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Numerical Applications Test 48

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Numerical Applications Test 48
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  • Question 1
    1 / -0
    The average of statistical data is called :
    Solution
    The average of statistical data is called MeanMean or Arithmatic meanArithmatic\ mean

    Hence, OpAOp-A is correct.
  • Question 2
    1 / -0
    Two men and 55  boys together can finish a piece of work in  44 days, while 33 men and 6 6  boys can finish in 33  days. Find the time taken :
    (i) by one man alone to finish the work
    (ii) by one boy alone to finish the work.
    Solution
    Let the amount of work done by a Man alone in a day be M M and by a Boy alone be B B

    Given, 22 men and 55 boys together can finish a piece of work in 4 4 days

    Amount of work they can do in one day 2M+5B2M + 5B =14 = \dfrac {1}{4} -- (1)

    Similarly, 3M+6B=13 3M + 6 B = \dfrac {1}{3} --- (2)

    Solving equations (1) (1) and (2) (2) , we get B=136 B = \dfrac {1}{36}

    Substituting B B in equation 2 2 , we get M=118 M = \dfrac {1}{18}

    Hence, one man alone will take 18 18 days to do the work and one boy will take 36 36 days.

  • Question 3
    1 / -0
    A contractor undertakes to dig a canal, 66 kilometers long in 3535 days and employed 9090 men. He finds that after 2020 days only 22 km of canal have been completed . How many more men must be employed to finish the work in time?
    Solution
    Total time given =35=35 days.

    To dig a canal of 22 km in 2020 days, 9090 men were required.

    So, to dig a canal of 11 km in 11 day, 90×202\dfrac{90\times20}{2} men will be required.

    So, to dig a canal of 44 km in 1515 days 900×415\dfrac{900\times4}{15} =240=240 men will be required.

    As 9090 men were already employed, so 150150 more men are required.
  • Question 4
    1 / -0
    Pooja and Ritu can do a piece of work in 171717\displaystyle \frac{1}{7} days. If one day work of Pooja be three fourth of one day work of Ritu; find in how many days each alone will do the work.
    Solution
    Let Pooja's one day work be xx
    And Ritu's one day work be yy
    Accprding to first condition,
    Pooja and Ritu can do work in 171717 \dfrac{1}{7} days i.e 1207\dfrac{120}{7} days
    x+y=11207\therefore x + y = \dfrac {1}{\dfrac{120}{7}} 
    x+y=7120\therefore x + y = \dfrac{7}{120}  .........Equation 11
    According to second condition,
    x=34yx = \dfrac{3}{4} y
    4x3y=0\therefore 4x - 3y = 0   .......... Equation 22
    Multiplying eqution 11 by 33 and solving equation 11 and 22 simultaneosly, we get
    3x+3y=7403x + 3y = \dfrac{7}{40}
    4x3y=04x - 3y = 0
    7x=740\therefore 7x = \dfrac{7}{40}
    x=140\therefore x = \dfrac{1}{40}
    Putting the value of xx in equation 11 , we get
    140+y=7120\dfrac{1}{40} + y = \dfrac{7}{120}
    y=4120=130\therefore y = \dfrac{4}{120} = \dfrac{1}{30}
    \therefore Pooja's one day work is x=140x = \dfrac{1}{40} 
    i.e. her whole work will be completed in 4040 days
    Similarly Pooja's work will be completed in 3030 days
    Answer : Pooja can do the work alone in 4040 days and Ritu in 3030 days
  • Question 5
    1 / -0
    A cabinet of ministers consists of 11 ministers, one minister being the chief minister. A meeting is to be held in a room having a round table and 11 chairsround it, one of them being meant for the chairman. The number of ways in which the ministers can take their chairs, the chief minister occupying the chairman's place, is
    Solution
    These are circular permutations
    so,
    =(n1)!=(n-1)! = 10!10!
  • Question 6
    1 / -0
    1616 men can complete a work in 1212 days, 2424 children can complete the same work in 1818 days, 1212 men and 88 children started working and after 88 days, 33 more children joined them. How many days will they now take to complete the remaining work ?
    Solution
    One day work of 16 men is 112\dfrac { 1 }{ 12 } .

    One day work of 24 children is 118\dfrac { 1 }{ 18 } .

    One men's one day work is 112×16\dfrac { 1 }{ 12\times 16 } .

    One child's one day work is 124×18\dfrac { 1 }{ 24\times 18 } .

    One day work of 12 men and 8 children is 1212×16+824×18\dfrac { 12 }{ 12\times 16 } +\dfrac { 8 }{ 24\times 18 } .

    One day work = 3554×8\dfrac { 35 }{ 54\times 8 } .

    Work done in 8 days = 8×3554×8=35548\times \dfrac { 35 }{ 54\times 8 } =\dfrac { 35 }{ 54 } .

    Remaining work = 13554=19541-\dfrac { 35 }{ 54 } =\dfrac { 19 }{ 54 } .

    The remaining work is 12×\times one men's one day work + 11×\times one child's one day work.

    Hence, 12×112×16+11×24×1812\times \dfrac { 1 }{ 12\times 16 } +11\times 24\times 18

    One day work multiplied by number of days to complete remaining work =  remaining work

    Therefore, the work is done in 4 days.

    (116+1124×18)x=1954\left( \dfrac { 1 }{ 16 } +\dfrac { 11 }{ 24 } \times 18 \right) x=\dfrac { 19 }{ 54 }

    \Rightarrow x=4x=4

    Thus they take 4 days to complete the remaining work.

    Hence option B is correct.
  • Question 7
    1 / -0
    A particular work can be completed by 66 men and 66 women in 2424 days, whereas the same work can be completed by 88 men and 1212 women in 1515 days. Find the time taken by 44 men and 66 women to complete the same work .
    Solution
    From the question, we have
    (66 men and 66 women) 11 day work =124=\dfrac{1}{24}

    (1212 men and 1212 women)11 day work=112=\dfrac{1}{12}   ...     (1)

    (88 men and 1212 women)11 day work=115=\dfrac{1}{15}     ...    (2)
    Subtracting (1) and (2), we get
    44 men work =112=\dfrac{1}{12}  115-\dfrac{1}{15} =160=\dfrac{1}{60}  

    22 men work =1120=\dfrac{1}{120}  

    44 men and 66 women work =1=1 day work of 66 men and 66 women  1-1 day work of 22 men.
    =124=\dfrac{1}{24} 1120-\dfrac{1}{120} =4120=\dfrac{4}{120}  =130=\dfrac{1}{30}  

    The time taken by 44 men and 66 women to complete the same work =30= 30 days.

    Hence, option B.
  • Question 8
    1 / -0
    AA and BB can do a piece of work in 55 days, BB and CC can do it in 77 days. AA and CC can do it in 44 days. Who among these will take the least time if put to do it alone ?
    Solution
    Let  the  work  be  W.Let \, \,the \, \,work \, \,be \, \,W.
    A  and  B  can  do  a  piece  of   work  in  5  daysA \, \, and\, \, B\, \, can\, \, do\, \, a \, \, piece \, \,of\, \,  work\, \, in \, \, 5 \, \,days  5A+5B=W  ..........(1)\therefore \, \,5A+5B=W \, \,..........(1)

    B  and  C  can  do  it  in  7  daysB \, \, and \, \, C \, \, can \, \,do \, \, it \, \, in \, \,7 \, \, days,  7B+7C=W  ......(2)\therefore, \, \,7B+7C=W \, \,......(2)

    A  and  C  can  do  it  in  4  daysA \, \, and \, \, C \, \, can \, \,do \, \, it \, \, in \, \,4 \, \, days,  4A+4C=W  .......(3)\therefore, \, \,4A+4C=W \, \,.......(3)

    Solving  (1)  (2)  and  (3)  We  get,Solving \, \, (1) \, \,(2) \, \,and \, \,(3) \, \,We \, \,get,

    Speed  of  A=43280  work/daySpeed \, \, of \, \, A=\dfrac{43}{280} \, \, work/day

    Speed  of  B=13280  work/daySpeed \, \, of \, \, B=\dfrac{13}{280} \, \, work/day

    Speed  of  C=27280  work/daySpeed \, \, of \, \, C=\dfrac{27}{280} \, \, work/day

    Since  speed  of  A  is  maximum  among  all  three,Since \, \,speed \, \,of \, \,A \, \,is \, \,maximum \, \,among \, \,all \, \,three, \,
    Hence  A  will  take  least  time  to  complete  the  work.Hence \, \,A \, \, will \, \,take \, \,least \, \,time \, \,to \, \,complete \, \,the \, \,work.
  • Question 9
    1 / -0
    A and B can do a job in 1212 days, B and C can do it in 1616 days. After A has worked for 55 days and B has worked for 77 days, C can finish the rest in 1313 days. In how many days, can C do the work alone?
    Solution
    AA and BB can do a job in 1212 days.
    BB and CC can do a job in 1616 days.
    AA worked for 55 days and BB worked for 77 days.
    Part of work done by AA and BB in 55 days=512\dfrac{5}{12}
    \therefore Part of work left after AA left=1512=12512=7121-\dfrac{5}{12}=\dfrac{12-5}{12}=\dfrac{7}{12}
    After AA left work BB worked for 22 more days.
    Part of work done by BB and CC in 22 days=216=18\dfrac{2}{16}=\dfrac{1}{8}
    \therefore part of work left after BB left=71218=14324=1124\dfrac{ 7 }{ 12 } -\dfrac{ 1 }{ 8 } =\dfrac{ 14-3 }{ 24 } =\dfrac{ 11 }{ 24 }
    Now 1124\dfrac{ 11 }{ 24 } part o work CC has to complete in 1111 days
    Since CC already had worked for 22 dayswith BB
    (part)complete work is done by CC in 11 1124 \dfrac { 11  }{ \dfrac { 11 }{ 24 }  }
    =2424days.
  • Question 10
    1 / -0
    33 men and 44 boys can do a piece of work in 1414 days, while 44 men and 66 boys do the same work in 1010 days. How long would it take one man to finish the same work ?
    Solution
    Let one man's one day's work be xx
    Let one boy's one day's work be yy
    According to first condition,
    3x+4y=1143x+4y = \dfrac{1}{14}     .......Equation 11
    According to second condition,
    4x+6Y=1104x+6Y = \dfrac{1}{10}     .......Equation 22
    Multiplying Equation 11 by 44 and Equation 22 by 33,
    12x+16y=41412x + 16y = \dfrac{4}{14}     .....Equation 33
    12x+18y=31012x + 18y = \dfrac{3}{10}     .....Equation 44
    Now solving Equation 33 and 44 simultaneously by substracting Equation 33 by Equation 44,
     we get,
    2y=414310\therefore -2y = \dfrac{4}{14} -\dfrac{3}{10}

    2y=4042140\therefore -2y = \dfrac{40-42}{140}

    y=1140\therefore y = \dfrac{1}{140}
    Putting the value of y in equation 11, we get,
    3x+4140=1143x + \dfrac{4}{140} = \dfrac{1}{14}

    x=170\therefore x = \dfrac{1}{70}

    Hence one man's one day  work is 170\dfrac{1}{70}

    \therefore one man completes the whole work in 7070 days.
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