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Comparing Quantities Test - 55

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Comparing Quantities Test - 55
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  • Question 1
    1 / -0
    A loss of 2020% is incurred when 66 articles are sold for a rupee. To gain 2020% how many articles should be sold for a rupee?
    Solution
    Given than, SP=1,LossSP = 1, Loss% =20= 20%
    We know that, SP=0.8×CPSP = 0.8\times CP
    CP=54CP = \dfrac {5}{4}
    To gain 2020%, SP=1.2×CP=120100×54=32SP=32SP = 1.2\times CP = \dfrac {120}{100}\times \dfrac {5}{4} = \dfrac {3}{2} \Rightarrow SP = \dfrac {3}{2}
    For Rs.3/2Rs. 3/2 number of articles sold =6= 6
    For Rs.1Rs. 1 number of articles to be sold =6×23=4= 6\times \dfrac {2}{3} = 4 articles.
  • Question 2
    1 / -0
    A, B, C enter into partnership. A invests some money at the beginning, B invests double the amount after 66 months and C invests thrice the amount after 88 months. If the annual profit is Rs27,00027,000. C's share is (in Rs)
    Solution
    Let the amount A invest is Rsxx.
    Time period of investment of A is 1212 months.
    Amount B invest is Rs22x.
    Time period of investment of B is 66 months.
    Amount C investment is Rs33x.
    Time period of investment of C is 44 months.
    Profit is divided in the ratio = Amount of investment \times Time period of investment
    PA:PB:PC=x×12:2x×6:3x×4P_A : P_B : P_C = x \times 12 : 2x \times 6 : 3x \times 4
    =12x:12x:12x12 x : 12 x : 12 x
    =1:1:11 : 1 : 1
    C's share of profit = 13×27000=9000\dfrac{1}{3} \times 27000 = 9000
  • Question 3
    1 / -0
    A house and store were sold for Rs. 12,000\text{Rs. } 12,000 each. The house was sold at a loss of 20%20\% of the cost and the store at a gain of 20%20\% of the cost. The entire transaction resulted in
    Solution
    12,000=H15H H=15,00012,000 = H - \dfrac {1}{5}H  \therefore H = 15,000
    12,000=S+15S S=10,00012,000 = S + \dfrac {1}{5}S  \therefore S = 10,000
    H+S=25,000 H + S = 25,000.
    Therefore, the sale resulted in a loss of Rs.1000Rs. 1000.
  • Question 4
    1 / -0
    Select the ratio that is equivalent to 12:1412:14
    Solution
    Given ratio is 12:1412:14
    It can be rewritten as 1214=6×27×2\dfrac{12}{14}=\dfrac{6\times 2}{7\times 2}
    Cancelling 22 from numerator and denominator, we get
    1214=67\dfrac{12}{14}=\dfrac{6}{7} i.e 6:76:7
    Hence, the answer is 6:76:7.
  • Question 5
    1 / -0
    A shop keeper earns a profit of 1212% on selling a book at 1010% discount on the printed price. The ratio of cost price to the printed price of the book is
    Solution
    Using the formula
    100±(Profit Loss)100Discount=M.PC.P\dfrac {100\pm (Profit\ Loss)}{100 - Discount} = \dfrac {M.P}{C.P}
    M.PC.P=11210010=11290=5645\dfrac {M.P}{C.P} = \dfrac {112}{100 - 10} = \dfrac {112}{90} = \dfrac {56}{45}
    Hence, C.PM.P=4556\dfrac {C.P}{M.P} = \dfrac {45}{56}.
  • Question 6
    1 / -0
    Mr,A owns a home worth Rs.10,00010,000. He sells it to Mr.B at a 10%10\% profit based on the worth of the house. Mr.B sells the house back to Mr.A at a 10%10\% loss. Then:
    Solution
    $$\text{C.P. of B = 110% of Rs. 10000 = Rs. 11000}$$
    $$\text{Loss of B = 10%}$$
    $$\text{S.P. of B = 90% of Rs. 11000 = Rs. 9900}$$
    Thus, C.P. of A = Rs. 9900\text{Thus, C.P. of A = Rs. 9900}
    $$\text{So, A gets [(10% of Rs. 10000) + (10000 - 9900)] = Rs. 1100}$$
  • Question 7
    1 / -0
    The cost CC of sending a parcel post package weighing PP gms, PP an integer is Rs.10Rs. 10 for the first grams and Rs.3Rs. 3 for each additional grams. The formula for the cost is
    Solution
    For P1P - 1 gms the charge is 33 rupees per grams. For the first gram the charge is 1010 rupees. C=10+3(P1)\therefore C = 10 + 3(P - 1); or
    The charge for each gram is 33 rupees with an additional charge of 77 rupees for the first gram. C=3P+7=3P3+10=10+3(P1)\therefore C = 3P + 7 = 3P - 3 + 10 = 10 + 3(P - 1).
  • Question 8
    1 / -0
    55% more is gained by selling a watch for Rs.350Rs. 350 than by selling it for Rs.340Rs. 340. The cost price of the watch is
    Solution
    Let cost price be Rs.xRs. x
    Then 55% of x=(350340)x = (350 - 340)
    55% of x=10x = 10
    x=10×1005=200x = 10\times \dfrac {100}{5} = 200
    Hence, cost price is Rs.200Rs. 200.
  • Question 9
    1 / -0
    Tanishka deposited Rs. 50,00050,000 in a bank for two years with the interest rate of 5.5%5.5\% p.a. How much interest would she earn?
    Solution
    Interest=50000×2×5.5100=5500 Interest = \cfrac{ 50000 \times 2 \times 5.5 }{100} = 5500
  • Question 10
    1 / -0
    A sum of money invested at 14%\% per annum amounted to Rs. 10400 in 5 years. Calculate the principal.
    Solution
    We have the formula for amount in simple interest as:

    A=P(1+rt)A=P(1+rt)

    Given, A=10400A=10400, t=5t=5, r=14%r=14\%

     10400=P(1+5(0.14))\therefore \ 10400=P(1+5(0.14))

    10400=P(1+0.7)=P(1.7)10400=P(1+0.7)=P(1.7)

    P=104001.7P=\dfrac{10400}{1.7}

    =6117.647.6=6117.647.6

    Hence, the principal amount is Rs. 6117.656117.65.
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