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Permutations and Combinations Test 32

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Permutations and Combinations Test 32
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  • Question 1
    1 / -0
    Directions for questions 1 to 3: Find the related word/ letters/numbers from given alternatives.
    12:72::8:?
    Solution
    12 $$\times \, \frac{12}{2}$$ = 72 similarly, 8 $$\times \, \frac{8}{2}$$ = 32
  • Question 2
    1 / -0
    If $$12\times 16 = 188$$ and $$14\times 18 = 248$$, then find the value of $$16\times 20 = ?$$
    Solution
    $$12\times 16 = 192 - 4 = 188$$
    $$14\times 18 = 252 - 4 = 248$$
    $$16\times 20 = 320 - 4 = 316$$.
  • Question 3
    1 / -0
    Select the missing number from the given alternatives.

    Solution
    48 $$\div$$ 2 = 24;
    24 $$\times$$ 3 = 72: 72 $$\div$$ 2 = 36;
    36 $$\div$$ 3 = 108; 108 $$\div$$ 2 = 54.
  • Question 4
    1 / -0
    Choose the correct answer from the alternatives given.
    If $$1^2+2^2$$ +.... + $$x^2$$ = $$\frac{x(x + 1)(2x + 1)}{6}$$then $$1^2$$ 
    + $$3^2$$+ $$5^2$$ + .... + $$19^2$$ is equal to
    Solution
    $$(1^2+3^2+5^2+...........+19^2)$$
    = $$\displaystyle\, $$(1^2$$ \, + \, $$2^2$$ \, + \, $$3^2$$ \, + \, ........... \, + \, 20^2)  \, - \, (2^2 \, + \, 4^2 \, + \, 6^2 \, + \, ........... \, + \, 20^2)$$
    = $$\displaystyle\, (1^2 \, + \, 2^2 \, + \, 3^2 \, + \, .......... \, + \, 20^2) - 4(1 \, + \, 2^2 \, + \, 3^2 \, + \, .......... \, + \, 10^2)$$
    Using the formula,
    $$\displaystyle\, = \frac{20 \times  21 \times  41}{6} - 4 \left ( \frac{10 \times  11 \times  21}{6} \right ) = 2870 - 1540 = 1330$$

  • Question 5
    1 / -0
    How many different $$4$$-person committees can be chosen form the $$100$$ members of the Senate ?
    Solution
    Total number of members $$=100$$.
    Members has to be selected $$=4$$.
    $$\therefore$$ different committees $$=^{100}C_{4}=\large{\frac{100!}{4!\times96!}}$$ $$=3,921,225$$.
  • Question 6
    1 / -0
    Choose the correct answer alternatives given.
    Select the missing number from the given alternatives.

    Solution
    80 = 8 $$\times$$ (8 + 2)
    143 = 11 $$\times$$ (11 + 2)
    323 = ? $$\times$$ (? + 2)
    $$\therefore$$ ? = 17
  • Question 7
    1 / -0
    Arrange these numbers in ascending order. 
    $$756, 567, 657, 676$$ 
    Solution
    $$\Rightarrow$$  Numbers are said to be in ascending order when they are arranged from the smallest to the largest number.
    $$\Rightarrow$$  The numbers which we have to arrange in ascending order are $$756,\,567,\,657$$ and $$676$$
    $$\Rightarrow$$  $$567<657<676<756$$
    $$\therefore$$  Ascending order $$=567,\,657,\,676,\,756$$
  • Question 8
    1 / -0
    The number of ways in which we can select 5 letters of the word INTERNATIONAL is equal to
    Solution
    We have thirteen letters.
    $$2l, 3N, 2T, 2A$$ and $$(E, O, L,R)8$$ (types)
    We can select 5(five) letters in the following manner:

    1. All different $${^8C}_5 = \dfrac{8.7.5}{1.2.3} = 56$$ ways

    2. 2 alike, 3 different $${^4C}_1 . {^7C}_3 = 435 = 140$$ ways (we have 4 sets of alike letters)

    3. 3 alike, 2 different $${^1C}_1 . {^7C}_2 = 1.21 = 21$$ ways (we have only one set of 3 alike)

    4. 3 alike and 2 alike $${^1C}_1 . {^3C}_1 = 1.3 = 3$$ ways

    5.Two sets of alike and one different 
    $${^4C}_2 . {^6C}_1 = 6.6 = 36$$ ways

    $$\therefore$$ Total number of selection is
    $$56 + 140 + 21 + 3 + 36 = 256$$ ways $$\Longrightarrow$$ (d)
  • Question 9
    1 / -0
    The area of the triangle with vertices at $$(-4, 1), (1, 2)(4, -3)$$ is
    Solution

  • Question 10
    1 / -0
    The area of the triangle whose vertices are (3,8), (-4,2) and (5,-1) is 
    Solution
    Let $$A(3,8), B(-4,2),C(5,-1)$$ be the vertices of the given $$\triangle ABC$$. 

    Then,

    $$(x_{1}=3,y_{1}=8),(x_{2}=-4,y_{2}=2),(x_{3}=5,y_{3}=-1)$$

    Area of $$\triangle ABC$$ = $$\dfrac{1}{2}|[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]|$$

    $$=\dfrac{1}{2}|3[2-(-1)]-4(-1-8)+5(8-2)|$$

    $$=\dfrac{1}{2}|9+36+30|=\dfrac{75}{2}=37.5 \ sq. units$$
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