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Simple Interest & Compound Interest Test - 4

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Simple Interest & Compound Interest Test - 4
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  • Question 1
    2 / -0.83

    A sum of Rs. 14,000 amounts to Rs. 22,400 in 12 years at the rate of simple interest. What is the rate of interest?

    Solution

    Amount = 22400, P = 14000, T = 12yrs
    SI = 22400 - 14000
    = 8400
    Rate = (8400*100)/(12*14000)
    = 5%
     

  • Question 2
    2 / -0.83

    A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

    Solution

  • Question 3
    2 / -0.83

    An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:

    Solution

    B is the correct option. Solution
    6 months interest will be 5%
    next 6 months interest will be 5%
    percentage effect = 5 + 5 +25/100
    = 10.25 Ans
     

  • Question 4
    2 / -0.83

    A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:

    Solution

    Let the rate of interest be x% per annum.
    For the loan to B:
    Simple interest = (5000)(2)(x)/100 = 100x
    For the loan to C:
    Simple interest = (3000)(4)(x)/100 = 120x
    Total simple interest received = 100x + 120x = 220x
    Given, 220x = 2200
    x = 10
    Therefore, the rate of interest per annum is 10%.

  • Question 5
    2 / -0.83

    What annual payment will discharge a debt of Rs. 6450 due in 5 years at 10% per annum?

    Solution

    Installment for first year = x

    Installment for second year = 1.10x

    Installment for third year = 1.20x

    Installment for third year = 1.30x

    Installment for final year = 1.40x

    Total amount to be paid = (1 + 1.10 + 1.20 + 1.30 + 1.40) ×x = 6450

    ∴x = 6450 / 6 = 1075

  • Question 6
    2 / -0.83

    A lends Rs. 1500 to B and a certain sum to C at the same time at 8% per annum simple interest. If after 4 years, A altogether receives Rs. 1400 as interest from B and C, then the sum lent to C is A.

    Solution

    A is the correct option.
    Simple interest on the total money lent at  
    8% FOR 4 YEARS = 1400
    Total money lent
    = 100/8 ×1400/4 =4375
    Therefore, money lent to C
    = 4375- 1500 = 2875

  • Question 7
    2 / -0.83

    If the simple interest on a certain sum of money after 25/8 years is 1/4 of the principal, what is the rate of interest per annum?

    Solution

    Take P=x
    given T=25/8
    &I=1/4x
    w.k.t I=PTR/100
    1/4x =x*25/8*R/100
    R = 8%

  • Question 8
    2 / -0.83

    A sum of Rs. 10 is lend to be returned in 11 monthly instalments of Rs. 1 each, interest being simple. The rate of interest charged is:

    Solution

    Given :

    A sum of Rs. 10 is lend to be returned in 11 monthly instalments of Rs . 1 each interest being simple  

    Formula used :

    Simple interest = P ×R ×T /100  

    Principal(P), Rate(R) and Time(T)

    Calculation :

    Let the rate of interest be R% per annum

    ⇒Amount to be paid (if paid at the end of 11 months)

    ⇒10 + [10  ×R  ×(11/12) / 100] = 10 + (11R/120)

    ⇒Total effective payment = (Rs. 1 + interest on Rs. 1 for 10 months) + (Rs. 1 + interest on Rs. 1 for 9 months) + .... +(Rs. 1 + interest on Rs. 1 for 1 months) + Rs. 1

    ⇒(1 + 1  ×R  ×(10/12) / 100)  + (1 + 1  ×R  ×(9/12) / 100) + .... + (1 + 1  ×R  ×(1/12) / 100) + 1

    ⇒(1 + 10R/1200) + (1 + 9R/1200) + .... + (1 + R/1200) + 1

    ⇒11 + R(1 + 2 + .... + 10)/1200

    ⇒11 + R(10  ×11 / 2)/1200

    ⇒11 + R(10  ×11 / 2)/1200

    ⇒11 + 11R/240  

    Now we have, 

    ⇒10 + 11R/120 = 11 + 11R/240

    ⇒11R/240 = 1

    ⇒R = 240/11 = 21 ×9/11%

    ∴Rate of interest is 21(9/11)%

  • Question 9
    2 / -0.83

    If a sum of Rs. 9 is lent to be paid back in 10 equal monthly installments of re. 1 each, then the rate of interest is

    Solution

    Let 's try to understand the problem step by step.

    A sum of Rs. 9 is lent to be paid back in 10 equal monthly installments of Re. 1 each.

    This means that the borrower is paying back Re. 1 per month for 10 months.

    Now, let 's calculate the interest paid in each installment.

    1. In the first month, the borrower still owes Rs. 9, so no interest is paid.
    2. In the second month, the borrower has already paid Re. 1, so he now owes Rs. 8. The interest paid would be on Rs. 8.
    3. In the third month, the borrower has paid Rs. 2, so he now owes Rs. 7. The interest paid would be on Rs. 7.
    4. This continues until the 10th month when the borrower has paid Rs. 9 and owes nothing.

    Now let 's calculate the total interest paid over the 10 months.

    Total Interest Paid = (Interest on Rs. 8) + (Interest on Rs. 7) + ... + (Interest on Re. 1)

    Let 's assume the rate of interest is "R "percent per month.

    Total Interest Paid = (8 * R) + (7 * R) + ... + (1 * R)

    Now, we know that the total amount paid is Rs. 10 (10 installments of Re. 1 each), and the total amount lent is Rs. 9. So, the total interest paid is Rs. 1.

    1 = (8 * R) + (7 * R) + ... + (1 * R)

    Now we can simplify the equation:

    1 = R * (8 + 7 + 6 + 5 + 4 + 3 + 2 + 1)

    1 = R * 36

    Now, let 's find the value of R (the rate of interest per month):

    R = 1/36

    Since we need to calculate the rate of interest in percentage, we multiply R by 100:

    R (%) = (1/36) * 100 = 2.78%

    Now, we have the monthly rate of interest. To find the annual rate of interest, we multiply the monthly rate by 12:

    Annual Rate of Interest = 2.78% * 12 = 33.33%

  • Question 10
    2 / -0.83

    Divide Rs. 2379 into 3 parts so that their amount after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is

    Solution

    The correct answer is D  as
    let assume 3 parts of amount are x,y,z
    x+y+z=2379
    SI for x= x*5*2/100 ⇒x/10
    Total amount after 2 years - x+(x/10)= 11x/10 ------------(1)
    SI for y= y*5*3/100=>15y/100
    Total amount after 3 years - y+(15y/100) = 115y/100 -------(2)
    SI for z= z*5*4/100=>z/5
    Total amount after 4 years - z+(z/5) = 6z/5 ---------------(3)

    As per the conditions "3 parts amounts after 2, 3 and 4 years respectively may be equal "(1) = (2) = (3)
    based on the above condition and x+y+z=2379, we can get x= 828, y= 792 and z= 759
    So answer is Rs.828

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