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Numerical Applications Test 25

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Numerical Applications Test 25
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  • Question 1
    1 / -0
    The sum of five numbers is 555555. The average of first two numbers is 7575 and the third number is 115115. What is the average of the last two numbers?
    Solution
    Let the five numbers be A,B,C,D,EA, B, C, D, E
    Given, A+B+C+D+E=555A + B + C + D + E =555
    A+B2=75A+B=150\displaystyle \frac{A+B}{2} = 75 \Rightarrow A + B = 150 and C=115C = 115
    150+115+D+E=555\therefore 150 + 115 + D + E = 555
    D+E=555265=290\Rightarrow D + E = 555 - 265 = 290
    \therefore Required Average =D+E2=2902=145= \displaystyle \frac{D + E}{2} = \frac{290}{2} = 145
  • Question 2
    1 / -0
    A tank can be filled by a tap in 2020 minutes and by another tap in 6060 minutes. Both the taps are kept open for 1010 minutes and then the first tap is shut off. After this, the tank will be completely filled in
    Solution
    Part of the tank filled by both the taps in 10 minutes =10×(120+ 160 )=10×( 3+160 )= 23= 10 \times \left( \displaystyle\frac { 1 }{ 20 } + \displaystyle\frac { 1 }{ 60 }  \right) = 10 \times \left( \displaystyle\frac { 3+1 }{ 60 }  \right) = \displaystyle\frac { 2 }{ 3 }
    Remaining part  =1 23= 13 = 1 - \displaystyle\frac { 2 }{ 3 } = \displaystyle\frac { 1 }{ 3 }
    160\displaystyle\frac { 1 }{ 60 }th part of the tank is filled by the second pipe in 1 min.
       13\therefore     \displaystyle\frac { 1 }{ 3 }rd of the tank will be filled by the second pipe in  60× 13=20 60 \times \displaystyle\frac { 1 }{ 3 } = 20  min.
  • Question 3
    1 / -0
    An electric pump can fill a tank in 33 hours. Because of a leak in the tank, it took 3123\displaystyle\frac { 1 }{ 2 } hours to fill the tank. In how much time can the leak drain all the water of the tank?
    Solution
    Part of the tank filled by the pump in 1 hour =13= \displaystyle\frac { 1 }{ 3 }
    Net part of the tank filled by the pump and the leak in 1 hour  = 1 72 = 27 = \displaystyle\frac { 1 }{ \displaystyle\frac { 7 }{ 2 }  } = \displaystyle\frac { 2 }{ 7 }

    \therefore  Part of the tank emptied by the leak in 1 hour  =  13 27= 121 = \displaystyle\frac { 1 }{ 3 } - \displaystyle\frac { 2 }{ 7 } = \displaystyle\frac { 1 }{ 21 }
    \therefore  The tank will be emptied by the leak in 2121 hours.
  • Question 4
    1 / -0
    AA and BB can do a job in 1212 days and BB and CC can do it in 1616 days. After AA has been working for 55 days and BB for 77 days, CC finishes rest of the work in 1313 days, in how many days can C do the work alone ?
    Solution
    A's 5 days work++ B's 7 days work ++ C's 13 days work ==1
    \therefore (A+B)s(A+B)'s 5 days work ++ (B+C)s(B+C)'s 2 days work ++ C's 11 days work == 1
    \therefore 512+216\dfrac{5}{12}+\dfrac{2}{16}++ C's 11 days work=1=1
    \therefore C's 11 days work==1(512+2161-(\dfrac{5}{12}+\dfrac{2}{16}) == 1124\dfrac{11}{24}
    \therefore C's 1 days work ==(1124×111 (\dfrac{11}{24} \times \dfrac{1}{11}) =124=\dfrac{1}{24}
    \therefore C alone can finish the work in 24 days.

  • Question 5
    1 / -0
    AA can do a piece of work in 77 days of 99 hr each and BB can do it in 66 days of 77 hr each. How long will they take to do it working together 425\displaystyle \frac{42}{5} hr a day?
    Solution
    A can do a piece of work in 7 days of 9 hr each or in 63 hrs.
    B can do it in 6 days of 7 hr each  or 42 hrs.
    So, in 1 hour, A can do 1/63 of the work and B can do 1/42 of the work.
    Together, they can finish 5/126 of the work in 1 hour or they work gets completed in 126/5 hrs.
    If they complete work in  42/5 hrs a day, they will take 3 days to complete the work working for 126/5 hours.
  • Question 6
    1 / -0
    Machines AA and BB produce 8,0008,000 clips in 44 hr and 66 hr respectively. If they work alternately for 11 hr A starting first, then the 8,0008,000 clips will be produced in:
    Solution

    (A+B)'s 2 hour's work=(14+16)=512=\left(\dfrac{1}{4}+\dfrac{1}{6} \right)=\dfrac{5}{12}

    (A+B)'s 4 hour's work=512×2=56=\dfrac{5}{12}\times 2=\dfrac{5}{6}

    Remaining work=156=16=1-\dfrac{5}{6}=\dfrac{1}{6}

    Now it is a's turn

    14\because \dfrac{1}{4} work by A in=1 hours

    16\therefore \dfrac{1}{6} work done by A in =4×16=23 hours=4\times \dfrac{1}{6}=\dfrac{2}{3}  hours

    \therefore total time taken=423=4.66  hours=4\dfrac{2}{3}=4.66 \  hours

  • Question 7
    1 / -0
    Ronald and Elan are working on an assignment Ronald takes 66 hours to type 3232 pages on a computer while Elan takes 55 hours to type 4040 pages. How much time will they take working together on two different computers to type an assignment of 110110 pages ?
    Solution

    Ronald work at a rate of 32 pages per 6 hrs=326=163pages/hr=\dfrac{32}{6}=\dfrac{16}{3} pages/hr

    Elan work at a rate of  40 pages per 5 hr=404=8 Pages/hr=\dfrac{40}{4}=8   Pages/hr

     

    If they  work together then they work at the rate of=163+8=403 pages/hr=\dfrac{16}{3}+8=\dfrac{40}{3}  pages/hr

    Let t time is required them to type 110 pages then

    403t=110\dfrac{40}{3}t=110

    40t=33040t=330

    t=33040=334=814=8 hr 15min.t=\dfrac{330}{40}=\dfrac{33}{4}=8\dfrac{1}{4}=8  hr  15 min.

  • Question 8
    1 / -0
    3636 men can complete a piece of work in 1818 days. In how many days will 2727 men complete the same work?
    Solution

    Step  - 1: Finding number of work units for the first case.{\textbf{Step  - 1: Finding number of work units for the first case}}{\text{.}}

                      Total 36 men work for a total of 18 days to complete it.{\text{Total 36 men work for a total of 18 days to complete it}}{\text{.}}

                      So total work units=36×18 =648{\text{So total work units}} = 36 \times 18  = 648

                      Thus, for the first case ,it’s a total of 648 work units.{\text{Thus, for the first case ,it's a total of 648 work units}}{\text{.}}

    Step  - 2: Finding number of days for the second case.{\textbf{Step  - 2: Finding number of days for the second case}}{\text{.}}

                      Let it takes x days for the workers to complete the task,so we can write{\text{Let it takes }}x{\text{ days for the workers to complete the task,so we can write}}

                      Number of work units=27 ×x=27x{\text{Number of work units}} = 27  \times x = 27x

                      Also, the work is same so the work units must be same{\text{Also, the work is same so the work units must be same}}

                      So, equating the work units for the two cases, we get{\text{So, equating the work units for the two cases, we get}}

                      648=27x648 = 27x

                      x=64827=24x = \dfrac{{648}}{{27}} = 24

    Hence option D is correct{\textbf{Hence option D is correct}}

  • Question 9
    1 / -0
    AA does 45\displaystyle \frac{4}{5}th of work in 2020 days. He then calls in BB and they together finish the remaining work in 33 days. How long BB alone would take to do the whole work ?
    Solution
    Let  the  total  work  be  x.Let \, \,the \, \,total \, \,work \, \,be \, \,x.
      The  speed  at  which  amount  of  work  A  does  in  20  days  is=x25\therefore \, \, The \, \, speed \, \, at \, \, which \, \, amount \, \,of \, \,work \, \,A \, \,does \, \,in \, \,20 \, \,days \, \,is=\dfrac{x}{25}
    Let  the  speed  of  B  be  B  work  per  day.Let \, \,the \, \,speed \, \,of \, \,B \, \,be \, \,B \, \,work \, \,per \, \,day.
    Since,   the  work  done  by  A  is  45xSince,  \, \, the \, \, work \, \,done \, \,by \, \,A \, \,is \, \,\dfrac{4}{5}x
      Remaining  work  is  x45x=x5\therefore \, \,Remaining \, \,work \, \,is \, \,x-\dfrac{4}{5}x=\dfrac{x}{5}
    If  A  and  B  work  together  then  we  have,If \, \,A \, \,and \, \,B \, \,work \, \,together \, \,then \, \,we \, \,have,

    x5x25+B=3\dfrac{\dfrac{x}{5}}{\dfrac{x}{25}+B}=3
      speed  of  B=2x75  work  per  day.\therefore \, \, speed \, \, of \, \, B=\dfrac{2x}{75} \, \,work \, \,per \, \,day.
    If  B  alone  works  then  the  work  will  be  completed  in  x2x75=  3712  days.If \, \,B \, \, alone \, \,works \, \, then \, \, the \, \,work \, \, will \, \,be \, \, completed \, \,in \, \,\dfrac{x}{\dfrac{2x}{75}}=\, \, 37 \dfrac{1}{2} \, \, days.

  • Question 10
    1 / -0
    AA can do a certain work in the same time in which BB and CC together can do it. If AA and BB together could do it in 1010 days and CC alone in 5050 days, then BB alone could do it in
    Solution
    (A+B)'s 1 day's work=110=\dfrac{1}{10}
    C's 1 day's work=150=\dfrac{1}{50}
    (A+B+C)'s 1 day's work=110+150=650=325.....(I)=\dfrac{1}{10}+\dfrac{1}{50}=\dfrac{6}{50}=\dfrac{3}{25}.....(I)
    According to the question
    A's 1 day's work=(B+C) 1day's work.........(II)
    From I and II we get
    2×2\times (A's  one  day's  work)=325=\dfrac{3}{25}
    A's one day's work=350=\dfrac{3}{50}
    \therefore B's 1 day's work=110350=250=125=\dfrac{1}{10}-\dfrac{3}{50}=\dfrac{2}{50}=\dfrac{1}{25}
    Hence B alone could do the work in 25 days.
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