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Comparing Quantities Test - 61

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Comparing Quantities Test - 61
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  • Question 1
    1 / -0
    What would be the C.I. obtained on an amount of Rs. $$12000$$ at the rate of $$9$$ p.c.p.a for $$3$$ years?
    Solution
    Given:
    $$P=12000$$
    $$R=9\%$$
    $$F=n=3\ year$$

    $$A=P \left (1+\dfrac {R}{100}\right)^n$$

    $$=12000 \left (1+\dfrac {9}{100}\right)^3$$

    $$=12000 \left (\dfrac {109}{100}\right)^3$$

    $$=12000\times \dfrac {109}{100}\times \dfrac {109}{100}\times \dfrac {109}{100}$$

    $$A=15540.348$$

    C.I.$$=A- P$$

    C.I.$$=15540.348-12000$$

    C.I$$=$$ Rs. $$3540.348$$

    Hence option (B) is the correct option
  • Question 2
    1 / -0
    The population of a town in $$2009$$ was $$125000$$. It increase $$10\%$$ per year. What is the population after $$3$$ years?
    Solution
    Taking $$P$$ as the population of the town in $$2009$$, $$n$$ is the number of years after which the population is to be calculated and $$R$$ is the rate at which the population increases, then:
    Population after $$3$$ years $$= P \left(1 + \dfrac{R}{100}\right)^{n} $$

    $$= 12500 \left(1 + \dfrac{10}{100}\right)^{3} $$

    $$= 12500 \left(1 + \dfrac{1}{10}\right)^{3} $$

    $$= 12500 \left( \dfrac{11}{10}\right)^{3} $$

    $$ = 12500 \times \dfrac{1331}{1000} $$

    $$ = 16,637.5$$ 
  • Question 3
    1 / -0
    If the cost of transporting $$150$$ kg of goods for $$120$$ km is Rs $$50.$$ what will be the cost of transporting $$180$$ kg of goods for $$300$$ km?
    Solution

    Cost of transporting $$150\ kg$$ over $$120\ km=50$$

    Cost of transporting $$1\ kg$$ over $$1\ km=\dfrac{50}{(150\times 130)}$$

    Cost of transporting $$180\ kg$$ over $$300\ km=\dfrac{50}{(150\times 130)}\times (180\times 300)=Rs.150$$

  • Question 4
    1 / -0
    If Rs $$2800$$ is $$\dfrac{2}{7}th$$ of the value of a house, the worth of the house (in Rs ) is?
    Solution
    let the worth of the house be $$x$$ 
    Given that $$\dfrac {2}{7}$$ of value of house is $$2800$$
    $$\dfrac {2}{7}x=2800$$
    $$x=2800\times \dfrac {7}{2}$$
    $$=1400\times 7$$
    $$Rs. 9800$$
    Worth of house $$=Rs. 9800$$

  • Question 5
    1 / -0
    The population of Mumbai increase $$5\%$$ per annum, its population was $$1000000$$ in $$2004$$. What was it's population after $$3$$ year?
    Solution
    Population after $$3$$ years $$= P\left(1 + \dfrac{r}{100}\right)^{n}$$

    $$= 1000000\left(1 + \dfrac{5}{100}\right)^{3}$$

    $$= 1000000\left(1 + \dfrac{1}{20}\right)^{3}$$

    $$= 1000000\left( \dfrac{21}{20}\right)^{3}$$

    $$= 1000000 \times \dfrac{9261}{8000} $$
    $$= 1157625$$

  • Question 6
    1 / -0
    Decrease $$300$$ by $$24$$%
    Solution
    $$24$$percent of $$300$$ is $$\dfrac{24}{100}\times 300=72$$
    Decrease 300 by $$24$$ percent$$=300-72=228$$
  • Question 7
    1 / -0
    Which of the following is the most beneficial customer? 
    Solution
    More the discount more it is beneficial Let $$x$$ be the cost price of article.
    a) $$2$$ successive discount $$\rightarrow \dfrac{20}{100}\times \left[\dfrac{20}{100}x\right]\rightarrow \left[\dfrac{x}{25}\right]$$
    b) $$40\%$$ at a time discount $$\rightarrow \dfrac{40}{100}x\rightarrow \left[\dfrac{2x}{5}\right]\rightarrow \left[\dfrac{10x}{25}\right]$$
    $$\therefore b \,discount >a\,discount$$

  • Question 8
    1 / -0
    At what prices should a shopkeeper mark a radio that costs him Rs 1200 in order that he may offer a discount of 20% on the marked price and still make a profit of 25%
    Solution
    Given: Cost price $$=$$ Rs. $$1200$$

    Profit $$\%=25$$

    $$\therefore $$ Selling price $$=125\%$$ of $$1200$$

                              $$=\dfrac{125}{100}\times 1200$$

                              $$=$$ Rs. $$1500$$

    Let the marked price be $$x$$

    and discount is of $$20\%$$          [Given]

    $$\Rightarrow 80\%$$ of $$x=1500$$

    $$\Rightarrow \dfrac{80}{100}\times x=1500$$

    $$\Rightarrow x=\dfrac{1500\times 100}{80}$$

    $$\Rightarrow x=1875$$

    Hence the marked price should be $$Rs.1875$$
  • Question 9
    1 / -0
    The owner of a toy shop charges his customer $$33$$% more then the cost price. If the customer paid Rs. $$4921$$ for a toy, then what was the cost price of the toy?
    Solution
    Let cost price of the toy $$=x$$ 

    Owner will change $$=x+ \dfrac{33}{100} \times x$$
    $$=494$$

    $$4921=\dfrac{133}{100}x$$

    $$x=\dfrac{4921 \times 100}{133}$$

    $$x=3700\ Rs$$
  • Question 10
    1 / -0
    A steno-typist deposits Rs. $$10,000$$ in a bank for a period of $$3$$ years at $$12$$% per annum compound interest. The interest accrued to her after maturity will be  
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