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Comparing Quantities Test - 47

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Comparing Quantities Test - 47
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  • Question 1
    1 / -0
    Arun buys an old machine for Rs. 2640 and spends Rs. 295 on its repairs. Find his gain or loss if he sells the machine for Rs. 3228. Also, express the gain or loss made as percent.
    Solution
    Total C.P. of a machine = $$Rs. 2640 + Rs. 295 = Rs.2935$$
    Given, S.P. = Rs. 3228; [which is less than total cost price]
    Therefore, gain = Rs. 3228 - Rs. 2935 = Rs. 293
    Also, gain% = $$\dfrac{gain}{total\space\ C.P.} \times 100$$%
    = $$\dfrac{293}{2935}\times 100 \approx 10$$%
  • Question 2
    1 / -0
    Spice prices of distributor A
    SpicePrice Per Kilogram
    Cinnamon$$ $7.00$$
    Nutmeg$$ $10.00$$
    Ginger$$ $5.00$$
    Cloves$$ $9.00$$
    If a customer buys $$2$$ kilograms each of cinnamon and ginger and then sells to another person, then what is his net profit if he earns a profit of $$ $3 $$ per $$100$$ grams? 
    Solution
    Given that a person buys 2 kg cinnamon and 2 kg ginger and sell both with profit of $ 3 per kg
    Than the total weight of cinnamon and ginger =2+2=4 kg =40000 gms
    He earn profit of $3 per 100 gms 
    Than he get profit =$$\dfrac{$3}{100}\times 40000=$120$$
  • Question 3
    1 / -0
    A hardware store sells lightbulbs in different quantities. The figure above shows the cost of various quantities. According to the figure, what is the cost of a single lightbulb?

    Solution
    From the graph we can see that cost of $$5$$ lightbulbs is $$ $9$$

    Hence, cost of $$1$$ lightbulb$$=\dfrac95=$1.80$$

  • Question 4
    1 / -0
    Martin is planning which crops to plant on his farm for the upcoming season. He has enough seed to plant $$4$$ acres of wheat and $$7$$ acres of soybeans, but the total area of farmland he owns is only $$9$$ acres. He earns $$ $90$$ per acre for every acre of wheat planted and $$ $120$$ for every acre of soybeans planted, and he must pay a $$10$$% tax on all money he earns from selling his crops. Calculate the maximum profit, in dollars that Martin can earn from planting wheat and soybeans this season.
    Solution
    In order to find the greatest profit, maximize the number of acres of soybeans Martin plants, since soybeans bring in more money per acre than does wheat. At most, MartIncan plant $$7$$ acres of soybeans. 
    Therefore, the most money he can make on soybeans is $$7\times 120=840$$.
    He then has $$9 - 7 = 2$$ acres left on which to plant wheat.
    The money he makes from this wheat is $$2\times 90=180$$.
    The total amount Martin makes before taxes is therefore $$840 + 180 = 1,020$$.
    The tax on this money equals $$1020\times 0.10=102$$.
    Subtract the amount Martin pays in taxes to get $$1,020 - 102 = 918$$ profit. 
    The correct answer is $$918$$.
  • Question 5
    1 / -0
    Ravi bought a motorcycle for Rs. 50000 and paid Rs. 2000 for its tyre. He sold it at a gain of 20%. What is Ravi's profit?
    Solution
    Total C.P. of a motorcycle = $$Rs.\ 50000 + Rs.\ 2000 = Rs.\ 52000$$
    Profit% = $$\dfrac{profit}{total\space\ C.P.} \times 100$$%
    Profit = $$\dfrac{profit \space\ percentage \times CP}{100}$$
    $$=\dfrac{20\times 52000}{100}= Rs. 10400$$
  • Question 6
    1 / -0
    A man bought a car for Rs. 75,000. He spent Rs. 9,000 on repairs and spent Rs. 1500 on other things. He sold the car for Rs. 1,20,000. Find his gain percent (approximately)
    Solution
    Total C.P. of a car = Rs. 75,000 + Rs. 9,000 + Rs. 1,500 = Rs.85,500
    Given, S.P. = Rs. 1,20,000; [which is less than total cost price]

    $$\therefore$$ Gain = Rs. 1,20,000 - Rs. 85,500 = Rs. 34,500

    Also, gain% = $$\dfrac{gain}{total\space\ C.P.} \times 100$$

    = $$\dfrac{34500}{85500}\times 100 = 40$$% (approximately)
  • Question 7
    1 / -0
    A television set was bought for Rs. $$10,500$$ and sold at Rs. $$9500$$. Find the loss.
    Solution
    Given, C.P. $$ =$$ Rs. $$ 10500$$, S.P. $$ =$$ Rs. $$ 9500$$
    We know that, loss $$=$$ C.P. $$-$$ S.P.
    $$= 10500 - 9500$$
    $$=$$ Rs. $$1000$$
    Thus, the loss is Rs. $$1000$$.
  • Question 8
    1 / -0
    An article when sold at Rs.$$25$$ yields a profit of $$25$$%. The article is sold at $$10$$% discount with a marked price of Rs.$$30$$. What would be the profit earned?
    Solution
    Let's assume Cost Price (CP) of article = $$x$$ Rs.

    Given that when SP $$= 25$$ Rs, then Profit Percentage $$= 25\%$$

    $$\Rightarrow SP=CP+CP\times\dfrac{Profit\ Percentage}{100}$$

    $$\Rightarrow 25=x+x\times\dfrac{25}{100}$$

    $$\Rightarrow 25=x+\dfrac{x}{4}=\dfrac{5x}{4}$$

    $$\therefore x=20\ Rs$$

    Now, article is sold at 10% discount at marked price of $$30$$ Rs.

    $$\therefore SP=MP-MP\times\dfrac{Discount\ Percentage}{100}=30-30\times\dfrac{10}{100}=27\ Rs$$

    $$Profit = SP-CP=27-20=7\ Rs$$

    $$Profit\ Percentage=\dfrac{Profit\times100}{CP}=\dfrac{7\times100}{20}=35\%$$

  • Question 9
    1 / -0
    If S.P. of an article is $$\dfrac {4}{3}$$ of its C.P. What is the profit percentage in the transaction?
    Solution
    Let's assume $$Cost\ Price = x$$

    Then, $$Selling\ Price = \dfrac{4}{3}x$$

    $$Profit = Selling\ Price-Cost\ Price=\dfrac{4}{3}x-x=\dfrac{x}{3}$$

    $$Profit\ Percentage=\dfrac{Profit\times100}{Cost\ Price}=\dfrac{\dfrac{x}{3}\times100}{x}=\dfrac{100}{3}=33\dfrac{1}{3}\%$$

  • Question 10
    1 / -0
    What is the fractional equivalent of 12.5%?
    Solution

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