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1. What is the difference between the compound interests on $₹5000$ for $1\dfrac{1}{2}$ years at $4\%$ per annum compounded yearly and half-yearly? | |

A. $₹2.04$ | B. $₹3.06$ |

C. $₹8.30$ | D. $₹4.80$ |

Discuss |

answer with explanation

Answer: Option A

Explanation:

Interest for first $6$ months in both cases, is $\dfrac{1}{2}(4\%\text{ of }5000)=100.$

In the next $6$ months, $\dfrac{1}{2}(4\%\text{ of }100)=2$ will be extra interest in half-year compounding.

In the next $6$ months, $\dfrac{1}{2}(4\%\text{ of }2)=0.04$ will be extra interest for half-year compounding.

Therefore, required difference $=2+0.4=2.04$

When interest is compounded yearly,

interest for first year $=4\%(5000)=200$

interest for next $6$ months $=\dfrac{200}{2}+\dfrac{1}{2}[4\%(200)]=104$

compound interest $=200+104=304$

$2\%(5000)=100\\2\%(100)=2\\2\%(2)=0.04$

When interest is compounded half-yearly,

compound interest for $1\dfrac{1}{2}$ years

$=3(100+2)+0.04=306.4$

(formula)

Required difference $=306.4-304=2.04$

When interest is compounded yearly,

interest for first year $=4\%(5000)=200$

interest for next $6$ months $=\dfrac{200}{2}+\dfrac{1}{2}[4\%(200)]=104$

compound interest $=200+104=304$

When interest is compounded half-yearly,

amount after $1\dfrac{1}{2}$ years $=5000\left(1+\dfrac{4/2}{100}\right)^{2×\frac{3}{2}}=5000\left(\dfrac{51}{50}\right)^3=5306.04$

compound interest $=5306.04-5000=306.4$

Required difference $=306.4-304=2.04$

When interest is compounded yearly,

interest for first year $=\dfrac{5000×4×1}{100}=200$

amount after first year $=5000+200=5200$

interest for next $6$ months $=\dfrac{5200×4×\dfrac{1}{2}}{100}=104$

compound interest $=200+104=304$

When interest is compounded half-yearly,

interest for $6$ months $=\dfrac{5000×4×\dfrac{1}{2}}{100}=100$

amount after $6$ months $=5000+100=5100$

interest for next $6$ months $=\dfrac{5100×4×\dfrac{1}{2}}{100}=102$

amount after $1$ year $=5100+102=5202$

interest for next $6$ months $=\dfrac{5202×4×\dfrac{1}{2}}{100}=104.04$

compound interest $=100+102+104.04=306.04$

Required difference $=306.4-304=2.04$

2. A bank offers $5\%$ compound interest calculated on half-yearly basis. A customer deposits $₹1600$ each on $1\text{st}$ January and $1\text{st}$ July of a year. At the end of the year, the amount he would have gained by way of interest is: | |

A. $₹121$ | B. $₹120$ |

C. $₹123$ | D. $₹122$ |

Discuss |

answer with explanation

Answer: Option A

Explanation:

$\dfrac{5}{2}\%(1600)=40\\\dfrac{5}{2}\%(40)=1$

Compound interest on $1600$ deposited on $1$st January

$=2×40+1=81$

(formula)

Compound interest on $1600$ deposited on $1$st July

$=\dfrac{1}{2}(5\%\text{ of }1600)=40$

Required gain $=81+40=121$

Case $1:$ $₹1600$ deposited on $1\text{st}$ Jan

Interest for $6$ months $=\dfrac{1}{2}[5\%(1600)]=40$

Interest for next $6$ months $=40+\dfrac{1}{2}[5\%(40)]=41$

Total interest $=40+41=81$

Case $2:$ $₹1600$ deposited on $1\text{st}$ Jul

Total interest $=\dfrac{1}{2}[5\%(1600)]=40$

Required gain $=81+40=121$

Amount after $1$ year on $₹1600$ (deposited on $1\text{st}$ Jan)

$=1600\left(1+\dfrac{5/2}{100}\right)^{2×1}=1600\left(\dfrac{41}{40}\right)^2=1681$

Compound interest $=1681-1600=81$

Amount after $\dfrac{1}{2}$ year on $₹1600 ($deposited on $1\text{st}$ Jul)

$=1600\left(1+\dfrac{5/2}{100}\right)^{2×\frac{1}{2}}=1600\left(\dfrac{41}{40}\right)=1640$

Compound interest $=1640-1600=40$

Required gain $=81+40=121$

3. There is $80\%$ increase in an amount in $8$ years at simple interest. What will be the compound interest of $₹14,000$ after $3$ years at the same rate? | |

A. $₹3714$ | B. $₹3794$ |

C. $₹4612$ | D. $₹4634$ |

Discuss |

answer with explanation

Answer: Option D

Explanation:

$80\%$ increase in $8$ years at simple interest. That is, $10\%$ each year. Therefore, rate of interest is $10\%.$

$10\%(14000)=1400\\10\%(1400)=140\\10\%(140)=14$

Required compound interest

$=3(1400+140)+14=4634$

(formula)

If principal is $100,$ simple interest for $8$ years $=80$ (because, $80\%$ increase is there due to the simple interest). Therefore,

$\text{R}=\dfrac{100×80}{100×8}=10$

Now let's find out the compound interest on $₹14,000$ for $3$ years at $10\%$

Amount after $3$ years

$=14000\left(1+\dfrac{10}{100}\right)^3\\=14000\left(\dfrac{110}{100}\right)^3=18634$

Compound interest $=18634-14000=4634$

4. The compound interest on $₹30,000$ at $7\%$ per annum is $₹4347.$ The period (in years) is: | |

A. $2$ | B. $3$ |

C. $1$ | D. $3.5$ |

Discuss |

answer with explanation

Answer: Option A

Explanation:

Let the period be $n$ years

Amount after $n$ years $=30000+4347=34347$

$30000\left(1+\dfrac{7}{100}\right)^\text{n}=34347\\\Rightarrow 30000\left(\dfrac{107}{100}\right)^\text{n}=34347\\\Rightarrow \left(\dfrac{107}{100}\right)^\text{n}=\dfrac{34347}{30000}= \dfrac{11449}{10000}=\left(\dfrac{107}{100}\right)^2\\\Rightarrow n=2$

Interest for first year $=7\%(30000)=2100$

$2100×2$ is close to $4347$ and therefore total years will be near to $2$

Interest for second year $=2100+7\%(2100)=2247$

$2100+2247=4347$

Therefore, required period is $2$ years.

5. The difference between simple and compound interest(compounded annually) on a certain sum of money for $2$ years at $4\%$ per annum is $₹1.$ What is the sum? | |

A. $₹600$ | B. $₹645$ |

C. $₹525$ | D. $₹625$ |

Discuss |

answer with explanation

Answer: Option D

Explanation:

Refer formula

$\text{P}\left(\dfrac{4}{100}\right)^2=1\\\Rightarrow \text{P}\left(\dfrac{1}{25}\right)^2=1\\\Rightarrow \dfrac{\text{P}}{25^2}=1\\\Rightarrow \text{P}=625$

Let the sum be $x$

Amount after $2$ years (when interest is compounded annually)

$=x\left(1+\dfrac{4}{100}\right)^2=x\left(\dfrac{26}{25}\right)^2$

Compound interest

$=x\left(\dfrac{26}{25}\right)^2-x=x\left[\dfrac{676}{625}-1\right]=\dfrac{51x}{625}$

Simple interest $=\dfrac{x×4×2}{100}=\dfrac{2x}{25}$

Difference between compound interest and simple interest is $1.$ Therefore,

$\Rightarrow \dfrac{51x}{625}-\dfrac{2x}{25}=1\\\Rightarrow \dfrac{51x-50x}{625}=1\\\Rightarrow x=625$

Let the sum be $x$

Compound interest

$=2[4\%(x)]+4\%[4\%(x)]$

(formula)

Simple interest $=2[4\%(x)]$

$2[4\%(x)]+4\%[4\%(x)]-2[4\%(x)]=1\\\Rightarrow 4\%[4\%(x)]=1\\\Rightarrow 4\%(x)=25\\\Rightarrow x=625$

aashish gupta

2015-03-27 06:26:58

sir , can we have sample question of other sections - logical reasoning , dATA INTERPRETATION .

sree jothi

2015-03-24 10:41:47

the difference between the S.I and C.I compounded every six months at rate of 30% per annum, at the end of 1 1/2 years of rs.5670.what is the sum?(rs.40,000 , Rs.60,000, Rs.64,000, Rs.72,000,Rs.80,000)

Dev

2015-04-05 19:08:42

Time = 1 1/2 year = 3/2 year

R = 30

Let P is the amount

Simple Interest = PRT/100 = P * 30 * (3/2) /100= 9P/20

Compound Interest = P[1 + (R/2)/100]^{2n} - P

= P[1 + (30/2)/100]^{(2*3/2)} - P

= P(1.15)^{3} - P

= P[(1.15)^{3}-1]

P[(1.15)^{3}-1] - 9P/20 = 5670

P[(1.15)^{3}-1- 9/20] = 5670

P = 80000

R = 30

Let P is the amount

Simple Interest = PRT/100 = P * 30 * (3/2) /100= 9P/20

Compound Interest = P[1 + (R/2)/100]

= P[1 + (30/2)/100]

= P(1.15)

= P[(1.15)

P[(1.15)

P[(1.15)

P = 80000

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