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Comparing Quantities Test - 65

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Comparing Quantities Test - 65
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The present population of a city is 3,02,500. If the population increases by 10% every year, what was the population of the city two years ago?
    Solution
    Let the population before two years ago be $$x$$

    Given the population increases by $$10\%$$

    So increase in population after one year is $$\dfrac{10}{100}\times x=0.1 x$$

    So population after one year is $$x+0.1 x=1.1 x$$

    Increase in population after another one year is $$\dfrac{10}{100} \times 1.1 x=0.11 x$$

    So population at present is $$1.1 x+0.11 x=1.21 x$$

    But given population at present is $$3,02,500$$
    $$\implies 1.21 x=302500$$

    $$\implies x=\dfrac{302500}{1.21}=2,50,000$$

    Population of city two years ago is $$2,50,000$$
  • Question 2
    1 / -0
    The price of $$282$$ apples is Rs. $$1974$$. What will be the price of $$24$$ dozens of apples?
    Solution

  • Question 3
    1 / -0
    If 16 shirts were sold for 20 notes of Rs 100, double its notes of Rs 50 and 200 notes of Rs 20, then what is the selling price of a shirt ?
    Solution
    Total amount obtained for 16 shirts
    $$= 20 \times 100 + 40 \times 50 + 200 \times 20$$
    $$= 8000 \ Rs$$

    So, Selling Price of a shirt $$ = \dfrac{8000}{16} = 500 \ Rs$$

    Thus, D is the correct answer.
  • Question 4
    1 / -0
    What is the cost price of a pen if it is sold for Rs 60 gaining 20% profit?
  • Question 5
    1 / -0
    Ramesh bought two boxes for Rs 1300. He sold one box at a profit of 20 % and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.
    Solution

  • Question 6
    1 / -0
    XYZ Ltd. incurs a net loss of Rs. 5, 000 during are accounting years. Depreciation for the relevant year amount to Rs. 1000, preliminary expenses written off during the accounting years is Rs. 3, 000 and a loss of Rs. 4, 000 is due to sale of plant and mach ____________.
  • Question 7
    1 / -0
    The compound interest on $$₹\ 10000$$ at $$10 \%$$ per annum for $$3$$ years, compounded annually, is
    Solution

    Given:

    Present value $$= ₹\ 10000$$

    Interest rate $$= 10 \%$$ per annum

    Time $$=3$$ years

    To find the amount we have the formula,

    Amount $$(A) = P (1+(r/100))^n$$

    where $$P$$ is present value, $$r$$ is rate of interest, $$n$$ is time in years

    Now substituting the values in above formula we get,

    $$\therefore A = 10000 (1 +10/100)^3$$

    $$\Rightarrow A = 10000 (11/10)^3$$

    $$\Rightarrow A = 121 (10) (11)$$

    $$\Rightarrow A = 1331 (10)$$
    $$\Rightarrow A = ₹\ 13310$$

    ∴ Compound interest $$= A – P$$

    $$= 13310 – 10000= ₹\ 3310$$

  • Question 8
    1 / -0

    Directions For Questions

    Select the correct answer in each of the following:

    ...view full instructions

    What percent of  $$90$$  is  $$120$$? 
    Solution
    Let us consider a number x

    So, $$x\%$$ of $$90$$  is  $$120$$ 
    $$x\% \times 90 = 120$$ 
    $$x/100 \times 90 = 120$$ 
    $$90x = 120 \times 100$$ 
    $$x = 12000/90 = 133 (1/3)\%$$
  • Question 9
    1 / -0

    Directions For Questions

    Select the correct answer in each of the following:

    ...view full instructions

    What percent of $$10kg$$  is $$250g$$?
    Solution
    We need to convert kg into grams by multiplying by $$1000$$ we get,

    $$10kg= 10 \times 1000 = 10000g$$ 
    Now, let us consider a number x

    $$x\%$$ of  $$10000$$ is $$250$$ 
    $$x\% \times 10000 = 250$$
    $$100x = 250$$
     $$x = 250/100 = 2.5\% $$
  • Question 10
    1 / -0

    Directions For Questions

    Select the correct answer in each of the following:

    ...view full instructions

    $$40\,\%$$ of  ? = $$240$$ 
    Solution
    Let us consider a number x

    $$40\%$$  of  $$ x = 240$$ 
    $$40/100 \times x = 240$$
    $$x = 240 \times 100/40$$ 
    $$x = 600 $$
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