Q). A person on a tour has Rs. 4,200 for expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by Rs. 70. Find the original duration of the tour.

(Q 33 A – 30/2/1 – CBSE 2026 Question Paper)

Ans:

Step 1: Let the original duration of the tour be T days.

∴ Daily expense =

=

Step 2: If Tour is extended by 3 days, then new duration = (T + 3) days

∴ New daily expense =

Step 3: Given condition that the new daily expense is reduced by 70

∴

∴

∴

∴

∴

∴ 60 (3) = T (T + 3)

∴ 180 = T ² + 3 T

∴ T ² + 3 T – 180 = 0

By mid-term splitting: T ² + 15 T – 12 T – 180 = 0

∴ T (T + 15) – 12 (T + 15) = 0

∴  (T + 15) (T – 12) = 0

∴ T = – 15 and T = 12

Step 4: Here, since duration can not be negative,

hence we drop T = – 15 and accept T = 12

∴ T = 12

Therefore, original duration of the tour is 12 days.

Check: If Original duration is 12 days, then daily expense = 4200 / 12 = Rs. 350
If this duration is increased by 3 days, then new duration = 12 + 3 = 15 days
∴ New daily expense = 4200 / 15 = Rs. 280
∴ Now difference in daily expense = 350 – 280 = Rs. 70
∵ This meets the given condition in the question, ∴ our answer is correct.

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