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.

$1/2$ 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.

$text{Solution:}$

$text{The amount for 2 years}$

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

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

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

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

$=18000(11/10)^2$

$=18000times11/10times11/10$

$=21780$

$text{The interest after 2 years}$

$text{SI=A-P}$

$=21780-18000$

$=3780$

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

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

$=(21780times10times1/2)/100$

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

$=1089$

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

$text{A = P + CI}$

$= 21780 + 1089 = 22869$

$text{Answer:}$

$text{A = ₹ 22869, CI = ₹ 4869}.$

$1/2$ 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?

$text{Solution:}$

$text{P = 60000, R = 12%, and}$

$text{ T = 6 months = }6/12 text{ year}$

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

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

$=3600$

$text{A = P + SI}$

$=60000 + 3600$

$=63600$

$text{Answer:}$

$text{Required amount ₹ 63600}$ .

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

$text{Solution:}$

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

$text{For 1 year}$

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

$=(80000times10times1)/100$

$=8000$

$text{Amount = P + SI}$

$=80000+8000$

$=88000$

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

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

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

$=4400$

$text{Amount = P + SI}$

$= 88000 + 4400$

$=92400$

$text{Answer:}$

$text{A ₹ 92400}$

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.

$text{Solution:}$

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

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

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

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

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

$=8000(21/20)^2$

$=8000times21/20times21/20$

$=20times441$

$=8820$

$text{Answer:}$

$text{Amount ₹ 8820}$

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?

$text{Solution:}$

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

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

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

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

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

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

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

$text{P}=54000times400/441$

$text{P}=21600000/441$

$=21600000/441$

$=48979.59$

$text{=48980 (approx)}$

$text{Answer:}$

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

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.

$text{Solution:}$

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

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

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

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

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

$=42000(23/25)^1$

$=42000times23/25$

$=1680times23$

$=38640$

$text{Answer:}$

$text{Value after 1 year ₹ 38640}.$

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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

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10th Maths Chapter 1

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NCERT Class 8th Solution of Exercise 8.1

NCERT Class 8th Solution of Exercies 8.2

Important formulatext{Important formula}

i) A = P(1+R100)ntext{ A = P}(1+text{R}/100)^text{n}ii) CI = A - Ptext{ CI = A - P}

EXERCISE 8.3

Comments

Popular posts from this blog

10th Maths Chapter 1

CBSE 10th and 12th Result

RSKMP 5th & 8th Result

$text{Important formula}$

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

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