Installments on Simple Interest and Compound Interest Case

Installments on Simple Interest and Compound Interest Case

Miscellaneous instances of Installments on Simple Interest and Compound Interest

: To determine the installment whenever interest is charged on SI

A phone that is mobile readily available for ?2500 or ?520 down re payment accompanied by 4 month-to-month equal installments. In the event that interest rate is 24%p.a. SI, determine the installment.

Installments on Simple Interest and Compound Interest Sol: that is one question that is basic. You must simply make use of the formula that is above determine the actual quantity of installment.

Consequently, x = P (1 + nr/100)/ (n + n(n-1)/2 * r/100))

Right Here P = 2500 – 520 = 1980

Thus, x = 1980(1 + 15 * 12/ 1200)/ (4 + 4* 3* 12/ 2 * 12 * 100)

= ?520

Installments on Simple Interest and Compound Interest Case 2: To determine the installment whenever interest is charged on CI

Exactly What yearly repayment will discharge a financial obligation of ?7620 due in 36 months at 16 2/3% p.a. Compounded interest?

Installments on Simple Interest and Compound Interest Sol: once again, we're going to utilize the following formula,

P (1 + r/100) n = X (1 + r/100) n-1 + X (1 + r/100) n-2 + X (1 + r/100) n-3 +…. + X (1 + r/100)

7620(1+ 50/300) 3 = x (1 + 50/300) 2 + x (1 + 50/300) + x

12100.2778 = x (1.36111 + 1.1667 + 1)

X = ?3430

Installments on Simple Interest and Compound Interest Case 3: To determine loan quantity whenever interest charged is Compound Interest

Ram borrowed cash and came back it in 3 equal quarterly installments of ?17576 each. Just just just What amount he'd lent if the interest rate ended up being 16 p.a. Compounded quarterly?

Installments on Simple Interest and Compound Interest Sol: in this instance, we're going to make use of present value technique even as we have to discover the initial amount lent by Ram.

Since, P = X/ (1 + r/100) n ………X/ (1 + r/100) 2 + X/ (1 + r/100)

Consequently, P = 17576/ (1 + 4/100) 3 + 17576/ (1 + 4/100) 2 + 17576/ (1 + 4/100)

= 17576 (0.8889 + 0.92455 + 0.96153)

= 17576 * 2774988

= 48773.1972

Installments on Simple Interest and Compound Interest Case 4: Gopal borrows ?1,00,000 from a bank at 10per cent p.a. Easy interest and clears your debt in 5 years. In the event that installments compensated at the conclusion of the very first, 2nd, 3rd and 4th years to clear your debt are ?10,000, ?20,000, ?30,000 and ?40,000 correspondingly, exactly exactly just what quantity must certanly be compensated by the end for the fifth 12 months to clear your debt?

Installments on Simple Interest and Compound Interest Sol: Total principal amount left after 5 th 12 months = 100000 – (10000 + 20000 + 30000 + 40000) = 100000 – 100000 = 0

Consequently, only interest component has got to be paid within the final installment.

Ergo, Interest when it comes to year that is first (100000 * 10 * 1) /100 =?10000

Interest for the year that is second (100000 – 10000) * 10/ 100 = ?9000

Interest when it comes to year that is third (100000 – 10000 – 20000) * 10/ 100 = ?7000

Interest when it comes to year that is fourth (100000 – 10000 – 20000 – 30000) * 10/ 100 = ?4000

Hence, Amount that need to paid into the installment that is fifth (10000 + 9000 + 7000 + 4000) = ?30000

Installments on Simple Interest and Compound Interest Case 5: a quantity of ?12820 due in three years, thus is completely paid back in three annual installments beginning after 12 months. 1st installment is ? the next installment in addition to 2nd installment is 2 /3 regarding the 3rd installment. If interest is 10% p.a. Get the installment that is first.

Installments on Simple Interest and Compound Interest Sol: allow installment that is third x.

Since, 2nd installment is 2 /3 associated with the 3rd, it should be 2 /3x. Last but not least, 1 st installment should be ? * 2 /3 *x

Now proceeding within the comparable fashion as we did early in the day and utilising the ingredient interest formula to determine the installment quantity.

P (1 + r/100) n = X (1 + r/100) n-1 + X (1 + r/100) n-2 + X (1 + r/100) n-3 +…. + X (1 + r/100)

12820 (1 + 10/100) 3 = ?X (1 + 10/100) 2 + ?X (1 + 10/100) 1 + X

12820(1.1) 3 = x (?(1.1) 2 + ?(1.1) + 1)

­­­17063.42 = x(0.40333 + 0.55 + 1)

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17063.42 = x* 1.953333

­­­­X = ?8735.53

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