OR

Interest = Amount – Principal => S. I. = A – P

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,

R_{2} = \( \frac{(n2-1)}{n1-1} \)xR_{1} and
T_{2} = \( \frac{(n2-1)}{n1-1} \)xT_{1}

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

**Explanation :**(2) Using Rule 1,

\( \frac{150*100}{4} * \frac{2}{1} \) = ₹7500

**Ans . **

2

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(1) Using Rule 1,

4200 = \( \frac{29400*6*R}{100} \)

=> R = \(\frac{4200}{294*6}\) = \(\frac{50}{21}\) = 2\(\frac{8}{21}\)%

**Ans . **

2

**Explanation :**(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

**Explanation :**(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

**Explanation :**Using Rule 1, Time = \( \frac{S.I. * 100}{P * R} \) =\( \frac{1080*100}{3000*12} \) = 3 years

**Ans . **

3

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(3) Using Rule 1,

Annual interest

= 365 × 2 = ₹730

Principal = \( \frac{SI*100}{Time*Rate} \)

= \( \frac{730*100}{1*5} \) = ₹14600

**Ans . **

3

**Explanation :**(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

**Explanation :**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

**Explanation :**(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

**Explanation :**(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

**Explanation :**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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(1) Principal = \( \frac{S.I. * 100}{R * T} \)

=\( \frac{60*100}{5*6} \) = Rs.200

**Ans . **

1

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

= 0.08 per annum

**Ans . **

1

**Explanation :**Rate = \( \frac{S.I. * 100}{P * T} \)

\( \frac{280*100}{400*10} \)

= 7% per annum

**Ans . **

3

**Explanation :**Rate = \( \frac{S.I. * 100}{P * T} \)

= \( \frac{\frac{1}{100}*100}{1*\frac{1}{12}} \)= 12% per annum

**Ans . **

3

**Explanation :**(3)S.I.= \( \frac{Principal * Rate * Time}{100} \)

=Rs. \( (4000*\frac{18}{12}*\frac{12}{100}) \)

= Rs.720

**Ans . **

2

**Explanation :**(2)T= \( \frac{S.I. * 100}{P * R} \)

=\( \frac{1080*100}{3000*12} \)

= 3 yers

**Ans . **

3

**Explanation :**(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

**Explanation :**P= \( \frac{S.I. * 100}{R * T} \)

= \( \frac{R * 100}{R * 2} \)

= Rs.50

**Ans . **

3

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(1)Using Rule 3,

R% = \( \frac{(\frac{41}{40}-1)*100}{\frac{1}{4}} \)%

= \( \frac{1}{40}*4 *100 \)%

= 10%

**Ans . **

2

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(3)Using Rule 3,

R% = \( \frac{(2-1)*100}{12} \)%

= \(\frac{25}{3}\)%

= \(8\frac{1}{3}\)%

**Ans . **

3

**Explanation :**(3)Using Rule 3,

R% = \( \frac{(\frac{7}{4}-1)*100}{12} \)%

= \(\frac{75}{4}\)%

= \(18\frac{3}{4}\)%

**Ans . **

4

**Explanation :**(4)Using Rule 3,

R_{1}= \( \frac{(2-1)*100}{5} \)%

= 20%

R_{2}= \( \frac{(3-1)*100}{12} \)%

= \(16\frac{2}{3}\)%

Lowest rate of interest = = \(16\frac{2}{3}\)%

**Ans . **

3

**Explanation :**(3)Using Rule 3,

T = \( \frac{(n-1)}{R} \)% years

= \(\frac{2-1}{\frac{25}{4}}*100\) years

= 16 years

**Ans . **

3

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(2)Using Rule 3,

R% = \( \frac{(2-1)*100}{8} \)%

= 12.5%

**Ans . **

2

**Explanation :**(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

**Explanation :**(1)Using Rule 3,

T = \( (\frac{(n-1)}{R}) \)*100 %

= \(\frac{3-1}{\frac{25}{4}}*100\)

= 16 years

**Ans . **

2

**Explanation :**(2) Using Rule 1,

Rate = R% per annum

Therefore, Time = \(\frac{R}{2}\) years

Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)

R = \( \frac{8}{25} * \frac{100}{\frac{R}{2}}\)

R^{2}= \( \frac{8*200}{25} \) = 64

R = 8% per annum

**Ans . **

3

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**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

**Explanation :**(3) Using Rule 13,

Here, P_{1}= Rs. 3000,

R_{1}= R,

T_{1}=\(\frac{5}{2}\) years,

P_{2}= Rs. 3200,

R_{2}= R,

T_{2}=\(\frac{5}{2}\) 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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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, R^{2}= \(\frac{400}{9}\)

R= \(\frac{20}{3}\)=\(6\frac{2}{3}\)%

**Ans . **

4

**Explanation :**(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

**Explanation :**(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

**Explanation :**(1) Using Rule 5,

Here , n = \(\frac{16}{25}\) , R=T

Now , R*R = \(\frac{16}{25}*100\)

R^{2}= \(\frac{1600}{25}\)

R= \(\frac{40}{5}\)=8%

**Ans . **

3

**Explanation :**(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

**Explanation :**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

**Explanation :**(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

**Explanation :**(2)Using Rule 5,

Here, n = \(\frac{1}{16}\), R = T

RT = n × 100

R^{2}= \(\frac{100}{16}\)

R = \(\frac{10}{4}\)

R = \(2\frac{1}{2}\)%

**Ans . **

1

**Explanation :**(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

**Explanation :**Using Rule 5,

n = \(\frac{1}{5}\), R = T

RT = n × 100

R^{2}= \(\frac{1}{4}*100\)

R^{2}= 25

R = 5%

**Ans . **

3

**Explanation :**Using Rule 13,

Here, P_{1}= P, R_{1}= 7.5%, T_{1}= 4 years.

P_{2}= P, R_{2}= 7.5%, T_{2}= 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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(3)Using Rule 5,

Here, n = \(\frac{1}{9}\) , R = T

RT = n × 100

R^{2}=\(\frac{1}{9}\)*100

R^{2}= \(\frac{100}{9}\)

R =\(\frac{10}{3}\)

R = 3\(\frac{1}{3}\)%

**Ans . **

2

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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, \(\frac{y}{z} = \frac{x}{y}\)

y^{2}= zx

**Ans . **

1

**Explanation :**(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

**Explanation :**(3)Using Rule 13.

Here, P_{1}= Rs. P, R_{1}= 12%, T_{1}= 4 years

P_{2}= Rs. P, R_{2}= 15%, T_{2}= 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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(2) Using Rule 1,

Time = \( \frac{S.I. * 100}{P * R} \)

= \( \frac{3}{10} * \frac{100}{10} \) = 3 years

**Ans . **

1

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**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

**Explanation :**(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

**Explanation :**(1)Using Rule 13,

P_{1}= P, R_{1}= R, T_{1}= 2

P_{2}= P, R_{2}= R + 3, T_{2}= 2

S.I.= 72

72 = \( \frac{P*(R+3)*2-P*R*2}{100} \)

7200 = 6P

P = 1200

**Ans . **

4

**Explanation :**(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

**Explanation :**(1) \( \frac{P*1*2}{100} \) = 24

P = \( \frac{2400}{2} \) = ₹1200

**Ans . **

3

**Explanation :**(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

**Explanation :**(1)Using Rule 13.

P_{1}= P, R_{1}= R, T_{1}= 2.

P_{2}= P, R_{2}= R + 3, T_{2}= 2.

S.I.= 300

300 = \( \frac{P*(R+3)*2-P*R*2}{100} \)

300 = \( \frac{6P}{100} \)

p = ₹5000

**Ans . **

4

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**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

**Explanation :**(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

**Explanation :**(2) Using Rule 1.

Let the principal be x .

Therefore, I_{1}= \( \frac{x*10*1}{2*100} = \frac{x}{20} \)

I_{2}= \( \frac{x*9*1}{3*100} = \frac{3x}{20} \)

I_{3}= \( \frac{x}{6} * \frac{12*1}{100} = \frac{x}{50} \)

Therefore, I_{1}+ I_{2}+ I_{3}= \( (\frac{x}{20}+\frac{3x}{100}+\frac{x}{50}) \)

=\( \frac{5x+3x+2x}{100} = \frac{x}{10}\)

Therefore, Average annual rate = 10%

**Ans . **

3

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(1) Using Rule 1.

Principal = Rs. (60000 – 10000)

= Rs. 50000

Therefore, S.I. = \( \frac{50000*15*2}{100} \)

= Rs. 15000

**Ans . **

2

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(4) I = I

_{1}+ I_{2}

I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)

I = \( \frac{R}{100} \)(P_{1}T_{1}+ P_{2}T_{2})

or, R = \( \frac{100I}{P1T1+P2T2} \)

Here, I = 900

P_{1}= 1200

T_{1}= 5 years

P_{2}= 1500

T_{2}= 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

**Explanation :**(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

**Explanation :**(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

**Explanation :**I.= \( \frac{P * R * T}{100} \)

Given : I = \( \frac{P}{16} \)

and T = R

So, on substitution we get

\( \frac{P}{16} = \frac{P*R*R}{100} \)

R^{2}= \(\frac{100}{16}\)

R = \( \frac{10}{4} \) % = \( \frac{5}{2} \)% = \(2\frac{1}{2}\)%

**Ans . **

2

**Explanation :**(2) Let the sum borrowed at 6% be x = P

_{1}

Then the sum borrowed at 10% = (16000 – x ) = P_{2}

Time is one year in both cases

R_{1}= 6%

R_{2}= 10%

I = I_{1}+ I_{2}

I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)

I = \( \frac{R}{100} \)(P_{1}T_{1}+ P_{2}T_{2})

P_{1}T_{1}+ P_{2}T_{2}= \(\frac{100I}{T}\) On substitution we get,

(x × 6) + (16000 – x)10 = \(\frac{100*1120}{1} \)

160000 – 4x =112000

4x = 48000

x = 12000

**Ans . **

3

**Explanation :**I = I

_{1}+ I_{2}

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

**Explanation :**(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 = A_{1}+ A_{2}+ A_{3}+ A_{4}

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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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

**Explanation :**(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.