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Permutations and Combinations Test 20

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Permutations and Combinations Test 20
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Weekly Quiz Competition
  • Question 1
    1 / -0
    In  the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?) 

    Solution

  • Question 2
    1 / -0
    In the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?)

    Solution
    Sum of outer digits ×3\times3 is the answer 
    (4+6+3+5)×3=18×3=54\Rightarrow (4+6+3+5)\times3=18\times3=54
    (2+3+6+5)×4=16×4=64\Rightarrow (2+3+6+5)\times4=16\times4=64
    (3+5+7+9)×5=120\Rightarrow (3+5+7+9)\times5=120
  • Question 3
    1 / -0
    In of the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?) Choose the correct answer for the term marked by the symbol (?)

    Solution
    16×4=64=82\rightarrow 16 \times4=64=8^2
    3×27=81=92\rightarrow 3 \times27=81=9^2
    18×8=144=122\rightarrow 18 \times8=144=12^2
  • Question 4
    1 / -0
    In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?) 
    6,9,18,21,30,?6,\,9,\,18,\,21,\,30,\,?
    Solution
    6,9,18,21,306,\,9,\,18,\,21,\,30
    Alternate series difference is common
      30+3=33\therefore\;30+3=33
  • Question 5
    1 / -0
    In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?) 
    0,3,9,18,30,?0,\,3,\,9,\,18,\,30,\,?
    Solution
    The difference between the consecutive numbers increases by 33.
    30=33-0 = 3
    93=69 - 3 = 6
    189=918-9 = 9
    3018=1230-18 = 12.
    Pattern is adding multiples of 3
    Next term is 30+15=4530+15=45
  • Question 6
    1 / -0
    There are 8 buses running from Kota to Jaipur and 10 buses running from Jaipur to Delhi. In how many ways a Person can travel from Kota to Delhi via Jaipur by bus?
    Solution
    Let E1E_1 be the event of travelling from Kota to Jaipur & E2E_2 be the event of travelling from Jaipur to Delhi by the person.
     E1E_1 can happen in 8 ways and E2E_2 can happen in 10 ways. 
    Since both the events E1E_1 and E2E_2 are to be happened 12 in order, simultaneously,the number of ways =8×10=80=8\times10=80
  • Question 7
    1 / -0
    In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)

    Solution
    Multiply opp digits & substract:
      (8×9)(2×3)=726=66\;\rightarrow (8\times9)-(2\times3)=72-6=66
            (7×8)(3×4)=5612=44\;\;\;\;\rightarrow (7\times8)-(3\times4)=56-12=44
            (9×11)(6×5)=9930=69\;\;\;\;\rightarrow (9\times11)-(6\times5)=99-30=69
  • Question 8
    1 / -0
    In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)

    Solution
    12+1=2,  32+1=10,  52+1=26\rightarrow 1^2+1=2,\;3^2+1=10,\;5^2+1=26
    22+1=5,  42+1=17,  62+1=37\rightarrow 2^2+1=5,\;4^2+1=17,\;6^2+1=37
    52+1=26,  72+1=50,  92+1=82\rightarrow 5^2+1=26,\;7^2+1=50,\;9^2+1=82
  • Question 9
    1 / -0
    - nmmn - mmnm - mnnm -
    Solution
    The series is n n m m / n n m m / n n m m / n n m m 
  • Question 10
    1 / -0
    If the area of a triangle is 6868 sq. units and the vertices are (6,7),(4,1)(6, 7), (-4, 1) and (a,9)(a, -9) then the value of aa is 
    Solution
    Area of a triangle with vertices (x1,y1)({ x }_{ 1 },{ y }_{ 1 }) ; $$({ x }_{ 2 },{ y

    }_{ 2 }) and  and ({ x }_{ 3 },{ y }_{ 3 })$$  is:
    $$A= \left| \dfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

    3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 }  \right| $$

    Hence, substituting the points (x1,y1)=(6,7)({ x }_{ 1 },{ y }_{ 1 }) = (6,7) ; $$({ x }_{ 2

    },{ y }_{ 2 }) = (-4,1)  and  and ({ x }_{ 3 },{ y }_{ 3 }) = (a,-9)$$ in the formula for area, we get:

    Area of triangle $$ = \left| \dfrac {  (6)(1+9)+(-4)(-9-7) + a(7-1) }{ 2 } 

    \right|  = 68 $$
    60+64+6a2  =68 \left| \dfrac { 60 + 64 + 6a }{ 2 }  \right|  = 68
    124+6a2 =68 \dfrac{124 + 6a}{2}  = 68
     124+6a=136  124 + 6a = 136
     6a=12  6a = 12
        a=2 \implies a = 2

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