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Comparing Quantities Test - 40

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Comparing Quantities Test - 40
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  • Question 1
    1 / -0
    If the cost of $$7$$ kg of rice is Rs. $$168$$, what is the cost of $$105$$ kg of rice?
    Solution

    Cost of $$1$$ kg of rice $$=\dfrac{168}{7}$$
                                 $$=Rs\ 24$$

    Thus cost of $$105$$kg rice is given as $$Rs\ (24\times 105)$$ that is equal to $$Rs\ 2520$$
  • Question 2
    1 / -0
    A shopkeeper makes a profit of $$20$$ $$\%$$ even after giving a discount of $$10$$ $$\%$$ on the advertised price of a T. V. set. If he makes a profit of Rs. $$450$$, the advertised price (in rupees) is
    Solution
    Let the advertised price be Rs. $$100$$
    $$\therefore$$ S.P.of the T.V. set
    $$=100-10=$$ Rs. $$90$$
    Since he gains $$20\%$$
    $$\therefore$$ C.P.of the T.V. set
    $$\displaystyle=\frac{100\times90}{120}=$$ Rs. $$75$$
    $$\therefore$$ Profit $$=90-75=$$ Rs. $$15$$
    If profit is Rs. $$15$$, then advertised Price $$=$$ Rs. $$100$$
    $$\therefore$$ Profit is Rs. $$450$$ advertised Price
    $$\displaystyle=\frac{100}{15}\times450=$$ Rs. $$3000$$
  • Question 3
    1 / -0
    A dealer sold three-fourth of the articles at a gain of $$20\%$$ and the remaining at CP. Find the gain earned by him in the whole transaction.
    Solution
    Let the CP of whole be x.

    CP of $$\displaystyle\frac{3}{4}th=\frac{3x}{4}$$

    CP of $$\displaystyle\frac{1}{4}th=\frac{x}{4}$$

    Total $$\displaystyle SP=\left(120\%\:of\:\frac{3x}{4}+\frac{x}{4}\right)=\left(\frac{120}{100}\times\frac{3x}{4}+\frac{x}{4}\right)=\frac{23x}{20}$$

    Gain $$\displaystyle=SP-CP=\frac{23x}{20}-x=\frac{23x-20x}{20}=\frac{3x}{20}$$

    Gain% $$\displaystyle=\left(\frac{3x}{20}\times\frac{1}{x}\times100\right)=15\%$$
  • Question 4
    1 / -0
    If the list price of a book is reduced by Rs 5, a person can buy 5 more books for Rs 300. Then the original cost of the book is
    Solution
    Let the original list price of the book be=Rs x
    Original number of books purchased=$$\frac { 300 }{ x } $$
    When the price of the book is reduced by 5
    New Number of books purchased will be=$$\frac { 300 }{ x-5 } $$
    New number of books purchased-Original number of books purchased=5
    $$\therefore \frac { 300 }{ x-5 } -\frac { 300 }{ x } =5$$
    $$\Rightarrow \frac { 300x-300x+1500 }{ x\left( x-5 \right)  } =5$$
    $$\Rightarrow { x }^{ 2 }-5x=300$$
     $$\Rightarrow { x }^{ 2 }-20x+15x-300=0$$
     $$\Rightarrow { x }\left( x-20 \right) +15\left( x-20 \right) =0$$
     $$\Rightarrow \left( x-20 \right) \left( x+15 \right) =0$$
     $$\Rightarrow x-20=0\quad or\quad x+15=0$$
    $$\Rightarrow x=20\quad or\quad x=-15$$
     $$\therefore x=20$$
    the original cost of book is Rs 20
  • Question 5
    1 / -0
    Successive discounts of $$\displaystyle12\frac{1}{2}\%$$ and $$\displaystyle7\frac{1}{2}\%$$ are given on the marked price of the table. If the customer pays Rs 2590, then what is the marked price of the table?
    Solution
    b'Marked price $$=$$ Rs 100. Price after 1st discount $$\displaystyle=Rs\:87\frac{1}{2}$$

    Price after 2nd discount $$\displaystyle=Rs\:92\frac{1}{2}\%$$ of $$\displaystyle87\frac{1}{2}$$

    But customer pays Rs 2590.

    $$\displaystyle\therefore M.P.=(2590\times100)\div\left(92\frac{1}{2}\%\:of\:87\frac{1}{2}\right)=Rs\:3,200$$.'
  • Question 6
    1 / -0
    The best source of revenue for the State Government is
    Solution
    Among the following, Sales Tax is the best source of income for the State Government.
    State government tend to obtain the largest portion of tax revenues from the sales and gross receipts tax.
  • Question 7
    1 / -0
    By selling a suitcase for Rs. 570, a man loses 5%. At what price must he sell it to gain 5%?
    Solution
    Selling Price=Rs.570
    Loss =5%
    $$\therefore C.P =\frac{100}{100-loss\%}\times S.P$$
    $$\Rightarrow \frac{100}{95}\times 570=Rs.600$$
    Now C.P of the suitcase=600
    Gain%=5%
    $$\therefore S.P=\frac{100+gain \%}{100} \times C.P$$
    $$=\frac{105}{100}\times 600=Rs.630$$
    Hence he sell Rs.630 to gain 5%.

  • Question 8
    1 / -0
    Find the compound interest on Rs 1,200 for 2 years at 10% per annum
    Solution
    Compound Interest $$C.I=A-P$$
    Formula for amount when compound interest is applied : $$A=P\left (1+\dfrac {r}{100}\right )^n$$
    Given $$P=1,200; n=2; r=10$$
    $$\therefore A=1,200\left (1+\dfrac {10}{100}\right )^2=1,200\left (\dfrac {11}{10}\right )^2=\dfrac {1,200\times 121}{100}=Rs\  1,452$$
    $$CI=1,452-1,200=Rs\  252$$
  • Question 9
    1 / -0
    A person purchased 10 dozen pens at the rate of Rs 4 per dozen. On checking, he found that 10 pens were not working. In order to earn 25% he should sell the remaining pens, each at
  • Question 10
    1 / -0
    Sita's salary was reduced by 10%. In order to reach her salary back to the original amount it must be raised by
    Solution

    Given that, sita's salary was reduced by 1010 percent,

    we have to find order to reach her salary back to the original amount it must be raised.


    Let, Sita’s salary $$=100$$

    Reduced 1010, then remaining salary $$=100-10$$ of $$100$$

                                                               $$=100-\dfrac{100-10}{100}$$

                                                               $$=Rs. 90$$

    Order to reach her salary back to the original amount it must be raised by $$10$$

         $$=\dfrac{10\times 100}{90}$$

         $$=\dfrac{100}{9}$$

         $$=11\dfrac {1}{9}\%$$

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