{"id":6004,"date":"2025-06-09T12:03:13","date_gmt":"2025-06-09T12:03:13","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=6004"},"modified":"2025-06-09T12:03:13","modified_gmt":"2025-06-09T12:03:13","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-interest-compound","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-interest-compound\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: COMPOUND INTEREST"},"content":{"rendered":"

COMPOUND INTEREST<\/u><\/h1>\n

Compound Interest:<\/strong> Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly <\/em>or half-yearly <\/em>or quarterly <\/em>to settle the previous account.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 In such cases, the amount after first unit of time becomes the principal for the second unit,the amount after second unit becomes the principal for the third unit and so on.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 After a specified period, the difference between the amount and the money <\/em><\/strong>borrowed is called<\/em><\/strong> the <\/em><\/strong>Compound Interest (abbreviated <\/em>as C.I.) for that period.<\/em><\/strong><\/p>\n

IMPORTANT FACTS AND FORMULAE<\/u><\/h2>\n

Let Principal <\/strong>= P, Rate <\/strong>= <\/strong>R% per annum, Time = n years. <\/strong><\/p>\n

    \n
  1. When interest is compound Annually:<\/strong><\/li>\n<\/ol>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Amount = P(1+R\/100)n<\/sup><\/p>\n

      \n
    1. When interest is compounded Half-yearly:<\/strong><\/li>\n<\/ol>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Amount = P[1+(R\/2)\/100]2n<\/sup>\u00ad<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 III. When interest is compounded Quarterly:<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Amount = P[ 1+(R\/4)\/100]4n<\/sup>\u00ad<\/p>\n

        \n
      1. When interest is compounded AnnuaI1y but time is in fraction, <\/strong>say 3(2\/5) years.<\/strong><\/li>\n<\/ol>\n

        Amount = P(1+R\/100)3 <\/sup>x (1+(2R\/5)\/100)<\/p>\n

          \n
        1. When Rates are different for different years, say Rl%, R2%, R3% for 1st, 2nd and 3rd year respectively.<\/li>\n<\/ol>\n

          Then, Amount = P(1+R1<\/sub>\/100)(1+R2<\/sub>\/100)(1+R3<\/sub>\/100)<\/p>\n

            \n
          1. Present worth of Rs.x due n <\/em>years hence is given by :<\/strong><\/li>\n<\/ol>\n

            Present Worth = x\/(1+(R\/100))n<\/sup><\/p>\n

            SOLVED EXAMPLES<\/h2>\n

            Ex.1. Find compound interest <\/em>on Rs. 7500 <\/em>at 4% per <\/em>annum for <\/em>2 years, compounded <\/em>annually<\/strong>.<\/strong><\/p>\n

            \u00a0 Sol.<\/strong><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Amount = Rs [7500*(1+(4\/100)2<\/sup>] = Rs (7500 * (26\/25) * (26\/25)<\/u>) <\/u>= Rs. 8112. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 therefore, C.I. = Rs. (8112 – 7500) = Rs. 612.<\/p>\n

            Ex. 2. Find compound interest <\/em>on Rs. 8000 <\/em>at 15% per <\/em>annum for <\/em>2 years <\/em>4 months, compounded <\/em><\/strong>annually<\/em><\/strong>.<\/em><\/p>\n

            \u00a0\u00a0 Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Time = 2\u00a0 years\u00a0 4\u00a0 months = 2(4\/12) years = 2(1\/3) years.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Amount = Rs’. [8000 X (1+\u00ad(15\/100))2 <\/sup>X (1+((1\/3)*15)\/100)]<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 =Rs. [8000 * (23\/20) * (23\/20) * (21\/20)]<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 = Rs. 11109.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 .<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 :. C.I. = Rs. (11109 – 8000) = Rs. 3109.<\/p>\n

            \u00a0\u00a0\u00a0 Ex. 3. Find the compound interest <\/em>on Rs. 10,000 in <\/em>2 years <\/em>at 4% per <\/em>annum, <\/strong>the<\/em><\/strong><\/p>\n

            \u00a0\u00a0\u00a0 interest <\/em><\/strong>being compounded half-yearly.<\/em><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (S.S.C. 2000)<\/strong><\/p>\n

            \u00a0\u00a0\u00a0 Sol.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0Principal = Rs. 10000; Rate = 2%\u00a0 per half-year;\u00a0 Time = 2 years = 4 half-years.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Amount =<\/p>\n

            Rs [10000 * (1+(2\/100))4<\/sup>] = Rs(10000 * (51\/50) * (51\/50) * (51\/50) * (51\/50))\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Rs. 10824.32.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 :. C.I. = Rs. (10824.32 – 10000) = Rs. 824.32.<\/p>\n

            \u00a0\u00a0 Ex. 4. Find the compound interest <\/em>on Rs. 16,000 <\/em>at 20% per <\/em>annum for <\/em>9 months,<\/em><\/strong><\/p>\n

            \u00a0\u00a0\u00a0 compounded quarterly.<\/em><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n

            \u00a0\u00a0 Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Principal = Rs. 16000; Time = 9 months =3 quarters;<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Rate = 20% per annum = 5% per quarter.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Amount = Rs. [16000 x (1+(5\/100))3<\/sup>] = Rs. 18522.<\/p>\n

              \n
            1. = Rs. (18522 – 16000) = Rs. 2522.<\/li>\n<\/ol>\n

              Ex. 5. If the simple interest <\/em>on a sum <\/em>of money <\/em>at 5% per <\/em>annum for <\/em>3 years is Rs. 1200, find the compound interest <\/em>on the same sum for the <\/em>same period at the same r<\/em>ate.<\/strong><\/p>\n

              Sol.<\/strong><\/p>\n

              Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. \u00a0\u00a0\u00a0\u00a0\u00a0 . .<\/p>\n

              So principal=RS [100*1200]\/3*5=RS 8000<\/p>\n

              Amount = Rs. 8000 x [1 +5\/100]^3 – = Rs. 9261.<\/p>\n

              .. C.I. = Rs. (9261 – 8000) = Rs. 1261.<\/p>\n

              Ex. 6. In what time will Rs. 1000 become Rs. <\/em>1331 at 10% per <\/em>annum compounded annually?<\/em>\u00a0 (S.S.C. 2004)<\/strong><\/p>\n

              Sol.<\/strong><\/p>\n

              Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,<\/p>\n

              [ 1000 (1+ (10\/100))n\u00a0 <\/sup>] = 1331 or (11\/10)n =\u00a0 <\/sup>(1331\/1000) = (11\/10)3<\/sup><\/p>\n

              n = 3 years.<\/p>\n

              Ex. 7. If Rs. 600 amounts <\/em>to Rs. 683.20 in <\/em>two years compounded <\/em>annually, find the<\/strong><\/p>\n

              rate of interest per <\/em>annum.<\/strong><\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0 Sol. Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Let the rate be R% per annum.. ‘Then,<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0 [ 500 (1+(R\/100)2 <\/sup>] = 583.20 or [ 1+ (R\/100)]2\u00a0 =\u00a0 <\/sup>5832\/5000 = 11664\/10000<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0 [ 1+ (R\/100)]2 <\/sup>= (108\/100)2<\/sup>\u00a0 or\u00a0 1 + (R\/100) = 108\/100\u00a0 or\u00a0 R = 8<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So, rate = 8% p.a.<\/p>\n

              Ex. 8. If the compound interest <\/em>on a certain sum <\/em>at 16 (2\/3)% to\u00a0 <\/em>3 years is Rs.1270,<\/em><\/strong><\/p>\n

              find the simple interest <\/em>on the same sum <\/em>at the same <\/em>rate and f or the <\/em>same period.<\/strong><\/p>\n

              Sol.<\/strong> Let the sum be Rs. x. Then,<\/p>\n

              \u00a0\u00a0 \u00a0C.I. = [ x * (1 + (( 50\/(3*100))3<\/sup> – x ] = ((343x \/ 216) – x) = 127x \/ 216<\/p>\n

              127x \/216 = 1270\u00a0 or\u00a0 x = (1270 * 216) \/ 127\u00a0 =\u00a0 2160.<\/p>\n

              Thus, the sum is Rs. 2160<\/p>\n

              S.I.<\/strong> = Rs\u00a0 ( 2160 * (50\/3) * 3 * (1 \/100 ) ) = Rs. 1080. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

              Ex. 9. The difference <\/em>between the compound interest <\/em>and simple interest <\/em>on a<\/strong><\/p>\n

              certain sum <\/em>at 10% per <\/em>annum for <\/em>2 years is Rs. <\/em>631. Find the sum.<\/em><\/strong><\/p>\n

              \u00a0Sol.<\/strong> Let the sum be Rs. x. Then,<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C.I. =\u00a0 x ( 1 + ( 10 \/100 ))2 <\/sup>– x\u00a0 =\u00a0 21x \/ 100 , \u00a0<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S.I.\u00a0 = (( x * 10 * 2) \/ 100) =\u00a0 x \/ 5<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (C.I) – (S.I) = ((21x \/ 100 ) – (x \/ 5 )) =\u00a0 x \/ 100<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( x \/ 100 )\u00a0 =\u00a0 632\u00a0 \uf0f3\u00a0 x\u00a0 =\u00a0 63100.<\/p>\n

              Hence, the sum is Rs.63,100.<\/p>\n

              Ex. 10. The difference <\/em>between the compound interest <\/em>and the simple <\/em>interest accrued <\/em>on an amount of Rs. 18,000 in <\/em>2 years was Rs. 405. What <\/em>was the <\/em>rate of interest p.c.p.a. <\/em>?\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Bank P.O. 2003)<\/strong><\/p>\n

              Sol<\/strong>. Let the rate <\/em>be R% p.a. then,<\/p>\n

              [ 18000 ( 1 + ( R \/ 100 )2 <\/sup>) – 18000 ] – ((18000 * R *\u00a0 2) \/ 100 ) = 405<\/p>\n

              18000 [ ( 100 + (R \/ 100 )2 <\/sup>\u00a0\/ 10000)\u00a0 –\u00a0 1 – (2R \/ 100 ) ]\u00a0 =\u00a0 405<\/p>\n

              \u00a018000[( (100 + R ) 2<\/sup> – 10000 – 200R) \/ 10000 ]\u00a0 =\u00a0 405<\/p>\n

              \u00a09R2<\/sup> \/ 5\u00a0 =\u00a0 405\u00a0 \uf0f3 R2<\/sup>\u00a0 =((405 * 5 ) \/ 9) = 225<\/p>\n

              \u00a0R = 15.<\/p>\n

              \u00a0Rate = 15%.<\/p>\n

              Ex. 11. Divide Rs. 1301 between A and B, so that the amount <\/em>of A after <\/em>7 years is equal to the amount <\/em>of B after <\/em>9 years, the interest being compounded at <\/em>4% per\u00a0 annum.<\/em><\/strong><\/p>\n

              Sol. Let the two parts be Rs. x and Rs. (1301 – x).<\/em><\/p>\n

              x(1+4\/100)7 <\/sup>=(1301-x)(1+4\/100)9<\/sup><\/em><\/p>\n

              x\/(1301-x)=(1+4\/100)2<\/sup>=(26\/25*26\/25)<\/em><\/p>\n

              625x=676(1301-x)<\/em><\/p>\n

              1301x=676*1301<\/em><\/p>\n

              x=676.<\/em><\/p>\n

              So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625.<\/em><\/p>\n

              Ex.12. a certain sum amounts to rs.7350 in 2 years and to rs.8575 in 3 years.find the sum and rate percent.<\/p>\n

              S.I on rs.7350 for 1 year=rs.(8575-7350)=rs.1225.<\/em><\/p>\n

              Rate=(100*1225\/7350*1)%=16 2\/3%<\/em><\/p>\n

              Let the sum be rs.x.then,<\/em><\/p>\n

              X(1+50\/3*100)2<\/sup>=7350<\/em><\/p>\n

              X*7\/6*7\/6=7350<\/em><\/p>\n

              X=(7350*36\/49)=5400.<\/em><\/p>\n

              Sum=rs.5400.<\/em><\/p>\n

              Ex.13.a sum of money amounts to rs.6690 after 3 years and to rs.10,035 after 6 years on compound interest.find the sum.<\/p>\n

              Sol.<\/em><\/strong> Let the sum be rs.P.then<\/em><\/p>\n

              P(1+R\/100)3<\/sup>=6690\u2026(i) and \u00a0 P(1+R\/100)6<\/sup>=10035\u2026(ii)<\/em><\/p>\n

              On dividing,we get (1+R\/100)3<\/sup>=10025\/6690=3\/2.<\/em><\/p>\n

              Substituting this value in (i),we get:<\/em><\/p>\n

              P*3\/2=6690 or P=(6690*2\/3)=4460<\/em><\/p>\n

              Hence,the sum is rs.4460.<\/em><\/p>\n

              Ex.14. a sum of money doubles itself at compound interest in 15 years.in how many years will it beco,e eight times?<\/p>\n

              P(1+R\/100)15<\/sup>=2P<\/em><\/p>\n

              (1+R\/100)15<\/sup>=2P\/P=2<\/em><\/p>\n

              LET P(1+R\/100)n<\/sup>=8P<\/em><\/p>\n

              (1+R\/100)n<\/sup>=8=23<\/sup>={(1+R\/100)15<\/sup>}3<\/sup>[USING (I)]<\/em><\/p>\n

              (1+R\/100)N<\/sup>=(1+R\/100)45<\/sup><\/em><\/p>\n

              n=45.<\/em><\/p>\n

              Thus,the required time=45 years.<\/em><\/p>\n

              Ex.15.What annual payment will discharge a debt of Rs.7620 due in 3years at <\/strong><\/p>\n

              16 2\/3% per annum interest?<\/strong><\/p>\n

              Sol.<\/strong> Let each installment beRs.x.<\/p>\n

              \u00a0\u00a0\u00a0\u00a0 Then,(P.W. of Rs.x due 1 year hence)+(P>W of Rs.x due 2 years hence)+(P.W of Rs. X due 3\u00a0\u00a0\u00a0<\/p>\n

              \u00a0\u00a0\u00a0\u00a0 years hence)=7620.<\/p>\n

              \\ x\/(1+(50\/3*100))+ x\/(1+(50\/3*100))2<\/sup> + x\/(1+(50\/3*100))3<\/sup>=7620<\/p>\n

              \u00db(6x\/7)+(936x\/49)+(216x\/343)=7620.<\/p>\n

              \u00db294x+252x+216x=7620*343.<\/p>\n

              \u00db x=(7620*343\/762)=3430.<\/p>\n

              \\Amount of each installment=Rs.3430.<\/p>\n

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