JEE Advanced 2022 — Paper 1 — MATH-I solution

JEE Advanced2022Paper 1MATH-I

Let $l_{1}, l_{2}, \dots, l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_{1}$ , and let $w_{1}, w_{2}, \dots, w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_{2}$ , where $d_{1} d_{2} = 10$ . For each $i = 1, 2, \dots, 100$ , let $R_{i}$ be a rectangle with length $l_{i}$ , width $w_{i}$ and area $A_{i}$ . If $A_{51} - A_{50} = 1000$ , then the value of $A_{100} - A_{90}$ is ____________.

Your answer

Solution

Let $l_{1} = a$ , $w_{1} = b$ . Then $l_{i} = a + (i - 1) d_{1}$ , $w_{i} = b + (i - 1) d_{2}$ and $A_{i} = l_{i} w_{i}$ .

Compute $A_{51} - A_{50} = (a + 50 d_{1}) (b + 50 d_{2}) - (a + 49 d_{1}) (b + 49 d_{2}) = b d_{1} + a d_{2} + 99 d_{1} d_{2}$ .

Given this equals $1000$ and $d_{1} d_{2} = 10$ , we get $b d_{1} + a d_{2} = 1000 - 990 = 10$ .

Similarly, $A_{100} - A_{90} = (a + 99 d_{1}) (b + 99 d_{2}) - (a + 89 d_{1}) (b + 89 d_{2}) = 10 (b d_{1} + a d_{2}) + (9 9^{2} - 8 9^{2}) d_{1} d_{2}$ .

$= 10 (10) + (188) (10) (10) = 100 + 18800 = 18900$ .

$∴ A_{100} - A_{90} = 18900$ .

← Back to full paper

Practice more MATH-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →