12,10,13}\{6, 4, a, 8, b, 12, 10, 13\}, so omitting aa or repeating 88 from the first line leads to a wrong equation. Always use the data set consistent with the solution working.

  • Using the variance formula incorrectly as σ2=∑xi2n−xˉ\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x} instead of subtracting xˉ2\bar{x}^2. This changes the second equation completely. Always subtract the square of the mean.

  • Forgetting the identity (a+b)2=a2+b2+2ab(a+b)^2 = a^2 + b^2 + 2ab when finding abab. If you use a+ba+b and a2+b2a^2+b^2 directly without this relation, the product cannot be obtained correctly. Use the identity before substituting values.

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