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 mobile is designed for ?2500 or ?520 down re payment followed closely 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: this is certainly one question that is basic. You need to simply make use of the formula that is above determine the total amount of installment.

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

Here P = 2500 – 520 = 1980

Ergo, 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

Just 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’ll utilize the formula that is following

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 exactly What amount he’d lent in the event that interest rate ended up being 16 p.a. Compounded quarterly?

Installments on Simple Interest and Compound Interest Sol: in this situation, we’ll make use of current value technique once we have to discover the initial sum 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 the bank at 10per cent p.a. Easy interest and clears your debt in 5 years. In the event that installments compensated at the conclusion associated with very first, 2nd, 3rd and 4th years to clear the debt are ?10,000, ?20,000, ?30,000 and ?40,000 correspondingly, just just what quantity ought to be compensated at the conclusion for the year that is fifth clear your debt?

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

Consequently, only interest component needs to be compensated into the installment that is last.

Ergo, Interest when it comes to very first 12 months = (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 for the year that is fourth (100000 – 10000 – 20000 – 30000) * 10/ 100 = ?4000

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

Installments on Simple Interest and Compound Interest Case 5: a sum of ?12820 due in three years, thus is completely paid back in three yearly installments beginning after one year. The initial installment is ? the next installment in addition to 2nd installment is 2 /3 regarding the installment that is third. If interest rate is 10% p.a. Get the very first installment.

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

Since, 2nd installment is 2 /3 associated with 3rd, it should be 2 /3x. And lastly, 1 st installment would be ? * 2 /3 *x

Now continuing into the fashion that is similar we did early in the day and making use of the element interest formula to calculate 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 title loans interest rate florida (?(1.1) 2 + ?(1.1) + 1)

­­­17063.42 = x(0.40333 + 0.55 + 1)

17063.42 = x* 1.953333

­­­­X = ?8735.53