1. Raju spent Rs. 6000 to produce 150 jack fruits. He marked each fruit 10% above its cost price and sold each fruit at 5 % discount. What is the total profit Raju earns?
   310
   270
   239
   275
   125
   Option B
   Cost price of a single fruit = 6000 /150 = Rs. 40.
   Marked price = 40 X 110 /100 =Rs.44.
   Selling price = 44 x (100 – 5)/ 100 = 44 x 95 / 100 = Rs. 41.80
   Therefore total profit = 41.80 x 150 – 6000 = 6270 -6000 = Rs.270.
   

    

2. Three friends planned to start a new project with Rs. 5, 00, 000 investment. Initially friend 1 and 2 gave 1,00,000 and 2,00,000 respectively. Two months after friend 3 gave Rs. 25,000 and gets Rs. 20000 as his share profit. Then find the sum of the shares of friends 1 and 2?
   4,12,000
   2,20,000
   1,20,000
   3,20,000
   1,76,000
   Option D
   Investment of Friend 1 = 1,00,000 x12
   Investment of Friend 2 = 2,00,000 x12
   Investment of Friend 3 = 25000 x 9
   Ratio of their investment = 12,00,000 : 24,00,000 : 2,25,000
   = 16 : 32 : 3
   Profit of 3 = (3 / 51) x Total investment = Rs.20,000
   Total investment = (20000 x 51) / 3 = 3,40,000
   So the sum of profit of friend 1 and 2
   = (16 + 32) / 51 x 3, 40 ,000=Rs.3,20,000.
   

    

3. The bank manager forms a secret 2 – digit code from the numbers 0 -9. But he set code as the first digit will not be 0 and the second number will not be even number. Then what are the possible ways to set the code?
   45
   40
   55
   65
   35
   Option B
   First number will not be zero implies there are 8 possible way for digit one . And for second digit we have 5 possibilities(1, 3 ,5 7, 9 ) as should not use the even numbers .So the possible number of ways = 8 x 5 =40.

    

4. Two boats travel in upstream, the speed of boat 1 is 10 km/hr , boat 2 is 20 km/hr. Find the distance between them after 2 hours?
   25
   10
   20
   30
   15
   Option C
   Speed of the stream = 10 + 20 / 2 = 15 kmph
   Boat 1 upstream speed = 10 – 15 = -5 kmph
   Boat 2 upstream speed = 20 – 15 = 5 kmph
   Negative sign indicates the boat cannot move upwards , so it’s drifting back.
   Distance of boat 1after 2 hrs = -5 x 2 = -10 km
   Distance of boat 2 after 2 hrs = 5 x 2 = 10 km
   So the gap = 10 – (-10) = 20 km
   

    

5. A vessel has 50 litre of juice. From the vessel , 5 litres juice was taken out and replaced with an equal quantity of water . If 5 litres of newly formed mixture is taken out of the jar, then what is the total quantity of juice left in the vessel?
   40.5
   45.9
   44.8
   42.5
   46.5
   Option A
   Final quantity left in the jar = 50 ( 1-5/50)2 =40.5 lt

    

6. 9 litres oil are drawn from a drum full of sesame oil and it is then filled with the same quantity of sunflower oil. 9 litres are drawn again filled with the same quantity of sunflower oil. The quantity of sesame oil now left in the drum to that of sunflower oil in it is 16:9.Find the capacity of the drum?
   55
   65
   25
   15
   45
   Option E
   Let the capacity of the drum is x lt.
   So ( 1-9/x)2 = 16/25
   1-9/x= 4/5
   =>x=45lt
   

    

7. Arun purchased 30 kg of wheat at the rate 15 kg and mixed with 10kg of rice purchased at the rate of 12. He wants to earn a profit of 10/kg by selling it.What should be the selling price of the mix per kg?
   24.25
   23.76
   22.15
   24.50
   24.75
   Option A
   Selling price of the mix per kg= ( 30*15+ 10*12)/( 30+10) + 10
   =450+ 120/ 40 + 10
   = Rs. 24.25
   

    

8. The average age of some boys and 5 girls is 8 years. the sum of the ages of 5 girls is 24 years and average age of boys is 20 years. find the number of boys.
   2
   1
   5
   3
   7
   Option B
   Let the number of boys = n
   So ( n+ 5) * 8 = 24+ 20n
    n= 1
   

    

9. The difference between the present age of friend 1 and friend 2 is 14 years. Seven years ago the ratio of their ages was 10:14 respectively. What is friend 1’s present age?
   100
   102
   105
   104
   107
   Option C
   Let friend 2 present age = 14x+7 years
   Friend 1 age = 10x+ 7 years
   14x-10x= 28
    x= 7
    so friend 2 age = 98+7 = 105 years.
   

    

10. Hema invested an amount for 2 years at 20 percent per annum at simple interest .Had the interest been compounded annually. she would have earned Rs.500 more as interest.What was the amount invested?
    12,500
    10,800
    11,800
    14,500
    15,900
    Option A
    Difference between CI and SI = 500
    500 = p*20*20/ 100 *100
    =>p=Rs. 12,500
    

     

1. There are 3 members in a family – husband, mate and their daughter. Husband’s age is thrice his daughter’s age and mate is 3 years younger than his husband. what’s the individual ratio of ages of daughter, husband and mate if their average age is 70?
   30: 90:87
   20:34:90
   10:40:90
   35:45:80
   15|:25:65
   Option A
   Let daughter’s age = x
   Husband’s age = 3x
   mate’s age = 3x – 3
   ⇒ x + 3x + (3x – 3) = 70 × 3
   ⇒ 7x =210
   ⇒ X = 30
   Daughter’s age = 30 years
   Husband’s age = 30 × 3 = 90 years
   mate age = 30 × 3 – 3 = 87 years
   Required ratio = 30 : 90 : 87
   

    

2. A spherical tank of surface area 616 sq m crammed with water planned to hide with a lid of the biggest circle size lying on the tank . Then what’s the diameter of the circular lid?
   12
   16
   14
   15
   11
   Option C
   Surface area of spherical vessel = 616 cm2
   ∴ 4πr2 = 616
   r2 = 616/ 4π =49

   ⇒ r = 7 cm
   ∴ Diameter of the lid of largest circle lying on vessel = 2r = 2 × 7 = 14 cm
   

    

3. If the average of six numbers is 30 and the average of first three numbers is 20 and the last two number is 19,What is the fourth number from the starting?
   23
   78
   87
   82
   90
   Option D
   Let the five numbers be a, b, c, d, e, f
   Given (a + b + c) / 3 = 20
    a+ b + c = 60
   (e + f)/2 = 19
    e+ f = 38
   a+ b+ c + d + e +f = 30 X 6 = 180
   60 + d + 38 = 180
    d = 180 – 60 – 38 = 82
   so the fourth number is 82.
   

    

4. Three students on their flash lights and turn off after every 12sec, 15sec , 20 sec respectively. If they all turn off simultaneously at 09:00: 00 hrs. If they continue this play then at what time they are expected to turn off their lights?
   8.25
   9.01
   8.09
   9.04
   8.00
   Option B
   Interval of change the light = l c m ( 12 , 15 , 20 )
   = 60 sec = 1 min
   Hence next at 9: 01 :00 they will turn off the light simultaneously.
   

    

5. In a population 20% of women are 200 less than three fifth of the total women population .Then how many number of women are there?
   500
   400
   200
   600
   700
   Option A
   Let the population of the women be x.
   Then (3/5) x – 200 = ( 20% of x )
   3 x/5 – ( 20 x / 100 ) = 200
   x = 500
   

    

6. A father decided to give his retirement amount Rs. 7,40,000 to his sons A, B , C in the proportion 1/3 : 2/ 5 : 3/6. Then what is the average amount of first two sons got from their father?
   1,20,00
   2,00,000
   2,20,000
   3,00,000
   4,00,000
   Option C
   Given proportion = 1/3 : 2/5 : 3 /6 = 10 : 12 : 15
   A ‘s share = (10/ 37 ) x 7,40,000 = Rs. 2,00,000
   B’ s share = (12 / 37) x 7, 40 , 000 = Rs. 2, 40, 000
   Hence average =( 200000 + 240000 )/ 2 = Rs. 2, 20 , 000.
   

    

7. Three years back the age of brother and sister was 5: 6 respectively. Five years after the ratio of their age become 7 :8 .Then what is the age of the brother ?
   10
   20
   30
   15
   25
   Option B
   Let the brother age be 5x and the sister age be 6x.
   Five years after brother age 5x + 3 + 5 = 5x + 8
   Five years after sister age 6x + 3+ 5 = 6x + 8
   So 5x + 8 / 6x + 8 = 7 / 8
    8 ( 5x + 8) = 7 ( 6x + 8)
    X =4 yrs
   Age of the brother is = 5 x 4 = 20 years
   

    

8. Two friends planned to buy a land into a partnership. Friend A invests Rs 45,000 and friend B invests Rs. 20,000. After 6 months Friend C has joined with and ready to give certain amount . If the car company gave the total discount Rs 20,000 and so Friend B got Rs.2000 back. Then What is the investment of Friend C?
   16,20,000
   15,90,000
   14,35,000
   13,90,000
   17,90,000
   Option A
   Let the investment of Friend A, B, C be x, y , z.
   The discount ratio is d1,d2,d3.
   x : y: z = 45,000 x 12 : 20,000×12 : z = 5,40,000 : 2,40,000 :z ( 12 months)
   total investment : B investment :: total discount : B discount
   7,80,000 + z : 2,40,000 :: 20,000: 2000
   So , 2,40,000/ ( 7,80,000 + z) = 2000 / 20000
   =>z=16,20,000
   

    

9. Three boats travelling at 10 km /h , 15 km/h and 20 km/h respectively in still water And they travel face to face directly to each other. At present they are 50 km apart. How many kilometers are they one minute before they clash each other?
   3.8
   4.8
   5.9
   5.8
   6.5
   Option C
   One minute before clash distance = Total – ( total distance of three boats before a minute)
   = 50 – ( 10 x 59 /60 – 15 x 59 /60 – 20 x 59 /60)
   = 50 – ( 9.8 + 14.6 + 19.7 )
   = 5.9 km
   

    

10. Vinay wants to travel from Delhi to Banglore which is 600 kilometer. If he opt for train , it takes 12 hrs more than the flight. If the speed of the train is doubled ,it takes ten hours less than the flight. what’s the speed of the flight?
    300
    200
    400
    250
    342
    Option A
    Speed of the train = 600 / 12 = 50 km / hr
    If it speed doubled = 50 x 2 = 100 km/hr
    So 100 / 10 = 10 times increased.
    Then original time 12-10 = 2 hrs
    Therefore speed = 600 / 2 = 300 kmph .
    

     

1. I. 10x²+9x+2=0
   II. 6y²– 27y-15=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option D
   10x²+9x+2=0
   10x² + 5x+ 4x+2=0
   5x(2x +1) + 2(2x +1)=0
   (2x+1) ( 5x+2)=0
   x= -1/2 , -2/5
   6y²– 27y-15=0
   6y² -30y+ 3y -15=0
   6y( y-5) + 3(y-5)=0
   (y-5) ( 6y+3)=0
   y= 5, -3/6
   Hence y≥x
   

    

2. I. 35x²+27x+4=0
   II. 9y²– 3y-2=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option E
   35x²+27x+4=0
   35x² + 20x+ 7x+4=0
   5x(7x +4) + 1(7x +4)=0
   (7x+4) ( 5x+1)=0
   x= -4/7 , -1/5
   9y²– 3y-2=0
   9y² -6y+ 3y -2=0
   3y( 3y-2) + 1(3y-2)=0
   (3y-2) ( 3y+1)=0
   y= 2/3,- 1/3
   Hence the relation cannot be determined
   

    

3. I. 35x²-4x-15=0
   II. 10y²+129y-13=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option A
   35x²-4x-15=0
   35x² – 25x+ 21x-15=0
   5x(7x -5) + 3(7x -5)=0
   (7x-5) ( 5x+3)=0
   x= 5/7 , -3/5
   10y²+129y-13=0
   10y² -y+ 130y -13=0
   y( 10y-1) + 13(10y-1)=0
   (10y-1) ( y+13)=0
   y= 1/10,- 13
   Hence x>y
   

    

4. I. 10x² +79x+63=0
   II. 20y²+3y-2=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option C
   10x²+79x+63=0
   10x² +70x+9x+63=0
   10x(x +7) + 9(x +7)=0
   (x+7) ( 10x+9)=0
   x= -7 , -9/10
   20y²+3y-2=0
   20y² -5y+ 8y -2=0
   5y( 4y-1) + 2(4y-1)=0
   (4y-1) ( 5y+2)=0
   y= 1/4, -2/5
   Hence y>x
   

    

5. I. 2x²-31x= -120
   II. 8y²-74y=-143
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option A
   2x²-31x= -120
   2x² -16x-15x+120=0
   2x(x -8) -15(x -8)=0
   (x-8) ( 2x-15)=0
   x= 8 , 15/2
   8y²-74y=-143
   8y² -22y-52y+143 =
   2y( 4y-11) -13(4y-11)=0
   (4y-11) ( 2y-13)=0
   y= 11/4, 13/2
   Hence x>y
   

    

6. I. 8x²+41x+5=0
   II. 49y²-49y-18=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option C
   8x²+41x+5=0
   8x² +40x+x+5=0
   8x(x +5) + 1(x +5)=0
   (x+5) ( 8x+1)=0
   x= -5 , -1/8
   49y²-49y-18=0
   49y² -63y+ 14y -18=0
   7y( 7y-9) + 2(7y-9)=0
   (7y-9) ( 7y+2)=0
   y= 9/7, -2/7
   Hence y>x ( 9/7> -1/8) and( -2/7 > -5)
   

    

7. I. 40x²-59x+21=0
   II. 14y²+45y-14=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option A
   40x²-59x+21=0
   40x² -24x-35x+21=0
   8x(5x -3) -7(5x -3)=0
   (5x-3) ( 8x-7)=0
   x= 3/5 , 7/8
   14y²+45y-14=0
   14y² +49y- 4y -14=0
   7y( 2y+7) – 2(2y+7)=0
   (2y+7) ( 7y-2)=0
   y= -7/2, 2/7
   Hence x>y
   

    

8. I. 40x²-13x+1=0
   II. 10y²-11y-6=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option E
   40x²-13x+1=0
   40x² -8x-5x+1=0
   8x(5x -1) -1(5x -1)=0
   (5x-1) ( 8x-1)=0
   x= 1/5 , 1/8
   10y²-11y-6=0
   10y² -15y+4y -6=0
   5y( 2y-3) +2(2y-3)=0
   (2y-3) ( 5y+2)=0
   y= 3/2, -2/5
   Hence the relation cannot be determined
   

    

9. I. 10x²-9x+2=0
   II. 6y²+ 27y-15=0
   
   x > y
   x ≥ y
   x < y
   x ≤ y
   x = y or Cannot be determined.
   Option B
   10x²-9x+2=0
   10x² – 5x- 4x+2=0
   5x(2x -1) – 2(2x -1)=0
   (2x-1) ( 5x-2)=0
   x= 1/2 , 2/5
   6y²+ 27y-15=0
   6y² +30y-3y -15=0
   6y( y+5) – 3(y+5)=0
   (y+5) ( 6y-3)=0
   y= -5, +3/6
   Hence x≥y
   

    

10. I. 35x²+51x+18=0
    II. 9y² =1
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option C
    35x²+51x+18=0
    35x² + 30x+ 21x+18=0
    5x(7x +6) + 3(7x +6)=0
    (7x+6) ( 5x+3)=0
    x= -6/7 , -3/5
    9y² =1
    9y²-1=0
    (3y-1) ( 3y+1)=0
    y= 1/3,- 1/3
    Hence y>x
    

     

11. I. 35x²+16x-3=0
    II. 10y²+136y-56=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option E
    35x²+16x-3=0
    35x² -5x+ 21x-3=0
    5x(7x -1) + 3(7x -1)=0
    (7x-1) ( 5x+3)=0
    x= 1/7 , -3/5
    10y²+136y-56=0
    10y² -4y+ 140y -56=0
    y( 10y-4) + 14(10y-4)=0
    (10y-4) ( y+14)=0
    y= 4/10,- 14
    Hence the relation cannot be determined
    

     

12. I. 10x²-136x-14=0
    II. 25y²-10y-8=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option A
    10x²-136x-14=0
    10x² -140x+x-14=0
    10x(x -14) + 1(x -14)=0
    (x-14) ( 10x+1)=0
    x= +14 , -1/10
    25y²-10y-8=0
    25y² -20y+ 10y -8=0
    5y( 5y-4) + 2(5y-4)=0
    (5y-4) ( 5y+2)=0
    y= 5/4, -2/5
    Hence x>y
    

     

13. I. 11x²+120x= 11
    II. 8y²+74y+143=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option E
    11x²+120x= 11
    11x² +121x-x-11=0
    11x(x +11) -1(x +11)=0
    (x+11) ( 11x-1)=0
    x=- 11 , 1/11
    8y²+74y+143=0
    8y² +22y+52y+143 =
    2y( 4y+11) +13(4y+11)=0
    (4y+11) ( 2y+13)=0
    y= -11/4, -13/2
    Hence the relation cannot be determined
    

     

14. I. 8x²+51x +18=0
    II. 49y²-1=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option C
    8x²+51x+18=0
    8x² +48x+3x+18=0
    8x(x +6) + 3(x +6)=0
    (x+6) ( 8x+3)=0
    x= -6 , -3/8
    49y²-1=0
    (7y-1) ( 7y+1)=0
    y= 1/7, -1/7
    Hence y>x
    

     

15. 40x²-13x+1=0
    18y²+77y-18=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option B
    40x²-13x+1=0
    40x² -8x-5x+1=0
    8x(5x -1) -1(5x -1)=0
    (5x-1) ( 8x-1)=0
    x= 1/5 , 1/8
    18y²+77y-18=0
    18y² +81y- 4y -18=0
    9y( 2y+9) – 2(2y+9)=0
    (2y+9) ( 9y-2)=0
    y= -9/2, 2/9
    Hence x ≥ y
    

     

16. I. 16x²+60x+14=0
    II. 8y²– 77y-30=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option C
    16x²+60x+14=0
    16x² + 56x+ 4x+14=0
    8x(2x +7) + 2(2x +7)=0
    (2x+7) ( 8x+2)=0
    x= -7/2 , -2/8
    8y²– 77y-30=0
    8y² -80y+ 3y -30=0
    8y( y-10) + 3(y-10)=0
    (y-10) ( 8y+3)=0
    y= 10, -3/8
    Hence y>x
    

     

17. I. 35x²+32x+5=0
    II. 9y²– 20y+24=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option C
    35x²+32x+5=0
    35x² + 25x+ 7x+5=0
    5x(7x +5) + 1(7x +5)=0
    (7x+5) ( 5x+1)=0
    x= -5/7 , -1/5
    9y²– 36y+32=0
    9y² -12y-24y +32=0
    3y( 3y-4) -8(3y-4)=0
    (3y-4) ( 3y-8)=0
    y= 4/3, 8/3
    Hence y>x
    

     

18. I. 35x²-18x-81=0
    II. 10y²+184y-114=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option A
    35x²-18x-81=0
    35x² – 45x+ 63x-81=0
    5x(7x -9) + 9(7x -9)=0
    (7x-9) ( 5x+9)=0
    x= 9/7 , -9/5
    10y²+184y-114=0
    10y² -6y+ 190y -114=0
    y( 10y-6) + 19(10y-6)=0
    (10y-6) ( y+19)=0
    y= 6/10,- 19
    Hence x>y
    

     

19. I. 10x²+79x+63=0
    II.²+10y-2=0
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option C
    10x²+79x+63=0
    10x² +70x+9x+63=0
    10x(x +7) + 9(x +7)=0
    (x+7) ( 10x+9)=0
    x= -7 , -9/10
    48y²+10y-2=0
    48y² -6y+ 16y -2=0
    6y( 8y-1) + 2(8y-1)=0
    (8y-1) ( 6y+2)=0
    y= 1/8, -2/6
    Hence y>x
    

     

20. I. 2x²-18x= -17
    II. ²-97y=-117
    
    x > y
    x ≥ y
    x < y
    x ≤ y
    x = y or Cannot be determined.
    Option E
    2x²-35x= -17
    2x² -34x-1x+17=0
    2x(x -17) -1(x -17)=0
    (x-17) ( 2x-1)=0
    x= 17 , 1/2
    20y²-97y=-117
    20y² -45y-52y+117 =
    5y( 4y-9) -13(4y-9)=0
    (4y-9) ( 5y-13)=0
    y= 9/4, 13/5
    Hence the relation cannot be determined
    

     

1. A person sold his painting for Rs.1000 and earned profit , is twice the loss by selling it for Rs.800. Find out the selling price of the painting to get 50% profit?
   300
   400
   500
   600
   700
   Option A
   Let CP= Rs.x
   profit = 2 loss
   ie) 1000- x= 2( x-800)
   =>3x=1800
   =>x= CP= Rs.300
   

    

2. A vendor purchased 50 kg of tomatoes from Farm 1 at Rs. 20 per kg and 30 kg of tomatoes for Rs. 22 per kg. He mixed two farms tomatoes for sale..At what rate the vendor should be sold to get 10% profit?
   24
   27
   34
   22
   23
   Option E
   CP of two farms tomatoes= Rs.( 50 *20 + 30 *22)
   = Rs. 1660
   CP of 1 kg tomatoes= 1660/(50+30) =1660/80 = Rs.20.75
   Profit % = 10%
   =>SP of 1 kg tomatoes= 20.75 * 110/100
   = Rs.23 ( approx)
   

    

3. A show room offers 10 % on a marked price on silk saree. But they charges 8% GST on the discount price..If the customer pays Rs.4000 for the saree including GST. Find the marked price of the saree.
   6000
   5000
   4000
   5800
   4800
   Option B
   Let the marked price be Rs.x
   S.P = x*90/100 * 108/100 = Rs.4860
   => x= 4860*100 *100/ 90*108
   => x= Rs.5000
   

    

4. A man bought a bicycle in a discount of 20% on marked price.The bicycle seller got 4% profit. If the cost price of the cycle is Rs.5000 find its marked price?
   6500
   4400
   3466
   7500
   5500
   Option A
   Let the marked price be Rs. x.
   given 80% of x= Rs.5200
   =>100% of x= 5200/80 *100 = Rs. 6500
   

    

5. In a vegetable shop vegetables are selling for 5% loss on cost price. But the shopkeeper uses 15% less weight for all the vegetables. Find out whether he gets profit or loss?
   12.5
   13.7
   12.0
   11.8
   14.8
   Option D
   let CP = Rs.100
   SP at 10% loss= 100-5/100 * 100 = Rs.95
   Actual CP= 100 * ( 100-15)/100 =Rs.85
   So profit percentage = ( 95-85)/85 *100 = Rs.11.8%

    

6. The average profit of First six months of an year is Rs. 4000 and the average profit of next 3 months is Rs.3500. Find the average profit of all the 9 months.
   3833.3
   4589.6
   3487.9
   4321.8
   3277.9
   Option A
   Required profit average=( 6* 4000 + 3* 3500)/ 9
   = (24000+10500)/9
   = Rs.3833.3
   

    

7. In a water tanker drinking water at the rate of Rs.5 per liter and salt water at Rs. 3 per liter are mixed in the ratio 4:1 and sells at Rs. 6 per liter. Find the profit percentage earned ?
   12.8
   56.9
   30.4
   43.8
   15.9
   Option C
   CP= (5*4 + 3*1)/ (4+1) = Rs. 23/5 = Rs.4.6
   SP= Rs. 6
   Profit =Rs.( 6-4.6)=Rs.1.4
   Profit %= 1.4/4.6 *100 = 30.4%
   

    

8. How many liters of buffalo milk worth Rs. 20 per liter to be added with 50 liters of cow milk worth Rs. 30 per liter so that by selling the milk at Rs. 28 per liter there will be a profit 12%.
   10
   40
   20
   50
   60
   Option D
   Let the quantity of buffalo milk = X liter
   CP of the mixture = 28* 100/112 =Rs.25
   So, (20 *x+ 30*50)/ ( x+50) = 25
    x =50 liters
   

    

9. Car 1 has sold at a profit of 5% and Car 2 is sold at a profit of 15%.Find the ratio of two selling prices.
   11:12
   20:12
   21:22
   21:15
   21:23
   Option E
   Ratio of two selling price = 105/100 * 100 : 115/100 * 100
   = 21:23
   

    

10. In a police selection ,It is required to get 216 points out of maximum marks to select. Ravi gets 186 marks and is considered failed by 10 % marks. What are the maximum points to get for the selection?
    100
    300
    400
    500
    800
    Option B
    Let the maximum marks be x.
    given,
    10% of x = 216-186
    =>x/10 = 30
    =>x= 300
    

     

1. Ramu gave Rs. 5000 to Raju for 2 years and Rs.3000 to Kiran for 4 years on simple interest at the same rate of interest and received Rs.4400 in total from both as interest. Find the rate of interest per annum.
   20
   30
   40
   50
   60
   Option A
   Let the rate of interest =R%.
   Then 5000 * 2* R / 100 + 3000 * 4* R /100 = 4400
   100R+120R= 4400
   =>220R= 4400
   =>R= 20%
   

    

2. A and B borrows Rs.5000 and Rs.3000 respectively at the same rate of interest for 2 ½ years. If B paid Rs.200 more interest than A, Find the rate of interest.
   2
   4
   6
   8
   5
   Option B
   From the question,
   5000 * 5*R/ 100*2 – 3000 * 5 * R / 100*2 = 200 ( 2 ½ = 5/2)
   =>10000 R/ 200 =200
   =>R= 4%
   

    

3. The simple interest on a sum of money 3 years is Rs. 240. And the compound interest on the same sum at the same rate for 2 years is Rs.170.What is the rate of interest?
   22.2
   23.7
   25.6
   21.6
   22.9
   Option A
   Simple interest for 1 year = 270/3= 90 =PR/100
   For two years= 90*2= Rs.180
   Difference for 2 years= 180- 160 = Rs.20
   Difference between C.I and S.I = PR2 /1002
   Rate % = 20/90 * 100
   =22.2%
   

    

4. A person deposits some amount in a bank. After some years a sum of money is doubled itself at 2 ½ % simple interest for a year. Then find the time .
   30
   35
   40
   45
   55
   Option C
   Let Principal=P.
   =>Amount = 2P
   Simple interest = 2P-P= Rs.P
   SI= P*R*T/100
   P= P*5 *T/2*100
   =>200 = 5T
   T= 200/5= 40 years.
   

    

5. A person gave Rs.4000 for 3 years to his friend and gave Rs. 5000 for 4 years to his brother at the same rate of simple interest. Finally he receives Rs.1600 as interest from both ,What is the rate of interest per annum?
   6
   7
   2
   5
   4
   Option D
   Let the rate of interest = x%
   4000 * x * 3/ 100 + 5000 * x*4 /100 = 1600
   120x + 200x = 1600
   =>320x= 1600
   =>x= 5%
   

    

6. A person borrowed some amount at the rate of 12% per annum at simple interest .After 5 years 4 months , he paid Rs.720 as an interest .Find the amount borrowed by him?
   Rs.1235
   Rs.1125
   Rs.1254
   Rs.1200
   Rs.1300
   Option B
   R= 12%, T= 6 years 8months.
   5 years 4 months =5 4 /12 = 5 1/3 = 16/ 3 years.
   Interest =PRT/ 100
   720 =P *12*16/ 100 *3
   =>P=Rs. 1125
   

    

7. A deposits Rs. 20,000 as fixed deposits at a bank at the rate of 10% per annum simple interest. But for some reasons he has to withdraw the entire money after 2 years . So bank has given lower interest to him. But he got Rs.4000 less amount what he would wants to get after 5 years, then find the rate of interest given by the bank ?
   23
   24
   25
   32
   21
   Option C
   Principal= Rs. 20,000
   Rate%= 10%
   SI for 5 years= 20000*10 *5/100 =Rs.10,000
   After three years interest received = 10000-4000= Rs.6000
   Rate%= 6000/ 12000 * 100/2 = 25%
   

    

8. Sonu deposits Rs.2500 in 2 years. She receives 4% compound interest for the first year. Second year the bank hikes its interest rate as 8%.Find the difference of the amount she received in the two years.
   200
   123
   543
   300
   400
   Option A
   P= Rs.2500
   R1= 4%, R2=8%
   Amount for first year = 2500 ( 1+ 4/100) =Rs.2600
   Second year= 2600 (1+8/100)= Rs.2800
   Difference = 2800- 2600 = Rs. 200
   

    

9. Man invests ½ of his retirement amount at 2% in one bank and 1/5 amount at 5% in the second bank and 1/8 th amount in the third bank at 8% .If simple interest of 2 years amount he got is Rs.10,000, Find the total amount he invested?
   Rs.176,099
   Rs.1,66,667
   Rs.1,76,900
   Rs.1,23,765
   Rs.1,89,002
   Option B
   Let the sum be Rs.100.
   SI = (100/2 * 2/100 * 2) +( 100/5 * 5/100 *2) + ( 100/8 * 8/100* 2)
   = 6
   Actual Sum = 10000/6 * 100 =Rs.1,66,667
   

    

10. Arun invests sum amount in the compound interest at 25 % for 5 years. At the end of five years he got Rs.2500 as an interest amount. If he invests in simple interest then what will be the interest amount on same sum for same period and rate
    Rs.2300
    Rs.2000
    Rs.1200
    Rs.2040
    RS.1600
    Option B
    C.I = P ( ( 1+25/100)2 – P)
    = 25P/16
    Given C.I = Rs2500
    P=Rs. 1600
    S.I = 1600 * 25*5/100
    =>S.I= Rs.2000
    

     

1. Machines A and B Produce 100 products in 10 days, Machines B and C together can produce the same products in 12 days and machines A and C
   together can produce that products in 15 days. In how many days can A produce the same products alone?
   24
   22
   21
   23
   27
   Option A
   1 day work of A and B together = 1/10 units
   1 day work of B and C together = 1/12 units
   1 day work of A and C together = 1/15 units
   1 day work of A + B and B + C and A+ C together = 1/10 + 1/12 + 1/15 units
   ie)( 2A+2B+ 2C) = 15/60 = ¼ units
   =>1 day work of (A+ B+C)= 1/8
   1 day work of A= 1 day work of ( A+B+C) – 1 day work of ( B+C)
   = 1/8 – 1/12
   = 1/24
    A can produce in 24 days
   

    

2. Two friends Ram and Sam can complete a project in 14 days and 35 days respectively.
   In how many days they will complete the project if they work together?
   5
   6
   10
   11
   12
   Option C
   1 day Work of Ram = 1/14
   1 day work of Sam= 1/35
   1 day work of Ram and Sam = 1/14 + 1/35 =7 /70 1/10
   =>Ram and Sam together can complete in 10 days
   

    

3. A can stitch a shirt in 12 hours. B and C can stitch a shirt in 9 hours while A and C together can do it in 6 hours.
   How long will B alone do the work?
   36
   33
   34
   35
   31
   Option A
   A -> 1/12 units
   B+C-> 1/9 units
   A+C-> 1/6 units
   C-> 1/6 – 1/12 = 1/12 units
   B-> 1/9 – 1/12= 1/36 units
   B can alone stitch in 36 hours
   

    

4. Worker 1 can complete a piece of work in 8 days. Worker 2 can complete in 10 days. Then Worker 3 also joined with them and they completed the same work in 4 days .
   Find how many days worker C alone complete the work?
   35
   36
   42
   44
   40
   Option E
   Worker 1 one day work = 1/8 units
   Worker 2 one day work = 1/10
   Workers 1+ 2+ 3 one day work = ¼
   Worker 3 one day work = 1/4 -1/8-1/10= 1/40
   =>Worker 3 can complete in 40 days alone.
   

    

5. Mano can construct a shop in 30 days. He works at it for 10 days and then Somu finishes in 40 days.
   In how many days taken to complete if they work together?
   10
   20
   30
   40
   50
   Option B
   Mano’s 1 day work = 1/30 units
    10 days work = 1/3 unit
    Remaining work = 1- 1/3 = 2/3 done by Somu.
    So somu’s one day work = 2/3/40 = 1/60 units
    1 day work of Mano and Somu = 1/30 + 1/60 =3/60=1/20
    So Mano and Somu can complete a work in 20 days together.
   

    

6. A project work could be finished in 27 days by some workers. but it was finished 9 days before the deadline
   as 10 more workers are joined. Find the number workers .
   21
   19
   18
   20
   22
   Option D
   let x be the number of workers.
   x workers can complete the work in 27 days.
   x+10 can complete the work in 18 days ( 27-9).
   ie) 27 * x = (x+ 10) * 18
   =>9x = 180
   => x=20
   

    

7. X can complete a piece of work in 20 days and Y can complete a work in 40 days . They began the work 10 days before finishing the work
   X leaves from the work. Find in how many days the work was completed?
   10
   12
   14
   16
   18
   Option A
   one day,
   A work ->1/20
   B work -> 1/40
   Let t be the total time to finish the work.
   =>t-10/20 + t/40 =1
   =>t/20 + t/40 = 1+1/2 = 3/2
   =>3t/20= 3/2
   =>t=10 days
   

    

8. A can do a work in 10 days and B in 15 days. If they work on it together for some days and 5/12 of work remains then
   find out how many days did they work together?
   3/2
   5/3
   7/2
   11/9
   7/9
   Option C
   1 day work of A= 1/10
   1 day work of B= 1/15.
   Let x be the days of work A and B together .
   x days work of A and B= x/10 + x/15
   x days work completed= 1-5/12=7/12
   x/10 + x/15 =7/12
   x/6=7/12
   =>x=7/2 days.
   

    

9. A takes four times as much time as B or five times as much time as C to finish a piece of work. Working together, they can finish the work in 22 days. B can do the work alone in:
   6
   5
   3
   8
   4
   Option B
   Let A takes x days to complete the work.
   B takes x/4 days to complete the work.
   C takes x/5 days to complete the work.
   A+B+C can complete the work = 1/x+4/x+5/x = 1/2
   =>10/x= 1/2
   =>x=20
   Hence B can take = 20/4 = 5 days to finish the work.
   

    

10. First Air cooler can cool the room in 10 minutes while second takes 15 minutes to cool under similar conditions.
    If both air coolers are switched on at same instance then how long will it take to cool the room approximately ?
    5
    7
    6
    4
    8
    Option C
    First cooler ->10 minutes
    second cooler-> 15 minutes
    Together = (a x b)/(a + b)
    = (15 x 10)/(15 + 10)
    = 150/25
    = 6 minutes
    

     

1. In a jewelry jewels will be rejected if it is 0.05% manufacturing defect. Then how many grams they should examine to reject 2 grams of jewels?
   4000
   3000
   2000
   1500
   4500
   Option A
   Let x be the grams to examine.
   0.05% of x = 2grams.
   0.05 /100 X x = 2
   x = 2 x 100/0.05 => x = 4000.
   So 4000 grams to examine to reject 2grams.
   

    

2. Sunil planned to deposits his money in a bank at 20% compound interest .How many years will take the deposits amount tripled?
   5
   6
   7
   8
   4
   Option C
   C.I >3P ( p- principal amt) ( tripled)
   => P ( 1+ r/100)n > 3P
   => ( 1+ r/100)n > 3
   => ( 1+ 20/100)n > 3
   => ( 120/100)n > 3
   => ( 6/5)n > 3
   When n= 7 ie) (6/5)7 > 3 ..Therefore 7 years after he will get the tripled amount.
   

    

3. A girl decided to read a book of 1000 pages in 20 days, reading 5hrs/day. How many days will she take to read a book of 1500 days with a rate of 2hrs/day?
   50 days
   75 days
   65 days
   70 days
   60 days
   Option B
   1 day= 5hrs=> 20 days = 20 x 5=100hrs
   ie) she can read 1000 pages in 100 hrs.
   so for 1 hr = 1000/100 = 10 pages she can complete.
   Then 2 hrs , 20 pages she can complete.
   Per day 20 pages , then for 1500 pages = 1500/20 = 75 days to complete.
   

    

4. 27.92% of 425.02-24.03% of 299.97= (?)
   Option
   Substituting (?) with “x”
   x=28%×425-24%×300
   x =119 – 72
   x=47
   

    

5. If 5 years before the ratio of ages of daughter and father is 3 : 15. After 5 years from now the ratio will become 8: 20. Then Find out the
   age of the father?
   25 years
   40 years
   35 years
   45 years
   30 years
   Option C
   Let the age of daughter before 5 years= 3x
   Let the age of father before 5 years = 15x
   Then , (3x + 10/15x+10) = 8/20
   =>x=2
   So age of father before 5 years= 15x= 15*2 = 30 years
   present age = 30+ 5= 35 years

    

6. Seetha scored 60 % of marks in Hindi, 70% marks in Science and ” M” marks in Maths.If Maximum marks of each subject is 150 and he scored 70% marks.
   Then find the marks he scored in Maths?
   
   110
   120
   130
   125
   135
   Option B
   Marks in Hindi = 60/100 * 150 = 90
   Marks in Science = 70/100 * 150 = 105
   Total marks in 3 subjects = 70/100 * 450 =315
   so 90 + 105 +M = 315
   => M= 120

    

7. Raju bought a LED T.V and got diwali discount 10% on marked price. The showroom got 4% profit .If Raju paid Rs. 40,000,
   then find the marked price of the T.V?
   
   Rs.45,000
   Rs.42,006
   Rs.46,000
   Rs.46,222
   Rs.43,000
   Option D
   selling price= 104/ 100 * 40, 000 = Rs. 41,600
   Let the marked price be x.
   So,90 % of x = Rs.41,600
   So 100 % of x= (41,600/ 90) * 100 =Rs.46,222

    

8. From a group of 6 soldiers and 4 police a committee of 5 people is formed for emergency rescue operation.
   The committee is to be formed having atleat 3 soldiers.Find the number of possible ways?
   
   160
   170
   186
   179
   125
   Option C
   (6C3 * 4C2) + ( 6C4 * 4C1 ) + 6C5 = 120 +60 + 6 = 186

    

9. A circular frame is inserted in a square frame of area 16 metre square.
   So what is the area of the part of square frame outside the circle?
   
   2.45
   4.56
   3.43
   5.89
   1.25
   Option C
   Let the side of the square frame = a meter
   area = a 2 = 16
   => a= 4 m
   =>diameter of a circular frame = 4m
   => radius= 2m
   area = 22/ 7 * 2 * 2 =12.57
   So remaing outer area= 16- 12.57=3.43m 2
   

    

10. Inlet pipe 1 is 6 times more efficient than inlet pipe 2 to fill the tank.If inlet pipe 1 can fill the tank in 32 min,
    then how much time is taken by both pipes to fill the tank when both the pipes are opened?
    24 min
    36min
    20min
    16 min
    28 min
    Option E
    Let efficiency of pipe 1= 7x
    efficiency of pipe 2= x
    Let both the pipes together can fill the tank in t min
    =>7x * 32 = ( 7x+ x) t
    =>t = 28 min

     

Directions(1-5): What approximate value will come in place of question mark ‘?’ in the following questions.

1. (1224)^1/2 + (485)^1/2 = ?*(360.98)^1/2
   4
   5
   6
   3
   7
   Option D
   (1224)^1/2 + (485)^1/2 = ?*(360.98)^1/2
   =>(1225)^1/2 + (484)^1/2 = ?*(361)^1/2
   =>35 +22 = ?+19
   =>? = 3
   

    

2. (220.82 – 149.21)/(19.9% of 89) = ?
   5
   6
   8
   7
   4
   Option E
   (220.82 – 149.21)/(19.9% of 89) = ?
   =>(221 – 149)/(20% of 90) = ?
   =>? = 72/18
   =>? = 4
   

    

3. (784.1 + 511.9 – 100.01)/3.99 = ?
   285
   300
   320
   410
   451
   Option B
   (784.1 + 511.9 – 100.01)/3.99 = ?
   =>(784 + 512 – 100)/4 = ?
   =>? = 1196/4
   =>? = 300
   

    

4. 59.980% of 599.98 + 34.99% of 499.99 = ?
   500
   320
   440
   535
   480
   Option D
   59.980% of 599.98 + 34.99% of 499.99 = ?
   =>60% of 600 + 35% of 500 = ?
   =>? = 360+175
   =>? = 535
   

    

5. 11.84*125.12/5.008+799.9 = ?
   1200
   1300
   1100
   1400
   1000
   Option C
   11.84*125.12/5.008+799.9 = ?
   =>12*125/5+800 = ?
   =>300+800 = ?
   =>? = 1100
   

    

6. Two alloys contain 40% and 35% of gold in it resp. If 24kg of first type alloy is mixed with 16 kg of second type of alloy then find the percentage of gold in the resulting mixture.
   49%
   44%
   38%
   40%
   30%
   Option C
   First alloy quantity = 0.4*24 = 9.6 kg
   Second alloy quantity = 0.35*16 = 5.6 kg
   % of gold in resulting mixture = {(9.6+5.6)/(24+16)}*100 = 38%
   

    

7. The monthly incomes of Ram and Shyam is 14:15 resp. and their monthly expenditure is Rs.4600 and Rs.4500 resp. If saving of Shyam is 12.5% more than the savings of Ram. Find the monthly income of Shyam.
   Rs.13500
   Rs.13300
   Rs.15000
   Rs.12000
   Rs.11200
   Option A
   (14x – 4600)/(15x – 4500) = 8/9
   =>x = 900
   Monthly income of Shyam = 15*900 = Rs.13500
   

    

8. A man travelled certain distance in three equal parts with different speeds of 12 km/hr., 16km/hr. and 20km/hr. resp. Find his average speed for the overall journey.
   11.02 km/hr.
   17.17 km/hr.
   18.84 km/hr.
   15.32 km/hr.
   10.21 km/hr.
   Option D
   Total time taken by the man = [x/12 + x/16 + x/20]
   Average speed = 3x/[ x/12 + x/16 + x/20] = 15.32 km/hr.
   

    

9. A man invested Rs.20000 at simple interest for 2 years at the rate of r% and gets an interest of Rs.2000. If he invested total amount (Principal and interest) in a scheme offering simple interest at the rate of (r+3)%. Find the total interest obtained by the man from scheme after 2 years?
   Rs.3450
   Rs.3245
   Rs.3520
   Rs.3151
   Rs.3320
   Option C
   Interest for one year = 2000/2 = Rs.1000
   Rate = 1000/20000*100 = 5%
   Interest earned = 22000*8%*2 = Rs.3520
   

    

10. Two dices are thrown, find the probability of getting an even number after multiplying numbers obtained from each dice.
    1/2
    3/4
    4/7
    7/9
    5/7
    Option B
    Probability = ½*1/2+1/2*1/2+1/2*1/2
    = ¾
    

     

1. 144, 173, 140, 169, 136, ?
   150
   160
   178
   165
   190
   Option D
   144 + 29 = 173
   173 – 33 = 140
   140 + 29 = 169
   169 – 33 = 136
   136 + 29 = 165
   

    

2. 14, 24, 43, 71, 108, ?
   160
   174
   154
   145
   130
   Option C
   14 + 10 = 24
   24 + 19 (=10 + 9) = 43
   43 + 28 (= 19 + 9) = 71
   71 + 37 (= 28 + 9) = 108
   108 + 46 (=37 + 9) = 154
   

    

3. 25, 48, 94, 186, 370, ?
   700
   738
   715
   725
   740
   Option B
   25 × 2 – 2 = 50 – 2 = 48
   48 × 2 – 2 = 96 – 2 = 94
   94 × 2 – 2 = 188 – 2 = 186
   186 × 2 – 2 = 372 – 2 = 370
   370 × 2 – 2 = 740 – 2 = 738
   

    

4. 11, 10, 18, 51, 200, ?
   900
   995
   845
   974
   920
   Option B
   11 × 1 – 1 = 10
   10 × 2 – 2 = 18
   18 × 3 – 3 = 51
   51 × 4 – 4 = 200
   200 × 5 – 5 = 995
   

    

5. 958, 833, 733, 658, 608, ?
   500
   450
   583
   620
   550
   Option C
   958 – 833 = 125
   833 – 733 = 100
   733 – 658 = 75
   658 – 608 = 50
   ? = 608 – 25 = 583
   

    

6. 112, 119, 140, 175, 224, ?
   220
   215
   287
   230
   240
   Option C
   112 + 1 × 7 = 119
   119 + 3 × 7 = 119 + 21 = 140
   140 + 5 × 7 = 140 + 35 = 175
   175 + 7 × 7 = 175 + 49 = 224
   224 + 9 × 7 = 224 + 63 = 287
   

    

7. 3, 100, 297, 594, 991, ?
   1520
   1362
   1488
   1122
   1200
   Option C
   3 + 97 = 100
   100 + 197 = 297
   297 + 297 = 594
   594 + 397 = 991
   991 + 497 = 1488
   

    

8. 12, 35, 81, 173, 357, ?
   725
   700
   770
   720
   745
   Option A
   12 + 1 × 23 = 35
   35 + 2 × 23 = 35 + 46 = 81
   81 + 2 × 46 = 81 + 92 = 173
   173 + 2 × 92 = 173 + 184 = 357
   357 + 2 × 184 = 357 + 368 = 725
   

    

9. 13, 20, 39, 78, 145, ?
   200
   212
   248
   230
   220
   Option C
   13 + 7 = 20
   20 + 19 (= 7 + 12) = 39
   39 + 39 (=19 + 20) = 78
   78 + 67 (= 39 + 28) = 145
   145 + 103 (= 67 + 36) = 248
   

    

10. 1, 7, 49, 343, ?
    2301
    2200
    2401
    2313
    2512
    Option C
    1 × 7 = 7
    7 × 7 = 49
    49 × 7 = 343
    343 × 7 = 2401
    

     

Directions(1-5):Answer the questions based on the following information.
The graph given below shows the total number of units produced by five companies P,Q,R,S and T and percentage of units sold by respective company in 2010.

1. What is the difference between the number of units which remain unsold by company S and T?
   5000
   5200
   4600
   4000
   4800
   Option C
   Required Difference = 6600 -2000 = 4600

    

2. Number of units sold by company R is how much percent more or less than the number of units sold by company S?
   14.2%
   17.5%
   19.7%
   22.2%
   10.9%
   Option B
   Percentage = [(6000 – 4950)/6000 ]*100 = 17.5%

    

3. What is the average number of unsold units of companies P,R and T together?
   4200
   3215
   3500
   4750
   3550
   Option D
   Average = (3600+4050+6600)/3 = 4750

    

4. Company R has three types of unsold products (P,Q,R), out of which 60% were of the type P while type Q and R were in the ratio 5:4 resp. Find the number of type Q products of company R which remain unsold.
   985
   900
   850
   800
   788
   Option B
   Total number of products of type Q and type R were unsold = 0.40*4050 = 1620
   Number of type Q which remain unsold = 5/9*1620 = 900
   

    

5. What is the total number of units sold by company P and Q together in 2010?
   7500
   5400
   7000
   6400
   6200
   Option D
   Total number of units = 2400+4000 = 6400

    

Directions(6-10): Answer the questions based on the given information.
The graph given below shows the number of students who play Cricket and Badminton from five different schools. One student play one game.



6. Find the average number of students who play Badminton in school A,B and D.
   200
   180
   190
   240
   210
   Option D
   Required Average = (260+180+280)/3 = 240

    

7. What is the difference between the average number of students who play Cricket in school A and B together and the average number of students who play Badminton in school C and D?
   20
   25
   32
   30
   45
   Option D
   Difference = (240+280)/2 – (210+250)/2 = 30

    

8. If the ratio of the number of boys to girls who play Cricket in school D is 2:3, Find the number of boys who play Cricket in school D?
   80
   100
   90
   85
   96
   Option E
   Number = (240/5)*2 = 96

    

9. Find the ratio of the number of students who play Cricket in school C to the number of students who play Badminton in B.
   4:5
   5:6
   7:5
   2:3
   8:9
   Option E
   Ratio = 160:180 = 8:9

    

10. The number of students who play Cricket in E is how much percent less than the number of students who play Badminton in A and C together?
    20%
    30%
    40%
    50%
    60%
    Option C
    Difference = (260+240)- 300 = 200
    Percentage = 200/500*100 = 40%