Math, asked by rashi20104070, 1 year ago

18. Find the time in which simple interest on Rs. 12900 at 8 1/3 % per annum will be Rs. 3225.

Answers

Answered by SimrenLalwani

As we know that:

C.I = Amount – Principle

= P(1+R100)n – P

= P[(1+R100)n−1]

164 = y[(1+5100)2−1]

164 = y[(1.05)2−1]

y = 1640.1025

y = 1600

Therefore, the required sum is Rs 1600.

Q2) Find the principal if the interest compounded annually at the rate of 10% for two years is Rs 210.

Solution:

Let the sum be Rs y.

Now we know that:

C.I = Amount – Principle

210 = P(1+R100)n – P

210 = P[(1+R100)n−1]

210 = y[(1+10100)2−1]

210 = y[(1.10)2−1]

y = 2100.21

y = 1000

Therefore, the required amount is Rs 1000.

Q3) A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.

Solution:

Let the sum be Rs y.

Next,

Amount = P(1+R100)n

A = P[(1+R100)n]

756.25 = y[(1+10100)2]

756.25 = y[(1.10)2]

y = 756.251.21

y = 625

Therefore, the required sum is Rs 625.

Q4) What sum will amount to Rs 4913 in 18 months, if the rate of interest is 1212% per annum, compounded half-yearly?

Solution:

Let the sum be Rs y.

Given:

Amount = Rs 4913

Rate of interest = 12.5 %

n = 18 months = 1.5 years (time)

Now we know that:

Amount = P(1+R200)2n

4913 = P(1+R200)2n

4913 = x(1+12.5200)3

4913 = P[(1.0625)3]

y = 49131.1995

y = 4096

Therefore, the required sum is Rs 4096.

Q5) The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.

Solution: It is given that

C.I – S.I = Rs 283.50

Rate of Interest = 15 %

n = 3 years(time)

suppose the sum is Rs y.

now we know that:

Amount = P(1+R100)n

= y(1+5100)3

= y(1.15)3 …..(1)

and also,

S.I = PRT100 = y(15)(3)100 = 0.45y

Amount = S.I + P = 1.45y …..(2)

Thus, we have:

y(1.15)3 – 1.45y = 283.50 [From (1) and (2)]

1.523y – 1.45y = 283.50

0.070875y = 283.50

y = 283.500.070875

= 4000

Therefore, the sum is Rs 4000.

Q6) Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1290 as interest compounded annually, find the sum she borrowed.

Solution:

Let us suppose that the money borrowed by Rachana is Rs y.

So, we have:

C.I = P(1+R100)n – P

1290 = y[(1+15100)2−1]

1290 = y[0.3225]

y = 12900.3225

= 4000

Thus, Rachana borrowed Rs 4000.

Q7) The interest on a sum of Rs 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.

Solution:

Let us assume the time period to be t years.

And, we have:

C.I = P(1+R100)t−P

163.20 = 2000(1+4100)t−2000

2163.20 = 2000(1.04)t

(1.04)t=2163.202000

(1.04)t=1.0816

(1.04)t=(1.04)2

when we compare both the sides, we get:

t = 2 years

Thus, the required time is two years.

Q8) In how much time would Rs 5000 amount to Rs 6655 at 10% per annum compound interest?

Solution:

Let us assume the time period to be t years.

So, then we have:

C.I = P(1+R100)t−P

6655 = 5000(1+10100)t−5000

11655 = 5000(1.10)t

(1.1)t=116555000

(1.1)t=2.331

(1.1)t=(1.1)3

On comparing both the sides, we get:

t = 3 years

Therefore, the required time is three years.

Q9) In what time will Rs 4400 become Rs 4576 at 8% per annum interest compounded half-yearly?

Solution:

Let us assume the time period be t years.

R = 8 % = 4 % (Half-yearly – 6 months)

So, we have:

Amount = P(1+R100)t

4576 = 4400(1+4100)t

4576 = 4400(1.04)t

(1.04)t=45764400

(1.04)t=1.04

(1.04)t=(1.04)1

On comparing both the sides, we get:

t = 1 year

Therefore, the required time is 6 months or half year.

Q10) The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.

Solution: It is given that

C.I – S.I = Rs 20

[P(1+4100)2−P]−P×4×2100=20

P[(1.04)2−P]−0.08P=20

0.0816P – 0.08P = 20

0.0016P = 20

P = 200.0016

Principal = Rs.12500

Therefore, the required sum is Rs 12500.

Q11) In what time will Rs 1000 amount to Rs 1331 at 10% per annum, compound interest?

Solution:

Let us assume the time to be t years.

So then,

Amount = P(1+10100)t

1331 = 1000(1+10100)t

(1.1)t=13311000

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Answered by wifilethbridge