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## Arithmetic: Compound Interest Test -6

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*Arithmetic: Compound Interest Test -6*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

The simple interest on Rs. 84000 for 3 years is Rs. 30240. On the same amount, for the same period and the same rate, what will be the compound interest?

A | Rs 30013.95 |

B | Rs 31013.95 |

C | Rs 32013.95 |

D | Rs 34013.95 |

Question 1 Explanation:

Principal value, P = Rs. 84000

Rate =R,

Time T = 3 years

We know:

SI = PRT/100

Therefore, R = (SI × 100)/ P x T

= 30240 × 100/84000 × 3

= 12% per annum

Therefore, the compound interest is

= P (1+ r/100 )

= 84000 (1 + 12/100)³ - 84000 = 84000 × 1.12 × 1.12 × 1.12 – 84000

= 118013.952 – 84000 = Rs. 34013.95

Rate =R,

Time T = 3 years

We know:

SI = PRT/100

Therefore, R = (SI × 100)/ P x T

= 30240 × 100/84000 × 3

= 12% per annum

Therefore, the compound interest is

= P (1+ r/100 )

^{n}– P= 84000 (1 + 12/100)³ - 84000 = 84000 × 1.12 × 1.12 × 1.12 – 84000

= 118013.952 – 84000 = Rs. 34013.95

Question 2 |

On a certain sum of amount, the difference between compound interest (compounded annually) and simple interest for 2 years at 10% per annum is Rs. 28. If the compound interest is reckoned half yearly, then the difference between two interests is:

A | 44 |

B | 28.5 |

C | 44.45 |

D | 43.42 |

Question 2 Explanation:

Difference between compound interest and simple interest can be written as

CI – SI = {P [(1+r/100)

= P [(1+r/100)

But the difference is given i.e. Rs. 28

So, 28= P [(1+r/100)

= 28 = P[(1+10/100)

On solving this equation we get principal value

= P = 2800

Now question tells that if the Interest reckoned half yearly

So amount will be

A= P (1+R/100)

A = 2800(1+5/100)

A= 3403.41

Therefore the compound interest will be 3403.41 – 2800= 603.41

And the simple interest will be S.I= (2800 x 10 x 2)/100= 560

And the difference is 603.41 – 560 =43.417 = Rs. 43.42

Hence, the right option is D

CI – SI = {P [(1+r/100)

^{T}-1} – PRT/100= P [(1+r/100)

^{T}-1-(RT/100)]But the difference is given i.e. Rs. 28

So, 28= P [(1+r/100)

^{T}-1-RT/100]= 28 = P[(1+10/100)

^{2 }-1-10 X 2/100]On solving this equation we get principal value

= P = 2800

Now question tells that if the Interest reckoned half yearly

So amount will be

A= P (1+R/100)

^{2T}A = 2800(1+5/100)

^{4}A= 3403.41

Therefore the compound interest will be 3403.41 – 2800= 603.41

And the simple interest will be S.I= (2800 x 10 x 2)/100= 560

And the difference is 603.41 – 560 =43.417 = Rs. 43.42

Hence, the right option is D

Question 3 |

Pankaj took a sum of Rs 4500 from Richa. He promised Richa that he will give back her money at the end of the year but she gave an option to him that he can pay her in two equal annual installments.Pankaj agreed on her suggestion .If the rate of interest taken by Richa was 10% per annum, compounded annually, find the amount of each instalment given be Pankaj.

A | 2390 |

B | 3429 |

C | 2560 |

D | none |

Question 3 Explanation:

In case of instalments we will apply the formula:

P= A/(1+r/100)

Where n is the number of years, A is the installment

4500 = A/(1+10/100)

4500 = 100A /121 + 10A/11

A= (4500 X 121)/210 =Rs. 2592.85

Therefore, the right answer is D

P= A/(1+r/100)

^{n}……………. + A/(1+r/100)Where n is the number of years, A is the installment

4500 = A/(1+10/100)

^{2}+ A/(1+10/100)4500 = 100A /121 + 10A/11

A= (4500 X 121)/210 =Rs. 2592.85

Therefore, the right answer is D

Question 4 |

The population of Dehradun was 15000 two years ago. If it had increased by 2% and 3% in the last two years, find the present population of Dehradun

A | 15760 |

B | 15758 |

C | 15768 |

D | 15759 |

Question 4 Explanation:

This is a simple question of compound growth.

When the population is increasing by a different growth rate,

the population at the end of year one becomes the base for year two.

Total Population= P(1+R/100) x (1+R/100)

=> 15000(1+2/100)

=> 15000 x 102/100 x 103/100 = 15759

Hence, the right answer is option D

When the population is increasing by a different growth rate,

the population at the end of year one becomes the base for year two.

Total Population= P(1+R/100) x (1+R/100)

=> 15000(1+2/100)

^{1}x (1+3/100)^{1}=> 15000 x 102/100 x 103/100 = 15759

Hence, the right answer is option D

Question 5 |

Richa bought a new BMW car. The value of the car is Rs. 6000000. If rate of depreciation is 10% for first year and 5% for the second and remains constant, then what will be the value of car at end of second year?

A | Rs. 5130300 |

B | Rs. 5131300 |

C | Rs. 5130000 |

D | none |

Question 5 Explanation:

Rate of depreciation is given by over the period of time n years is given the formula:

P(1-R/100)

Keep in mind that this is formula for negative compound growth, that is all

Now we have:

P = 6000000

Rate 10% and 5%

T= total of 2 years, one year each at a certain rate.

So P(1-R/100)

6000000(1-10/100)(1-5/100)

= 6000000 x (90/100) x (95/100)

= 600 x 90 x 95 =Rs. 5130000

P(1-R/100)

^{t1}(1-R/100)^{t2}…………………….(1-R/100)^{tn}Keep in mind that this is formula for negative compound growth, that is all

Now we have:

P = 6000000

Rate 10% and 5%

T= total of 2 years, one year each at a certain rate.

So P(1-R/100)

^{t1}(1-R/100)^{t2}6000000(1-10/100)(1-5/100)

= 6000000 x (90/100) x (95/100)

= 600 x 90 x 95 =Rs. 5130000

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