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Comparing Quantities Test - 22

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Comparing Quantities Test - 22
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  • Question 1
    1 / -0
    Mark again the correct answer in each of the following:
    $$(3/4)$$ as rate percent is
    Solution
    Because,
    $$=(3/4)\times 100$$
    $$=(300/4)$$
    $$=75\%$$
  • Question 2
    1 / -0
    The per cent that represents the unshaded region in the figure.

    Solution
    By counting , 
    Unshaded parts $$= 40$$
             Total parts $$= 100$$
    Unshaded fraction $$=\dfrac{\text {unshaded parts}}{\text{total parts}}= \dfrac{40}{100} = 40\%$$
  • Question 3
    1 / -0
    A man loses $$12\frac{1}{2} $$ % of his money and after spending 70% of the remainder, he was left with Rs. 210. The money that he had at the beginning is
    Solution
    Let Rs. $$x$$ be the amount at the beginning.
    Loss $$ = 12.5\%$$
    Remainder $$ = \cfrac{87.5x}{100}$$

    $$210 = \cfrac{30}{100} \times \dfrac{87.5x}{100}$$

    $$\therefore x=800$$
  • Question 4
    1 / -0
    A person purchases $$20$$ litres of juice at $$Rs. 2.20$$ per litre  and diluted it with water to  make the contents $$22$$ litres.  In order to earn $$10\%$$ profit he should sell the juice at
    Solution
    Volume of juice purchased $$= 20$$ litres
    Rate at which juice purchased $$= Rs.2.20$$ per litre

    $$\Rightarrow$$ Cost price of juice $$= 20 \times 2.2 = Rs..44$$
    In order to gain $$10\%$$ profit, 
    Selling price $$= \dfrac{110}{100} \times 44 = 48.4$$

    Final volume of juice after dilution $$= 22$$ litres
    $$\therefore$$ Rate of selling $$= \dfrac{48.4}{22} = Rs. 2.20$$
  • Question 5
    1 / -0
    How many $$\text{kg}$$ of sugar costing $$\text{Rs.}\ 5.75$$ per $$\text{kg}$$ should be mixed with $$75\ \text{kg}$$ of cheaper sugar costing $$\text{Rs.}\ 4.50$$ per $$\text{kg}$$ so that the mixture is worth $$\text{Rs.}\ 5.50$$ per $$\text{kg}$$ ?
    Solution
    Let say $$x\ \text{kg}$$ sugar costing $$\text{Rs.}\ 5.75$$ per $$\text{kg}$$ is added.

    Total cost $$= \text{Rs.}\ (5.75x + 75) \times 4.50$$
                     $$=\text{Rs.}\ (5.75x+337.5)$$
    Total sugar $$= (x + 75)\ \text{kg}$$ 

    Average cost of sugar mixture $$=\dfrac{\text{Total cost of sugar}}{\text{Total sugar}}$$

    $$5.5 = \dfrac{\left(5.75x + 337.5\right)}{\left(x + 75\right)}$$

    $$5.5x + 412.5 = 5.75x + 337.5$$

    $$75 = 0.25x$$

    $$x =\dfrac{75}{0.25}$$

    $$\therefore\,x = 300\ \text{kg}$$
  • Question 6
    1 / -0
    A shopkeeper professes to sell the goods at cost price, but he uses a weight of 900 grams for a Kilogram. Then his gain percent is
    Solution
    Shopkeeper uses weight of 900 grams for a kilogram or 1000 grams.
    $$\Rightarrow$$  $$\text{Profit} = 1000-900=100$$
    $$\text{Profit}\ \%=\dfrac{100}{900}\times 100=\dfrac{100}{9}\%$$
  • Question 7
    1 / -0
    A certain sum of money Q was deposited for 5 years and 4 months at 4.5%  simple interest and mounted to Rs. 248. Then the value of Q is
    Solution
    We know that,
    Amount = principle + simple interest
    $$A=P+\cfrac{PRT}{100}$$
    $$A=P[1+\cfrac{RT}{100}]$$            ...... (1)
    Where, $$A$$ is amount.
                $$P$$ is principle.
                $$R$$ is rate of interest.
                $$T$$ is time in year.

    Given, $$A=248$$
                $$P=Q$$
                $$R=4.5$$%
                $$T=5+\cfrac{4}{12}=\cfrac{16}{3}$$
    Substitute these values in equation (1) and find the value of $$Q$$.
    $$248=Q[1+\cfrac{4.5\times\frac{16}{3}}{100}]$$
    $$248=Q[1+\cfrac{24}{100}]$$
    $$248=1.24Q$$
    $$\therefore  Q=200$$
  • Question 8
    1 / -0
    A began business with Rs. 4,500 and was joined afterward by B with Rs. 5,400. If the profit at the end of the year was divided in the ratio of 2:1, then the time of joining B was after:
    Solution
    $$\Rightarrow$$   Let B remained in the business for $$x$$ months. Then,
    $$\Rightarrow$$   $$\dfrac{A}{B}=\dfrac{4500\times 12}{5400\times x}=\dfrac{54000}{5400x}=\dfrac{10}{x}$$

    $$\therefore$$  $$\dfrac{10}{x}=\dfrac{2}{1}$$

    $$\Rightarrow$$  $$2x=10$$
    $$\therefore$$      $$x=5$$
    $$\Rightarrow$$    Thus, B remained in the business for 5 months.
    $$\Rightarrow$$    So, B joined the business after $$7\, months.$$
  • Question 9
    1 / -0
    A VCR and TV were bought for Rs 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.
    Solution
    Cost Price (CP) of VCR is Rs $$8000$$

    Cost Price (CP) of TV is Rs $$8000$$

    Total CP$$=$$Rs $$16,000$$

    Loss on$$ VCR=CP\times Loss$$%

    $$=8000\times\dfrac{4}{100}$$

    $$=320$$

    Profit on $$TV=CP\times Profit$$%

    $$=8000\times\dfrac{8}{100}$$

    $$=640$$

    Total profit$$=640-320$$

                     $$=320$$

    Percentage profit $$=\dfrac{profit}{CP}\times100$$

                                  $$=\dfrac{320}{16000}\times100$$

                                  $$=2$$%
  • Question 10
    1 / -0
    Rakesh took a loan for 7 years at the rate of 6% p.a. S.I. If the total interest paid was Rs. 2100, the principal was
    Solution
    Given that,
    Rakesh took a loan for $$7$$ years at $$6\%$$ per annum simple interest.
    He paide $$Rs.\ 2100$$ interest on the amount.

    To find out,
    The principal amount.

    We know that, $$S.I.=\left(\dfrac{P\times R\times T}{100}\right)$$

    Here, $$S.I.=Rs.\ 2100,\ R=6\%$$ and $$T=7\ years$$

    Hence, $$2100=\left(\dfrac{P\times 6\times 7}{100}\right)$$

    $$\Rightarrow P=\left(\dfrac{2100\times 100}{6\times 7}\right)$$

    $$\Rightarrow P=\left(\dfrac{2100\times 100}{42}\right)$$

    $$\Rightarrow P=50\times 100$$

    $$\therefore\  P=Rs.\ 5000$$

    Hence, the principal amount was $$Rs.\ 5000$$.
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