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

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

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

Right 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 calculate the installment whenever interest is charged on CI

Just just just What yearly repayment will discharge a debt of ?7620 due in three years at 16 2/3% p.a. Compounded interest?

Installments on Simple Interest and Compound Interest Sol: once again, we’ll utilize the after 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 calculate loan quantity whenever interest charged is Compound Interest

Ram borrowed cash and came back it in 3 equal quarterly installments of ?17576 each. Exactly exactly just 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 cases like this, we’re going to utilize value that is present even as we have 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 a bank at 10% p.a. Easy interest and clears your debt in 5 years. In the event that installments compensated by the end of this initial, second, 3rd and 4th years to clear your debt are ?10,000, ?20,000, ?30,000 and ?40,000 correspondingly, exactly just just what quantity must certanly be compensated by the end associated with 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

Therefore, only interest component needs to be compensated within the installment that is last.

Thus, Interest for the year that is first (100000 * 10 * 1) /100 =?10000

Interest when it comes to 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 36 months, hence is completely paid back in three installments that are annual after 12 months. The very first installment is ? the next installment and also the 2nd installment is 2 /3 for the installment that is third. 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 regarding the 3rd, it will likely be 2 /3x. Last but not least, 1 st installment would be ? * 2 /3 *x

Now continuing within the fashion that is similar we did earlier and with 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

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