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Comparing Quantities : Exercise 8.3 (Mathematics NCERT Class 8th)


Q.1 Calculate the amount and compound interest on
(a) Rs 10,800 for 3 years at 12{1 \over 2}% per annum compounded annually.
(b) Rs 18,000 for 2{1 \over 2} years at 10% per annum compounded annually.
(c) Rs 62,500 for 1{1 \over 2} years at 8% per annum compounded half yearly.
(d) Rs 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) Rs 10,000 for 1 year at 8% per annum compounded half yearly.
Sol. (a) Here, principal (P) = Rs 10,800, Rate(R) = 12{1 \over 2}% = {{25} \over 2}% (annual), Number of years (n) = 3.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= Rs \left[ {10800{{\left( {{\rm{1 + }}{{25} \over {{\rm{200}}}}} \right)}^3}} \right]
= Rs \left[ {10800{{\left( {{{225} \over {{\rm{200}}}}} \right)}^3}} \right]
= Rs \left( {10800 \times {{225} \over {200}} \times {{225} \over {200}} \times {{225} \over {200}}} \right)
= Rs 15377.34
Compound Interest, C.I. = A – P = Rs (15377.34 – 10800) = Rs 4,577.34

(b) Here, principal (P) = Rs 18,000, Rate(R) = 10%, Number of years (n) = 2{1 \over 2} years.
Calculation for first 2 years:
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= Rs \left[ {18000{{\left( {{\rm{1 + }}{1 \over {{\rm{10}}}}} \right)}^3}} \right]
= Rs \left( {18000 \times {{11} \over {10}} \times {{11} \over {10}}} \right)
= Rs 21,780
Calculation for next 6 months:
S.I. = Rs \left[ {{{21780 \times {1 \over 2} \times 10} \over {100}}} \right]= Rs 1089
Therefore, interest for the first two years = Rs (21780 - 18000) = Rs 3780
And interest for next {1 \over 2}year = Rs 1,089
Hence, total C.I. = Rs 18000 + Rs 4869 = Rs 22,869

(c) Here, principal (P) = Rs 62,500, Rate(R) = 8% per annum or 4% per half year, Number of years (n) = 1{1 \over 2}.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= Rs \left[ {62500{{\left( {{\rm{1 + }}{4 \over {{\rm{100}}}}} \right)}^3}} \right]
= Rs \left( {62500 \times {{26} \over {25}} \times {{26} \over {25}} \times {{26} \over {25}}} \right)
= Rs 70304
Compound Interest, C.I. = A – P = Rs (70304 – 62500) = Rs 7,804

(d) Here, principal (P) = Rs 8,000, Rate(R) = 9% per annum or {9 \over 2}% per half year, Number of years (n) = 1.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= Rs \left[ {8000{{\left( {{\rm{1 + }}{9 \over {{\rm{200}}}}} \right)}^2}} \right]
= Rs \left[ {8000{{\left( {{{209} \over {{\rm{200}}}}} \right)}^2}} \right]
= Rs 8736.20
Compound Interest, C.I. = A – P = Rs (8736.20 – 8000) = Rs 736.20

(e) Here, principal (P) = Rs 10,000, Rate(R) = 8% per annum or 4% per half year, Number of years (n) =1.
We know that, there are 2 half years in 1 year.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= Rs \left[ {10000{{\left( {{\rm{1 + }}{4 \over {{\rm{100}}}}} \right)}^2}} \right]
= Rs \left( {10000 \times {{26} \over {25}} \times {{26} \over {25}}} \right)
= Rs 10,816
Compound Interest, C.I. = A – P = Rs (10816 – 10000) = Rs 816

Q.2 Kamala borrowed Rs 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 \over {12}}years).
Sol. Here, principal (P) = Rs 26,400, Rate(R) = 15% per annum, Number of years (n) =2{4 \over {12}}years.
Calculation for first 2 years:
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= Rs \left[ {26400{{\left( {{\rm{1 + }}{{15} \over {{\rm{100}}}}} \right)}^2}} \right]
= Rs \left( {26400 \times {{23} \over {20}} \times {{23} \over {20}}} \right)
= Rs 34,914
Calculation for next {1 \over 3} years:
S.I. = Rs\left( {{{34914 \times {1 \over 3} \times 15} \over {100}}} \right)= Rs 1,745.70
Now, interest for the first two years = Rs (34914 – 26400) = Rs 8,514
And interest for the next {1 \over 3}year = Rs 1,745.70
Total C.I. = Rs (8514 + 1745.70) = Rs 10,259
Hence, Amount = P + C.I. = Rs 26400 + Rs 10259.70 = Rs 36,659.70

Q.3 Fabina borrows Rs 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?
Sol. Here, principal (P) = Rs 12,500, Rate(R) = 12% per annum, Number of years (n) =3 years.
S. I. for Fabina = {{{\rm{P \times R \times T}}} \over {{\rm{100}}}}= {{{\rm{12500 \times 12 \times 3}}} \over {{\rm{100}}}}= Rs 4500
For Radha:
Principal (P) = Rs 12,500, Rate(R) = 10% per annum, Number of years (n) =3 years
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= \left[ {12500{{\left( {{\rm{1 + }}{{10} \over {{\rm{100}}}}} \right)}^3}} \right]
= \left[ {12500{{\left( {{\rm{1 + }}{1 \over {{\rm{10}}}}} \right)}^3}} \right]
= \left( {12500 \times {{11} \over {10}} \times {{11} \over {10}} \times {{11} \over {10}}} \right)
= Rs 16,637.50
Compound Interest, C.I. = A – P = Rs (16637.50 – 12500) =Rs 4,137.50
The difference between Fabina’s and Radha’s interest = Rs (4500 – 4137.50) = Rs 362.50
Therefore, Fabina will have to pay Rs 362.50 more.

Q.4 I borrowed Rs 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?
Sol. Here, principal (P) = Rs 12,000, Rate(R) = 6% per annum, Number of years (n) =2 years.
S. I. = {{{\rm{P \times R \times T}}} \over {{\rm{100}}}}= {{{\rm{12000 \times 6 \times 2}}} \over {{\rm{100}}}}= Rs 1,440
Calculation for Compound Interest (C.I.):
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
= \left[ {12000{{\left( {{\rm{1 + }}{6 \over {{\rm{100}}}}} \right)}^3}} \right]
= \left[ {12000{{\left( {{\rm{1 + }}{3 \over {{\rm{50}}}}} \right)}^3}} \right]
= \left( {12000 \times {{53} \over {50}} \times {{53} \over {50}}} \right)
= Rs 13,483.20
Therefore, Compound Interest = A – P = Rs 13483.20 – Rs 12000 = Rs 1,483.20
Now, C.I. – S.I. = Rs 1,483.20 – Rs 1,440 = Rs 43.20
Therefore, extra amount to be paid is Rs 43.20

Q.5 Vasudevan invested Rs 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?
Sol. (i) Here, principal (P) = Rs 60,000, Rate(R) = 12% per annum = 6% per half year, Number of years (n) =6 months = 1 half year.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
=Rs \left[ {60000{{\left( {{\rm{1 + }}{6 \over {{\rm{100}}}}} \right)}^3}} \right]
= Rs \left( {60000 \times {{106} \over {100}}} \right)
= Rs 63,600

(ii) We know that, 2 half years in 1 year. Therefore, n = 2.
A =Rs \left[ {60000{{\left( {{\rm{1 + }}{6 \over {{\rm{100}}}}} \right)}^2}} \right]
= Rs \left( {60000 \times {{106} \over {100}} \times {{106} \over {100}}} \right)
= Rs 67,416

Q.6 Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1{1 \over 2} years if the interest is
(i) compounded annually.
(ii) compounded half yearly.
Sol. (i) Here, principal (P) = Rs 80,000, Rate(R) = 10% per annum, Number of years (n) =6 months = 1{1 \over 2} half year.
Calculation for 1st year:
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
=Rs \left[ {80000{{\left( {{\rm{1 + }}{{10} \over {{\rm{100}}}}} \right)}^1}} \right]
= Rs \left( {80000 \times {{11} \over {10}}} \right)
= Rs 88,000
Calculation for next {1 \over 2}year taking Rs 88,000 as principal:
S.I. = {{{\rm{P \times R \times T}}} \over {{\rm{100}}}}= Rs {{{\rm{88000 \times 10 \times }}{1 \over 2}} \over {{\rm{100}}}}= Rs 4,400
Now, interest for next first year = Rs 88,000 – Rs 80,000 = Rs 8,000
And interest for next {1 \over 2}year = Rs 4,400
Here, C.I. = Rs 8,000 + Rs 4,4000 = Rs 12,400
Hence, Total C.I. = Rs (80000 + 12400) = Rs 92,400

(ii) For the interest compounded half yearly:
Rate(R) = 10% per annum = 5% per half yearly.
There will be three half years in 1{1 \over 2}years.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
A = Rs \left[ {80000{{\left( {{\rm{1 + }}{5 \over {{\rm{100}}}}} \right)}^3}} \right]
= Rs \left( {80000 \times {{21} \over {20}} \times {{21} \over {20}} \times {{21} \over {20}}} \right)
= Rs 92,610
Thus, the difference between amounts = Rs 92,610 – Rs 92,400 = Rs 210

Q.7 Maria invested Rs 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
Sol. (i) Here, principal (P) = Rs 8,000, Rate(R) = 5% per annum, Number of years (n) =2 years.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
A = Rs \left[ {8000{{\left( {{\rm{1 + }}{5 \over {{\rm{100}}}}} \right)}^2}} \right]
= Rs \left( {8000 \times {{21} \over {20}} \times {{21} \over {20}}} \right)
= Rs 8,820

(ii) Now, calculating S.I. for next year taking Rs 8,820
S.I. = Rs {{{\rm{8820 \times 5 \times 1}}} \over {{\rm{100}}}}= Rs 441

Q.8 Find the amount and the compound interest on Rs 10,000 for1{1 \over 2} years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
Sol. Here, principal (P) = Rs 10,000, Rate(R) = 10% per annum = 5% per half year, Number of years (n) = 1{1 \over 2}years.
In 1{1 \over 2} there will be 3 half years.
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
A = Rs \left[ {10000{{\left( {{\rm{1 + }}{5 \over {{\rm{100}}}}} \right)}^3}} \right]
= Rs \left( {10000 \times {{21} \over {20}} \times {{21} \over {20}} \times {{21} \over {20}}} \right)
= Rs 11,576.25
Now, Compound Interest C.I. = A – P
= Rs 11576.25 – Rs 10000
= Rs 1,576.25
Now, calculating amount for the first year
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
A = Rs \left[ {10000{{\left( {{\rm{1 + }}{{10} \over {{\rm{100}}}}} \right)}^1}} \right]
= Rs \left( {10000 \times {{11} \over {10}}} \right)
= Rs 11,000
Now, taking Rs 11,000 as principal, S.I. for next half year will be:
S.I. = \left( {{{11000 \times 10 \times {1 \over 2}} \over {100}}} \right)
= Rs 550
Hence, interest for the first year = Rs 11000 – Rs 10000 = Rs 1000
Therefore, Compound Interest C.I. = Rs 1000 + Rs 550 = Rs 1,550
Thus, the interest would be more when compounded half year as compared to the interest when compounded annually.

Q.9 Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12{1 \over 2}% 2 per annum, interest being compounded half yearly.
Sol. Here, principal (P) = Rs 4,096, Rate(R) = 12{1 \over 2}% per annum = {{25} \over 4}% per half year, Number of months = 18.
In 18 months, there will be 3 half years, therefore,
Amount, A = {\rm{P}}{\left( {{\rm{1 + }}{{\rm{R}} \over {{\rm{100}}}}} \right)^{\rm{n}}}
A = Rs \left[ {4096{{\left( {{\rm{1 + }}{{25} \over {{\rm{400}}}}} \right)}^3}} \right]
= Rs \left( {4096 \times {{17} \over {16}} \times {{17} \over {16}} \times {{17} \over {16}}} \right)
= Rs 4,913
Therefore, the required amount is Rs 4,913.

Q.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?
Sol. (i) Given, population in 2003 = 54,000.
Hence, 54000 = Population in 2001 x {\left( {{\rm{1 + }}{5 \over {{\rm{100}}}}} \right)^2}
Population in 2001= \left( {54000 \times {{20} \over {21}} \times {{20} \over {21}}} \right)
= Rs 48979.59
Hence, the population in the year 2001 was approximately 48,980.

(ii) Population in 2005 = 54000 x {\left( {{\rm{1 + }}{5 \over {{\rm{100}}}}} \right)^2}
= \left( {54000 \times {{21} \over {20}} \times {{21} \over {20}}} \right)
= 59,535
Hence, the population in the year 2005 would be 59,535.

Q.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.
Sol. Given, the initial count of bacteria is 5,06,000.
Now, bacteria at the end of 2 hours = 506000 x {\left( {{\rm{1 + }}{{2.5} \over {{\rm{100}}}}} \right)^2}
= \left( {506000 \times {{41} \over {40}} \times {{41} \over {40}}} \right)
= 5,31,616.25
Hence, the count of bacteria at the end of 2 hours will be approximately 5,31,616

Q.12 A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Sol. Here, principal (P) = Rs 42,000, rate = 8% per annum
Depreciation = *% of Rs 42,000 per year
= Rs \left( {{{42000 \times 8 \times 1} \over {100}}} \right)
= Rs 3,360
Therefore, value after one year = Rs 42000 – Rs 3360 = Rs 38,640



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