# Rd Sharma Solutions Class 8 Chapter 14

## Rd Sharma Solutions Class 8 Chapter 14 Compound Interest

Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 14, Compound Interest. Here students can easily find Exercise wise solution for chapter 14, Compound Interest. Students will find proper solutions for Exercise 14.1, 14.2, 14.3, 14.4 and 14.5 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

Compound Interest Exercise 14.1 Solution :

Question no – (1)

Solution :

As per the question,

Principal = 3000 Rs

rate = 5%

n = 2 years

Amount = P(1 + r/100)^n

= 3000 (1 + 5/100)²

= 3000 (100+5/100) ²

= 3000 × 105/100 × 105/1000

= 3307.5

Compound Interest

= 3307.5 – 3000.00

= 307.5 Rs.

Therefore, the compound interest will be Rs. 307.5

Question no – (2)

Solution :

According to the question,

Principal = 400 Rs

rate = 5%

n = 2 years

Amount = p (1 + r/100)^n = 4000 (1 + 5/100)²

= 4000 (100+5/100) ² = 4000 (105/100) ²

= 400 × 105/100 × 105/100 = 4410

Compound Interest,

= 4410 – 4000

= 410

Thus, the compound interest will be 410 Rs.

Question no – (3)

Solution :

As per the given question,

Principal = 8000 Rs

rate = 15%

n = 3

Amount = p (1 + r/100)^n

= 8000 (1 + 15/100)³

= 8000 (100 + 15/100)³

= 8000 × 105/100 × 105/1000 × 105/100

= 12167 Rs.

Compound Interest,

= 12167 – 8000

= 4167 Rs.

Therefore, the compound interest that Rohit will gets Rs. 4167

Question no – (4)

Solution :

According to the question,

Principal = 1000 Rs

rate = 8%

n = 1 1/2 = 3/2 years

Amount = p (1 + r/100)^2n

= 1000 (1 + 8/100)^2 3/2

= 1000 (100 + 8/100)³

= 1000 × 108/100 × 108/100 × 108/100

= 1124.86

Compound Interest,

= 1124.86 – 1000

= 124.86 Rs

Therefore, the compound interest on Rs 1000 will be Rs 124.86.

Question no – (6)

Solution :

According to the question,

Principal = 16000 Rs

rate = 12 1/2% = 25/2%

n = 3 years

Amount = P (1 + r/100)^n

= 16000 (1 + 25/2/100)³

= 16000 (1 + 25/200)³

= 16000 (200 + 25 /200)

= 16000 × 225/200 × 225/200 × 225/200

= 22781.25 Rs

Compound Interest,

= 22781.75 – 16000

= 6781. 25 Rs.

Therefore, the total compound interest payable by Swati is Rs. 6781.25

Question no – (7)

Solution :

According to the question,

P = 64000 Rs

r = 10%

n = 1 1/2y = 3/2y

A = p (1 + r/100) ^2n

= 64000 (1 + 10/100)^2 2/3

= 64000 (100 + 10/100)³

= 64000 × 100/100 × 110/100 × 100/100

= 78088 Rs

Compound interest = 78088 – 64000

= 10088 Rs

Thus, the compound interest payable by Roma is Rs. 10088

Question no – (8)

Solution :

As per the question,

Principal = 20000

Rate = r = 18%

Time = n = 2 y

For Simple Interest,

Simple Interest = PRT/100

= 20000 × 18 × 2 /100

= 7200 Rs

Now, For Compound Interest,

Amount = P (1 + r/100)^n

= 2000 (1 + 18/100)²

= 20000 (100 + 18/100)²

= 20000(100 + 18/100)²

= 20000 ×118/100 × 118/100

= 27848 Rs

Compound Interest

= 27848 – 20000

= 7848 Rs

His gain

= 7848 – 7200

= 648 Rs

Question no – (9)

Solution :

According to the question,

Principal = 800

rate = 20%

n = 9 month = 9/12 years = 3/4 × 4 quarter

Amount = P (1 + r/100)^n

= 8000 (1 + 20/100)³

= 8000 (100 + 20/100)³

= 8000 × 120/100 × 120/100

= 9261 Rs

Compound Interest,

= 9261 – 8000

= 1261 Rs

Therefore, the compound interest on Rs 8000 will be Rs. 1261.

Question no – (10)

Solution :

As per the given question,

Simple Interest = 200

Rate = r = 10%

n = Time = 2 years

For Simple Interest,

According to question,

Simple Interest = PRT /100

or, P = SI × 100/ RT

= 200 × 100/10 × 2

= 1000 Rs

For Compound Interest,

Amount = P (1 + r/100)

= 1000 (1 + 10/100)²

= 1000 (100 + 10/100)²

= 1000 × 110/100 × 110/100

= 1210 Rs

Compound Interest

= 1210 – 1000

= 210 Rs

Therefore, the compound interest will be Rs. 210

Question no – (11)

Solution :

As per the given question,

Principal = 64000 Rs

rate =10%

n = 1 year

Compound Interest for quarterly we know that,

Amount = P (1 + r/100)^4n

= 64000 (1 + 10/100)⁴

= 64000 (100 + 10/100)⁴

= 64000 × 110/100 × 110/100 × 110/100 × 110/100

= 70644.03 Rs

Compound Interest,

= 70644.03 – 64000

= 6644.03 Rs

Therefore, the compound interest on Rs 64000 will be Rs 6644.03

Question no – (12)

Solution :

According to the given question,

Principal = 7500 Rs

rate = 12/4% = 3% [quarterly]

n = 9 month = 9/12 years = 3/4 × 4 quarterly = 3 quarter

Amount = P (1 + r/100)^n = 7500 (1 + 3/100) ³

= 7500 (100 + /100)³

= 7500 × 103/100 v 103/100 × 103/100

= 8195.45 Rs

Therefore, Ramesh will receive Rs 8195.45 after 9 months.

Question no – (13)

Solution :

As per the given question,

Principal = 9600 Rs

rate = 5 1/2% = 11/2%

n = 3 years

Amount = P (1 + r/100)^n

= 9600 (1 + 11/2/100)³

= 9600 (1 + 11/200)³

= 9600 (200 +11/200)³

= 9600 × 211/100 × 211/200 × 211/200

= 11272.72 Rs

Compound Interest = Amount – Principal

= 1127.72 – 9600

= 1672.72 Rs

Therefore, compound interest Anil will pay after 3 years will be Rs 1672.72

Question no – (14)

Solution :

According to the question,

Principal = 12000 Rs

rate = 5%

n = 3 years

Amount = P (1 + r/100)³ = 12000(1 + 5/100)³

= 12000 (100 + 5/100)³

= 12000 × 105/100 × 105/100 × 105/100

= 13891.50 Rs

Compound Interest,

= 1389.50 – 12000.00

= 1891.50 Rs

Hence, Surabhi has to pay Rs 1891.50 to the company after 3 years.

Question no – (15)

Solution :

As per the given question,

Principal = 40000

rate = 7%

n = 3 years

Amount = P(1 + r/100)^n

= 4000 (1 + 7/100)³

= 12000 (100 + 7/100)³

= 40000 × 107/100 × 107/100 × 107/100

= 45796 Rs

Compound Interest,

= 45769 – 40000

= 5796 Rs

Therefore, Daljit will pays Rs. 5796 after 2 years.

Compound Interest Exercise 14.2 Solution :

Question no – (1)

Solution :

(i) Principal = 3000

rate = 5%

n = 2 years

Amount = p (1 + 5/100)^n = 3000 (100 + 5/100)²

= 3000 × 105/100 × 105/100

= 3307.5 Rs.

Compound Interest,

= 3307.5 – 3000

= 307.5 Rs.

Therefore, the amount will be 3307.5 Rs and compound Interest will be 307.5 Rs.

(ii) Principal = 3000 Rs

rate = 18%

n = 2 years

Amount = P (1 + r/100)^n = 3000 (1 + 18/100)²

= 3000 × 118/100 × 18/100

= 4177.2 Rs.

Compound Interest,

= 4177.2 – 3000

= 1177.2 Rs.

Therefore, the amount will be 4177.2 Rs and compound Interest will be 1177.2 Rs.

(iii) Principal = 5000 Rs.

rate = 10%

n = 2y

Amount = P (1 + r/100)^n = 5000 (1 + 10/100)²

= 5000 (100 + 10/100)²

= 5000 × 110/100 × 110/100

= 6050 Rs.

Compound Interest,

= 6050 – 5000 Rs.

= 1050 Rs.

Therefore, the amount will be 6050 Rs and compound Interest will be 1050 Rs.

(iv) Principal = 2000 Rs

rate = 4%

n = 3

Amount = P (1 + r/100)^n = 2000 (1 + 4/100)³

= 2000 × (100 + 4/100)³

= 2000 × 104/100 × 104/100 × 104/10

= 2249.68 Rs

Compound Interest,

= 2249.68 – 2000

= 249 .68 Rs

Therefore, the amount will be 2249.68 Rs and compound Interest will be 249 .68 Rs.

(v) In then given question,

Principal = 12800

rate = 7 1/2% = 7.5%

n = 3 years

∴ Amount = P (1 + r/100)^n

= 12800 (1+7.5/100)³

= 12800 (1.075)³

= 15901.40 Rs

Compound Interest,

= 15901.40 – 12800.00

= 3101.40 Rs

Therefore, the amount will be 15901.40 Rs and compound Interest will be 3101.40 Rs.

(vi) Principal = 10000

rate = 20%

n = 2 year

Amount = P (1 + r/100)^2n = 10000 (1 + 20/200)⁴

= 10000 (10 + 2/20) = 10000(12/20)⁴

= 10000 × 12/20 × 12/20

= 14641 Rs

Compound Interest,

= 14641 – 10000

= 4641 Rs

Therefore, the amount will be 14641 Rs and compound Interest will be 4641 Rs.

(vii) Principal = 160000 Rs

rate = 10 paise per rupee per annum compounded half-yearly.

n = 2 years

Amount = P (1 + r/100)^2n = 1000 (1 + 10/100)⁴

= 10000 (100 + 10/100)⁴

= 10000 (110/100) ⁴

= 10000 × 11 × 11 × 11 × 11/10 × 10 × 10 × 10

= 19448.10 Rs

Compound Interest,

= 19448.10 – 16000

= 3448.10 Rs

Therefore, the amount will be 19448.10 Rs and compound Interest will be 3448.10 Rs.

Question no – (2)

Solution :

According to the question,

Principal = 2400 Rs

rate = 20%

n = 3 years

Amount = P (1 + r/10)^n = 2400 (1 + 20/100)³

= 2400 (100+20/100)³

= 2400 × 120/100 × 120 × 100 × 120/100

= 4147.20 Rs

Therefore, the required amount will be Rs 4147.20.

Question no – (3)

Solution :

As per the given question,

Principal = 16000 Rs

rate = 12 1/2% = 25/2 = 12.5%

n = 3 years

Amount = P (1 + r/1000)^n

= 16000 (1 + 72.5/100) ³

= 16000 (1.125)³

= 22781.25 Rs

Hence, the amount which is payable Rasheed to Rahman will be 22781.25 Rs.

Question no – (4)

Solution :

In the given question,

Principal = 1000

rate = 10%

n = 2 years

Amount = P (1 + r/100)^n

= 1000 (1 + 10/100)²

= 1000 (1.1)²

= 1210 Rs

Therefore, the amount that Meera has to pay back will be 1210 Rs.

Question no – (6)

Solution :

As per the given question,

Principal = 16000 Rs

Rate = r = 17 1/2% = 17.5%

Time  = n = 2 years

For Simple Interest,

S.I. = PRT/100 = 1600 × 10 × 2/100

= 5600 Rs

For Compound Interest,

Amount = P (1 + r/100)^n

= 1600 (1 + 17.5/100) ²

= 16000 (1.175)²

= 22090 Rs

Compound Interest,

= 22090 – 16000

= 6090 Rs

His gain at end of 2 years,

= 6090 – 5600

= 490 Rs

Therefore, Amit will gain Rs 490 at the end of 2 years.

Question no – (7)

Solution :

According to the given question,

Principal = 4096 Rs

rate = 12 1/2% = 25/2 = 12.5%

n = 18 months = 18/12 = 3/2 years

Amount = P (1 + r/200)^2n

= 4096 (1 + 12.5/200)^2 2/3

= 4096 (1.0625)³

= 4913 Rs

Therefore, the required amount will be 4913 Rs.

Question no – (8)

Solution :

As per the question,

Principal = 8000 Rs

rate  = 10%

n = 1 1/2 = 2/3

Amount = P (1 + r/200)^2n

= 8000 (1 + 10/200)^2 3/2

= 8000 (1.05)³

= 9261 Rs

Compound Interest,

= 9261 – 8000 Rs

= 1261 Rs

Therefore, the amount will be 9261 and comp[und interest will be 1261 Rs.

Question no – (9)

Solution :

As per the given question,

Principal = 57600 Rs

rate = 12 1/2% = 12.5%

n = 1 1/2 = 3/2

Amount = P (1 + r/200)^2n

= 5770 (1 + 12.5/200)^2 3/2

= 57600 (1.0625)³

= 69089.06 Rs

Hence, the amount that she pays will be 69089.06 Rs.

Question no – (10)

Solution :

According to the given question,

Principal = 64000 Rs

rate = 5%

n = 1 1/2 = 3/2

Amount = P (1 + r/200)

= 6400 (1 + 5/200)^2.3/2

= 6400 (1 + 5/200)³

= 6400 (1.025)³

= 68921 Rs

Compound Interest,

= 68921 – 64000

= 4921 Rs

Therefore, Abha will pay 4921 Rs interest after one year and a half.

Question no – (11)

Solution :

According to the given question,

Principal = 10000 Rs

rate = 20%

n = 2 years

For compounded annually,

Amount = P (1 + r/100)^n

= 1000 (1 + 20/100)²

= 10000 (1.2)²

= 14400 Rs

Now, for compounded half-yearly,

Amount = P (1 + r/200)^2n = 1000 (1 + 20/200) ^2.2

= 10000 (1.1)⁴

= 14641 Rs

Difference,

= 14641 – 14400

= 241 Rs

Question no – (12)

Solution :

As per the given question,

Given in the question,

Principal = 245760 Rs

Rate = 12.5%

n = 2 years

For compounded annually,

Amount = P (1 + r/100)^n

= 245760 (1 + 12.5/100)²

= 245760 (1.2656)

= 311040 Rs

Now, for compounded semi annually,

Amount = P (1 + r/200)^2n

= 245760 (1 + 12.5/200)^2.2

= 245760 (1.0625)⁴

= 313203.75 Rs

His gain after 2 years,

= 313203.75 – 311040.00

= 2163.75 Rs.

Question no – (13)

Solution :

From the question,

Principal = 8192 Rs

rate = 12 1/2 = 12.5%

n = 18 months = 18/12= 3/2 years

Amount = P (1 + r/200)^2n

= 8192 (1 + 12.5/200)^2 3/2

= 8192 (1.0625)³

= 9826 Rs

Therefore, the amount that David would receive will be 9826 Rs.

Question no – (17)

Solution :

According to the question,

Principal = 15625 Rs

rate = 16%

n = 2 1/4 years

Amount = P (1 + r/100)² (1+ r/200)

= 15625 (1 + 16/100)² (1 + 16/200)

= 15625 × (1.16)² × (1.04)

= 21866 Rs.

Therefore, payment Ramu have to make will be Rs. 21866

Question no – (20)

Solution :

Let, the sum is = P

rate = 20%

n = 2 years

For compounded yearly,

A₁ = P (1 + r/100) ^n

= P (1 + 20/100)²

= P (1.20)²

= 1.44P

Compound Interest,

= 1.44p – P

= 0.44p

Now, for compounded half-yearly,

A₂ = P (1 + r/200) ^2n

= P (1 + 20/200)^2×2

= P (1.1) ⁴

= 1.4641p

Compound Interest,

= 1.4641p – P

= 0.4641P

According to question,

0.4641P – 0.44P = 482

or, 0.0241P = 482

or, P = 482/0.024

or, P = 20,000 Rs

Therefore, the required sum will be Rs. 20,000

Question no – (21)

Solution :

As per the given question,

Simple Interest = 5200 Rs.

R = r = 6 1/2% = 6.5%

T = n = 2 years

For Simple Interest,

S.I = PRT/100

or, Principal = Simple Interest × 100/R.T

= 5200 × 100/6.5 × 2

= 40000 Rs.

Now, for Compound Interest,

Amount = P (1 + r/100)^n

= 4000 (1 + 6.5/100)²

= 40000 (1.06)²

= 45369 Rs.

Compound Interest,

= 45369 – 40000

= 5369 Rs.

Therefore, the compound interest will be 5369 Rs.

Question no – (22)

Solution :

According to the question,

Simple Interest = 1200 Rs

R = r = 5%

T = n = 3 y

For Simple Interest,

Simple Interest = PRT/100

or, P = Simple Interest × 100/R.T

= 1200 × 100/ 5 × 3

= 8000

Now for Compound Interest,

A = P (1 + r/100)^n

= 8000 (1 + 5/100)³

= 8000 (1.05)³

= 9261

Compound Interest,

= 9261 – 8000

= 1261

Therefore, the compound interest is Rs. 1261

Compound Interest Exercise 14.3 Solution :

Question no – (1)

Solution :

As per the given question,

Let, sum = p

rate = 5%

n = 2

Compound Interest = 164 Rs

According to question,

Compound Interest = Amount – Principal

or, 164 = P (1+ R/100)^n – p

or, 164 = p [(1 + 5/100)² – 1]

or, 164 = p [(1.05) ² – 1] = p × 0.1025

or, p = 164/0.1025

= 1600 Rs

Therefore, the sum will be 1600 Rs.

Question no – (2)

Solution :

Let, Principal = P

rate = 10%

n = 2 years

Compound Interest = 216

Now according to question,

Compound Interest = Amount – Principal

= P (1 + r/100)^n – p

= p [(1 + 10/100) ²- 1]

or, 216 = p [(1 .1)² – 1]

or, P × 0.21 = 216

or, p = 216/0.21

= P = 1000 Rs

Therefore, the required principal will be 1000 Rs.

Question no – (3)

Solution :

According to the given question,

Let, sum is = p

Amount = 756.25 Rs

rate = 10%

n = 2 years

Amount = p (1 + r/100) ²

= p (1 + 10/100) ² p × (1.1)

= p × 1.21

or, 756.25 = p × 1.21

or, p = 756.25/1.21

= P = 625 Rs

Therefore, the required sum will be 625 Rs.

Question no – (4)

Solution :

As per the question Let, sum is = p

rate = 12 1/2% = 12.5%

n = 18 months = 18/12 years = 3/2

Amount = P (1 + r/200)^2n

= p (1 + 12.5/200)^2 3/2

= p (1 + 1.0625)³

or, 4913 = 1 (1.0625)³ = p × 1.995

or, P = 4913/1.995

= P = 4096 Rs

Therefore, the required sum will be 4096 Rs.

Question no – (5)

Solution :

As per the question Let, sum is = P

Rate = r = 15%

Time = n = 3 years

Compound Interest – Simple Interest = 283.50 Rs

For the Simple Interest,

Simple Interest = PRT/100

= P × 15 × 3 /100

= 0.45P

A = 0.45P + P

= 1.45P

For Compound Interest,

A = P (1 + r/100)^n

= p (1 + 15/100)³

= p (1.15)³ = 1.523P

According to question,

1.523 – 1.45 = 283.50

or, 0.070875P = 283.50

or, P = 283.50/0.070875

= 4000 Rs

Therefore, the required sum will be 4000 Rs.

Question no – (6)

Solution :

Let, the Sum is = P

rate = 15%

n = 2

Compound Interest = 1290 Rs

Compound Interest = Amount – Principal

= P (1 + r/100)^n – p

= P [(1 + 15/100)² – 1]

or, 1290 = P (0.3225)

or, P = 0.3225/1290

= P = 4000 Rs.

Therefore, Rachana will borrowed 4000 Rs.

Question no – (7)

Solution :

Let, the time period = n

Compound Interest = 163.20 (as per the question)

rate = 4%

principal = 2000 Rs

Compound Interest = Amount – Principal

= p (1 + r/100)^n – p

= p [(1 + r/100)^n – 1]

or, 163.20 = 2000 (1.04)^n

or, (1.04)^n = 163.20/2000 = 1.0816

or, (1.04) = (1.04)²

or, n = 2 years

Therefore, the period will be 2 years.

Question no – (8)

Solution :

Let, time needs = n year.

Principal = 5000

Compound Interest = 6655

rate = 10%

Compound Interest = Amount – Principal

= [p (1 + r/100)^n – p]

= p [91 + r/100)^n – 1]

or, 6655 = 5000 [(1 + 10/100)^n -1]

or, 6655 = 5000 (1.10)^n

or, (11/10)^n = (11/10) ³

or, n = 3 years

Therefore, the required time will be 3 years

Question no – (9)

Solution :

Let, time period = n years.

rate = 8%

Amount = 4576 Rs

principal = 4400 Rs

Amount = p (1 + r/200)^2n

= 4400 (1 + 8/200)^2n

or, 4576/4400 = (1 + 4/100)^2n

or, 1.04 = (1.04)^2n

or, (1.04)^2n = (1.04)

or, 2n = 1

or, n = 1/2 year

Therefore, the required time will be 1/2 year.

Question no – (10)

Solution :

Let, principal = P

rate = 4%

n = 2

For Simple Interest,

S.I = PRT/100 = P × 4 × 2/100

= 8p/100 = 0.08p

= 2/25 P

For Compound Interest,

Compound Interest = A – P [P (1 + r/100)^n – p]

= P [(1 + 4/100)² – 1]

or, Compound Interest = P [(1 + 1/25)² – 1]

According to question –

Compound Interest – Simple Interest = 20

P (1 + 1/25)² – 1] – 2/25 P

= P = 20

or, p [(51/625) – 2/25] = 20

or, p [51 – 50/625] = 20

or, p /625 = 20

or, p = 20 × 625

= P = 12500 Rs

Therefore, the sum will be 12500 Rs.

Question no – (11)

Solution :

Let, time needs = n years

rate = 10%

principal = 1000

Amount = 1331 Rs

Now, according to question,

A = p (1 + r/100)^n

or, 1331 = 1000 (1 + 10/100)^n

or, 1331/1000 = (110/100)^n

or, (11/10) ^n = (11/10)³

or, n = 3 years

Therefore, the required time will be 3 years.

Question no – (12)

Solution :

Let, rate = r%

Amount = 774.40 (as per the question)

principal = 640 Rs

n = 2 years

According to question,

A = P (1 + r/100)^n

or, 774.40 = 640 (1 + r/100)^n

or, (1 + r/100)²

= 774.40/640

or, (1 + r/100) ² = 774/640 × 100

= 7744/6400= (88/80)²

or, 1 + r/100 = 88/80

or, r/100 = 88/80 – 1 = 88 – 80/80

= 8/80 = 1/10

or, r = 10%

Therefore, the rate percent will be 10%

Question no – (13)

Solution :

Let, the rate is = r%

n = 1 1/2 = 3/2

Amount = 2315.25

Principal = 2000 Rs

According to question,

A = P (1 + r/200)^2n = 2000 (1 + r/100)^2 2/3

= 200 (1 + r/100)³

or, 2315.25/2000 = (1 + r/200)

or, (1 + r/200)³ = 1.157 = (1.05)³

or, 1 + r/200 = 1.05

or, r/200 = 1.05 – 1 = 0.05

∴ r = 0.05 × 200 = 10%

Therefore, the rate percent per annum will be 10%.

Question no – (14)

Solution :

Let, the principle = P and Rate = R%

Simple Interest = 200

Compound Interest = 210

n = 2 = years

For Simple Interest,

Simple Interest = PRT/100

or, PR = SI × 100 / t

= 200 × 100/2

= 10000

For, Compound Interest,

Compound Interest = A – P

or, 210 = [p (1 + R/100)^n ] – p = p [(1 + R/100)² – 1]

or, 210 = p [(1 + R/100) ² – 1]

or, 210 [{1² + 2. R/100 1 + (R/100) ² – 1]

or, 210 [{1 + R/50 + R²/1000 – 1}]

= R/50 + R² /1000 = 200R + R² /1000

or, 210 × 1000 = P (R² + 200R)

or, 210 × PR = P (R² + 200R) [∵ P R = 10000]

or, R² = 210R – 200R = 10r

or, R = 10%

P = 10000/R = 10000/10

= P = 1000 Rs

Therefore, the sum will be Rs.1000 and the rate of interest will be 10%

Question no – (15)

Solution :

Let, Rate = r%

n = 1 1/2 = 3/2

Amount = 2315.25

Principal = 2000

According to question,

A = P (1 + r/200)^2n = 2000 (1 + r/100)^2 2/3

= 200 (1 + r/100)³

or, 2315.25/2000 = (1 + r/200)

or, (1 + r/200)³ = 1.157 = (1.05)³

or, 1 + r/200 = 1.05

or, r/200 = 1.05 – 1 = 0.05

r = 0.05 × 200 = 10%

Therefore, the rate percent per annum will be 10%.

Question no – (16)

Solution :

Let, rate = r % and principle = p

Amount = 2P

n = 3 years

According to question,

A = P (1 + r/100)^n

or, 2p = p (1 + r/100)³

or, 1 + r/100 = 3√2 = 1.2599

or, r/100= 1.2599 – 1 = 0.2599

or, r = 0.2599 × 100 = 25.99%

Therefore, the required rate will be 25.99%

Question no – (17)

Solution :

Let, rate = r% and principle = p

Amount = 4P

n = 2 years

According to question,

A = P (1 + r/200)^2n

or, 4p = p (1 + r/200)^2.2 = r (1 + r/200)⁴

or, (1 + r/200)⁴ = 4

or, 1 + r/200 = 4√4 = 1.4142

or, r/200 = 1.4142 – 1 = 0.4142

or, r = 0.4142 × 200 = 82.84%

Therefore, the required rate will be 82.84%

Question no – (18)

Solution :

Given in the question,

A = 5832

r = 8%

n = 2 year

A = P (1 + r/100)^n = P (1 + 8/100)²

or, 5832 = p (27/25)²

or, P = 8 5832 × 25 × 25/27 × 27

= P = 5000 Rs.

Question no – (19)

Solution :

Let, the sum = p

Time = n = 2

Rate = r = 7.5%

For, simple interest,

Simple Interest

= PRT/100 = P × 7.5× 2/100

= 0.15P

For Compound Interest,

Compound Interest = A – P = [(p (1 + r/100)^n – p]

= [(1 + 7.5/100)² – 1]

= p (1.155625 – 1)

= 0.155625 p

According to question,

Compound Interest – Simple Interest = 360

or, 0.155625p – 0.15p = 360

or, 0.005625p = 360

or, p 360/0.005625 = 64000 Rs

Therefore the sum will be 64000 Rs.

Question no – (20)

Solution :

Let, the sum = p

Rate = r = 6 2/3% = 20/3%

Time = n = 3

For Simple Interest

Simple Interest = PRT/100

= P × 20 × 3/3 × 100

= p/5

For, Compound Interest,

Compound Interest = A – P

= [P (1 + r/100)^n – P]

= p[(1+20/30)³ – 1}

= P (4096/3375 – 1)

According to question,

Compound Interest – Simple Interest = 46

P (4096/3375 – 1) – p/5 = 46

or, p (4096/3375 – 1 – 1/5) = 46

or, p (4096 – 3375 – 675/3375) = 46

or, p × 46/3375 = 46

or, p = 3375

Therefore, the sum will be 3375 Rs.

Question no – (21)

Solution :

From the question,

A = 13230 Rs

p = 12000 Rs

R = 5%

n = ?

According to question,

A = P (1 + r/100)^n = 12000 (1 + 5/100)^n

or, 13230/12000 = (1.05)^n

or, (1.0.5)^n = (1.05)^n

or, n = 2

Therefore, the value of n will be 2.

Question no – (22)

Solution :

Let, Rate = r%

From the given question,

principal = 4000 Rs

Compound Interest = 410

n = 2

Compound Interest = A – P = [p (1 + r/100)^n – p]

or, 410 = 4000 (1 + r/100)² – 4000

or, 4000 + 410 = 4000 (1 + r/100)² – 4000

or, (1 + r/100)² = 4410/4000 = (1.05)²

or, 1 + r/100 = 1.05

or, r/100 = 1.05 – 1 = 0.05

or, = 0.05 × 100 = 5%

Therefore, the rate percent will be 5%

Question no – (23)

Solution :

Let, Principle = p

A = 10404 Rs (as per the given question)

n = 2

r = 2%

A = P (1 + r/100)^n

or, 10404 = p (1 + 2/100)²

= P × (1.02)²

= p × 1.0404

or, p = 10404/1.0404

= 10000 Rs

Therefore, the deposited sum will be 10,000 Rs.

Question no – (25)

Solution :

Let, Rate = r %

In the question,

A = 1102.50

p = 1000

n = 2 years

A = P (1 + r/100)^n = 1000 (1 + r/100)²

or, 1102.50 = (1 + r/100)²

or, 1.1025 = (1 + r/100)²

or, (1 + r/100)² = (1.05)²

or, 1 + r/100 = 1.05

or, r/100 = 1.05- 1 = 0.05

or, r = 0.05 × 100 = 5%

Therefore, the required rate percent will be 5%

Question no – (26)

Solution :

Let, time = n years

rate % = 10%

C.I = 378

Principal = 1800 Rs

C.I = A – P = P (1 + r/)^n – p

or, 378 = 1800 (1 + 10/100)^n – 1800

or, 1800 × (1.1)^n = 378 + 1800 = 2178

or, (1.1)^n = 2178/1800 = 1.21

or, (1.1)^n = (1,1)²

or, n = 2 years

Therefore, the required time will be 2 years.

Question no – (27)

Solution :

Let, sum = p

r = 6 3/4% = 27/4% (as per the question)

n = 2

A = 45582.25

A = p (1 + r/100)^n = p (1 + 27/2100)²

or, 45582.25 = p (1.0675)²

or, p = 45582.25/1.13955625

= 40000 Rs.

Therefore, the sum of money will be 40,000 Rs.

Question no – (28)

Solution :

Let, sum = p

Amount = 453690 Rs (As per the given question)

n = 2

Rate = 6.5%

Amount = (1 + r/100)²

= p (1 + 6.5/100)² = p (1.065)²

or, 453690 = p × 1.134225

or, p = 453690/1.134225

= P = 400000 Rs

Therefore, the required sum will be 4,00,000 Rs.

Compound Interest Exercise 14.4 Solution :

Question no – (1)

Solution :

As per the given question,

Population = 2800

Rate = 5%

n = 2 years

Population after 2 years = p (1 + R/100)^n

= 2800 (1+ 5/100)²

= 28000 (1.05)²

= 30870 Rs

Therefore, the population will be Rs 30870 after 2 years.

Question no – (2)

Solution :

According to the given question,

Population = 125000

R₁ = 5.5%

R₂ = 3.5%

n = 3 years

Net death rate = R = (5.5 – 3.5) = 2%

Population after 3 years,

= p (1 + r/100)^n = 125000 (1 + 2/100)²

= 125000 × (1.02)³

= 132651

Therefore, the population of city after 3 years will be 132621.

Question no – (3)

Solution :

As per the given question,

P = 25000

n = 3y

R₁ = 4%

R₂ = 5%

R3 = 8%

Population after 3 years,

= p (1 + R₁/100) (1 + R₂ /100) (1 + R3/100)

= 25000 (1 + 4/100) (1 + 5/100) (1 + 3/100)

= 25000 × 1.04 × 1.05 × 108

= 29484

Therefore, its population will be 29484 after 3 years.

Question no – (4)

Solution :

Given in the question,

P = 50000

R₁ = 4%

R₂ = 5%

R3 = 3%

n = 3

Population after 3 years,

= p (1 + R₁/100) (1 + R₂/100) (1 + R3/100)

= 50000 (1+ 4/100) (1 + 5/100) (1 + 3/100)

= 50000 × (1.04) × (1.05) × (1.03)

= 56238

Therefore, the present population will be 56238.

Question no – (5)

Solution :

As per the given question,

Population after 3 year = 9261

n = 3 years

R = 5%

Let, p = ?

According to question,

p (1 + r/100)^n = 9261

or, p (1 + 5/100)³ = 9261

or, p × (1.05)³ = 9261

or, p × 1.157625 = 9261

or, p = 9261/1.157625 = 8000

Therefore, 3 years ago the population was 8000.

Question no – (6)

Solution :

Let, Rate = r %

n = 3 years (given in the question)

p = 40000

After 3 years growth of production = 46305

According to question,

p (1 + r/100)6n = 46305

or, 40000 (1 + r/100)³ = 46305

or, (1 + r/10) ³ = 46305/40000= (21/20) ³

or, 1 + r/100 = 21/20

or, r/100 = 21/20 – 1 = 1/20

or, r = 5%

Therefore, the annual rate of growth of the production of scooters is 5%

Question no – (7)

Solution :

Present population = 196830

r = 3%

n = 3 years

Let, population before 3 years age = p

According to question,

p (1 + r/100)n = 196830

or, p (1 + 3/100)³ = 196830

or, p (1.08)³ = 196830

or, p = 196830/(1.08)³

= 196830/1.259712

= 156250

Therefore, 3 years ago the population was 156250.

Question no – (9)

Solution :

According to the question,

P = 13125000

R₁ = 10%

R₂ = 8%

R3 = 12%

Population after 3 years hours,

= p (1 + R₁/100) (1 – R₂/100) (1 + R3/100)

= 13125000 (1 + 10/100) (1 – 8/100) (1 + 12/100)

= 13125000 × 1.01 × 0.92 × 1.12

= 14876400

Therefore, the count of bacteria after 3 hours will be 14876400.

Question no – (10)

Solution :

As per the given question,

p = 72000

R₁ = 7%

R₂ = –10%

Population at the end of ‘2000’,

= p (1 + R₁/100) (1 – R₂/100)

= 72000 (1 + 7/100) (1 – 10/100)

= 72000 × 1.07 × 0.9 = 69336

Therefore, the population at the end of the year 2000 is 69336.

Question no – (11)

Solution :

According to the given question,

P = 64000

R₁ = – 25%

R₂ = – 25%

R3 = 25%

No of workers after 4th year,

= p (1 – R₁/100) (1 – R₂/100) (1 – R3/100)

= 6400 (1 – 25/100) (1 – 25/100) (1 + 25/100)

= 6400 × 0.75 × 0.75 × 1.25

= 4500

Therefore, 4500 workers were working during the fourth year.

Question no – (12)

Solution :

As per the question we know,

P = 100000

R₁ = 5%

R₂ = + 10

R3 = 12

Aman’s profit after 3 years,

= p (1 – R₁/100) (1 – R₂/100) (1 + R3/100)

= 1000000 (1 – 5/100) (1 + 10/100) (1 + 12/100)

= 1000000 × 0.95 × 1.01 × 1.12 = 117040

Net profit,

= 117040 – 100000

= 17040

Therefore, the net profit will be 17040 Rs.

Question no – (13)

Solution :

As per the question present population of the town,

= P = 175760

R = 40%

n = 3

Population after 3 years,

= p (1 + r/100)³ = (1 + 40/100)³

or, 175760 = P × (1.04)³ = 1.124864P

or, p = 175760/1.124864 = 156250

Therefore, the population 3 years ago was 156250.

Question no – (14)

Solution :

According to the given question,

P = 8000

R₁ = 15%

R₂ = 5%

Production after 3 years,

= P (1 + R₁/100)² (1 – R²/100)

= 8000 (1 + 15/100)² (1 – 5/100)

= 8000 (1.15)² (0.95) = 10051

Thus, the production after 3 years is 10051.

Question no – (15)

Solution :

As per the given question,

P = 6760000

R = 4%

(i) Population in 2001 – where = n = 2y

= p (1 + R/100)n = 6760000 (1 + 4/100)²

= 676000 (1.04)²

= 7311616

(ii) Population in 1997 – where n = 2 years

= p (1- R/100)n = 6760000 (1 – 4/100) ²

= 6760000 (0.96)² = 62500016

Question no – (16)

Solution :

According to the given question,

P₁ = 2500000

R = 5%

R = 10%

n = 1y

Profit after 1 year,

(P₂) = p (1 + 100)n

= 2500000 (1 + 5/100)

= 2500000 × 1.05 = 2625000

Profit after next year,

= P₂ (1 + r/100)n 2625000 (1 + 5/100)

= 2625000 × 1.10

= 2887500

Total profit,

= 2887500 – 2500000

= 387500 Rs

Therefore, his total profit will be 387500 Rs.

Compound Interest Exercise 14.5 Solution :

Question no – (1)

Solution :

As per the given question,

P = 16000

R = 5%

n = 2 year

Value after 2 years,

= p (1 + r/100)n

= 16000 (1 + 5/100)

= 16000 × (0.95) ²

= 14440 Rs

Therefore, its value after 2 years will be 14440 Rs.

Question no – (2)

Solution :

From the given question,

P = 100000

R = 10%

n = 2 year

Value of machine after 2 years,

= p (1 – R/100)n – 1000000 (1 – 10/100)²

= 100000 × (0.90)²

= 81000

Depreciation,

= 100000 – 81000

= 19000 Rs.

Therefore, the total depreciation will be 19000 Rs.

Question no – (3)

Solution :

According to the given question,

P = 640000

r = 5%

n = 2 years

6 months = 6/12 y = 1/2y

Value after 2 years,

= p (1 + r/200)2n

= 640000 (1 + 5/200)2.2

= 640000 (1.025)⁴

= 706440. 25

Therefore, the value of the plot after 2 years is Rs. 706440. 25.

Question no – (4)

Solution :

According to the question,

P = 30000

R = 25

n = 3y

Value after 3 years,

= P (1- r/100)n = p (1 – 25/100)³

= 30000 (0.75)³

= 12656.25

Therefore, the value of the house after 3 years is Rs. 12656.25

Question no – (5)

Solution :

As per the given question,

P = 9680

R = 12%

n = -2y

Price = p (1 – R/100)n

= 9680 (1 – 12/100) – 2

= 9680 (0 . 88) – 2

= 125000

Therefore, It was purchased at Rs. 125000.

Question no – (6)

Solution :

Given in the question,

P = 9680 Rs

R = 12%

n = – 2y

Price = p (1 – R/100)n

= 9680 (1 – 12/100) – 2

= 9680 (0 . 88) – 2

= 125000 Rs

It was purchased at Rs. 125000.

Question no – (8)

Solution :

According to the question,

P = 50000

R₁ = -4%

R₂ = 5%

R3 = 10%

∴ Final investment,

= p (1 + R₁/100) (1 + R₂/100) (1 + R3/100)

= 500000 (1 – 4/100) (1 + 5/100) (1 + 10/100)

= 50000 × (0.96) × (1.05) × (1.01)

= 554400 Rs

Net profit,

= 554400 – 500000

= 54400 Rs

Therefore, the net profit will be Rs. 54400

Next Chapter Solution :

Updated: June 14, 2023 — 6:30 am