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 designed for ?2500 or ?520 down re payment followed closely by 4 month-to-month equal installments. If the interest is 24%p.a. SI, calculate the installment.

Installments on Simple Interest and Compound Interest Sol: this might be one question that is basic. You need to just utilize the formula that is above calculate the actual quantity of installment.

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

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 exactly What payment that is annual discharge a debt of ?7620 due in 36 months at 16 2/3% p.a. Compounded interest?

Installments on Simple Interest and Compound Interest Sol: Again, we are going to 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 determine loan quantity whenever interest charged is Compound Interest

Ram borrowed money and came back it in 3 equal quarterly installments of ?17576 each. 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 even as we need certainly to discover the initial sum borrowed 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 end regarding 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 what quantity should really be compensated by the end associated with 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 needs to be compensated when you look at 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 for the year that is third (100000 – 10000 – 20000) * 10/ 100 = ?7000

Interest when it comes to 4th 12 months = (100000 – 10000 – 20000 – 30000) * 10/ 100 = ?4000

Hence, Amount that need to paid when you look at 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 installments that are annual after 12 months. The initial installment is ? the next installment in addition to 2nd installment is 2 /₃ associated with 3rd installment. If interest is 10% p.a. Get the very first installment.

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

Since, 2nd installment is 2 /₃ regarding the third, it’ll be 2 /₃x. Last but not least, 1 st installment will likely to be ? * 2 /₃ *x

Now continuing into the comparable fashion as we did earlier 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)

17063.42 = x* 1.953333

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­­­­X = ?8735.53