Simple Interest
Exam Duration: 45 Mins Total Questions : 25
What will be the amount a man would get , if invests Rs.12000at 9\(\frac{1}{2}\)% per annum simple interestfor 2 yr.
- (a)
Rs. 13280
- (b)
Rs.16280
- (c)
Rs.12280
- (d)
Rs.14280
Here, P =Rs.12000, R = \(\frac{19}{2}\)% per annum and T = 2 yr
SI = ( \(\frac{P \times T \times R}{100}\) )
= Rs.( 12000 \(\times\)2\(\times\) \(\frac{19}{2}\)\(\times\) \(\frac{1}{100}\) )
= Rs. 2280
Amount = ( P + SI )
= Rs. ( 12000 + 2280 )
= Rs. 14280
What sum of money will fetch a simple interest of Rs.3900 for 2\(\frac{1}{2}\) yr at the rate of 13% per annum?
- (a)
Rs.12600
- (b)
Rs.11800
- (c)
Rs.10400
- (d)
Rs.12000
Let the sum be Rs. x
Then, 3900 = ( x \(\times\) \(\frac{5}{2}\)\(\times\) 13\(\times\) \( \frac{1}{100}\))
\(\Rightarrow\) x = \(\frac{3900\times2\times100}{5\times13}\)
= 12000
Thus, the principalis Rs.12000.
A sum doubles after 10 yr, if it is invested at simple interest. In how many years will it become four times the original sum invested?
- (a)
32 yr
- (b)
24 yr
- (c)
30 yr
- (d)
28 yr
Let the original sum be Rs.x.
Then, SI = Rs.x.
R = \(( \frac{100\times SI}{P \times T} )\)
= \((\frac{100\times x}{x \times 10})\)
= 10 % per annum
Now, P= Rs.x, SI = Rs. (4x-x) = Rs.3x
R = 10% annum
T = \((\frac{100\times SI}{P\times R})\)
= \((\frac{100\times3x}{x\times10})\)
= 30 yr
A certain sum amounts Rs.18600 in 4 yr and Rs.20400 in 6 yr under simple Interest. Find the Sum.
- (a)
Rs.16780
- (b)
Rs. 15450
- (c)
Rs.16000
- (d)
Rs.15000
Amount in 6 yr = Rs. 20400
Amount in 4 yr = Rs. 18600
SI for 2 yr = Rs.(20400 - 18600)
= Rs.1800
SI for 4 yr = Rs.(1800\(\times\)2)
= Rs.3600
Principal = (Amount in 4 yr )-( SI in 4 yr )
= Rs.( 18600-3600)
= Rs. 15000
A certain sum becomes \( \frac{10}{9}\) times itself in one year. Find the rate of simple interest .
- (a)
11\(\frac{1}{2}\)%
- (b)
11\(\frac{1}{9}\)%
- (c)
12\(\frac{1}{2}\)%
- (d)
12\(\frac{1}{9}\)%
Let the sum - Rs.x, Amount = Rs.\(\frac{10x}{9}\), T= 1 yr
SI = ( Amount - sum)
= Rs. \((\frac{10x}{9}-x)\)
= Rs. \((\frac{x}{9})\)
R = \((\frac{100\times SI}{P\times T})\)
= \((100\times\frac{x}{9}\times\frac{1}{x})\)
= \((\frac{100}{9})\)
= 11 \(\frac{1}{9}\)% per annum.
A sum of Rs. x is lent for simple interest at 9.5% per annum for 4 yr. What is the total amount?
- (a)
Rs.1.38x
- (b)
Rs.1.24x
- (c)
Rs.1.16x
- (d)
Rs.1.42x
P= Rs.x, R = 9.5% per annum, T = 4 yr
A = ( P + SI )
= x + ( x \(\times\) 4\(\times\)9.5\(\times\) \(\frac{1}{100}\) )
Rs. 1.38 x
'A' lends Rs.4300 at 8\(\frac{1}{2}\)% per annum simple interest for 2\(\frac{1}{2}\)yr .'B' lends Rs.645 at 12 \(\frac{1}{2}\)% per annum simple interest for 12 yr. Who gets more interest?
- (a)
A
- (b)
B
- (c)
Both get the same
- (d)
Data insufficient
SI earned by A = Rs.(4300\(\times\) \(\frac{5}{2}\)\(\times\) \(\frac{17}{2}\)\(\times\) \(\frac{1}{100}\) )
= Rs. 913.75
Si earned by B = Rs.( 645\(\times\)12\(\times\) \(\frac{25}{2}\)\(\times\)\(\frac{1}{100}\) )
= Rs. 967. 50
Thus, the SI earned by B is more.
A man borrows Rs.12000 at 8 % simple interest for 2 yr. He repays Rs.12000 at the end of 2 yr and balance at the end of third year .How much did he repay at the end of third year?
- (a)
Rs.2145.40
- (b)
Rs.2073.60
- (c)
Rs.2042.20
- (d)
Rs.2130.90
SI for 2 yr Rs. = \((\frac{1200\times2\times8}{100})\)
= Rs. 1920
Amount to repayed after 2 yr
= Rs. (12000+1920)
= Rs.13920
But the man repays Rs.12000 at the end of 2 yr.
\(\therefore \)Principal remaining = Rs.(13920-1200)
= Rs.1920
Time = 1 yr, R = 8 % per annum
Amount to be repaid = Rs.(1920+ \(\frac{1920\times1\times8}{100}\))
= Rs. 2073.60
At simple interest of 5% , 6% and 8% for three consecutive year, the interest earned is Rs.760.Find the principal.
- (a)
Rs. 4600
- (b)
Rs.3200
- (c)
Rs.4000
- (d)
Rs.3600
Let the prinicipal be x. Then,
\((\frac{x\times1\times5}{100})\)+ \((\frac{x\times1\times6}{100})\)+ \((\frac{x\times1\times8}{100})\) = 760
\(\Rightarrow\) \(\frac{5x}{100}\) + \( \frac{6x}{100}\) + \(\frac{8x}{100}\) = 760
\(\Rightarrow\) 19x = 760 \times 100
\(\Rightarrow\) x = \((\frac{760\times100}{19})\)
= Rs.4000
A man borrowed Rs. 5000 at 6\(\frac{1}{4}\) per annum simple interest for 6 yr. Instead of clearing the loan with interest at the end of 6 yr, if he returns the loan with interest at the end of 4 yr, how much does the man save ?
- (a)
Rs.625
- (b)
Rs.450
- (c)
Rs.575
- (d)
Rs.700
P = Rs. 5000 , R = \(\frac{25}{4}\) % per annum , T = 6 yr
Simple Interest for 6 yr
= Rs (5000\(\times\)6\(\times\)\(\frac{25}{4}\)\(\times\)\(\frac{1}{100}\))
= Rs. 1875
Simple Interest for 4 yr
= Rs.(5000\(\times\)4\(\times\)\(\frac{25}{4}\)\(\times\)\(\frac{1}{100}\))
= Rs. 1250
Savings = Rs(1875-1250)
= Rs. 625
A man borrowed Rs.12000 for 4 yr at 7\(\frac{3}{4} \)% per annum and a year later he again borrowed another Rs.11000 for 3 yr at the same time . How much should be pay at the need to settle the loans?
- (a)
Rs.30510
- (b)
Rs.28610
- (c)
Rs.31450
- (d)
Rs.29590
In the first case
P = Rs. 12000 , T = 4 yr and R = \( \frac{31}{4}\) % per annum
SI = \((\frac{PTR}{100})\)
= Rs.(12000\(\times\)4\(\times\)\( \frac{31}{4}\)\(\times\)\(\frac{1}{100}\))
= Rs.3720
Amount = (P+SI)
= Rs. (12000+3720)
= Rs.15720
In the second case
P = Rs.12000, T =3 yr and r = \(\frac{31}{4}\)% per annum
SI = \( (\frac{PTR}{100})\)
= Rs.(12000\(\times\)3\(\times\)\(\frac{31}{4}\)\(\times\)\(\frac{1}{100}\))
= Rs.2790
Amount = (P+SI)
= Rs.(12000+2790)
= Rs.14790
\(\therefore\) Total money to be paid at thee nd
= Rs.(15720+14790)
= Rs.30510
Ramesh took a loan of Rs.6000 at 6% per annum simple interest for 3 yr. After one year the rate was increased to 6\(\frac{1}{2}\)% per annum. How much shall Ramesh have to pay at the end to clear his loan?
- (a)
Rs. 7480
- (b)
Rs.7390
- (c)
Rs.7140
- (d)
Rs.7260
SI for the first year = Rs. \((\frac{6000\times6\times1}{100})\)
= Rs.360
SI for the next two year = Rs.(6000\(\times\)\(\frac{13}{2}\)\(\times\)\(\frac{1}{100}\))
= Rs.780
\(\therefore\) Amount to be paid at the end
= Rs.(6000+360+780)
= Rs. 30510
A person borrowed Rs.2000 at 5% per annum simple interest and immediately lent it at 6% per annum simple interest .At the end of 2\(\frac{1}{2}\) yr he collected the amount and settled his loan. What was his profit.
- (a)
Rs.45
- (b)
Rs.40
- (c)
Rs.50
- (d)
Rs.56
For borrowed money P = Rs.2000, R = 5 % per annum and T = \(\frac{5}{2} \)yr
SI = Rs.(2000\(\times\)\(\frac{5}{2} \)\(\times\)5\(\times\)\(\frac{1}{100}\))
= Rs. 250
For lent money
P = Rs.2000, R% = 6% per annum and T= \(\frac{5}{2} \) yr
SI = Rs.(2000\(\times\)\(\frac{5}{2} \)\(\times\)6\(\times\)\(\frac{1}{100}\))
= Rs. 300
His profit = Rs. (300-250)
= Rs. 50
What sum of money lent out at simple interest will amount to Rs. 3400 in 3 yr at 1% per month?
- (a)
Rs.2400
- (b)
Rs.1800
- (c)
Rs.1600
- (d)
Rs.2500
A = Rs.3400 , T = 3 yr,
R = 1% per month = 12% per annum
Let the principal be Rs.x.
SI = \((\frac{PTR}{100})\)
= Rs.\((\frac{x\times3\times12}{100})\)
= Rs.\(( \frac{36x}{100})\)
\(\Rightarrow\) \( \frac{36x}{100}\)= 3400
\(\Rightarrow\) x = \(\frac{3400\times100}{136}\)
= Rs.2500
Hence , the sum = Rs. 2500
A man borrowed Rs.8000 at 6% per annum interest for 5 yr. After 3 yr he returned Rs.7000 .How much amount should he return at the end to settle the loan?
- (a)
Rs. 2732.80
- (b)
Rs.2612.20
- (c)
Rs.2824.40
- (d)
Rs.2190.50
P = Rs.8000, R = 6% per annum and T = 3 yr.
SI for 3 yr = Rs.(8000\(\times\)3\(\times\)6\(\times\)\(\frac{1}{100}\))
= Rs. 1440
Amount after 3 yr = Rs. (8000+1440)
= Rs. 9440
Money returned after 3 yr = Rs. 7000
Money left = Rs.(9440-7000)
= Rs. 2440
Time left = (5-3)yr
= 2 yr
SI = \((\frac{PTR}{100})\)
= Rs . \((\frac{2400\times2\times6}{100})\)
= Rs. 292.80
\(\therefore\) Amount to be returned at th eend
= Rs. (2440+292.80)
= Rs. 2732.80
A person invests money in three different schemes for 5 yr, 10 yr and 15 yr at 8%, 10 % and 12% simple interest respectively. At the completion of each scheme, he gets the same interest. The ratio of his investments is
- (a)
45:9:5
- (b)
45:18:10
- (c)
25:18:10
- (d)
25:9:5
Let , the required ratio be x : 1: y. Then, SI on Rs. x for 5 yr at 8% per annum = SI on Rs.1 for 10 yr at 10% per annum
\(\Rightarrow\) \( \frac{x\times5\times8}{100}\) = \( \frac{1\times10\times10}{100}\)
\(\Rightarrow\) x = \(\frac{100}{40}\)
= \( \frac{5}{2}\)
SI on Rs.y for 15 yr at 12% per annum
\(\Rightarrow\) \( \frac{1\times10\times10}{100}\) = \( \frac{y\times15\times12}{100}\)
\(\Rightarrow\) y = \( \frac{100}{180}\)
= \( \frac{5}{9}\)
Required ratio = x:1:y
= \(\frac{5}{2}\) :1 : \(\frac{5}{9}\)
= 45:18:10
Simple interest on Rs.1680 for 4 yr at 7\(\frac{1}{2}\) % per annum is equal to the simple interest on Rs.1200 at 7% per annum for a certain period of time. The period of time is
- (a)
7 yr
- (b)
6yr
- (c)
5\(\frac{1}{3}\)yr
- (d)
7\(\frac{1}{4}\)yr
Let the required period of time be x yr. Then,
1680\(\times\)4\(\times\)\(\frac{15}{2}\)\(\times\)\(\frac{1}{100}\) = 100\(\times\)x\(\times\)7\(\times\)\(\frac{1}{100}\)
\(\Rightarrow\) x = \( \frac{1680\times2\times15}{1200\times7}\)
= 6 yr
Hence the required time period is 6 yr.
If the simple interest for 5 yr be equal to 40 % of the principal , it will be equal to the principal after.
- (a)
12 yr 3 months
- (b)
12 yr 6 months
- (c)
12 yr 4 months
- (d)
12 yr 9 months
Let , the principal be Rs.x and the rate be R% per annum. Then,
x\(\times\)\(\frac{R}{100}\)\(\times\)5 = \(\frac{40}{100}\)\(\times\)x
\(\Rightarrow\) R = 8% per annum
Let , the required time be t yr.
x\(\times\)t\(\times\)8\(\times\) \(\frac{1}{100}\) = x
\(\Rightarrow\) t = \(\frac{100}{8}\)yr
= 12 \(\frac{1}{2}\)yr
= 12 yr 6 months
Anurag lends 20% of the sum at 10 % per annum, 50% of rest annum at 10 % per annum and the rest at 8% per annum simple interest. What would be the rate of interest per annum calculated on the whole sum?
- (a)
10.4%
- (b)
10.8%
- (c)
10.6%
- (d)
10.2%
Let ,the whole sum be Rs. 100.
Sum at 15 % per annum = Rs. 20
Sum at 10 % per annum = Rs. 40
Sum at 8% per annum = Rs.40
SI on Rs.100 for 1 yr
= (20\(\times\)1\(\times\)\(\frac{15}{100}\))+(40\(\times\)1\(\times\)\(\frac{10}{100}\))+(40\(\times\)1\(\times\)\(\frac{8}{100}\))
= Rs.(3+4+3.2)
= Rs.10.2
\(\therefore\) Required rate = 10.2% per annum
Sudha borrowed some money at the rate of 6% per annum for the first 2 yr, 8% per annum for the next 4 yr and 12% per annum for the period beyond 6 yr. If the total interest paid by him at the end of 14 yr is Rs.6300, how much money did she borrow?
- (a)
Rs.4500
- (b)
Rs.4800
- (c)
Rs.4200
- (d)
Rs.4900
Let , the sum be Rs.x. Then,
(x\(\times\)2\(\times\)\(\frac{6}{100}\))+(x\(\times\)4\(\times\)\(\frac{8}{100}\))+(x\(\times\)8\(\times\)\(\frac{12}{100}\))
= 6300
\(\Rightarrow\) \( \frac{x}{100}\)(12+32+96) = 6300
\(\Rightarrow\) \(\frac{140x}{100}\) = 6300
\(\Rightarrow\) x = \(\frac{6300\times100}{140}\)
= 4500
Hence , the money borrowed is Rs. 4500.
An automobile financier claims to be lending money at simple interest , but he includes the interest every 4 months for calculating the principal, if he is charging an interest of 12 %, the effective rate of interest is approximately.
- (a)
13.25%
- (b)
12.5%
- (c)
12.75%
- (d)
13.5%
Let the sum be Rs.100. Then
SI for first 4 months = Rs.(100\(\times\)\(\frac{4}{12}\)\(\times\)12\(\times\)\(\frac{1}{100}\))
= Rs. 4
SI for next 4 months = Rs.(104\(\times\)\(\frac{4}{12}\)\(\times\)12\(\times\)\(\frac{1}{100}\))
= Rs. 4.16
SI for first last 4 months = Rs. (108.16\(\times\)\(\frac{4}{12}\)\(\times\)12\(\times\)\(\frac{1}{100}\))
= Rs.4.3264
So, amount at the end of 1 year
= Rs.(100+4+4.16+4.3264)
= Rs. 112.4864
\(\therefore \) Effective rate = (112.4864-100)
= Rs.12.4864%
= 12.5%
A sum was put at simple interest at a certain rate for 3 yr.Had it been put at 1% higher rate, it would have fetched Rs. 5100 more . The sum is
- (a)
Rs. 170000
- (b)
Rs.150000
- (c)
Rs.125000
- (d)
Rs.12000
Simple Interst for 1 yr = \(\frac{5100}{3}\)
= Rs.1700
1 % of sum = 1700
Sum = \(\frac{1700\times100}{1}\)
= Rs.17000
The present worth of bill due 7 months , hence is Rs.1200. If the bill were dure to at th eend of 2\frac{1}{2} yr, its present worth would be Rs.1016.The rate percent of bill is
- (a)
16%
- (b)
8%
- (c)
10%
- (d)
18%
Let , the rate of interest be r%
Then, A = 1200 + \(\frac{1200\times r\times7}{12\times100}\)
= 1200+7r
Again , 1200+7r = 1016
= \(\frac{1016\times r\times2.5}{100}\)
\(\Rightarrow \) 1200+ 7r = 1016+25.4r
18.4r = 184
r = \(\frac{184}{18.4}\)
= 10%
A invests two equal amounts earning 10 % and 12% of interst annually . If the interest on them earned is Rs.1650 in an year, then the sum invested in each ( in rupees ) is
- (a)
17000
- (b)
15000
- (c)
8500
- (d)
7500
Let , the investment e x each.
Interest = Rs. 1650
\(\frac{10x}{100}\) + \(\frac{12x}{100}\)= 1650
\(\Rightarrow\) x = Rs. 7500
A sum of Rs.5000 was lent partly at 6% and partly at 9% simple interest .If the interest received after one year is Rs. 390 , find the ration in which the money was lent.
- (a)
2:3
- (b)
3:4
- (c)
1:2
- (d)
3:2
Let , th eamount lent at 6% be Rs.x and amount lent at 9% is (5000-x)
\(\frac{6\times x\times1}{100}\) + \(\frac{(5000-x)\times 9\times1}{100}\)= 390
\(\Rightarrow\) 6x+45000-9x = 39000
\(\Rightarrow\) 3x = 45000-39000
= 6000
\(\Rightarrow\) x = 2000 and 5000 - x = 3000
The ratio of two amounts = 2000: 3000
= 2 : 3