CAT 2025

Slot 3

QA

Question & Solution

Let the thousands, hundreds, tens, and units digits of the number be $$a$$, $$b$$, $$c$$, and $$d$$, respectively.

The sum of the digits in the thousands, hundreds, and tens places is 15;  $$a+b+c = 15$$ .....(1)

The sum of the digits in the hundreds, tens, and units places is 16;  $$b+c+d = 16$$ .....(2)

Solving equations (1) and (2), we get $$d = a+1$$.

The digit in the tens place is 6 more than the digit in the units place;  $$c=d+6 = a+1+6 = a+7$$ .....(3)

Substituting the value of $$c$$ in terms of $$a$$ in equation (1), we get;  $$a+b+a+7 = 15$$  or  $$b=8-2a$$

Thus, the number is:  $$\underline{a}\text{ }\underline{8-2a}\text{ }\underline{a+7}\text{ }\underline{a+1}$$

We know that since the number is a four digit number, $$a\geq 1$$. But all the four digits have to be less than or equal to $$9$$ but greater than or equal to $$0$$, thus,

$$8-2a\geq 0$$  gives  $$a\leq 4$$  and  $$a+7\leq 9$$  gives  $$a\leq 2$$.

Thus, $$a$$ can be $$1$$ or $$2$$.

The two possible values for the number, therefore, are, $$1682$$ and $$2493$$, when we put $$a=1$$ and $$a=2$$, respectively.

The difference between the only two possible values is, $$2493 - 1682 = 811$$

Option A is the correct answer.