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Comparing Quantities Test - 63

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Comparing Quantities Test - 63
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  • Question 1
    1 / -0
    A shopkeeper purchased $$20$$ dozen bananas for $$\text{Rs }\,208$$, Four dozen of the bananas were rotten. At what price per dozen must he sell the remaining bananas so that he is neither at a loss nor at a profit?
    Solution
    Cost of $$20$$ dozen bananas $$=\text{Rs }208$$
    Remaining bananas $$=$$ Total bananas $$-$$ rotten bananas
    $$=20-4=16$$ dozens
    let the selling price of 1 dozen bananas be $$x$$
    as per the question, $$16$$ dozen bananas are required to be sold at $$\text{Rs }208$$
    $$\implies 16x=208$$
    $$\implies x=13$$
    Hence, the remaining bananas are to be sold at $$\text{Rs 13}$$ per dozen 
  • Question 2
    1 / -0
    The compound interest on $$Rs.5,000$$ at $$4\%$$ per annum for $$2$$ years compounded annually will be
    Solution
    Given, $$P=5000$$
    $$R=4\%$$
    $$T=2\ years$$

    Amount $$A=P\Big(1+\dfrac{R}{100}\Big)^T$$

    $$A=5000\Big(1+\dfrac{4}{100}\Big)^2$$

        $$=5000\times\Big(\dfrac{104}{100}\Big)^2$$

        $$=Rs. 5408$$

    Compound interest $$=A-P$$

                                      $$=5408-5000=Rs. 408$$
  • Question 3
    1 / -0
    $$10$$ cylindrical pillars of building have to be painted.If the diameter of each pillar is $$50$$m and the height is $$4$$m.Find the cost of painting at the rate of Rs.$$7$$ per square metre.
    Solution
    Diameter of the pillar$$=50$$m
    Radius$$=25$$m
    Height$$=4$$m
    $$\therefore$$ Curved surface area of a pillar$$=2\pi rh=2\times\dfrac{22}{7}\times 25\times 4$$ 
    Curved surface area of $$10$$ pillars$$=2\times\dfrac{22}{7}\times 25\times 4\times 10$$ 
    Cost of painting$$=$$Rs.$$7$$per sq.m
    $$\therefore$$ Total cost of painting $$10$$ pillars$$=2\times\dfrac{22}{7}\times 25\times 4\times 10\times 7$$
    $$=$$Rs.$$44,000$$

  • Question 4
    1 / -0
    The marked  price of a book is Rs. $$100$$. The shopkeeper gave $$25\%$$ discount on it. Then the sale price of the book is
    Solution

  • Question 5
    1 / -0
    To gain $$25\% $$ after allowing a discount of $$10\% ,$$ the shopkeeper must mark the price of the article which costs him $$Rs 360$$ as :
    Solution
    The correct answer is option (a).
    Suppose, the marked price of the article $$=Rs. x$$.
    Cost price of the article $$=Rs. 360$$
    As per question,
    $$x-x\times \dfrac{10}{100}-\dfrac{25\times 360}{100}=360$$
    or $$x-\dfrac{x}{10}-90=360$$
    or $$\dfrac{9x}{10}-90=360$$
    or $$\dfrac{9x}{10}=360+90$$
    So, $$x=Rs. 500$$
  • Question 6
    1 / -0
    If a quarter kg of potato costs $$60$$ paise, how many paise does $$200$$ gm cost?
    Solution
    Given,

    Let the Required cost be $$x$$ paise.  

    Therefore, $$250 : 200 :: 60 : x$$

    $$\Rightarrow  (250 \times  x) = (200 \times 60)$$

    $$\Rightarrow x= 48$$
  • Question 7
    1 / -0
    $$120\%$$ of $$45$$
    Solution
    Given,

    $$120\%$$ of $$45$$

    $$=\dfrac{120}{100} \times 45$$

    $$=\dfrac{12}{2} \times 9$$

    $$=6 \times 9$$

    $$=54$$
  • Question 8
    1 / -0
    The population of a particular area of a city is $$5000$$. It increases by $$10 \%$$ in $$1^{st}$$ year. It decreases by $$20 \%$$ in the $$2^{nd}$$ year because of some reason. In the $$3^{rd}$$ year, the population increases by $$30 \%$$. What will be the population of area at the end of $$3$$ years?
    Solution
    Population of a city at the starting of first year $$=5000$$
    It increases by $$10\%$$ in $$1^{st}$$ year.
    i.e., Population at the end of $$1^{st}$$ year $$=$$ Population at the starting of $$2^{nd}$$ year 
    $$=5000+10\%\space of\space5000\\=5000+\dfrac{10}{100}\times5000\\=5500$$
    It decreases by $$20\%$$ in $$2^{nd}$$ year.
    i.e., Population at the end of $$2^{nd}$$ year $$=$$ Population at the starting of $$3^{rd}$$ year 
    $$=5500-20\%\space of\space5500\\=5500-\dfrac{20}{100}\times5500\\=5500-1100\\=4400$$
    It increases by $$30\%$$ in $$3^{rd}$$ year.
    i.e., Population at the end of $$3^{rd}$$ year
     $$=4400+30\%\space of\space4400\\=4400+\dfrac{30}{100}\times4400\\=4400+1320\\=5720$$

    So, $$\text{C}$$ is the correct option.
  • Question 9
    1 / -0
    On selling an article for Rs. 42, there is a percent profit of 5%. If the same article is sold for Rs. 50. What is the percent profit?
    Solution

  • Question 10
    1 / -0
    The cost price of a car is $$Rs. \,4,00,000$$. If its price decreases by $$10 \%$$ every year, then when will be the cost of car after $$3$$ years?
    Solution
    Cost price of car $$(P)=\mathrm{Rs.}\space4,00,000$$
    Rate of depreciation $$(R)=10\%$$
    Time $$(T)=3\space\text{years}$$
    So, Cost of car after $$3\space\text{years}=P(1-\frac{R}{100})^T$$
    $$=4,00,000(1-\frac{10}{100})^3$$
    $$=4,00,000(1-0.1)^3$$
    $$=4,00,000(0.9)^3$$
    $$=4,00,000\times0.729$$
    $$=\mathrm{Rs.}\space2,91,600$$
    So, $$\text{C}$$ is the correct option.
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