TheCATExam
Sign in

CAT 2020 Slot 1 — QA Question 7

Basics of AverageEasy

The mean of all 4-digit even natural numbers of the form ‘aabb’, where a > 0, is

Answer & solution

  • A

    5050

  • B

    4864

  • C

    4466

  • 5544

Solution

Easy

A mean is the sum over the count, but here a symmetry shortcut is faster: the value aabb‾=1100a+11b\overline{aabb}=1100a+11b is linear in aa and bb separately, and a,ba,b range over their allowed sets independently. So the mean equals 1100⋅(mean a)+11⋅(mean b)1100\cdot(\text{mean }a)+11\cdot(\text{mean }b).

1

Identify the allowed digits. "Even" fixes the units digit, so b∈{0,2,4,6,8}b\in\{0,2,4,6,8\}; and a>0a>0 gives a∈{1,2,…,9}a\in\{1,2,\dots,9\}. Each (a,b)(a,b) pair is a valid number, and the digits vary independently.

a∈{1,2,…,9}(9 values)b∈{0,2,4,6,8}(5 values)\begin{aligned} &a\in\{1,2,\dots,9\}\quad(9\text{ values})\\ &b\in\{0,2,4,6,8\}\quad(5\text{ values}) \end{aligned}
2

Average each digit separately. Because the number is linear in aa and bb, the mean of aabb‾=1100a+11b\overline{aabb}=1100a+11b is 1100 aˉ+11 bˉ1100\,\bar a+11\,\bar b.

aˉ=1+2+⋯+99=459=5bˉ=0+2+4+6+85=205=4\begin{aligned} &\bar a=\frac{1+2+\cdots+9}{9}=\frac{45}{9}=5\\ &\bar b=\frac{0+2+4+6+8}{5}=\frac{20}{5}=4 \end{aligned}
3

Combine.

mean=1100 aˉ+11 bˉ=1100⋅5+11⋅4=5500+44=5544\begin{aligned} &\text{mean}=1100\,\bar a+11\,\bar b=1100\cdot 5+11\cdot 4=5500+44=5544 \end{aligned}
55445544
CAT 2020 Slot 1 QA Q7: The mean of all 4-digit even natural numbers of the form ‘aabb’, where a > 0, is — Solution | TheCATExam