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Comparing Quantities Test - 33

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Comparing Quantities Test - 33
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  • Question 1
    1 / -0
    A discount of $$15\%$$ on one article is the same as a discount of $$20\%$$ on another article. The cost of the two articles can be
    Solution
    Let the cost of the two articles be $$x$$ and $$y$$. 
    Then, $$15\%$$ of $$x=20\%$$ of $$y$$
    $$\Rightarrow \dfrac{15}{100}\times x=\dfrac{20}{100}\times y$$
    $$\Rightarrow \dfrac{x}{y}=\dfrac{20}{100}\times \dfrac{100}{15}$$
    $$\Rightarrow \dfrac {x}{y}=\dfrac {20}{15}=\dfrac {4}{3}$$
    $$\therefore$$ $$x$$ and $$y$$ should be in the ratio $$4: 3$$.
    Then the cost price $$=$$ Rs. $$80$$ and Rs. $$60$$.
  • Question 2
    1 / -0
    If SP of an article is $$\displaystyle\frac{4}{3}$$ of its CP then the profit in the transaction is 
    Solution
    Suppose CP of article = Rs. 100
    SP = $$100\,\times\, \displaystyle\frac{4}{3}\,=\, Rs. \frac{400}{3}$$
    Profit = SP - CP = $$\displaystyle\frac{400}{3}\, -\, 100\, =\, Rs. \frac{100}{3}$$

    Profit % = $$\displaystyle\frac{\frac{100}{3}}{100}\, \times\, 100\, =\, 33\frac{1}{3}$$ %
  • Question 3
    1 / -0
    What would be the printed price of a watch purchased at Rs. $$380$$, so that after giving $$5\%$$ discount, there is $$25\%$$ profit?
    Solution

    Given, C.P. $$=$$ Rs. $$ 380$$, Profit $$= 25\%$$
    $$\therefore$$ S.P. $$=$$ Rs. $$ \left (\dfrac {380\times 125}{100}\right )=$$ Rs. $$ 475$$
    Discount $$=$$ $$5\%$$
    $$\therefore$$ M.P. $$=$$ Rs. $$ \left (\dfrac {475\times 100}{95}\right )=$$ Rs. $$ 500$$

  • Question 4
    1 / -0
    A shopkeeper earns a profit of $$15$$% after selling a book at $$20$$% discount on the printed price. The ratio of the cost price and the printed price of the book is 
    Solution
    Let the printed price of the book be Rs. $$100$$ 
    After a discount of $$20\%$$, S.P. $$=$$ Rs. $$80 $$
    Profit earned $$= 15\%$$
    Thus, Cost Price of Book $$=$$ $$\displaystyle \frac{100}{115}\times 80$$ =$$ \displaystyle \frac{1600}{23}$$
    Hence, (C.P) : (printed price) $$=$$ $$\displaystyle \frac{1600}{23}:100$$
    $$=$$ $$16:23$$
  • Question 5
    1 / -0
    Successive discounts of $$12\dfrac {1}{2}$$% and $$7\dfrac {1}{2}$$% are given on the marked price of a cupboard. If the customer pays Rs $$2590$$, then what is the marked price?
    Solution

    Let the M.P. of the cupboard be Rs $$x$$. 

    Then, $$(100-12.5)\%$$ of $$(100-7.5)\%$$ of Rs. $$x=2590$$

    $$\Rightarrow \dfrac {87.5}{100}\times \dfrac {92.5}{100}\times x=2590$$
    $$\Rightarrow x=\dfrac {2590\times 100\times 100}{87.5\times 92.5}=$$ Rs. $$3200$$

  • Question 6
    1 / -0
    A shopkeeper fixes the MP of an item $$35$$% above its CP.  The percentage of discount allowed to gain$$ 8$$% is:
    Solution
    Let $$C.P.$$ of article = Rs. $$100 $$
    $$M.P.\,=\, 100\, \times\, \displaystyle\frac{135}{100}\,=\, Rs. 135$$
    $$SP$$ required = $$\displaystyle100\times\frac{108}{100}\,=\, Rs. 108$$
    On Rs. $$135$$ discount allowed
    = Rs. $$135 -$$ Rs. $$108 =$$ Rs. $$27$$
    Hence on Rs. $$100$$ discount allowed is
    $$\displaystyle\frac{27}{135}\, \times\, 100\,=\, 20\%$$
  • Question 7
    1 / -0
    An article when sold at a gain of $$5$$% yield Rs. $$15$$ more than when sold at a loss of $$5$$ $$\%$$. Its cost price would be:
    Solution
    Suppose the CP of the article $$=$$ Rs. $$100$$
    I case :
    Profit $$= 5\%$$
    SP $$=$$ $$100 + 5$$ $$=$$ Rs. $$105$$
    II case :
    Loss $$= 5\%$$
    SP $$=$$ CP $$-$$ loss $$=$$ $$100 - 5$$ $$=$$ Rs. $$95$$
    Difference between two SP's
    $$=$$ $$105 - 95 $$ $$=$$ Rs. $$10$$
    If the difference $$=$$ Rs. $$10$$, then CP $$=$$ Rs. $$100$$
    If the difference $$=$$ Rs. $$15$$, then CP $$= \frac{100}{10}\times 15=$$ Rs. $$150$$
  • Question 8
    1 / -0
    If $$5$$% more is gained by selling an article for Rs. $$350$$ than by selling it for Rs. $$340$$, the cost of the article is:
    Solution

    Let the C.P is =Rs. x

    When S.P is Rs.340 then gain %$$=\dfrac{340-x}{x}\times 100$$

    When S.P is rs.350 then gain %$$=\dfrac{350-x}{x}\times 100$$

    According to the question

    $$\Rightarrow \left[\dfrac{350-x}{x}\times 100 \right]-\left[\dfrac{340-x}{x}\times 100 \right]=5$$

    $$\Rightarrow \dfrac{100}{x}[350-x-340+x]=5$$

    $$\Rightarrow \dfrac{100}{x}[10]=5$$

    $$\Rightarrow x=\dfrac{1000}{5}=200$$

    Hence C.P of article is Rs.200.

  • Question 9
    1 / -0
    If the discount is $$10$$%, an item bought for Rs.$$9$$ is priced at:
    Solution

    let the M.P $$=$$ Rs. $$x$$

    Then $$S.P=x-10\%$$ of $$x$$

    $$\Rightarrow x-\dfrac{10x}{100}$$

    $$\Rightarrow x-\dfrac{x}{10}$$

    $$\Rightarrow \dfrac{9x}{10}$$
    Then $$\dfrac{9x}{10}=9$$

    $$\Rightarrow 9x=10\times 9$$

    $$\Rightarrow x=10$$

  • Question 10
    1 / -0
    By selling an article at $$\left (\dfrac {2}{5}\right)^{th}$$ of the marked price, there is a loss of $$25\%$$. The ratio of the marked price and the cost price of the article is
    Solution

    Let M.P. $$=$$ Rs. $$100$$

    Then S.P. $$=\dfrac{2}{5}  $$ of M.P. $$=\dfrac{2}{5}\times 100=$$ Rs. $$40$$

    Loss $$=25\%$$

    Then, C.P. $$=\dfrac{100}{100- \text{loss} \%}\times$$ S.P.

    $$\Rightarrow \dfrac{100}{100-25}\times 40$$

    $$\Rightarrow \dfrac{100}{75}\times 40=$$ Rs. $$\dfrac{160}{3}$$

    M.P. : C.P. $$=100:\dfrac{160}{3}=300 : 160$$

    $$=15:8$$

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