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Comparing Quantities Test - 65

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Comparing Quantities Test - 65
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  • Question 1
    1 / -0
    The maximum retail price (MRP) of a product is $$55\%$$ above its manufacturing cost. The product is sold through a retailer, who earns $$23\%$$ profit on his purchase price. What is the profit percentage for the manufacturer who sells his product to the retailer? The retailer gives $$10\%$$ discount on MRP.
    Solution
    The manufacturer sells the product to retailer and then retailer sells to the buyer.

    Assume manufacturing cost $$=100$$ and manufacturing profit $$=x\%$$.

    As maximum retail price (MRP) of a product is $$55\%$$ above its manufacturing cost.

    MRP $$=100(1+0.55)=155$$

    The retailer gives $$10\%$$ discount on MRP. So, buyer price is $$90\%$$ of MRP.

    Buyer price $$=155\times 0.9$$

    Manufacturer makes $$x\%$$ profit and then retailer makes $$23\%$$ profit.

    So, $$100(1+x)\times (1+0.23)=155\times 0.9$$

    $$100+100x=113.41$$

    Therefore, $$x=0.134$$

    Hence, manufacturer profit $$=13.4\%$$ 
  • Question 2
    1 / -0
    Two solutions A and B are mixed in the ratio 2 : 3 by volume in litre. The C.P. of solution A is Rs. 350 per litre and that of B is Rs. 400 per litre. If one-fifth of solution is sold at Rs. 460 per litre and remaining of it at Rs. 550 per litre, find the profit percent on the whole.
    Solution
    Given that,
    Two solutions $$A$$ and $$B$$ are mixed in the ratio $$2 : 3$$ by volume in litre.
    The $$C.P.$$ of solution $$A$$ is $$Rs.\  350$$ per litre and that of $$B$$ is $$Rs.\  400$$ per litre.
    One-fifth of the mixed solution is sold at $$Rs.\ 460$$ per litre and remaining at $$Rs.\ 550$$ per litre.

    To find out,
    The profit percentage on the whole mixture.

    Let the quantity of solution $$A$$ be $$2x$$ litres.
    Hence, the quantity of solution $$B$$ will be $$3x$$ litres.

    The cost of one litre of solution $$A = Rs.\ 350  $$
    Hence, the cost of $$2x$$ litres of $$A=2x\times 350$$
    $$=Rs.\ 700x$$

    Also, the cost of one litre of solution $$B = Rs.\ 400  $$
    Hence, the cost of $$3x$$ litres of $$B=3x\times 400$$
    $$=Rs.\ 1200x$$

    Now, the total solution will be $$2x+3x=5x$$ litres.
    And, the cost of $$5x$$ litres of the solution $$=Rs.\ 700x+1200x$$
    $$=Rs.\ 1900x$$

    One-fifth of the solution is $$x$$ litres and the remaining is $$4x$$ litres.
    Now, one-fifth of the solution is sold at $$Rs.\ 460$$ per litre.
    Hence, selling price of $$x$$ litres of solution $$=x\times 460$$
    $$=Rs.\ 460x$$

    Also, the remaining solution is sold at $$Rs.\ 550$$ per litre.
    Hence, the selling price of $$4x$$ litres of solution $$=4x\times 550$$
    $$=Rs.\ 2200x$$

    So, the selling price of the total $$5x$$ litres of solution $$=Rs. 460x+2200x$$
    $$=Rs.\ 2660x$$

    So, for complete $$5x$$ litres of solution, we have: 
    $$C.P.=Rs.\ 1900x$$ and $$S.P.=Rs.\ 2660x$$

    We know that, $$profit\%=\dfrac{S.P.-C.P.}{C.P.}\times 100$$

    Hence, $$profit\%=\dfrac{2660x-1900x}{1900x}\times 100$$

    $$\Rightarrow profit\%=\dfrac{760x}{1900x}\times 100$$

    $$\Rightarrow profit\%=\dfrac{760}{19}$$

    $$=40$$

    Hence, the profit percent on the whole is $$40\%$$.
  • Question 3
    1 / -0
    A shopkeeper makes a profit of $$20\%$$ on selling a transistor radio for $$\text{Rs } 840$$. For how much should he sell another transistor radio, whose cost price is $$\text{Rs }25$$ more than the first one, to make the same percentage profit ?
    Solution

    Given, for first transistor
    S.P $$=$$ Rs. $$840 $$ 
    Profit $$ = 20 $$ $$\%$$
    Now, with this S.P. and for Profit $$\%$$ $$ = \dfrac {\text {S.P.} - \text {C.P.}}{\text {C.P.}} \times 100 = 20 \%$$ 
    $$ \Rightarrow  \dfrac {840 - \text {C.P.}}{\text {C.P.}} \times 100 = 20 $$
    $$ \Rightarrow  840 - \text {C.P.} = 0.20 \text {C.P.} $$
    $$\Rightarrow  1.20 \text {C.P.} = 840 $$
    $$ \Rightarrow \text { C.P. }=$$ Rs. $$  700 $$
    And for another transistor
    $$ \text {C.P} =$$ Rs. $$ 700 +$$ Rs. $$25 =$$ Rs. $$ 725 $$ 
    Profit $$ = 20 $$ $$\%$$
    So, Profit $$\%$$ $$ = \dfrac {\text {S.P. }- \text {C.P.}}{\text {C.P.}} \times 100 =20 $$
    $$ = \dfrac {\text {S.P. }- 725.}{725} \times 100 =20 $$
    $$ \Rightarrow \text { S.P.} - 725 =145 $$
    $$ \Rightarrow \text { S.P.} =$$ Rs. $$ 870 $$ 

  • Question 4
    1 / -0
    Interest rate per conversion period is calculated by 
    Solution

  • Question 5
    1 / -0
    If the selling price of $$16$$ watches is equivalent to the cost price of $$20$$ watches then the gain per cent is
    Solution

  • Question 6
    1 / -0
    $$A$$ sells an article to $$B$$ for Rs. $$45000$$ losing $$10\%$$ in the transaction. $$B$$ sells it to $$C$$ at a price which would have given a profit of $$10\%$$ to $$A$$. By what percent does $$B$$ gain ?
    Solution
    Let cost price for $$A$$ is Rs. $$x$$
    Given that Selling price for $$A$$ $$=$$ Rs. $$45000$$ and Loss percentage $$=10\%$$
    We know that $$\text{Selling price}=\text{Cost price}-\text{loss}$$
    $$\Rightarrow 45000=x-10\% \ \text{of} \ x = x-\dfrac{x}{10}$$
    $$\Rightarrow 45000=\dfrac{9x}{10}$$
    $$\Rightarrow  x=$$ Rs. $$ 50000$$
    Now, cost price for $$B =$$ Selling price for $$A$$ $$= $$ Rs. $$45000$$
    $$B$$ gets $$10\%$$ profit on $$A$$'s cost price.
    Therefore, $$ \text{Selling price for B}=\text{Cost price for A}+\text{Profit on A's cost price}$$
    $$\Rightarrow$$ SP for $$B=50000+10\%\ \text{of} \ 50000=55000$$
    So, for $$B$$:
    CP $$= 45000 $$, SP $$= 55000$$
    Hence, $$\text{Profit} =$$ SP $$-$$ CP $$=55000-45000=$$ Rs. $$10000$$
    Therefore, $$ \text{Profit percentage} = \dfrac{\text{Profit}\times 100}{CP}=\dfrac{10000\times 100}{45000}=\dfrac{1000}{45}=22.22\%=22\% (\text{Approx})$$
  • Question 7
    1 / -0
    Ajay has purchased a second hand car by taking a loan of Rs$$5000$$ from bank at a rate of interest of $$8\%$$ for $$4$$ years. Calculate the total amount he will be paying to bank a loan of Rs$$5000$$ at the end of $$4$$ years
  • Question 8
    1 / -0
    Rate of interest is decided by
  • Question 9
    1 / -0
    A person invested part of Rs.$$45000$$ at $$4$$% and the rest at $$6$$%. If his annual income from both are equal, then what is the average cost of interest?
  • Question 10
    1 / -0
    A manufacturer makes 1200 articles at Rs 2.50 each. He marks each of them at such a price that if only 800 articles are sold, he would make a profit of 40% of his outlay. But, however, he is able to sell 840 articles at their marked price and the remaining at 60% of their marked price. Calculate manufacturer's total profit as per cent (in the integer value).
    Solution
    Total cost price (CP) $$=Number\ of\ articles\times\ Cost\ of\ each\ article=1200\times 2.50=3000\ Rs$$

    Let marked price of articles $$=x\ Rs$$

    So, Selling price of 800 articles (SP) $$=800x\ Rs$$

    Manufacturer gets 40% profit after selling 800 articles.

    We know that, $$SP=CP+Profit$$

    $$\Rightarrow 800x=3000+40\%\ of\ 3000$$

    $$\Rightarrow 800x=4200$$

    $$\therefore x=\dfrac{42}{8}=5.25$$

    Now, Since Manufacturer solds $$840$$ articles at marked price and remaining $$(1200-840=360)$$ at $$60\%$$ of marked price.

    So, In this case, Total Selling price (SP') $$=(840\times 5.25)+(360\times (60\%\ of\ 5.25))$$

    $$\Rightarrow SP' = 4410+1134=5544\ Rs$$

    So, $$Profit=SP'-CP=5544-3000=2544\ Rs$$

    $$\therefore Profit\ percentage=\dfrac{profit\ times 100}{CP}=\dfrac{2544\times 100}{3000}=84.8=85\%\ (Nearest Integer)$$
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