1. The average speed of a car is one-third the speed of a bus. The bus covers 1125 km in 15 hours. How much distance will the car cover in 36 minutes?
   10
   15
   20
   25
   30
   Option B
   The bus covers 1125 km in 15 hrs ⇒ speed of train = distance / time = 1125/15 = 75km/hr
   Now speed of car = 1/3 (speed of train) = 1/3 (75) = 25 km/hr = (25/60)km/min = 5/ 12 Km/min
   Hence distance traveled by car in 36 min= speed × time = 5 × 36/12 = 5 × 3 = 15km

    

2. Francise purchases 60 kg. of grains for Rs.700. The cost price of Dal was Rs.15 per kg and that of wheat was Rs.10 per kg. Find the quantity of Dal and wheat purchased.
   20
   30
   40
   50
   60
   Option A
   Let the quantity of Dal bought be x kg and that of wheat be (60-x) kg.
   15x + (60-x)10 = 700
   15x + 600 – 10x = 700
   5x + 600 = 700
   5x = 100
   x= 20
   

    

3. There were 11 soldiers in a camp. If the number of soldiers increased by 4, the expenses of the food increased by Rs. 45 per day while the average expenditure per head decreased by Rs. 13. Find the original expenditure of the food.
   660
   450
   550
   430
   200
   Option A
   Let the average expenditure per head = x
   No. of students = 11
   11x + 45 = 15 (x-13)
   11x 45 = 15x – 195
   4x = 240
   x = 60
   Original expenditure of food = 11x 60=Rs.660
   

    

4. The mobile is sold at a profit of 6.5%. Then its CP and SP are increased by Rs. 120 and Rs. 150 respectively and the profit increases to 7.5%. Find the mobile’s new cost price.
   1200
   1400
   1600
   2220
   1800
   Option D
   Initially SP = 1.065CP
   CP’ = CP + 120 & SP’ = SP + 150
   Finally SP’ = 1.075CP’
   Hence,
   SP+150 = 1.075(CP+120)
   1.065CP+150 = 1.075(CP+120)
   0.01CP = 150-129
   0.01CP = 21
   CP = 2100
   CP’ = 2220
   

    

5. Akil and Arun started a business in partnership investing Rs. 10,000 and Rs. 7,500 respectively. After 3 months, Arjun joined them with Rs. 10,000. What will be Arun’s share in the total profit of Rs 12,000 earned at the end of 1 year from the starting of the business?
   Rs. 4600
   Rs. 3400
   Rs. 3000
   Rs. 4500
   Rs. 3600
   Option E
   ratio= (10,000 x 12) : (7500 x 12) : (10,000 x 9) = 4 : 3 : 3
   Hence, Arun’s share = (3/10) x 12000 = Rs. 3600
   

    

6. The average ages of the group of students is 20 years and one new student joined , then the average age of the group is increased by 1 years. If the initial number of students from the group is 28, what is the age of new student?
   45
   46
   47
   48
   49
   Option E
   Total age of the group = 20 * 28 = 560
   New total age of the group = 29 * 21 = 609
   New student age = 609 – 560 = 49 years
   

    

7. Ibrahim invests Rs.4800 at simple interest for 4 years and Ishan invests Rs.3200 at simple interest for 5 years. If both of them invest their money for same rate of interest, the difference between the interest amounts received by them is Rs.160, find the rate of interest per annum?
   3
   4
   5
   6
   7
   Option C
   SI = P * N * R/100
   4800 * 4 * R/100 – 3200 * 5 * R/100 = 160
   192R – 160R = 160
   32R = 160
   R = 5%
   

    

8. 24 boys were appointed to complete the work in 40 days. But after 20 days, it was found that only one third of the work only completed. In how many extra boys needed to be complete the work in time?
   25
   24
   23
   22
   21
   Option B
   (boys1 * day 1)/ work 1 = (boys2 * day 2) / work2
   (24 * 20)/(1/3) = (24 + x) * 20/(2/3)
   48 = 24 + x
   x = 24
   

    

9. The ratio of milk and water in a can is 3: 2. When 20 liters and added 5 liters of water, the ratio becomes 8: 7. Find the initial quantity of the milk and water in the can?
   40
   50
   60
   70
   80
   Option C
   Milk in 20 liters = 3/5 * 20 = 12 liters
   Water in 20 liters = 2/5 * 20 = 8 liters
   (3x – 12)/(2x – 8 + 5) = 8/7
   21x – 84 = 16x – 24
   x = 12
   Initial quantity of the mixture = 5x = 5 * 12 = 60 liters
   

    

10. The ratio of the number of men to women in the hall A is 3: 2 and the ratio of the number of men to women in the hall B is 4: 5. If the ratio of the total number of people in the hall A to B is 5: 9 and the difference between the number of people in hall A and B is 16. What is the ratio of the number of men in hall A to number of women in hall B?
    3: 7
    3: 1
    9: 5
    3: 5
    3: 2
    Option D
    The ratio of the total number of people in hall A to B = 5: 9 (5x, 9x)
    9x – 5x = 16
    4x = 16
    x = 4
    Total number of people in hall A = 5 * 4 = 20
    Total number of people in hall B = 9 * 4 = 36
    Number of men in A = 3/5 * 20 = 12
    Number of women in B = 5/9 * 36 = 20
    Required ratio = 12: 20
    = 3: 5
    

     

1. 33.99% of 120 + 59.87% of 650 +150.32% of 90= ?
   567
   345
   569
   128
   450
   Option A
   42+390+135=567

    

2. 69.885 of 79.56 +1800/ 8.99- 19.87* 11.56=?
   14
   16
   18
   20
   22
   Option B
   56+200+240=16

    

3. √528.99 * 4.5 + 79.87% of 134.56 =? + 15% of 180
   126
   146
   196
   180
   175
   Option C
   115 + 108 =? + 27
   ?=196
   

    

4. (63.6 ÷ 125.6) * (314.9 ÷ 15.8) * ? =40.4% of 150
   5
   6
   7
   8
   9
   Option B
   (32/63) * (315/16) * ?=60
   2 * 5 * ? = 60
   ?= 6
   

    

5. √13688.88 ÷ 12.88 + ? % of 319.88 = 130.9 * √9
   120
   130
   140
   150
   160
   Option A
   √13689 ÷ 13 + ? % of 320 = 131 * √9
   9 + ? % of 320=393
   ?=120
   

    

6. 3239.67 ÷ 29.5 * 15 =? * 16
   145
   101
   111
   156
   178
   Option B
   ? * 16 = (3240/30) * 15
   ? * 16 = 3240/2
   ? = 101(approx)
   

    

7. 60.80% of 984.67 – 150.9046 =? – 63% of 698
   890
   880
   780
   680
   580
   Option B
   61% of 985 – 151.046 =? – 63% of 698
   60/100 * 985 – 151 =? – 63/100 * 700
   591 – 151 =? – 441
   ? = 881 = 880 (Approx.)
   

    

8. (448.068/32) ÷ (18.923/778.69) * (228.054/76) =?
   45
   51
   64
   58
   45
   Option D
   ? = (448/32) * (779/19) * (228/76)
   ? = 14 + 41 + 3
   ? = 58
   

    

9. 1235.0067 + 656.900009 + 90.00763 + 452.987 + 6753.099 =?
   5688
   6888
   5888
   8988
   9188
   Option E
   1235 + 657 + 90 + 453 + 6753 =?
   9188 =?
   

    

10. √7395.90 * √11024.89 + √626.666 =? – √731.009
    9087
    9082
    9088
    9056
    9045
    Option B
    √7396 * √11025 + √625 =? – √729
    86 * 105 + 25 =? – 27
    9030 + 25 + 27 =?
    9082 =?
    

     

1. Ratio of the cost price to marked price of the Book is 4: 5. the shop offer a discount of 15% and the marked price of the article is Rs.720. What is the profit percentage of the Book?
   3.25%
   5.25%
   4.25%
   5.25%
   6.25%
   Option E
   CP = 4/5 * 720 = Rs.576
   SP = 720 * 85/100 = Rs.612
   Profit percentage = (612 – 576)/576 * 100
   = 6.25%
   

    

2. Kayal and Mayan together can complete a work in 30 days. If Mayan alone complete 20% of the work in 16 days. In how many days Kayal alone complete 75% of the work?
   36
   38
   40
   42
   44
   Option A
   K+ M = 1/30
   M = 5/1 * 16 = 80 days
   K= 1/30 – 1/80
   K = (8 – 3)/240 = 5/240
   K = 1/48
   K complete the 75% of the work in = 48 * ¾ = 36 days
   

    

3. Kiran and Mithun invests in the ratio of 2: 3 and the investment period of x months and y months respectively. If the total profit of the business is 3500 and Mithun’s share is Rs.1500, then what is the ratio of x: y?
   2/1
   3/4
   2/5
   4/7
   2/9
   Option A
   profit = 3500 – 1500 = 2000
   (2 * x)/(3 * y) = 2000/1500
   2x/3y = 4/3
   x/y = 2/1
   

    

4. Person S invests Rs.8000 in scheme A which offer simple interest at 15% per annum for x years. After x years S received the total amount is Rs.17600. Find the value of x
   5
   6
   7
   8
   9
   Option D
   SI = (P * N * R)/100
   SI = 17600 – 8000 = Rs.9600
   9600 = (8000 * 15 * N)/100
   N = 8 years
   

    

5. Ratio of the speed of Ship in still water to river is 4: 1. If the ship rowing the speed of 20 kmph in still water, then what is the ratio of the speed of ship rowing against the current and along the current?
   3: 5
   3: 7
   4: 5
   3: 8
   3: 10
   Option A
   Speed of river = ¼ * 20 = 5 kmph
   Speed of the ship along the river = 20 + 5 = 25 kmph
   Speed of the boat against the river = 20 – 5 = 15 kmph
   Required ratio = 15: 25
   = 3: 5
   

    

6. A container of 120 liters contains oil and ghee. If 60 % of oil and 30 % of ghee is taken out, then the vessel will be half empty. Find the ratio of initial quantity of oil and ghee in the container?
   5:6
   2 : 1
   1:6
   7:4
   2:0
   Option B
   O+ G = 120 -à (1)
   (60/100) * O + (30/100) * G = 60
   6O + 3G = 600 —> (2)
   By solving the equation (1) and (2), we get,
   O = 80 liters, G = 40 liters
   Required ratio = 80 : 40 = 2 : 1
   

    

7. How many 6 letter words with or without meaning can be formed out of the letters of the word, ‘INCUBATORS’, if repetition of letters is not allowed?
   169006
   350861
   151200
   713058
   601427
   Option C
   The word contains 10 different letters.
   Required number of words = 10p6
   = > (10 * 9 * 8 * 7 * 6 * 5) = 151200
   

    

8. A guy drives 360 km from point A to B in 5 hours and backs to point B in 4 hours. If M is the average speed of the entire trip, then the average speed for the journey from B to A exceeds M by?
   10 km/hr
   56km/hr
   89km/hr
   45km/hr
   16km/hr
   Option A
   The speed from A to B = 360/5 = 72 km/hr
   The speed from B to A = 360/4 = 90 km/hr
   The average speed of the entire trip = 2xy/(x+ y)
   = > [(2 * 72 * 90) / 162]
   = > 80 km/hr
   Required difference = 90 – 80 = 10 km/hr
   

    

9. A purse has the dollars of 10p, 50p and 1 rupee is in the ratio of 5: 4: 3. If there is Rs. 22 in all, then find the total number of 50p coins?
   90
   14
   54
   98
   16
   Option E
   A purse has the coins of 10p, 50p and 1 rupee is in the ratio = 5: 4: 3 (5x, 4x, 3x)
   22 = [(5x * 10) / 100] + [(4x * 50) / 100] + [(3x * 1)]
   22 = (50x/100) + (200x/100) + 3x
   22 = (x/2) + 2x + 3x
   44 = 11x
   x = 4
   The total number of 50p coins = 4x = 16
   

    

10. A express train crosses a pole in 25 sec and the same train crosses a platform of length 150 m in 40 sec. Find the length of the train?
    567m
    605m
    845m
    250 m
    831m
    Option D
    Let the length of express be x,
    Here, the speed of the train is equal. So,
    (x / 25) = (x + 150) / 40
    8x = 5x + 750
    3x = 750
    x = 250
    The length of the train (x) = 250 m
    

     

Directions (1-5): Study the following information carefully and answer the questions that follow:
The following pie chart shows the percentage distribution of total number of players who play hockey in 6 different clubs total number of players who play hockey is 8000.
The following table shows the ratio of total number of players who play skating and hockey in different clubs.

1. Find the ratio between the total number of players who play skating in club P to that of club S.
   130:187
   120:111
   111:123
   141:160
   117:119
   Option A
   The total number of players who play skating in club P
   = >8000 * (18/100) * (13/9) = 2080
   The total number of players who play skating in club S
   = >8000 * (22/100) * (17/10) = 2992
   Required ratio = 2080 : 2992 = 130 : 187
   

    

2. Find the difference between the total number of players who play Hockey in club Q and R together to that of total number of players who play skating in club Q and T together.
   509
   504
   505
   409
   550
   Option B
   The total number of players who play Hockey in club Q and R together
   = >8000 * (36/100) = 2880
   The total number of players who play skating in club Q and T together
   = >8000 * (16/100) * (17/20) + 8000 * (14/100) * (23/20)
   = >1088 + 1288 = 2376
   Required difference = 2880 – 2376 = 504
   

    

3. Total number of players who play Hockey in club P and S together is approximately what percentage of total number of players who play Skating in club R and U together?
   112
   114
   115
   117
   119
   Option D
   Total number of players who play Hockey in club P and S together
   = > 8000 * (40/100) = 3200
   Total number of players who play Skating in club R and U together
   = >8000 * (20/100) * (31/40) + 8000 * (10/100) * (15/8)
   = >1240 + 1500 = 2740
   Required % = (3200 / 2740) * 100 = 116.788 % = 117 %
   

    

4. Find the average number of players who play skating in all the given institutions except club T.
   1280
   1480
   1780
   1680
   1980
   Option C
   The total number of players who play skating in all the given institutions except club T
   = > (8000 * (18/100) * (13/9) + 8000 * (16/100) * (17/20) + 8000 * (20/100) * (31/40) + 8000 * (22/100) * (17/10) + 8000 * (10/100) * (15/8))
   = > (2080 + 1088 + 1240 + 2992 + 1500) = 8900
   Required average = 8900 / 5 = 1780
   

    

5. Find the difference between the average number of players who play hockey in club T and U together to that of the average number of players who play skating in club Q and T together.
   228
   823
   123
   512
   713
   Option A
   The average number of players who play hockey in club T and U together
   = >8000 * (24/100) * (1/2) = 960
   The average number of players who play skating in club Q and T together
   = >[8000 * (16/100) * (17/20) + 8000 * (14/100) * (23/20)] / 2
   = > (1088 + 1288) / 2 = 1188
   Required difference = 1188 – 960 = 228
   

    

Directions (6-10): Study the following table carefully and answer the questions that follow:



6. What is the ratio of the total number of farmers in city A, C and E together to that of the total number of police in city B, C and D together?
   38: 31
   38: 41
   38: 39
   38: 30
   38: 37
   Option E
   Required ratio = (120 + 160 + 100): (120 + 100 + 150)
   = 380: 370 = 38: 37
   

    

7. If the number of male farmers in city B is 26 and total number of males in city B is 48% of the total number of farmers in city D. What percentage of female police in city B from the total people (Farmers and Police)?
   22
   23
   24
   25
   26
   Option D
   Total number of male in city B = 48/100 * (200)
   = 96
   Number of male Police in city B = 96 – 26 = 70
   Number of female police in city B = 120 – 70 = 50
   Required percentage = 50/200 * 100 = 25%
   

    

8. What is the difference between the number of police in city D and C is what percentage more/less than the number of farmers in city E?
   50
   55
   60
   65
   70
   Option A
   Difference between police in city D and C = 150 – 100 = 50
   Required percentage = (100 – 50)/100 * 100
   = 50%
   

    

9. What is the average number of Farmers in all the given cities together?
   131
   132
   133
   134
   135
   Option B
   Required average = (120 + 80 + 160 + 200 + 100)/5
   = 660/5 = 132
   

    

10. What is the ratio of the difference between the police in the city D and B to that of the difference between the farmers in the city A and C?
    3: 10
    3: 11
    3: 7
    3: 4
    3: 5
    Option D
    Required ratio = (150 – 120): (160 – 120)
    = 30: 40 = 3: 4
    

     

1. 10²-x-2=0
   6y²+33y+15=0
   x<y
   x<=y
   x>y
   x>=y
   the relation cannot be determined
   Option E
   10x²-x-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+33y+15=0
   6y² +30y+ 3y +15=0
   6y( y+5) + 3(y+5)=0
   (y+5) ( 6y+3)=0
   y= -5, -3/6
   Hence the relation cannot be determined
   

    

2. 35x²-13x-4=0
   9y² +9y-2=0
   x<y
   x<=y
   x>y
   x>=y
   the relation cannot be determined
   Option C
   35x²-13x-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 +9y-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 x>y
   

    

3. 35x²-46x+15=0
   10y²+131y+13=0
   x<y
   x<=y
   x>y
   x>=y
   the relation cannot be determined
   Option C
   35x²-46x+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+131y+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. 10x²+59x+45=0
   20y²-13y+2=0
   x<y
   x<=y
   x>y
   x>=y
   the relation cannot be determined
   Option A
   10x²+59x+45=0
   10x² +50x+9x+45=0
   10x(x +5) + 9(x +5)=0
   (x+5) ( 10x+9)=0
   x= -5 , -9/10
   20y²-13y+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. 2x²-31x= -120
   8y²-74y=-143
   x<y
   x<=y
   x>y
   x>=y
   the relation cannot be determined
   Option C
   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. 8x²-41x-5=0
   49y²-49y-18=0
   x<y
   x<=y
   x>y
   x>=y
   the relation 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 x>y
   

    

7. 40x²-59x+21=0
   14y²+53y+14=0
   x<y
   x<=y
   x>y
   x>=y
   the relation cannot be determined
   Option C
   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+53y+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. 40x²-13x+1=0
   10y²+19y+6=0
   x<y
   x<=y
   x>y
   x>=y
   the relation cannot be determined
   Option C
   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+19y+6=0
   10y² +15y+4y -6=0
   5y( 2y+3) +2(2y+3)=0
   (2y=3) ( 5y+2)=0
   y= -3/2, -2/5
   x>y
   

    

9. 10x²-9x+2=0
   6y²+ 27y-15=0
   x<y
   x<=y
   x>y
   x>=y
   the relation 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 x≥y
   

    

10. 35x²+51x+18=0
    16y² =1
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option A
    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
    16y² =1
    16y²-1=0
    (4y-1) ( 4y+1)=0
    y= 1/4,- 1/4
    Hence y>x
    

     

Find the wrong term:-

1. 13.., 77.., 202.., 418.., 763…, 1273
   763
   13
   77
   202
   418
   Option A
   13 + 43 = 77
   77 + 53 = 202
   202 + 63 = 418
   418 + 73 = 761
   761 + 83 = 1273
   The wrong term is 763
   

    

2. 53760…, 6720.., 960.., 160.., 80…, 8
   6720
   960
   160
   80
   8
   Option D
   53760 ÷ 8 = 6720
   6720 ÷ 7 = 960
   960 ÷ 6 = 160
   160 ÷ 5 = 32
   32 ÷ 4 = 8
   The wrong term is 80
   

    

3. 80.., 120.., 300.., 1040.., 4725.., 25987.5
   80
   120
   300
   1040
   4725
   Option D
   80 * 1.5 = 120
   120 * 2.5 = 300
   300 * 3.5 = 1050
   1050 * 4.5 = 4725
   4725 * 5.5 = 25987.5
   The wrong term is 1040
   

    

4. 325.., 362.., 401.., 442.., 484.., 530
   325
   362
   401
   442
   484
   Option E
   182 + 1 = 325
   192 + 1 = 362
   202 + 1 = 401
   212 + 1 = 442
   222 + 1 = 485
   232 + 1 = 530
   The wrong term is 484
   

    

5. 19.., 20.., 45.., 129.., 520.., 2605
   19
   20
   45
   129
   520
   Option C
   19 * 1 + 1 = 20
   20 * 2 + 2 = 42
   42 * 3 + 3 = 129
   129 * 4 + 4 = 520
   520 * 5 + 5 = 2605
   The wrong term is 45
   

    

Find the missing numbers:-

6. 56.., 92.., 67.., 83.., 74.., ?
   76
   78
   80
   82
   90
   Option B
   56 + 62=92
   92 – 52 =67
   67 + 42 =83
   83 – 32 =74
   74 + 22 =78
   

    

7. 5040.., 840.., 168.., ?.., 14.., 7
   40
   42
   44
   46
   48
   Option B
   5040 ÷ 6=840
   840 ÷ 5=168
   168 ÷ 4=42
   42 ÷ 3=14
   14 ÷ 2=7
   

    

8. 240.., 40.., ?.., 50.., 150.., 75
   100
   200
   300
   400
   500
   Option B
   240 ÷ 6 =40
   40 * 5 =200
   200 ÷ 4 =50
   50 * 3 =150
   150 ÷ 2 =75
   

    

9. 12.., 6.., 12.., 42.., 210.., ?
   1245
   1365
   1355
   1555
   1755
   Option B
   12 * 0.5=6
   6 * 2 =12
   12 * 3.5=42
   42 * 5=210
   210 * 6.5=1365
   

    

10. 71.., 77.., 89.., 109.., 139.., ?
    181
    171
    197
    121
    131
    Option A
    71 + (22 + 2) =77
    77 + (32 + 3) =89
    89 + (42 + 4) =109
    109 + (52 + 5)=139
    139 + (62 + 6)=181
    

     

Directions (1-5): Study the following information carefully and answer the questions that follow:
There are three garments A, B and C sold shirt and pants in January 2019.
Number of shirt sold in garment C is 28 less than the number of pants sold in garment B. Number of shirt sold in garment B is 240, which is 20% more than the number of pants sold in the same garment.
Total number of items sold in garment A is 50% of the total number of items sold in garment B. Number of pants sold in garment A is half of the number of shirt sold in garment B.
The ratio of the number of shirt sold in garment A to that of the number of pants sold in garment C is 25: 27.

1. Number of shirt sold in garment A is what percentage more/less than number of pants sold in garment B?
   
   10
   20
   30
   40
   50
   Option E
   Number of shirt sold in garment B = 240
   Number of pants sold in garment B = 240 * (100/120) = 200
   Number of shirt sold in garment C = 200 – 28 = 172
   Total number of items sold in garment B = 240 + 200 = 440
   Total number of items sold in garment A = 440 * 50/100 = 220
   Number of pants sold in garment A = ½ * 240 = 120
   Number of shirt sold in garment A = 220 – 120 = 100
   Number of pants sold in garment C = 100 * (27/25) = 108
   Required percentage = (200 – 100)/200 * 100
   = 100/200 * 100 = 50% less
   

    

2. What is the ratio of the total number of pants sold in all the garments together to that of the total number of items sold in garment C?
   117: 27
   117: 70
   113: 70
   117: 10
   17: 70
   Option B
   Required ratio = (120 + 240 + 108): 280
   = 468: 280
   = 117: 70
   

    

3. What is the difference between the total number of items sold in garment A to that of C?
   50
   60
   70
   80
   90
   Option B
   Required diff = 280 – 220 = 60

    

4. If garment D sold number of pants is 20% more than that of garment A and the ratio of the number of pants sold in garment D to that of shirt is 2: 3. What is the total number of items sold in garment D?
   340
   300
   240
   360
   120
   Option D
   Number of pants sold in garment D = 120 * (120/100) = 144
   Number of shirt sold in garment D = 144 * (3/2) = 216
   Required total = 144 + 216 = 360
   

    

5. If 20% of the shirt sold in garment B is defective, find the number of non – defective shirt sold in garment B?
   132
   223
   256
   190
   192
   Option E
   Required number of shirt = 240 * 80/100 = 192

    

Directions (6-10): Study the following information carefully and answer the questions that follow:

The pie chart shows the distribution of total number of senior professors in different departments.
Number of senior professors in Department X = 18


6. What is the percentage of number of senior professors in department U?
   15
   25
   35
   45
   55
   Option A
   Required percentage = 54˚/360˚ * 100
   = 15%
   

    

7. Find the total number of junior professors in department T, U and X together?
   80
   70
   60
   90
   100
   Option D
   Required total = (36 + 30 + 24) = 90

    

8. If total number of female professors in department Y is 40 and the number of junior male professors in department Y is 20, then find the total female senior professors in the same department?
   23
   24
   25
   26
   27
   Option B
   Total number of professors = 54 + 36 = 90
   Total number of female professors = 40
   Total number of male professors = 90 – 40 = 50
   Number of senior male professors = 50 – 20 = 30
   Number of senior female professors = 54 – 30 = 24
   

    

9. If the number of senior professors in department S is 25% more than department V and the ratio of senior to junior professors in department S is 3: 4. Number of junior professors in department S is what percent more/less than the number of junior professors in department U?
   100
   120
   130
   140
   110
   Option A
   Number of senior professors in department S = 125/100 * 36 = 45
   Number of junior professors in department S = 45/3 * 4 = 60
   Required percentage = (60 – 30)/30 * 100 = 30/30 * 100 = 100%
   

    

10. Which of the following department has maximum number of professors (Senior + Junior)?
    X
    Y
    T
    U
    V
    Option B
    From department T = 45 + 36 = 81
    From department U = 27 + 30 = 57
    From department V = 36 + 32 = 68
    From Department X = 18 + 24 = 42
    From department Y = 54 + 36 = 90
    Department Y has maximum number of professors.
    

     

1. Two cycles are travelling towards each other .what is the distance between them just before 30 minutes they meet ?
   I ) They were originally 100 km apart
   II) Speed of a cycle A is 30 km / hr and speed of cycle B is 40 km /hr apart.
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option B
   Using the statement I we cannot answer the question because no speed is given.
   But using II we can.
   Since the cycles are going to meet 30 minutes , which means , the relative speed = 30 + 40 = 70 km/hr.
   So the gap = speed x time .So we can find the distance .Hence option B is correct.
   

    

2. Saral wins in a very faculty election by defeating his solely opponent Varun. Next year the varsity population magnified 50%, and now Varun defeats Saral. Notice the supporters of Varun wins the second election.
   I ) Majority of votes by which Varun won are 1000 more than the majority of votes by which Saral has won in the first election.
   II) Supporters of Saral increased by 30 % and Supporters of Varun increased by 75%
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option C
   Clearly both statements are given the required answer, Hence option C is correct.

    

3. Average weight of 10 school children are 60 kg. Children X with average weight Y is joined the group so the total average weight is increased. Find the new average weight of children.
   I) X + Y = 65
   II) Y is prime.
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option D
   Clearly Y is greater than 60. Because average only increased if the value will be greater than 60.
   Also X denote s the number of children , so it will be the positive value. So by first statement we can get ( 1, 64), ( 2, 63) , ( 3, 62 ), ( 4 , 61) since y > 60.
   But by statement II Y is a prime number . So ( 4, 61 ) is correct.
   So 4 children with average weight 61are added in the group.
   Hence D is correct.
   

    

4. Who among Rani, Jaya, Akila and Neema is the tallest?
   I. Jaya is taller than Neema.
   II. Rani is not as tall as either Akila or Jaya.
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option E
   From I: J >N
   From II :R < A or J
   From the given data we can’t say Who is tallest or shortest .
   

    

5. In a garland how many roses are there between sunflower and lily in a row of 35 roses?
   I) sunflower is 6 places away from lily, which is 20th from the left end.
   II) Sunflower is 13th from the left end and lily is 8th from the right end.
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option B
   From second statement, we get
   (12 roses) sunflower (12 roses) lily (7 roses).
   

    

6. When is mybirthday this year ?
   I. It is between october 16 and 18, october 16 being Wednesday.
   II. It is not on Friday.
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option A
   From Statement I, we conclude that My birthday is on october 17 , which is Thursday, this year. So, only I is needed.
   

    

7. Emirates plains are always punctual in Dubai. How long, at the most, will Mr. mohamed have to wait for the bus ?
   I. Mr. mohamed has come to the airport stand at 9 A.M.
   II. There is airport at 10 A.M. and Possibly another airport even earlier.
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option C
   From both the given statements, we find that Mr.mohamed reached the airport stand at 9 A.M. and a airport is sure to arrive at 10 A.M. So, Mr. airport has to wait for at the most one hour.
   

    

8. How many iphones were sold in 2019?
   I. Last year 100 phones were sold.
   II. The number of phones sold this year was 2.5 times that of last year.
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option C
   Clearly both statements are given the required answer, Hence option C is correct.

    

9. How is Bheema related to Akil?
   I. Dinesh is the father of Akil and Bheema.
   II. Akil is Bheema’s brother.
   
   If Statement I is sufficient for the answer.
   If Statement II is sufficient for the answer.
   If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
   If Statement I or II is sufficient for the answer.
   both Statement I and II are insufficient for the answer.
   Option A
   Using the statement I we can answer the question as bheema is a brother of akil.
   

    

10. What is G’s age?
    I. G, V and K are all of the same age.
    II. Total age of V, K and A is 64 and A is as old as V and K together.
    If Statement I is sufficient for the answer.
    If Statement II is sufficient for the answer.
    If Statement I and II are sufficient for the answer but neither of the two alone is sufficient
    If Statement I or II is sufficient for the answer.
    both Statement I and II are insufficient for the answer.
    Option C
    As given in statements I and II, we have
    G = V = K,
    V + K + A = 64
    and A = V + k.
    take V + K = A in second,
    we get 2A = 64 or A = 32.
    Thus, V + K = 32 and V = K.
    So, V = K = 16. Thus, G = 16.
    

     

1. 13.928 % of 210 + 43.0851 % of 350 + 70.6428% of 140 = ?
   160
   180
   260
   280
   360
   Option D
   1/7 * 210 + 3/7 * 350 + 5/7 * 140 = ?
   30 + 150 + 100 = ?
   280 = ?
   

    

2. 2.99 * ? + 116.99 = 672.009– 232.213
   160
   170
   180
   190
   200
   Option C
   23 * ? + 117 = 672 – 232
   23 * ? + 117 = 4489 – 232
   ? * 23 = 4140
   ? = 180
   

    

3. (20.89 ÷ 797.88) * (49.009 ÷ 735.34) * ? = 10
   5400
   5500
   5600
   5700
   5800
   Option D
   (21 ÷ 798) * (49 ÷ 735) * ? = 10
   1/38 * 1/15 * ? = 10
   ? = 5700
   

    

4. 33.11 % of 80 + 69.780 % of 310 + 127 = ?
   350
   360
   370
   380
   390
   Option C
   26 + 217 + 127 = ?
   370 = ?
   

    

5. √1155.55 + √899.99 – (6858.9)(1/3) =?
   15
   25
   35
   45
   55
   Option D
   √1156 + √900 – (6859)(1/3) =?
   34 + 30 – 19 = ?
   45 = ?
   

    

6. (17.06)^2 – (8.911)^2 + (325.17 + 549.85) ÷ ? = 233
   15
   25
   35
   45
   55
   Option C
   17^2 – 9^2 + (325 + 550)/x = 233
   289 – 81 + (875/x) = 233
   (875/x) = 233 – 289 + 81
   (875/x) = 25
   x = 875/25 = 35
   

    

7. 18.7892 % of 898.99 + 13.8199 % of 1047 = 20 % of ?
   1610
   1590
   1790
   1280
   1450
   Option B
   (19/100) * 900 + (14/100) * 1050 = (20/100) * x
   171 + 147 = (x/5)
   318 * 5 = x
   x = 1590
   

    

8. 26.89 × 348.519 ÷ 5.0011 – ? – 129.52 = 12
   1890
   1780
   `1860
   1750
   2380
   Option D
   (27 * 350)/5 – x – 130 = 12
   1890 – x – 130 = 12
   1890 – 130 – 12 = x
   x = 1748 = 1750
   

    

9. 300.078 = 22.99 × 256.089 ÷ 8.11 – ? – 159.75
   273
   274
   275
   276
   278
   Option D
   300 = 23 * (256/8) – x – 160
   x = 736 – 160 – 300 = 276
   

    

10. (4.67/9.192) of 179.81 × 60.12 ÷ 14.93 – ? = 122
    278
    279
    280
    281
    282
    Option A
    (5/9) * 180 * (60/15) – 122 = x
    x = 400 – 122 = 278
    

     

11. 25.5 % of 350.175 – 31.55 % of 599.27 =? – 157.94 – (27.21)2
    678
    786
    876
    768
    788
    Option B
    26 % of 350 – 32 % of 600 = x – 158 – 272
    (26/100)*350 – (32/100)*600 = x – 158 – 729
    91 – 192 + 158 + 729 = x
    X = 786
    

     

12. 5 2/3 of 300.178 + 3 1/8 of 430.167 =? % of 749.85
    410
    411
    412
    413
    414
    Option C
    6 of 300 + 3 of 430 = x% of 750
    (300*6) + (3*430) = (x/100)*750
    1800 + 1290 = 15x/2
    (3090*2)/15 = x
    X = 412
    

     

13. (554.3672 + 958.6325 – 432.72) ÷ 14.85 =?
    72
    73
    74
    76
    75
    Option A
    (554 + 959 – 433)/15 = x
    X = 1080/15 = 72
    

     

14. [((148.9 ÷ 2.99) ÷ 10.97)2 + (64.13 × 9.05)1/2] = (?)2
    5
    6
    7
    8
    9
    Option C
    [((150 ÷ 3) ÷ 10)2 + (64 × 9)1/2] = x2
    (150/30)2 + (8*3) = x2
    25 + 24 = x2
    X2 = 49
    X = 7
    

     

15. ? ÷ 4.009 + 71.903 × 15 = 4823.612 + 1598.23
    23134
    13465
    32415
    12354
    21368
    Option E
    x ÷ 4 + 72 × 15 = 4824 + 1598
    (x/4) + 1080 = 6422
    (x/4) = 5342
    X = 5342*4 = 21368
    

     

1. Raju and Shiva are partners in a business. Raju invests Rs. 30,000 for 8 months and shiva invests Rs.40,000 for 10 months. Out of a profit of Rs.48,000. Raju ’s share is
   
   Rs.13,000
   Rs.12,000
   Rs.18,000
   Rs.17,000
   Rs.15,000
   Option
   Ratio of their shares =(30000×8):(40000×10)
   = 3 : 5
   Raju’s share = Rs.(48,000 ×3/8) = Rs.18,000.

    

2. A started a business investing Rs.35,000. B joined him after six months with an amount of Rs.52,500 and C joined them with Rs.70,000 after another six months. The amount of profit earned should be distributed in what ratio among A, B and C respectively, 3 years after A started the business?
   12 : 15 : 17
   11 : 15 : 16
   17 : 13 : 16
   12 : 15 : 16
   10 : 15 : 16
   Option D
   A : B : C = (35,000×36) : (52,500×30) : (70,000×24)
   = 12 : 15 : 16

    

3. Prem and Ram started a business investing Rs.80,000 and Rs.10,000 respectively. In what ratio the profit earned after 2 years be divided between Prem and Ram respectively?
   7:1
   13:1
   12:1
   8:1
   10:1
   Option D
   Prem : Ram = 80000 : 10000
   =8:1

    

4. Person 1 started a business by investing Rs.50,000. After six months, Person 2 joined her with a capital of Rs.80,000. After 3 years, they earned a profit of Rs.24,500. What was Person 1’s share in the profit?
   Rs.15500
   Rs.10500
   Rs.14500
   Rs.17500
   Rs.13500
   Option B
   person 1 : person 2 = 50000 × 36 : 80000 × 30
   = 3 : 4
   Person 1’s share = Rs.24500 × 3 7
   = Rs.10500
   

    

5. man 1 and 2 started a partnership business investing some amount in the ratio of 3 : 5. man 3 joined them after six months with an amount equal to that of man 2. In what proportion should the profit at the end of one year be distributed amount 1, 2 and 3?
   36 : 60 : 30
   26 : 60 : 30
   46 : 60 : 40
   6 : 60 : 30
   18 : 60 : 30
   Option A
   Let the initial investments of 1 and 2 be 3x and 5x.
   1: 2 : 3 =(3x ×12):(5x ×12):(5x×6) = 36 : 60 : 30

    

6. A and B started a business investing Rs.20,000 and Rs.35,000 respectively. Out of a total profit of Rs.11,000. B’s share is
   Rs.7,000
   Rs.6,000
   Rs.5,000
   Rs.8,000
   Rs.9,000
   Option A
   Ratio of their shares = 20,000 : 35000 =4 : 7
   B’s share = Rs.(11,000×7/11) = Rs.7,000

    

7. Kushal started a business investing Rs.18000. After five months, Sanjay joined with a capital of Rs.16000. If at the end of the year, they earn a profit of Rs.13,940, then what will be the shares of Sanjay in the profit?
   
   Rs.5,760
   Rs.6,760
   Rs.4,760
   Rs.9,760
   Rs.8,760
   Option C
   Kushal : Sanjay =(18000×12):(16000 ×7) = 108 : 56 = 27 : 14
   Sanjay’s share = Rs.(13,940 ×14/41) =Rs.4,760.
   

    

8. Mano, Praju and Aamin invested Rs.8000, Rs.4000 and Rs.8000 respectively in a business. Aamin left after six months .If after eight months, there was a gain of Rs.8010, then what will be the share of Praju?
   
   Rs.890
   Rs.800
   Rs.750
   Rs.1780
   Rs.900
   Option D
   Murugan : Prasanna : Arun =(8000×6):(4000×8):(8000×8)
   = 48 : 32 : 64 = 3 : 2 : 4
   Kamal’s share = Rs.8010×2/9) = Rs.1780
   

    

9. A, B and C enter into a partnership. They invest Rs.20,000, Rs.40,000 and Rs.60,000 respectively. At the end of the first year, B withdraws Rs.20,000 while at the end of the second year, C withdraws Rs.40,000. In what ratio will the profit be shared at the end of 3 years?
   3 : 4 : 7
   3 : 5 : 7
   3 : 6 : 7
   5 : 4 : 7
   8 : 4 : 7
   Option A
   A : B: C =(20000×36) : (40000×12+20000×24) : (60000×24+20000×12)
   = 3 : 4 : 7

    

10. Ankitha starts a business with Rs.3500 and after 5 months, Bindhu joins with Ankitha as his partner. After a year, the profit is divided in the ratio 2 : 3. What is Bindhu’s contribution in the capital?
    
    10000
    12000
    15000
    16000
    9000
    Option E
    Let B’s capital be Rs.x.
    Then, 3500×12 ⁄ 7x=2⁄3
    3500×12×3=2×7x
    ‹=›14x = 126000
    x = 9000