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Comparing Quantities Test - 41

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Comparing Quantities Test - 41
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  • Question 1
    1 / -0
    Four dealers advertise the same list price for a TV set. Which one of the following discount series is more advantageous to the customer?
  • Question 2
    1 / -0
    Which discount series is profitable to the buyer first -: 25%, 12%,3%, or second -: 18%, 1.7%,5%?
    Solution
    First discount series is better as the total discount which a customer will get is more.
    in first total discount will be=$$25\%$$+$$12\%$$+$$3\%$$=$$40\%$$
    in second discount will be=$$18\%$$+$$1.7\%$$+$$5\%$$=$$24.7\%$$
    so first series is better.
  • Question 3
    1 / -0
    A dealer earned a profit of $$5\%$$ by selling a radio for Rs.$$714$$, What is cost price of the radio?
    Solution
    Let the cost price of the radio be Rs.$$x$$.
    And, profit$$=5$$%
    Now, selling price $$=Rs.(x+5\%\ of\ x)=Rs.\cfrac{105}{100}x$$
    $$\therefore$$  $$\cfrac{105}{100}x=714$$
    $$\Rightarrow x=714\times \cfrac{100}{105}=680$$
    Hence, the cost price of the radio is Rs.$$680$$
  • Question 4
    1 / -0
    Mr. Rai bought a machine that was tested at Rs 320. He was given successive discounts of 20% and 10%, then the price he paid was
    Solution
    List price of the machine=Rs320
    after the first discount of 20% price will be$$=320-\frac { 20 }{ 100 } \times 320=256$$
    now a discount of 10% was given on Rs256
    price of the machine will be,$$=256-\frac { 10 }{ 100 } \times 256=230.40$$
    The price he paid was Rs230.40
  • Question 5
    1 / -0
    A man borrowed $$Rs,500$$ at a rate of $$6\%$$ per annum. At the end of 3$$\displaystyle \frac{1}{2}$$ years, he paid back $$Rs. 350$$ and gave his radio for the balance amount. The price of the radio is
    Solution
    Given principal $$P=Rs.500$$
    Rate of interest $$R=6\%$$
    Time $$T=3\dfrac{1}{2}$$ years$$=\dfrac{7}{2}$$ years

    Interest $$I=\dfrac{PTR}{100}$$

    $$\therefore \displaystyle I = \frac{\displaystyle 500 \times \frac{7}{2} \times 6}{100} = Rs. 105$$

    Amount $$= P + I$$
                  $$= Rs.(500 + 105)$$ 
                  $$= Rs. 605$$
    So,  price of radio $$=Rs.( 605 - 350) = Rs. 255$$

    Hence, the price of the radio is $$Rs.255$$
  • Question 6
    1 / -0
    A water filter is available for 500 down payment followed by two instalments of 600 each. If the total interest paid is 100, then what is the cash price of the water filter?
    Solution
    Total amount paid $$ = Rs  500 + 2(600) = Rs 1700 $$
    In this $$ Rs  100 $$ is interest , so actual price $$ = Rs 1700 - 100 = Rs 1600 $$
  • Question 7
    1 / -0
    $$X$$ sells an article to $$Y$$ at $$15\%$$ profit. $$Y$$ sells it to $$Z$$ at $$10\%$$ profit. What is $$X's$$ cost price, if $$Y$$ makes a profit of $$23?$$
    Solution
    Let the $$CP$$ of the article be $$Rs. 100$$.

    Then for $$X\ to\ Y$$:
     Given, $$ Profit = 15 $$ %
    $$\Rightarrow \dfrac {SP-CP}{CP} \times 100 = 15 $$

    $$\Rightarrow \dfrac { S_x-100}{100} \times 100 = 15 $$

    $$\Rightarrow S_X = Rs 115 $$

    Now, for $$Y\ to\ Z$$,
    $$ CP = Rs 115 $$
    Given, $$ Profit = 10 $$ %

    $$\Rightarrow \dfrac {SP-CP}{CP} \times 100 = 10 $$

    $$\Rightarrow \dfrac { S_Y- 115}{115} \times 100 = 10 $$

    $$\Rightarrow S_Y = Rs 126.5 $$

    So, profit for Y $$ = 126.5 - 115 = Rs 11.5 $$  when CP  of article for X $$ = Rs 100 $$

    But given profit $$ = Rs 23 $$

    So, actual CP $$ = Rs. \dfrac {23}{11.5} \times 100 = Rs. 200 $$

  • Question 8
    1 / -0
    A dealer purchases $$15$$ articles for $$\text{Rs }25$$ and sells $$12$$ articles for $$\text{Rs }30.$$ Find the profit percentage.
    Solution
    C.P. of $$15$$ articles is $$\text{Rs }25$$
    $$\therefore$$  C.P. of each article $$ = \dfrac {20}{15} = \dfrac {5}{3} $$
    S.P. of $$12$$ articles is $$\text{Rs }30$$
    $$\therefore$$  S.P. of each article $$ = \dfrac {30}{12} = \dfrac {5}{2} $$

    Now, Profit $$\% = \dfrac {S.P. - C.P.}{C.P.} \times 100 = \frac {\dfrac {5}{2}- \dfrac {5}{3}}{ \dfrac {5}{3}} \times 100 = 50 \%$$

  • Question 9
    1 / -0
    A trader marked a watch 40% above the cost price and then gave a discount of 10% He made a net profit of Rs 468 after paying a tax of 10% on the gross profit What is the cost price of the watch?
    Solution
    Let the cost price of watch $$ = $$Rs$$100$$
    Marker price $$=100+40=$$Rs$$ 140$$
    Then selling price =$$\dfrac{90}{100}\times 140=126$$ Rs
    So gain$$ =126-100=26$$ Rs
    Net gain$$= 26-10\%$$ of $$26=26\times 0.9=23.4$$ 
    Thus for the cost price of Rs $$100$$ net gain is Rs $$23.4$$
    According to the given condition,
     cost price =$$\dfrac{468}{23.4}\times 100=$$ Rs $$2000$$
  • Question 10
    1 / -0
    The compound interest on $$Rs. 10,000$$ in $$\displaystyle2\frac{1}{2}$$ years at $$4\%$$ per annum is
    Solution
    Given principal$$=Rs.10,000$$
    Time$$=2.5$$ years
    Rate of interest$$=4\%$$
    Compound interest for first year$$=\dfrac{PTR}{100}=Rs.\dfrac { 10000\times 1\times 4 }{ 100 } =Rs.400$$
    Amount at the end of first year$$=Rs.(10000+400)=Rs.10400$$

    Amount for first year $$=$$ Principal for second year$$=Rs.10400$$
    Compound interest for second year$$=\dfrac{PTR}{100}=Rs.\dfrac { 10400\times 1\times 4 }{ 100 } =Rs.416$$

    For next $$6$$ months,
    Principal $$=Rs.(10400+416)=Rs.10816$$
    Time$$=6\ months=\dfrac{1}{2}\ years$$
    Compound interest for next six months will be$$=\dfrac{PTR}{100}=Rs.\dfrac { 10816\times 1\times 4 }{ 100\times 2 } =Rs.216.32$$
    Total compound interest $$=Rs.(400+416+216.32)=Rs.1032.32$$

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