Data Sufficiency

What is the median of a set of 10 positive integers?

(1) The numbers are in an AP with the first term being 10
(2) The arithmetic mean of these 10 numbers is 45

The OA is the option C.

I don't know how to solve this DS question. I'd be thankful for some help here. Please.

Hello M7MBA.

First of all, I think there is a mistake. I think it should say 10 positive numbers, not positive integers.

The question is "What is the median of a set of 10 positive integers?". Since there are 10 elements (even number), the median is equal to the arithmetic mean of x_5 and x_6, where all numbers are ordered from the least to the greatest. Hence, we want to find $$M=\frac{x_5+x_6}{2}$$

First Statement

(1) The numbers are in an AP with the first term being 10

Here, we know that x_1=10 and x_n=10+(n-1)d, where d is a common difference. Hence $$x_5=10+4d\ \ \ and\ \ \ \ x_6=10+5d$$ Therefore $$M=\frac{10+4d\ +10+5d}{2}=\frac{20+9d}{2}$$ Since we don't know the value of d, then we can't determine the median. NOT SUFFICIENT.

Second Statement

(2) The arithmetic mean of these 10 numbers is 45

Now, we know that $$\frac{x_1+x_2+\cdots+x_9+x_{10}}{10}=45\ \ \Rightarrow\ \ \ x_1+x_2+\cdots+x_9+x_{10}=450.$$ We can not determine the median. NOT SUFFICIENT.

First Statement + Second Statement

(1) The numbers are in an AP with the first term being 10
(2) The arithmetic mean of these 10 numbers is 45

Now we have $$450=x_1+x_2+\cdots+x_9+x_{10}=10+\left(10+d\right)+\cdots+\left(10+8d\right)+\left(10+9d\right)$$ $$450=100+45d\ \ \ \Rightarrow\ \ 45d=350\ \ \ \Rightarrow\ \ d=\frac{70}{9}.$$ Hence we get from the first statement that $$M=\frac{20+9d}{2}\ =\frac{20+9\ \frac{70}{9}}{2}=\frac{20+70}{2}=\frac{90}{2}=45.$$ Therefore, we get the median of the set.

So, using both statements together is sufficient. SUFFICIENT.

The correct answer is the option C.

I hope it helps.