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Numerical Applications Test 27

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Numerical Applications Test 27
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  • Question 1
    1 / -0
    $$12$$ children take $$16$$ days to complete a work which can be completed by $$8$$ adults in $$12$$ days. $$16$$ adults started working and after $$3$$ days, $$10$$ adults left and $$4$$ children joined them. How many days will they take to complete the remaining work ?
    Solution
    $$12$$ children take $$16$$ days to complete a work
    $$8$$ adults take $$12$$ days to complete the same job.
    Let the productivity of children be $$c$$ and the productivity of adult be $$m$$
    Thus,
    $$12c\times 16=8m \times 12$$
    $$=>c=\displaystyle \frac{1}{2}m$$

    Work done by $$8$$ men in $$12$$ days $$8 m \times 12$$ $$=96m$$
    Work done by $$16$$ men in $$3$$ days $$=16 m \times 3$$$$=48m$$
    Work left $$=96m - 48 m =48m$$
    After $$3$$ days $$10$$ adults left and $$4$$ children joined them
    Thus, $$6$$ men and $$4$$ children have to complete the remaining job.
    Let the time taken by them be $$t$$ days. 
    Thus
     $$(6m+4c)t=48m$$
    $$=>t=\displaystyle \frac{48m}{6m+4\times \frac{1}{2}m}$$

    $$=>t=\displaystyle \frac{48m}{6m+2m}$$

    $$=>t=6$$

  • Question 2
    1 / -0
    Pipe A fills a tank of $$700$$ litres capacity at the rate of $$40$$ litres a minute. Another pipe B fills the same tank at the rate of $$30$$ litres a minute. A pipe at the bottom of the tank drains the tank at the rate of $$20$$ litres a minute. If pipe A is kept open for a minute and then closed and pipe B is kept open for a minute and then closed and then pipe C is kept open for a minute and then closed and the cycle repeated, how long will it take for the empty tank to overflow?
    Solution
    In 3 Minutes Volume Supplied 
    $$\Rightarrow 40L+30L-20L=50L$$

    So, In 1 Minute Volume Supplied will be 
    $$\Rightarrow \dfrac{50L}{3}$$

    Time Taken to Fill The tank Completely

    $$\Rightarrow \dfrac{700L}{\dfrac{50L}{3}}=42mins$$

    Therefore Answer is $$42mins$$
  • Question 3
    1 / -0
    Two pipes A and B can fill a tank in $$36$$ hours and $$45$$ hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
    Solution
    Part filled by $$A$$ in $$1$$ hour $$=\dfrac{1}{36}$$
    Part filled by $$B$$ in $$1$$ hour $$=\dfrac1{45}$$;
    Part filled by $$(A + B)$$ in $$1$$ hour $$=\dfrac1{36}+\dfrac1{45}=\dfrac9{180}=\dfrac1{20}$$
    Hence, both the pipes together will fill the tank in $$20$$ hours.
  • Question 4
    1 / -0
    Two pipes A and B can fill a cistern in $$\displaystyle 37\frac{1}{2}$$ min and $$45$$ min respectively. Both pipes were opened and the cistern was filled in just  $$\displaystyle \frac{1}{2}$$ an hour if the pipe B is turned off after what time it is opened? 
    Solution
    Let B be turned off after x minutes. Then,
    Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.
    $$\therefore x\left(\dfrac{2}{75}+\dfrac{1}{45} \right)+(30-x)\dfrac{2}{75}=1$$
                       $$\Rightarrow \dfrac{11x}{225}+\dfrac{60-2x}{75}=1$$
                       $$\Rightarrow 11x+180-6x=225$$
                        $$\Rightarrow 5x=225-180$$
                        $$\Rightarrow 5x=45$$
                       $$\Rightarrow x=9$$
  • Question 5
    1 / -0
    A cistern can be filled by a tap in $$4$$ hours while it can be emptied by another tap in $$9$$ hours. If both the taps are opened simultaneously then after how much time will the cistern get filled ?
    Solution
    A cistern can be filled by a tap in $$4$$ hours while it can be emptied by another tap in $$9$$ hours.
    Let tap $$A$$ fill the cistern and tap $$B$$ empty the cistern.
    Work done by tap $$A$$ $$ =\dfrac{1}{4}$$

    Work done by tap $$B$$ $$ =\dfrac{1}{9}$$
    Work done to fill the tap when both are opened $$=\dfrac{1}{4}-\dfrac{1}{9}$$
                                                                                    
                                                                                     $$=\dfrac{9-4}{36}$$

                                                                                     $$=\dfrac{5}{36}$$

    Thus, time taken to completely fill the cistern $$=\dfrac{36}{5}$$ $$=7.2$$ hrs
  • Question 6
    1 / -0

    $$A$$ works twice as fast as $$B.$$ If $$B$$ can complete a work in $$12$$ days independently, the number of days in which $$A$$ and $$B$$ can together finish the work in 

    Solution
    Part of work done by $$B$$ in $$1$$ day $$=\dfrac1{12}$$
    Since, $$A$$ works twice as $$B$$
    So, Part of work done by $$A$$ in $$1$$ day $$=\dfrac2{12}=\dfrac16$$
    $$\therefore $$  Part of work done by $$A+B$$ in $$1$$ day $$=\dfrac16+\dfrac1{12}=\dfrac14$$
    So, $$A$$ and $$B$$ together can finish the work in $$4$$ days.
    Hence, A is the correct option.
  • Question 7
    1 / -0
    A large tanker can be filled by two pipes $$A$$ and $$B$$ in $$60$$ min and $$40$$ min. respectively. How many minutes will it take to fill the tanker from empty state if $$B$$ is used for half the time and $$A$$ and $$B$$ fill it together for the other half ?
    Solution

    Part fill by (A+B) in 1 minute$$=\dfrac{1}{60}+\dfrac{1}{40}=\dfrac{5}{120}=\dfrac{1}{24}$$

    Let the tank is filled in x minute then

    $$\dfrac{x}{2}\times \left(\dfrac{1}{24}+\dfrac{1}{40}  \right)=1$$

    $$\dfrac{x}{2}\times \dfrac{1}{15}=1$$

    $$\dfrac{x}{30}=1$$

    $$x=30  \  min$$

     

     

  • Question 8
    1 / -0
    A water tank can be filled by a tap in $$6$$ hours while it can be emptied by another tap in $$8$$ hours. If both the taps are opened simultaneously then after how much time will the tank get filled ?
    Solution
    Let L be the capacity of tank in liters.

    Tank will be filled at the rate $$\dfrac{L}{6}\,$$ liters\hour.

    While, The tank will get emptied at the rate $$\dfrac{L}{8}\,$$ liters\hour.

    $$\therefore$$ Effective rate of filling the tank is $$\dfrac{L}{6}-\dfrac{L}{8}=\dfrac{L}{24}$$ liters/hour.

    $$\therefore$$ The tank will get filled in $$L \times \dfrac{24}{L}=24$$ hours.
  • Question 9
    1 / -0
    Two pipes $$A$$ and $$B$$ together can fill a cistern in $$4$$ hours. Had they been opened separately, then $$B$$ would have taken $$6$$ hours more than $$A$$ to fill the cistern. How much time will be taken by $$A$$ to fill the cistern separately ?
    Solution
    Let the cistern be filled by pipe $$A$$ alone in $$x$$ hours

    Then, pipe $$B$$ will fill it in $$(x+6)$$ hours.

    $$\therefore \displaystyle \frac{1}{x}+\frac{1}{x+6}=\frac{1}{4}$$

    $$\Rightarrow \displaystyle \frac{x+6+x}{x(x+6)}=\frac{1}{4}$$

    $$\Rightarrow x^2-2x-24=0$$
    $$\Rightarrow x^2-6x+4x-24=0$$
    $$\Rightarrow x(x-6)+4(x-6)=0$$
    $$\Rightarrow (x-6)(x+4)=0$$
    If we neglect the negative value then $$x=6$$
    Hence the cistern be filled by pipe $$A$$ alone in $$6$$ hrs.

  • Question 10
    1 / -0
    A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank five hour faster than the first pipe and 4 hour slower than the third. The time required by the first pipe is:
    Solution


    Suppose first pipe alone takes $$x$$ hours to fill the tank

    Then ,second and third pipe takes $$(x-5)$$ and $$(x-9)$$ hours to fill the tank.

    $$\therefore \displaystyle \frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9}$$

    $$\Rightarrow \displaystyle \frac{x-5+x}{x(x-5)}=\frac{1}{x-9}$$

    $$\Rightarrow (2x-5)(x-9)=x(x-5)$$

    $$\Rightarrow x^2-18x+45=0$$

    $$\Rightarrow x^2-15x-3x+45=0$$

    $$\Rightarrow x(x-15)-3(x-15)=0$$

    $$\Rightarrow (x-3)(x-15)=0$$

    $$x=15   hrs.$$  


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