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OP Malhotra Class 9 ICSE Maths Solutions Chapter 2 Compound Interest
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 Math Book, Chapter 2, Compound Interest. Here students can easily find step by step solutions of all the problems for Compound Interest, Exercise 2a, 2b, 2c and 2d Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Compound Interest Exercise 2(a) Solution :
Question no – (1)
Solution :
Given, Principle = ₹10000
Rate% p.a. = 12%
Number of years = 2
Now, Amount = p(1 + r/100)2
= 10,000 (1 + 12/100)2
= 10,000 (100 + 12/100)2
= 10,000 × 112 × 112/100 × 100
= 12544
∴ Compound interest.
= (12544 – 10,000)Rs
= 2544 Rs
Hence, the compound interest will be 2544 Rs.
Question no – (2)
Solution :
In the given question,
Principle = ₹5000,
Rate% p.a. = 10%
Number of years = 2
Amount after 2 years.
= 5000 (1 + 10/100)2
= 5000 (100 + 10/100)2
= 5000 × 110 × 110/100 × 100 × 100
= 1210 Rs × 5
= 6050 Rs
Compound interest,
= (6050 – 5000) Rs
= 1050 Rs
Therefore, the compound interest will be 1050 Rs.
Question no – (3)
Solution :
As per the question,
Principle = Rs 2800
Rate% p.a. = 10%
Number of years = 1 1/2
∴ Amount,
= 2800 (1 + 10/100)1/2
= 2800 (1 + 10/100)3/2
= 2800 (100 + 10/100)3/2
= 2800 × (110/100)3/2
= 2800 × (11/10)2×3/2
= 2800 × 11 × 11 × 11/10 × 10 × 10
= 3726.80 Rs
∴ Compound interest,
= (3726.8 – 2800) Rs
= 926.8 Rs
Thus, the compound interest will be 926.8 Rs
Question no – (4)
Solution :
Given, Principle = Rs 2000
Rate% p.a. = 20%
Number of years = 2
Amount,
= 2000 (1 + 20/100)2
= 2000 (100 + 20/100)2
= 2000 × 120 × 120/100 × 100
= 1440 Rs × 2
= 2880Rs
∴ Compound interest,
= (2880 – 2000) Rs
= 880 Rs
Thus, the Compound interest will be 880 Rs.
Question no – (5)
Solution :
Here, Principal = 20480 Rs
rate = 6 1/4% 25/4
Time = 2 year 73 days
= 3 year × 73 days
= 2 year × 73/365
= 2 1/5 year
= 1155 year
After 1 year amount will be,
= 20480 + (20480 × 25 × 1/4 × 100)
= 20480 + 1280
= 21760 Rs
After 2nd year amount will be,
= 21760 + (21760 × 25 × 1/4 × 100)
= (21760 + 1360)
= 23120 Rs
After 1/5 year amount will be,
= (23120 + 23120 × 25 × 1/4 × 100 × 5)
= (23120 + 289) Rs
= 23409 Rs
∴ Compound interest,
= (23409 – 2-480) Rs
= 2929
Therefore, the compound interest will be 2929 Rs.
Question no – (6)
Solution :
As per the question,
Principal = Rs 15625
Rate = 4% per annum
Time = 3 years
Amount × (1 + 4/100)3
= 15625 × (100 + 4/100)3
= 15625 × 104 × 104 × 104/100 × 100 × 100
= 18756 Rs
Therefore, the required amount will be 18756 Rs
Question no – (7)
Solution :
Anurag will have to pay after 3 years,
= 8000 (1 + 5/100)3
= 8000 × (105/100)3
= 8000 × 21 × 21 × 21/20 × 20 × 20
= 9261 Rs
Hence, after 3 years Anurag will pay Rs 9261
Question no – (8)
Solution :
Amount in 2nd year,
= 93750. (1 + 9.6/100)2
= 93750 × (109.6/100)2
= 112614 Rs
∴ Amount in 3rd year,
= 93750 × (1 + 9.6/100)3
= 93750 × (1 + 9.6/100)3
= 12342.94 Rs
∴ Compound Interest in 3rd year,
= (123424.94 – 112614)
= 10810.94 Rs
Question no – (9)
Solution :
(i) What is the sum due at the end of the first year?
At the end of 1st year amount
= 9600 + (9604 × 10/100)
= (9600 + 960) Rs
= 10560 Rs
(ii) What is the sum due at the end of the second year?
Sum of end of 2nd year
= 10560 + (10560 × 10/100) Rs
= 10560 + 1056
= 11616 Rs
(iii) Find the compound interest earned in the first 2 years.
1st 2 year compound interest earned
= (11616 – 9600) Rs
= 2016 Rs
(iv) Find the compound interest at the end of 3 years.
At end of 3 year Amount
= 11616 + (11616 × 10/100) Rs
= 11616 + 1161.6
= 12777.6 Rs
Question no – (10)
Solution :
(i) Find the compound interest after one year.
Amount after one year = 10,000 (1 + 10/100)1
= 10,000 × 110/100
= 11000 Rs
∴ Compound Interest = (11000 – 10,000) Rs
= 1000 Rs
(ii) Find the compound interest for 2 years.
2 years amount = 11000 + 11000 ∴ 10/100
= 11000 + 1100
= 12100 Rs
Interest = (12100 – 10000)
= 2100
(iii) Find the sum of money required to clear the debt at the end of 2 years.
= Sum of money 1200 to clear the debt at the end of 2 years.
(iv) Find the difference between the compound interest and the simple interest at the same rate for 2 years.
S.I of 2 years = (10,000 × 10 × 2/100) Rs
= 2000 Rs
C.I of 2 years = 2100 Rs
∴ Difference if C.I and S.I
= (2100 – 2000) Rs
= 100 Rs
Question no – (11)
Solution :
Here, rate = 12%/2 = 6%
time = 1 year = 2 half/years
∴ Amount = 5000 (1 + 6/100)²
= 5000 (106/100)²
= 5000 × 53/50 × 53/50
= 561 Rs
∴ Compound Interest,
= (5618 – 5000) Rs
= 618 Rs.
Therefore, the compound interest will be 618 Rs.
Question no – (12)
Solution :
According to the question,
rate = 10% = 10/2 = 5%
time = 1 1/2% = 3/2 year, 1 and half year
∴ Amount,
= 16000 (1 + 5/100)3
= 16000 × (105/100)3
= 16000 × 21/20 × 21/20 × 21/20
= 185222 Rs
∴ Compound Interest,
= (18522 – 16000) Rs
= 2522 Rs.
Thus, the amount is Rs 185222 and the and the compound interest is Rs. 2522
Question no – (13)
Solution :
Amount,
= 43,000 (1 + 7/100) (1 + 2/100)
= 40,000 × 107/100 × 108/100
= 46224 Rs
∴ Compound Interest after 2 year,
= (46224 – 40,000) Rs
= 6,224 Rs
Therefore, the due amount will be 46224 Rs and C.I will be 6,224 Rs.
Question no – (14)
Solution :
Let, Principal = P
I = PRT/100 …(we know)
= PRT = I × 100
= P = 50 × 100/5 × 2
= 500
∴ Amount in C.I
= 500 (1 + 5/100)2
= 500 × 105 × 105/100 × 100
= 551.25 Rs
∴ Compound Interest,
= (551.25 – 500) Rs
= 51.25 Rs
Thus, the compound interest will be 51.25 Rs
Question no – (15)
Solution :
(i) the interest for the 1st year;
Amount in 1 st year
= (46875 + 46875 × 4/100)
= 46875 + 1875
= 48750 Rs
∴ Compound Interest
= (48750 – 46875) Rs
= 1875 Rs
(ii) the amount standing to his credit at the end of the 2nd year;
∴ 2nd year amount
= 48750 + (48750 × 4/100)
= 48750 + 1950
= 50700 Rs
(iii) the interest for the 3rd year.
3rd year Interest
= 50700 × 4/100
= 2028 Rs
Compound Interest Exercise 2(b) Solution :
Question no – (1)
Solution :
As per the given question,
Principal = Rs 625
Rate of interest = 4%p.a
Time = 2 year
∴ Amount,
= 625 (1 + 4/100)²
= 625 (104/100)²
= 625 × 104 × 104/100 × 100
= 676 Rs
∴ Compound interest,
= (625 – 625) Rs
= 51 Rs
Therefore, the amount will be 676 Rs ands C.I will be 51 Rs
Question no – (2)
Solution :
Form the question we know,
Principal = Rs 8000
Rate of interest = 15%p.a
Time = 3 year
Now Amount,
= 8000 (1 + 15/100)3
= 8000 (1115/100)3
= 8000 × (23/20)3
= 8000 × 23 × 23 × 23/20 × 20 × 20
= 12167 Rs
∴ Compound interest,
= (12167 – 8000)
= 4167 Rs
Thus, the amount will be 12167 Rs and the compound interest will be 4167 Rs.
Question no – (3)
Solution :
Given in the question,
Principal = Rs 1000
Rate of interest = 10%p.a
Time = 3 year
Now Amount,
= 1000 (1 + 10/100)3
= 1000 (110/100)3
= 1000 × 11 × 11 × 11/10 × 10 × 10
= 1331 Rs
∴ Compound Interest,
= (1331 – 1000)
= 331 Rs
Hence, Thus, the amount is Rs 1331 and the compound interest is 331 Rs.
Question no – (4)
Solution :
Given, Principal = Rs 8000
Rate of interest = 10% half yearly
Time = 1 1/3 year
Now Amount,
= 8000 (1 + 10 10/2 × 100)
= 8000 (1 + 5/100)3
= 80000 (105/100)3
= 8000 × 105 × 105 × 105/100 × 100 × 100
= 9261 Rs
∴ Compound Interest,
= (9261 – 8000)
= 1261 Rs
Therefore, the amount will be 9261 and the compound interest will be 1261 Rs.
Question no – (5)
Solution :
Given in the question,
Principal = Rs 700
Rate of interest = 20% half yearly
Time = 1 1/2 year
∴ Amount,
= 700 (1 + 10/100)3
= 700 (110/100)3
= 700 × (11/10)3
= 700 × 11 × 11 × 11/10 × 10 × 10
= 931.7 Rs
∴ Compound Interest,
= (7000 – 931.7)
= 231.7 Rs
Thus, the amount will be 931.7 and the compound interest will be 231.7 Rs.
Question no – (6)
Solution :
According to the question,
Principal = 40960
rate = 12.5 % = 12 1/2% = 25/2%
time = 1 1/2 year
Now Amount
= 40960 (1 + 25/22.2 × 100)3
= 40960 (1 + 1/16)3
= 40960 (17/16)3
= 40960 × 17 × 17 × 17/16 × 16 × 16
= 49130 Rs
∴ Compound Interest,
= (49130 – 40960) Rs
= 8170 Rs
Therefore, Amar will pay 8170 Rs after 1 1/2 years.
Question no – (7)
Solution :
First, Sudhir Amount,
= 2000 (1 + 10/100)1
= 2000 × 110/100 × 1
= 2200 Rs
∴ Interest,
= (2200 – 2000) Rs
= 200 Rs
Now, Prashant Amount,
= 2000 (1 + 5/100)2
= 2000 (1 + 1/20)2
= 2000 (1 + 1/20)2
= 2000 × 21 × 21/20 × 20
= 2205 Rs
∴ Compound interest,
= (2205 – 2000) Rs
= 205 Rs
Now, the difference,
= (200 – 205)
= 5 Rs
Hence, the difference in the interest receive by them will be 5 Rs.
Question no – (8)
Solution :
As per the question,
Amount = Rs 25000
Time = 2 years
Rate = 4 and 5 per cent per year
∴ Amount,
= 25000 (1 + 4/100) (1 + 5/100)
= 25000 × 104/100 × 105/100
= 273000 Rs
Thus, the amount will be 273000 Rs.
Question no – (9)
Solution :
First, the Amount,
= 40,000 (1 + 5/100) (1 + 10/100) (1 + 15/100)
= 40,000 × 105/100 × 110/100 × 115/100
= 53130 Rs
Now, the Profit,
= (53130 – 40,000) Rs
= 13130 Rs
Therefore, the Total Profit will be 13130 Rs.
Question no – (10)
Solution :
[Given that (1.0575)40 = 9.35869]
= Here, time = (2008 – 1988)
= 20 year
= 40 half/year
= R = 11.5
= 5.75 and half yearly
Investment Principal = 5000
Amount P (1 + R/100)2
= 500 (1 + 5.75/100)40
= 500 (1 + 70.0575)40
= 46793.45 Rs
Therefore, the Maturity Amount will be 46793.45 Rs.
Question no – (11)
Solution :
According to the question,
Rate = 2 1/2%
= 5/2%
Time = 3 year
∴ After 3 year population will be,
= P (1 + r/100)3
= 64000 (1 + 5/2 × 100)3
= 64000 (1 + 1/40)3
= 68921
Hence, the number of inhabitants at the end of 3 years will be 68921.
Question no – (12)
Solution :
As per the given question,
Principal = 1250 Rs
R = 20%
n = 3 month
∴ After 3 month no of stray days,
= P(1 – R/100)n
= 1250 (1 – 20/100)³
= 1250 (80/100)³
= 1250 × 4/5 × 4/5 ×4/5
= 640
Therefore, the number of stray dogs in the city will be 640.
Question no – (13)
Solution :
Here, P = 8000 blood donors.
R = 20% per half year
x = 20% per half year = 3/2 year = 3 half year
∴ After 3 year increased donors,
= P (1 + R/100)x
= 8000 (1 + 20/100)3
= 8000 (1 + 1/5)3
= 8000 (6/5)3
= 8000 × 6/5 × 6/5 × 6/5
= 13824 donors
∴ Total number of new registrants will be,
= (14824 – 8000)
= 5824
Therefore, the number of registrants will be 5824.
Compound Interest Exercise 2(c) Solution :
Question no – (1)
Solution :
As we know that, Amount = P (1 + R/100)x
∴ 1323 = P (1 + 5/100)2
= P (105/100)2
= P (21/20)2
= P = 1323 × 20 × 20/21 × 21
= 1200 Rs
Thus, Mohan need invest 1200 Rs.
Question no – (2)
Solution :
Let, Sum is P
∴ Amount = P (1 + R/100)x
= 1352 = P(1 + 4/100)2
= P (1 + 2/100)2
= P (1 + 1/25)2
= p (1 + 26/25)2
= p = 1352 × 25 × 25/26 × 26
= 1250 Rs
Therefore, the required sum will be 1250 Rs.
Question no – (3)
Solution :
Here, Amount = 9768 Rs
r1 = 10%
r2 = 11%
∴ Amount = P(1 + r1/100) (1 + r2/100)
= 9768 = p(1 + 10/100) (1 + 11/100)
= 9768 = P (110/100) (111/100)
= P = 9768 × 10/11 × 100/111
= 8000 Rs
Therefore, the Principal will be 8000 Rs.
Question no – (4)
Solution :
Let, Principal = 100 Rs
t = 2 year
∴ S.I – C.I = 15 Rs
R = 5%
∴ Simple Interest,
= PRT/100
= 100 × 5 × 5/100
= 10 Rs
Now, Amount C.I,
= P (1 + R/100)x
= 100 (1 + 5/100)
= 100 (105/100)2
= 100 (25/20)2
= 100 × 21 × 21/20 × 20
= 441/4
Now, Compound interest,
= 441/4 – 100
= 441 – 400/4
= 41/4
∴ Compound Interest – Simple Interest,
= 41/4 – 10
= 41 – 40/4
= ¼
Now, if the difference is 1 then Principal,
= 100 × 4/1
= 400 Rs
∴ If the difference is 15 then principal,
= (400 × 15) Rs
= 6000 Rs
Question no – (5)
Solution :
Let, principal = 100
= R = 6 2/3% = 20/3%
t = 3 years
= S.I = 100 × 20 × 3/100 × 3
= 20 Rs
Amount = P (1 + R/100)x
= 100 (1 + 20/3 × 100)
= 100 (1 + 1/15)3
= 100 × (16/15)3
= 409600/3375
= 16384/135
Compound interest = 16384/135 – 100
= 16384 – 13500/135
= 2884/135 Rs
∴ C.I – S.I
= 12884/135 – 20
= 2884 – 2700/135
= 184/135 Rs
When difference is 184/135,
Principal is 100 × 135/184
and, when difference is 184, the principal is,
= 100 × 135/184 × 184
= 13500 Rs
Question no – (6)
Solution :
Let, Principal = 100
t = 2 year
R = 5%
S.I = 100 × 5 × 2/100
= 10 Rs
Amount in C.I = 100 (1 + 5/100)2
= 100 (1 + 1/20)2
= 100 (20 + 1/20)2
= 100 × (21/20)2
= 100 × 21 × 21/20 × 20
= 441/4 Rs
∴ C.I = 441/4 – 100
= 441- 400/4
= 41/4 Rs
∴ C.I – C.I = 41/4 – 10
= 41 – 40/4
= 1/4
When difference is 1/4 the n principal is 100.
When difference 1 the principal is,
= 100 × 4/1
= 400
When difference is 50 then principal is,
= (400 × 50)
= 20000 Rs
Question no – (7)
Solution :
(i) Principal = ₹ 196, Amount = ₹ 225, time, = 2 years.
Here, P = 196 Rs
= A = 225 Rs
t = 2 year
∴ A = P (1 + R/100)t
= A/P (1 + R/100)t
= 225/196 (1 + R/100)2
= (15/14)2 = (1 + R/100)2
= 1 + R/100 = 15/14
= R/100 = 15/14 – 1
= 15 – 14/14
= R = 1 × 100/14
= 100/14 = 50/7
= 7 1/7%
(ii) Principal = ₹ 3136, Compound interest = ₹ 345, Time = 2 years.
= P = 3136 Rs
C.I = 345 Rs
t = 2 year
Amount = (3136 + 345)
= 3481 Rs
A = P(1 + R/100)t
= A/P (1 + R/100)t
= 3481/3136 = (1 + R/100)2
= (59/56)2 = (1 + R/100)2
= 1 + R/100 = 59/56 – 1
= 59 – 56/56
= 3/56
= R = 3 × 100/56
= 75/14%
= 5 5/14%
Question no – (8)
Solution :
Here, P = 1000
A = 2000
t = 5 year = 10 half years
∴ Amount = P(1 + R/100)t
= A/P = (1 + R/100)t
= 2000/1000 = (1 + R/100)10
= 2/1 = (1 + R/100)10
= 1 + R/100 = 10√2
= R/100 = 10√2 – 1
= R/100 = 0.072
= R = 7.2
∴ Final Rate,
= 7.2 × 2
= 14.4%
Therefore, the rate of interest will be 14.4%.
Question no – (9)
Solution :
Here, P = 8000
A = 9261 Rs
R = 5%
∴ Amount = p (1 + R/100)2
= A/P = (1 + R/100)2
= 9261/8000 = (1 + 5/100)t
= (1 + 1/20)t
= 9261/8000 (1 + 1/20)t
= (21/20)3 = (21/20)t
∴ t = 3 years
Therefore, the required time will be 3 years.
Question no – (10)
Solution :
Here, Amount (A) = p (1 + R/100)x
A = P (1 + R/100)x
= A/P = (1 + R/100)x
= 3993/3000 = (1 + 10/100)x
= (110/100)x
= (1331/1000) = (11/10)x
= (11/10)3 (11/10)x
∴ = x = 3
∴ Time is 3 years,
= 3/2 year
= 1 1/2 years
∴ Rate = 20%
= 20/2
= 10% half year.
Question no – (11)
Solution :
Here, Amount in 3 year = 614.55 Rs
Amount in 2 year = 578.40 Rs
∴ 1 year Interest
= (614.55 – 578.40)
= 36.15 Rs
∴ Rate of interest
= 36.15 × 100/578.40 × 1
= 25/4
= 6 1/4%
Therefore, the rate of interest will be 6 1/4%
Compound Interest Exercise 2(d) Solution :
Question no – (1)
Solution :
After 3 years male it,
= P (1 – R/100)t
= 18000 (1 – 10/100)3
= 18000 (100 – 10/100)3
= 18000 = 110 × 110 × 110/100 × 100 × 100
= 18000 × 90 × 90 × 90/100 × 100 × 100
= 13122 Rs
Thus, the cost of a machine will be 13122 Rs.
Question no – (2)
Solution :
As per the given question,
The population of a town increases by 20% every year
If its present population is 2,16,000
(i) after 2 years,
= After 2 years population will be
= P (1 + R/100)2
= 2,16,000 (1 + 20/100)2
= 216000 (120/100)2
= 216000 (120/100)2
= 216000 × 12/10 × 12/10
= 311040
(ii) 2 years ago.
= 2 years ago population was,
= 2,16,000 (1 – 20/100)2
= 2,16,000 (1 – 1/5)2
= 2,16,000 (5 – 1/5)2
= 2, 16,000 (5 – 1/5)2
= 2,16,000 × 4/5 × 4/5
= 1500000
Question no – (3)
Solution :
(i) In 1979, value was,
= 18400 (1 – 8/100)
t = yr
R = 8%
= 184000 (1 – 2/25)
= 18400 × 23/25
= 16928 Rs
(ii) in 1981 value was,
= 18400 (1 + 8/100)
= 18400 (108/100)
= 18400 × 108/144 |
= 19872
Question no – (4)
Solution :
Here,
T = 3 years
R = 12 1/2% = 25/2%
P = 15360
∴ After 3 years value,
= P (1 – R/100)1/2
= 15360 (1 0 25/2 × 100)3
= 15360 (1 – 1/8)3
= 15360 × 7 × 7 × 7/8 × 8 × 8
= 10290 Rs
Thus, value after 3 years is 10290 Rs.
Question no – (5)
Solution :
According to the question,
A new car is purchased for Rs 2,50,000
Its value depreciates at rate of 10% in first year,
8% in 2nd year and then 6% every year
After 4 years value will be,
= 250000 (1 – R1/100) (1 – R2/100) (1 – R3/100)
= 250000 (1 – 10/100) (1 – 8/100) (1 – 6/100) (1 – 6/100)
= 250000 × 90/100 × 92/100 × 94/100 × 94/100
= 182905.20 Rs
Thus, value of the car after 4 years will be 182905.20 Rs.
Question no – (6)
Solution :
As per the question we know,
Principal = 10,000
R1 = 10%
R2 = 10%
R3 = 10%
∴ After 3 hour, of bacteria,
= 10,000 (1 + R1/100) (1 – R2/100) (1 + R3/100)
= 10,000 (1 + 10/100) (1 – 10/100) (1 + 109/100)
= 10,000 × 110/100 × 90/100 × 110/100
= 10890
Thus, the bacteria count at the end of 3 hours will be 10,890
Question no – (7)
Solution :
As we know Amount = P (1 + R/100)t
∴ Amount = (1 + R/100)t
= 48400/40,000 = (1 + R/100)2
= (11/10)2 = (1 + R/100)2
= 1 + R/100 = 11/10
= R/100 = 11/10 – 1
= 11 – 10/10
= R/100 = 1/10
= R = 10%
Therefore, Rate of growth will be 10%
Question no – (8)
Solution :
Let, time = t year
Amount = p (1 + R/100)t
= A/P = (1 – R/100)t
= 364500/500000 = (1 – R/100)t
= 3645/5000 = (1 – 10/100)t
= (90/100)t
= 729/1000 = (90/100)t
= (9/10)3
= (90/10)t
= 3
Therefore, the required time will be 3 years.
Question no – (9)
Solution :
After 3 year value of that will be,
= P (1 + R/100)2
= 10,00000 (1 + 20/100)x
= 10,00000 (1 + 1/5)3
= 1000000 (6/5)3
= 1000000 × 6/5 × 6/5 × 6/5
= 1728000 Rs
∴ After 3 years value of car will be,
= p (1 – R/100)x
= 3,20,000 (1 – 15/100)3
= 3,20,000 (100 – 15/100)3
= 3,20,000 (85/100)3
= 3, 20,000 × (17/20)3
= 320, 000 × 17 × 17 × 17/20 × 20 × 20
= 1965 Rs
∴ Total costing for buy,
= (10,000000 + 320,000)
= 13,20,000
Selling price,
= (1720000 + 196520) Rs
= 1916520Rs
∴ Gain,
= (1924520 – 1320,000)
= 604520 Rs
Question no – (10)
Solution :
In the beginning of fourth year workers will,
= P (1 – R1/100) (1 – R2/100) (1 + R3/100)
= 2000 (1 – 5/100) (1 – 5/100) (1 + 10/100)
= 800 × 95/100 × 95/100 × 110/100
= 7942
Thus, 7942 workers were working during the fourth year.
Next Chapter Solution :
👉 Chapter 3 👈
