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Numerical Applications Test 4

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Numerical Applications Test 4
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  • Question 1
    1 / -0
    Mean of a set of observations is the value which
    Solution
    Mean is the value which is the representative of the whole group. As, this takes into account all the values present in the group and averages them.
  • Question 2
    1 / -0
    In a cricket test match the scores of ten players are: 8585, 3232, 00, 5454, 2929, 101101, 7373, 6464, 2929 and 3636. Find the mean of the runs.
    Solution
    Runs scored by 1010 players =85,32,0,54,29,101,73,64,29,36= 85,32,0,54,29,101,73,64,29,36
    Mean =SumNumber of players= \cfrac{\text{Sum}}{\text{Number of players}}
    Mean =85+32+0+54+29+101+73+64+29+3610= \cfrac{85 + 32 + 0 + 54 + 29 + 101 + 73 + 64 + 29 + 36}{10}
    Mean =50310= \cfrac{503}{10}
    Mean =50.3= 50.3
  • Question 3
    1 / -0
    The heights of 66 boys in a group are 142142 cm, 154154 cm, 146146 cm, 145145 cm, 151151 cm and 150150 cm. Find the mean height per boy
    Solution
    Mean height =Sum of the height of the all boysNo. of boys=\cfrac{\text{Sum of the height of the all boys}}{\text{No. of boys}}
                          =142+154+146+145+151+1506=\cfrac{142+154+146+145+151+150}{6}
                          =8886=148=\cfrac{888}{6}=148 cm
    Hence, the mean height is 148148 cm.
  • Question 4
    1 / -0
    Find the mean of first ten odd natural numbers.
    Solution

    Step  - 1: Listing first 10 odd natural numbers{\textbf{Step  - 1: Listing first 10 odd natural numbers}}

                        First 10 odd natural numbers  =  1, 3, 5, 7, 9, 11, 13, 15, 17, 19{\text{First 10 odd natural numbers  =  1, 3, 5, 7, 9, 11, 13, 15, 17, 19}}

    Step  - 2: Calculating mean{\textbf{Step  - 2: Calculating mean}}

                        We know that, mean of n numbers  =  Sum of n numbersn{\text{We know that, mean of n numbers  =  }}\dfrac{{{\text{Sum of n numbers}}}}{{\text{n}}}

                         Mean of first 10 odd natural numbers,\therefore {\text{ Mean of first 10 odd natural numbers,}}

                        M  =  1  +  3  +  5  +  7  +  9  +  11  +  13  +  15  +  17  +  1910{\text{M  =  }}\dfrac{{{\text{1  +  3  +  5  +  7  +  9  +  11  +  13  +  15  +  17  +  19}}}}{{{\text{10}}}}

                         M  =  10010  =  10 \Rightarrow {\text{ M  =  }}\dfrac{{{\text{100}}}}{{{\text{10}}}}{\text{  =  10}}

    Thus, the mean of first 10 odd natural numbers is 10{\textbf{Thus, the mean of first 10 odd natural numbers is 10}}

  • Question 5
    1 / -0
    In a monthly test the marks obtained in mathematics by 1616 students of a class are as follows
    0,0,2,2,3,3,3,4,5,5,5,5,6,6,7,80, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8
    The arithmetic mean of the marks obtained is
    Solution
    Since, Mean =0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+816= \cfrac {0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8}{16}
          \Rightarrow Mean =6416= \cfrac {64}{16}
          \Rightarrow Mean = 4=  4
          \therefore Option BB is correct.
  • Question 6
    1 / -0
    A contractor undertook to do a certain piece of work in 99 days. He employed certain number of men, but 66 of them being absent from the very first day, the rest could finish the work in 1515 days. The number of men originally employed were
    Solution
    Let there be x x men at the begining less men, more days
    \therefore  15:9::x:(x6)15 : 9 :: x : (x - 6)  \Rightarrow  15(x6)=9x 15(x - 6) = 9x
    \Rightarrow  6x=906x = 90  \Rightarrow x= 15 x = 15
  • Question 7
    1 / -0
    Out of the following four choices which does not show the coinciding of the hour hand and minute hand :
    Solution
    Clearly, at 9:59:05 the hour hand and the minute hand does not coincide.
    Answer is Option C
  • Question 8
    1 / -0
    If in a non leap year it is Tuesday on 28th February then the day which will be on 28th march is ------
  • Question 9
    1 / -0
    A fort had provision of food for 150150 men for 4545 days. After 1010 days, 2525 men left the fort, The number of days for which the remaining food will last, is
    Solution
    After  1010 days: 150150 men had food for 3535 days.
    Suppose  125125  men had food for xx days. Less men, More days; More men less days.
    \therefore  125:150::35:x125 : 150 : : 35 : x
    125×x=150×35\Rightarrow\, 125\, \times\, x\, =\, 150\, \times\, 35
    x=150×35125x=42\Rightarrow\, x\, =\, \displaystyle \frac{150\, \times\, 35}{125}\, \Rightarrow\, x\, =\, 42
  • Question 10
    1 / -0
    If the arithmetic mean of 6,8,5,7,x6, 8, 5, 7, x and 44 is 7,7, then xx is
    Solution
    Step - 1: Solving for x.\textbf{Step - 1: Solving for x.}
                    Observation given = 6,8,5,7,x,4\text{Observation given = }{6, 8, 5, 7, x, 4}
                    Arithmetic mean = 7\text{Arithmetic mean = 7}
                    Arithmetic mean = Sum of observations.Number of observations.\text{Arithmetic mean = }\dfrac{\text{Sum of observations.}}{\text{Number of observations.}}
                    6+8+5+7+x+46=7\Rightarrow \dfrac{6 + 8 + 5 + 7 + x + 4}{6} = 7
                    30+x=7×630 + x = 7 \times 6
                    x=4230x = 42 - 30
                    x=12x = 12
    Hence, the value of x is 12. (Option A)\textbf{Hence, the value of x is 12. (Option A)}
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