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 :
👉 Chapter 20 👈
