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Numerical Applications Test 47

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Numerical Applications Test 47
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  • Question 1
    1 / -0
    A group of students comprises of $$5$$ boys and $$n$$ girls. If the number of ways, in which a team of $$3$$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is $$1750$$, then $$n$$ is equal to
    Solution
    Number of ways of selecting a team with 1 boy and 2 girls$$=^{5}C_{1}. ^{n}C_{2} $$

    Number of ways of selecting a team with 2 boy and 1 girls$$=^{5}C_{2} . ^{n}C_{1} $$

    Given $$^{5}C_{1}. ^{n}C_{2} + ^{5}C_{2} . ^{n}C_{1} = 1750$$

    $$5\cdot \dfrac {n(n-1)}{2!}+10\cdot n=1750$$

    $$n^{2} + 3n = 700$$

    $$\therefore n = 25$$.
  • Question 2
    1 / -0
    The number of four -digit numbers strictly grater than $$4321$$ that can be formed using the digits $$0,1,2,3,4,5$$ (repetition of digits is allowed) is:
    Solution
    (1) The number of four-digit numbers Strating with 5 is equal to $$6^3 = 216$$
    (2) Starting with 44 and 55 is equal to $$36\times 2 = 72$$
    (3) Starting with 433, 434 and 435 is equal to $$6\times 3 = 18$$
    (4) Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
    so total number are
    $$716 + 72 + 18 + 4 = 310$$
  • Question 3
    1 / -0
    Let $$ x,y,z$$ be three observations. The mean of these observation is 
    Solution
    We know that
    $$\text{Mean}=\dfrac{\text{sum of observations}}{\text{number of observations}}$$

    $$\therefore \text{ Mean}=\dfrac{x+y+z}{3}$$
  • Question 4
    1 / -0
    The mean of the first $$6$$ odd natural numbers is
    Solution
    First 6 odd natural numbers are:
    $$1, 3, 5, 7, 9, 11$$
    $$1 + 3 + 5 + 7 + 9 + 11$$
    Mean $$= \dfrac{1 + 3 + 5 + 7 + 9 + 11}{6}$$
    $$= \dfrac{36}{6} = 6$$
  • Question 5
    1 / -0
    Four alternative options are given for each of the following statement . Select the correct  option .
    The mean of score $$ 10 , 15 , 12 , 15 , 15  $$ is
  • Question 6
    1 / -0
    Four alternative options are given for each of the following statement . Select the correct  option .
    The mean of first three multiples of $$ 5 $$ is
  • Question 7
    1 / -0
    How many ways $$ 5 $$ persons can sit around the round table :
    Solution
    Total number of ways to sitting $$ 5 $$ persons around the round table.
    $$ =(5-1)! = 4! $$
    $$ = 4 \times 3 \times 2 \times  1 = 24 $$
    hence option (B) is correct.
  • Question 8
    1 / -0
    A student got 85, 87 and 83 marks in Mathematics, Physics and Chemistry respectively. Mean of obtained marks.
    Solution
    Given marks: $$85,\ 87,\ 83$$

    Mean of marks $$= \dfrac{\mathrm{Sum\ of\ marks}}{\mathrm{No.\ of\ subjects}}$$

                              $$= \dfrac{85+87+83}{3} $$

                              $$= \dfrac{255}{3}$$

                              $$ = 85$$
  • Question 9
    1 / -0
    53, 75, 42, 70 are marks obtained by 4 students is statistics. Arithmetic mean of their marks is:
    Solution
    Given marks = 53, 75, 42, 70
    Arithmetic mean $$= \frac{Sum\ of\ observations}{No.\ of\ observations}$$
                                 $$= \frac{53+75+42+70}{4}$$
                                  $$= \frac{240}{4} = 60$$
  • Question 10
    1 / -0
    The mean of $$15, 0, 10, 5$$ will be :
    Solution
    The given number are $$15,0,10,5$$
    The mean of a set of numbers is the sum divided by the number of terms.
    Then, mean $$=\dfrac{15+0+10+5}{4}=\dfrac{30}{4}\\=7.5$$

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