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Comparing Quantities Test - 31

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Comparing Quantities Test - 31
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  • Question 1
    1 / -0
    A cycle agent buys $$30$$ bicycles, of which 8 are first grade and the rest are second grade, for Rs. $$3150$$. Find at what price he must sell the first grade bicycles so that if he sells the second grade bicycles at three quarters of the price, he may make a profit of $$40\%$$ on his out lay?
    Solution
    Let the S.P.of one $$A$$ grade cycle $$=$$ Rs. $$n$$
    Then, the S.P. of one $$B$$ grade cycle $$=$$ Rs. $$\dfrac {3n}{4}$$
    C.P. of $$30$$ cycles $$=$$ Rs. $$3150$$, profit $$= 40\%$$.
    $$\therefore$$ S.P. of $$30$$ cycles $$=\dfrac {140}{100}\times  3150=$$ Rs. $$3150\times 1.4$$
    According to the question, we have
    $$8n+22\times \dfrac {3n}{4}=3150\times 1.4$$
    $$\Rightarrow 32n+66n=17640$$
    $$\Rightarrow 98n=17640$$
    $$\Rightarrow n=180$$
    Therefore, he should sell the first grade bicycle at Rs. $$180$$ each.
  • Question 2
    1 / -0
    Mohan bought $$20$$ dining tables for Rs. $$12000$$ and sold them at a profit equal to the selling price of $$4$$ dining tables. The selling price of each dining table is
    Solution
    Profit $$=$$ (S.P. of $$20$$ tables) $$-$$ (C.P. of $$20$$ tables)
    $$\Rightarrow$$ S.P. of $$4$$ tables $$=$$ S.P. of $$20$$ tables $$-$$ C.P. of $$20$$ tab1es
    $$\Rightarrow$$ S.P. of $$16$$ tables $$=$$ C.P. of $$20$$ tables $$=$$ Rs. $$ 12000$$
    $$\Rightarrow$$ S.P. of $$1$$ dining table $$=$$ Rs. $$ \dfrac {12000}{16}=$$ Rs. $$ 750$$
  • Question 3
    1 / -0
    Sanjay sold a bicycle to Salman at $$46\%$$ profit. Salman spent Rs. $$40$$ on repairs and sold it to Sunil for Rs. $$1500$$. In this deal, Salman made neither profit nor loss. What is the cost price for Sanjay?
    Solution

    Let the C.P. for Sanjay be Rs. $$x$$
    Then, S.P. for Sanjay $$= x + 46\%$$ of $$x$$
    $$=x+\dfrac {46}{100}x=\dfrac {146x}{100}$$
    $$\therefore$$ C.P. of Salman $$=\dfrac {146x}{100}+40$$
    $$\because$$ Salman made neither profit nor loss.
    $$\dfrac {146x}{100}+40=1500$$
    $$\Rightarrow \dfrac {146x}{100}=1460$$

    $$\Rightarrow x=\dfrac {1460\times 100}{146}=$$ Rs. $$1000$$.

  • Question 4
    1 / -0
    A shopkeeper sells an article at a loss of $$12\dfrac {1}{2}\%$$, Had he sold it for Rs. $$51.80$$ more than he would have earned a profit of $$6\%$$. The cost price of the article is
    Solution
    Let the C.P. of the article be Rs. $$x$$
    According to the question, we have
    S.P. at $$6\%$$ profit $$-$$ S.P. at $$12\dfrac {1}{2}\%$$ loss $$=$$ Rs. $$51.80$$
    $$\Rightarrow \dfrac {x\times (100+6)}{100}-\dfrac {x\times (100-12.5)}{100}=51.80$$
    $$\Rightarrow \dfrac {106x}{100}-\dfrac {87.5x}{100}=51.80$$
    $$\Rightarrow \dfrac {18.5x}{100}=51.80 \Rightarrow x=\dfrac {51.80\times 100}{18.5}=280$$
  • Question 5
    1 / -0
    A man bought two goats for Rs. $$1008$$. He sold one at a loss of $$20\%$$ and the other a profit of $$44\%$$. If each goat was sold for the same price, the cost price of the goat which was sold at a loss was
    Solution

    Let the C.P. of a goat be Rs. $$x$$ 

    Then, C.P. of the other goat $$=$$ Rs. $$(1008 - x)$$
    $$\because$$ S.P. of the both the goats is the same,
    Therefore, $$x\times \dfrac {(100-20)}{100}=(1008-x)\times \dfrac {144}{100}$$
    $$\Rightarrow 80x=1008\times 144-144x$$
    $$\Rightarrow 224x=1008\times 144$$
    $$\Rightarrow x=$$ Rs. $$ \dfrac {1008\times 144}{224}=$$ Rs. $$648$$

  • Question 6
    1 / -0
    A dealer buys an old cooler listed at Rs. $$950$$ and gets a discount of $$10\%$$. He spends Rs. $$45$$ for its repair. If he sells the cooler at a profit of $$25\%$$, then the selling price of the cooler is
    Solution

    C.P. $$=$$ Rs. $$950 - 10\%$$ of Rs. $$950 $$

    $$ \Rightarrow 950-\dfrac{10}{100}\times 950$$

    $$\Rightarrow 950-95=$$ Rs. $$ 855$$
    Repairs cost $$=$$ Rs. $$ 45$$
    $$\therefore$$ Net C.P. $$=$$ Rs. $$855+$$ Rs. $$45=$$ Rs. $$900, \text{Profit}=25\%$$

    S.P. $$=\dfrac{100+\text{profit}\%}{100}\times \text{C.P.}$$

    $$\therefore$$ S.P. $$=$$ Rs. $$\left (\dfrac {900\times 125}{100}\right )=$$ Rs. $$1125$$

  • Question 7
    1 / -0
    By selling a transistor for Rs. $$572$$, a shopkeeper earns a profit equivalent to $$30\%$$ of the cost price of the transistor. What is the cost price of the transistor?
    Solution

    Let the C.P. of the transistor be Rs. $$x$$
    Profit $$=30\%$$ of Rs. $$x$$ $$=$$ Rs. $$\dfrac {3x}{10}$$
    S.P. $$=$$ Rs. $$ 572$$

    C.P. $$+$$ Profit $$=$$ S.P.
    Therefore, $$ x+\dfrac {3x}{10}=572$$

    $$\Rightarrow \dfrac {13x}{10}=572$$
    $$\Rightarrow x=$$ Rs. $$ \dfrac {572\times 10}{13}=$$ Rs. $$ 440$$

  • Question 8
    1 / -0
    A tradesman sold an article at a loss of $$20\%$$. If the selling price had been increased by Rs. $$100$$, there  would have been a gain of $$5\%$$. The cost price of the article was
    Solution

    Let the C.P. of the article be Rs. $$x$$

    Then, S.P. of the article at a loss of $$20\%$$
    $$=\dfrac {80}{100}\times x=$$ Rs. $$ \dfrac {80x}{100}$$
    S.P. of the article at a profit of $$5\%$$
    $$=\dfrac {105}{100}\times x=$$ Rs. $$ \dfrac {105x}{100}$$
    According to the question, we have
    $$\dfrac {80x}{100}+100=\dfrac {105x}{100}$$

    $$\Rightarrow \dfrac {105x}{100}-\dfrac {80x}{100}=100$$
    $$\Rightarrow x=\dfrac {100\times 100}{25}=$$ Rs. $$400$$

  • Question 9
    1 / -0
    A person sells a table at a profit of $$20\%$$. If he had bought it at $$10\%$$ less cost and sold for Rs. $$105$$ more, he would have gained $$35\%$$. The cost price of the table is
    Solution
    Let the C.P. of the table be Rs. x.
    Then, S.P. of the table at 20% profit$$=\dfrac {120}{100}\times Rs. x=Rs \dfrac {6x}{5}$$
    C.P. of the table at 10% loss $$=\dfrac {90}{100}\times Rs. x$$$$=Rs \dfrac {9x}{10}$$
    S.P. of the table at 35% profit now$$=\dfrac {135}{100}\times Rs \dfrac {9x}{10}$$
    Then according to the question
     $$\dfrac {6x}{5}+105=\dfrac {135}{100}\times \dfrac {9x}{10}$$

    $$\Rightarrow \dfrac {6x}{5}+105=\dfrac {243}{200}x$$

    $$\Rightarrow \dfrac {243}{200}x-\dfrac {6}{5}x=105$$

    $$\Rightarrow \left (\dfrac {243-240}{200}\right )x=105$$

    $$\Rightarrow x=\dfrac {105\times 200}{3}=Rs. 7000$$.
  • Question 10
    1 / -0
    A man bought two old scooters for Rs. $$18000$$. By selling one at a profit of $$25\%$$ and the other at a loss of $$20\%$$, he neither gains nor loses. Find the cost price of each scooter.
    Solution
    Let the C.P. of the $$1^{st}$$ scooter $$=$$ Rs. $$ x$$. 
    Then, C.P. of the $$2^{nd}$$ scooter $$=$$ Rs. $$ (18000 - x)$$
    S.P. of $$1^{st}$$ scooter $$=$$ Rs. $$ x\times \dfrac {125}{100}=$$ Rs. $$ \dfrac {5x}{4}$$
    S.P. of $$2^{nd}$$ scooter $$=$$ Rs. $$ (18000-x)\times \dfrac {80}{100}$$ $$=$$ Rs. $$ \dfrac {4}{5}(18000-x)$$
    Since Mohan neither gains nor loses, his total S.P. $$=$$ Total C.P.
    $$\Rightarrow \dfrac {5x}{4}+\dfrac {4}{5}(18000-x)=18000$$
    $$\Rightarrow \dfrac {5x}{4}-\dfrac {4x}{5}=18000-14400=3600$$
    $$\Rightarrow \dfrac {25x-16x}{20}=3600\Rightarrow 9x=3600\times 20$$
    $$\Rightarrow x=\dfrac {3600\times 20}{9}=8000$$
    $$\therefore$$ C.P. of $$1^{st}$$ scooter $$=$$ Rs. $$ 8000$$
    C.P. of $$2^{nd}$$ scooter $$=$$ Rs. $$ 10,000$$.
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