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 = ?

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