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11. A sum of $₹14,000$ amounts to $₹22,400$ in $12$ years in simple interest. What is the rate of interest? | |

A. $7\%$ | B. $6\%$ |

C. $5\%$ | D. $4\%$ |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Simple interest(SI) for $12$ years $=22400-14000=8400$

$\text{R}=\dfrac{100×\text{SI}}{\text{PT}}=\dfrac{100×8400}{14000×12}=5$

12. A sum of $₹725$ is lent in the beginning of a year at a certain rate of interest. After $8$ months, a sum of $₹362.50$ more is lent but at the rate twice the former. At the end of the year, $₹33.50$ is earned as interest from both the loans. What was the original rate of interest? | |

A. $5\%$ | B. None of these |

C. $3.47\%$ | D. $4.5\%$ |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Suppose $₹725$ is lent out at rate of R% for $1$ year. Then, at the end of 8 months, $₹362.50$ more is lent out at rate of 2R% for the remaining $4$ months($1/3$ year)

Total simple interest $=33.50$

$\Rightarrow \dfrac{725×\text{R}×1}{100}+\dfrac{362.50×\text{2R}×\dfrac{1}{3}}{100}=33.50\\\Rightarrow \dfrac{725\text{R}}{100}+\dfrac{725\text{R}}{100}×\dfrac{1}{3}=33.50\\\Rightarrow \dfrac{725\text{R}}{100}×\dfrac{4}{3}=33.50\\\Rightarrow \dfrac{29\text{R}}{3}=33.50\\\Rightarrow \text{R}=3.47$

$₹725$ is lent out at rate of R% for $1$ year $\cdots(1)$

$₹362.50$ is lent out at rate of 2R% for $4$ months

$\equiv$ $₹362.50$ is lent out at rate of R% for $8$ months

$\equiv$ $₹725$ is lent out at rate of R% for $4$ months $\cdots(2)$

From $(1)$ and $(2),$

Simple interest on $725$ for $1$ year $4$ months at R% $=33.50$

$\text{R}=\dfrac{100×33.5}{725×\dfrac{16}{12}}=\dfrac{33.5}{29×\dfrac{1}{3}}\\=\dfrac{33.5×3}{29}=3.47$

13. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of $10\%,$ the effective rate of interest after one year becomes: | |

A. $10.5\%$ | B. $10.25\%$ |

C. $10\%$ | D. None of these |

Discuss |

answer with explanation

Answer: Option B

Explanation:

Suppose the automobile financier lends $100$

Simple interest for first $6$ months $=5$

hint

For $1$ year, simple interest is $10\%$ of $100,$ which is $10$ and for $6$ months, simple interest is half of $10,$ which is $5.$

Using formula, we get the same as $\dfrac{100×10×\dfrac{1}{2}}{100}=5$

After $6$ months, he adds the simple interest to principal. Therefore, principal after $6$ months $=100+5=105$

Simple interest for next $6$ months $=5.25$

hint

For $1$ year, simple interest is $10\%$ of $105,$ which is $10.5$ and for $6$ months, simple interest is half of $10.5$ which is $5.25$

Using formula, we get the same as $\dfrac{105×10×\dfrac{1}{2}}{100}=5.25$

Total simple interest for $1$ year $=5+5.25=10.25$

Hence, effective rate of interest $=10.25\%$

hint

$100$ gives $10.25$ as simple interest for $1$ year and therefore interest rate is $10.25\%.$

Using formula, we get the same as $\dfrac{100×10.25}{100×1}=10.25$

Suppose the automobile financier lends $100$

Simple interest for first $6$ months $=5$

So, additional amount he gets (when compared to the normal simple interest on $₹100$ for $1$ year at $10\%$) is the simple interest on this $5$ for next $6$ months at $10\%$

simple interest on $5$ for $6$ months at $10\%$

$\equiv$ simple interest on $5$ for $1$ year at $5\%$

$\equiv$ simple interest on $100$ for $1$ year at $\dfrac{5}{20}\%$

So, effective rate of interest

$=10\%+\dfrac{5}{20}\%=10\%+0.25\%=10.25\%$

Refer formula

Suppose the automobile financier lends $100$

Amount after $1$ year $=100\left(1+\dfrac{10/2}{100}\right)^{2×1}=100\left(\dfrac{21}{20}\right)^2=110.25$

Total simple interest for $1$ year $=110.25-100=10.25$

Therefore, effective rate of interest $=10.25\%$

14. A lent $₹5000$ to B for $2$ years and $₹3000$ 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. What is the rate of interest per annum? | |

A. $7\%$ | B. $5\%$ |

C. $10\%$ | D. $8\%$ |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Let required rate be R%

Simple interest on $₹5000$ for $2$ years + simple interest on $₹3000$ for $4$ years $=₹2200$

$\Rightarrow \dfrac{5000×\text{R}×2}{100}+\dfrac{3000×\text{R}×4}{100}=2200\\\Rightarrow \text{100R+120R}=2200\\\Rightarrow \text{220R}=2200\\\Rightarrow \text{R}=10$

SI on $₹5000$ for $2$ years + SI on $₹3000$ for $4$ years $=₹2200$

$\Rightarrow$ SI on $₹50$ for $1$ year + SI on $₹30$ for $2$ years $=₹11$

$\Rightarrow$ SI on $₹30$ for $\dfrac{5}{3}$ years + SI on $₹30$ on $2$ years $=₹11$

$\Rightarrow$ SI on $₹30$ for $\dfrac{11}{3}$ years $=₹11$

$\Rightarrow$ SI on $₹30$ for $1$ year $=₹3$

$\Rightarrow$ SI on $₹100$ for $1$ year $=₹10$

Therefore, required rate is $10\%$

15. What annual payment will discharge a debt of $₹6450$ due in $4$ years at simple interest of $5\%$ per annum? | |

A. $₹1800$ | B. $₹1400$ |

C. $₹1500$ | D. $₹1600$ |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Refer formula

Required annual payment $=\dfrac{100×6450}{100×4+\dfrac{5×4(4-1)}{2}}\\=\dfrac{100×6450}{400+30}=1500$

Amount needs to be repaid in $4$ years is $₹6450.$ Suppose $x$ is paid annually to repay this debt.

Amount paid after first year $=x$

Interest on $x$ for remaining $3$ years

$=\dfrac{x×5×3}{100}$

Amount paid after second year $=x$

Interest on $x$ for remaining $2$ years

$=\dfrac{x×5×2}{100}$

Amount paid after third year $=x$

Interest on $x$ for remaining $1$ year

$=\dfrac{x×5×1}{100}$

Amount paid after fourth year $=x$ and this closes the entire debt.

Therefore

$x+\dfrac{x×5×3}{100}+x+\dfrac{x×5×2}{100}+x+\dfrac{x×5}{100}+x=6450\\4x+\dfrac{5x(3+2+1)}{100}=6450\\\Rightarrow 4x+\dfrac{3x}{10}=6450\\\Rightarrow 40x+3x=64500\\\Rightarrow 43x=64500\\\Rightarrow x=1500$

rishi

2015-06-02 14:59:35

In the 12th solution from where does 4 come. What is the calculation for this

usha

2015-06-11 17:32:07

in one year after 8 months 362.5 more is added for remaning months (i.e) for 1 year 12months so 12-8 = 4 months

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shashi

2015-03-11 06:11:10

In 15th problem, 1500*4 will be 6000 only...and we r supposed to repay 6450 with rate of 15% p.a.......plzz...explain.

shashank

2015-05-20 19:32:59

look sashi...1500 is amount over which interest will be charge differently for the 4 years...for first year it will be 15 % so with 1500 interest of rs 225...then for 2nd year interst is 10 % ....so it will be 150..similary for 3 rd and 4th......so if we add up the intersets we get 225 + 150+ 75 = 450.....hope u understood this

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vamshi

2015-07-16 17:22:33

why is the interest different for each year ? or is it the interest amount earned over 3 yrs 1st then 2 yrs then 1yr

If it is like that bank earns the interest on our installment. At the same time there is interest imposed on the remaining amount of money we need to pay. Why is it not adding up?

Still it is out of logic that an installment of 1500/- every year for four years will discharge a debt of 6450 because it just doesn't add up. Am i wrong

OR is it like this that the amount 6450/- is principal + interest

If it is like that bank earns the interest on our installment. At the same time there is interest imposed on the remaining amount of money we need to pay. Why is it not adding up?

Still it is out of logic that an installment of 1500/- every year for four years will discharge a debt of 6450 because it just doesn't add up. Am i wrong

OR is it like this that the amount 6450/- is principal + interest

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srilakshmi

2014-08-21 15:21:16

In 11th problem why the calculation of simple interest is for 4 years instead of 12 years ????

kamlesh

2015-02-11 20:28:51

it is written by mistake 4 instead of 12.

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