ARITHMETICINTEREST & INSTALLMENTSMATHS

QUESTIONS ON INSTALLMENTS [PART-I]

QUESTIONS ON INSTALLMENTS [PART-I]

CASE-I

TO FIND RATE OF SIMPLE INTEREST WHEN MONTHLY INSTALLMENT IS GIVEN

QUERY 1

A sum of Rs 10 is lent to be returned in 11 monthly installments of Re 1 each, interest being simple. The rate of interest is?

A) 100/11%
B) 10%
C) 11%
D) 240/11%

RONNIE BANSAL
When it’s a monthly installment it’s better to find the one month equivalent principal.

As the sum of Rs 10 has be returned in 11 monthly equal installments, the payment of the principal to be made will end on the completion of 10 months, Re 1 being for every month.

And the principal for the first month will be Rs 10, for the second month Rs 9, for the third month Rs 8, and so on.

Thus the one month equivalent sum = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55

As this sum of Rs 10 has to be paid in 11 monthly installments of Re 1 each, the amount (principal + interest) will be Rs 11, and the interest to be paid = 11 – 10 = 1

Now the principal (equivalent to one month) = 55;
interest = 1
time = 1 month = 1/12 years (as the principal has been converted equivalent to one month)

So rate = 100*12/55*1 (rate = 100*interest/principal*time)
=> 1200/55
=> 240/11 So option ‘D’ is correct.

QUERY 2

A refrigerator of MRP Rs 5000 is available in Rs 1000 cash down payment and 9 monthly equal installment of Rs 500 each. Find rate of interest per annum.

A) 100/3%
B) 50/3%
C) 100/7%
D) 33%

MAHA GUPTA
When it’s a monthly installment it’s better to find the one month equivalent principal.

Now the principal got as loan = MRP – Cash Down Payment = 5,000 – 1,000 = 4,000

As the sum of Rs 4,000 has be returned in 9 monthly equal installments, the payment of the principal to be made will end on the completion of 8 months, Rs 500 being for every month.

And the principal for the first month will be Rs 4,000, for the second month Rs 3,500, for the third month Rs 3,000, and so on.

Thus the one month equivalent sum = 4,000 + 3,500 + 3,000 + 2,500 + 2,000 + 1,500 + 1,000 + 500 = 18,000

As this sum of Rs 4,000 has to be paid in 9 monthly installments of Re 500 each, the amount (principal + interest) will be Rs 4,500 and the interest to be paid = 4,500 – 4,000 = 500

Now the principal (equivalent to one month) = 18,000;
interest = 500
time = 1 month = 1/12 years (as the principal has been converted equivalent to one month)

So rate = (100*500)/(18,000*1/12)                                —-(rate = 100*interest/principal*time)
=> 50,000*12/18,000
=> 100/3  (option ‘A’)

QUERY 3

A computer is available for Rs. 39000 cash or Rs. 17000 as cash down payment followed by five monthly installments of Rs. 4800 each. What is the rate of interest per annum under the installment plan?

A) 35.71%
B) 37.71%
C) 36.71%
D) 38.71%

MAHA GUPTA
When it’s a monthly installment it’s better to find the one month equivalent principal.

Now the principal got as loan = MRP – Cash Down Payment = 39,000 – 17,000 = 22,000. Obviously the principal for the first month will be Rs 22,000, for the second month 17,200, for the third month Rs 12,400, and so on; as the monthly installment is of Rs 4,800.

Thus the one month equivalent sum = 22,000 + 17,200 + 12,400 + 7,600 + 2,800 = 62,000

As this sum of Rs 22,000 has to be paid in 5 monthly installments of Rs 4,800 each, the amount (principal + interest) will be Rs 24,000 and the interest to be paid = 24,000 – 22,000 = 2,000

Now the principal (equivalent to one month) = 62,000;
interest = 2,000
time = 1 month = 1/12 years (as the principal has been converted equivalent to one month)

So rate = (100*2,000)/(62,000*1/12)                                        —-(rate = 100*interest/principal*time)
=> 2,00,000*12/62,000
=> 1,200/31 = 38.71 (option ‘D’)

 

CASE-II

TO FIND MONTHLY INSTALLMENT WHEN RATE OF SIMPLE INTEREST IS GIVEN

QUERY 4

A mobile phone is available for Rs 2500 cash or Rs 520 cash down payments followed by 4 monthly equal installments. If the rate of interest charged is 25% per annum simple interest, calculate the installment.

A) 520
B) 480
C) 550
D) 600

MAHA GUPTA
When it’s a monthly installment it’s better to find the one month equivalent principal.

Now price to be paid in installments = MRP – Cash Down Payment = 2,500 – 520 = 1,980.
Number of equal monthly installments = 4
Rate of interest = 25% per annum

Now let the amount of each installment = Rs x
Therefore interest paid in installment scheme = 4x – 1980           —- (i)

Thus the one month equivalent sum = 1980 + (1980 – x) + (1980 – 2x) + (1980 – 3x) = Rs (7,920 – 6x)

Now the principal (equivalent to one month) = 7,920 – 6x
rate = 25% per annum
time = 1 month = 1/12 years (as the principal has been converted equivalent to one month)
Therefore Interest = [(7920 – 6x)*25*1]/100*12
= (7920*25 – 25x)/1200                                                                       —- (ii)

From (i) and (ii)
(7920*25 – 25x)/1200 = 4x – 1980
=> x = 520 (option ‘A’)

 

CASE-III

TO FIND YEARLY INSTALLMENT WHEN RATE OF SIMPLE INTEREST IS GIVEN

QUERY 5

Reena borrowed Rs 6450 at 5% simple interest repayable in 4 equal installments. What will be the annual installment payable by him ?

A) 1710
B) 1800
C) 1910
D) 1860

MAHA GUPTA
A similar type of question is in R S Aggrawal also, and guys are found to be confused therefore. In R S Aggrawal the language of the question is different; here it is:

WHAT ANNUAL PAYMENT WILL DISCHARGE A DEBT OF Rs 6450 DUE IN 4 YEARS AT 5% SIMPLE INTEREST?

Well, now see the difference in the words of these 2 questions. In the question above 6450 is the sum that was borrowed, whereas in R S Aggrawal 6450 is the sum which is due in 4 years i.e. equal to the addition of the borrowed sum and the interest. If you want to solve the this question as per the method of R S Aggrawal you’ll have to find the amount due after 4 years first, which of course will be 6450*(120100) = 7740

Now see what the amount of installment is:
Formula of Installment: INSTALLMENT = 13148008_1016565988430729_493913047_o

where A= amount due, n = number of years, r = rate of interest.

Hence the installment =100*7740[100*4 + {4*(4 – 1)5}/2] = 774000430 = 1800 (option ‘B’)

QUERY 6

Rahul borrows Rs 1,00,000 from a bank at 10% p.a. simple interest and clears the debt in five years. If the installments paid at the end of the first, second, third and fourth years to clear the debt are Rs 10,000, Rs 20,000, Rs 30,000 and Rs 40,000 respectively, what amount should be paid at the end of the fifth year to clear the debt?

A) Rs 20,000
B) Rs 24,500
C) Rs 30,000
D) Rs 35,900

MAHA GUPTA
Always remember, in the case of simple interest, installment amount will always be reduced from principal and the interest will be calculated on the remaining principal.

Now, principal left after the payment of the above installments = 1,00,000 – (10,000 + 20,000 + 30,000 + 40,000) = 1,00,000 – 1,00,000  = 0

This means there is no principal remains to be paid, only the interest will be calculated that will be given as the last installment to clear the debt.

Now, interest for the 1st year = 10% of 1,00,000 = (1,00,000*1*10)/100 = 10,000

Interest for the 2nd year = 10% of (1,00,000 – amount of first installment) = 10% of (1,00,000 – 10,000) = (90,000*1*10)/100 = 9,000

Interest for the 3rd year = 10% of (90,000 – amount of second installment) = 10% of (90,000 – 20,000) = (70,000*1*10)/100 = 7,000

Interest for the 4th year = 10% of (70,000 – amount of third installment) = 10% of (70,000 – 30,000) = (40,000*1*10)/100 = 4,000

Interest for the 5th year = 10% of (40,000 – amount of fourth installment) = 10% of (40,000 – 40,000) = 0

We see there is no principal left to be paid, therefore balance of the debt will be only the interest i.e. 10,000 + 9,000 + 7,000 + 4,000 = 30,000 (option ‘C’)

 

CASE-IV

TO FIND MONTHLY INSTALLMENT WHEN RATE OF COMPOUND INTEREST IS GIVEN

QUERY 7

A builder borrows Rs. 2550 to be paid back at the compound rate of 4% per annum by the end of 2 years in 2 equal yearly installments. How much will each installment be ?

A) 1252
B) 1352
C) 1350
D) 1150

Amit jha
2550 is the Principal
Rate of interest 4%
Therefore, the Amount payable after 1 year = 2550 + 10% of 2550 = 2550 + 102 = 2652

Now let X be the amount of equal installment
Therefore, loan left after giving 1st installment will be = Total Amount payable – installment = 2652 – X

So, amount of the installment after the completion of the second year = amount of loan left plus interest thereon = (2652 – X)*104%
= (2652 – X)*(104/100)

But the amount of installment is X
Thus (2652 – X)(104/100) = X
=> X = 1352 (option ‘B’)

QUERY 8

A man borrows Rs 21000 at 10% compound interest. How much he has to pay equally at the end of each year, to settle his loan in two years?

A) 12000
B) 12100
C) 12200
D) 12300

MAHA GUPTA
21,000 is the Principal
Rate of interest 10%
Therefore, the Amount payable after 1 year = 21,000 + 10% of 21,000 = 21,000 + 2,100 = 23,100

Now let X be the amount of equal installment
Therefore, loan left after giving 1st installment will be = Total Amount payable – installment = 23,100 – X

So, amount of the installment after the completion of the second year = amount of loan left plus interest thereon = (23,100 – X)*110%
= (23,100 – X)*(110/100)
= (23,100 – X)*(11/10)

But the amount of installment is X
Thus, (23,100 – X)*(11/10) = X
=> X = 12,100 (option ‘B’)

QUERY 9

A man buys a scooter on making a cash down payment of Rs 16224 and promises to pay two more yearly installment of equivalent amount in next two years. If the rate of interest is 4% per annum compounded yearly, the cash value of the scooter is:

A) 40000
B) 46824
C) 46000
D) 50000

MAHA GUPTA
When the rate of interest is compound, A = P(1 + r/100)t, where A = Amount, P = principal, r = rate of interest, t = term

So P = A/(1 + r/100)t

According to the question each installment is worth Rs 16624, means this will be the AMOUNT i.e. PRINCIPAL + INTEREST for each year.

So PRINCIPAL for the first installment =16224/(1 + 4/100)1 = 16224/(26/25)1

= 16224(25/26) = 15600

Likewise PRINCIPAL for the second installment = 16224/(26/25)²
= 16224(25/26)² = 15000

But the buyer has already paid Rs 16224 as cash down payment.

Therefore CASH VALUE of the scooter = Cash down payment + Principal for the 1st installment + Principal for the 2nd installment = 16224 +15600 + 15000 = 46824 (option ‘B’)

QUERY 10

A man borrows money from a finance company and has to pay it back in two equal half yearly installments of Rs 4945 each. If the interest is charged at the rate of 15% per annum compounded semi-annually, find the principal and total interest paid.

A) Rs 8879,        Rs 1011
B) Rs 8800,       Rs 1000
C) Rs 8897,        Rs 1200
D) Rs 7988,        Rs 1159

MAHA GUPTA
When the rate of interest is compound, A = P(1 + r/100)t, where A = Amount, P = principal, r = rate of interest, t = term

So P = A/(1 + r/100)t

According to the question each installment is worth Rs 4945, means this will be the AMOUNT i.e. PRINCIPAL + INTEREST for every six months.

So PRINCIPAL for the first installment              (Rate of interest will be half of the given as the interest is to be compounded semi-annually)
= 4945/(1 + 152 /100)1                                                                                                   
= 4945(1 + 15/200)
= 4945/(215/200)
= 4945*(200/215)
= 4945*(40/43) = 4600

Likewise PRINCIPAL for the second installment         (Rate of interest will be half of the given as the interest is to be compounded semi-annually)
= 4945/(1 + 152 /100)²
= 4945/(1 + 15/200)²
=4945/(215/200)²
=4945*(200/215)²
= 4945*(40/43)² = 4279.07

Thus, the total money borrowed (principal) = 4600 + 4279.07 = 8879.07 = 8879 (approx.) [option ‘A’]

And, the total interest paid = 4925*2 –  8879 = 1011 (option ‘A’)

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

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