Simple Interest (S.I.)= \( \frac{Principal * Rate * Time}{100} \)
or,
S.I.= \( \frac{P * R * T}{100} \)
P= \( \frac{S.I. * 100}{R * T} \),
R= \( \frac{S.I. * 100}{P * T} \),
T= \( \frac{S.I. * 100}{P * R} \),
If there are distinct rates of interest for distincttime periods i.e.
Rate for 1st t1 years -> R1%
Rate for 2nd t2 years -> R2%
Rate for 3rd t3 years -> R3%
or,
Then, Total S.I. for 3 years = \( \frac{P(R1T1 * R2T2 * R3T3)}{100} \)
If a certain sum becomes ‘n’ times of itself in
T years on Simple Interest, then the rate per cent per annum
is,
R% = \( \frac{(n-1)}{T} \)x100% and
T = \( \frac{(n-1)}{R} \)x100%
If a certain sum becomes n1 times of itself at
R1% rate and n2 times of itself at R2% rate, then,
R2 = \( \frac{(n2-1)}{n1-1} \)xR1 and
T2 = \( \frac{(n2-1)}{n1-1} \)xT1
If Simple Interest (S.I.) becomes ‘n’ times of
principal i.e.
S.I. = P × n then.
RT = n × 100
If an Amount (A) becomes ‘n’ times of certain
sum (P) i.e.
A = Pn then,
RT = (n – 1) × 100
If the difference between two simple interests
is ‘x’ calculated at different annual rates and times, then
principal (P) is
P = \( \frac{(x*100)}{dr * dt} \)
where dr = Difference in rate & dt = Difference in time
If a sum amounts to x1 in t years and then
this sum amounts to x2 in t yrs. Then the sum is given by
P = \( \frac{((da)*100)}{ (ci) * time} \)
where da = Difference in amount & ci = Change in time
If a sum with simple interest rate, amounts
to ‘A’ in t1 years and ‘B’ in same t2 years, then,
R% = \( \frac{(B-A)*100}{A.t2-B.t1} \) and
P = \( \frac{A.t2-B.t1}{t2-t1} \)
If a sum is to be deposited in equal
instalments, then,
Equal instalment = \( \frac{A*200}{T[200+(T-1)r]} \)
where T = no. of years, A = amount, r = Rate of Interest.
To find the rate of interest under current
deposit plan,
r = \( \frac{S.I. * 2400}{n(n+1)*dA00} \)
where n = no. of months & dA = deposited ammount
If certain sum P amounts to Rs. A1 in t1
years at rate of R% and the same sum amounts to Rs. A2 in
t2 years at same rate of interest R%. Then,
(i) R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
(ii) P = \( \frac{A2T1-A1T2}{T1-T2} \)
The difference between the S.I. for a certain
sum P1 deposited for time T1 at R1 rate of interest and
another sum P2 deposited for time T2 at R2 rate of interest
is
S.I. = \( \frac{P2R2T2-P1R1T1}{100} \)
Ans .
2
(2) Using Rule 1,
\( \frac{150*100}{4} * \frac{2}{1} \) = ₹7500
Ans .
2
(3) Using Rule 1,
Principal (P) = ₹1600
T = 2 years 3 months = \( (2+\frac{3}{12})yrs. = (2+\frac{1}{3})yrs. = \frac{9}{4}yrs. \)
S.I = ₹252
R = % rate of interest per annum
=> R = \( \frac{100*S.I.}{P*t} \)
= \( \frac{100*252}{1600*\frac{9}{4}} \)
Rate of interest = 7% per annum.
Ans .
4
(4) If the principal be x and rate
of interest be r% per annum,
then
SI after 1 year = 920 – 880 = ₹40
Therefore, SI after 2 years = ₹80
=> 880 = x + 80
=> x = ₹(880 – 80) = ₹800
Aliter : Using Rule 12,
P = \( \frac{A2T1-A1T2}{T1-T2} \)
= \( \frac{920*2-880*3}{2-3} \)
= \( \frac{1840-2640}{-1} \)
= \( \frac{-800}{-1} \)
= ₹800
Ans .
1
(1) Using Rule 1,
If rate of interest be R% p.a. then,
S.I.= \( \frac{Principal * Rate * Time}{100} \)
Therefore, \( \frac{600*2*R}{100} + \frac{1500*4*R}{100}\)
= 900
=> 120R + 60R = 900
=> 180R = 900
=> R = \( \frac{900}{180} = 5%\)
Ans .
4
(4) Using Rule 1,
Let the rate of interest per annumbe r%
According to the question,
\( \frac{5000*2*R}{100} +\frac{300*2*R}{100} = 2000 \)
=> 100R + 120R = 2200
=> 220R = 2200
=> R = \( \frac{2200}{220} =10% \)
Ans .
1
(1) Simple interest for 2 years
= ₹(568 – 520) = ₹48
Therefore, Interest for 5 years
=₹\( \frac{48}{2} \)*5 = ₹120
Principal = ₹(520 – 120) = ₹400
Aliter : Using Rule 12,
P = \( \frac{A2T1-A1T2}{T1-T2} \)
= \( \frac{568*5-520*7}{5-7} \)
= \( \frac{2840-3640}{-2} \)
= \( \frac{-800}{-2} \)
=₹ 400
Ans .
3
(3) Using Rule 1,
Simple interest gained from ₹500
= \( \frac{500*12*4}{100} \) = ₹240
Let the other Principal be x.
S.I. gained = ₹(480 – 240)= ₹240
Therefore, \( \frac{x*10*4}{100} \)=240
=> x=\( \frac{240*100}{40} \) = ₹600
Ans .
3
Difference in rate
= \( (8-7\frac{3}{4}) \) % = \( \frac{1}{4} \)%
Let the capital be ₹x.
Therefore, \( \frac{1}{4} \)% of x = 61.50
=> x = 61.50 × 100 × 4
= ₹24600
Ans .
4
(4) Using Rule 1,
Let the sum lent to C be x
According to the question,
\( \frac{2500*7*4}{100} + \frac{x*7*4}{100} = 1120 \)
or 2500 × 28 + 28x = 112000
or 2500 + x = 4000
or x = 4000 – 2500 = 1500
Ans .
4
(4) S.I. for 1\( \frac{1}{2} \)years
= (873 – 756) = 117
S.I. for 2 years
= ₹\( (117*\frac{2}{3}*2) = ₹156 \)
Therefore, Principal = 756 – 156 = ₹600
Now, P = 600, T = 2,
S.I. = 156
Therefore, R= \( \frac{100*S.I.}{P*T} \)
\( \frac{100*156}{600*2} =13 \)%
Aliter : Using Rule 12,
Rate of interest = \( \frac{A1-A2}{A2T1-A1T2} \)X100
= \( \frac{756-873}{873*2-756\frac{7}{2}} \)X100
= \( \frac{-117}{1746-2646} \)X100
= \( \frac{-117}{-900} \)X100
=13%
Ans .
3
(3) Using Rule 1,
P = \( \frac{A * 100}{100 + R * T} \)
=\( \frac{7000 * 100}{1000+\frac{10}{3}*5} \)
=\( \frac{7000*100*3}{350} \)
= ₹6000
Ans .
4
(4) Using Rule 1,
Let the principal be x.
S.I. = \( \frac{Principal * Rate * Time}{100} \)
=> 5400 = \( \frac{x * 12 * 3}{100} \)
=> \( x = \frac{5400 * 100}{12 * 3} \) = ₹15000
Ans .
3
(3) Principal + S.I. for \(\frac{5}{2}\)years = ₹1012 ...(i)
Principal + S.I. for 4 years = ₹1067.20 ...(ii)
Subtracting equation (i) from (ii)
S.I. for \(\frac{3}{2}\)years = ₹55.20
Therefore, S.I. for \(\frac{5}{2}\)years = 55.20 x \(\frac{2}{3}\) x \(\frac{5}{2}\) = ₹92
Therefore, Principal
= ₹(1012 – 92) = ₹920
Therefore, Rate = \(\frac{92*100}{920*\frac{5}{2}}\)
= \(\frac{2*92*100}{920*5}\) = 4%
Aliter : Using Rule 12,
Rate of interest = \( \frac{A1-A2}{A2T1-A1T2} \)X100
= \( \frac{1012-1067.50}{1067.50*\frac{5}{2}-1012*4} \)X100
= \( \frac{-55.2}{2668-4048} \)X100
= \( \frac{-55.2}{-1380} \)X100
=4%
Ans .
2
(2) Principal + SI for 2 years = ₹720 .... (i)
Principal + SI for 7 years = ₹1020 .....(ii)
Subtracting equation (i) from (ii) get,
SI for 5 years
= ₹(1020 – 720) = ₹300
Therefore, SI for 2 years
= ₹300 × \(\frac{2}{5}\) = ₹120
Therefore, Principal
= ₹(720 – 120) = ₹600
Aliter : Using Rule 12,
P = \( \frac{1020*2-720*7}{2-7} \)
= \( \frac{2040-5040}{-5} \)
= \( \frac{-3000}{-5} \)
= ₹600
Ans .
4
(4) Using Rule 1,
The sum of money will give ₹365
as simple interest in a year.
=> S.I. = \(\frac{PRT}{100}\)
=> 365 = \(\frac{P*5*1}{100}\)
=> P = \(\frac{365*100}{5}\) = ₹7300
Ans .
3
(3) Using Rule 1,
Let the sum be x.
Using formula, I = \(\frac{PRT}{100}\) we have
\( \frac{x*\frac{15}{12}*\frac{15}{2}}{100} \) - \( \frac{x*\frac{8}{12}*\frac{25}{2}}{100} \)
= 32.50
=> \( \frac{25x}{2400} \) = 32.50
=> x = \( \frac{32.50*2400}{25} \) = 3120
Therefore, Required sum = ₹3120
Ans .
1
(1) Let each instalment be x
Then,
\((x+\frac{x*5*1}{100})+(x+\frac{x*5*2}{100})+(x+\frac{x*5*3}{100})+x\)=6450
=> \((x+\frac{x}{20})+(x+\frac{x}{10})+(x+\frac{3x}{20})+x\)=6450
=> \((\frac{21x}{20})+(\frac{11x}{10})+(\frac{23x}{20})+x\)=6450
=> \( \frac{21x+22x+23x+20x}{20} \)=6450
=> \( \frac{86x}{20} \) = 6450
=> x = \( \frac{6450*20}{86} \) = ₹1500
Aliter : Using Rule 10,
Equal instalment
= \( \frac{6450*200}{4[200+(4-1)*5]} \)
= \(\frac{6450*200}{4(215)}\)
= \(\frac{6450*50}{215}\)
= ₹1500
Ans .
1
(1) Using Rule 1,
Interest = ₹(81–72)= ₹9
Let the time be t years.
Then, 9 = \( \frac{72*25*t}{4*100} \)
=> t = \(\frac{9*400}{72*25}\)=2 years
Ans .
1
Using Rule 1,
Time from 11 May to 10 September,
1987
= 21 + 30 + 31 + 31 + 10
= 123 days
Therefore, 123 days = \( \frac{123}{365} \) year
Therefore, S.I. = \( \frac{7300*123*5}{365*100}\) = ₹123
Ans .
3
(3) Using Rule 1,
Case I :
S.I. = \( \frac{5000*2*4}{100}\) = ₹400
Case II :
S.I. = \( \frac{5000*25*2}{100*4}\) = ₹625
Therefore, ₹(625-400) = ₹225
Ans .
3
(3) Using Rule 1,
Let the sum lent at 4% = Rs.x
Therefore, Amount at 5%= (16000 – x )
According to the question,
\( \frac{x*4*1}{100} + \frac{(16000-x)*5*1}{100}\)
= 700
=> 4x + 80000 – 5x = 70000
=> x = 80000 – 70000
= Rs. 10000
Ans .
4
(4) Using Rule 1,
After 10 years,
SI = \( \frac{1000*5*10}{100} \)= ₹ 500
Principal for 11th year
= 1000 + 500 = 1500
SI = ₹(2000 – 1500) = ₹500
Therefore, T = \( \frac{SI*100}{P*R} = \frac{500*100}{1500*5} = \frac{20}{3}\) yeras =6\(\frac{2}{3}\) years
Therefore, Total time = 10 + 6\( \frac{2}{3}\) = 16\( \frac{2}{3}\) years
Ans .
4
(4)
P + S.I. for 5 years = 5200 ..(i)
P + SI for 7 years = 5680 ...(ii)
On subtracting equation (i) from (ii
SI for 2 years = 480
Therefore, SI for 1 year = ₹240
Therefore, From equation (i),
P + 5 × 240 = 5200
=> P = 5200 – 1200 = ₹4000
Therefore, R = \( \frac{SI*100}{T*P} \)= \( \frac{240*100}{1*4000} \) = 6%
Aliter: Using Rule 12,
R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
= \( \frac{5200-5680}{5680*5-5200*7} \)X100
= \( \frac{-480}{28400-36400} \)X100
= \( \frac{-480}{-8000} \)X100
=6%
Ans .
3
(3) Using Rule 1,
S.I. = 956 – 800 = Rs. 156
Therefore, Rate = \(\frac{SI*100}{Principal*Time}\) = \(\frac{156*100}{800*3}\) = 6.5% per annum
Therefore, New rate = 10.5%
Therefore, S.I. = \(\frac{Principal*Time*Rate}{100}\) = \(\frac{800*3*10.5}{100}\) = ₹252
Therefore, ammount = 800+252 = ₹1052
Ans .
1
(1) Using Rule 1,
Let the rate of interest be R per
cent per annum.
Therefore, \( \frac{400*2*R}{100} + \frac{550*4*R}{100} + \frac{1200*6*R}{100} \) = 1020
=> 8R+22R+72R = 1020
=> 102R = 1020
=> R=\(\frac{1020}{102}\)=10%
Ans .
1
(1) Using Rule 1,
4200 = \( \frac{29400*6*R}{100} \)
=> R = \(\frac{4200}{294*6}\) = \(\frac{50}{21}\) = 2\(\frac{8}{21}\)%
Ans .
2
(2) Using Rule 1,
Let the amount lent at 4% be x
\ Amount lent at 5% = (60000 – x )
According to the question,
\(\frac{(60000-x)*5*1}{100}\) + \(\frac{x*4*1}{100}\)
=> 2560
=> 300000-5x + 4x = 256000
=> x = 300000-256000=₹44000
Ans .
4
(4) Principal + interest for 8 years= ₹2900... (i)
Principal + interest for 10 years = ₹3000 ... (ii)
Subtracting equation (i) from (ii)
Interest for 2 years = 100
Therefore, Interest for 8 years = \(\frac{100}{2}\)*8 = ₹400
From equation (i),
Principal = ₹(2900 – 400) = ₹2500
Therefore, Rate = \( \frac{SI*100}{Time*Principal} \) = \( \frac{400*100}{8*2500} \)=2%
Aliter : Using Rule 12,
R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
= \( \frac{2900-3000}{3000*8-2900*10} \)X100
R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
= \( \frac{-100}{24000-29000} \)X100
= \( \frac{-100}{-5000} \)X100
= 2%
Ans .
1
Using Rule 1,
Time = \( \frac{S.I. * 100}{P * R} \) =\( \frac{1080*100}{3000*12} \) = 3 years
Ans .
3
(3) Interest for 1 year
= ₹(925 – 850) = ₹75
Therefore, If a sum becomes a1 in t1 years
and a2 in t2 years then rate of
interest = \( \frac{100(a2-a1)}{(a1t2-a2t1)} \)%
=\( \frac{100(925-850)}{850*4-3*925} \) = \(\frac{7500}{625}\) = 12%
Therefore, Principal = \( \frac{SI*100}{Time*Rate} \) = \( \frac{75*100}{1*12} \) = ₹625
Aliter : Using Rule 12,
P = \( \frac{A2T1-A1T2}{T1-T2} \)
= \( \frac{925*3-850*4}{3-4} \)
= \( \frac{2775-3400}{-1} \)
= \( \frac{-625}{-1} \)
= ₹625
Ans .
2
(2) Using Rule 1,
S.I. = 2641.20 – 1860
= 781.2
Time = \( \frac{S.I. * 100}{P * R} \) =\( \frac{781.2*100}{1860*12} \) = 3.5 = 3\(\frac{1}{2}\) years
Ans .
2
(2) Using Rule 18 of ‘percentage’ chapter,
Present population = 10000 \( (1-\frac{20}{100}) \)2
= 10000 \( * \frac{4}{5} * \frac{4}{5} \) = 6400
Ans .
3
(3) Using Rule 1,
Annual interest
= 365 × 2 = ₹730
Principal = \( \frac{SI*100}{Time*Rate} \)
= \( \frac{730*100}{1*5} \) = ₹14600
Ans .
3
(4) If principal = x and rate = r%
per annum, then
1380 = x + \(\frac{x*3*r}{100}\) .....(i)
1500 = x + \(\frac{x*5*r}{100}\) ....(ii)
S.T. for to years = 1500-1380 = ₹120
Therefore, \(\frac{x*2*r}{100}\) =120
Therefore, \(\frac{xr}{100}\) = 60
Therefore, From equation(i)
1380 = x + 60 × 3
=> x = 1380 – 180 = ₹1200
From equation (iii)
\( \frac{1200*r}{100} \) = 60
=> r = \( \frac{6000}{1200} \) = 5% per annum
Aliter : Using rule 12
R = \( \frac{A1-A2}{A2T1-A1T2} \)X100%
= \( \frac{1380-1500}{1500*3-1380*5} \)X100
= \( \frac{-120}{4500-6900} \)X100%
= \( \frac{-120}{-2400} \)X100
= 5%
Ans .
3
S.I. for 1 year = 14250 – 12900 = Rs. 1350
S.I. for 4 years = 1350 × 4 =Rs. 5400
Therefore, Principal = 12900 – 5400 =Rs. 7500
Therefore, Rate = \( \frac{SI*100}{Principal*Time} \)
= \( \frac{5400*100}{7500*4}\)
=18% per annum
Aliter : Using Rule 12,
R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
= \( \frac{12900-14250}{14250*4-12900*5} \)X100
= \( \frac{-1350}{57000-64500} \)X100
= \( \frac{1350}{7500} \)X100
= 18%
Ans .
1
(1) Using Rule 1,
Required time = t years
S.I. = \( \frac{Principal * Rate * Time}{100} \)
Therefore, \( \frac{6000*5*4}{1000} \) = \( \frac{8000*3*t}{100} \)
=> 6000*4*5=8000*3*t
Therefore, t = \( \frac{6000*4*5}{8000*3} \)=5 years
Ans .
2
(2) Using Rule 1,
Principal = \( \frac{S.I. * 100}{R * T} \)
= \( \frac{1*100}{\frac{1}{365}*5} \) = \( \frac{365*100}{5} \)
= Rs. 7300
Ans .
1
S.I. for 5 years
= Rs. (1020 – 720) = Rs. 300
Therefore, S.I. for 2 years
= \( \frac{300}{5} * 2\) = Rs.120
Therefore, Principal = Rs. (720 – 120)
= Rs. 600
Ans .
4
(4) Using Rule 1,
Number of days from 5th January to 31st May = 26 + 28 + 31 + 30 + 31 = 146
Therefore, S.I. = \( \frac{P*T*R}{100} \) = \( \frac{36000*146*9.5}{365*100} \) = Rs.1368
Ans .
3
(3)\( \frac{Principal}{Interest} \) = \( \frac{10}{3} \)
\( \frac{Interest}{Principal} \) = \( \frac{3}{10} \)
Time = \( \frac{SI*100}{P*R} \)
\( \frac{3}{10} * \frac{100}{6} \) = 5 years
Ans .
1
(1) Principal = \( \frac{S.I. * 100}{R * T} \)
=\( \frac{60*100}{5*6} \) = Rs.200
Ans .
1
(1) According to the question,
S.I. for 2 years 6 months
= Rs. (5500 – 4000)
=> S.I. for \( \frac{5}{2} \) years = Rs.1500
Therefore, S.I. for 1 year = \( \frac{1500*2}{5} \)
= Rs.600
Rate = \( \frac{S.I.*100}{Principal*Time} \)
= \( \frac{1200*100}{2800*2} \) = \( \frac{150}{7} \)
= 21 \( \frac{3}{7}% \)per annum
Ans .
2
(2) Principal = \( \frac{S.I. * 100}{R * T} \)
= \( \frac{840*100}{8*5} \)
= Rs.2100
Case II,
S.I. = Rs. 840
Principal = Rs. 2100
Time = 5 years
Rate = \( \frac{S.I. * 100}{P * T} \)
=\( \frac{840 * 100}{2100*5} \)
= 8 % per annum
Ans .
2
(2) Let first part be x.
Therefore, Second part
= Rs. (2800 – x)
According to the question,
(S.I.)= \( \frac{Principal * Rate * Time}{100} \)
Therefore, \( \frac{x*5*9}{100}\) = \( \frac{(2800-x)*6*10}{100} \)
3x = 4*2800-4x
7x = 4*2800
x = \( \frac{4-2800}{7} \)
=Rs.1600
Ans .
2
(2) According to the question,
\( \frac{S.I.}{Principal} = \frac{2}{5} \)
Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{2}{5} * \frac{100}{5} \) = 8% per annum
Ans .
1
Rate = \( \frac{S.I. * 100}{P * T} \)
\( \frac{280*100}{400*10} \)
= 7% per annum
Ans .
3
Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{\frac{1}{100}*100}{1*\frac{1}{12}} \)= 12% per annum
Ans .
3
(3)S.I.= \( \frac{Principal * Rate * Time}{100} \)
=Rs. \( (4000*\frac{18}{12}*\frac{12}{100}) \)
= Rs.720
Ans .
2
(2)T= \( \frac{S.I. * 100}{P * R} \)
=\( \frac{1080*100}{3000*12} \)
= 3 yers
Ans .
3
(3) Let the principal be Rs. x.
According to the question,
x + S.I. for 2 years = Rs. 5182 ...(i)
x + S.I. for 3 years = Rs. 5832 ...(ii)
By equation (ii) – (i),
S.I. for 1 year
= Rs. (5832 – 5182)
= Rs. 650
Therefore, S.I. for 2 years
= Rs. (2 × 650) = Rs. 1300
Therefore, Principal
= Rs. (5182 – 1300)
= Rs. 3882
Ans .
2
P= \( \frac{S.I. * 100}{R * T} \)
= \( \frac{R * 100}{R * 2} \)
= Rs.50
Ans .
3
(3)S.I.= \( \frac{Principal * Rate * Time}{100} \)
= \( \frac{2000*2*5}{100} \) = Rs.250
Therefore, Required amount
= Rs. (2000 + 200)
= Rs. 2200
Ans .
1
(1) S.I. = Amount – Principal
= Rs. (6900 – 6000)
= Rs. 900
Therefore, Rate =\( \frac{S.I. * 100}{P * T} \)
= \( \frac{900 * 100}{6000*3} \)= 5% per annum
Ans .
1
(1)Using Rule 3,
R% = \( \frac{(\frac{7}{6}-1)*100}{3} \)%
= \( \frac{1}{18} *100\)%
= \(\frac{50}{9}\)%
= \( 5\frac{5}{9} \)%
Ans .
1
(1)Using Rule 3,
R% = \( \frac{(\frac{41}{40}-1)*100}{\frac{1}{4}} \)%
= \( \frac{1}{40}*4 *100 \)%
= 10%
Ans .
2
(1)Using Rule 3,
R% = \( \frac{(3-1)*100}{20} \)%
= 10%
Now , T = \( \frac{(n-1)}{R} \)x100%
= \(\frac{2-1}{10}*100\)%
= 10 years
Ans .
4
(4) Using Rule 1,
Let P be the principal and R%
rate of interest.
Therefore, S.I. = \( \frac{Principal * Rate * 10}{100} \) = \( \frac{Principal * Rate}{10} \)
According to the question,
\( \frac{PR}{10}\) = \( (P+\frac{PR}{10}) * \frac{2}{5}\)
\( \frac{R}{10}\) = \( (1+\frac{R}{10}) * \frac{2}{5}\)
\( \frac{R}{10}\) = \( \frac{R}{25} + \frac{2}{5}\)
\( \frac{R}{10}\) - \( \frac{R}{25} = \frac{2}{5}\)
\( \frac{5R-2R}{50}\) = \frac{2}{5}\)
\( \frac{3R}{50}\) = \frac{2}{5}\)
R = \( \frac{50*2}{3*5} = \frac{20}{3} = 6\frac{2}{3}%
Ans .
2
(2) Using Rule 1,
SI = (7200–6000)
= 1200
Therefore, SI = \( \frac{PRT}{100} \)
1200 = \( \frac{600*R*4}{100} \)
R = \( \frac{1200*100}{6000*4} \)= 5%
New rate of R = 5×1.5 = 7.5%
Then, SI = \( \frac{6000*7.5*5}{100} \) = ₹2250
Therefore, Amount = (6000 + 2250) = ₹8250
Ans .
3
(3)Using Rule 3,
R% = \( \frac{(3-1)*100}{15} \)%
= \(\frac{40}{3}\)%
Now , T = \( \frac{(n-1)}{R} \) years
= \(\frac{2-1}{\frac{40}{3}}*100\)
= 30 years
Ans .
3
(3)Using Rule 3,
R% = \( \frac{(2-1)*100}{12} \)%
= \(\frac{25}{3}\)%
= \(8\frac{1}{3}\)%
Ans .
3
(3)Using Rule 3,
R% = \( \frac{(\frac{7}{4}-1)*100}{12} \)%
= \(\frac{75}{4}\)%
= \(18\frac{3}{4}\)%
Ans .
4
(4)Using Rule 3,
R1 = \( \frac{(2-1)*100}{5} \)%
= 20%
R2 = \( \frac{(3-1)*100}{12} \)%
= \(16\frac{2}{3}\)%
Lowest rate of interest = = \(16\frac{2}{3}\)%
Ans .
3
(3)Using Rule 3,
T = \( \frac{(n-1)}{R} \)% years
= \(\frac{2-1}{\frac{25}{4}}*100\) years
= 16 years
Ans .
3
(3)Using Rule 3,
R% = \( \frac{(2-1)*100}{10} \)%
= 10%
Now , T = \( \frac{(n-1)}{R} \)*100 years
= \(\frac{3-1}{10}*100\)
= 20 years
Ans .
4
(4)Using Rule 3,
T = \( \frac{(n-1)}{R} \)% years
= \(\frac{2-1}{15}*100\)
= \(\frac{20}{3}\) years
= \(6\frac{2}{3}\) years
Ans .
3
(3)Using Rule 3,
T = \( \frac{(n-1)}{R} \)% years
= \(\frac{2-1}{12}*100\)
= \(\frac{25}{3}\) years
= \(8\frac{1}{3}\) years
= 8 years, 4 months.
Ans .
2
(2)Using Rule 3,
R% = \( \frac{(2-1)*100}{8} \)%
= 12.5%
Ans .
2
(2)Using Rule 3,
R% = \(\frac{n-1}{T}*100\) %
=( \frac{(2-1)*100}{16} \)%
= \(6\frac{1}{4}\)%
Now , T = \( \frac{(n-1)}{R} \)*100
= \(\frac{3-1}{\frac{25}{4}}*100\)
= 32 years
Ans .
1
(1)Using Rule 3,
T = \( (\frac{(n-1)}{R}) \)*100 %
= \(\frac{3-1}{\frac{25}{4}}*100\)
= 16 years
Ans .
2
(2) Using Rule 1,
Rate = R% per annum
Therefore, Time = \(\frac{R}{2}\) years
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
R = \( \frac{8}{25} * \frac{100}{\frac{R}{2}}\)
R2 = \( \frac{8*200}{25} \) = 64
R = 8% per annum
Ans .
3
(3) Case I,
Interest = Principal
Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{100}{7} \)% per annum
Case II,
Interest = 3 × Principal
Time = \( \frac{S.I. * 100}{P * R} \)
= \( \frac{3*100}{\frac{100}{7}} \)
= 3 * 7 = 21 years
Ans .
4
(4)Difference in rates = 8 – 5 = 3%
Since, 3% => 2320 – 2200 = 120
Therefore, 5% => \( \frac{120}{3} \) *5 = 200
Therefore, Principal = Rs. (2200 – 200) = Rs. 2000
Therefore, Time = \( \frac{S.I. * 100}{P * R} \)
= \( \frac{200 * 100}{2000*5} \) = 2 years
Ans .
2
(2) Let principal be Rs. x.
Therefore, Amount = Rs. 2x
Therefore, Interest = Rs. (2x – x)
= Rs. x
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{x*100}{x*15} \) = \( \frac{20}{3} \)
= \(6\frac{2}{3}\)% per annum
Ans .
2
(2) Principal = Rs. x
Interest = Rs. x
Time = 6 years
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{x*100}{x*16} \) = \( \frac{50}{3} \)
Case II,
Interest = \(\frac{x*12*50}{100*3}\) = Rs.2x
i.e., Amount is thrice the principal.
Ans .
3
(3) Principal = Rs. x (let)
Therefore, Amount = Rs. 5x
Interest = Rs. (5x – x) = Rs. 4x
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{4x * 100}{x*8} \)= 5% per annum
Ans .
4
(4) Let principal be Rs. x.
Therefore, Amount = Rs. 2x
Interest = Rs. (2x – x) = Rs. x
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{x * 100}{x*8} \) = \(12\frac{1}{2}\)% per annum
Ans .
3
(3) According to the question,
Principal = Rs. x.
Interest = Rs. x.
Time = \(\frac{50}{3}\)years
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{x * 100}{x*\frac{50}{3}} \) = 6% per annum
Ans .
3
(3)Using Rule 5,
Here, n = \( \frac{2}{5} \)and R = 8%
RT = n*100
T = \(\frac{n*100}{R}\)
T = \(\frac{2}{5}*\frac{100}{8}\)
T = 5 years
Ans .
2
(2)Using Rule 5,
Here, n = \( \frac{1}{5} \)and T = 4 years
R = \(\frac{n*100}{T}\)
R = \(\frac{1}{5}*\frac{100}{4}\)
R = 5 %
Ans .
1
(1)Using Rule 5,
Here, n = \( \frac{9}{25} \)and T = 6 years
R = \(\frac{n*100}{T}\)
R = \(\frac{9}{25}*\frac{100}{6}\)
R = 6%
Ans .
1
(1)Using Rule 5,
Here, n = \( \frac{1}{4} \)and T = 5 years
R = \(\frac{n*100}{T}\)
R = \(\frac{1}{4}*\frac{100}{5}\)
R = 5%
Ans .
2
(2)Using Rule 5,
Here, n = \( \frac{3}{8} \)and T = \(\frac{25}{4}\) years
R = \(\frac{n*100}{T}\)
R = \(\frac{3}{8}*\frac{100}{\frac{25}{4}}\)
R = 6%
Ans .
2
(2) Using Rule 1,
S.I. = \( \frac{Principal * Rate * Time}{100} \)
Therefore, 1200 + \( \frac{1200*7*r}{12*100} \) = Amount (A)
=> 1200 + 7r = A .........(i)
and, 1016 + \( \frac{1016*5*r}{2*100} \) = A
Therefore, 1016 + 25.4r = A ...(ii)
\ 1016 + 25.4r = 1200 + 7r
25.4r – 7r = 1200 – 1016
18.4r = 184
r = \(\frac{184}{18.4}\) = 10% per annum
Ans .
2
Using Rule 5,
Here, S.I. =\(\frac{2}{5}\)amount
S.I. =\(\frac{2}{5}\) (P+S.I.)
S.I. =\(\frac{2}{5}\)S.I. +\(\frac{2}{5}\) P
\(\frac{3}{5}\) = \(\frac{2}{5}\)P
S.I. =\(\frac{2}{3}\)P
Here, n = \( \frac{2}{3} \)and T = 10 years
R = \(\frac{n*100}{T}\)
R = \(\frac{2}{3}*\frac{100}{10}\)
R = \(6\frac{2}{2}\)%
Ans .
3
(3) Using Rule 13,
Here, P1 = Rs. 3000,
R1 = R,
T1 =\(\frac{5}{2}\) years,
P2 = Rs. 3200,
R2 = R,
T2 =\(\frac{5}{2}\) years,
Difference S.I. = Rs. 40
40 = \( \frac{3200*R*\frac{5}{2}-3000R*\frac{5}{2}}{100} \)
4000 = 8000R - 7500R
R = 8%
Ans .
4
(4)Using Rule 13,
P1 = P, R1 = 5%, T1 = 3years.
P2 = P, R2 = 5%, T2 = 4 years.
S.I.= 42
42 = \(\frac{20P-15P}{100}\)
P = 42 × 20
P = ₹840
Ans .
3
(3)Using Rule 13,
P1 = Rs. 1500, R1, T1 = 3 years.
P2 = Rs. 1500, R2, T2 = 3 years.
S.I. = Rs. 13.50
13.50 = \(\frac{1500*R2*3-1500*R1*3}{100}\)
\(\frac{1350}{100}\) = \(\frac{4500(R2-R1)}{100}\)
R2-R1=\(\frac{1350}{4500}\)=\(\frac{27}{90}\)=\(\frac{3}{10}\)=0.3%
Ans .
2
(2) Using Rule 1,
We know that
S.I. = \(\frac{PRT}{100}\)
According to question,
S.I.= \(\frac{4}{9}\)P
& R = T (numerically)
Therefore, \(\frac{4}{9}\)P = \(\frac{P*R*R}{100}\)
Therefore, R2 = \(\frac{400}{9}\)
R= \(\frac{20}{3}\)=\(6\frac{2}{3}\)%
Ans .
4
(4)Using Rule 13,
P1 = P, R1 = 4%, T1
= 8 months = \(\frac{8}{12}\)years
P2 = P, R2 = 5%, T2
= 15 month = \(\frac{15}{12}\) years
S.I. = 129
129 = \(\frac{P*5*\frac{15}{12}-P*4*\frac{8}{12}}{100}\)
12900 = \( \frac{75P-32P}{12} \)
12900 = \(\frac{43P}{12}\)
P = ₹3600
Ans .
4
(4) Using Rule 1,
Let the sum lent in each case be
x.
Then, \( \frac{x*9*2}{100} + \frac{x*10*2}{100} \) = 760
\( \frac{x*2}{100}(9+10) \) = 760
\( \frac{2*19x}{100} \)
x = \(\frac{760*100}{2*19}\) = ₹2000
Ans .
1
(1) Using Rule 5,
Here , n = \(\frac{16}{25}\) , R=T
Now , R*R = \(\frac{16}{25}*100\)
R2 = \(\frac{1600}{25}\)
R= \(\frac{40}{5}\)=8%
Ans .
3
(3) Using Rule 1,
Let the sum lent out at 12.5% be x
Therefore, Sum lent out at 10% = 1500 – x
Now, \( \frac{(1500-x)*10*5}{100} \) = \( \frac{x*12.5*4}{100} \)
50(1500-x)=50x
2x=1500
x=\( \frac{1500}{2} \) = ₹750
Ans .
1
Using Rule 5,
Here, n = \(\frac{30}{100} = \frac{3}{10} \), T = 6 years.
RT = n × 100
R × 6 = \(\frac{3}{10}\)*100
R = 5%
As,S.I. = P
S.I. = \( \frac{P * R * T}{100} \)
100 = RT
100 = 5 × T
This is possible only when T = 20.
Ans .
3
(3) Using Rule 1,
Let the period of time be T years.
Then, \( \frac{400*5*T}{100} = \frac{500*4*6.25}{100} \)
T = \( \frac{500*4*6.25}{400*5} \) = \(\frac{25}{4}\)
= 6\(\frac{1}{4}\) years
Ans .
2
(2)Using Rule 5,
Here, n = \(\frac{1}{16}\), R = T
RT = n × 100
R2 = \(\frac{100}{16}\)
R = \(\frac{10}{4}\)
R = \(2\frac{1}{2}\)%
Ans .
1
(1) Using Rule 1,
Let the larger part of the sum be x
Therefore, Smaller part = (12000 – x)
According to the question,
\( \frac{x*3*12}{100} = \frac{(12000-x)*9*16}{2*100} \)
36 x = (12000 – x ) 72
x = (12000 – x ) × 2
x + 2x = 24000
3x = 24000
x = \(\frac{24000}{3}\)
= ₹8000
Ans .
2
Using Rule 5,
n = \(\frac{1}{5}\), R = T
RT = n × 100
R2 = \(\frac{1}{4}*100\)
R2 = 25
R = 5%
Ans .
3
Using Rule 13,
Here, P1 = P, R1 = 7.5%, T1 = 4 years.
P2 = P, R2 = 7.5%, T2 = 5 years.
S.I. = Rs. 150
S.I. = \( \frac{P2R2T2-P1R1T1}{100} \)
150 = \( \frac{P*7.5*5-P*7.5*4}{100} \)
15000 = 7.5P
P = \(\frac{15000}{7.5}\)
P = ₹2000
Ans .
1
(1) Using Rule 1,
Let first part be x and second
part be(1750 –x )
According to the question,
x × \(\frac{8}{100}\) = (1750 – x ) * \(\frac{6}{100}\)
8x + 6x = 1750 × 6
14x = 1750 × 6
x = \(\frac{1750*6}{14}\)= 750
Therefore, Interest = 8% of 750
= 750 * \( \frac{8}{100} \) = 60
Ans .
3
(3) Using Rule 1,
Let the period of time be T years.
Therefore, 800 + \( \frac{800*12*T}{100} \) = 910 + \(\frac{910*10*T}{100}\)
800 + 96 T = 910 + 91T
96 T – 91 T = 910 – 800
5T = 110
T = \(\frac{110}{5}\) = 22 years.
Ans .
3
(3)Using Rule 5,
Here, n = \(\frac{1}{9}\) , R = T
RT = n × 100
R2 =\(\frac{1}{9}\)*100
R2 = \(\frac{100}{9}\)
R =\(\frac{10}{3}\)
R = 3\(\frac{1}{3}\)%
Ans .
2
(2) 411, Using Rule 1,
Let 'r' be the rate of interest
190 = \(\frac{500*4*r}{100}+\frac{600*3*r}{100}\)
20r + 18r = 190 v
38r = 190
r = \( \frac{190}{38} \) = 5%
Ans .
2
(2)Using Rule 7,
Here, P = Rs. 500, x = Rs. 2.50,
Difference in time = 2 years.
Difference in rate = ?
500 = \(\frac{2.50*100}{(diff. in rate)*2}\)
Different in rate = 0.25%
Ans .
3
(3) Using Rule 1,
Let the principal be x.
Time = \( \frac{S.I. * 100}{P * R} \)
= \(\frac{x*100*3}{x*50}\) = 6 years
Ans .
2
(2) Using Rule 1,
\( \frac{P*R*1}{100} = \frac{P*5*2}{100} \)
[Since, Capital is same in both cases]
r × 1 = 5 × 2
r = 10%
Ans .
1
(1) Using Rule 1,
S.I. = \( \frac{Principal * Rate * Time}{100} \)
Therefore, \( \frac{4000*3*x}{100} \) = \( \frac{5000*2*12}{100} \)
x = \( \frac{5*2*12}{4*3} \).
= 10% per annum
Ans .
4
(4) Using Rule 1,
S.I. = \( \frac{P * R * T}{100} \)
Therefore, y = \( \frac{x*T*R}{100} \)
and z = \( \frac{y*T*R}{100} \)
So, \(\frac{y}{z} = \frac{x}{y}\)
y2= zx
Ans .
1
(1) Using Rule 1,
Amount lent at 8% rate of interest = ₹x
Therefore, Amount lent at \(\frac{4}{3}\)% rate of interest = (20,000 – x)
Therefore, S.I. = \( \frac{Principal * Rate * Time}{100} \)
\( \frac{x*8*1}{100} + \frac{(20000-x)*\frac{4}{3}*1}{100} \) = 800
\( \frac{2x}{25} + \frac{20000-x}{75} \) = 800
\( \frac{6x+20000-x}{75} \) = 800
5x + 20,000 = 75 × 800 = 60,000
5x = 60,000-20,000 = 40,000
x = \(\frac{40000}{5}\)= 8000
Ans .
3
(3)Using Rule 13.
Here, P1 = Rs. P, R1 = 12%, T1 = 4 years
P2 = Rs. P, R2 = 15%, T2 = 5 years
S.I. = Rs. 1350
S.I.= \( \frac{P2R2T2-P1R1T1}{100} \)
1350 = \( \frac{P*15*5-P*12*4}{100} \)
135000 = 75 P – 48P
135000 = 75 P
P = Rs. 5000
Ans .
3
(3) Using Rule 1,
True discount = \( \frac{Amount*Rate*Time}{100+(Rate*Time)} \)
\(\frac{2400*5*4}{100+(5*4)}\)
= \(\frac{2400*5*4}{120}\) = Rs.400
S.I. = \(\frac{2400*5*4}{100}\) = Rs.480
Required difference = Rs. (480 – 400) = Rs. 80
Ans .
4
(4) Using Rule 1,
Let the sum lent at the rate of
interest 5% per annum is x and
at the rate of interest 8% per
annum is (1550 – x)
According to the question,
\( \frac{x*5*3}{100} + \frac{(1500-x)*8*3}{100} \) = 300
\( \frac{15x}{100} + \frac{37200-24x}{100} \) = 300
15x + 37200 – 24x = 300 × 100
9x = 7200
Therefore, x = ₹800 and,
1550 – x = 1550 – 800 = ₹750
Therefore, Ratio of money lent at 5% to that at 8% = 800 : 750 = 16 : 15
Ans .
2
(2) Using Rule 1,
Let the sum of x be lent at the
rate of 4% and (5000 – x) at the
rate of 5%
\( \frac{x*4*2}{100} + \frac{(5000-x)*5*2}{100} \) = 440
8x + 50000 – 10x = 44000
2x = 50000 – 44000 = 6000
x = ₹3000
Therefore, ₹(5000 – x)
= ₹(5000 – 3000) = ₹2000
Now, Required ratio
= 3000 : 2000 = 3 : 2
Ans .
4
(4) Required ratio = 5 : \(\frac{2}{5}\) = 25 : 2
\( \frac{loan amount }{Interest Amount} = \frac{5}{2} \)
Interest Rate = \(\frac{2}{5}\)
Since, \( [ \frac{P+I}{I} = \frac{5}{2}=> \frac{P}{I}+1 = \frac{5}{2} => \frac{P}{I} = \frac{3}{2} => I = \frac{2}{5} ] \)
\( \frac{loan amount }{Interest Rate} = \frac{5}{\frac{2}{5}} \)
\( \frac{25}{2} \) or 25:2
Ans .
1
(1) Using Rule 1,
P1 : P2 : P3 = \( \frac{1}{r1t1} : \frac{1}{r2t2} : \frac{1}{r3t3} \)
= \( \frac{1}{6*10} : \frac{1}{10*12} : \frac{1}{12*15} \)
\( \frac{1}{60} : \frac{1}{120} : \frac{1}{180} \)
= 6:3:2
Ans .
3
(3) Using Rule 1,
Case-I,
Interest = 5x – 4x = x
Therefore, x = \(\frac{4x*R*T}{100}\)
T =\( \frac{25}{R} \) years
Case-II,
T = \( \frac{25}{R} \)+ 3 = \( \frac{25+3R}{R} \) years
SI = 7 y – 5y = 2y
Therefore, 2y = \( \frac{5y*R*(25+3R)}{R*100} \)
40 = 25 + 3R
3R = 40 –25 = 15 %
R = \( \frac{15}{3} \) =5%
Ans .
4
(4) Using Rule 1,
\( \frac{Principal}{Amount} = \frac{10}{12} \)
\( \frac{Amount}{Principal} = \frac{Principal+Interest}{Principal} \)
= \(\frac{12}{10}\)
1 + \(\frac{Interest}{Principal} = \frac{12}{10} \)
\(\frac{Interest}{Principal} = \frac{2}{10} = \frac{1}{5} \)
Therefore, Rate = \(\frac{1}{5}\)*100 = 20%
Ans .
2
(2) Using Rule 1,
Time = \( \frac{S.I. * 100}{P * R} \)
= \( \frac{3}{10} * \frac{100}{10} \) = 3 years
Ans .
1
(1) Using Rule 1,
First part = Rs. x and second part
= (12000 – x )
Therefore, \( \frac{x*3*12}{100} = \frac{(12000-x)*9*16}{200} \)
\(\frac{x}{12000-x} = \frac{9*16*100}{3*12*200} \)
\( \frac{2}{1} \)= 2 : 1
Ans .
1
(1) Using Rule 1,
Principal : Interest = 25 : 1
Interest : Principal = 1 : 25
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \(\frac{1}{25}\) × 100 = 4% per annum
Ans .
2
(2) Using Rule 1,
\( \frac{Principal}{Interest} = \frac{10}{3} \)
\( \frac{Interest}{Principal} = \frac{3}{10} \)
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \(\frac{3}{10}*\frac{100}{5}\) = 6% per annum
Ans .
3
(3) Principal lent at 8% S.I. = Rs. x.
Therefore, Principal lent at 10% S.I. = Rs. (4000 – x)
S.I. = \( \frac{Principal * Rate * Time}{100} \)
\(\frac{x*8}{100}+\frac{(4000-x)*10}{100}\) = 352
8x + 40000 – 10x = 35200
2x = 40000 – 35200 = 4800
x = \( \frac{4800}{2} \) = Rs. 2400
Ans .
3
(3) Using Rule 1,
Interest = ₹. (480–400) = ₹80
Therefore, 80 = \(\frac{400*r*4}{100}\)
r = 5
Now, r = 7% (2% increase)
Therefore, S.I. = \( \frac{400*7*4}{100} \) = 112
Therefore, Amount = ₹(400+112) = ₹512
Ans .
1
1) Using Rule 1,
Let his capital be x.
According to the question,
\( \frac{x*11.5}{100} - \frac{x*10}{100} = 55.50 \)
or (11.5 – 10)x = 5550
or 1.5x = 5550
or x = \( \frac{5550}{1.5} \) = ₹3700
Ans .
1
(1) Using Rule 1,
Change in SI
= \( (\frac{25}{2}-10) \)% = \(\frac{5}{2}\)%
Therefore, \(\frac{5}{2}\)% of principal = ₹1250
Therefore, Principal = ₹ \( \frac{1250*2*100}{5} \) = ₹50000
Ans .
1
(1)Using Rule 13,
P1 = P, R1 = R, T1 = 2
P2 = P, R2 = R + 3, T2 = 2
S.I.= 72
72 = \( \frac{P*(R+3)*2-P*R*2}{100} \)
7200 = 6P
P = 1200
Ans .
4
(4) If the sum lent be Rs. x, then
\( \frac{x*2.5*3}{100} \) = 540
x = \(\frac{540*100}{2.5*3}\) = ₹7200
Ans .
1
(1) \( \frac{P*1*2}{100} \) = 24
P = \( \frac{2400}{2} \) = ₹1200
Ans .
3
(3) If the capital after tax deduction
be x, then
x × (4 – 3.75) % = 48
\(\frac{x*0.25}{100}\)= 48
\(\frac{x*25}{10000}\)= 48
\(\frac{x}{400}\)= 48
x = 48 × 400 = ₹19200
Therefore, Required capital
= \( \frac{19200*100}{96} \)
= ₹20000
Ans .
1
(1)Using Rule 13.
P1 = P, R1 = R, T1 = 2.
P2 = P, R2 = R + 3, T2 = 2.
S.I.= 300
300 = \( \frac{P*(R+3)*2-P*R*2}{100} \)
300 = \( \frac{6P}{100} \)
p = ₹5000
Ans .
4
(4) Using Rule 1,
S.I. = 3264 – 2400 = ₹864
Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{864*100}{2400*4} \) = 9% per annum
New rate = 10% per annum
Therefore, S.I. = \( \frac{2400*10*4}{100} \) = ₹960
Therefore, Amount = 2400 + 960 = ₹3360
Ans .
4
(4) Using Rule 1,
S.I. = ₹(920 – 800) = ₹120
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{120*100}{800*3} \)= 5% per annum
New rate = 8% per annum
Therefore, S.I. = \( \frac{800*3*8}{100} \) = ₹192
Therefore, Amount = (800 + 192) = ₹992
Ans .
1
(1) Using Rule 1,
Case I,
S.I. = 920 – 800 = ₹120
Rate = \( \frac{S.I. * 100}{P * T} \)
= \(\frac{120*100}{800*3}\) = 5% per annum
Case II,
Rate = 8% per annum
S.I. = \( \frac{800*8*3}{100} \) = ₹192
Therefore, Amount = Principal + S.I.
= (800 + 192) = ₹992
Ans .
1
(1) Using Rule 1,
S.I. = 2352 – 2100 = 252
Rate = \( \frac{S.I. * 100}{P * T} \).
= \( \frac{252 * 100}{2100*2} \) = 6% per annum
New rate = 5%
Therefore, S.I. = \(\frac{252*5}{6}\) = ₹210
Ans .
3
(3) Using Rule 1,
S.I. = 956 – 800 = Rs. 156
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{156 * 100}{800*3} \)
= 6.5%
New rate = (6.5 + 4)%
= 10.5%
Thereforel, S.I. = \( \frac{Principal * Rate * Time}{100} \)
= \( \frac{800*3*10.5}{100} \)
= Rs. 252
Therefore, Amount = Rs.(800 + 252)
= Rs.1052
Ans .
4
(4) Using Rule 1,
Amount deposited in bank = Rs. x (let)
Difference of rates = 5 – \( \frac{7}{2} \)= \(\frac{3}{2}\) % per annum
Therefore, S.I. = \( \frac{Principal * Rate * Time}{100} \)
\( \frac{x*1*3}{100} \) = 105
x = \(\frac{105*200}{3}\) = Rs. 7000
Ans .
1
(1) Using Rule 1,
Let x be lent at 8%, then (10000 – x) is lent at 10%.
Accordingly,
\( \frac{10000*9.2*t}{100} = \frac{x*8*t}{100} + \frac{(10000-x)*10*t}{100} \)
\( \frac{92000t}{100} = \frac{8xt}{100} + \frac{(10000-x)10t}{100} \)
92000t = 8xt + (10000 – x) 10t
92000 = 8x + 100000 – 10x
2x = 8000
x = 4000
Therefore, First part = ₹4000
Second part = ₹6000
Ans .
1
(1) Let x be lent on 8%.
Therefore, (1000 – x ) is lent on 10%.
Interest = 9.2% of 1000 = 92
Therefore, 92 = \( \frac{x*8}{100} + (\frac{1000-x}{100})*10 \)
8x + 10000 – 10x = 9200
– 2x = 9200 – 10000
x = \(\frac{800}{2}\) = ₹400 = first part
Therefore, Second part = 600
Ans .
1
(1) Interest = (7000 + 630 × 8) – 12000
= (7000 + 5040) – 12000
= 12040 – 12000 = 40
Total Principal
= 5000 + 4370 + 3740 + 3110
+ 2480 + 1850 + 1220 + 590
= 22360
Rate = \( \frac{40*100*12}{22360*1} \) = 2.1 per cent
Ans .
4
(4) Let the sum be ₹100.
For initial six months, Interest
= 100 * \( \frac{6}{100} * \frac{6}{12} \) = 3 %
Now, sum = 100 + 3 = ₹103
For another six months,
Interest = 103 * \( \frac{6}{100} * \frac{6}{12} \) = 3.09
Therefore, Rate of interest per annum = 3 + 3.09 = 6.09%
Ans .
3
(3) Let the person have 100.
Then SI for 1 year = ₹\( (\frac{40*15*1}{100} + \frac{30*10*1}{100} + \frac{30*18*1}{100} ) \)
= ₹(6+3+5.4) = ₹14.4
Therefore, Rate of interest on whole sum = 14.4%
Ans .
4
(4) SI earned after two years
= \( \frac{15600*10*2}{100} \) = ₹3120
Therefore, Principal for next two years
= ₹(15600 + 3120) = ₹18720
SI earned at the end of fourth
year = \(\frac{18720*10*1}{100}\) = ₹1872
Ans .
1
(1) Let x be lent at 10% per annum.
Therefore, (1500 – x ) is lent at 7% per
annum.
Now,
\( \frac{x*10*3}{100} + \frac{(1500-x)*7*3}{100} \) = 396
30x + 31500 – 21x
= 39600
9x = 39600 – 31500
x = \( \frac{8100}{9} \) = 900
Ans .
2
Using Rule 10,
Here, A = 848,
T = 4 years, r = 4%
Equal instalment = \( \frac{848*200}{4[200+(4-1)4]} \)
= \( \frac{848*200}{4*212} \) = ₹200
Ans .
3
(3) Using Rule 1.
Remaining amount
= (50000 – (8000 + 24000)) = 18000
Let 18000 be lent at the rate of r% p.a.
According to the question,
\( \frac{8000*11*1}{2*100} + \frac{24000*6*1}{100} + \frac{18000*r*1}{100} \) = 3680
440 + 1440 + 180r = 3680
1880 + 180r = 3680
180r = 3680 – 1880 = 1800
r = \(\frac{1800}{180}\)= 10%
Ans .
2
(2) Using Rule 1.
Let the principal be x .
Therefore, I1 = \( \frac{x*10*1}{2*100} = \frac{x}{20} \)
I2 = \( \frac{x*9*1}{3*100} = \frac{3x}{20} \)
I3 = \( \frac{x}{6} * \frac{12*1}{100} = \frac{x}{50} \)
Therefore, I1 + I2 + I3 = \( (\frac{x}{20}+\frac{3x}{100}+\frac{x}{50}) \)
=\( \frac{5x+3x+2x}{100} = \frac{x}{10}\)
Therefore, Average annual rate = 10%
Ans .
3
(3) Using Rule 1.
If the principal be x, then
Simple interest = (770 – x)
Therefore, Principal = \( \frac{S.I. * 100}{R * T} \)
x = \( \frac{(700-x)*100}{4*10} \)
2x =(770 – x) × 5
2x + 5x = 770 × 5
7x = 770 × 5
Therefore, x = \(\frac{770*5}{7}\) = ₹550
Ans .
4
(4) Using Rule 1.
S.I. on ₹12000 = \( \frac{12000*8*1}{100} \) = ₹960
Desired gain on ₹20000
= 20000 * \(\frac{10}{100}\)= ₹2000
Therefore, S.I. on ₹8000 = 2000 – 960 = ₹1040
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
=\( \frac{1040*100}{8000} \) = 13% per annum
Ans .
2
(2) Using Rule 1.
S.I. after five years
= \( \frac{Principal * Rate * Time}{100} \) = \( \frac{12000*5*10}{100} \) = 6000
Interest earned
= (6000 – 3320) = 2680
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{2680 * 100}{12000*3} \) = \( \frac{67}{9} = 7\frac{4}{9}\)%
Ans .
4
(4) Using Rule 1.
Case I
Let principal be x then Amount = 3x
S.I. = 2x
Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
= \(\frac{2x*100}{x*8}\)=25%
Case II
Time = \( \frac{S.I. * 100}{P * R} \)
= \(\frac{3x*100}{x*25}\)= 12 years
Ans .
2
(2) Using Rule 1.
Required percent
= \( \frac{1}{4}*3 + \frac{2}{3}*5 + (1-\frac{1}{4}-\frac{2}{3})*11 \)
=\( \frac{3}{4}+\frac{10}{3}+\frac{11}{12} =\frac{9+40+11}{12} \) =5%
Ans .
1
(1) Using Rule 1.
120 = \( \frac{300*4*r}{100} + \frac{400*3*r}{100} \)
24r = 120
r = \frac{120}{24} = 5% per annum
Ans .
3
(3) Using Rule 1.
If the sum of money be
x, then
=\( \frac{x*6*3}{100}+\frac{x*5*9}{100}+\frac{x*3*13}{100} \) = 8160
18x + 45x + 39x = 816000
102x = 816000
x = \(\frac{816000}{102}\)
= 8000
Ans .
3
(3) Using Rule 1.
If each amount lent be x, then
=\( \frac{x*7*4}{100}+\frac{x*5*4}{100} \) = 960
\( \frac{48x}{100} \) = 960
x = \(\frac{960*100}{48}\)
x = ₹2000
Ans .
3
(3) Using Rule 1.
Let the money lent to Tom be Rs. x.
Simple interest = \( \frac{Principal * Rate * Time}{100} \)
Therefore, \( \frac{500*8*4}{100} + \frac{x*8*4}{100} \) = 210
160 + \( \frac{32x}{100} \) = 210
\( \frac{32x}{100} \) = 210 – 160 = 50
x = \( \frac{50*100}{32} \) = Rs. 156.25
Ans .
1
(1) Using Rule 1.
Rate = \( \frac{20}{3} \)% per annum
Therefore, S.I.= \( \frac{Principal * Rate * Time}{100} \)
= \( \frac{2600*20*T}{3*100} \)
Therefore, Required Time = 3 years
Ans .
1
(1) Using Rule 1.
Principal = Rs. (60000 – 10000)
= Rs. 50000
Therefore, S.I. = \( \frac{50000*15*2}{100} \)
= Rs. 15000
Ans .
2
(2) Using Rule 1.
Let the loans taken by A, B and
C be Rs. x, Rs. y and Rs. z respectively.
Therefore, x + y + z = Rs. 7930
S.I. = \( \frac{Principal * Rate * Time}{100} \)
According to the question,
x+\(\frac{x*2*5}{100}\)= y+\(\frac{y*3*5}{100}\)= z+\(\frac{z*5*4}{100}\)
\(\frac{100x+10x}{100} = \frac{100y+15y}{100} = \frac{100z+20z}{100} \)
110x = 115y = 120z
22x = 23y = 24z
\(\frac{22x}{6072} = \frac{23y}{6072} = \frac{24z}{6072} \)
\(\frac{x}{276} = \frac{y}{264} = \frac{z}{253} \)
x:y:z = 276:264:253
Sum of terms of ratio = 276 + 264 + 253 = 793
Therefore, A's Loan = \(\frac{276}{793}\)*7930
= Rs.2760
Ans .
2
(2) Using Rule 1.
Remaining amount
= Rs. (16000 – 4000)
= Rs. 12000
\Therefore, S.I. = \( \frac{Principal * Rate * Time}{100} \)
= \( \frac{12000*15*12}{12*100} \) = Rs.1800
Therefore, Total amount paid
= Rs. (16000 + 1800)
= Rs. 17800
Ans .
4
(4)Using Rule 1.
S.I. after 1 year = \( \frac{Principal * Rate * Time}{100} \)
= \( \frac{x*5}{100} \) = Rs. \( \frac{x}{20} \)
Principal for 2nd year
= Rs. \( 2x+\frac{x}{20} \) = Rs. \( \frac{41x}{20} \)
S.I. after second year
= Rs. \( \frac{41x}{20}*\frac{5}{100} \)
= Rs. \( \frac{41x}{400} \)
Principal for third year
= Rs. \( 3x+\frac{41x}{400} \)
= Rs. \( \frac{1200x+41x}{400} \)
= Rs. \( \frac{1241x}{400} \)
Therefore, S.I. after 3rd year
= Rs. \( \frac{1241x}{400} * \frac{5}{100} \)
= Rs. \( \frac{1241x}{8000} \)
Therefore, Required amount
= Rs. \( (3x+\frac{1241x}{8000}) \)
= Rs. \( \frac{24000x+1241x}{8000} \)
= Rs. \( \frac{25241x}{8000}\)
Ans .
3
(3) Using Rule 1.
S.I. = \( \frac{Principal * Rate * Time}{100} \)
= \( \frac{100000*6*6}{100} \)
= Rs.36000
Total pocket money = 6 × 2500 = Rs. 15000
Total expenses of trust = 6 × 500 = Rs. 3000
Total expenses = Rs. (15000 + 3000) = Rs. 18000
Therefore, Amount to be received by the boy
= Rs. (100000 + 36000 – 18000)
= Rs. 118000
Ans .
1
(1) Let amounts be equal in T
years.
S.I. = \( \frac{Principal * Rate * Time}{100} \)
Therefore, P + \( \frac{P*x*T}{100} \) = Q + \( \frac{Q*y*T}{100} \)
\(\frac{PxT}{100}-\frac{QyT}{100}\) = Q-P
T\( \frac{Px-Qy}{100} \)= Q-P
T=100\( (\frac{Q-P}{Px-Qy}) \)
Ans .
4
(4) Let the principal be Rs. 100
Interest = Rs. 10
Actual principal = Rs. 90
Q Interest on Rs. 90 = Rs. 10
Therefore, Interest on Rs. 100
= \(\frac{10}{90}\)*100
= \(\frac{100}{9}\) = 11\(\frac{1}{9}\)%
Ans .
2
(2) Let the principal be Rs. P.
S.I. = \( \frac{Principal * Rate * Time}{100} \)
= \( \frac{P*5*5}{100} \) = Rs. \( \frac{P}{4} \)
Amount = P + \( \frac{P}{4} \) = Rs. \( \frac{5P}{4} \)
According to the question,
\( \frac{5P}{4} * \frac{2}{100} \) = 5
\( \frac{P}{4} \) = 5
P = 40*5
Rs.200
Ans .
1
(1) Principal = Rs. 1950, Rate =10% per annum
S. I. = \(\frac{Principal*Rate*Time}{100}\)
=\(\frac{1950*1*10}{100}\)
= Rs.195
Therefore, Amount = Rs.(1950+195)= Rs.2145
Therefore, 2200-2145=55
Therefore, There is gain of 55 Rs.
Ans .
1
(1) Using Rule 1.
160 = \( \frac{500*4*r}{100} + \frac{400*3*r}{100} \)
32r = 160
r = \frac{160}{32} = 5% per annum
Ans .
1
(1) Let amount of loan per head
be Rs. x.
S.I. = \( \frac{Principal * Rate * Time}{100} \)
Therefore, \( \frac{x*7*8}{100}- \frac{x*5*5}{100} \) = 62000
\( \frac{56x}{100}- \frac{25x}{100} \) = 62000
31x = 6200000
x = Rs.200000
Ans .
1
(1) Let the amount in Post office
be Rs. x.
Therefore, Amount in Bank
= Rs. (320000 – x)
S.I. = \( \frac{Principal * Rate * Time}{100} \)
Therefore, \( \frac{x*9}{100} + \frac{(320000-x)*8}{100} \) = 027600
9x + 2560000 – 8x = 2760000
x = 2760000 – 2560000
= Rs. 200000
Therefore, Amount in bank
= Rs. (320000 – 200000)
= Rs. 120000
Ans .
3
(3) Sum given to Anil = Rs. x
Sum given to Sunil =Rs. (x + 2000)
S.I. = \( \frac{Principal * Rate * Time}{100} \)
\( \frac{(x+2000)*12*3}{100} - \frac{x*10*3}{100} \) = 1020
36x + 72000 – 30x = 102000
6x = 102000 – 72000 = 30000
x = Rs. 5000
Therefore, Sum given to Sunil = Rs. 7000
Ans .
3
(3) S.I.= \( \frac{P * R * T}{100} \)
Let the required rate of interest
be R% per annum.
\( \frac{30000*5*10}{100} - \frac{30000*3*R}{100} \) = 7800
15000 – 900R = 7800
900R = 15000 – 7800 = 7200
R = \(\frac{7200}{900}\) = 8% per annum.
Ans .
3
(3) Case-I,
Interest = 8000 – 6000
= Rs. 2000
Rate = \( \frac{S.I. * 100}{P * T} \) = \( \frac{2000 * 100}{6000*4} \)
=\(\frac{25}{3}\)%
Therefore, Case-II,
Time = \(\frac{175*100}{525*\frac{25}{3}}\) = 4 years
Ans .
1
(1) P = ₹1460 ,
R = 10%
1992 is a leap year
Therefore, T = (24 + 31 + 25) days = 80
days.
I = \(\frac{PRD}{36500}\)
I = \( \frac{1460*10*80}{36500} \)
I = ₹32.
Note :We have excluded 5th February
but included 25th
Ans .
2
(2) Here, P = ₹15000
R = 15%
T = 2 years
A = \(P\frac{100+RT}{100})\)
= \(15000\frac{100+15*2}{100})\)
= 15000 * \(\frac{130}{100}\)
A = ₹19500
Ans .
3
(3) Here, P = ₹5000
A = ₹6000
T = 4 years
So, I = A – P
= ₹(6000 – 5000) = ₹1000
R = \( \frac{100I}{PT} \)
R = \( \frac{100*1000}{5000*4} \)
R = 5%.
Ans .
4
(4) I = I1 + I2
I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)
I = \( \frac{R}{100} \)(P1T1 + P2T2)
or, R = \( \frac{100I}{P1T1+P2T2} \)
Here, I = 900
P1 = 1200
T1 = 5 years
P2 = 1500
T2 = 2 years
R = \( \frac{100*900}{(1200*5)*(1500*2)} \)
R = \( \frac{90000}{9000} \)
R = 10%.
Note : In case of more than two
investment, sum the products of
principal and time of each case.
Ans .
1
(1) A = P + I
So, P remains same in both cases.
Only amount of interest is different
in two cases because the time
period are different.
P + Interest for 5 years = ₹1800
and P + Interest for 3 years = ₹1680
On subtraction we get,
Interest for 2 years = ₹120
Now, we solve for the case of 3 years .
Interest for 3 years = ₹120 * \( \frac{3}{2} \) = ₹180
And amount after 3 years = ₹1680
Principal (P) = A – I = ₹(1680 – 180) = ₹1500.
R = \( \frac{100I}{PT} \)
R = \( \frac{100*180}{1500*3} \)
R = 4%.
Note : Alternatively, we could
have solved for 5 years too and
got the same answer.
Ans .
2
(2) A sum doubles itself when
amount of interest becomes equal
to the principal.
So, I = P
Given, R = 5%
T = \( \frac{100I}{PR} \)
On substitution we get,
T = \( \frac{100*P}{P*5} \)
T = 20 years.
Ans .
4
I.= \( \frac{P * R * T}{100} \)
Given : I = \( \frac{P}{16} \)
and T = R
So, on substitution we get
\( \frac{P}{16} = \frac{P*R*R}{100} \)
R2 = \(\frac{100}{16}\)
R = \( \frac{10}{4} \) % = \( \frac{5}{2} \)% = \(2\frac{1}{2}\)%
Ans .
2
(2) Let the sum borrowed at 6%
be x = P1
Then the sum borrowed at 10%
= (16000 – x ) = P2
Time is one year in both cases
R1 = 6%
R2 = 10%
I = I1 + I2
I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)
I = \( \frac{R}{100} \)(P1T1 + P2T2)
P1T1 + P2T2 = \(\frac{100I}{T}\)
On substitution we get,
(x × 6) + (16000 – x)10 = \(\frac{100*1120}{1} \)
160000 – 4x =112000
4x = 48000
x = 12000
Ans .
3
I = I1 + I2
I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)
or T = \(\frac{100I}{P1R1*P2R2}\)
= \(\frac{100*390}{(1500*4)*(1400*5)}\)
= \( \frac{39000}{13000} \)
T = 3 years
Ans .
4
(4) Let each equal annual instalment
be x.
First instalment is paid after 1
year and hence will remain with
the lender for the remaining (4 – 1) = 3 years.
Similarly, second instalment will
remain with the lender for 2
years, third instalment for 1 year
and the final fourth instalment remain
x as such.
A = A1 + A2 + A3 + A4
A = \(P(\frac{100+RT}{100})\)
A = x\([\frac{100+5*3}{100}+\frac{100+5*2}{100}+\frac{100+5*1}{100}+\frac{100+5*0}{100}]\)
12900 = x\([\frac{115+110+105+100}{100}]\)
12900 = \( \frac{430}{100} \)x
x = \( \frac{12900*100}{430} \)
x = ₹3000
Ans .
1
(1) Principal + S.I. for 3 \( \frac{1}{2} \)years = ₹7360.50 ....... (i)
Principal + S.I. for 2 years = ₹6780 ...... (ii)
On subtracting equation (ii) from (i),
S.I. for 1 \(\frac{1}{2}\) years = ₹580.50
Therefore, S.I. for 2 years = ₹\((\frac{580.50*2*2}{3})\) = ₹774
Therefore Principal = ₹(6780 – 774 ) = ₹6006
And, rate of interest
=\( \frac{774*100}{6006*2} \)
= 6.4% per annum.
Ans .
2
(2) Interest = ₹(6678 – 5600) = ₹1078
Rate = \( \frac{S.I. * 100}{P * T} \)
= \( \frac{1078*100*2}{5600*7} \)
= \( 5\frac{1}{2} \)% per annum
Therefore, S.I. on ₹9600 for 5 \( \frac{1}{4} \) years
= ₹\( (\frac{9600}{100}*\frac{21}{4}*\frac{11}{2}) \) = ₹2772
therefore, Amount = ₹(9600 + 2772) = ₹12372
Ans .
3
(3) Let the sum be 100.
Number of days from January 1
to August 8 = 31 + 28 + 31 + 30
+ 31 + 30 + 31 + 7 = 219 days
= \( \frac{219}{365} \) year = \( \frac{3}{5} \)year
S.I. on ₹100 for \(\frac{3}{5}\) year at 7% = ₹\((\frac{100*3*7}{100*5})\)= ₹\(\frac{21}{5}\)
If required money is ₹\( \frac{21}{5} \), sum = ₹100
If required money is Rs. 63, sum = \((100*\frac{5}{21}*63)\)
= ₹1500
Ans .
4
(4) Number of monthly instalments = 15
Monthly instalment = 800
Time (T) = \( \frac{15}{12} = 1\frac{1}{4} \)
Therefore Total amount paid = (800 × 15) = 12,000
Interest = (12,000 – 10,000) = 2,000
Therefore, Rate of return
= \( \frac{100*2000*1*4}{10000*5} \)= 16%
Ans .
1
(1) Monthly instalment = ₹500
Total loan = ₹4000
Therefore, Number of instalments = \(\frac{4000}{500}\)=8
Once the payment starts, outstanding
balances will go on diminishing.
Hence, from point of view of interest,
principal = 4000 + 3500
+ 3000 + 2,500 + 2000 + 1500
+ 1,000 + 500 = ₹18,000
Therefore, Interest on ₹18,000 for 1 month at 6% P.a.
= \( \frac{18000*6*1}{100*12} \) = ₹90
Average rate of interest
= \( \frac{I*100}{PT} \)
T= 8 months = \( \frac{8}{12} \) years
= \( \frac{90*100*13}{4000*8} \) = \(\frac{27}{8}\)% = \( 3\frac{3}{8} \)%
Ans .
2
(2) Let the first part be x. Then
second part = (6800 – x)
Interest on first part for 3\(\frac{1}{3}\)years at 6%
= \( \frac{x*6*\frac{10}{3}}{100} \) = \( \frac{x}{5} \)
Interest on second part for 3\( \frac{1}{2} \)years at 4%
= \( \frac{(6800-x)*4*\frac{7}{2}}{100} \)
= ₹\( \frac{(6800-x)*7}{50} \)
According to the problem,
\( \frac{x}{5} = \frac{(6800-x)*7}{50} \)
10x = (6800 – x) 7
10x = 47600 – 7x
17x = 47600
x = 2800
Hence, first part = ₹2800
and second part
= ₹(6800 – 2800) = ₹4000.