### Comparing Quantities Using Proportion Class 8th Mathematics AP Board Solution

##### Question 1.Sudhakar borrows Rs. 15000 from a bank to renovate his house. He borrows the money at 9% p.a. simple interest over 8 years. What are his monthly repayments?Answer:Principal (P) = Rs. 15000Time period (T) = 8Rate of interest (R) = 9%  I = Rs. 10800∴ Interest for 8 years is Rs. 10800Amount to be paid at the end of 8 years = Principal + interestAmount = 15000+10800= 25800∴ Amount to be paid at the end of 8 years is Rs. 25800 ⇒ monthly repayment = = 268.75∴ Sudhakar pays Rs. 268.75 monthly.Question 2.A TV was bought at a price of Rs. 21000. After 1 year the value of the TV was depreciated by 5% (Depreciation means reduction of the value due to use and age of the item). Find the value of the TV after 1 year.Answer:Cost price of TV = Rs. 21000Depreciation = 5%Depreciation after 1 year = 5% of 21000 ∴ Depreciation after 1 year = Rs. 1050Value of TV after 1 year = cost price – depreciation= 21000 – 1050= Rs. 19950∴The value of TV after 1 year is Rs. 19,950Question 3.Find the amount and the compound interest on Rs. 8000 at 5% per annum, for 2 years compounded annually.Answer:Principal (P) = Rs. 8000Time period (n) = 2Rate of interest (R) = 5%  = 8000×1.052= 8820∴ Amount = Rs. 8820  = (8000×1.052)-8000= 8820-8000= 820∴ Compound Interest = Rs. 820Question 4.Find the amount and the compound interest on Rs. 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.Answer:Principal (P) for 1st year = Rs. 6500Rate of interest (R) for first year = 5%Interest on first year = 5% of 6500= = 325∴ Interest for 1st year = Rs. 325Principal (P) for second year = Principal (P) for 1st year + Interest for 1st yearPrincipal (P) for second year = Rs. 6500 + Rs. 325 = Rs. 6825Rate of interest (R) for second year = 6%Interest on second year = 6% of 6825= = 409.5∴ Interest for 2nd year = Rs. 409.5Total interest = Rs. 325+Rs. 409.5 = Rs. 734.5Amount at second year = Principal (P) for 2nd year + Interest for 2nd year= 6825+409.5= 7234.5∴ Amount at second year = Rs. 7234.5Question 5.Prathibha borrows Rs. 47000 from a finance company to buy her first car. The rate of simple interest is 17% and she borrows the money over a 5 year period.Find:(a) How much amount Prathibha should repay the finance company at the end of five years.(b) her equal monthly repayments.Answer:(a) Principal (P) = Rs. 47000Time period (T) = 5Rate of interest (R) = 17%  I = Rs. 39950∴ Interest for 5 years is Rs. 39950Amount at the end of 5 years = Principal + interestAmount = 47000+39950= 86950∴ Amount at the end of 5 years is Rs. 86950(b) ⇒ monthly repayment = = 1449.17∴ prathibha’s monthly repayment isRs. 1449.17.Question 6.The population of Hyderabad was 68,09,000 in the year 2011. If it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015.Answer:The population of Hyderabad (P) = 68,09,000Time period (n) = 2015-2011 = 4Rate of interest (R) = 4.7%  = 6809000×1.0474= 81821994∴ The population of Hyderabad at the end of 2015 = 8,18,21,994Question 7.Find Compound interest paid when a sum of Rs. 10000 is invested for 1 year and 3 months at 8 % per annum compounded annually.Answer:Principal (P) = Rs. 10000Rate of interest (R) = 8.5%Time period (T) = 1year 3 monthsFor T = 1 year,  I = Rs. 850∴ Interest for 1 year is Rs. 850Amount at the end of 1 year = Principal + interestAmount = 10000+850= 10850∴ Amount at the end of 1 year is Rs. 10850For T = 3 months =   I = Rs. 230.56∴ Interest for year is Rs. 230.56Amount at the end of year = Principal + interestAmount = 10850+230.56= 11080.56∴ Amount at the end of year is Rs. 11080.56Total interest for 1.3 year = Interest for 1 year + Interest for year= 850+230.56= 1080.56∴ The compound interest paid is Rs. 1080.56Question 8.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 years, if the interest is compounded annually and compounded half yearly.Answer:For compounded AnnuallyPrincipal (P) = Rs. 80000Time period (n) = 1Rate of interest (R) = 10%  = 80000×1.053= 88000∴ Amount for 1 year = Rs. 88000Interest for remaining 6 months = = 4400∴ Amount for 1.5 years = Rs. 88000+4400 = Rs. 92400 = 92400-80000∴ Compound Interest = Rs. 12400For compounded half yearlyPrincipal (P) = Rs. 80000Time period (n) = 3Rate of interest (R) for half year = 10%× = 5%  = 80000×1.053= 92610∴ Amount = Rs. 92610 = 92610-80000∴ Compound Interest = Rs. 12610∴ the difference in amounts = amount for For compounded half yearly - amount for For compounded annually= 92610-92400= Rs. 210∴ The difference in amounts is Rs. 210Question 9.I borrowed Rs. 12000 from Prasad at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compounded annually, what extra amount would I have to pay?Answer:Principal (P) = Rs. 12000Rate of interest (R) = 6%Time period (T) = 2years  I = Rs. 1440∴ Interest for 2 years is Rs. 1440This sum to be borrowed at 6% per annum compounded annually,Principal (P) = Rs. 12000Rate of interest (R) = 6%Time period (n) = 2years  = 12000×1.062= 13483.2∴ Amount = Rs. 13483.2 = 13483.2-12000∴ Compound Interest = Rs. 1483.2∴ The difference in interest = 1483.2 - 1440= Rs. 43.2∴ The difference in interest is Rs. 43.2Question 10.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, 06,000Answer:Principal (P) = 506000Rate of interest (R) = 2.5%Time period (n) = 2hours  = 506000×1.0252= 531616.25∴ The number of the bacteria at the end of 2 hours is 531616(approximately)Question 11.Kamala borrowed Rs. 26400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?Answer:Principal (P) = 26400Rate of interest (R) = 15%Time period (T) = 2 years and 4 monthsAmount for 2 years,  = 26400×1.152= Rs. 34914∴ Amount for 2 year = Rs. 34914Interest for remaining 4 months = = 1745.70∴ Total amount for 2 years and 4 months = Rs. 34914 + 1745.70= Rs. 36659.70∴ The total amount to clear the loan is Rs. 36659.70Question 12.Bharathi borrows an amount of Rs. 12500 at 12% per annum for 3 years at a simple interest and Madhuri borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?Answer:Principal (P) = Rs. 12500Rate of interest (R) = 12%Time period (T) = 3yearsInterest paid by bharathi,  I = Rs. 4500∴ Interest paid by bharathi is Rs. 4500Amount paid by madhuri,Principal (P) = Rs. 12500Rate of interest (R) = 10%Time period (n) = 3years  = 12500×1.13= Rs. 16637.5∴ Amount paid by madhuri isRs. 16637.5Interest = A-P= Rs. 16637.5 – 12500= Rs. 4137.5∴ Interest paid by madhuri is Rs. 4137.5On comparing the interests paid by bharathi and madhuri,4500-4137.5 = 362.5∴ Bharathi paid Rs.362.5 more than by madhuri.Question 13.Machinery worth Rs. 10000 depreciated by 5%. Find its value after 1 year.Answer:Principal (P) = 10000Depreciation (R) = 5%Time period (n) = 1 year   = Rs. 9500∴The value of machinery after 1 year is Rs. 9500Question 14.Find the population of a city after 2 years which is at present 12 lakh, if the rate of increase is 4%.Answer:Present population (P) = 12 lakhRate of interest (R) = 4%Time period (n) = 2years  = 1200000×1.042= 1297920∴ The population of a city after 2 years is 1297920Question 15.Calculate compound interest on Rs. 1000 over a period of 1 year at 10% per annum, if interest is compounded quarterly?Answer:Principal (P) = 1000Rate of interest (R) = 10%Time period (n) = 1 yearFor quarterly, n = 4Rate of interest (R) for quarterly =    = 1000 = 1103.81∴ Amount = Rs. 1103.81Interest = A-P= 1103.81-1000= 103.81∴ compound interest = Rs. 103.81

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