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Numerical Applications Test 14

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Numerical Applications Test 14
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Weekly Quiz Competition
  • Question 1
    1 / -0
    A garrison of '$$n$$' men had enough food to last for $$30$$ days. After $$10$$ days, $$50$$ more men joined them. If the food now lasted for $$16$$ days, what is the value of $$n$$?
    Solution
    After $$10$$ days, the food for n men is there for $$20$$ days.
    This food can be eaten by $$(n + 50)$$ men in $$16$$ days.
    $$\therefore 20n = 16(n + 50)$$
    $$\therefore n = 200$$
  • Question 2
    1 / -0
    A bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise coins in the ratio $$3 : 8 : 10$$. What is the number of $$50$$-paise coins?
    Solution
    Here bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise.
    Coin ratio is $$ 3 : 8 : 10$$.
    Value ratio $$= 3 \times 1: 8\times \dfrac {1}{2} : 10 \times \dfrac {1}{10}$$
    $$= 3 : 4 : 1$$
    Number of $$50$$ paise coins $$= \dfrac {4}{8} \times$$ Rs. $$112 =$$ Rs. $$56$$
    Therefore, number of $$50$$ paise coins $$= 2\times 56 = 112$$.
  • Question 3
    1 / -0
    If $$15$$ men can do a piece of work in $$20$$ days. In how many days can $$25$$ men finish the same work?
    Solution
    Using direct proportion method
    15 men do work in 20 days
    Therefore 25 men do work in $$15 \ast  \dfrac{20}{25} = 12$$
  • Question 4
    1 / -0
    Two pipes A and B can fill a tank in $$24$$ and $$30$$ minutes respectively. Both are turned on together. But at the end of $$8$$ minutes, the first is turned off. The time taken to fill the tank B is
    Solution
    Work done by A and B in $$8\ min$$.
    $$= 8\times \left (\dfrac {1}{24} + \dfrac {1}{30}\right )$$
    $$= 8\times \left (\dfrac {5 + 4}{120}\right ) = \dfrac {9}{15} = \dfrac {3}{5}$$.
    Remaining work $$1 - \dfrac {3}{5} = \dfrac {2}{5}th$$
    $$\dfrac {2}{5}$$ of the tank is filled by B in $$\dfrac {2}{5} \times 30 = 12\ min$$.
  • Question 5
    1 / -0
    If the wages of $$45$$ women amount to $$Rs. 15525$$ in $$48$$ days, how many men must work $$16$$ days to receive $$Rs. 5750$$, the daily wages of a man being double that of a woman?
    Solution
    45 women work for 48 days and get =  Rs.15525
    45 women work for 1 day and get =  Rs 15525/48
    1 woman's payment for 1 day =  15525 / (48 × 45)
    Daily wage of 1 woman  =  Rs. 115 / 16
    Now,
    Daily wage of 1 man= 2 × daily wages of 1 woman
    2×(115 / 16)= Rs 115 / 8 
    Required number of men = Rs 5750 × (8 / 115) / 16 = 25 men
    So option B is the correct answer.
  • Question 6
    1 / -0
    Two pipes A and B can fill a tank in $$4$$ hours and $$5$$ hours respectively. If they are turned up alternatively for one hour each, the time taken to fill the tank is
    Solution
    Tank filled in $$2\ hours = \dfrac {1}{4} + \dfrac {1}{5} = \dfrac {9}{20}$$
    Tank filled in $$4\ hours = 2\times \dfrac {9}{20} = \dfrac {9}{10}$$.
    Now $$\dfrac {1}{10}th$$ of the tank filled by A in $$\left (\dfrac {1}{10}\times 4\times 60\right )min$$. i.e., $$24$$ min. Total time $$= 4\ hrs. 24 min$$.
  • Question 7
    1 / -0
    A can do a piece of work in $$20$$ days B in $$15$$ days and C in $$12$$ days. How soon can the work be done if A is assisted by B on one day and by C on the next alternately?
    Solution
    $$(A + B)'s 1\ day$$ work $$= \left (\dfrac {1}{20} + \dfrac {1}{15}\right )$$ of the work, i.e., $$\dfrac {7}{60}$$
    $$(A + C)^{s} 1\ days$$ work $$= \left (\dfrac {1}{20} + \dfrac {1}{12}\right )$$, i.e. $$\dfrac {8}{60}$$ of the work.
    $$2$$ days work done $$= \dfrac {7}{60} + \dfrac {8}{60} = \dfrac {1}{4}$$.
    Since $$\dfrac {1}{4}$$ of work is done in $$2$$ days, so whole of the work is done in $$8$$ days.
  • Question 8
    1 / -0
    A man do a piece of work in $$40$$ days. He worked at it for $$5$$ days, then B finished it in $$21$$ days. The number of days that A and B will take together to finish the same work is
    Solution
    Work done by A and B in $$8\ min$$.
    $$= 8\times \left (\dfrac {1}{24} + \dfrac {1}{30}\right )$$
    $$= 8\times \left (\dfrac {5 + 4}{120}\right ) = \dfrac {9}{15} = \dfrac {3}{5}$$.
    Remaining work $$1 - \dfrac {3}{5} = \dfrac {2}{5}th$$
    $$\dfrac {2}{5}$$ of the tank is filled by B in $$\dfrac {2}{5} \times 30 = 12\ min$$.
  • Question 9
    1 / -0
    A square pyramid can contain $$16\ m^3$$ of water. The height of the pyramid is $$3\ m$$. Calculate the length of base of the square pyramid.
    Solution
    Volume of square pyramid  $$=16{ m }^{ 3 }$$
    $$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times h=16{ m }^{ 3 }\Rightarrow \dfrac { 1 }{ 3 } \times { a }^{ 2 }\times 3=16\Rightarrow a=\sqrt { 16 } =4cm$$
    $$\therefore $$  Length of base $$= 4m$$
  • Question 10
    1 / -0
    Which of the following relations is correct ?
    Solution
    We know that
    $$Speed =\dfrac{Distance}{Time}$$

    Hence, this is the answer.
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