{"id":6002,"date":"2025-06-06T13:01:43","date_gmt":"2025-06-06T13:01:43","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=6002"},"modified":"2025-06-06T13:01:43","modified_gmt":"2025-06-06T13:01:43","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-interest-simple","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-interest-simple\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: SIMPLE INTEREST"},"content":{"rendered":"

\u00adSIMPLE INTEREST<\/strong><\/h1>\n

\u00a0<\/strong>\u00a0IMPORTANT FACTS AND FORMULAE<\/p>\n

1.. Principal<\/strong>: The money borrowed or lent out for a certain period is called the<\/p>\n

principal <\/strong>or the sum.<\/strong><\/p>\n

    \n
  1. 2. Interest<\/strong>: Extra money paid for using other’s money is called<\/li>\n
  2. Simple Interest (S.I.)<\/strong> : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.<\/strong><\/li>\n<\/ol>\n

    Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,<\/p>\n

      \n
    • I. = (P*R*T )\/100<\/li>\n<\/ul>\n

      \u00a0 (ii)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P=(100*S.I)\/(R*T) ;R=(100*S.I)\/(P*T) and T=(100*S.I)\/(P*R)<\/p>\n

      SOLVED EXAMPLES<\/u><\/strong><\/h2>\n

      Ex. 1<\/strong>. Find the simple interest on Rs. 68,000 at 16 2\/3% per annum for 9 months<\/strong>\u00a0 \u00a0 <\/strong>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/strong>\u00a0<\/p>\n

      Sol.<\/strong>\u00a0 P = Rs.68000,R = 50\/3% p.a and T = 9\/12 years\u00a0 = 3\/4years.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0 \\\u00a0 S.I. = (P*R*T)\/100 = Rs.(68,000*(50\/3)*(3\/4)*(1\/100))= Rs.8500<\/p>\n

      Ex. 2. Find the simple interest on Rs. 3000 at 6 1\/4% per annum for the period from<\/strong><\/p>\n

      4th Feb., 2005 to 18th April, 2005.<\/strong><\/p>\n

      Sol.<\/strong> Time = (24+31+18)days = 73 days = 73\/365 years = 1\/5 years.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P = Rs.3000 and R = 6 \u00bc %p.a = 25\/4%p.a<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\\S.I. = Rs.(3,000*(25\/4)*(1\/5)*(1\/100))= Rs.37.50.<\/p>\n

      Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted\u00a0<\/p>\n

      Ex. 3. A sum at simple interests at 13 \u00bd % per annum amounts to Rs.2502.50 after 4 years find the sum.<\/strong><\/p>\n

      Sol.<\/strong>\u00a0 Let sum be Rs. x then , S.I.=Rs.(x*(27\/2) *4*(1\/100) ) = Rs.27x\/50<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\amount = (Rs. x+(27x\/50)) = Rs.77x\/50<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\ 77x\/50 = 2502.50 \u00db x = 2502.50 * 50<\/u>\u00a0\u00a0\u00a0 = 1625<\/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 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 77<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence , sum = Rs.1625.<\/p>\n

      \u00a0<\/strong>Ex. 4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere<\/strong><\/p>\n

      interest rate is increased by 8%, it would amount to bow mucb ?<\/strong><\/p>\n

      Sol.<\/strong> S.l. = Rs. (920 – 800) = Rs. 120; p = Rs. 800, T = 3 yrs. _<\/p>\n

      \u00a0\u00a0\u00a0\u00a0 . R = ((100 x 120)\/(800*3) ) % = 5%.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 New rate = (5 + 3)% = 8%.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 New S.l. = Rs. (800*8*3)\/100 = Rs. 192.<\/p>\n

      :\u00a0\u00a0\u00a0\u00a0\u00a0 New amount = Rs.(800+192) = Rs. 992.<\/p>\n

      Ex. 5. Adam borrowed some money at the rate of 6% p.a. for the first two years , at\u00a0 the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. 1\u00a3 he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?<\/strong><\/p>\n

      \u00a0<\/strong>Sol<\/strong>. Let the sum borrowed be x. Then,<\/p>\n

      (x*2*6)\/100 + (x*9*3)\/100 + (x*14*4)\/100 = 11400<\/p>\n

      \u00db \u00a0(3x\/25 + 27x\/100 + 14x \/ 25) = 11400\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00db 95x\/100 = 11400 \u00db x = (11400*100)\/95 = 12000.<\/p>\n

      Hence , sum\u00a0 borrowed = Rs.12,000.<\/p>\n

      Ex. 6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 \u00bd years. Find the sum and rate of interests.<\/strong><\/p>\n

      \u00a0<\/strong>Sol..<\/strong> S.I. for 1 \u00bd years = Rs.(1164-1008) = Rs.156.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S.l. for 2 years = Rs.(156*(2\/3)*2)=Rs.208<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Principal = Rs. (1008 – 208) = Rs. 800.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now, P = 800, T = 2 and S.l. = 208.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Rate =(100* 208)\/(800*2)% = 13%<\/p>\n

      Ex. 7. At what rate percent per annum will a sum of money double in 16 years.<\/strong><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong>. Let principal = P. Then, S.l. = P and T = 16 yrs.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\Rate = (100 x P)\/(P*16)% = 6 \u00bc % p.a.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

      Ex. 8. The simple interest on a sum of money is 4\/9 of the principal .Find the rate percent and time, if both are numerically equal.<\/strong><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol<\/strong>. Let sum = Rs. x. Then, S.l. = Rs. 4x\/9<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Let rate = R% and time = R years.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, (x*R*R)\/100=4x\/9 or R2 <\/sup>=400\/9 or R = 20\/3 = 6 2\/3.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\Rate = 6 2\/3 % \u00a0\u00a0<\/sup>and Time = 6 2\/3 years = 6 years 8 months.<\/p>\n

      Ex. 9. The simple interest on a certain sum of money for 2 l\/2 years at 12% per<\/strong><\/p>\n

      annum\u00a0 is Rs. 40 less tban the simple interest on the same sum for 3 \u00bd\u00a0 years at 10% per annum. Find the sum.<\/strong><\/p>\n

      \u00a0<\/em><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>Sol.<\/strong>\u00a0 Let the sum be Rs. x <\/em>Then, ((x*10*7)\/(100*2)) \u2013 ( (x*12*5)\/(100*2)) = 40\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 \u00db (7x\/20)-(3x\/10)=40\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00dbx <\/em>= (40 * 20) = 800.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, the sum is Rs. 800.<\/p>\n

      \u00a0<\/strong>Ex. 10. A sum was put at simple interest at a certain rate for 3 years. Had it been<\/strong><\/p>\n

      put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.<\/strong><\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/strong>\u00a0\u00a0\u00a0 \u00a0 Sol. <\/strong>Let sum = P and original rate = R.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, [ (P*(R+2)*3)\/100] \u2013 [ (P*R*3)\/100] = 360.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00db 3PR + 6P – 3PR = 36000 \u00db 6P=36000 \u00db P=6000<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, sum = Rs. 6000.<\/p>\n

      \u00a0<\/strong>Ex. 11. What annual instalment will discharge a debt of Rs. 1092 due in 3 years<\/strong><\/p>\n

      at 12% simple interest?<\/strong><\/p>\n

      \u00a0\u00a0\u00a0Sol <\/strong>. Let each Instalment be Rs. x <\/em><\/p>\n

      \u00a0<\/em>Then, ( x+ ((x*12*1)\/100)) + (x+ ((x*12*2)\/100) ) + x <\/em>= 1092<\/p>\n

      \u00a0 \u00db ((28x\/25) <\/em>+ (31x\/25) + x) <\/em>= 1092\u00a0\u00a0 \u00db (28x+31x+25x)=(1092*25)<\/p>\n

      \u00a0 \u00db x= (1092*25)\/84 = Rs.325.\u00a0\u00a0<\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0\\ Each instalment = Rs. 325.<\/p>\n

      Ex. 12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at<\/strong><\/p>\n

      6%. If the total annual income is Rs. 106, find the money lent at each rate.<\/strong><\/p>\n

      \u00a0\u00a0<\/strong>\u00a0\u00a0\u00a0 Sol.<\/strong> Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 – x).<\/em><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0 \\((x*8*1)\/100) + ((1550-x)*6*1)\/100=106<\/p>\n

      \u00a0\u00a0\u00a0\u00a0 \u00db8x + 9300 \u20136x=10600 \u00db 2x = 1300\u00a0 \u00db x = 650.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0 \\ Money lent at 8% = Rs. 650. Money lent at 6% = Rs. (1550 – 650) = Rs. 900.<\/p>\n

      \u00adRS Aggarwal’s “Quantitative Aptitude” is a widely acclaimed resource for competitive exam preparation, offering comprehensive coverage of various mathematical topics, including Simple Interest. To access the content on Simple Interest, you might consider obtaining the book through authorized channels.<\/p>\n

      Authorized Purchase Options:<\/strong><\/p>\n