Simple interest questions

(D) 16 years

Let \(x\) be the principal amount ' \(y\) ' be the time to double the money.

Then interest will also be 'x'.

\(\therefore \quad x=\frac{x \times 25 \times y}{4 \times 100}\)

\(400=25 y \quad \therefore y =16\) years

From the given data we can write ,
\(P - 420 = \frac{(P \times 5 \times 8)}{100}\)
\(100P - 42000 = 40P\)
\(60 P = 42000\)
\(P = 700\)Rs.

According to the question, If principal be Rs. x, then
S.I = Rs.x
Time= \(\frac{({SI}\times{100}) } {({principal}\times{rate}) }\)
=\(\frac{{x}\times{100} } {{x}\times{\frac{25}{2}} }\)
=\(\frac{200}{25}\) =8 years

A certain sum of money amounts to Rs.9,766 in 3 years at simple interest at R% per annum and to Rs.10,849 in years at the same rate of simple interest.

Simple interest of 1.5 years = 10849 – 9766 = Rs. 1083

Simple interest of 1 years =\( 1083 \times \frac{2}{3}= Rs. 722 \)

Simple interest of 3 years \( = Rs. 722 \times 3= 2166 \)

Amount of 3 years = Rs.\( 9766 \)

Principal \( = 9766 - 2166= 7600 \)

As we know,

Simple Interest \(=\frac{\text { Principal } \times \text { Rate } \times \text { time }}{100}\)

\(\Rightarrow 722=\frac{7600 \times R \times 1}{100}\)

\( ⇒ R=\frac{722}{76}=9.5\)%

Let Principal = P

Rate of Interest = R%

Required Ratio = \(\frac{(\frac{PR×5}{100})}{(\frac{PR×15}{100})} =\frac{5}{15}=\frac{1}{3}=1:3\)

Let the principal be Re. 1
\(\therefore \mathrm{S.I.}=\frac{41}{40}-1=\frac{1}{40}\)
Now, rate \(=\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Time }}\)
\(=\frac{\frac{1}{40} \times 100}{1 \times \frac{1}{4}}=\frac{100 \times 4}{40}=10 \%\)

\( R= \frac{100}{30}=\frac{10}{3}\) = 3 \( \frac{1}{3}\) %

\( \frac{100}{8}= 12.5 \)

\(\therefore\) Rate of interest \(=12.5 \% \mathrm{p} a\)

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

\(12.80=\frac{1600\left(R_{2}-R_{1}\right) \times 5}{100}\)

\(R{1}-R{2}=0.16 \) %

SI is also P since the amount is doubled in 5 years

\( R=\frac{SI\times100}{P\times5} \)

R=20%

Let the partial amount at 5% be x and the partial amount at 8% be (1550-x) .
Interest on x at 5% for 3 years + interest on (1550-x) at 8% for 3 years = 300
\(\frac{x \times 5 \times 3}{100} +\frac{(1550-x) \times 8 \times 3}{100} =300 \)
\(\frac{x \times 5}{100} +\frac{(1550-x) \times 8}{100} = 100 \)
\(5x + 8(1550 −x) = 10000\)
\(5x + 12400 −8x = 10000 \)
\(x = 800\)
Required Ratio = \(x : (1550-x) = 800 : (1550-800) = 800 : 750 = 16 : 15 \)

From the given data we can write ,
\(P - 340 = \frac{(P \times 4 \times 8)}{100}\)
\(100P - 34000 = 32P\)
\(68 P = 34000\)
\(P = 500\)Rs.

Let the principal amount be \( P \).

According to the problem, the amount after \(4\) years is \( \frac{9}{5}P \).

Using the simple interest formula:

\[ A = P + I \]

\[ I = A - P = \frac{9}{5}P - P = \frac{9}{5}P - \frac{5}{5}P = \frac{4}{5}P \]

\[ I = \frac{P \times r \times t}{100} \]

\[ \frac{4}{5}P = \frac{P \times r \times 4}{100} \]

\[ \frac{4}{5} = \frac{r \times 4}{100} \]

\[ r = \frac{4}{5} \times \frac{100}{4} = \frac{100}{5} = 20 \]

Thus, the rate percent per annum is \( \textbf{20%} \).

Let the sum borrowed be x.

Then, \(\frac{(x\times2\times6)}{100}\) + \(\frac{(x\times9\times3)}{100}\) +\(\frac{ (x\times14\times4)}{100}\) = 11400

\(\frac{3x}{25}\)+ \(\frac{27x}{100}\) + \(\frac{14x}{25}\) = 11400

\(\frac{95x}{25}\) = 11400

x = \(\frac{11400\times100}{95}\)

Hence , sum borrowed = Rs.12,000.

Given,

\(\mathrm{t}=4\) years

\(\therefore S . I a t 5 \%=\frac{x \times 5 \times 4}{100}=\frac{20 x}{100}\)

\(\therefore S . I\) at \(7.5 \%=\frac{(2450-x) \times 7.5 \times 4}{100}=\frac{30(2450-x)}{100}\)

Given, Total simple interest \(=\) Rs. 615

\(\frac{20 x}{100}+\frac{30(2450-x)}{100}=615\)

\(\Rightarrow 20 x+73500-30 x=61500\)

\(\Rightarrow 10 x=12000\)

\(\Rightarrow x=\operatorname{Rs} .1200\)

Simple interest= 900 - 750 = Rs.150
Principal= Rs.750
Time taken = 2 years
Rate = \( 100 \times \frac{150}{750 \times 2} = 10 \) %
New rate = 10 + 5 = 15 %
New simple interest = \(750 \times 15 \times \frac{2}{100} = Rs. 225 \)

Amount after \(2\) year on Rs.\(3000\) at \(10\)% per annum when interest is compounded annually

Amount = \(P\:(1+(\frac{R}{100})^{2} \)

Amount = \(3000\:(1+\frac{10}{100})^{2} \)

Amount = \(3000\:(\frac{110}{100})^{2} \)

Amount = \(33\times 110 \)

Compound Interest = \(3630 - 3000 = 630\)

Given that simple interest on a certain sum of money for \(4\) years at \(5\)% per annum is half of the compound interest.

i.e., simple interest = \(\frac{630}{2}= 315 \)

SI = \(\frac{PRT}{100}\)

P = \(\frac{100\times SI}{R\times T}\)

P = \(\frac{100\times 315}{5\times 4}\)

P = \(5 \times 315 = 1575\)

Let the sum borrowed is P,

Then, according to the question

\( [ \frac{P × 6× 3}{ 100}] + [ \frac{P × 9 × 5}{100}] + [ \frac{P × 13 × 3}{100}] = 8160 \)

\( \frac{18P + 45P + 39P}{100} = 8160 \)

\( \frac{102P}{ 100} = 8160 \)

∴ P = 8000 Hence, Swathi borrowed Rs. 8000

• Take the sum of money as 100

  At the end of third years it becomes \(100\times\frac{8}{5}\)=160

• here interest=160-100=60

  1 year interest=\(\frac{60}{3}=20\)

• rate of interest=\(\frac{20}{100}\times100=20\%\)

\(S.I = 1150 – 1000 = 150 \)
\(\frac{PNR}{100} = 150\)

\(\frac{(1000\times3\times R)}{ 100}= 150\)

\(R = 5\)

R is increased by \(3\)%,

New R \(= 5+3 = 8\)%
New S.I \(= \frac{(1000\times3\times8)}{100} = 240\)

Amount \(= 1240\)

The amount of money is doubled, so to find in how many years, \(\frac{100}{R}\) </br> \(R = 8\tfrac{1}{3} = \frac{25}{3}\) </br>

The number of years \(= \frac{100}{\frac{25}{3}}\) </br>

\(= \frac{100\times3}{25}\) </br> \(= 4\times3 = 12\) years

Let the principle amount be P and the rate of interest per annum be R%.

Given: The sum of money becomes\( \frac{5}{4} \) of itself in \( \frac{1}{3} \) years.

∴ Simple interest in this time period = \( (\frac{5P}{4}) – P = \frac{P}{4} \)

We know that. Simple interest \( = \frac{P × R × T}{100} \)

\(\therefore \frac{P}{4}=\frac{P \times R \times \frac{1}{3}}{100}\)

\(\Rightarrow R=\frac{300}{4} \%=75 \%\)

Hence, the rate of interest per annum is 75%.

Let percentage rate of interest be r
Let previously, sum was P, after 15 years, sum = 2P
∴ Interest = \(2P – P = P\)
∵ \(SI = \frac{PRT}{100}\)
∴ \(P = \frac{(P × r × 15)}{100}\)
⇒ \(15r = 100\)
⇒ \(r = 6\frac{2}{3}\%\)

\(\frac{Principal}{Intrest} = \frac{10}{3}\)

\(\frac{Interest}{Principal} = \frac{3}{10}\)

\(S.I = \frac{Time\times Principal\times Rate}{100}\)

\(Rate = \frac{S.I \times 100}{Principal\times Time}\)

\(Rate = \frac{3 \times 100}{10\times 5}\)

\(Rate = 6% \)

Let us write ratio \(25: 18\) as \(50: 36\)

So, the ratio of principal and one year simple interest will be \(50: 9\)

Rate of interest \(=\frac{9}{50} \times 100=18 \%\)

Hence, option D is correct.

The interest rate for Rs 800 for 4 years which gives 960-800 = 160
will be = \(\frac{160\times100}{800\times4} = 5\)%
New rate = 5% + 2% = 7%
\(\therefore New Interest = \frac{800\times4\times7}{100} = Rs 224\)
\(\therefore New Amount = 800+220 = Rs 1024\)

13.50 = \(\frac {1500 \times 3 \times r_1}{100}\) - \(\frac {1500 \times 3 \times r_2}{100}\)

13.50 = \(\frac {4500}{100}\)\((r_1 - r_2\))

\((r_1 - r_2\)) \( =\frac {135}{450} \)

\( => 0.3 \)%

Rs.7920 in 4 years
Rs.8880 in 6 years
960 interest increases in 2 years
If for 4 years (960 x 2 = 1920)
7920-1920 = 6000

SI = (P × r × t)/100

Where, P = principal, r = rate% per annum, t = time in years

Let the money lent at 10% be ‘a’.

Given, total money lent = Rs. 2000

Money lent at 20% = 2000 – a

SI at 10% for 3 years = (a × 10 × 3)/100

SI at 20% for 3 years = ((2000 – a) × 20 × 3)/100

Given, total interest received after 3 years is Rs. 750.

\(\Rightarrow \frac{30 a}{100}+\frac{120000-60 a}{100}=750\)

\(\Rightarrow 120000-30 a=75000\)

\(\Rightarrow 45000=30 a\)

\(\Rightarrow \mathrm{a}=\operatorname{Rs} .1500\)

Sum of money lent at \(20 \%=2000-1500=\) Rs. 500

Ratio of money lent at \(10 \%\) to that at \(20 \%=1500 / 500=3: 1\)

Let the the sum of money(P) be Rs.x.

Time(T) = \(3 \frac{1}{8}\) Years = \(\frac{25}{8}\) Years

Simple interest (SI) = \(\frac{x}{4}\)

Rate of interest per annum(R) = \(\frac{100 \times SI}{PT} = \frac{100 \times \frac{x}{4}}{x \times \frac{25}{8}}= \frac {100 \times x \times 8}{4 \times x \times 25}=8\)%

Let the money lent at \(7 \%\) and \(9 \%\) be Rs. \(x\) and Rs. \((8000-x)\) respecitively.

\(SI =\left( x \times 1 \times \frac{7}{100}\right)+(8000- x ) \times 1 \times \frac{9}{100}=620\)

\(\Rightarrow \frac{7 x}{100}+720-\frac{9 x}{100}=620\)

\(\Rightarrow \frac{2 x}{100}=100\)

\(\Rightarrow x=5000\)

Money lent are is in the ratio \(=5000:(8000-5000)=5: 3\)

Hence, option (A) is correct.

Simple Interest for 3 years = (Rs.12005 - Rs.9800) = Rs.2205

Simple Interest for 1 years =\( 2205\div 3 = Rs.735 \)

Simple Interest for 5 years = \(735\times 5=Rs.3675\)

Principal(P) = (Rs.9800- Rs.3675 ) = Rs.6125

\(R = \frac{100×SI}{PT}=\frac{100\times 3675}{6125 \times 5} =12 \% \)

SI=\( \frac{P\times R\times N }{100} \)

\( \frac{2000\times 10\times 1 }{100} \)=200

Given, Rate of interest for first 6 years = 3% p.a

Rate of interest for next 9 years = 5% p.a

Rate of interest beyond 3 years = 13% p.a

Simple interest for 11 years = Rs 81600

Concept

Simple interest, SI = PNR100 P → princiapl R → rate of interest T → time

Calculation

Let the money borrowed be Rs. P SI for 11 years = SI for first 6 years at 3% rate + SI for next 9 years at 5% rate + SI for 3 years at 13% rate

\( SI for 11 years = \frac{18P}{100}+ \frac{45P}{100}+ \frac{39P}{100} \)= Rs 81600

\( ⇒ \frac{102P}{100} \) = 81600

Or, P = 80000

Therefore the principal is Rs 80000

Time \(=5-2=3\) years
Simple interest for 3 years \(=12800-9920=2880 \mathrm{RS}\).
Simple interest for 2 years \(=\frac{2880}{3} \times 2=\) Rs. 1920
Principal = Amount \(-\) Interest \(=9920-1920=\) Rs. 8000
We know that:
Simple interest \(=\frac{P \times R \times T}{100}\) \[ \Rightarrow 2880=\frac{8000 \times 3 \times R}{100} \] \[ \Rightarrow R=\frac{2880 \times 100}{8000 \times 3} \] \[ \Rightarrow \mathrm{R}=12 \% \] Therefore, required rate of interest per annum is \(12 \%\).

\( R = \frac{I\times100}{PN} \)

=\( \frac{X\times100}{5X\times4} \)

=\( \frac{25x}{5x} \)

R=5.

• Rate of interest (R) = \(\frac{I \times 100}{PN}\)

• Ratio of Principal and Simple interest = 15 : 1

\(\Rightarrow\) Principal = 15 & Simple interest = 1

\(\therefore\) Rate of interest (R) = \(\frac{1 \times 100}{15}\) = 6\(\tfrac{2}{3}\)%

Principal = P.
S.I = P
Rate = \(\frac{100}{16} = 6 \frac{1}{4} \)%

SI = PNR/100

If it doubles then SI = P
\(P= \frac{PNR}{100}\)

Given \(N = 12\)
\(R= \frac{100}{12} = \frac{25}{3}\)

\(I=P \times i \times t \)
here , i = rate of interest per annum

The simple interest \(I_1= \frac{6000 \times 5 \times 4}{100}=1200\) Rs
From the question it is clear that ,\(I_1 =I_2\)
\(\therefore 1200 = \frac{8000 \times 3 \times t}{100}\)
\(1200=240 t\)
\(t = 5\) years

• Let the ratio be x

• So, principal is 10x and the simple interest is 7x

• Now, according to question

\( 7x = \frac{10x × R × 5}{100} \)

\( R = \frac{7 × 100 }{50} \).

• ∴ Rate percent = 14%.

If amout= P

S.I = P- 140

\( S.I = \frac{P \times N \times R}{100} \)

\( S.I = \frac{P \times 6 \times 5}{100} \)

\( P - 140 = \frac{30 P}{100} \)

\( 100P - 14000 = 30 P \)

\(\Rightarrow P = 200 \)

Let the sum of money be Rs.x

Amount after 5 years = 33x/20

T = 5 years

R = ?

\[\textrm{Simple Interest, SI = }(\frac{33x}{20}-x) =\frac{13x}{20}\] \[\textrm{Rate of interest R =} \frac{100×SI}{PT}=\frac{100×\frac{13x}{20}}{x×5} \] \[=\frac{100×13x}{x×5×20}
=13 \]