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Quiz: Simple Interest

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Number of Questions: 20

Question: 6 -

David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B? 

Options:

  1. Rs. 6500

  2. Rs. 5000

  3. Cannot be determined

  4. Rs. 8000

    5. Answer:

    Rs. 5000
    Solution:

    Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
    (x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
    10x + 12y + 15z = 320000
    Now, z = 240% of y = 12/5 y
    And, z = 150% of x = 3/2 x
    x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
    16y  + 12y + 36y = 320000
    y = 5000
    Sum invested in scheme B = Rs. 5000.

Question: 7 -

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in scheme B? 

Options:

  1. Rs. 6400

  2. Rs. 6500

  3. Rs. 7500

  4. Rs. 7200

    5. Answer:

    Rs. 6400
    Solution:

    Let the sum invested in scheme A be Rs. x and that in scheme B be Rs. (13900 - x). Then,
    (x * 14 * 2)/100 + [(13900 - x) * 11 * 2]/100 = 3508
    28x - 22x = 350800 - (13900 * 22)
    6x = 45000 => x = 7500
    So, sum invested in scheme B = (13900 - 7500) = Rs. 6400.

Question: 8 -

Divide Rs. 2379 into 3 parts so that their amounts 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?

Options:

  1. Rs. 792

  2. Rs. 759

  3. Rs. 828

  4. Rs. 818

    5. Answer:

    Rs. 828
    Solution:

    Let the parts be x, y and [2379 - (x + y)]

    x + (x * 2 * 5/100) = y + (y * 3 * 5/100) = z + (z * 4 * 5/100)
    11x/10 = 23y/20 = 6z/5 = k
    x = 10k/11, y = 20k/23, z = 5k/6
    But x + y + z = 2379
    10k/11 + 20k/23 + 5k/6 = 2379
    k = (2379 * 11 * 23 * 6)/3965 = (3 * 11 * 23 * 6)/5
    x = [10/11 * (3 * 11 * 23 * 6)/5] = 828
    Hence, the first part is Rs. 828.

Question: 9 -

An amount of Rs. 100000 is invested in two types of shares. The first yields an interest of 9% p.a and the second, 11% p.a. If the total interest at the end of one year is 9 3/4 %, then the amount invested in each share was? 

Options:

  1. Rs. 52500; Rs. 47500

  2. Rs. 82500; Rs. 17500

  3. Rs. 62500; Rs. 37500

  4. Rs. 72500; Rs. 27500

    5. Answer:

    Rs. 62500; Rs. 37500
    Solution:

    Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x). Then,
    (x * 9 * 1)/100 + [(100000 - x) * 11 * 1]/100 = (100000 * 39/4 * 1/100)
    (9x + 1100000 - 11x)/100 = 39000/4 = 9750
    x = 62500
    Sum invested at 9% = Rs. 62500
    Sum invested at 11% = Rs. (100000 - 62500) = Rs. 37500.

Question: 10 -

A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum? 

Options:

  1. Rs.515

  2. none of these

  3. Rs.500

  4. Rs.525

    5. Answer:

    none of these
    Solution:

    840 = P [1 + (10*3)/100]

    P = 646