SSC CGL SIMPLE INTEREST AND COMPOUND INTEREST MCQ ENGLISH

Correct Answer : (B)

Explanation : Given,Principal = Rs. 2000Rate =r=5%Time =n=2 yearsAs we know,Amount = Principal ×(1+r100)nAt the end of the first year,Amount =2000×105100=2000×2120= Rs. 2100At the end of the second year,Principal for the second year =2000+2100=4100Amount =4100×105100=4100×2120= Rs. 4305So, her investments be at the end of the 2nd year is Rs. 4305.

Correct Answer : (A)

Explanation : Given:Jyoti Invested =2 yearsSita invested =3 yearsWe know that,Amount =P(1+r100)tNow,Amount received by Jyoti : Amount received by Sita =(1+2100)2:(1+2100)3⇒(102100)2:(102100)3⇒100:102⇒50:51∴ Amount received by Jyoti : Amount received by Sita is 50:51 .

Correct Answer : (D)

Explanation : Given,P1= Vipul’s investment = ₹ 15000P2= Manish’s investment = ₹ 20000R1=20%R2=30%T1=T2=3 yearsVipul amount after 3 year,15000×120100×120100×120100Vipul amount = ₹ 25,920CI=A−PCI of Vipul for 3 years =25,920−15,000= ₹ 10,920Manish amount after 3 years,20000×130100×130100×130100Manish amount = ₹ 43,940CI of Manish for 3 years =43,940−20,000= ₹ 23,940Total interest =23,940+10,920= ₹ 34,860

Correct Answer : (D)

Explanation : Given:The sum of money gets tripled in 14 years.We know that,I=(P×t×r)100Where, P= Principle, t= Time of investment, r= Rate of interestLet us assume the money invested =P and rate of SI=r% per annumIf the money gets tripled, then interest received =(3P−P)=2PAgain for the money to get quadrupled, interest =(4P−P)=3PAccording to the question,2P=(P×r×14)100⇒r=1007Now, let’s assume the time required to quadruple the money =t yearsAccording to the question,3P=(P×1007×t)100⇒t=21 years∴ It will take 21 years to quadruple the money.

Correct Answer : (C)

Explanation : Let the equal amount borrowed be Rs. ‘P’.Time = 4 years∵ Simple interest = (Principal × Rate × Time)100Simple interest at 5% = (P×5×4)100 = Rs. 0.2PSimple interest at 4% = (P×4×4)100 = Rs. 0.16PTotal interest after 4 years = 0.2P + 0.16P = Rs. 0.36PBut, total interest = Rs. 4050.36P = 405⇒ P = 4050.36 = Rs. 1125∴ Amount borrowed = Rs. 1125

Correct Answer : (C)

Explanation :

Correct Answer : (A)

Explanation : Let the principal amount = Rs. 100xRate of interest =r%Time period =2 yearsSimple Interest =P×R×T100=1400⇒100x×r×2100=1400⇒2rx=1400⇒x=14002r=700rCompound Interest =P[(1+R100)T−1]=1449⇒100x[(1+r100)2−1]=1449⇒100x[(1+r21002+2r100)−1]=1449⇒(100×700r)[r210000+2r100]=1449⇒7r+1400=1449⇒7r=1449−1400=49⇒r=497=7%

Correct Answer : (C)

Explanation : Given:Rate of interest =2.5% per monthAmount paid after 6 months = Rs. 13110Amount, A=P+SISimple interest, SI=P×R×T100Where P→ Principal, R→ rate of interest, T→ timeSuppose the sum borrowed be Rs. xSI=x×2.5×6100=0.15xA=x+0.15x=1.15x1.15x=13110⇒x=11400⇒ Amount of interest =0.15×11400= Rs. 1710

Correct Answer : (D)

Explanation : Given,Principal = Rs. 480Amount = Rs. 520Time =20 months =2012 yearsAs we know,Simple interest =P×R×T( in years )100Simple interest for 2012 years = Amount − principal =520−480⇒520−480=480×R×20100×12⇒40=8R⇒R=5%∴ The interest rate per annum is 5% .

Correct Answer : (C)

Explanation : Given,Amount =2× Principal, When time =3 yearsAs we know,Amount =P+P×R×T100P= Principal, R= Rate, T= Time2P=P+(3×P×R100)⇒P=3PR100⇒R=P×1003P⇒R=100352P=P+P×1003×T100⇒51P=P×1003×T100⇒T=51×3⇒T=153

Correct Answer : (C)

Explanation : Given,Amount (A)=3600Principal (P)=2400A=P(1+r100)n⇒3600=2400(1+r100)6⇒(1+r100)6=32For 12 years,⇒A=2400(1+r100)12⇒A=2400[(1+r100)6]2⇒A=2400×(32)2= Rs. 5400∴ The amount after 12 years at the same rate of interest will be Rs. 5400 .

Correct Answer : (D)

Explanation : Given:Principal = Rs. 27570R1=18%R2=12%R3=10%T1=1 yearT2=1 yearT3=1 yearWe know that,Simple Interest ={P×(R1×T1+R2×T2+R3×T3)}100Where,P− PrincipalR− Rate of InterestT− Time periodNow, Substitute the values in the formula,Simple Interest ={27570×(18×1+12×1+10×1)}100= Rs. 11028Total amount =P+ Simple Interest=27570+11028= Rs. 38598

Correct Answer : (B)

Explanation : Let principal =P∴ Amount =3PInterest =3P−P=2PAccording to the question,2P=P×R×20100⇒R=10%Let after t years it will become double.Thus, Interest =2P−P=P⇒P=P×10×t100⇒ t=10 years

Correct Answer : (B)

Explanation : Given:Time period =20 yearsFormula:SI =(P×R×T)100A=P+SIThe sum of money is 3PSI =3P−P⇒2P=(P×R×20)100⇒2=R5⇒R=10%∴ The rate of interest is 10% .

Correct Answer : (A)

Explanation : Given:Rate =5%Simple interest =40We know that,Simple interest =(P×R×T)100Amount =P(1+R100)nNow,Let the sum = PSimple Interest =40Rate of interest =5%Time =2 yearsSimple interest =(P×R×T)100⇒40=P×5×2100⇒P=(40×100)(5×2)⇒P=400Compound interest of 400 ,Amount =400(1+5100)2⇒400×105100×105100⇒400×2120×2120⇒441Compound Interest =441−400= Rs. 41

Correct Answer : (C)

Explanation : Given,Principal (P)=12000Rate (R)=15%Time (T)=3 yearsAs we know,Simple interest (SI)=P×R×T100=12000×15×3100= ₹ 5400

Correct Answer : (B)

Explanation : Given:Simple and compound interest on a principal at a certain rate of interest for 2 years is Rs. 1400 and Rs. 1470 respectivelyFormula Used:S.I=P×R×T100C.I=P[1+R100]n−1}where P= principal, R= interest rate per annum T and n= time periodLet the principal be ‘ P ‘⇒1400=P×R×2100 ⋅⋅⋅⋅⋅⋅⋅ (i)also,⇒1470=P(1+R210000)+(2R100−1)⇒1470=PR(R10000+150) ⋅⋅⋅⋅⋅⋅⋅ (ii)dividing equation (ii) by (i), we get⇒2120=(R10000+150)×50⇒2120=R200+1⇒R=10% p.a.so, P=1400×10020=7000∴ Required value of principal (In Rs. )=7000

Correct Answer : (D)

Explanation : Given:Rate of depreciation =25%3rd year value −4 th year value = Rs. 17,199We know that:Depreciated value after ‘n’ years =p[1−(τ100)]nLet the present value be ‘p’∵ 3rd year value th year value Rs. 17,199⇒p[1−(25100)]3−p[1−(25100)]4=17,199⇒p[1−(14)]3−p[1−(14)]4=17,199⇒p(34)3−p(34)4=17,199⇒27p64−81p256=17,199⇒(108p−81p)256=17,199⇒27p256=17,199⇒p=(17,199×256)27⇒p=637×256⇒p=1,63,072∴ Value of machine after 1 year =1,63,072[1−(25100)]1=1,63,072[1−(14)]=1,63,072(34)= Rs. 1,22,304

Correct Answer : (C)

Explanation : Principal is Rs. 5000, Rate is 5%Amount is Rs. 6250A=P+{(P×r×t)100}Here, A= Amount, P= Principalr= Rate, t= Time Rs. 6250= Rs. [5000+{(5000×5×t)100}]⇒ Rs. [5000+(50×5×t)]= Rs. 6250⇒5000+250t=6250⇒250t=6250−5000⇒t=1250250 ⇒t=5

Correct Answer : (A)

Explanation : C.I. when interest compounded yearly, = Rs. [5000×(1+4100)×(1+12×4100)]= Rs. (5000×2625×5150)= Rs. 5304. C.I. when interest is compounded half-yearly, = Rs. [5000×(1+2100)3]= Rs. (5000×5150×5150×5150)= Rs. 5306.04 = Rs. [5000×(1+2100)3]= Rs. (5000×5150×5150×5150)= Rs. 5306.04 ∴ Difference =Rs⋅(5306.04−5304)= Rs. 2.04

Correct Answer : (D)

Explanation :

Correct Answer : (B)

Explanation : Given,P1= ₹ 20000P2= ₹ 40000R1=10%R2=15%T1=T2=3 yearsFor principal P1 ,SI1=P1R1T1100SI1=20000×10×3100SI1= ₹ 6000For principal P2 ,SI2=P2R2T2100SI2=40000×15×3100SI2= ₹ 18000Total SI=SI1+SI2=6000+18000= ₹ 24000

Correct Answer : (B)

Explanation : Given:Time period =3 yearsInterest amount = Rs. 2160Now,Let the loan amount be Rs. x .We know first-year interest is 5% and that keeps increasing by 1% with every passing year.Total interest earned =5%+6%+7%=18%So, according to the question,18% of x= Rs. 2,160⇒x=(2160×100)18⇒ Rs. 12,000∴ The loan amount was Rs. 12,000 .

Correct Answer : (A)

Explanation : Given:Two equal annual installments = Rs. 5547 each, the rate of interest (R) was 712% p.pa compounded yearlyAs we know,P=[ Installment(1+R100)1+Installment (1+R100)2]⇒P=[5547(4340)1+5547(4340)2]⇒P=5547(4043+16001849)⇒P=5547(33201849)⇒P= Rs. 9960Total Installment =5547×2= Rs. 11094Total Interest = Rs (11094−9960)= Rs. 1134∴ Total interest is Rs. 1134 .

Correct Answer : (C)

Explanation : Given,Principal (P)= Rs. 8500Rate of interest (r)=4%Time (n)=2 yearsAs we know,Compound interest =P(1+r100)n−P∴ Compound interest =8500(1+4100)2−8500=8500×(104100)2−8500=8500×(2625)2−8500=8500×676625−8500=8500×(676625−1)=8500×(676−625625)=8500×51625= Rs. 693.6

Correct Answer : (A)

Explanation : Given:Amount lent for 3 years = Rs. 5000Difference between compound interest = Rs. 2986We know that,Compound interest =P(1+R100)n−PWhere, P= principal, R= rate of interest and n= time periodAccording to the question,Compound interest in 3 years =5000×(1+10100)3−5000= Rs. 1655Compound interest in 4 years =2986+1655= Rs. 4641⇒A(1+10100)4−A=4641⇒A(1+110)4−A=4641⇒A[(1110)4−1]=4641⇒A[(14,64110,000)−1]=4641⇒A(4,64110,000)=4641⇒A=10000∴ The amount invested for 4 years is Rs. 10000 .

Correct Answer : (C)

Explanation : Given:Rate =5% p.a.Compound interest for second year = Rs. 210Let the principal amount be Rs. ‘ P ‘.Compound interest after ‘ n ‘ years = Principal × [(1+ Rate 100)n−1]C.I. after 2 years −C.I. after 1 year = Rs. 210[P×{(1+5100)2−1}]−[P×{(1+5100)−1}]=210⇒P×(1.052−1−0.05)=210⇒P×(1.1025−1−0.05)=210⇒P=2100.0525⇒P= Rs. 4000∴ Required principal amount = Rs. 4000

Correct Answer : (A)

Explanation : Let the sum be Rs. x. Then,C.I. =[x(1+4100)2−x]=(676625x−x)=51625xS.l. =(x×4×2100)=2×25∴51×625−2×25=1⇒x=625

Correct Answer : (C)

Explanation : Given:After 2 years sum amounts to Rs. 3200 and after 4 years it amounts to Rs. 12800Initial amount invested on simple interest for 3 years at a rate of 20% per annumIn compound interest, if the time interval is same then the amount will be compounded annually in the same ratioFormula used:Simple interest (SI)= Principal(P)×Rate(R)×Time(T)100And in compound interest (Cl) ,Amount (A)=P(1+R100)TCl=A−PLet the initial amount be PFor compound interest,3200=P(1+R100)2 ⋅⋅⋅⋅ ⋅ (i)12800=P(1+R100)4 ⋅⋅⋅⋅ ⋅ (ii)From equation2equation1 we get,4=(1+R100)2 ⋅⋅⋅⋅ ⋅ (iii)⇒2=1+R100⇒R=100Put the value of equation (iii) in (i)⇒3200=P×4⇒P=800∴ Sum invested by Ashish = Rs. 800If this amount invested on simple interest at 20% per annum for 3 yearsSI=(800×20×3100)=480∴ Interest received = Rs. 480

Correct Answer : (C)

Explanation : Simple interest gained on the principal amount. SI=prt100 where ‘ p ‘ is the principal amount, ‘ r ‘ is rate of interest, and ‘ t ‘ is the time period. Time period is 3 years 5 months, and 15 days ⇒3 years 5 months and 15 days =3+512+15365 =3+0.4166+0.0410 =3.457 years Simple interest for the sum of Rs. 2400 for 3.457 years at the rate of 9% . ⇒SI=prt100 =2400×9×3.457100 =746.712 Total amount A=p+SI . So, substituting the value of p as 2400 and the simple interest as 746.712 in the equation A=p+SI ⇒A=p+SI =2400+746.712 =3146.712 The amount is Rs. 3146.712 and the simple interest is Rs. 746.712.