Mixed Quantitative Aptitude Questions Set 150

Banya
  1. A person invests money in 3 different schemes for 3 years, 5 years and 6 years at 10%, 12% and 15% simple interest respectively. At the completion of each scheme, he gets the same interest. The ratio of his investments is:
    4:3:3
    3:2:1
    6:3:2
    5:2:4
    1:1:2
    Option C
    (3×10%)of A = (5 × 12%) of B = (6 × 15%) of C
    (A, B, C are the investments)
    0.3 A = 0.6 B = 0.9 C
    A:B:C: = 6:3:2

     

  2. A sells a horse to B for Rs. 9720, thereby losing 19 per cent, B sells it to C at a price which would have given A 17 per cent profit. Find B’s gain.
    4220
    4400
    4000
    4141
    4320
    Option E
    CP for A = 9720*100/(100-19) = 12000
    SP with 17% profit for A = 12000*(100+17)/100 = 14040
    B’s gain = 14040 – 9720 = 4320

     

  3. In an election between 2 candidates, 75% of the voters cast their votes, out of which 2% votes were declared invalid. A candidate got 18522 votes which were 75% of the valid votes. The total number of voters enrolled in the election was:
    28200
    31300
    24500
    33600
    30000
    Option D
    Let total number of voters = x
    Voters who cast their votes = 0.75x
    Valid votes polled = 0.98*0.75x
    Valid votes polled for a candidate
    => 18522 = 0.98*0.75*0.75x
    =>x = 33600

     

  4. The length of each side of a rhombus is equal to the length of the side of square whose diagonal is 80√2. If the length of the diagonals of rhombus are in ratio 3:4, then its area (in cm^2) is:
    6144
    6424
    6901
    5580
    6000
    Option A
    Length of each side of rhombus = 80 cm
    Let the diagonals = 3x, 4x
    (3x/2 )^2 + (4x/2 )^2 = 6400
    X^2 = 6400 ×4/25 = 1024
    => x = 32
    Area of rhombus = 1/2× 96 × 128 = 6144 cm^2

     

  5. Two vessels A and B contains milk and water mixed in the ratio 8:5 and 5:2 respectively. The ratio in which these two mixture be mixed to get a new mixture containing 69(3/13)% milk is:
    3:4
    1:3
    2:7
    7:8
    8:9
    Option C
    8/13 ==== 5/7
    ====9/13
    2/91 ====1/13
    2:7

     

    6. Directions(6-10): What approximate value will come in place of question mark ‘?’ in the following questions.

  6. ? = 𝟓𝟔. 𝟗𝟏𝟓𝟔 𝐨𝐟 𝟐𝟖. 𝟎𝟓𝟔 ÷ 𝟕𝟔. 𝟎𝟕𝟓𝟒 × 𝟓. 𝟗𝟕𝟒3
    125
    115
    100
    126
    110
    Option D
    ? = 𝟓𝟔. 𝟗𝟏𝟓𝟔 𝐨𝐟 𝟐𝟖. 𝟎𝟓𝟔 ÷ 𝟕𝟔. 𝟎𝟕𝟓𝟒 × 𝟓. 𝟗𝟕𝟒3
    =>? = 57 × 28 × 1/76 × 6
    =>? = 126

     

  7. 𝟓𝟔𝟕𝟕. 𝟏𝟑𝟐𝟏 + 𝟒𝟗𝟏𝟑. 𝟗𝟏𝟑𝟑 − 𝟑𝟕𝟗𝟖. 𝟗𝟐 = ? +𝟐𝟎. 𝟎𝟎𝟓% 𝐨𝐟 𝟑𝟗𝟔𝟎. 𝟏𝟑𝟐1
    6790
    5670
    6160
    6000
    5909
    Option D
    𝟓𝟔𝟕𝟕. 𝟏𝟑𝟐𝟏 + 𝟒𝟗𝟏𝟑. 𝟗𝟏𝟑𝟑 − 𝟑𝟕𝟗𝟖. 𝟗𝟐 = ? +𝟐𝟎. 𝟎𝟎𝟓% 𝐨𝐟 𝟑𝟗𝟔𝟎. 𝟏𝟑𝟐1
    =>5677 + 4914 – 3799 = ? + 20/100 × 3960
    =>6792 = ? + 792
    =>? = 6000

     

  8. 𝟔𝟓𝟗. 𝟗𝟕 × (? )^𝟐 = (𝟔𝟒. 𝟗𝟐)^𝟐 + 𝟐𝟒. 𝟗𝟗𝟕% 𝐨𝐟 𝟔𝟖𝟔𝟎. 𝟎𝟎𝟏3
    2
    5
    3
    1
    6
    Option C
    𝟔𝟓𝟗. 𝟗𝟕 × (? )^𝟐 = (𝟔𝟒. 𝟗𝟐)^𝟐 + 𝟐𝟒. 𝟗𝟗𝟕% 𝐨𝐟 𝟔𝟖𝟔𝟎. 𝟎𝟎𝟏3
    => 660 ×(? )^2 = (65)² + 25/100 × 6860
    =>660 × (? )^2= 4225 + 1715
    => (? )^2= 5940/660
    =>? = √9
    =>? = 3

     

  9. 𝟐𝟑. 𝟖𝟑% 𝐨𝐟 𝟔𝟐𝟓. 𝟎𝟐 − 𝟏𝟎𝟎. 𝟎𝟏 =? % 𝐨𝐟 𝟑𝟓𝟗𝟗. 𝟗𝟗 + 𝟗𝟖.𝟏𝟑 ÷ 𝟔. 𝟗𝟗𝟗𝟗
    2
    1
    3
    5
    4
    Option B
    𝟐𝟑. 𝟖𝟑% 𝐨𝐟 𝟔𝟐𝟓. 𝟎𝟐 − 𝟏𝟎𝟎. 𝟎𝟏 =? % 𝐨𝐟 𝟑𝟓𝟗𝟗. 𝟗𝟗 + 𝟗𝟖.𝟏𝟑 ÷ 𝟔. 𝟗𝟗𝟗𝟗
    => 24/100 × 625 – 100 = ?/100 × 3600 + 98/7
    ⇒ 50 = ? × 36 + 14
    ⇒ ? = (50 – 14)/36
    ⇒ ? = 1

     

  10. 𝟏𝟏.𝟗𝟗𝟒% 𝐨𝐟 𝟓𝟎𝟎. 𝟎𝟑 + 𝟏𝟔.𝟎𝟏×?/𝟐𝟎.𝟎𝟒 = 𝟏𝟓𝟎. 𝟎𝟏𝟐𝟑 + (𝟐𝟓. 𝟗𝟓𝟑𝟏 × 𝟑𝟓. 𝟏𝟐𝟏)
    1122
    1250
    1225
    1420
    1330
    Option B
    𝟏𝟏. 𝟗𝟗𝟒% 𝐨𝐟 𝟓𝟎𝟎. 𝟎𝟑 + 𝟏𝟔.𝟎𝟏×?/𝟐𝟎.𝟎𝟒 = 𝟏𝟓𝟎. 𝟎𝟏𝟐𝟑 + (𝟐𝟓. 𝟗𝟓𝟑𝟏 × 𝟑𝟓. 𝟏𝟐𝟏)
    12/100 × 500 + 16 × ?/20 = 150 + (26 × 35)
    =>60 + 4 × ?/ 5 = 150 + 910
    => 4 × ? /5 = 1060 – 60
    =>? = 1000 × 5/4
    => ? = 1250

     


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