# Installments on Simple Interest and Compound Interest Case

Installments on Simple Interest and Compound Interest Case

## Miscellaneous situations 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, calculate the installment.

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

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

Right 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

Just just What annual 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 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 calculate 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 had been 16 p.a. Compounded quarterly?

Installments on Simple Interest and Compound Interest Sol: in this situation, we shall utilize value that is present once we want to get 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 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 of this very first, second, third and 4th years to clear your debt are ?10,000, ?20,000, ?30,000 and ?40,000 correspondingly, just exactly exactly what quantity should really be compensated at the conclusion associated with the 5th 12 months to 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 paid into the final installment.

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

Hence, Amount that need to paid into the 5th installment = (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 12 months. The initial installment is ? the next installment and also the 2nd installment is 2 /3 of this installment that is third online installment loans. If interest rate is 10% p.a. Get the installment that is first.

## Installments on Simple Interest and Compound Interest Sol: allow 3rd installment be x.

Since, 2nd installment is 2 /3 associated with 3rd, it’ll be 2 /3x. Last but not least, 1 st installment are going to be ? * 2 /3 *x

Now continuing into the fashion that is similar we did earlier in the day and making use of the element 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)

17063.42 = x* 1.953333

­­­­X = ?8735.53