Quantitative Aptitude: Simple Interest/Compound Interest Set 04

Simple Interest/Compound Interest Questions for IBPS PO, SBI PO, NIACL, NICL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams

  1. Find the compound interest on Rs36,000 at a rate in which Rs216 becomes Rs343 in 3years and the time is 2years?
    A) Rs12000
    B) Rs12500
    C) Rs13000
    D) Rs13500
    E) Rs14200
    View Answer
    Option C
    Solution:

    first we find the rate
    3√216 : 3√343
    6 : 7
    (+1)
    1/6*100= 16(2/3) %
    Now R = 16(2/3) %=1/6
    6…………….7
    6…………….7
    36 49
     (13) =13000
  2. If a principal becomes triple in 3years on C.I. then find in how many years it will be 27 fold?
    A) 39years
    B) 9years
    C) 18years
    D) 27years
    E) 10years
    View Answer
    Option B
    Solution:

    in C.I principal increase like
    1….3….9…..27
    …3…..3….3
    = 9years
  3. If a principal becomes amount of rs14500 at 14(2/7)% rate of interest in 3years at simple interest. Find the S.I on principal?
    A) Rs4250
    B) Rs4300
    C) Rs4400
    D) Rs4350
    E) Rs4270
    View Answer
    Option D
    Solution:
    R = 14(2/7)% = 1/7
    S.I remains same in all years so…
    (P)7 + 1+1+1= 10(A)
    10-7 = 3S.I
    10 =14500
    1=1450
    3=4350
  4. If the difference between C.I and S.I is rs256 at 20% rate of interest in 3years. Find the amount on C.I?
    A) Rs4320
    B) Rs2500
    C) Rs3456
    D) Rs3200
    E) Rs3478
    View Answer
    Option C
    Solution:

    S.I in 3years = 20*3= 60%
    C.I in 3years = 5……………6
    .                       5……………6
    .                       5………..….6
    .                    125            216
    .                            (91)
    91/125*100= 72.8
    Difference = 72.8 – 60= 12.8
    12.8%= 256
    100% = 2000
    Now P = 2000
    Means in C.I …. 125 =2000
    1 = 16
    216 = 3456
  5. A sum becomes 8000 in 3years and 10000 in 6years at C.I. Find the sum ?
    A) Rs6400
    B) Rs6500
    C) Rs6000
    D) Rs7000
    E) Rs7200
    View Answer
    Option A
    Solution:

    x : y = y : z
    x : 8000 = 8000 : 10000
    x = 6400
  6. Find the C.I on rs9000 at 15% rate of interest for 3years?
    A) Rs4645.87
    B) Rs4680.87
    C) Rs4685.87
    D) Rs4687.87
    E) Rs4356.77
    View Answer
    Option D
    Solution:

    15% of 9000 = 1350
    1350……………1350………………1350
    .                      202.5…………….202.5
    .                                               202.5
    .                                               30.37
    = 4687.87
  7. Find the compound interest on 18000 at 20% rate of interest in 1(1/2) years, if compounded half yearly ?
    A) Rs5958
    B) Rs4916
    C) Rs5780
    D) Rs3500
    E) Rs6724
    View Answer
    Option A
    Solution:

    in half yearly we make rate half and time double.
    So R =20/2 = 10%
    T = 3/2 * 2= 3years
    So 10% of 18000 = 1800
    1800………………1800………………….1800
    .                           180……………………180
    .                                                         180
    .                                                           18
    = 5400 + 540 + 18 = 5958
  8. Find the difference between S.I and C.I on Rs 5000 if rate of interest for first year is 10% and 2nd year is 15% and 3rd year is 20%?
    A) Rs300
    B) Rs320
    C) Rs330
    D) Rs340
    E) Rs360
    View Answer
    Option D
    Solution:

    S.I = 10+15+20 = 45%
    C.I ..     10…….11
    .            20……23
    .              5…….6
    .         1000…..1518
    .               (518)
    = 518/1000*100 = 51.8 %
    = difference = 51.8 – 45 = 6.8%
    = 6.8% of 5000 = 340
  9. If the principal become 6 fold on S.I in 10 years then find in how many years it will be 12 fold?
    A) 24years
    B) 22years
    C) 12years
    D) 20years
    E) 25years
    View Answer
    Option B
    Solution:

    P …………………..6P
    6P – P = 5P interest
    5P = 10years
    P = 2years
    11 P = 22years
  10. If the compound interest on a sum at 25% rate of interest is Rs900 then find the S.I of 3years at same rate?
    A) Rs1000
    B) Rs1100
    C) Rs1300
    D) Rs1200
    E) Rs1500
    View Answer
    Option D
    Solution:

    S.I = 25*3 = 75%
    C.I = 25% =1/4
    4…………..5
    4……………5
    16…………25
    25 -16 = 9
    9 = 900
    16 = 1600 = principal
    So 75% of 1600 =1200

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25 Thoughts to “Quantitative Aptitude: Simple Interest/Compound Interest Set 04”

  1. vijeta mona

    question no 4 ,
    plz explain the method

    C.I in 3years = 5……………6
    . 5……………6
    . 5………..….6
    . 125 216
    . (91)

    1. AVI™ (Sinchen loveR)

      skip this method
      use basic method
      si 20*3 = 60
      ci 20% for 3 year when u calculated value comes = 72.8
      differ 72.8-60=12.8
      12.8 is value of 256
      256/12.8*100=2000
      this is P =2000
      now u calculate Ci on 2000 for 20% in 3 year u find answer

      1. _/_

        no
        ryt hai qs
        solution bhi ryt hai

        1. _/_

          concept hota hai one less hota hai

          1. _/_

            solution diya hai achche se pdo aa jayga

          2. Alice

            try this

            R=(n-1)*100/T where n=no of times

            find out R frm here then

            T=(n-1)*100/R put value of R frm abv

        2. $Ec$tATiC$

          bro…here principal means amount …;)

      2. Ak$hita $inghal

        are dese quiz enough for maths?

        1. AVI™ (Sinchen loveR)

          yes

          bt u have to practice more

      3. AVI™ (Sinchen loveR)

        1 to 12
        mtlb difference 12-1 =11 so here 11 comes

        1 to 6

        difference 6-1=5
        10/5=2
        11*2= 22 ans everything is correct here question and solution i hope u understand

        1. AVI™ (Sinchen loveR)

          always wlc bro

    2. ??????????|||√乇刀り乇イイム™

      CI For 3 year there are 2-3 approaches:
      1st =3r.3r^2.r^3
      2nd=effective method
      3rd=percentile
      you can use any approach ,
      By Effective method = 20+20(20*20/100)=44
      again consider 44 as first value and 20 is another =44+22(44*22/100)=72.8

    3. $Ec$tATiC$

      p =100
      =>si =60 (3yrs)
      =>ci =72.8 (3yrs)
      =>12.8 = 256 =>100 =2000
      now 72.8 =1456
      =>totl =2000+1456 =3456

    1. AVI™ (Sinchen loveR)

      SI = 2250
      CI = 2590
      Diff = 340
      no prob its easy

    2. ??????????|||√乇刀り乇イイム™

      See –
      For Si =10+15+20=45%
      For CI=
      use effective method For first 2 year = 10+15+(10*15/100)=25+1.50=26.50 ,
      again use the same, For remaining =
      26.5+20+(26.5*20/100)=46.5+5.30=51.80
      Diff of CI and SI==51.80-45=6.80%
      6.80% of 5000=340

  2. AVI™ (Sinchen loveR)

    all done but i have a doubt on Q 10 i think this question is incomplete
    any mod plz help to understand this question

  3. ??????????|||√乇刀り乇イイム™

    Re-check Qo no 10.
    Info is not properly mentioned /

  4. Marry..

    all done bt qus 10-time is not given for ci

    1. Vijay Yaduvansi

      same problem

  5. Alice

    Q8> however i gt the answer correct frm diffrent method this method seems new to me..pls explain @Shubhra ma’am (hw 10..11
    20…23
    5…6)

    C.I .. 10…….11
    . 20……23
    . 5…….6
    . 1000…..1518
    . (518)
    = 518/1000*100 = 51.8 %
    = difference = 51.8 – 45 = 6.8%
    = 6.8% of 5000 = 340

  6. Rakesh19

    pls explain 5 th q.. how to cal

  7. ~lonely~hanker~nitrous oxide~

    tq mam

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