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Comparing Quantities Test - 25

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Comparing Quantities Test - 25
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  • Question 1
    1 / -0
    An agent makes a profit of 20% even after a discount of 10% on the advertised price. If he makes a profit of Rs. 900 on the sale of a scooter, then the advertised price is
    Solution
    The advertising price is basically the marked price.
    Agent makes a profit of $$20\%$$ after giving a discount of $$10\%$$

    Marked Price $$\times$$ $$(1$$ $$-$$ Discount$$\%) = $$Cost Price $$\times (1 + $$Profit$$\%)$$

    $$\therefore MP\times \dfrac{90}{100} = CP \times \dfrac{120}{100} = SP$$ 

    Profit $$= SP - CP$$
    $$\therefore 900 = \dfrac{20}{100} \times CP$$
         $$CP = 4500$$

    $$\Rightarrow$$ Advertised Price $$= \dfrac{120}{90} \times 4500 =$$ Rs. $$6000$$

    Answer $$=$$ Rs $$6000$$
  • Question 2
    1 / -0
    A sum of money becomes Rs. 13380 after 3 years and Rs. 20070 after 6 years on compound interest. The sum is
    Solution
    Let the sum be Rs. $$x$$. Then               ...(i)
    $$x (1+\cfrac{R}{100})^3 = 13380$$
    and $$x (1+\cfrac{R}{100})^6 = 20070$$         ...(ii)
    Dividing equation (ii) by (i), we get
    $$(1+\cfrac{R}{100})^3 = (\cfrac{200070}{13380}) = (\cfrac{3}{2})$$
    $$\therefore x\cfrac{3}{2} = 13380$$
    $$\Rightarrow x = (13380\times \cfrac{2}{3}) = 8920$$
    Hence, the sum is Rs. 8920
  • Question 3
    1 / -0
    A shopkeeper sells a badminton racket marked at Rs. 30  at 15% discount and gives a  shuttle cock costing Rs. 1.50  free with each badminton racket. He then makes a profit of 20%.  His cost price per badminton racket is
    Solution
    $$\Rightarrow$$  Marked price = $$Rs.30$$
    $$\Rightarrow$$  Selling price = $$Rs.[(\dfrac{85}{100}\times 30)-1.50]$$

    $$\Rightarrow$$  Selling price = $$Rs.(25.50-1.50)=Rs.24$$.

    $$\Rightarrow$$  Let Cost price be $$Rs.x$$
    $$\Rightarrow$$  Then, $$120\%$$ of $$x=24$$

    $$\Rightarrow$$  $$x=\dfrac{24\times 100}{120}=Rs.20$$.
  • Question 4
    1 / -0
    A shopkeeper purchases 11 pens for Rs. 10 and sells them at the rate of Rs. 11, then the profit is
  • Question 5
    1 / -0
    If the cost of a pen is Rs. 12.60 and the gain was $$10\%$$ of the marked price, then the marked price of the pen was
    Solution
    Given cost price 12.60
    Profit $$= 10\%$$
    Selling price = Cost price + Profit
    Profit $$=12.60 \times \cfrac{10}{100}=1.26$$
    Selling price $$= 12.60+1.26=13.86$$, which is approximately 14. 
    So, the correct answer is option B.
  • Question 6
    1 / -0
    A tree increases annually by one-fourth of its height. What will be it's height after $$2$$ years, if it stands $$64$$ cm high today?
    Solution
    Initial height $$=64\ cm$$
    Every year it increases one-fourth of it's height.
    Increase $$\%= (\cfrac{1}{4}\times 100\%)=25\%$$

    Height after $$2$$ years $$= 64\times \left(1+\cfrac{25}{100}\right)^2\ cm$$
                                      $$= 64\times \left(1+\cfrac{1}{4}\right)^2\ cm$$
                                       $$= 64\times \left(\cfrac{5}{4}\right)^2\ cm$$
                                      $$= (64\times \cfrac{5}{4}\times \cfrac{5}{4})\ cm$$ 
                                      $$= 100\ cm$$ 

    Hence, the height of the tree after $$2$$ years will be $$100\ cm$$
  • Question 7
    1 / -0
    A shopkeeper mixes 80 kg sugar worth of Rs. 6.75 per kg with 120 kg of sugar worth of Rs. 8 per kg. He earns a profit of 20% by selling the mixture. He sells it at the rate.
    Solution
    Total cost $$=80\times 6.75+120\times 8=Rs. 1500$$
    Selling price $$=1500\times \dfrac {120}{100}=Rs. 1800$$
    S.P. per kg $$=\dfrac {1800}{200}=Rs.9$$
  • Question 8
    1 / -0
    If the population of a town is  64,000 and its annual increase is 10%, then its correct population at the end of 3 years will be
    Solution
    Population (P) $$= 64,000$$
    Annual increase rate $$(r) = 10\%$$
    Hence, population after $$n=3$$ years
    $$= P(1+\cfrac{r}{100})^n$$
    $$= 64,000(1+\cfrac{10}{100})^3$$
    $$= 64,000\times (\cfrac{11}{10})^3$$
    $$= 64,000\times \cfrac{11\times 11\times 11}{1000}$$
    $$= 85,184$$
  • Question 9
    1 / -0
    The current population of a town is 10,000. If the population increases by $$10\%$$ every year, then the  population of the town after three years will be
    Solution
    $$A=P(1+\cfrac{R}{100})^n$$
    $$A=10000(1+0.1)^3=13310$$
  • Question 10
    1 / -0
    If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 12000, the compound interest on the same sum for the same period at the same rate, is
    Solution
    Clearly, Rate = $$5\; \%$$, Time = $$3$$ years and S.I. = $$\text{Rs.}\;12000$$
    Sum $$= \text{Rs.} \left (\cfrac{100\times 12000}{3\times 5}\right ) = \text{Rs.}\; 80000$$
    Amount $$= \text{Rs. } \left [80000\times \left (1+\cfrac{5}{100}\right )^3\right ]$$
                  $$= \text{Rs. } 92610$$
    We know that C.I. = Amount - Sum
    So,
    C.I. $$ = 92610-80000$$
          $$= \text{Rs. } 12610$$
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