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Basics of Financial Mathematics Test 17

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Basics of Financial Mathematics Test 17
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  • Question 1
    1 / -0
    Find the compound interest on Rs. 64006400 for 22 years, compounded annually at 152%\dfrac{15}{2}\% per annum.
    Solution
    \Rightarrow  P=Rs.6400,n=2years,R=152%P=Rs.6400,\, n=2\,years,\, R=\dfrac{15}{2}\%
    \Rightarrow  A=P(1+R100)nA=P(1+\dfrac{R}{100})^n

    \Rightarrow  A=6400(1+152×100)2A=6400(1+\dfrac{15}{2\times 100})^2

    \Rightarrow  A=6400×215200×215200A=6400\times \dfrac{215}{200}\times \dfrac{215}{200}
    \Rightarrow  A=Rs.7,396A=Rs.7,396
    \Rightarrow  Compoundinterest=APCompound\, interest = A-P
    \Rightarrow  CompoundInterest=Rs.7,396Rs.6,400=Rs.996Compound\, Interest =Rs. 7,396-Rs.6,400=Rs.996
  • Question 2
    1 / -0
    If money is invested at r percent interest, compounded annually, the amount of the investment will double in approximately 70/r years. If Pat's parents invested $5000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of the investment 18 years later, when Pat is ready for college?
    Solution
    \Rightarrow  We have given that rr = Percent interest.
    \Rightarrow  We are also given that an investment with double in approximately 70/r70/r years.
    \Rightarrow  We are told to invest $5,000\$5,000 at 8%8\% for 18 years.
    \Rightarrow  Put r=8r=8
    \Rightarrow  So, 708\dfrac{70}{8} is about 9years9\,years, meaning investment will double in 9years9\,years.
    \Rightarrow  In the first 9 years, $5,000\$5,000 doubles to $10,000\$10,000
    \Rightarrow  In the next 9 years, $10,000\$10,000 doubles to $20,000\$20,000
    \therefore   The approximate total amount of investment 18 years later, when Pat is ready for college is $20,000\$20,000.
  • Question 3
    1 / -0
    The present population of town is 1250012500 and it is increasing at the rate of 8%8\% per annum. What will be the population of town after two years?
    Solution
    Given, present population of town =12500,R=8%=12500, R=8\%
    Therefore, population after 22 years =12500×{1+8100}2=12500\times\begin{Bmatrix}1+\dfrac{8}{100}\end{Bmatrix}^2
    =12500×{108100}2=12500\times\begin{Bmatrix}\dfrac{108}{100}\end{Bmatrix}^2
    =14580=14580
  • Question 4
    1 / -0
    Find the interest for the principal Rs. 41104110 at the rate of 6%6\% for 22 years.
    Solution
    Simple interest, I=I = PRT100\dfrac{PRT}{100}
    Where, PP is the principal =4110=4110
    RR is the rate =6%=6\%
    TT is the time period =2=2 years
    I=I = 4110×6×2100\dfrac{4110\times 6\times 2}{100}
    I=I = Rs. 493.20493.20
  • Question 5
    1 / -0
    John earned Rs. 100100 as simple interest on Rs. 600600 for 66 months. Find the annual rate of interest.
    Solution
    Given, Simple interest == Rs. 100100, Principal == Rs. 600600 and period 66 months
    We know the formula,
    Rate of interest, R=R = I×100PT\dfrac{I\times 100}{PT}
    R=R = 100×100600×0.5\dfrac{100\times 100}{600\times 0.5}
    R=33.33%R = 33.33\%
    Therefore, the annual rate of interest is 33.33%33.33\%.
  • Question 6
    1 / -0
    Identify the two statements by which the principal sum can be calculated?
    P : The sum amounts to Rs. 690690 in 33 yrs at Simple Interest.
    Q : The sum amounts to Rs. 750750 in 55 yrs at Simple interest.
    R : The rate of interest is 5%5\% per annum.
    Solution
    Any two of three will give the answer because if amount, rate and number of years are given, we can easily find out the principal sum.
  • Question 7
    1 / -0
    Priya invested a certain amount of money and got back an amount of Rs. 1500015000. If the bank paid an interest of Rs. 50005000, what is the amount Priya invested?
    Solution
    Amount == Rs. 1500015000
    Simple interest == Rs. 50005000
    Principal == Amount - simple interest
    =150005000= 15000 - 5000
    Principal =10000= 10000
    Therefore, priya invested Rs. 1000010000.
  • Question 8
    1 / -0
    In a culture, the present bacteria count was found as 67,60,00067,60,000. It was found that the number was increased by 4%4\% per hour. Find the number of hours before which the bacteria count was 62,50,00062,50,000.
    Solution
    Let the number of years be nn
    67,60,000=62,50,000[1+4100]n\Rightarrow 67,60,000=62,50,000 \left [1+\dfrac{4}{100}\right]^n
    676625=[1+125]n\Rightarrow \dfrac{676}{625}=[1+\dfrac{1}{25}]^n
    [2625]2=[2625]n\Rightarrow \left [\dfrac{26}{25}\right]^2=\left [\dfrac{26}{25}\right]^n
    n=2n=2 hrs
    Thus, the number of hours are 22.
  • Question 9
    1 / -0
    The present population of a city is 1852200018522000. If it has been increased at the rate of 5%5\% per annum, find the population(approx) after 33 years.
    Solution
    Given, Principal amount 1852200018522000, rate of interest =5%=5\% and period =3=3 years
    Total Population =18522000[1+5100]3=18522000\left [1+\dfrac{5}{100}\right]^3 
                                =18522000[1+120]3=18522000\left [1+\dfrac{1}{20}\right]^3
                               =21441530=21441530(approx)
  • Question 10
    1 / -0
    Find rate, when principal == Rs. 30003000; interest == Rs. 400400; time =3= 3 years.
    Solution
    Given, Principal == Rs. 30003000, interest == Rs. 400400, time =3=3 years
    We know the formula,
    Rate of interest, R=R = I×100PT\dfrac{I\times 100}{PT}
    R=R = 400×1003000×3\dfrac{400\times 100}{3000\times 3}
    R=4.44%R = 4.44\%
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