Quantitative Aptitude

Simple and Compound Interest

3,394 Questions
All Simple interest and its inverse forms 768 Compound interest with annual compounding 742 Simple-interest growth and split sums 626 Difference between compound and simple interest 479 Half-yearly and quarterly compounding 252 Changing rates and fractional years 201 Installments (CI based) 190 Ratio of principal to interest 118 Population growth and depreciation 16

Simple and Compound Interest Questions

Multiple choice

Consider a bond which offers 10% p.a. rate of interest to be compounded half yearly. What is the effective rate of interest per annum?

  1. 10%

  2. 10.25%

  3. 5%

  4. 20%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Suppose Rs. 100 is invested. The rate per annum is 10%, i.e. 5% per half year. The amount after first half year will be 100 + 5%, i.e. Rs. 105 and amount after second half year, i.e. one year will be 105 + 5%, i.e. Rs. 110.25. Thus, effective rate of interest per annum is 110.25 - 100 = 10.25%.

Multiple choice

Ravi borrowed a sum of money from the bank at an interest of 10.5% per annum and for 2.5 years. If he paid an interest of Rs.1863.75, which of the following could be the total sum borrowed by Ravi?

  1. Rs. 5140

  2. Rs. 7600

  3. Rs. 7100

  4. Rs. 6330

  5. Rs. 7231
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rate of interest = 10.5 = 21/2% p.a. Simple Interest (S.I) = Rs.1863.75 T = 2.5 = 5/2 years Principal (P) =  (S.I x 100) / (R x T) = (1863.75 x 100 x 2 x 2) / (21 x 5) = Rs. 7100

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths
If the interest is calculated at $6\%$ p.a. and is compounded at the end of March and September every year, The interest earned up to $31^{st}$ March and then after completing all the entries, find the amount that the account holder would have received had he closed the account on $20^{th}$ Oct. the same year. A page from the passbook of a Saving Book account in a particular year is given below:
Date Particulars Debit (Rs.) Credit (Rs.) Balance (Rs.)
Jan. $13$ By Cash $5,000.00$ $5,000.00$
Feb. $13$ To self $500.00$
March $24$ By Cheque $2,000.00$
March $31$ By Interest
May $20$ By Cash $800.00$
July $7$ To Cheque $1,400.00$
July $18$ By Cash $1,600.00$
Sept. $15$ To Cheque $3,200.00$
Sept. $26$ By Cheque $2,350.00$
  1. Rs. $4517.86$
  2. Rs. $3890.1$
  3. Rs. $4329.39$
  4. Rs. $6898.10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In bank gives interest on minimum balance between 10 Th to last date of month in S.B A\c

Date       Particulars          Debit(Rs)      Credit (Rs)     Balance (Rs )
Jan 13      By cash                                    5000.00       5000.00
Feb 13      To self                  500.00                             4500.00
March 24  By cheque                              2000.00        6500.00
March 31  By interest                                     45.00       6545.00
May 20     By cash                                        800.00      7345.00
July 7        To cheque         1400.00                               5945.00
July 18      BY cash                                        1600.00     7545.00
Sept 15     To cheque          3200.00                              4345.00
Sept 16     By cheque                                   2350.00     6695.00
Sept 30    By interest                                      203.10      6898.10
Oct 20      To A\C closed    6898.10                                  NIL  

Then product for the month Feb and March=$4500\times2=Rs 9000$
Then interest =$\dfrac{9000\times 6}{12\times 100}= 45$ Rs
Then product for month April and May=$6545\times 2=Rs 13090$
  Or product for the month June=Rs 7345
Product for the month July =Rs 5945
Product for the month August =Rs 7545
product for the month Sept=Rs 6695
Then total product up to month Sept=13090+7345+5945+7545+6695=40620 Rs
Then interest up to Sept =$\dfrac{40620\times 6}{12\times 100}= 203.10$Rs
The a|c closed on 20 Th Oct Then no interest paid for the month Oct
Then amount paid Rs. 6898.10.
    

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

Mark invests Rs. $6500$ in a savings account his annual interest rate is $7\%$ compounded annually. What is the approximate balance of his savings account after $2\dfrac{1}{2}$?

  1. $6500$
  2. $5500$
  3. $7700$
  4. $8200$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given: $P =$ Rs. $6500, r = 7\%, n = 1$ and $t =$ $2\dfrac{1}{2}$ years
$\Rightarrow A = 6500\left (1+\dfrac{0.07}{1}\right)^{1\times 2.5}$
$\Rightarrow A = 6500\times 1.07^{2.5}$
$\Rightarrow A = 6500\times 1.184294$
$\Rightarrow A =$ Rs. $7697.91$ $\text{approx}$ $7700$

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

If you have a bank account whose principal is Rs. $5000$, and your bank compounds the interest twice a year at an interest rate of $12\%,$ how much money do you have in your account at the year's end?

  1. $4272$
  2. $5272$
  3. $6272$
  4. $7272$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: $P = 5000, r = 12\%, n = 2$ years
We know the formula $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Substituting the given values int he formula, we get

$A = 5000\left [\left (1+\dfrac{12}{100}\right)^2\right]$
$A =$ Rs. $6272$

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

Mari deposited Rs. $20000$ in a savings bank account. She would be paid interest at $12\%$ per annum compounded annually. Find the interest to her credit at the end of second year.

  1. $1088$
  2. $3088$
  3. $5088$
  4. $7088$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $P = $ Rs. $20000$, $r = 12\%$, $n = 1$, $t = 2$ years
$A =$ $P\left (1+\dfrac{r}{n}\right)^{nt}$
$A =$ $20000\times (1+0.12)^{1\times 2}$
$A = 20000 \times  1.2544$
$A = Rs. 25088$
Amount $=$ Principal $+$ Interest
Interest $= A - P$
$= 25088 - 20000$
$=$ Rs. $5088$

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

What will a deposit of Rs. $4,500$ at $10\%$ in a savings account compounded yearly interest be worth if left in the bank for $9$ years?

  1. $2110.77$
  2. $4110.77$
  3. $6110.77$
  4. $8110.77$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $P =$ Rs. $4500$, $r = 10\%$, $n = 1$, $t = 9$ years
$A =$ $P(1+0.1)^{1.9}$
$A = 4500 \times 2.357948$
$A =$ Rs. $10610.77$
Amount $=$ Principal $+$ Interest
Interest $= A - P$
$= 10610.77 - 4500$
$=$ Rs. $6110.77$

Multiple choice maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

A certain sum of money at simple interest amounts to $Rs. 1012$ in $2\dfrac {1}{2}$ years and to $Rs. 1067.20$ in $4$ years. The rate of interest per annum is

  1. $2.5$%
  2. $3$%
  3. $4$%
  4. $5$%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the principal be $P$ and rate of interest be $r$%.
According to question,
$1012 = P + \dfrac {P\times r \times 5}{100\times 2} .... (1)$
Interest in $\dfrac {3}{2} years = 1067.20 - 1012 = Rs. 55.20$
$P = \dfrac {I\times 100}{R\times T} = \dfrac {55.20\times 100}{3} = \dfrac {3680}{r}$
Putting values in equation $(1)$,
$1012 = \dfrac {3680}{r} + \dfrac {3680\times r\times 5}{r\times 100\times 2}$
$1012 = \dfrac {3680}{r} + 92$
$\dfrac {3680}{r} = 1012 - 92 = 920$
$r = \dfrac {3680}{920} = 4$% per annum.
Hence, the rate of interest is $4$% per annum.

Multiple choice maturity value of recurring deposits banking maths

Mr Bittu deposits a certain sum of money each month in a recurring deposit account. If the rate of interest is $8\%$ per annum and Mr Bittu gets Rs $8,088$ from the bank after $3$ years. Find the value of his monthly instalment.

  1. $400$
  2. $800$
  3. $200$
  4. $140$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the monthly installment be $P$

Given, maturity value $=8088$, $n=3$ years $=3\times 12=36$ months, $r=8\%$
Interest $=\dfrac {Pn(n+1)r}{2400}$
$=\dfrac {P\times 36\times 37 \times 8}{2400}$
$=\dfrac {111P}{25}$
We know, amount $=Pn+\dfrac {111P}{25}$
$8088=36P+\dfrac {111P}{25}$
$=\dfrac {1011P}{25}$
$\therefore 8088= \dfrac {1011P}{25}$
$\therefore P=200$
Therefore, monthly installment  is Rs. $200$.

Multiple choice maturity value of recurring deposits banking maths

Divya has a recurring deposit in a bank at $ 5\%$ per annum simple interest. If she pay monthly installment of Rs. $2000$ for annually. Find her Maturity value.

  1. $6049.931$
  2. $6249.931$
  3. $6149.931$
  4. $6549.931$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\dfrac{-1}{3}}}$ 
R $=$ Monthly Installment $= 2000 $
i $=$ Rate of Interest $5\%$
n $=$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{2000 \times [(1+5/400)^{n} - 1]}{1-(1+5/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{2000\times\dfrac{5}{400}}{1-(\dfrac{405}{400})^{-1/3}}$
$M = 6049.931$

Multiple choice maturity value of recurring deposits banking maths

Sheela has a recurring deposit in a bank at $2\%$ per annum simple interest. If she pay monthly installments Rs.$200$ for annually. Find her Maturity value.

  1. $621.99$
  2. $611.99$
  3. $631.99$
  4. $601.99$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 200$ 
$i =$ Rate of Interest/ 400 $= \dfrac {2}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{200 \times [(1+2/400)^{n} - 1]}{1-(1+2/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{200\times\dfrac{2}{400}}{1-(\dfrac{402}{400})^{-1/3}}$
$M = 601.99$

Multiple choice maturity value of recurring deposits banking maths

Rita has a recurring deposit in a bank at $4\%$ per annum simple interest. If she pay monthly installment s Rs.$1000$ for annually. Find her Maturity value.

  1. $3019.97$
  2. $4019.97$
  3. $2019.97$
  4. $5019.97$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R $= Monthly Installment $= 1000 $
$i =$ Rate of Interest/ 400 $= \dfrac {4}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{1000 \times [(1+4/400)^{n} - 1]}{1-(1+4/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{1000\times\dfrac{4}{400}}{1-(\dfrac{404}{400})^{-1/3}}$
$M = 3019.97$

Multiple choice maturity value of recurring deposits banking maths

Nithya deposited Rs. $100$ per month for $12$ months in a bank's recurring deposit account. If the bank pays interest at the rate of $10\%$ per annum, find the amount she gets on maturity.

  1. $1205$
  2. $1265$
  3. $1345$
  4. $1450$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: $P = 100, T(n) = 12$ months, $R = 10\%$
Equivalent principal for $12$ months $=$ $100\times \dfrac{n(n+1)}{2}$
$=$ $100\times\dfrac{12(13)}{2}$
$= 50 \times  12 \times  13$
Interest $=$ $\dfrac{PRT}{100}$
$=$ $\dfrac{50\times 12 \times 13\times 10\times 1}{100\times 12}$
$=$ $65$
Maturity amount $=$ P$\times$T $+$ Interest
$=$ $100 \times 12 + 65$
$=$ Rs. $1265$

Multiple choice maturity value of recurring deposits banking maths

Mohan deposited Rs.$80$ per month in a cumulative deposit account for six years. Find the amount payable to him on maturity, if the rate of interest is $6\%$ per annum.

  1. $6811.20$
  2. $2000$
  3. $4811.20$
  4. $3811.20$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the monthly installment be $P$

Given, monthly installment Rs.$=80$, $r=6\%$, $n=6\times 12=72$
We know, Interest $=\dfrac {Pn(n+1r)}{2400}$
$=\dfrac {80 \times 72 \times 73\times 6}{2400}$
$=\dfrac {5256}{5}$
Required amount $=Pn+{5256}{5}$
$=80 \times 72 +\dfrac {5256}{5}$
$=28800+\dfrac {5256}{5}$
$=6811.20$

Multiple choice maturity value of recurring deposits banking maths

Priya has a recurring deposit in a bank at $10\%$ per annum simple interest. If she pay monthly installment s Rs.$700$ for annually. Find her Maturity value.

  1. $2034.904$
  2. $2134.904$
  3. $3134.904$
  4. $4134.904$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 700 $
$i =$ Rate of Interest/ 400 $= \dfrac {10}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{700 \times [(1+10/400)^{n} - 1]}{1-(1+12/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{700\times\dfrac{10}{400}}{1-(\dfrac{410}{400})^{-1/3}}$
$M = 2134.904$