We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.More informationAgree
menu ×
Google
Custom Search
cancel
search
FacebookTwitterLinkedIn×
share
page location
×
ad

Solved Examples(Set 9) - Simple Interest

41. If the simple interest on a certain sum of money for $4$ years is one–fifth of the sum, then the rate of interest per annum is
A. $7\%$B. $5\%$
C. $4\%$D. $6\%$
Discuss
answer with explanation

Answer: Option B

Explanation:

Solution 1

Let the sum be be $x$

Then, simple interest $=\dfrac{x}{5}$

T $=4$

$\text{R}=\dfrac{100×\text{SI}}{\text{PT}}=\dfrac{100×\dfrac{x}{5}}{x×4}=5$

Solution 2

Total simple interest of $4$ years is one–fifth of the sum, that is, $20\%$ of the sum.

rate of interest per annum $=\dfrac{20}{4}\%=5\%$

42. If a certain sum is invested for T years at simple interest of $10\%$ per annum, it amounts to $₹400.$ If the rate of interest was $4\%$ per annum, it would have amounted to $₹200.$ Find T.
A. $45$ yearsB. $40$ years
C. $50$ YearsD. $60$ years
Discuss
answer with explanation

Answer: Option C

Explanation:

Solution 1

Refer formula

$\text{T}=\dfrac{\text{P}_1-\text{P}_2}{\text{P}_2\text{R}_1-\text{P}_1\text{R}_2}×100\\=\dfrac{400-200}{200×10-400×4}×100=50$

Solution 2

Let the sum be $x$

Simple interest on $x$ at $10\%$ per annum in $\text{T}$ years $=400-x$

Simple interest on $x$ at $4\%$ per annum in $\text{T}$ years $=200-x$

simple interest $\propto$ R (because P and T are constants here). Therefore,

$400-x:200-x=10:4\\\Rightarrow 400-x:200-x=5:2\\\Rightarrow 800-2x=1000-5x\\\Rightarrow x=\dfrac{200}{3}$

Simple interest on $\dfrac{200}{3}$ at $10\%$ per annum in $\text{T}$ years $=400-\dfrac{200}{3}=\dfrac{1000}{3}$

Therefore, T $=\dfrac{100×\dfrac{1000}{3}}{\dfrac{200}{3}×10}=50$

Solution 3

Let the sum be $x$

$x$ amounts to $400$ at $10\%$ per annum in $\text{T}$ years.
Simple interest $=400-x$

$400-x=\dfrac{x×10×\text{T}}{100}~\cdots(1)$

$x$ amounts to $200$ at $4\%$ per annum in $\text{T}$ years.
Simple interest $=200-x$

$200-x=\dfrac{x×4×\text{T}}{100}~\cdots(2)$

Dividing $(1)$ by $(2)$

$\dfrac{400-x}{200-x}=\dfrac{5}{2}\\\Rightarrow 800-2x=1000-5x\\\Rightarrow 3x=200\\\Rightarrow x=\dfrac{200}{3}$

Substituting value of $x$ in $(1)$

$400-\dfrac{200}{3}=\dfrac{\left(\dfrac{200}{3}\right)×10×\text{T}}{100}\\\Rightarrow 400-\dfrac{200}{3}=\dfrac{20\text{T}}{3}\\\Rightarrow 1200-200=20\text{T}\\\Rightarrow 20\text{T}=1000\\\Rightarrow\text{T}=50$

43. Arun lent a sum of money at simple interest for $6$ years. If the same amount was paid at $4\%$ higher, Arun would have got $₹120$ more. Find the sum.
A. $₹600$B. $₹500$
C. $₹200$D. $₹400$
Discuss
answer with explanation

Answer: Option B

Explanation:

Simple interest at $4\%$ is $₹120$

$\text{P}=\dfrac{100×\text{SI}}{\text{RT}}=\dfrac{100×120}{4×6}=500$

44. Mr.Mani invested an amount of $₹12000$ at a simple interest rate of $10\%$ per annum and another amount at a simple interest rate of $20\%$ per annum. The total interest earned at the end of one year on the total amount invested is $14\%$ per annum. Find the total amount invested.
A. $₹25000$B. $₹15000$
C. $₹10000$D. $₹20000$
Discuss
answer with explanation

Answer: Option D

Explanation:

Solution 1

Refer formula

$14=\dfrac{12000×10+\text{P}_2×20}{12000+\text{P}_2}\\\Rightarrow 12000×14+14\text{P}_2=120000+20\text{P}_2\\\Rightarrow 6\text{P}_2=48000\\\Rightarrow \text{P}_2=8000$

Total amount invested $=12000+8000=20000$

Solution 2

Let the other amount invested be $x$

Total simple interest $=\dfrac{12000×10×1}{100}+\dfrac{x×20×1}{100}=1200+\dfrac{x}{5}$

But $1200+\dfrac{x}{5}$ is the simple interest on $12000+x$ at $14\%$ per annum for $1$ year. Therefore,

$\Rightarrow 1200+\dfrac{x}{5}=\dfrac{(12000+x)×14×1}{100}\\\Rightarrow 120000+20x=14×12000+14x\\\Rightarrow 6x=48000\\\Rightarrow x=8000$

Total amount invested $=12000+x=12000+8000=20000$

45. A lent $₹4000$ to B for $2$ years and $₹2000$ to C for $4$ years on simple interest at the same rate of interest and received $₹2200$ in all from both of them as interest. The rate of interest per annum is:
A. $15\%$B. $12\%$
C. $13.75\%$D. $14\%$
Discuss
answer with explanation

Answer: Option C

Explanation:

Solution 1

Let the rate of interest per annum be R%

Simple interest on $₹4000$ for $2$ years at R% + simple interest on $₹2000$ for $4$ years at R% $=2200$

$\Rightarrow \dfrac{4000×\text{R}×2}{100}+\dfrac{2000×\text{R}×4}{100}=2200\\\Rightarrow 160\text{R}=2200\\\Rightarrow \text{R}=\dfrac{2200}{160}=\dfrac{55}{4}=13.75$

Solution 2

$₹4000$ for $2$ years and $₹2000$ for $4$ years give same simple interest.

Therefore, simple interest on $₹4000$ for $2$ years $=\dfrac{2200}{2}=1100$

R $=\dfrac{100×1100}{4000×2}=13.75$

Add Your Comment

(use Q&A for new questions)
?
Name
cancel
preview