Advertisements
Advertisements
प्रश्न
Find the interest and the amount on:
₹ 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.
Advertisements
उत्तर
Given: Principal (P) = ₹5000
Rate (R) = 8% p.a.
Time (T) = 23 December 2011 to 29 July 2012
| Dec. | Jan. | Feb. | March | April | May | June | July |
| 8 | 31 | 29 | 31 | 30 | 31 | 30 | 29 |
Total of 219 days =`219/365`years
∴ S.I. = `(P xx R xx T)/100`
= `(5000 xx 8 xx 219)/(100 xx 365)`
= `(5000 xx 8 xx cancel(219)^3)/(100 xx cancel(365)^5)`
= `(5000 xx 8 xx 3)/(500)`
= `(cancel(5000)^10 xx 8 xx 3)/cancel(500)`
= 10 × 8 × 3
= ₹ 240
∴ Amount = P + I
= ₹ 5000 + 240
= ₹ 5240
APPEARS IN
संबंधित प्रश्न
The simple interest on a certain sum of money is `3/8` of the sum in `6 1/4` years. Find the rate percent charged.
Aravind borrowed a sum of ₹ 8,000 from Akash at 7% per annum. Find the interest and amount to be paid at the end of two years
In what time will ₹ 17800 amount to ₹ 19936 at 6% per annum?
A sum of ₹ 48,000 was lent out at simple interest and at the end of 2 years and 3 months the total amount was ₹ 55,560. Find the rate of interest per year
The value of a machine depreciates at 10% per year. If the present value is ₹ 1,62,000, what is the worth of the machine after two years?
If a principal is getting doubled after 4 years, then calculate the rate of interest. (Hint: Let P = ₹ 100)
Given the principal = Rs 40,000, rate of interest = 8% p.a. compounded annually. Find amount if period is 2 years.
The difference of interest for 2 years and 3 years on a sum of ₹ 2100 at 8% per annum is ______.
The simple interest on a sum of ₹ P for T years at R% per annum is given by the formula: Simple Interest = `(T xx P xx R)/100`.
Radhika borrowed ₹ 12000 from her friends. Out of which ₹ 4000 were borrowed at 18% and the remaining at 15% rate of interest per annum. What is the total interest after 3 years?
