5 Mark Question

Question 15 Marks

Find the time when:
A sum of money invested at 8% per annum amounts to Rs. 12122 in 2 years. What will it amount to in 2 years 8 months at 9% per annum?

Answer

Amount = Rs. 12122
Let principal (P) = Rs. 100
Rate (r) = 8% p.a.
Time (t) = 2 years
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{100\times8\times2}{100}$
$=\text{Rs. }16$
And amount = P × S.I. = Rs. 100 + 16
$= \text{Rs. }116$
If amount is Rs. 116, then principal
$= \text{Rs. }100$
And if amount is Rs. 1, then principal
$=\text{Rs. }\frac{100}{116}$
And if amount is Rs. 12122, then principal
$=\text{Rs. }\frac{100\times12122}{116}$
$=\text{Rs. }10450$
Now, rate (r) = 9%
And time (t) = 2 years, 8 months
$=2\frac23\text{ years}=\frac83\text{ years}$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\text{Rs. }\frac{10540\times9\times8}{100\times3}$
$=\text{Rs. }2508$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }10450+2508$
$=\text{Rs. }12958$

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Question 25 Marks

Find the time when:
A sum of money put at 11% per annum amounts to Rs. 4491 in 2 years 3 months, what will it amount to in 3 years at the same rate?

Answer

Amount = Rs. 4491
Let principal (P) = Rs. 100
Rate (r) =11% p.a.
$\text{Time(t)}=2\text{ years}\ 3\text{ months}=2\frac14=\frac94\text{ years}$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{100\times11\times9}{100\times4}$
$=\text{Rs. }\frac{99}{4}$
$\therefore\text{Amount}\text{ A}=\text{P}+\text{S.I.}$
$=\text{Rs. }100+\text{Rs. }\frac{99}{4}=\text{Rs. }\frac{499}{4}$
If amount is $\text{Rs. }\frac{499}{4},$ then principal
$= \text{Rs. }100$
If amount is Rs. 1, then principal
$=\text{Rs. }\frac{100\times4}{499}$
$=\text{Rs. }3600$
Now, interest for 3 years at the rate of 11%
$=\text{Rs. }=\frac{3600\times11\times3}{100}=\text{Rs. }1188$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }3600+\text{Rs. }1188$
$=\text{Rs. }4788$

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Question 35 Marks

Find the time when:
Divide Rs. 3600 into two parts such that if one part be lent at 9% per annum and the other at 10% per annum, the total annual income is Rs. 333.

Answer

Total sum = Rs. 3600
Let first part = Rs. x
Then second part = Rs. (3600 - x)
Interest on first part for 1 year at 9% p.a.
$=\frac{\text{x}\times9\times1}{100}=\frac{9\text{x}}{100}$
And interest on second part for 1 years at 10% p.a.
$=\frac{(3600-\text{x})\times10\times1}{100}$
$=\frac{10(3600-\text{x})}{100}$
But total interst = Rs. 333
$\therefore\frac{9\text{x}}{100}+\frac{10(3600-\text{x})}{100}=333$
$\Rightarrow 9\text{x} + 10 (3600 - \text{x}) = 33300 $
$\Rightarrow 9\text{x} + 36000 - 10\text{x} = 33300 $
$\Rightarrow -\text{x} = 33300 - 36000 $
$\Rightarrow-\text{x} = - 2700 $
$\Rightarrow \text{x} = 2700$
First part = Rs. 2700
and second part = Rs. 3600 - 2700 = Rs. 900

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Question 45 Marks

Find the time when:
Divide Rs. 3000 into two parts so that the simple interest on the first part for 4 years at 8% per annum is equal to the simple interest on the second part 2 years at 9% per annum.

Answer

Total sum = Rs. 3000
Let first part = Rs. x
Then second part = Rs. (3000 - x)
Now, interest on first part at the rate of
$8\%\text{ for 4 years} =\frac{\text{x}\times8\times4}{100}=\frac{32}{100}\text{x}$
And interest on the second part for 2 years
at $9\%=\frac{(3000-\text{x})\times9\times2}{100}$
$=\frac{18(3000-\text{x})}{100}$
But in both the case, interest is same.
$\therefore\frac{32}{100}\text{x}=\frac{18(3000-\text{x})}{100}$
$\Rightarrow32\text{x}=18(3000-\text{x)}$
$\Rightarrow\text{32x}=54000-18\text{x}$
$\Rightarrow32\text{x}+\text{18x}=54000$
$\Rightarrow50\text{x}=54000$
$\Rightarrow\text{x}=\frac{54000}{50}=1080$
$\therefore$ First part = Rs. 1080
And second part = Rs. (3000 - 1080)
= Rs. 1920

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Maths 5 Mark Question

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