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Open in App Solution Principal (P) = 4500FinalAmount(A)= 5040 Simple Interest (SI) = Final Amount Principal = $ 5040 4500= 540 Time (T) = 2 years Let the required rate of interest be R% Therefore, at 6% rate per annum the money 4500willbecome 5040 after 2 yearsSimilar questions Q. At what rate of simple interest Rs.4500 will become Rs.5040 after 2 years? Q. Frank paid the interest on the loan he bought 10 years ago is $4500 at a simple interest rate of 15%. What is the
amount he borrowed? Q. A sum of money doubles itself after 10 years at a certain rate of interest. What is the rate of simple interest? Q. There is 60% increase in an amount in 6 years at simple interest rate. What will be the compound interest of ₹12,000 after 3 years at the same rate? Q. There is 100% increase in an amount in 8 years,at simple interest. Find the compound interest on Rs.
8000 after 2 years at the same rate of interest. View More At what rate of interest will a sum of Rs. 4500 amounts to Rs. 6525 at simple interest for 5 years ?
Answer (Detailed Solution Below)Option 1 : 9% Given: Principal = Rs. 4500 Amount = Rs. 6525 Time = 5 years Concept used: Simple interest Calculation: \({\rm{si = }}\frac{{{\rm{P}} \times {\rm{R}} \times {\rm{T}}}}{{100}}\) Principal + interest = Amount Interest = Amount - Principal Interest = 6525 - 4500 = 2025 Interest for 5 years = Rs. 2025 Interest for 1 year = 2025/5 = Rs. 405 \({\rm{Rate = }}\frac{{{\rm{Interest}}\; \times \;{\rm{100}}}}{{{\rm{Principal}}\; \times \;{\rm{Time}}}}\) Rate = \(\frac{{405\; \times \;100}}{{4500}} = 9\% \) Stay updated with the Quantitative Aptitude questions & answers with Testbook. Know more about Interest and ace the concept of Simple Interest.
Rs Aggarwal Solutions for Class 7 Math Chapter 12 Simple Interest are provided here with simple step-by-step explanations. These solutions for Simple Interest are extremely popular among Class 7 students for Math Simple Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal Book of Class 7 Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal Solutions. All Rs Aggarwal Solutions for class Class 7 Math are prepared by experts and are 100% accurate. Page No 165:Answer:P=Rs. 6400, R=6%, T=2 yearsS.I. =P×R×T100=6400×6×2100 =Rs. 768Amount=P+S.I. =6400+768= Rs. 7168 Page No 165:Answer:P=Rs. 2650, R=8%, T=21 2 years =52 years S.I.= P×R×T100=2650×8×5100×2 =Rs. 530Amount=P+S.I. =2650+530=Rs. 3180 Page No 165:Answer:P=Rs.1500, R=12% , T=3+312=134 years S.I.=P×R×T100 =1500×12×13100×4 =Rs. 585Amount=P+S.I. =1500+585 =Rs. 2085 Page No 165:Answer:P= Rs . 9600R=712% T=5 months =512 yearsS.I .=P×R×T100 =9600×15×5100×2×12 =Rs. 300Amount= P+ S.I.=9600+300=Rs. 9900 Page No 165:Answer:P=Rs.5000 , R=9% , T=146 days=146365 yearsS.I.=P×R×T100=5000×9×146100×365 =Rs. 180Amount=P+S.I.=5000+180=Rs. 5180 Page No 165:Answer:P=Rs. 6400, S.I. = Rs. 1152, R=6% T =S.I.×100P×R=1152×1006400×6 =1152384 =3 years Page No 165:Answer:P=Rs. 9540 , S.I.=Rs. 1908, R=8%T = S.I.×100 P×R=1908×1009540×8 =104 =212 years Page No 165:Answer:P=Rs. 5000, A=Rs. 6450, R=12% S.I.=A−P =6450−5000 =Rs. 1450T =S.I×100P×R=1450×100 5000×12 =2912 =2512 =2 years 5 months Page No 166:Answer:P = Rs. 8250, S.I.=Rs. 1100, T=2 yearsR=S.I.×100P×T=1100×100 8250×2 =1100165=6.67% Page No 166:Answer:P= Rs. 5200 , S.I.=Rs. 975 [ T=2 12 years=52 years] R=S.I.×100P×T=975×100×25200×5 = 19526=7.5% Page No 166:Answer:P=Rs. 3560 , A=Rs. 4521. 20 , T=3 yearsS.I.= A−P =4521.20−3560 =Rs. 961.20 R =S.I.×100P×T=961.20×1003560×3 =96120×100100×3560×3 =9% Page No 166:Answer:P=Rs 6000, R=12%, T=3 years 8 months = 3812=4412 yearsS.I. =P×R×T100=6000×12×44100×12= Rs 2640A= P+S.I. = 6000+2640 = Rs 864 0 Page No 166:Answer:P=Rs. 12600 R=15% T=3 yearsS.I.=P×R×T100=12600×15×3100 =Rs. 5670A =Rs. 12600+Rs. 5670=Rs. 18270 Hari had to pay Rs. 18270 to the money lender, but he paid Rs. 7070 and a goat . ∴Cost of the goat =Rs. 18270−Rs. 7070 =Rs. 11200 Page No 166:Answer:Let the sum be Rs. P. S.I.=Rs . 829.50, T=3 years, R=10%Now, P=S.I×100R×T =829.50×10010×3 =82953= 2765Hence, the sum is Rs. 2765. Page No 166:Answer:Let the required sum be Rs. x. A=Rs. 3920, R=7 12%, T=3 yearsNow, Now, S.I.=P×R×T100=x×15×32×100= 9x40A=P+S.I. = x+9x40=40x+9x40 = 49x40 But the amount is Rs. 3920. =>49x40=3920=>x=3920×4049 =15680049=3200Hence, the required sum is Rs. 3200. Page No 166:Answer:Given: R=11%, T=2 years 3 months = 2+3 12=2712 yearsLet the required sum be Rs. x.S.I.=P×R×T100=x×11×279100×124=99x400A=P+S.I. =x+99x400=400x+99x400=499x400 But the amount is Rs. 4491. =>499x400=4491 =>x=4491×400499=1796400499=3600Hence, the required sum is Rs . 3600.∴ S.I.=P×R×T100=3600×11×3100=Rs. 1188∴Amount=P+S.I.=3600+1188 =Rs . 4788 Page No 166:Answer:Let the required sum be Rs. x.S.I.=P×R×T100=x×8× 2100=16x100 A=P+S.I. =x+16x100=100x+16x100= 116x100But the amount is Rs. 12122. =>116x100=12122 =>x=12122×100116=10450Now, S.I.=P×R×T100=10450×93× 328100×1241=Rs. 2508 ∴A=P+S.I. = Rs. 10450 + Rs. 2508 = Rs. 12958 Page No 166:Answer:P= Rs. 3600 A=Rs. 4734 T= 312=72 years S.I.= A−P =4734−3600 =Rs. 1134 R=S.I.×100P×T =1134×100×23600×7=9% Page No 166:Answer:P=Rs. 640, A=Rs . 768, T=2 years 6 months =52 years S.I.=A−P =768−640 =Rs. 128 R=S.I.×100P×T=128×100×2640×5=8% P=Rs. 850, R=8%, T=3 years∴S.I. =P×R×T100=850×8×3100=204010 =Rs. 204∴A=P+S.I. =850+204 =Rs. 1054 Page No 166:Answer:P=Rs. 5600, A=Rs. 6720, R=8% S.I.=A−P =6720−5600 =Rs. 112 0T=S.I.×100P×R=1120×1005600×8=1120448=212 years Page No 166:Answer:Let the sum be Rs. x . Amount=8x5 ∴S.I.=A−P=8x5−x =3x5Let the rate be R%. S.I.=P×R×T100=>3x5=x×R×5110020=>3x×20=R×x×5=>R=3×x×204x×5=12Hence, the rate of interest is 12% . Page No 166:Answer:Amount in 3 years=(Principal+ S.I. for 3 years)=Rs. 837 Amount in 2 years=(Principal+ S.I. for 2 years)=Rs. 783 On subtracting: S.I. for 1 year=(837−783)=Rs. 54S.I. for 2 years=(541× 2)=Rs. 108∴Sum=Amount for 2 years−S.I. for 2 years=783−108 =Rs. 675 P=Rs. 675, S.I.=Rs. 1 08 and T=2 yearsR=S.I.×100P×T =108×10050267527× 21 =8% Page No 166:Answer:Amount in 5 years=(Principal+ S. I. for 5 years)=Rs. 5475 Amount in 3 years =(Principal+ S.I. for 3 years)=Rs. 4745 On subtracting: S.I. for 2 years=(5475−4745) =Rs. 730S.I. for 3 years=(7302×3)=Rs. 1095∴Sum=Amount for 3 years−S.I. for 3 years =4745−1095 =Rs. 3650 P=Rs. 3650, S.I .=Rs. 1095, T=3 yearsR=S.I.×100P×T =1095×1003650×3 = 10% Page No 166:Answer:Let the first part be Rs. x. Second part =(3000−x) ∴ S.I. on x at 8% per annum for 4 years=x×8×42110 05025=8x25S.I. on (3000−x) at 9 % per annum =(3000−x)×9×2110050 =27000−9x50 ∴8x25=27000−9x50=>8x=(27000−9x)×251 502=>16x=27000−9x=>16x+9x=27000=>x=270001080251=1080 ∴First part = Rs.1080 Second part =(3000−1080)=Rs. 1920 Page No 166:Answer:Let the first part be Rs. x. Second part =(3600− x)∴S.I. on x at 9% per annum for 1 years=x×9×1100=9x100And, S.I. on (36 00−x) at 10% per annum =(3600−x)×1×101100 =3600−x10∴9x100+3600−x10=333=>9x+36000−10x100=333= >−x+36000=33300=>−x=33300−36000=>−x=−2700=>x=2700First part = Rs. 2700 Second part =(3600−2700)=Rs. 900 Page No 166:Answer:( a) Rs.125Principal =Rs. 6250Simple Interest=4% per annumTime =6 months =12 yearsSimple Interest=P×R×T100Simple Interest=6250×4×1100×2Simple Interest=2502=Rs. 125 Page No 166:Answer:(b) Rs.3500 Amount =Rs. 3605Time =219365 days= 219365 daysRate=5% per annumAmount = Sum +Sum×Rate×Time100Amount = Sum (1+Rate×Time100)Sum=3605 1+5100×219365=3605×3650037595Sum= Rs. 3500 Page No 166:Answer:(c) 8% Let the sum be Rs. x.Rate of interest = r%Time=212 years=52 yearsAmount=65× Sum Rate=?Amount =65×SumPrincipal + S.I. = AmountPrincipal+Principal×Rate×Time100=65×Principal =>x+xr×5100×2=65x=>x(1+5r100×2)=65x=> 1+r40=65=>r=40×15=>r=8So, the rate of interest is 8%. Page No 167:Answer:(b) 9 months 4.(b)Let the time be t years.Principal=Rs. 8000Amount=Rs. 836 0Rate =6% per annumAmount=Principal(1+Rate×Time100)83608000=1+6×t100 =>83608000−1=6t100=>t=(8360−80008000)×1006 =3608000 ×1006 =68×12 months = 9 months Page No 167:Answer:(b) 10% Let the sum be Rs. x and the rate be r%.A/Q:Amount =2x⇒ P+S.I.=2 x⇒P+P×R×T100=2x=>x(1+r×10100)=2x=>100+ 10r100=2=>10r=200−100⇒10r=100⇒r=10010⇒r=10 Page No 167:Answer:(c) Rs. 100x Simple Interest=Rs. xRate= x% per annum Time = x yearsSimple Interest=Principal×Rate×Time100=>x=Principal×x×x100=> Principal=Rs. 100x Page No 167:Answer:(b) 8% Time=5 yearsSimple interest=25P= >P×Rate×Time100=25P=>Rate×5100=25⇒Rate=2×100 5×5=>Rate=8% Page No 167:Answer:(c) 22 years R1=12%R2=10% P1=Rs.8000P2=Rs.9100Let their amounts be equal in T years.Amount1=S.I.1+P1 =P 1×R1×T100+P1 =8000×12×T 100+8000 = 960T+8000Amount2=S. I.2+P2 =P2×R2×T100 +P2 =9100×10×T100+9100 = 910T+9100Amount1=Amount2 ⇒960T+8000=910T+9100⇒960T-910T=9100-8000⇒50T=1100⇒T=22 Hence, after 22 years their amounts will be equal. Page No 167:Answer:(c) Rs. 768 Let the rate be R %.S.I.= A-P = 720-600 =Rs. 120Time=4 yearsR=100×SI P×T R=100×120600×4 =5Rate of interest=5%Now, R=(5+2)%=7%S.I.=P×R×T100 =600×7×4100 =Rs. 168Amount= SI+P =600+168 = Rs. 768 Page No 167:Answer:(d) y2 = zx y=S.I. on x=x×R×T100 ...(i)z=S.I. on y=y×R×T100 ...(ii)Dividing equation (i) by ( ii):⇒yz=x×R×T100×100y×R×T⇒yz=xy⇒y2=xz Page No 167:Answer:(a) 114 years Rate=10% per annumSimple Interest=0.125 × Principal=>Principal×Rate×Time100=0.125×Principal=>Time10 =0.125=>Time=1.25=114 years Page No 167:Answer:(b) Rs 2400 Page No 168:Answer:P=Rs 6300, R=8% , T=812 years ∴S.I.=P×R×T100=6300×8×82142100×126 31=Rs 336 Page No 168:Answer:Let the sum be Rs x.S.I.=P×R×T 100=x×10×211005=x5 Now, A=P+S. IA=x+x5=6x5But the amount is Rs 6600.∴6x5=6600=>x=66001100×56=5500Hence, the required sum is Rs 5500. Page No 168:Answer:P=Rs 3625 , A=Rs 4495, T=2 years∴S.I.=A―P =4495―3625 =Rs 870∴S.I.=P×R×T100 =>870=3625×R×2100=>R=870×1003625×2=>R=870007250=12% Page No 168:Answer:P=Rs. 3600, A=Rs. 4410, R=9% S.I.=A−P =4410−3600 =Rs. 810∴T=S.I.×100P×R=810×1003600×9=9036=212 years Page No 168:Answer:Let the sum be Rs x. Amount =Rs 2x ∴S.I.=(2x−x)=Rs xTime=12 years ∴ P=x, S.I.=x, T=12 years∴ R=100×S.I.P×T=100×xx× 12=8.3% Page No 168:Answer:Let the sum be Rs x. Amount =43xS.I.=A−P=43x−x = x3Let the rate be R%.S.I.=P×R×T100=>x3=x×R×6100=>R =x×100x×6×3=10018=5.55Hence, the rate of interest is 5.55%. Page No 168:Answer:(c) 9% Page No 168:Answer:(c) 3500 Page No 168:Answer:(a) 9 months P=Rs 6000, A=Rs 6360, R=8% S .I.=A−P =6360−6000 =Rs. 360T=S.I.×100P×T=360×1006000×8 = 68 years =68×12 months =728 =9 months Page No 168:Answer:(c) 12% Let the sum be Rs x. S.I.=Rs 35 Let the rate be R% per annum.S.I.=P×R×T100=x×R×5100= Rx20∴3x5=Rx20=>R=3x×205x=12% Page No 168:Answer:(d) Rs 100xLet the sum be Rs P.Time=x years Rate of interest, r=x% per annumS.I.=Rs xSimple Interest =P×R×T100 =>x =P×x×x100 =>P=x×100x×x∴P=(100x) Page No 168:Answer:(b) 10% Let the sum be Rs x.Amount=Rs 2xTime=10 years∴S.I.=A−P=2x−x =Rs x∴S.I.=P×R×T100=>x=x×R×10100=>R=x× 100x×10=>R =10 Page No 168:Answer:(i)P=100×S.I.R×T(ii)R=100×S. I.(P)×T(iii)10% Let the sum be Rs x. Amount= Rs 2x S.I.=A-P=2x-x=Rs xLet the rate of inetrset be R%.Time=10 yrsS.I.=P×R×T100⇒x =x×R×10100⇒R=10010 ⇒R =10(iv) 8% Let the sum be Rs x. Amount= Rs 65xS.I.=A-P =65x-x=Rs x5Let the rate of inetrset be R%.Time =2.5 yrsS.I.=P×R×T100⇒x5=x×R×2.5100 ⇒R=1002.5×5 ⇒R =8 Page No 168:Answer:(i) FalseS.I.=P×R×T100⇒x=x×R×x100⇒R=100×xx×x ⇒R=100x%(ii) True(iii) TrueLet the sum be Rs x.S.I.=P×R×T100S.I.=x×10×10100S.I.=Rs xAmount= S.I.+P=x+x=Rs 2x(iv) TrueP=Rs 1000 , R=5%, T=73 days=73365 yrsS.I.=P ×R×T100 =1000×5×73100×365 =Rs 10 View NCERT Solutions for all chapters of Class 7 At what rate percent per annum will a sum of Rs 3600 become Rs 4500 in 10 years at simple interest?The rate percent is 9%.
In what time will Rs 4500 amount to Rs 4770 at 3 per annum?At the time of 2 years, The amount Rs. 4500 will become Rs. 4770 at 3% per annum.
At what rate percent per annum will the simple interest on a sum of money be 2 5?∴ Rate is 4% per annum.
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At what rate percentage per annum simple interest will Rs 4500 amounts to Rs 5715 in 3 years?
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