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Comparing Quantities Test - 29

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Comparing Quantities Test - 29
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  • Question 1
    1 / -0
    I bought two tables for Rs. $$5400$$. I sold one at a loss of $$5\%$$ and the other at a gain of $$7\%$$. In the whole transaction, I neither gained nor lost. Find the costs of the tables separately.
    Solution
    Let the C.P. of one table $$=$$ Rs. $$x$$ 
    Then, C.P. of the other table $$=$$ Rs. $$(5400 - x)$$
    S.P. of the first table at a loss of $$5\%$$ $$=$$ Rs. $$\dfrac {95}{100}\times x$$

    S.P. of the other table at a gain of $$7\%$$ $$=$$ Rs. $$\dfrac {107}{100}\times (5400-x)$$
    Since there is no gain no loss.
    Total S.P. $$=$$ Total C.P.
    $$\Rightarrow \dfrac {95x}{100}+\dfrac {107\times (5400-x)}{100}=5400$$

    $$\Rightarrow \dfrac {95x}{100}-\dfrac {107x}{100}+\dfrac {107\times 5400}{100}=5400$$

    $$\Rightarrow \dfrac {-12x}{100}=\dfrac {5400\times 100-5400\times 107}{100}$$

    $$\Rightarrow \dfrac{-12x}{100}=\dfrac{540000-577800}{100}$$

    $$\Rightarrow -12x=-37800$$

    $$\Rightarrow x=\dfrac{37800}{12}=$$ Rs. $$3150$$

    $$\therefore$$ Cost of the two tables are Rs. $$3150$$ and $$(5400-3150) =$$ Rs. $$2250$$.
  • Question 2
    1 / -0
    Arun borrowed a sum of money from Jayant at the rate of $$8\%$$ simple interest for the first four years, $$10\%$$ p.a. for the next $$6$$ years and $$12\%$$ p.a. beyond $$10$$ years. If he pays a total of Rs. $$12160$$ as interest only at the end of $$15$$ years, how much money did he borrow?
  • Question 3
    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 4
    1 / -0
    A person lent a certain sum of money at $$4\%$$ simple interest and in $$5$$ years, the interest amounted to Rs. $$520$$ less than the sum lent. The sum lent was
    Solution
    Let the sum lent be $$Rs. x$$

    The interest in $$5$$ years $$=\dfrac{x\times 4\times 5}{100}$$

    Given: $$\dfrac{x\times4\times5}{100}=x-520$$

    $$\Rightarrow \dfrac{x}{5}=x-520$$

    $$\Rightarrow x-\dfrac{x}{5}=520$$

    $$\Rightarrow \dfrac{4x}{5}=520$$

    $$\Rightarrow 4x=520\times 5$$

    $$\Rightarrow x=\dfrac{520\times 5}{4}$$

    $$\Rightarrow\;x=$$ Rs. $$650$$

    Therefore, the sum lent is Rs. $$650$$
  • Question 5
    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 6
    1 / -0
    The cost price of $$20$$ articles is the same as selling price of $$x$$ articles. If the profit is $$25$$%, then the value of $$x$$ is
    Solution
    $$\textbf{Step 1:Calculate CP of 20 articles and SP of x Articles}$$
                   $$\text{Let the C.P. of 1 article be}$$ $$Rs. 1$$. 

                   $$\therefore \text{C.P. of x articles be }$$$$ Rs. x$$,

                   $$\Rightarrow \text{S.P. of x articles=C.P. of 20 articles=}$$ $$Rs. 20$$

                   $$\text{Profit for x Article= SP of x Article - CP of x Article}$$

                   $$\therefore \text{Profit =Rs. (20 - x)}$$. 

                   $$\Rightarrow \text{Profit} \% $$$$= 25 \%$$

                   $$\Rightarrow \dfrac{SP-CP}{CP} \times100=25$$

                   $$\Rightarrow \dfrac {(20-x)}{x}\times 100=25$$

                   $$\Rightarrow 2000-100x=25x$$

                   $$\Rightarrow 125x=2000$$

                   $$\Rightarrow x=16$$

    $$\textbf{Therefore,Value of x is 16}$$
  • Question 7
    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 8
    1 / -0
    A man gains $$10\%$$ by selling a certain article for a certain price. If he sells it at double the price, then the profit made is
    Solution

    Let the C.P. $$=$$ Rs. $$100$$,

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

    $$= \dfrac {110}{100}\times  100$$
    $$=$$ Rs. $$110$$
    New S.P. $$=2\times 110=$$ Rs. $$220$$
    $$\therefore$$ Profit $$=$$ Rs. $$220-$$ Rs. $$100=$$ Rs. $$120$$
    Profit $$\%$$ $$=\dfrac {120}{100}\times 100=120\%$$

  • Question 9
    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 10
    1 / -0
    An article passing through two hands is sold at a profit of $$38\%$$ at the original cost price. If the dealer makes a profit of $$20\%$$, then the profit per cent made by the second is:
    Solution
    Let the initial cost price of the article be Rs. 100, Profit = 20%
    S.P. of the 1st dealer $$=Rs \frac {100\times 120}{100}=Rs. 120$$
    Let the second profit % $$= x$$. Then,
    S.P. of the 2nd dealer $$=Rs \frac {120\times (100+x)}{100}$$
    Final S.P. of the article after passing through two hands at a profit= 38%
    $$=\frac {100\times (100+38)}{100}=Rs. 138$$
    Given, $$\frac {120\times (100+x)}{100}=138$$
    $$\Rightarrow 12000 + 120x = 13800$$
    $$\Rightarrow 120x = 1800 \Rightarrow x = 15$$%.
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