# A person on tour has Rs. 360 for his daily expenses. If he exceeds his tour programme by four days, he must cut down his daily expenses by Rs. 3 per day. Find the no. of days of his tour programme.

Let the original duration of the tour be x days.

Total expenditure on tour = Rs. 360

∴ Expenditure per day Person extends his tour by 4 days.

Duration of extended tour = ( x + 4) days

∴ New expenditure per day It is given that expenditure per day is cut down by Rs.3 ⇒ x 2 + 4x = 480

x 2 + 4x – 480 = 0

⇒ x 2 + 24x – 20x – 480 = 0

⇒ x ( x + 24) – 20 ( x + 24 ) = 0

( x – 20 ) ( x + 24) = 0

either ( x + 24 ) = 0 or ( x – 20) = 0

either x = – 24 or x = 20

But number of days can't be negative. So, x = 20.

Hence, original number of days of the tour was 20 days.

• 97

Let the no of days be n

Total money -> Rs 360

So, daily expense = 360/n

If the no. of days increase by 4, then new no. of days -> n + 4

And new daily expense = 360/(n + 4)

We are given that if no of days increase by 4, then daily expenses are reduced by 3

So, the equation thus formed is

360/n - 360/(n + 4) = 3

=> (360n + 1440 - 360n)/n2 + 4n = 3

By cross multiplication

3n2 + 12n = 1440

Dividing LHS and RHS by 3

n2 + 4n = 480

=> n2 + 4n - 480 = 0

=> n2 + 24n - 20n - 480 = 0

=> n(n + 24) - 20(n + 24) = 0

=> (n - 20) (n + 24) = 0

=> n = 20 or - 24

Since the no. of days cannot be in negative,

no. of days = 20

Hope this helps!!

• 14
What are you looking for?