Find what comes in the question mark.

1. 1… 4… 20… 120… 840… ?
   4590
   6720
   4320
   5560
   7690
   Option B
   1 * 4 = 4
   4 * 5 = 20
   20 * 6 = 120
   120 * 7 = 840
   840 * 8 = 6720
   

    

2. ?… 8776… 1096… 136… 16… 1
   73216
   70288
   70716
   75216
   70216
   Option E
   (8776 – 8)/8 = 1096
   (1096 – 8)/8 = 136
   (136 – 8)/8 = 16
   (16 – 8)/8 = 1
   8776 * 8 + 8 = 70216
   

    

3. 13…15… 27… ?… 143… 293
   45
   54
   63
   36
   69
   Option C
   13 + 12 + 13 =15
   15 + 22 + 23 = 27
   27 + 32 + 33 = 63
   63 + 42 + 43= 143
   143 + 52 + 53 = 293
   

    

4. 41… 52… 67… 89… 121… ?
   146
   156
   166
   176
   186
   Option C
   41 52 67 89 121 166
   11 15 22 32 45
   4 7 10 13
   3 3 3
   

    

5. 6… 3… 3… 4.5… ?… 22.5
   9
   11
   13
   15
   17
   Option A
   6 * 1/2 = 3
   3 * 2/2 = 3
   3 * 3/2 = 4.5
   4.5 * 4/2 = 9
   9 * 5/2 = 22.5
   

    

Find the wrong number in the following series:-


6. 22… 36… 54… 76… 102… 130
   22
   36
   54
   76
   130
   Option E
   22 36 54 76 102 132
   14 18 22 26 30

    

7. 37… 75… 227… 911… 4559… 27358
   27358
   4559
   911
   227
   75
   Option A
   37 * 2 + 1 = 75
   75 * 3 + 2 = 227
   227 * 4 + 3 = 911
   911 * 5 + 4 = 4559
   4559 * 6 + 5 = 27359
   

    

8. 107… 118… 131… 148… 168… 190
   190
   167
   148
   131
   118
   Option B
   107 + 11 = 118
   118 + 13 = 131
   131 + 17 = 148
   148 + 19 = 167
   167 + 23 = 190
   

    

9. 17… 18… 34… 54… 68… 161
   161
   68
   54
   34
   18
   Option A
   17 * 2 = 34 … 18 * 3 = 54
   34 * 2 = 68 … 54 * 3 = 162
   

    

10. 47… 54… 66… 80…. 97…. 112
    54
    66
    80
    97
    112
    Option C
    47 + 7 * 1 = 54
    54 + 6 * 2 = 66
    66 + 5 * 3 = 81
    81 + 4 * 4 = 97
    97 + 3 * 5 = 112
    

     

1. I. x² -4x -21 = 0
   II. y² = 64
   
   x
   x<=y
   x>y
   x>=y
   Hence the relation cannot be determined
   Option E
   x² -4x -21 = 0
   x² -7x + 3x -21 = 0
   x( x-7) + 3( x-7)=0
   (x-7) (x+3)=0
   x=7 or x= -3
   y 2 = 64
   y= ±8
   Hence the relation cannot be determined
   

    

2. I. x² +7x +10 = 0
   II. y² -5y + 6=0
   
   x>y
   x>=y
   x< y
   x<=y
   it cannot be determined
   Option C
   x² + 7x +10 = 0
   x² +2x + 5x + 10 = 0
   x( x+ 2) + 5( x+2)=0
   (x+2) (x+5)=0
   x=-2 or x= -5
   y² -5y + 6=0
   y² -2y -3y+ 6=0
   y(y-2) -3( y-2)=0
   (y-2) ( y-3) =0
   y= 2 or 3
   so x< y
   

    

3. I. x² -8x +16 = 0
   II. y² +4y + 4=0
   
   x> y
   x>=y
   x
   x<=y
   it cannot be determined
   Option A
   x² -8x +16 = 0
   (x-4)2= 0
   =>x=4
   y² +4y + 4=0
   ( y+ 2) 2 = 0
   y= -2
   Hence x> y
   

    

4. I. x² -12x +36 = 0
   II. y² +22y + 121=0
   
   x> y
   x>=y
   x
   x<=y
   it cannot be determined
   Option A
   x² -12x +36 = 0
   (x-6)2= 0
   =>x=6
   y² +22y+121=0
   ( y+ 11) 2 = 0
   y= -11
   Hence x> y
   

    

5. I. x² +8x +15 = 0
   II. y² -10y + 24=0
   
   x>y
   x>=y
   x< y
   x<=y
   it cannot be determined
   Option C
   x² + 8x +15 = 0
   x² +3x + 5x + 15 = 0
   x( x+ 3) + 5( x+3)=0
   (x+3) (x+5)=0
   x=-3 or x= -5
   y² -10y + 24=0
   y² -4y -6y+ 24=0
   y(y-4) -6( y-4)=0
   (y-4) ( y-6) =0
   y= 4 or 6
   so x< y
   

    

6. I. x² -2x +1 = 0
   II. y² +8y + 16=0
   
   x
   x<=y
   x> y
   x>=y
   it cannot be determined
   Option C
   x² -2x +1 = 0
   (x-1)2= 0
   =>x=1
   y² +8y + 16=0
   ( y+ 4) 2 = 0
   y= -4
   Hence x> y
   

    

7. I. x² -3x -10 = 0
   II. y² = 4
   
   x
   x<=y
   x>y
   x ≥ y
   it cannot be determined
   Option D
   x² -3x -10 = 0
   x² -5x + 2x -10 = 0
   x( x-5) + 2( x-5)=0
   (x-5) (x+2)=0
   x=5 or x= -2
   y 2 = 4
   y= ± 2
   Hence x ≥ y
   

    

8. I. x² +24x + 144=0
   II. y² -18y +81 = 0
   
   x
   x<=y
   x>y
   x>=y
   it cannot be determined
   Option A
   x² +24x + 144=0
   ( x+ 12) 2 = 0
   x= -12
   y² -18y +81 = 0
   (y-9)2= 0
   =>y=9
   Hence y> x
   

    

9. I.2x² +13x +15 = 0
   II. y² -6y -7=0
   
   x
   x<=y
   x>y
   x>=y
   it cannot be determined
   Option A
   2x² +13x +15 = 0
   2x² +3x +10x +15 = 0
   x( 2x + 3) + 5( 2x + 3)=0
   (2x+3) ( x+5)=0
   x= -3/2 or -5
   y² -6y -7=0
   y² -7y + y-7=0
   y( y-7) + 1( y-7)=0
   ( y-7) ( y+1)=0
   y= 7 or -1
   Hence y> x
   

    

10. I. x² -2x -8=0
    II. 4y² +14y +6 = 0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x² -2x -8=0
    x² -4x + 2x-8=0
    x( x-4) + 2( x-4)=0
    ( x-4) ( x+2)=0
    x= 4 or -2
    4y² +14y +6 = 0
    4y² +2y +12y +6 = 0
    y( 4y + 2) + 3( 4y + 2)=0
    (4y+2) ( y+3)=0
    y= -2/4 or -3
    Hence x>y
    

     

11. I. 3x² +16x +5 = 0
    II. 5y²+ 37y +60=0
    
    x
    x<=y
    x>y
    x≥y
    it cannot be determined
    Option D
    3x² +16x +5 = 0
    3x² +x +15x +5 = 0
    x( 3x + 1) + 5( 3x + 1)=0
    (3x+1) ( x+5)=0
    x= -1/3 or -5
    5y²+ 37y +60=0
    5y² + 25y +12y +60
    5y(y+5)+12(y+5)=0
    ( y+5)( 5y +12) =0
    y= -5 or -12/5
    Hence x≥y
    

     

12. I. 2x² -13x +20 = 0
    II. 3y²-10y +7=0
    
    x
    x<=y
    x>y
    x<=y
    it cannot be determined
    Option C
    2x² -13x +20 = 0
    2x² -8x -5x +20 = 0
    2x(x-4) – 5(x-4) = 0
    (x-4) ( 2x-5)=0
    x= 4 or 5/2
    3y² -10y +7 = 0
    3y² -3y -7y +7 = 0
    3y(y-1) – 7(y-1) = 0
    (y-1) ( 3y-7)=0
    y= 1 or 7/3
    Hence x>y
    

     

13. I. x² -12x +20 = 0
    II. 6y² -29y +20 = 0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x² -12x +20 = 0
    x² -10x -2x +20 = 0
    x(x-10) – 2(x-10) = 0
    (x-10) ( x-2)=0
    x= 10 or 2
    6y² -29y +20 = 0
    6y² -24y -5y +20 = 0
    6y(y-4) – 5(y-4) = 0
    (y-4) ( 6y-5)=0
    y= 4 or 5/6
    Hence x>y
    

     

14. I. x² -6x +8 = 0
    II. y² -6y +5 = 0
    
    x
    x<=y
    x>y
    x>=y
    Hence the relation cannot be determined
    Option E
    x² -6x +8 = 0
    x² -4x -2x +8 = 0
    x(x-4) – 2(x-4) = 0
    (x-4) ( x-2)=0
    x= 4 or 2
    y² -6y +5 = 0
    y² -y -5y +5 = 0
    y(y-1) – 5(y-1) = 0
    (y-1) ( y-5)=0
    y= 1 or 5
    Hence the relation cannot be determined
    

     

15. I. x² -12x +32 = 0
    II. y² -20y +100 = 0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x² -12x +32 = 0
    x² -8x -4x +32 = 0
    x(x-8) – 4(x-8) = 0
    (x-8) ( x-4)=0
    x= 8 or 4
    y² -20y +100 = 0
    (y-10) ^2=0
    y= 10
    Hence y> x
    

     

16. I. x² -x -30 = 0
    II. y 2 = 169
    
    x
    x<=y
    x>y
    x>=y
    Hence the relation cannot be determined
    Option E
    x² -x -30 = 0
    x² -6x + 5x -30 = 0
    x( x-6) + 5( x-6)=0
    (x-6) (x+5)=0
    x=6 or x= -5
    y 2 = 169
    y= ±13
    Hence the relation cannot be determined
    

     

17. I.x² + 3x -10 = 0
    II. y² -13y/2 + 3=0
    
    x< y
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x² + 3x -10 = 0
    x² -2x + 5x – 10 = 0
    x( x- 2) + 5( x-2)=0
    (x-2) (x+5)=0
    x=2 or x= -5
    y² -13y/2 + 3=0
    y² –y/2 -6y+ 3=0
    y(y-1/2) -6( y-1/2)=0
    (y-1/2) ( y-6) =0
    y= 1/2 or 6
    so x< y
    

     

18. I. x² -10x +25 = 0
    II. y² +14y + 49=0
    
    x
    x<=y
    x> y
    x>=y
    it cannot be determined
    Option C
    x² -10x +25 = 0
    (x-5)2= 0
    =>x=5
    y² +14y + 49=0
    ( y+ 7) 2 = 0
    y= -7
    Hence x> y
    

     

19. I. x² +30x +225 = 0
    II. y² -30y+225=0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x² +30x +225 = 0
    (x+15)2= 0
    =>x=-15
    y² -30y+225=0
    ( y- 15) 2 = 0
    y= 15
    Hence y>x
    

     

20. I. x² + 5x -36 = 0
    II. y² +3y + 2=0
    
    x
    x<=y
    x>y
    x>=y
    cannot be determined
    Option E
    x² + 5x -36 = 0
    x² -4x + 9x -36 = 0
    x( x-4) + 9( x-4)=0
    (x-4) (x+9)=0
    x=4 or x= -9
    y² +3y + 2=0
    y² +2y +y+ 2=0
    y(y+2) + 1( y+2)=0
    (y+2) ( y+1) =0
    y= -2 or -1
    so cannot be determined
    

     

21. I. x² +32x +256 = 0
    II. y² -28y + 196=0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x² +32x +256 = 0
    (x+16)2= 0
    =>x= -16
    y² -28y + 196=0
    ( y-14 ) 2 = 0
    y= 14
    Hence y>x
    

     

22. I. x² – 64=0
    II. y1/2 = 5
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x² – 64=0
    (x-8) (x+8)=0
    x=8 or x= -8
    y 1/2 = 5
    y= 25
    Hence y>x
    

     

23. I. x² -18x +77 = 0
    II. y² +8y +15=0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x² -18x +77 = 0
    x² -7x -11x +77 = 0
    x( x-7) -11( x-7)=0
    (x-7) ( x-11)=0
    x= 7 or 11
    y² +8y +15=0
    y² +5y +3 y+15=0
    y( y+5) + 3( y+5)=0
    ( y+5) ( y+3)=0
    y= -5or -3
    Hence x> y
    

     

24. I. x² -12x +11 = 0
    II. y² -5y -150=0
    
    x
    x<=y
    x>y
    x>=y
    Hence it cannot be determined
    Option E
    x² -12x +11 = 0
    x² -x -11x +11 = 0
    x( x-1) -11( x-1)=0
    (x-1) ( x-11)=0
    x= 1 or 11
    y² -5y -150=0
    y² +10y -15 y-150=0
    y( y+10) -15( y+10)=0
    ( y+10) ( y-15)=0
    y= -10 or +15
    Hence it cannot be determined
    

     

25. I. x² +3x +2 = 0
    II. y² + y =0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option B
    x² +3x +2 = 0
    x² +x +2x +2 = 0
    x( x+1) +2( x+1)=0
    (x+1) ( x+2)=0
    x= -1 or -2
    y² + y =0
    y ( y+1)=0
    y= 0 or -1
    Hence x ≤ y
    

     

26. I. x² -24x +119 = 0
    II. y²+ 6y + 8 =0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x² -24x +119 = 0
    x² -17x-7x +119 = 0
    x( x-7) -17( x-7)
    (x – 7)(x – 17) = 0
    x = 7 or 17
    y²+ 6y + 8 =0
    y² +4y +2y+8 = 0
    y( y+4) +2( y+4)
    (y+2)(y +4) = 0
    y= -2 or -4
    So, x >y
    

     

27. I. 6x² + x – 2 = 0
    II. 20y² – 13y + 2 = 0
    
    x
    x<=y
    x>y
    x>=y
    the relation cannot be determined.
    Option E
    6x² + x – 2 = 0
    (2x – 1)(3x + 2) = 0
    x = 1/2 or -2/3
    20y² – 13y + 2 = 0
    20y² – 8y -5y+ 2 = 0
    4y(5y -2) -1( 5y-2)=0
    (4y– 1)(5y – 2) = 0
    y = 1/4 or 2/5
    So the relation cannot be determined.
    

     

28. I. 3x² + 13x+ 14 = 0
    II. 2y4 – 3y² + 2 = 0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option B
    3x² + 13x+ 14 = 0
    3x² + 7x + 6x + 14 = 0
    (x + 2)(3x + 7) = 0
    x = -2, -7/3
    2y4 – 3y² + 2 = 0
    y4 -3y²/2 +1=0
    y4 + y²/2 -2y² +1=0
    (y² – 2)(y² – 1/2) = 0
    y= ±√ 2, , √1/2 or -√1/2
    So, y≥ x
    

     

29. I. x² – 2x – 48= 0
    II. 6y² + 35y + 25 = 0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option E
    x² – 2x – 48= 0
    x² +6x –8x-48 = 0
    (x – 8)(x + 6) = 0
    x = -6 or + 8
    6y² + 35y + 25 = 0
    6y² +15y +10y + 25 = 0
    (3y + 5)(2y + 5) = 0
    y = -5/3 or -5/2
    Hence it cannot be determined
    

     

30. I. x² -24x + 143 = 0
    II. y²– 6y + 8 = 0
    
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x² -24x + 143 = 0
    x² -13x-11x + 143 = 0
    (x – 11)(x – 13) = 0
    x = 11 or 1 3
    y²– 6y + 8 = 0
    y²– 2y -4y+ 8 = 0
    (y– 4)(y– 2) = 0
    y = 4 or 2
    So y < x
    

     

1. If the marked price of the Sofa is Rs.2800 and the Shop offers X % of discount. If he sold at the sofa at Rs.2400, then find the value of X?
   12
   13
   14
   15
   16
   Option C
   2800 * (100 – X)/100 = 2400

   100 – X = 600/7

   X = 100/7

   X = 14.28 %

    

2. A 72 litres of mixture of water and milk in the ratio of 4: 5. Then adding some amount of water and the volume of the mixture is increased by 88 liters. The quantity of the milk is how much percentage more or less than of water in the final mixture?
   17
   18
   19
   20
   10
   Option A
   Milk = 5/9 * 72 = 40

   Water = 4/9 * 72 = 32

   Added water quantity = 88 – 72 = 16

   Total quantity of water = 16 + 32 = 48

   Required percentage = [(48 – 40)/48] * 100 = 16.67 %

    

3. The income of Pattu is Rs. 8000 and he spends Rs. 3200 and save the remaining amount. What percent of salary he saves?
   50
   60
   70
   90
   75
   Option B
   Income = 8000

   Expenditure = 3200

   Savings = 8000 – 3200 = 4800

   Required percentage = 4800/8000 * 100 = 60 %

    

4. The ratio of income of Revathi and Gayathri is 4: 5. The sum of their expenditures is Rs. 40000 and the amount of the savings of Revathi is equal to the expenditure of Gayathri. Find the ratio of their incomes to the sum of their savings?
   2
   1.9
   1.8
   1.7
   1.6
   Option
   Income Expenditure Savings

   R … 4x … y … 4x – y

   G … 5x …. 4x – y …. 5x – (4x – y) = x + y

   Required ratio = (4x + 5x): (4x – y + x + y)

   = > 9x : 5x = 9 : 5

   = > (9/5) = 1.8

    

5. Ravi and Kavi shared a profit in the ratio of 5: 4. They invested for 7 months and 8 months respectively. Then find the ratio of their investments?
   1.9
   1.6
   1.4
   1.8
   1.6
   Option C
   Investment ratio * Period ratio = Ratio of profit

   Ratio of investment = Ratio of profit / Period ratio

   = > (5 / 7) : (4 / 8)

   = > 10 : 7

   = > (10/7) = 1.428

    

6. The sum Rs. 12000 invested at same rate of interest in simple and compound interest scheme and the difference between the interests is Rs. 270 for 2 years. Find the rate of interest?
   12
   13
   14
   15
   16
   Option D
   Difference = P * R * R / 100 * 100

   270 = 12000 * R * R / 100 * 100

   Rate of interest = 15%

    

7. What is the rate of interest on the amount Rs. 20000 invests in the compound interest scheme for 2 years and the amount obtained after 2 years is Rs. 26450?
   15
   16
   17
   18
   19
   Option A
   CA = P * (1 + R/100)n

   26450 = 20000 * (1 + R/100)2

   529/400 = (1 + R/100)2

   23/20 = (100 + R)/100

   115 = 100 + R

   R=15%

    

8. A plane covered a distance of 2100 km in 14 hours. Ratio of the speed of Helicopter to speed of the Plane is 4: 3. In how much distance will the Helicopter cover in 3 hours?
   300
   400
   500
   600
   700
   Option D
   Speed of P = 2100/14 = 150 kmph

   Speed of H = 4/3 * 150 = 200 kmph

   Distance covered by H = 200 * 3 = 600 km

    

9. A man rides a bicycle at the speed of 20 kmph and reaches the office at 15 minutes late. If he increases his speed by 5 kmph and reaches the office in time, then what is the distance between house and office?
   10
   15
   20
   25
   30
   Option D
   Distance = (20 * 25 * 15/60)/5

   = 25 km

    

10. The length of train X is 600 m . And it crosses train Y running in opposite direction in 36 seconds and train X also crosses a man standing in a platform in 36 seconds. What is the speed of train Y if the length of train Y is half of train X?
    10
    20
    30
    40
    50
    Option C
    Speed of train X = 600/36 * 18/5

    = 60 kmph

    (Speed of train Y + 60) * 5/18 = 900/36

    Speed of train Y= 30 kmph

     

11. What is the speed of the train if train crosses a lamp post in 18 seconds and crosses 125 m long tunnel in 36 seconds.
    23
    24
    25
    26
    27
    Option C
    Length of train = x

    Speed of train = y

    x = y * 5/18 * 18

    x = 5y

    (x + 125) = y * 5/18 * 36

    x + 125 = 10y

    5y + 125 = 10y

    y = 25 kmph

     

12. Pipe 1 can fill the tank in 20 hours and ratio of the efficiency of pipe 2 to B is 3: 2. Pipe 2 and pipe 3 together can fill the tank in 15 hours. In how many hour pipe 1, 2 and pipe 3 together can fill the tank?
    15
    10
    12
    14
    16
    Option B
    1 = 1/20

    2 = 3/2 * 20 = 30 hours

    2 + 3 = 1/15

    C = 1/15 – 1/30

    C = 1/30

    1 + 2 + 3 = 1/20 + 1/30 + 1/30

    = (3 + 2 + 2)/60

    = 7/60

    Required time = 60/7 hours

    From quantity II,

    Required time = 1/30 + 1/15

    = 3/30 = 1/10

    = 10 hours

     

13. Cistern A can fill the tank in 30 hours and Cistern B can fill the tank in 15 hours. In what time will the tank is completely fill, if both the cisters are opened together?
    5
    10
    15
    20
    25
    Option B
    Required time = 1/30 + 1/15

    = 3/30 = 1/10

    = 10 hours

     

14. Ratio of the ages of kanish and kannan is 4: 5 and after 4 years the age of ramesh is 22 years. If the average ages of three of them is 15 years, then what is the present age of Kannan?
    15
    20
    25
    30
    35
    Option A
    Ramesh’s present age = 22 – 4 = 18 years

    9x + 18 = 15 * 3

    9x = 27

    x = 3 years

    Kannan’s present age = 5 * 3 = 15 years

     

15. If the sum of the ages of two brothers is 45 years and the ratio of the ages of 1 to 2 after 21 years is 17: 12, then what is the present age of 2?
    15
    16
    17
    18
    19
    Option A
    1 + 2 = 45

    Sum of the ages of 1 and 2 after 21 years = 45 + 21 + 21 = 87 years

    2s age after 21 years = 12/29 * 87 = 36 years

    2’s present age = 36 – 21 = 15 years

     

Directions (1-5): The following table shows the total number of baskets manufactured by 5 different companies in a certain year.

1. Total number of baskets defaulted by the company C1 is approximately what percentage more/less than the total number of baskets defaulted by the company C5?
   
   5
   6
   7
   8
   9
   Option C
   Total number of baskets defaulted by the company C1
   = > 3200000*(2/100) = 64000
   Total number of baskets defaulted by the company C5
   = > 4000000*(1.5/100) = 60000
   Required % = [(64000 – 60000)/60000]*100 = 7 % more
   

    

2. Find the average number of baskets manufactured by all the given companies together?
   34.2
   32.4
   43
   43.1
   22
   Option A
   The average number of baskets manufactured by all the given companies together
   = > (32 + 26 + 48 + 16 + 40)/5
   = > 162/5 = 32.4
   

    

3. Find the difference between the total number of baskets sold by the company C2 to that of company C4?
   450000
   430000
   833200
   420000
   356700
   Option C
   The total number of baskets sold by the company C2
   = >2600000*(95/100)*(84/100) = 2074800
   The total number of baskets sold by the company C4
   = > 1600000*(97/100)*(80/100) = 1241600
   Required difference = 2074800 – 1241600 = 833200
   

    

4. Find the ratio between the total number of baskets defaulted by the company C4 to that of C3?
   56:89
   12:56
   34:76
   56:64
   65: 96
   Option E
   The total number of baskets defaulted by the company C4
   = > 26*(5/100) = 1.3 lakhs
   The total number of baskets defaulted by the company C3
   = > 48*(4/100) = 1.92 lakhs
   Required ratio = 1.3: 1.92 = 130: 192 = 65: 96
   

    

5. Total number of baskets sold by the company C1 is approximately what percentage of total number of baskets sold by the company C3?
   71
   72
   73
   74
   75
   Option E
   Total number of baskets sold by the company C1
   = > 3200000*(98/100)*(78/100) = 2446080
   Total number of baskets sold by the company C3
   = > 4800000*(96/100)*(72/100) = 3317760
   Required % = (2446080/3317760)*100 = 73.726 = 75 %
   

    

Directions (6-10): Following table shows the total number of people went for a different states in a certain week and the ratio of male and female also given.




6. Total number of people visited Tamilnadu and Goa together is approximately what percentage of total number of people visited Odissa and Rajasthan together?
   130 %
   125 %
   135 %
   145 %
   120 %
   Option C
   Total number of people visited Tamilnadu and Goa together
   = > 5200 + 3200 = 8400
   Total number of people visited Odissa and Rajasthan together
   = > 2600 + 3600 = 6200
   Required % = (8400/6200)*100 = 135 %

    

7. Find the ratio between the total number of male visited Maharashtra to that of total number of female visited Goa?
   5: 2
   1: 2
   9: 4
   7: 6
   5: 3
   Option A
   The total number of male visited Maharashtra
   = > 4500*(2/3) = 3000
   The total number of female visited Goa
   = > 3200*(3/8) = 1200
   Required ratio = 3000: 1200 = 5: 2

    

8. Find the difference between the total number of male visited Odissa and Rajasthan together to that of female visited the same places?
   400
   300
   500
   800
   200
   Option E
   The total number of male visited Odissa and Rajasthan together
   = > 2600*(5/13) + 3600*(5/9)
   = > 1000 + 2000 = 3000
   The total number of female visited Odissa and Rajasthan together
   = > 2600*(8/13) + 3600*(4/9)
   = > 1600 + 1600 = 3200
   Required difference = 3200 – 3000 = 200

    

9. Find the average number of female visited Tamilnadu and Rajasthan together?
   2270
   2650
   2830
   2750
   2380
   Option D
   The total number of female visited Tamilnadu and Rajasthan together
   = > 5200*(3/4) + 3600*(4/9)
   = > 3900 + 1600 = 5500
   Required average = 5500/2 = 2750

    

10. Total number of people visited Maharashtra is what percentage of total number of people visited Rajasthan?
    120 %
    125 %
    145 %
    140 %
    165 %
    Option B
    Required % = (4500/3600)*100 = 125 %

     

1. A CNC machine become 10% 20% & 40% more efficient when it is supported by three other supportive machine P,Q& R respectively. Find the net saving of fuel consumed by the CNC machine if it is supported by all the three supportive machine independently?
   58
   49
   66.8
   76.5
   64.3
   Option C
   Improvement in efficiency = Net fuel saving= Reduction in fuel consumed
   Consumption of fuel reduces by 10%,20%, and 40% successively.
   Let initial consumption of fuel be 100 Liter
   10% Improvement in efficiency = 10% Net fuel saving= 10% Reduction in fuel consumed
   Thus fuel used by main CNC machine after adding supportive machine P = 100-10 = 90 Liter
   20% Improvement in efficiency = 20% Net fuel saving= 20% Reduction in fuel consumed
   Thus fuel used by main CNC machine after adding supportive machine Q = 90- 18=72 Liter
   40% Improvement in efficiency = 40% Net fuel saving= 40% Reduction in fuel consumed
   Thus fuel used by main CNC machine after adding supportive machine R = 72-28.8=43.2 Liter
   Thus total fuel saved by main machine = 100-43.2 = 66.8 Liter
   = (66.8/100)*100 = 66.80%
   

    

2. Sara buys a Micro oven for Rs. 11000. After using it for an year she sell it at a loss of 15%. Then again she buys a new Oven at a price 15% higher than the previous one. She uses the new Oven for one year and sell it at a loss of 20% then find her average loss per year during this period?
   2100
   2090
   3010
   3000
   2200
   Option B
   Loss in first year = 15% 0f 11000
   =1650
   Cost of mobile in second year 115% of 11000
   = 11000 + 1650
   = 12650
   Loss in second year = 20% of 12650
   = 2530
   Total loss = 1650 +2530
   =4180
   Average loss= 4180/2 = 2090
   

    

3. A construction can be finished by X labour in 80 days. If 10 labours leave before the work started then it will take 20 extra days to complete the work. Find the number of days to complete the same work if 100 people are working.
   10
   20
   30
   40
   50
   Option D
   If X labour can be completed the work in 80 days
   Then 1 labor can complete the work in 80*X days ————–I
   After 10 labors leave the job total labors = (X-10) and work completed in 80+20= 100 days
   Now,
   (X-10) labours can complete the work in 100 days
   Then 1 labours can complete the work in 100*(x-10) days ————-II
   Equating I and II value
   80*X = 100*(X-10)
   X = 50 labours
   Thus 50 labours can complete the work in 80 days
   100 labours can complete the work in (80*50)/100 = 40 days
   

    

4. The average speed of a school van is 60km/h excluding its stoppage time and if stoppage time is included its average become 50km/hr. How many minutes does the school van stop in an hour?
   5
   10
   15
   20
   25
   Option B
   Let the distance travel by school van is 300 km (L.C.M of 60 & 50)
   Time taken when the speed is 60km/h = 300/60
   =5 hours
   Time taken when the speed is 50km/h = 300/50
   =6 hours
   In 6 hours bus travel only 5 hours and 1 hours stop on its stoppage
   Thus in one hour stoppage time =1/6 hour
   =(1/6*60) minutes
   = 10 minutes
   

    

5. The average salary of 25 executives is 55000 rupees. If 5 executive were dropped from it the average reduced by 5000. What is the average salary of the executives who were dropped?
   60000
   56000
   70000
   75000
   55000
   Option D
   Total salary of 25 = 25* 55000
   =1375000
   Average salary of remaining 20 = 55000 – 5000
   = 50000
   Total salary of remaining 20 = 50000 * 20
   = 1000000
   Sum of the salary or 5 who left = 1375000 – 1000000
   =375000
   Average salary or 5 who left 375000/5 =75000

    

6. 34484.78 ÷ 10.99 + (4.76/9) of 12149.56 + 55.87 % of 8999.5 = ?^2 – 451
   120
   121
   122
   123
   124
   Option E
   (34485/11) + (5/9)*12150 + (56/100)*9000 = x^2 – 451

   3135 + 6750 + 5040 + 451 = x^2

   15376 = x^2

   x = 124

    

7. If the radius of the spherical ball is 4.9 cm, what is the surface area of the ball?
   301.84 cm^2
   334.76 cm^2
   391.22 cm^2
   201.74 cm^2
   271.84 cm^2
   Option A
   Radius of the sphere = 4.9 cm

   Surface area of the square = 4 * 22/7 * r * r

   = 4 * 22/7 * 4.9 * 4.9 = 301.84 cm2

    

8. If a man sold the dress for Rs. 1200 and he earned X% of the profit. If he increase the selling price of the dress by Rs. 300 and he can makes the profit of (X + 30) %, then find the value of X?
   10
   20
   30
   40
   50
   Option B
   CP * (X + 100)/100 = 1200

   CP * (X + 130)/100 = 1500

   120000/(X + 100) = 150000/(X + 130)

   4X + 520 = 5X + 500

   X = 20 %

    

9. A blackboard is in rectangular shape.If the ratio of the length of the board to the side of the another square board is 5 : 8 and the ratio of the breadth of the board to the side of the square board is 1: 2. If the difference between the perimeter of the board and the square board is 56 cm, then what is the area of the rectanglular board?
   450
   470
   340
   360
   320
   Option E
   Ratio of length = 5: 8

   Ratio of the breadth = 1: 2

   Ratio of the length and breadth of the rectangle and side of the square = 5: 4: 8

   Perimeter of the rectangle = 2 * (l + b) = 2 * (9x) = 18x

   Perimeter of the square = 4 * a = 4 * 8x = 32x

   32x – 18x = 56

   x = 4 cm

   Area of the rectangle = l * b

   = (5 * 4) * (4 * 4)

   = 320 cm2

    

10. Ratio of the number of red, blue and yellow cards in the bag is 4: 3: 5. If 18 yellow cards taken out and 10 blue cards is added, then the ratio of red, blue and yellow cards become 6: 7: 3. What is the total number of cards in the bag at initially?
    72
    74
    76
    78
    80
    Option A
    Ratio of the red, blue and yellow cards = 4: 3: 5

    (5x – 18)/(3x + 10) = 3/7

    9x + 30 = 35x – 126

    26x = 156

    X = 6

    Total number of cards = 12x = 12 * 6 = 72

     

Find what comes in the question mark.

1. 0.8.. 3.8.. 12.6.. 44.8..?
   
   188.2
   180.2
   88.2
   198.2
   168.2
   Option A
   0.8 × 1 + 3 = 3.8
   3.8 × 2 + 5 = 12.6
   12.6 × 3 + 7 = 44.8
   44.8 × 4 + 9 = 188.2
   

    

2. 33 .. 110.. 453.. 2282.. ?
   
   13711
   12467
   12345
   13423
   15672
   Option A
   13 × 2 + 7 = 33
   33 × 3 + 11 = 110
   110 × 4 + 13 = 453
   453 × 5 + 17 = 2282
   2282 × 6 + 19 = 13711
   

    

3. 664.. 334.. 85.5.. ?.. 4.03125 ..
   
   16
   15
   25.25
   16.25
   15.25
   Option D
   664 + 4 ÷ 2 = 334
   334 + 8 ÷ 4 = 85.5
   85.5 + 12 ÷ 6 = 16.25
   16.25 + 16 ÷ 8 = 4.03125

    

4. 1 .. 3.. 7.. 13.. 22.. ? ..
   
   30
   34
   21
   31
   33
   Option
   1 + 1 + 1 = 3
   3 + 2 + 2 = 7
   7 + 3 + 3 = 13
   13 + 4 + 5 = 22
   22 + 5 + 7 = 34
   

    

5. 5 .. 41.. 321.. 1921.. ?..
   
   7681
   Option
   5 × 10 – 9 = 41
   41 × 8 – 7 = 321
   321 × 6 – 5 = 1921
   1921 × 4 – 3 = 7681
   

    

6. 384.. 381.. 372.. 345.. ?
   
   100
   101
   102
   103
   104
   Option C
   -3, -9, -27 ,-81 -243, ….
   

    

7. 12.. 14.. 32.. 102 ..?..
   
   Option
   clue: ×1 + 2, ×2 + 4, ×3 + 6, ×4 + 8, ×5 + 10
   

    

8. 841.. 961… 1089.. ?.. 1369 ..
   
   1225
   1125
   1325
   1432
   1200
   Option A
   clue: +120, +128, +136, +…. +160, ….
   

    

9. 1108.. 1117.. 1142.. 1191 .. ?..
   
   Option
   clue: +3^2, +5^2, +7^2, +11^2, +13^2,….
   

    

10. 529.. 841.. 961.. ? .. 1681
    
    1432
    1345
    1234
    1369
    1145
    Option
    clue:square of prime numbers from 23

     

1. 4^4 * 9^3 / 2^3 * 3^2 = ?
   2592
   2345
   2564
   2568
   2266
   Option A
   4^4 * 9^3 / 2^3 * 3^2 = ?
   => ( 2^2)4 *( 3^2) 3/ 2^3 * 3^2
   =>2 ^8 * 3 ^6 / 2^3 * 3^ 2
   =>2 ^8-3 * 3^ 6-2
   =>32* 81
   =2592
   

    

2. 120% of 4000 + ? = 125% of 2200
   2100
   2200
   2050
   2450
   2300
   Option C
   120% of 4000 + ? = 125% of 2200
   =>4800 – 2750
   =>2050
   

    

3. ½ of ? + 2/3 of 117 = 20% 0f 400
   3
   4
   6
   5
   7
   Option B
   ½ of ? + 2/3 of 117 = 20% of 400
   ½ of ? + 2*39 =80
   1/2of ? +78 =80
   So ½ of ? = 2
   ?= 4
   

    

4. 5 1/5 + 3 1/3 + 4 1/6 -1 4/9 =?
   11 23/90
   11 21/90
   11 25/90
   12 23/90
   14 23/90
   Option A
   5 1/5 + 3 1/3 + 4 1/6 -1 4/9 =?
   =>(5+3+4-1) +( 1/5 + 1/3 + 1/6 – 4/9)
   =>11 + (18+30+15-40)/90
   =>11+(23)/90
   =>11 23/90
   

    

5. ½ of ¼ of 1/5 of 1/6 of 1000=?
   4.6
   4.4
   4.2
   4.5
   4.7
   Option C
   ½ * ¼ * 1/5 * 1/6 *1000
   =>1000/240= 4.2
   

    

6. 16.67% of 36 + 33.33 % of 123*2 – √121 =?
   55
   66
   77
   88
   99
   Option C
   ?=1/6 of 36 + 1/3 of 123 *2 – 11 [16.67 % = 1/6, 33.33 % = 1/3]
   ?= 6+41*2-11
   ?= 6+82-11
   ?=77
   

    

7. 34*36 + √529 * √121 – 93= ?
   648
   548
   878
   748
   458
   Option D
   ? = (35 – 1) (35 + 1) + 23 *11 – 729 [a2 – b2 = (a + b) (a – b)]
   ? = 352 – 12 + 23 *11 – 729
   ? =1225 – 1+ 253 – 729
   ? =748
   

    

8. ?/√384 = √864/?
   22
   23
   24
   25
   26
   Option C
   ? * ? = √384 *√864
   ? * ? = √ (64x6x144x6)
   ? * ? =8x6x12 =576
   ? = 24
   

    

9. (6 x 136÷8 +132) / (628 ÷ 16 – 26.25) =?
   17
   18
   19
   20
   21
   Option B
   (6 x (136/8) + 132) / (628/16 – 26.25) =?
   (102 + 132) / (39.25 – 26.25) = ?
   234 / 13 =?
   18 = ?
   

    

10. 42.85 % of 140 + 62.5 % of 120 – 16.67 % of 36 =?
    123
    125
    127
    129
    131
    Option D
    3/7 of 140 + 5/8 of 120 – 1/6 of 36 =?
    [42.85 % = 3/7, 62.5 % = 5/8, 16.67 % = 1/6]
    ? = 60 + 75 – 6 = 129
    

     

Directions (1-5): Study the following information carefully and answer the questions given below.

Three rooms 1, 2 and 3 consume some units by using AC, fridge and other appliances.
Room 1: AC0 consumes 120 units which is 20% more than the other appliances. A total unit consumed by AC, fridge and other appliances is 300 units.
Room 2: Number of units consumed by AC is half the number of units consumed by fridge in room 1. Number of units consumed by fridge is half the number of units consumed by AC in room 1. Total number of units consumed by AC, fridge and other appliances is 250 units.
room 3: AC consumes 60 units. Average number of units consumed by AC, fridge and other appliances is 100 units. Ratio of the number of units consumed by fridge to other appliances is 1: 2.

1. What is the difference between the total units consumed by AC in all the three rooms together and the total units consumed by fridge in all the three rooms together?
   
   0
   1
   2
   3
   4
   Option A
   Room AC fridge other appliances
   1 ….. 120 …… 80( 300 –( 120+100) …… 100 ( 120/120 *100)
   2 …… 40 ( 80/2) ….. 60( 120/2) …… 150 ( 250-( 60+40)
   3 ……. 60 ……. 80 ( 300-60)*1/3 160 ( 300-60)*2/3

   Required difference = (120+40+60) – (80+60+80)
   = 220 – 220 = 0
   

    

2. 2) Number of units consumed by AC in roomB is what percentage of the number of units consumed by other appliances in roomC?
   15%
   25%
   35%
   45%
   55%
   Option B
   Required percentage = 40/160*100 = 25%

    

3. 3) What is the ratio between the number of units consumed by other appliances in room B and roomC?
   15:1
   25:16
   35:16
   15:26
   15:16
   Option E
   Required ratio = 150: 160= 15:16

    

4. 4) Find the difference between the average units consumed by AC in all the three rooms together and the average units consumed by other appliances in all the three rooms together
   63.33%
   83.33%
   73.33%
   43.33%
   53.33%
   Option A
   Required difference = [(100+150+160)/3- (120+40+60)/3]
   = 410/3 – 220/3 = 190/3 = 63.33%
   

    

5. 5) Number of units consumed by fridge in roomA is what percentage more/less than the number of units consumed by AC in roomB?
   45%
   60%
   80%
   100%
   90%
   Option D
   Required percentage = (80-40)/40*100 = 100%

    

Directions (6-9): Study the following information carefully and answer the questions given below.


Myntra selling pants which is available in 3 types i.e., jeans, cotton, and synthetic. The pants are sold in three-time slots i.e., 4 am- 5 am, 7 am – 8 am and 10 am-11 am. The price per pant for the jeans, cotton and synthetic type is Rs. 400, Rs. 450 and Rs. 500 . And the total number of pants of all types sold in all three timings in a day is 16000.

4 am-5 am: 15% of the pants are sold in this time slot. The ratio of the number of cotton type pant sold to the number of jeans type pant sold is 7:2. The number of synthetic type pants sold is 600 hence the number of jeans type pants sold is (A) times the number of synthetic type pants sold. The total revenue generated by the jeans type and cotton type pant is Rs. (B) lakh.

7 am – 8 am: The number of pants sold in this time slot is 225% more than the number of pants sold from 4 am-5 am. The number of synthetic type pants sold is 25% more than the number of jeans type pants sold and the number of cotton pants sold is 11/5th of the number of synthetic pants sold. The revenue generated by selling jeans type pants is (C) % less than the revenue generated by selling synthetic type pants.

10 am-11 am: The remaining pants are sold in this time slot and the ratio of the number of jeans, cotton and synthetic pants sold is 8:15:6 respectively. The ratio of revenue generated by selling jeans type pants to the revenue generated by selling synthetic type pants is (D).

6. What will come at the place of ‘A’?
   1/3
   2/3
   4/3
   5/3
   6/5
   Option B
   4am-5am:
   Total number of pants sold = 16000*(15/100) = 2400
   The sum of the number of a jeans and cotton-type pant sold = 2400-600 = 1800
   Number of cotton type pant sold = 1800*(7/9) = 1400
   Number of jeans type pant sold = 1800-1400 = 400
   Number of jeans type pants sold = (2/3)*600 = 400
   Hence the number of jeans type pants sold is (2/3) (A) times the number of synthetic type pants sold.

    

7. What will come at the place of ‘B’?
   
   8.3lakh
   5.3lakh
   7.9lakh
   6.4lakh
   4.5 lakh
   Option C
   4am-5am:
   Total number of pants sold = 16000*(15/100) = 2400
   The sum of the number of jeans and cotton type pant sold = 2400-600 = 1800
   Number of cotton type pant sold = 1800*(7/9) = 1400
   Number of jeans type pant sold = 1800-1400 = 400
   Number of jeans type pants sold = (2/3)*600 = 400
   Hence the number of jeans type pant sold is (2/3) (A) times the number of synthetic type pant sold.
   Revenue generated by selling jeans type and cotton type pants = 400*400 + 450*1400 = 160000 + 630000 = 790000 = 7.9lakh (B)

    

8. What will come at the place of ‘C’?
   
   32%
   36%
   44%
   56%
   78%
   Option B
   7am – 8am:
   The number of pants sold = 2400*(100 +225)/100) = 2400*(325/100) = 7800
   Let the number of jeans type pant sold = 100x and the number of synthetic type pant sold = 100x*(100+25/100) = 100*(125/100) = 125x
   The number of cotton type pant sold = 125x*(11/5) = 275x
   100x+125x+275x = 7800
   500x = 7800
   x = 15.6
   Number of jeans type pants sold = 100x = 100*15.6 = 1560
   Revenue generated by selling jeans type pants = 1560*400 = 624000
   Number of synthetic type pants sold = 125x = 125*15.6 = 1950
   Revenue generated by selling synthetic type pants = 1950*500 = 975000
   Number of cotton type pants sold = 275x = 275*15.6 = 4290
   Required percentage = {(975000-624000)/975000}*100 = (351000/975000)*100 = 36% (C)

    

9. What will come at the place of ‘D’?
   
   11:15
   12:15
   16:15
   13:17
   22:16
   Option C
   10 am-11 am:
   The number of pants sold = 16000-(2400+7800) = 16000-10200 = 5800
   Number of jeans type pants sold = 5800*(8/29) = 1600
   Revenue generated by selling jeans type pant sold = 1600*400 = 640000
   Number of cotton type pants sold = 5800*(15/29) = 3000
   Revenue generated by selling cotton type pants = 3000*450 = 1350000
   Number of synthetic type pants sold = 5800*(6/29) = 1200
   Revenue generated by selling synthetic type pants = 1200*500 = 600000
   Required ratio = (640000/600000) =16:15 (D)

    

1. The L.C.M of 4/2, 2/6, 8/9 is
   5/3
   6/3
   7/3
   8/3
   10/3
   Option D
   L.C.M= L.C.M of numerator/H.C.F of denominator
   =L.C.M of 4,2,8/H.C.F of 3,6,9=8/3
   

    

2. The ratio between two number is 6:5 if their HCF is 10. Then the numbers are.
   10,20
   40,50
   60,50
   50,70
   30,80
   Option C
   Let the number be 6x, 5x
   HCF of 6x,5x=x
   ie) x=10(given)
   Then the numbers are 6x,5x=60,50

    

3. Find the average of first 14 natural number.
   105
   109
   111
   134
   156
   Option A
   average of first n natural number=n(n+1)/2
   here n=14
   average=14(14+1)/2=7×15=105
   

    

4. The sum of ages of three sister born of the intervals of 2 years each is 42 years. What is the age of the youngest sister ?
   11
   12
   13
   14
   15
   Option B
   x+(x+2)+(x+4)=42 (given the intervals are 2 years)
   3x+6=42  x =36/2=12
    Age of the youngest sister=12 years
   

    

5. Cost of a mobile is Rs .5000 and selling price is Rs. 7000.What is the profit percent?
   30
   35
   40
   45
   50
   Option C
   Profit % = p/c.p x 100
   =2000/5000 x 100
   P%=40
   

    

6. Ganesh can do a task in 4days . Prem can do a task in 3 days. How long if will take if both Ganesh &Prem work together?
   2 d 3 hr
   4d 6 hr
   6 d 7 hr
   1 d 5 hr
   4 d 3 hr
   Option D
   Ganesh 1day work=1/4
   Prem 1day work =1/3
   (Ganesh+Prem) =1/4+1/3=7/12
   So Ganesh& Prem will complete in 12/7=1 5/7=1day 5hrs together.
   

    

7. If a boat travels with a speed 25km/hr in still water and the speed of stream is 5km/hrs . What would be the time taken by boat to go 40 km upstream?
   2
   4
   6
   8
   10
   Option A
   Upstream speed = x-y km/hr
   =25-5=20 km/hr
   T=distance/speed
   T=40/20=2hrs
   

    

8. what is the simple interest on Rs 60000 at 5 ½% per annum for 2 years?
   5900
   6600
   7700
   6800
   7800
   Option B
   S.I = P X R X T/100=60000 x 5 ½ x 2/ 100
   S.I=Rs 6600
   

    

9. Ram deposits Rs .20,000 for 2years with 5% compound interest. What amount will be get at last?
   22050
   45000
   13500
   26500
   24000
   Option A
   Amount= P x (1+r/100)n
   =20000 X (1+5/100)2
   =20000 x 105/100)2
   =20000x 105×105/10000
   =Rs .22050
   

    

10. A florist choose one rose at random from the basket which have rainbow colors of roses. what is the probability of choosing yellow color rose?
    1/8
    1/7
    1/9
    1/5
    1/6
    Option B
    S=(rainbow color roses)
    S=(V,I,B,G,Y,O,R)=N(S)=7
    P(Yellow rose)=P(S)/N(S)=1/7 (Only one yellow rose)
    

     

11. How many multiples of 2 are there in 10 to 80?
    42
    40
    38
    36
    48
    Option D
    Multiples of 2 =>last digits must be 0, 2, 4, 6, 8
    And first digit 1,2 ,3,4,5,6,7,
    So there are 7 choices for the first number and 5 choices for the second number.
    So there are 5x 7 =3 5 + 1= 36 (include 80) multiples of 2 in between 10 and 80.
    

     

12. In a shop two partners invested in the ratio 10:20 and they got the profits in the ratio 5:4.If the first partner invest the money for 10 months the how many months the second partner invested?
    12
    14
    16
    18
    20
    Option C
    Investment ratio: 10:20
    Profit ratio :5:4
    First partner invested 10 months.
    Second parner invested x (take) months.
    So 10 x 20 / 10 x x = 5/4
    20/x = 5/4
    x = 16 months

     

13. Two glasses of equal capacity contains sand and color powder.The ratio of the mixing is 2:3 and 5:4 respectively. These mixture are transferred into a big bowl.Then what is the ratio of sand and color powder in the bowl?
    1:4
    5:7
    2:9
    7:7
    4:5
    Option D
    Ratio= 2 + 5 : 3+4 = 7: 7

     

14. The population of mosquitoes in a certain colony doubles every two hours. If 4 hours ago there were 1,000 mosquitoe in the colony, in approximately how many hours will the colony population exceed 250,000 mosquitoe?
    6
    8
    10
    12
    14
    Option D
    4 hrs ago= 1000
    now = 4000
    12 hrs= 2,56,000 So approximately 12 hrs.

     

15. A boy can work equal to 2 days of girl in a day. One child can work one-third of the work done by a girl in 1 day. If an company owner hires 52 pairs of hands of boys, girls and children in the ratio of 6 : 5 : 2 and pays them in all Rs.1060 at the end of the work, then what will be the daily wages (in rupees) of a child?
    
    4
    5
    6
    7
    8
    Option B
    1 boy = 2 girl and 1 girl = 3 children
    Work done by 1 boy= work done by 2 girls= work done by 6 children
    Ratio of their efficiency = 6 : 3 : 1
    Let an estate-owner hires boys, girls and children as 6x, 5x and 2x respectively.
    Given,
    6x+5x+2x= 52
    Or x=4
    Number of boys = 24
    girls=20 and children = 8
    As wages are proportional to the amount of work done, then
    Wages divided into ratio of = = 36 : 15 : 2
    Share of children=1060/53 *2 = 40
    Share of 1 child = 40/8=5

     

1. 1/2 + 3 – 4 * 5 + [6 – 7 * 8/4]
   -22.5
   -21.6
   -23.6
   -34.7
   -24.5
   Option E
   = 1/2 + 3 – 4 * 5 + [6 – 7 * 2]

   = 1/2 + 3 – 4 * 5 + [6 – 14]

   = 1/2 + 3 – 4 * 5 – 8

   = 0.5 + 3 – 4 * 5 – 8

   = 0.5 + 3 – 20 – 8

   = 3.5 – 20 – 8

   = -24.5

    

2. √727.5601* 4.99 -370 + [15.8 * √624.61]
   156
   165
   174
   189
   145
   Option B
   = 27 * 5 – 370 + [16 * 25]

   = 27 * 5 – 370 + 400

   = 135 – 370 + 400

   = 535 – 370

   = 165

    

3. [(6*7 – 3+2 * 8) + 9/3] * 5 – 1 + 7
   234
   296
   288
   124
   156
   Option B
   = [(6*7 – 3+2 * 8) + 9/3] * 5 – 1 + 7

   = [(42 – 3 + 16) + 3] * 5 – 1 + 7

   = [55 + 3] * 5 – 1 + 7

   = 58 * 5 – 1 + 7

   = 290 – 1 + 7

   = 296

    

4. [(10.09)^2 * (√289.06)] – [14 – 6 * 7 + (√169.009) + (11.02)^2]
   1594
   1678
   1254
   1890
   1567
   Option A
   = [100 * 17] – [14 – 42 + (13 + 121)]

   = 1700 – [14 – 42 + 134]

   = 1700 – 106

   = 1594

    

5. [√{3√(8.09) + 6.66} – 3.59 * ( 6.79 + √( 225.1) * 5.556 ) + 13]
   -560
   -345
   -372
   -457
   -569
   Option C
   = [√8.96 – 3.89 * (7 + 15 * 6) + 13]

   = [3 – 4 * 97 + 13]

   = [3 – 388 + 13]

   = -372

    

6. 21.8% of 167.99 – 19% of 320 + 24% of 460 – 13% of 110
   45.78
   34.89
   72.26
   67.90
   78.90
   Option C
   = [20%+2%] of 168 – [ 20%-1%] of 320+ [ 20% + 2% + 2%] of 460 – [ 11% + 1% +1%] of 110

   = [33.6 + 3.36] – [64 – 3.2] + [92 + 9.2 + 9.2] – [12.1 + 1.1 + 1.1]

   = 36.96 – 60.8 + 110.4 – 14.30

   = 147.36 – 75.1

   = 72.26

    

7. 127.81x + √1680.99 = 12.0993 + 169.099
   13.5
   15.5
   14.5
   12.5
   15.6
   Option C
   128x + 41 = 123 + 169

   128x = 1728 + 169- 41

   128x = 1856

   x=14.5

    

8. x^3 – 149.796 * 1.5 = (4.91)^2 * 9.099 + 176.82% of (10)^3
   12
   13
   14
   15
   16
   Option B
   x3– 150 * 1.5 = 25 * 8 + 177.2% of 1000

   x3– 225 = 200 + 1772

   x3 = 1972 + 225= 2197

   x= 13

    

9. x% of 159.9 = √[(√676.089) + 110% of 111 + 21.09% of 100.0998]
   8.125%
   7.89%
   9.56%
   6.45%
   7.45%
   Option A
   x% of 160 = √[ 26 + 110% of 111 + 20% of 100 ]

   x% of 160 = √[ 26 + 122 + 21]

   x% of 160 = √169

   x=(13/160)*100 = 8.125%

    

10. 2^x = [(31.901 * 4.952)/3.91] * [10.24 % of √2499]
    14
    10
    12
    15
    17
    Option B
    2x = [32 * 25/4] * [10.24% of 50]

    = 200 * 10.24 * 50/100

    = 1024

    x = 10