8th Maths 8.3

Chapter 8

Comparing Quantities

NCERT Class 8th Solution of Exercise 8.1

NCERT Class 8th Solution of Exercies 8.2

`text{Important formula}`

i)`text{ A = P}(1+text{R}/100)^text{n}`

ii)`text{ CI = A - P}`

EXERCISE 8.3

1. Calculate the amount and compound interest on

(a) ₹ 10,800 for 3 years at 12`1/2` % per annum compounded annually.
(b) ₹ 18,000 for 2`1/2` years at 10% per annum compounded annually.

(c) ₹ 62,500 for 1`1/2` year at 8% per annum compounded half yearly.

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.

(You could use the year-by-year calculation using SI formula to verify). 

(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.

(a) ₹ 10,800 for 3 years at 12`1/2` % per annum compounded annually.

`text{Solution:}`

`text{P = ₹ 10800, n = 3 years, R}`` = 12 1/2 = 25/2%`

`text{A = P}(1+text{R}/100)^text{n}`

`=10800(1+25/(2times100))^3`

`=10800(1+1/(2times4))^3`

`=10800(1+1/8)^3`

`=10800((8+1)/8)^3`

`=10800(9/8)^3`

`=10800times(9times9times9)/(8times8times8)`

`=15377.34`

`text{CI = A - P}`

`=15377.34-10800`

`=4577.34`

`text{Answer:}`

`text{A= ₹ 15377.34, CI= ₹ 4577.34}`

(b) ₹ 18,000 for 2`1/2` years at 10% per annum compounded annually.

`text{Solution:}`

`text{The amount for 2 years}`

`text{A = P}(1+text{R}/100)^text{n}`

`=18000(1+10/100)^2`

`=18000(1+1/10)^2`

`=18000((10+1)/10)^2`

`=18000(11/10)^2`

`=18000times11/10times11/10`

`=21780`

`text{The interest after 2 years}`

`text{SI=A-P}`

`=21780-18000`

`=3780`

`text{Principal = ₹ 21780, for } 1/2 text{ years}`.

`text{SI}=(text{P}times text{R} times text{T})/100`

`=(21780times10times1/2)/100`

`=(21780times 10times1)/(2times100)`

`=1089`

`text{CI = 3780 + 1089 = 4869}`

`text{A = P + CI}`

`= 21780 + 1089 = 22869`  

`text{Answer:}`

`text{A = ₹ 22869, CI = ₹ 4869}.`

(c) ₹ 62,500 for 1`1/2` year at 8% per annum compounded half yearly.

`text{Solution:}`

`text{P = 62500, n }= 1 1/2 text{ years} = 3/2 text{ years }``text{= 3 half years}`

`text{R = 8% yearly} = 8/2 = text{ 4% half yearly}` 

`text{A = P}(1+text{R}/100)^text{n}`

`= 62500(1+4/100)^3`

`= 62500(1+1/25)^3`

`= 62500((25+1)/25)^3`

`= 62500(26/25)^3`

`= 62500times26/25times26/25times26/25`

`= 4times26times26times26`

`= 70304`

`text{CI = A - P}`

`= 70304 - 62500`

`= 7804`

`text{Answer:}`

`text{A = ₹ 70304, CI = ₹ 7804}`

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.

(You could use the year-by-year calculation using SI formula to verify).

`text{Solution:}`

`text{P = 8000, n = 1 year = 2 half years}`

`text{R = 9% yearly = } 9/2 text{ half yearly}`

`text{A = P}(1+text{R}/100)^text{n}`

`=8000(1+9/(2times100))^2`

`=8000(1+9/200)^2`

`=8000((200+9)/200)^2`

`=8000(209/200)^2`

`=8000times209/200times209/200`

`=8736.20`

`text{CI = A - P}`

`=8736.20 - 8000 = 736.20`

`text{Answer:}`

`text{A = ₹ 8736.20, CI = ₹ 736.20}`

(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.

`text{Solution:}`

`text{P = 10000, n = 1 year = 2 half years}`

`text{R = 8% yearly }= 8/2 = text{ 4 half yearly}`

`text{A = P}(1+text{R}/100)^text{n}`

`=10000(1+4/100)^2`

`=10000(1+1/25)^2`

`=10000((25+1)/25)^2`

`=10000(26/25)^2`

`=10000times26/25times26/25`

`=10816`

`text{CI = A - P}`

`=10816 - 10000 = 816`

`text{Answer:}`

`text{A = ₹ 10816, CI = ₹ 816}.`

2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd-year amount for `4/12` years).

`text{Solution:}`

`text{P = 26400 , R = 15%,  n = 2 year 4 months}`

`text{For 2 years}`

`text{A = P}(1+text{R}/100)^text{n}`

`=26400(1+15/100)^2`

`=26400(1+3/20)^2`

`=26400((20+3)/20)^2`

`=26400(23/20)^2`

`=26400times23/20times23/20`

`=34914`

`text{For 4 month}`

`text{SI }=(text{ P}times text{R}times text{T})/100`

`=(34914times 15times4)/(100times12)`

`=1745.70`

`text{A = 34914 + 1745.70}`

`=36659.70` 

`text{Answer:}`

`text{Amount after 2 years 4 months is ₹ 36659.70}.`

3. Fabina borrows t ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

`text{Solution:}`

`text{Fabina}`

`text{P = ₹ 12500, R = 12%, and  T = 3 years}`

`text{SI}=(text{P}times text{R} times text{T})/100`

`=(12500times12times3)/100`

`=125times12times3`

`=4500`

`text{Radha}`

`text{P = ₹ 12500, R = 10, and  n = 3 years}`

`text{A = P}(1+text{R}/100)^text{n}`

`=12500(1+10/100)^3`

`=12500(1+1/10)^3`

`=12500((10+1)/10)^3`

`=12500(11/10)^3`

`=12500times11/10times11/10times11/10`

`=16637.5`

`text{CI = A - P}`

`=16637.5 - 12500`

`=4137.5`

`text{Difference = Fabina -  Radha}`

`=4500 - 4137.5`

`=362.5`

`text{Answer:}`

`text{Fabina pays ₹ 362.5 more}`.

4. I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

`text{Solution:}`

`text{P = 12000, R = 6%, and T = 2 years}`

`text{SI}=(text{P}times text{R}times text{T})/100`

`=(12000times6times2)/100`

`=1440`

`text{P = 12000, R = 6%, and n = 2 years}`

`text{A = P}(1+text{R}/100)^text{n}`

`=12000(1+6/100)^2`

`=12000(1+3/50)^2`

`=12000((50+3)/50)^2`

`=12000(53/50)^2`

`=12000times53/50times53/50`

`=13483.2`

`text{CI = A - P}`

`=13483.2 - 12000`

`=1483.2`

`text{Extra amount = 1483.2 - 1440}`

`=43.20`

`text{Answer:}`

`text{Extra amount is ₹ 43.20}`.

5. Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

i) after 6 months?

ii) after 1 year?

i) `text{Solution:}`

`text{P = 60000, R = 12%, and}``text{ T = 6 months = }6/12 text{ year}`

`text{SI}= (text{P}times text{R} times text{T})/100`

`=(60000times12times6)/(100times12)`

`=3600`

`text{A = P + SI}`

`=60000 + 3600`

`=63600`

`text{Answer:}`

`text{Required amount ₹ 63600}`.

ii) `text{Solution:}`

`text{P = 60000, R }= 12/2% = 6%`

`text{n= 1 year = 2 half years}`

`text{A = P}(1+text{R}/100)^text{n}`

`= 60000(1+6/100)^2`

`= 60000(1+3/50)^2`

`=60000((50+3)/50)^2`

`=60000(53/50)^2`

`=60000times53/50times53/50`

`=67416`

`text{Answer:}`

`text{Required amount ₹ 67416.}`

6. Arif took a loan of t ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paid after 1`1/2` year if the interest is 

i) compounded annually.
ii) compounded half yearly

i) `text{Solution:}`

`text{P = 80000, R = 10%, n }= 1 1/2`

`text{For 1 year}`

`text{SI} = (text{P}times text{R} times text{T})/100`

`=(80000times10times1)/100`

`=8000`

`text{Amount = P + SI}`

`=80000+8000`

`=88000`

`text{For }1/2 text{ year}`

`text{SI} = (text{P}times text{R} times text{T})/100`

`=(88000times10times1)/(100times2)`

`=4400`

`text{Amount = P + SI}`

`= 88000 + 4400`

`=92400`

`text{Answer:}`

`text{A ₹ 92400}`

ii) `text{Solution:}`

`text{R = } 10/2 = 5% text{half yearly}`

`text{n = } 1 1/2 = 3/2 = 3 text{ half year}`

`text{A = P}(1+text{R}/100)^text{n}`

`=80000(1+5/100)^3`

`=80000(1+1/20)^3`

`=80000((20+1)/20)^3`

`=80000(21/20)^3`

`=80000times21/20times21/20times21/20`

`=10times9261`

`=92610`

`text{Answer:}`

`text{A ₹ 92610}.`

7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
i) The amount credited against her name at the end of the second year.
ii) The interest for the 3rd year.

i) `text{Solution:}`

`text{P = 8000, R = 5%, n = 2 years}`

`text{A = P}(1+text{R}/100)^text{n}`

`=8000(1+5/100)^2`

`=8000(1+1/20)^2`

`=8000((20+1)/20)^2`

`=8000(21/20)^2`

`=8000times21/20times21/20`

`=20times441`

`=8820`

`text{Answer:}`

`text{Amount ₹ 8820}`

ii) `text{Solution:}`

`text{P = 8000, R = 5%, n = 3 years}`

`text{A = P}(1+text{R}/100)^text{n}`

`=8000(1+5/100)^3`

`=8000(1+1/20)^3`

`=8000((20+1)/20)^3`

`=8000(21/20)^3`

`=8000times21/20times21/20times21/20`

`=9261`

`text{Interest for 3rd year = 9261 - 8820}`

`=441`

`text{Answer:}`

`text{Interest for 3rd year ₹ 441.}`

 

8. Find the amount and the compound interest on ₹ 10,000 for 1`1/2` year at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

`text{Solution:}`

`text{Interest is compounded half yearly}` 

`text{P = 10000, R = 10% yearly =5% half yearly}`

`text{n = }1 1/2 = 3/2 text{= 3 half year}`

`text{A = P}(1+text{R}/100)^text{n}`

`=10000(1+5/100)^3`

`=10000(1+1/20)^3`

`=10000((20+1)/20)^3`

`=10000(21/20)^3`

`=10000times21/20times21/20times21/20`

`=11576.25`

`text{CI = A - P}`

`=11576.25 - 10000`

`=1576.25`

`text{Interest is compounded annually}`

`text{P = 10000, R = 10% yearly}`

`text{ n = }1 1/2 text{ year}`

`text{SI} = (text{P}times text{R} times text{T})/100`

`=(10000times10times1)/100`

`=1000`

`text{Amount = P + SI}`

`= 10000 + 1000`

`=11000`

`text{Interest for }1/2 text{ year}`

`text{SI} = (text{P}times text{R} times text{T})/100`

`=(11000times10times1)/(100times2)`

`=550`

`text{Total interest = 1000 + 550}`

`= 1550`

`text{Difference between interest =1576.25-1550}`

`=26.25`

`text{Answer:}`

`text{A 11576.25, Interest 1576.25, and yes}`

9. Find the amount which Ram will get on ₹ 4096 if he gave it for 18 months at 12`1/2`%
per annum, interest being compounded half-yearly.

 

`text{Solution:}`

`text{A = P}(1+text{R}/100)^text{n}`

`text{P = 4096, R }= 12 1/2% text{ yearly }`` = 25/4% text{ half yearly}`

`text{ n = 18 month =}1 1/2 = 3/2 text{= 3 half year}`

`text{A = P}(1+text{R}/100)^text{n}`

`=4096(1+25/(100times4))^3`

`=4096(1+1/16)^3`

`=4096((16+1)/16)^3`

`=4096(17/16)^3`

`=4096times17/16times17/16times17/16`

`=4913`

`text{Answer:}`

`text{Amount ₹ 4913}.`

10. The population of a place increased to ₹ 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?

i) `text{Solution:}`

`text{P=54000, R = 5%, n = 2003-2001= 2 years.}`

`text{A = P}(1-text{R}/100)^text{n}`

`54000=text{P}(1+5/100)^2`

`54000=text{P}(1+1/20)^2`

`54000=text{P}((20+1)/20)^2`

`54000=text{P}(21/20)^2`

`54000=text{P}times20/21times20/21`

`text{P}=54000times400/441`

`text{P}=21600000/441`

`=21600000/441`

`=48979.59`

`text{=48980 (approx)}`

`text{Answer:}`

`text{Population in 2001 is 48980 (approx)}.`

ii) `text{Solution:}`

`text{P=54000, R = 5%, n = 2005-2003= 2 years.}`

`text{A = P}(1+text{R}/100)^text{n}`

`=54000(1+5/100)^2`

`=54000(1+1/20)^2`

`=54000((20+1)/20)^2`

`=54000(21/20)^2`

`=54000times21/20times21/20`

`=54000times441/400`

`=540times441/4`

`=238140/4`

`=59535`

`text{Answer:}`

`text{Population in 2005 is 59535}.`

11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour Find the bacteria at the end of 2 hours if the count was initially ₹ 5,06,000

`text{Solution:}`

`text{P = 506000, R = 2.5%, n = 2 hours}`

`text{A = P}(1+text{R}/100)^text{n}`

`=506000(1+2.5/100)^2`

`=506000(1+1/40)^2`

`=506000((40+1)/40)^2`

`=506000(41/40)^2`

`=506000times41/40times41/40`

`=1265times1681`

`text{=531616.25 (approx)}`

`text{Answer:}`

`text{531616 (approx) bacteria}`.

12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

`text{Solution:}`

`text{P = 42000, R = 8%,  n = 1 year}`

`text{A = P}(1-text{R}/100)^text{n}`

`=42000(1-8/100)^1`

`=42000(1-2/25)^1`

`=42000((25-2)/25)^1`

`=42000(23/25)^1`

`=42000times23/25`

`=1680times23`

`=38640`

`text{Answer:}`

`text{Value after 1 year ₹ 38640}.`