Mixed Quantitative Aptitude Questions Set 102

Directions(1-5): Find the missing term of the following questions.

  1. 520,513,494,457,396,?
    320
    333
    305
    314
    325
    Option C
    +1^3-2^3
    +2^3-3^3
    +3^3-4^3
    +4^3-5^3
    +5^3-6^3
    ? = 305

     

  2. 1812,1820,1804,1828,?,1836
    1796
    1700
    1770
    1754
    1784
    Option A
    +8*1
    -8*2
    +8*3
    -8*4
    +8*5
    ? = 1796

     

  3. 42,22,24,39,82,?
    215
    198
    222
    210
    200
    Option D
    *0.5+1
    *1+2
    *1.5+3
    *2+4
    *2.5+5
    ? = 210

     

  4. 74,74,111,222,?,1665
    520
    500
    525
    533
    555
    Option E
    *1
    *1.5
    *2
    *2.5
    *3
    ? =555

     

  5. 5,28,74,143,?,350
    222
    235
    242
    255
    274
    Option B
    +23*1
    +23*2
    +23*3
    +23*4
    +23*5
    ? = 235

     

  6. The average weight of (x+5) eggs is 10 grams. If 5 more eggs are added then the average gets doubled. Find the value of x if the 5 eggs which are added weigh 200 grams.
    4
    5
    2
    3
    6
    Option B
    [10(x+5)+200]/[x+10] = 2*10
    => x = 5

     

  7. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
    44%
    40%
    57%
    50%
    63%
    Option C
    votes received by the winning candidate = 11628
    Total votes =1136+7636+11628=20400
    Required percentage =11628/20400×100
    =969/17=57%

     

  8. Ayisha’s age is 1/6th of her father’s age. Ayisha’s father’s age will be twice Shankar’s age after 10 years. If Shankar’s eight birthdays was celebrated two years before, then what is Ayisha’s present age.
    6
    4
    7
    3
    5
    Option E
    Let Ayisha’s present age =x
    Then, her father’s age =6x
    Shankar’s age after 10 years =1/2(6x+10) =3x+5
    Therefore, Shankar’s age after 10 years =8+12=20
    3x+5=20
    =>x=15/3=5

     

  9. A train of length 250 m can cross a platform of length x m in seconds. Also it can cross a train of half of its length running with speed of 35 m/s in opposite direction in 5 seconds. Find the value of x.
    100
    125
    150
    135
    144
    Option C
    (250+x)= 10*s —-(1)
    (250+0.5*250) = 5*(s+35) —–(2)
    On solving (1) and (2), we get
    s= 40 m/s x = 10*40 – 250
    => x = 150 m

     

  10. A bag contains 6 apples, 8 bananas and (x+2) oranges. 2 fruits are drawn at random . Find the value of x if the probability that both fruits are oranges is 2/51.
    2
    4
    1
    5
    3
    Option A
    Probability of selecting 2 oranges = (x+2)C2/(16+x)C2 Now, (x+2)C2/(16+x)C2 = 2/51
    => (x+2)(x+1)/(x+16)(x+15) = 2/51
    => 49x^2 + 91x – 378 = 0
    => (x-2)(7x +27) = 0
    =>x = 2

     


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