Mixed Quantitative Aptitude Questions Set 150

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
   
   
    
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