Question

ꜱᴏʟᴠᴇ

❌ɴᴏ ꜱᴩᴀᴍᴍɪɴɢ ❌​

Answer 1

Let total time to catch the thief be n minutes

Total distance covered by thief = (100n) metres

Total distance covered by policeman = 100 + 110 + 120 + ...+ (n - 1) terms

After that please refer to the attachment.

Answer 2

$\huge\mathfrak\purple {solution}$

Let the total time taken be (n-1) min in which the police catch the thief .

As theif ran 1 min before police started running .

• Time taken by thief before he was caught = (n-1+1) min = n min

The total distance covered by thief at 100m/min = 100 × n

= 100n meter

Total distance covered by policeman in (n-1) min .

= 100 + 110 + 120 + ......(n+1)terms

$= \frac{(n - 1)}{2} (2 \times 100 + (n - 2)10)$

$= \frac{(n - 1)}{2} (2 \times 100 + (n - 2)10)$

$= \frac{(n - 1)}{2} (200 + (n - 2)10)$

According to the que ...

Total distance covered by thief in n min = total distance covered by police in (n-1 )min .

$= > 100n = \frac{(n - 1)}{2} (200 + (n - 2)10)$

$= > 100n = (n - 1)(100 + 5n - 10)$

$= > 100n = (n - 1)(90 + 5n)$

$= > 100n = 90n - 90 + 5 {n}^{2} - 5n$

$= > 5 {n}^{2} + 85n - 10n - 90 = 0$

$= > 5 {n}^{2} - 15n - 90 = 0$

$= > {n}^{2} - 3n - 18 = 0$

$= > {n}^{2} - 6n + 3n - 18 = 0$

$= > n(n - 6) + 3(n - 6) = 0$

$= > (n - 6)(n + 3) = 0$

=> n = 6 , -3

n = -3 rejected

# n = 6

Hence , time taken by policeman to catch the thief = n -1 = 6-1 = 5 min .