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