Installment Example-1

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Home > Word Problems calculators > Installment Word Problems example

Installment examples ( Enter your problem )

( Enter your problem )

2. HCF-LCM Word Problem
(Next method)

1. Example-1

Problem : 1 / 8 [ Installment ] Enter your problem 1. A briefcase is available for Rs 800 cash or for Rs 500 cash down payment and Rs 320 to be paid after 6 months. Find the rate of interest charged under the installment plan.


Solution:	Cash Price = `800` Down Payemnt = `500` Remaining Balance = `800 - 500 = 300` The installment to be paid at the end of `6` months = `320` `:.` the interest charged on Rs `300`, for a period of `6` months = Rs `320` - Rs `300 = 20` If R % is the rate of interest per annum, then `(300 × R × 6) / (100 × 12) = 20` `R = 13.3333` Thus, the rate of interest charged under the installment paln is = `13.3333` per annum

Problem : 2 / 8 [ Installment ] Enter your problem 2. A bicycle is sold for Rs 1800 cash or for Rs 600 cash down payment followed by 2 monthly installmnets of Rs 610 each. Compute the rate of the interest charged under the installment scheme.


Solution:	Cash Price = `1800` Down Payemnt = `600` Number of Installment = `2` Installment Amount = `610` Remaining Balance = `1800 - 600 = 1200` After `2` months Amount = `1200 + 1200 × 2 / 12 × R / 100 = 1200 + (2400 × R) / 1200 ` ` = 1200 + 2 × R ->(1)` `1^(st)` Installment = `610 + (610 × R × 1) / (12 × 100)` Total Installment `= 610 × 2 + (610 × R × (1)) / (12 × 100)` ` = 1220 + (610 × R × (1)) / 1200` ` = 1220 + 61/120 × R ->(2)` From (1) and (2) `1200 + 2 × R = 1220 + 61/120 × R` `2 × R - 61/120 × R = 1220 - 1200` `179/120 × R = 20` `R = 2400/179` `:. R = 2400/179` OR METHOD Cash Price = `1800` Down Payemnt = `600` Number of Installment = `2` Installment Amount = `610` Remaining Balance = `1800 - 600 = 1200` Total Installment = `610 × 2 = 1220` Interest = `1220 - 1200 = 20 ->(1)` Principal for `1^(st)` month = `1200` Principal for `2^(nd)` month = `590` Total = `1790` `:.` Installment = `(1790 × R × 1) / (100 × 12) ->(2)` From (1) and (2) `(1790 × R × 1) / (100 × 12) = 20` `R = (20 × 100 × 12) / 1790` `R = 2400/179`

Problem : 3 / 8 [ Installment ] Enter your problem 3. A washing machine is available at Rs 6400 cash or for Rs 1400 cash down payment and 5 monthly installments of Rs 1030 each. Calculate the rate of interest charged under the instalment plan.


Solution:	Cash Price = `6400` Down Payemnt = `1400` Number of Installment = `5` Installment Amount = `1030` Remaining Balance = `6400 - 1400 = 5000` After `5` months Amount = `5000 + 5000 × 5 / 12 × R / 100 = 5000 + (25000 × R) / 1200 ` ` = 5000 + 125/6 × R ->(1)` `1^(st)` Installment = `1030 + (1030 × R × 4) / (12 × 100)` `2^(nd)` Installment = `1030 + (1030 × R × 3) / (12 × 100)` `3^(rd)` Installment = `1030 + (1030 × R × 2) / (12 × 100)` `4^(th)` Installment = `1030 + (1030 × R × 1) / (12 × 100)` Total Installment `= 1030 × 5 + (1030 × R × (4 + 3 + 2 + 1)) / (12 × 100)` ` = 5150 + (1030 × R × (10)) / 1200` ` = 5150 + 103/12 × R ->(2)` From (1) and (2) `5000 + 125/6 × R = 5150 + 103/12 × R` `125/6 × R - 103/12 × R = 5150 - 5000` `49/4 × R = 150` `R = 600/49` `:. R = 600/49` OR METHOD Cash Price = `6400` Down Payemnt = `1400` Number of Installment = `5` Installment Amount = `1030` Remaining Balance = `6400 - 1400 = 5000` Total Installment = `1030 × 5 = 5150` Interest = `5150 - 5000 = 150 ->(1)` Principal for `1^(st)` month = `5000` Principal for `2^(nd)` month = `3970` Principal for `3^(rd)` month = `2940` Principal for `4^(th)` month = `1910` Principal for `5^(th)` month = `880` Total = `14700` `:.` Installment = `(14700 × R × 1) / (100 × 12) ->(2)` From (1) and (2) `(14700 × R × 1) / (100 × 12) = 150` `R = (150 × 100 × 12) / 14700` `R = 600/49`

Problem : 4 / 8 [ Installment ] Enter your problem 4. A computer is sold by a company for Rs 19200 cash or for Rs 4800 cash down payment together with 5 equal monthly installments. If the rate of interest charged by the company is 12 % per annum, find each installment.


Solution:	Cash Price = `19200` Down Payemnt = `4800` Number of Installment = `5` Rate = `12` Remaining Balance = `19200 - 4800 = 14400` At the end of `5` months, Amounts to = `14400 + (14400 × 5 × 12) /(12 × 100) = 14400 + 720` ` = 15120 ->(1)` Let Installment = `X` `1^(st)` Installment `= X + (12 × X × 4) / (12 × 100)` `2^(nd)` Installment `= X + (12 × X × 3) / (12 × 100)` `3^(rd)` Installment `= X + (12 × X × 2) / (12 × 100)` `4^(th)` Installment `= X + (12 × X × 1) / (12 × 100)` Total Installment `= X + ( X + (12 × X × 4) / (12 × 100) ) + ( X + (12 × X × 3) / (12 × 100) ) + ( X + (12 × X × 2) / (12 × 100) ) + ( X + (12 × X × 1) / (12 × 100) )` Total Installment `= 5 X + (12 × X ×(4 + 3 + 2 + 1))/ (12 × 100)` Total Installment `= 5 X + 1/10 × X` Total Installment `= 51/10 X ->(2)` From (1) and (2) `51/10 X = 15120` `X = (15120) / (51/10)` `X = 50400/17` `:.` Installment Amount = `50400/17` OR METHOD Cash Price = `19200` Down Payemnt = `4800` Number of Installment = `5` Rate = `12` Remaining Balance = `19200 - 4800 = 14400` Let Installment = `X` `:.` Interest Paid = `5 X - 14400 ->(1)` Principal for `1^(st)` month = `14400` Principal for `2^(nd)` month = `14400 - 1 X` Principal for `3^(rd)` month = `14400 - 2 X` Principal for `4^(th)` month = `14400 - 3 X` Principal for `5^(th)` month = `14400 - 4 X` Total `= 72000 - 10 X` `:.` Interest `= ((72000 - 10 X) × 12 × 1) / (100 × 12) ->(2)` `:.` From (1) and (2) `((72000 - 10 X) × 12 × 1) / (100 × 12) = 5 X - 14400` `(72000 - 10 X) = 500 X - 1440000` `500 X + 10 X = 72000 + 1440000` `510 X = 1512000` `:. X = 50400/17`

Problem : 5 / 8 [ Installment ] Enter your problem 5. A State Government announces sale of flats of Rs 555000 cash or Rs 42750 cash down payment and 3 equal Yearly installments. The Government also charges a nominal interest at the rate of 8 % per annum copuounded as installment plan in this welfare scheme. If one purchases a flat under this installment plan, calculate the value of each installment.


Solution:	Cash Price = `555000` Down Payemnt = `42750` Number of Installment = `3` Rate = `8` Remaining Balance = `555000 - 42750 = 512250` Let Installment = `X` Now present value of amount of X Rs paid at the end of 1st Year ` = X / ( 1 + 8/100 )` ` = X × 25/27` Similarly, for `2^(nd)` Year = `X × (25/27)^2` Similarly, for `3^(rd)` Year = `X × (25/27)^3` `:.` The total of the present values of the `3` Installment is Rs `( X × 25/27 + X × (25/27)^2 + X × (25/27)^3 )` and this must be equal to `512250` `:. ( X × 25/27 + X × (25/27)^2 + X × (25/27)^3 ) = 512250` `:. X ( 50725/19683 ) = 512250` `X = 403304670/2029` `:.` Installment Amount = `403304670/2029` Total Amount Paid `= 403304670/2029 × 3 = 1209914010/2029` Interest = Total Paid - Balance Amount `= 1209914010/2029 - 512250 = 170558760/2029`

Problem : 6 / 8 [ Installment ] Enter your problem 6. A TV set is available for Rs 19650 cash payment or for Rs 3100 cash down payment and 3 equal Yearly installments. If the shopkeeper charges interest at the rate of 10 % per annum compounded as installment plan, calculate the amount of each installment.


Solution:	Cash Price = `19650` Down Payemnt = `3100` Number of Installment = `3` Rate = `10` Remaining Balance = `19650 - 3100 = 16550` Let Installment = `X` Now present value of amount of X Rs paid at the end of 1st Year ` = X / ( 1 + 10/100 )` ` = X × 10/11` Similarly, for `2^(nd)` Year = `X × (10/11)^2` Similarly, for `3^(rd)` Year = `X × (10/11)^3` `:.` The total of the present values of the `3` Installment is Rs `( X × 10/11 + X × (10/11)^2 + X × (10/11)^3 )` and this must be equal to `16550` `:. ( X × 10/11 + X × (10/11)^2 + X × (10/11)^3 ) = 16550` `:. X ( 3310/1331 ) = 16550` `X = 6655` `:.` Installment Amount = `6655` Total Amount Paid `= 6655 × 3 = 19965` Interest = Total Paid - Balance Amount `= 19965 - 16550 = 3415`

Problem : 7 / 8 [ Installment ] Enter your problem 7. A man borrows money from a finance company and has to pay it back in 2 equal Half Yearly installments of Rs 4945 each. If the interest is charged by the finance company at the rate of 15 % per annum compounded as installment plan, find the principal and the total interest paid.


Solution:	InstallAmt = `4945` Number of Installment = `2` Rate = `7.5` The amount of `4945` Rs paid at the end of 1st Year has its principal equal to ` = 4945 / ( 1 + 7.5 / 100 )` ` = 4945 / ( 107.5 / 100 )` ` = 4945 × 0.9302` Similarly, for `2^(nd)` Year = `4945 × (0.9302)^2` `:.` The total of the present values of the `2` Installment is Rs `( 4945 × 0.9302 + 4945 × (0.9302)^2 )` ` = 8879.0698` Total Amount Paid `= 4945 × 2 = 9890` Interest = Total Paid - Balance Amount `= 9890 - 8879.0698 = 1010.9302`

Problem : 8 / 8 [ Installment ] Enter your problem 8. Ram borrowed a sum of money and returned it in 3 equal Quarterly installments of Rs 17576 each. Find the sum borrowed, if the rate of interest charged was 16 % per annum compounded as installment plan. Find also the total interest charged.


Solution:	InstallAmt = `17576` Number of Installment = `3` Rate = `4` The amount of `17576` Rs paid at the end of 1st Year has its principal equal to ` = 17576 / ( 1 + 4 / 100 )` ` = 17576 / ( 104 / 100 )` ` = 17576 × 0.9615` Similarly, for `2^(nd)` Year = `17576 × (0.9615)^2` Similarly, for `3^(rd)` Year = `17576 × (0.9615)^3` `:.` The total of the present values of the `3` Installment is Rs `( 17576 × 0.9615 + 17576 × (0.9615)^2 + 17576 × (0.9615)^3 )` ` = 48775` Total Amount Paid `= 17576 × 3 = 52728` Interest = Total Paid - Balance Amount `= 52728 - 48775 = 3953`

2. HCF-LCM Word Problem
(Next method)

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