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General Test - 13

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  • Question 1
    4 / -1

    Ajit, Anand, and Ayush enter into a partnership investing their Money in the ratio \(3:4:5\). What is the ratio of their time period of investments if Ayush receives Rs \(15000\) as a share from a profit of Rs \(51000\)? Given that Ajit invested twice the time Ayush invested.

    Solution

    Let the time period of their investments of Ajit, Anand, and Ayush be \(2x:y:x\) respectively.

    We know that,

    Profit \(=\)Capital \(\times\) Time

    Profit will be shared in the ratio \(6x:4y:5x\)

    Given that,

    \((\frac{5 x}{(11 x+4 y)})\times51000=15000\)

    \(\Rightarrow \frac{5 x}{(11 x+4 y)}=\frac{5}{17}\)

    \(\Rightarrow 85 x=55 x+20 y\)

    \(\Rightarrow 30 x=20 y\)

    \(\Rightarrow 3 x=2 y\)

    Let \(x=2 k\) and \(y=3 k\)

    \(\therefore\) Required ratio \(=2 x: y: x=4 k: 3 k: 2 k=4: 3: 2\)

  • Question 2
    4 / -1

    The compound interest on Rs. 8000 @ 10% per annum for 1.5 years, if the interest is calculated half-yearly is:

    Solution

    Given,

    Principal \((P)=\) Rs. \(8000\)

    Rate \((R)=10 \%\)

    Interest is calculated half-yearly. So, \(R=\frac{10}{2}=5 \%\)

    Time \(=1.5\) years \(=3\) half years. So, \(n=3\)

    As we know,

    Amount \((A)= P × (1 + \frac{R}{100})^{n}\)

    \(=8000 \times\left(1+\frac{5}{100}\right)^{3}\)

    \(=8000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\)

    \(\therefore\) Amount \(=\) Rs. \(9261\)

    Compound interest = Amount - Principal

    \(\therefore\) Compound interest \(=9261-8000=\) Rs. \(1261\)

  • Question 3
    4 / -1

    If a certain sum of money becomes 3 times of itself in 5 years at compound interest, In how much time it will become 81 times of itself?

    Solution
    Let sum at the start be \(x\)
    After 5 years it will become \(x \times 3=3 x\)
    After 10 years it will become \(3 x \times 3=9 x\)
    After 15 years it will become \(9 x \times 3=27 x\)
    After 20 years it will become \(27 x \times 3=81 x\)
    that is 81 times the initial amount of \({x}\).
    \(\therefore\) It will take 20 years to become 81 times
  • Question 4
    4 / -1

    The cost price of 50 Coca-Cola Bottles is equal to the selling price of 40 Coca-Cola Bottles. Find the Profit/Loss percentage.

    Solution

  • Question 5
    4 / -1

    Suresh loses 15% when he sells his computer for Rs. 17,000. If he sells it for Rs. 20,448, then his loss/profit percentage is:

    Solution

    Selling price = Rs.17000;Loss % = 15%

    Selling price = cost price ×\(\frac{(100 - loss \%)}{100}\)

    Cost price \(=17000 \times \frac{100 }{ 85}=\) Rs. 20000

    \(\therefore\) Profit percentage \(=\frac{(20448-20000)} { 20000} \times 100\) \(=2.24 \%\)

  • Question 6
    4 / -1

    Gaurav earns Rs. \(800\) per day. After some weeks, he earns Rs. \(960\) per day. What is the percentage increase in his daily earnings?

    Solution

    Increase in his earnings \(=\) Rs. \((960-800)\)= Rs. \(160\)

    Now,Percentage increase in his daily earnings \(=\) Rs. \((\frac{160}{800} \times 100)\)\(= 20\%\)

    ∴ The required percentage increase in his earnings is \(20\%\)

  • Question 7
    4 / -1

    The population of a village decreases every year at the rate of 3 %. If the present population is 28227 , what was the population of the village two years ago?

    Solution

    Present Population, P = 28227, Rate = 3 %, Time (n) = 2 years

    \(P=P_0\left(1-\frac{R}{100}\right)^n\),

    \(P_0 = \frac{28227 \times 100 \times 100}{97 \times 97}\)\( = 30000\)

  • Question 8
    4 / -1

    Two pipes A and B can fill an empty tank in 8 hours and 12 hours respectively. They have opened alternately for 1 hour each, starting with pipe A first. In how many hours will the empty tank be filled?

    Solution

    Given:

    Pipe A alone can fill the tank = 8 hrs

    Pipe B alone can fill the tank = 12 hrs

    Time ratio of Pipe A and B = 8 ∶ 12 = 2 ∶ 3

    Concept used:

    Efficiency is inversely proportional to time.

    Calculation:

    Efficiency ratio of pipe A and B = 3 ∶ 2

    Total work = 3 × 8 = 24

    Both pipes opened alternately for 1 hr.

    Total work in 2 hrs = 3 + 2 = 5

    Work in 8 hrs = 5 × 4 = 20

    Work in next 1 hr (9th) = 20 + 3 = 23

    Remaining work = 24 - 23 = 1

    B alone can work in 1 hr = 2

    B alone can work in\(\frac{1}{2}\)hr = 1

    So, the total work complete in\(9 \frac{1}{2}\)hrs.

  • Question 9
    4 / -1

    What should come in place of question mark (?) in the following question.

    \(77 \div 7 \times 4+(\frac{1}{5})\) of \(15 \times 4=?\)

    Solution

    Given: \(77 \div 7 \times 4+(\frac{1}{5})\) of \(15 \times 4=?\)

    \(\Rightarrow 77 \div 7 \times 4+3 \times 4=?\) =44+12 =56

  • Question 10
    4 / -1

    Simplify the following expression.

    \(4 \div\left[2+2 \div\left\{2+2 \div\left(2+\frac{2}{3}\right)\right\}\right]\)

    Solution

    Given: \(4 \div\left[2+2 \div\left\{2+2 \div\left(2+\frac{2}{3}\right)\right\}\right]\)

    \( \Rightarrow 4 \div\left[2+2 \div\left\{2+2 \times \frac{3}{8}\right\}\right] \)

    \( \Rightarrow 4 \times \frac{22}{60}=\frac{22}{15}\)

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