- Staff Selection Commission Mathematics 1999 to 2017 - Simple Interest

# Staff Selection Commission Mathematics - Simple Interest

## Important Points :

• Borrowed money is called Principal and it is denoted by ‘P’.

• Money is borrowed for certain time period, that time is called interest time and it is denoted by ‘T’ or ‘t’.

• The principal becomes Amount when interest is added to it Amount is represented as A.

• So, Amount = Principal + Interest => A = P + S. I.
OR
Interest = Amount – Principal => S. I. = A – P

• When Interest is payable half – yearly Rate will be half and time will be twice

• When Interest is payable quarterly Rate will be one-fourth and time will be four times.

• ## Rules:

RULE 1:

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= $$\frac{S.I. * 100}{P * R}$$,

RULE 2:

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 = $$\frac{P(R1T1 * R2T2 * R3T3)}{100}$$

RULE 3:

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%

RULE 4:

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

RULE 5:

If Simple Interest (S.I.) becomes ‘n’ times of principal i.e.
S.I. = P × n then.
RT = n × 100

RULE 6:

If an Amount (A) becomes ‘n’ times of certain sum (P) i.e.
A = Pn then,
RT = (n – 1) × 100

RULE 7:

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

RULE 8:

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

RULE 9:

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 = $$\frac{A.t2-B.t1}{t2-t1}$$

RULE 10:

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.

RULE 11:

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

RULE 12:

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 = $$\frac{A2T1-A1T2}{T1-T2}$$

RULE 13:

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. = $$\frac{P2R2T2-P1R1T1}{100}$$

## TYPE-I

Ans .

2

1. Explanation :

(2) Using Rule 1,

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

Ans .

2

1. 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

1. 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

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

1. 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

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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

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

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

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

1. 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

1. 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

1. 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

1. 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

1. 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

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

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

1. 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

1. 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

1. Explanation :

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

Ans .

3

1. 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

1. 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

1. 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

1. Explanation :

(3) Using Rule 1,
Annual interest
= 365 × 2 = ₹730
Principal = $$\frac{SI*100}{Time*Rate}$$
= $$\frac{730*100}{1*5}$$ = ₹14600

Ans .

3

1. 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

1. 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

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

1. 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

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

1. 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

1. 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

1. Explanation :

(1) Principal = $$\frac{S.I. * 100}{R * T}$$
=$$\frac{60*100}{5*6}$$ = Rs.200

Ans .

1

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

1. 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

1. 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

1. 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

2. = 0.08 per annum

Ans .

1

1. Explanation :

Rate = $$\frac{S.I. * 100}{P * T}$$
$$\frac{280*100}{400*10}$$
= 7% per annum

Ans .

3

1. Explanation :

Rate = $$\frac{S.I. * 100}{P * T}$$
= $$\frac{\frac{1}{100}*100}{1*\frac{1}{12}}$$= 12% per annum

Ans .

3

1. Explanation :

(3)S.I.= $$\frac{Principal * Rate * Time}{100}$$
=Rs. $$(4000*\frac{18}{12}*\frac{12}{100})$$
= Rs.720

Ans .

2

1. Explanation :

(2)T= $$\frac{S.I. * 100}{P * R}$$
=$$\frac{1080*100}{3000*12}$$
= 3 yers

Ans .

3

1. 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

1. Explanation :

P= $$\frac{S.I. * 100}{R * T}$$
= $$\frac{R * 100}{R * 2}$$
= Rs.50

Ans .

3

1. 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

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

## TYPE - II

Ans .

1

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

1. Explanation :

(1)Using Rule 3,
R% = $$\frac{(\frac{41}{40}-1)*100}{\frac{1}{4}}$$%
= $$\frac{1}{40}*4 *100$$%
= 10%

Ans .

2

1. 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

1. 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 1. 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

1. 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

1. Explanation :

(3)Using Rule 3,
R% = $$\frac{(2-1)*100}{12}$$%
= $$\frac{25}{3}$$%
= $$8\frac{1}{3}$$%

Ans .

3

1. Explanation :

(3)Using Rule 3,
R% = $$\frac{(\frac{7}{4}-1)*100}{12}$$%
= $$\frac{75}{4}$$%
= $$18\frac{3}{4}$$%

Ans .

4

1. Explanation :

(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

1. Explanation :

(3)Using Rule 3,
T = $$\frac{(n-1)}{R}$$% years
= $$\frac{2-1}{\frac{25}{4}}*100$$ years
= 16 years

Ans .

3

1. 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

1. 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

1. 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

1. Explanation :

(2)Using Rule 3,
R% = $$\frac{(2-1)*100}{8}$$%
= 12.5%

Ans .

2

1. 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

1. Explanation :

(1)Using Rule 3,
T = $$(\frac{(n-1)}{R})$$*100 %
= $$\frac{3-1}{\frac{25}{4}}*100$$
= 16 years

Ans .

2

1. Explanation :

(2) Using Rule 1,
Rate = R% per annum
Therefore, Time = $$\frac{R}{2}$$ years
Therefore, Rate = $$\frac{S.I. * 100}{P * T}$$
R = $$\frac{8}{25} * \frac{100}{\frac{R}{2}}$$
R2 = $$\frac{8*200}{25}$$ = 64
R = 8% per annum

Ans .

3

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

## TYPE - III

Ans .

3

1. 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

1. 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

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

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

1. 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

1. 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

1. 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

1. Explanation :

(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%

## TYPE-IV

Ans .

4

1. 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

1. 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

1. 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, R2 = $$\frac{400}{9}$$
R= $$\frac{20}{3}$$=$$6\frac{2}{3}$$%

Ans .

4

1. 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

1. 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

1. Explanation :

(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

1. 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

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

1. 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

1. Explanation :

(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. 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

1. Explanation :

Using Rule 5,
n = $$\frac{1}{5}$$, R = T
RT = n × 100
R2 = $$\frac{1}{4}*100$$
R2 = 25
R = 5%

Ans .

3

1. Explanation :

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. 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

1. 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

1. Explanation :

(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

1. 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

1. 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

1. 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

1. 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

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

1. 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, $$\frac{y}{z} = \frac{x}{y}$$
y2= zx

Ans .

1

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

1. Explanation :

(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

1. 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

## TYPE-V

Ans .

4

1. 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

1. 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

1. 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

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

1. 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

1. 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

1. Explanation :

(2) Using Rule 1,
Time = $$\frac{S.I. * 100}{P * R}$$
= $$\frac{3}{10} * \frac{100}{10}$$ = 3 years

Ans .

1

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

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

1. 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

1. 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

## TYPE-VI

Ans .

3

1. 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

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

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

1. Explanation :

(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

1. 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

1. Explanation :

(1) $$\frac{P*1*2}{100}$$ = 24
P = $$\frac{2400}{2}$$ = ₹1200

Ans .

3

1. 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

1. Explanation :

(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

1. 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

1. 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

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

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

1. 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

1. 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

## TYPE-VII

Ans .

1

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

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

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

1. 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

1. 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

1. 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

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

1. 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

1. 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

1. Explanation :

(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

1. 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

1. 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

1. 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

1. 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

1. 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

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

1. 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

1. 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

1. 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

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

1. Explanation :

(1) Using Rule 1.
Principal = Rs. (60000 – 10000)
= Rs. 50000
Therefore, S.I. = $$\frac{50000*15*2}{100}$$
= Rs. 15000

Ans .

2

1. 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

1. 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

1. 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

1. 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

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

1. 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

1. 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

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

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

## TEST YOURSELF

Ans .

1

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

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

1. 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

1. 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

1. 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

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

1. 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

1. 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

1. Explanation :

(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. 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

1. 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

1. Explanation :

I.= $$\frac{P * R * T}{100}$$
Given : I = $$\frac{P}{16}$$
and T = R
So, on substitution we get
$$\frac{P}{16} = \frac{P*R*R}{100}$$
R2 = $$\frac{100}{16}$$
R = $$\frac{10}{4}$$ % = $$\frac{5}{2}$$% = $$2\frac{1}{2}$$%

Ans .

2

1. Explanation :

(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

1. Explanation :

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

1. 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 = 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. 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

1. 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

1. 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

1. 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

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

1. 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.