JEE Mathematics 2023 Question with Solution

Given: The mean of $5$ observations is $5$ and the variance is $8$. Three observations are $1$, $3$, $5$. Let the remaining two observations be $a$ and $b$.

Find: The value of $a^3+b^3$.

Using the mean formula,

$$\frac{1+3+5+a+b}{5}=5$$

So,

$$a+b=16$$

Using the variance formula,

$$\sigma^2=\frac{\sum x_i^2}{5}-\left(\frac{\sum x_i}{5}\right)^2$$

Substitute the given values:

$$8=\frac{1^2+3^2+5^2+a^2+b^2}{5}-25$$

Hence,

$$a^2+b^2=130$$

Now from

$$(a+b)^2=a^2+b^2+2ab$$

we get

$$16^2=130+2ab$$ $$256=130+2ab$$ $$2ab=126$$ $$ab=63$$

Therefore $a$ and $b$ are $7$ and $9$. Now compute the sum of cubes:

$$a^3+b^3=7^3+9^3=343+729=1072$$

Therefore, the sum of cubes of the remaining two observations is $1072$. The correct option is A.

The solution also contains a conflicting label stating option C, but the worked value is $1072$, which matches option A.

Given: $a+b=16$ and $a^2+b^2=130$.

Find: $a^3+b^3$.

First find $ab$ using

$$(a+b)^2=a^2+b^2+2ab$$

So,

$$256=130+2ab$$ $$ab=63$$

Now use the identity

$$a^3+b^3=(a+b)^3-3ab(a+b)$$

Substitute the values:

$$a^3+b^3=16^3-3\cdot 63\cdot 16$$ $$=4096-3024=1072$$

Therefore, the required sum is $1072$.