1. 15 boys or 30 girls or 45 men can complete the work in 12 days. Find the number of days taken by 8 boys and 10 girls and 15 men to complete the work?
   12
   13
   10
   9
   8
   Option C
   15 boys= 30 girls

   1 boy = 2 girl

   15 boys = 45 men

   1 boy = 3 men

   8 boy + 10 girls + 15 men = 8 b + 5 b + 5 b = 18 boys

   boys ……. days

   15 …………… 12

   18 ………….. ?

   (15 * 12) = (18 * x)

   x = (15 * 12) / 18 = 10 days

    

2. Kiran and Kishore do a certain work in 42 days and 30 days respectively.Kishore started the work alone and then after 6 days Kiran joined him till the completion of the work. How long did the work last?
   10
   20
   30
   40
   50
   Option B
   Work done by Kishore in 6 days = (1/30) * 6 = 1/5

   Remaining work = 1 – 1/5 = 4/5

   (Kiran + Kishore)’s 1 day work = 1/42 + 1/30

   = > (42 + 30) / (42 * 30)

   = > 2/35

   Together can complete the work in 35/2 days

   So 4/5 work will be done by both

   = > (4/5) * (35/2)

   = > 14 days

   Total time taken = 14 + 6 = 20 days

    

3. P alone can do a piece of work in x days. Q alone can do the same work in 10 days more than P. R alone can do the same work in 10 days. If all of them working together and completed the work in 5 5/11 days, then find the value of ‘x’?
   20
   30
   40
   50
   60
   Option A
   Given,

   (1/x) + (1/(x + 10)) + 1/10 = 11/60

   (1/x) + (1/(x + 10)) = 11/60 – 1/10

   (2x + 10) / (x2 + 10x) = 1/12

   24x + 120 = x2 + 10x

   x2 – 14x – 120 = 0

   x = 20, -6 (Eliminate –ve value)

   x = 20 days

    

4. A mother is 2 times faster than his daughter. If the daughter can complete a piece of work in 16 days, then how long will it take for both mother and daughter to complete the same piece of work?
   4 1/2
   3 1/3
   5 1/3
   2 1/5
   1 1/7
   Option C
   Efficiency ratio of mother and daughter = 2 : 1

   The day ratio of mother and daughter = 1 : 2 (x, 2x)

   2x = 16

   x = 8

   mother and dauhter’s one day work,

   = > (1/8) + (1/16)

   = > 3/16

   mother and daughter can complete a piece of work in,

   = > 16/3 = 5 1/3 days

    

5. 46 female labours earned Rs. 172500 by working 25 days. How many male labours must work for 24 days to receive Rs. 230400 provided the daily wages of a man is twice that of a female labour?
   20
   22
   30
   32
   40
   Option D
   1 F’s 1 day wages = Total wages / (Number of days * total F)

   = > 172500 / (46 * 25) = 150

   1 M’s 1 day wages = 2 * 1 F’s 1 day wages = 2 * 150 = 300

   Total F = 230400 / (24 * 300) = 32 M

    

6. 12 Male workers can complete a work in 15 days and 15 Female workers can complete the same work in 24 days. Then find 15 males and 20 Females can complete the work?
   7 1/5
   5 1/5
   3 1/5
   4 1/5
   10 1/5
   Option A
   Total work = M * days

   12 m * 15 = 15 f * 24

   1 m = 2 f
   15 m + 20 f = 15 m + 10 m = 25 m

   Males days

   12 15

   25 ?

   (12 * 15) = (25 * x)

   x = (12 * 15) / 25 = 36 / 5 = 7 1/5 days

    

7. Lal is 20 % more efficient than Taj to complete the work. Both of them together complete the work in 5 5/11 days. Find the number of days taken by Lal alone to complete the work?
   10
   11
   12
   13
   14
   Option A
   Efficiency ratio = > 120: 100 = 6: 5

   Day ratio = > 5: 6 (5x, 6x)

   According to the question,

   (1/5x) + (1/6x) = (11/60)

   (6x + 5x) / (30×2) = (11/60)

   11x / 30×2 = (11/60)

   x = 2

   The number of days taken by Lal alone to complete the work = 10 days

    

8. A garments wanted 40 women to complete a project in 15 days. 40 women started working, after 9 days they notices that only three-fifth of the work gets completed. Then how many extra women can be employed to complete the remaining work on time?
   0
   1
   2
   3
   4
   Option A
   women Days Work

   40 15 1

   40 9 3/5

   ? 6 2/5

   (40 * 9) / (3/5) = [(40 + x) * 6] / (2/5)

   40 = 40 + x

   x = 0

   There is no extra women needed to complete the work.

    

9. A and B can complete the work alone in x days and (x + 10) days respectively. They work together to complete the work in 6 2/3 days. Find the efficiency ratio of A to that of B?
   2 : 5
   2 : 1
   3 : 1
   2 : 7
   4 : 1
   Option B
   A = x days, B = (x + 10) days

   Given,

   (1/x) + (1/(x + 10)) = 1/(6 2/3)

   [x + x + 10] / [x (x + 10)] = 3/20
   (2x + 10) / (x2 + 10x) = 3/20

   40x + 200 = 3×2 + 30x

   3×2 – 10x – 200 = 0

   x = 10, -6.66 (negative value will be eliminated)

   So, x = 10

   The efficiency ratio of A and B = 1/10 : 1/20 = 2 : 1

    

10. 35 workers can complete a piece of work in 18 days. After 8 days from the start of the work, some worker left. If the remaining work was completed by the remaining workers in 14 days, then find the number of workers left after 8 days from the start of the work?
    10
    15
    20
    12
    11
    Option A
    Total work = workers*days

    Total units of work = 35*18 = 630 work

    Work done in 8 days = 35*8 = 280 work

    Remaining work = 630 – 280 = 350 work

    Let the number of workers left after 8 days be x,

    According to the question,

    350 / 14 = 35 – x

    25 = 35 – x

    x = 10

    After 8 days from the start of the work, 10 worker left the job.

     

1. x³ – 4x² + 3x = 0
   7y³ – 23y² + 6y = 0
   
   x
   x<=y
   x>y
   x>=y
   none
   Option D
   x³ – 4x² + 3x = 0
   This gives x(x² – 4x + 3) = 0
   ⇒ x(x – 3)(x – 1) = 0
   ⇒ x = 0, 1 or 3
   7y³ – 23y² + 6y = 0
   ⇒y(7y² – 23y + 6) = 0
   ⇒y(7y – 2)(y – 3) = 0
   ⇒ y = 0, 2/7 or 3
   x>=y
   

    

2. x²+34x+289=0
   y²+28y+196=0
   
   x
   x<=y
   x>y
   x>=y
   none
   Option A

   x²+17x+17x+289=0
   So, X =-17

   Y = -14
   =>x
   

    

3. 4X² + 24X + 36 =0
   5Y² + 20Y + 20 =0
   
   x
   x<=y
   x>y
   x>=y
   none
   Option A
   4X² + 24X + 36 =0
   X² + 6X + 9 =0
   X² + 3X + 3x + 9 =0
   X = -3
   5Y² + 20Y + 20 =0
   Y² + 4Y + 4 =0
   Y² + 4Y + 4y + 4 =0
   Y= -2
   So, Y>X
   

    

4. x⁵ = 32
   y¹⁰ = 1024
   
   x
   x<=y
   x>y
   x>=y
   none
   Option D
   x = 2 and y = + /- 2

    

5. x² + x = 56
   y² – 17y + 72 = 0
   
   x
   x<=y
   x>y
   x>=y
   none
   Option A
   x² + x = 56
   x² + 8x – 7x -56 = 0
   x(x+8) -7 (x+8) =0
   (x+8) (x-7) =0
   x= -8 , 7
   y² – 17y + 72 = 0
   y² – 9y – 8y + 72 =0
   y(y-9) -8(y-9) = 0
   (y-9) (y-8) =0
   y= 9, 8
   x < y
   

    

6. 3x² +13x +14 = 0
   3y²+11y+10 = 0
   
   x
   x<=y
   x>y
   x>=y
   none
   Option B
   . 3x² +13x +14 = 0
   => (3x+7) (x+2)
   => x = -7/3, -2
   3y²+11y+10 = 0
   => (y+2) (3y+5)
   => y = -2 , -5/3
   So x ≤ y

    

7. 144x² – 23 = 2
   12y+√25=√100
   
   x
   x<=y
   x>y
   x>=y
   none
   Option B
   144x² = 23+2=25
   x²=25/144
   =>x=+or- 5/12
   12y+√25=√100
   12y= = 5
   y=5/12
   y ≥ x

    

8. x² – 32x + 255 = 0
   y² – 39y + 378= 0
   
   x
   x<=y
   x>y
   x>=y
   none
   Option A
   x² – 32x + 255 = 0
   x² – 15x – 17x + 255 = 0
   x(x-15)-17(x-15)=0
   (x-15)(x-17)=0
   x=15,17
   y² – 39y + 378 = 0
   y² – 18y – 21y + 378 = 0
   y(y-18)-21(y-18)=0
   (y-18)(y-21)=0
   y=18, 21
   y>x
   

    

9. x² – 15x + 84 = 4x
   y² – 15y + 156 = 10y
   
   x
   x<=y
   x>y
   x>=y
   none
   Option B
   x² – 15x + 84 = 4x
   ⇒ x² – 19x + 84 = 0
   ⇒ x² – 12x – 7x + 84 = 0
   ⇒ x( x – 12) – 7 ( x – 12) = 0
   ⇒ (x – 7) (x – 12) = 0
   ⇒ x = 7, 12

   y² – 15y + 156 = 10y
   ⇒ y² -25y + 156 = 0
   ⇒ y² – 12y – 13y + 156 = 0
   ⇒ y( y – 12) – 13 ( y – 12) = 0
   ⇒ (y – 13) (y – 12) = 0
   ⇒ y = 13, 12
   ∴ y ≥ x
   

    

10. 6x + 3y = 24
    x +2y = 8.5
    
    x
    x<=y
    x>y
    x>=y
    none
    Option A
    x = 2.5 y = 3 heyce, x < y

     

Directions (1-5): Find the wrong term in the series:

1. 40.. 72.. 106.. 148.. 196
   40
   72
   106
   148
   196
   Option B
   40 + 30 = 70
   70 + 36 = 106
   106 + 42 = 148
   148 + 48 = 196
   

    

2. 8..18.. 57.. 232..1160
   8
   18
   57
   232
   1160
   Option E
   8 * 2 + 2 = 18
   18 * 3 + 3 = 57
   57 * 4 + 4 = 232
   232 * 5 + 5 = 1165
   

    

3. 80.. 96… 121… 159…. 206
   80
   96
   121
   159
   206
   Option D
   80 + 4² = 96
   96 + 5² = 121
   121 + 6² = 157
   157 + 7² = 206
   

    

4. 316… 324… 309… 329… 306
   316
   324
   309
   329
   306
   Option C
   316 + 8 = 324
   324 – 12 = 312
   312 + 17 = 329
   329 – 23 = 306
   

    

5. 1333.. 1731.. 2201.. 2749.. 3378
   1333
   1731
   2201
   2749
   3378
   Option E
   11³ + 2 = 1333
   12³ + 3 = 1731
   13³ + 4 = 2201
   14³ + 5 = 2749
   15³ + 6 = 3381
   

    

Directions (6-10): The following series follow some particular pattern. Following that pattern find the number which will come in place of ‘?’.

6. 11.. 32..126.. 627.. 3758.. ?
   
   26090
   26302
   26301
   24560
   25671
   Option C
   Clue: *3 – 1, *4 – 2, *5 – 3, *6 – 4, *7 – 5,..
   answer 26301
   

    

7. 71.. 248.5… 869.75… ? .. 10654.4375
   
   6044.125
   2044.125
   1544.125
   2044.125
   3044.125
   Option E
   Clue:- *3.5
   answer 3044.125
   

    

8. 34.. 42.. 111… 397..? … 8781
   
   1713
   1535
   2763
   2761
   2289
   Option
   *1 + 23, *2 +33, *3 + 43, *4 + 53, *5 + 63,..
   answer 1713
   

    

9. 1127.. 1248.. 2976.. 3145.. 5889… ?
   
   5113
   5223
   6114
   6118
   5544
   Option C
   +11^2, +12^3, +13^2, +14^3, +15^2, +16^3,..
   answer 6114
   

    

10. 747… 765… 788… 821.. 869… 937… ?
    
    1100
    1200
    1400
    1030
    1500
    Option D
    747… 765… 788… 821.. 869… 937… 1030
    18 23 33 48 68 93
    5 10 15 20 25

     

1. The shopkeeper sold TV table for Rs. 2000 and incurred a loss. Had he sold the item for Rs. 3600, his gain would have been equal to three times of the amount of loss that he incurred. At what price should he sell the article to gain 20 %?
   Rs. 4000
   Rs. 2340
   Rs.1880
   Rs. 2890
   Rs. 2880
   Option E
   Loss = C.P – 2000

   Profit = 3600 – C.P

   Profit = 3*Loss

   3600 – C.P = 3 * [C.P – 2000]

   3600 – C.P = 3C.P – 6000

   9600 = 4C.P

   C.P = 9600/4 = Rs. 2400

   SP = 2400 * (120/100) = Rs. 2880

    

2. The sum of lengths of two express trains X and Y is 640 m. The speed of train X to that of train Y is in the ratio of 3 : 4. The time taken by train X and train Y to cross a pole is in the ratio of 4 : 5. Find the difference between the length of train X to that of Y?
   150 m
   160 m
   120 m
   130 m
   140 m
   Option B
   Let the length of train X be x m,

   So, the length of train Y = 640 – x

   The speed of train X to that of train Y is in the ratio = 3 : 4 (3y, 4y)

   Time = Length of train / Speed

   4 = x / 3y

   x = 12y -> (1)

   (640 – x) / 4y = 5

   640 – x = 20y

   640 – 12y = 20y

   640 = 32y

   y = 20

   The length of train x = x = 12 * 20 = 240 m

   The length of train Y = 640 – x = 640 – 240 = 400 m

   Required difference = 400 – 240 = 160 m

    

3. In still water the boat speed is 36 km/hr and the speed of stream is 12 km/hr. The time taken by boat to travel from X to Y downstream is 3 ½ hours less to travel from Y to Z upstream. If the distance between X to Y is 24 km more than the distance between Y to Z, then find the distance between X to Y?
   200 km
   216 km
   234 km
   116 km
   316 km
   Option B
   The speed of downstream = 36 + 12 = 48 km/hr

   The speed of upstream = 36 – 12 = 24 km/hr

   Let the distance between Y to Z be x km,

   The distance between X to Y = (x + 24) km

   Given,

   x / 24 – (x + 24) / 48 = 7/2

   (2x – x – 24) / 48 = 7/2

   x – 24 = 168

   x = 192

   The distance between X to Y = x + 24 = 192 + 24 = 216 km

    

4. Man 1 and 2 are 585 km away and both of them start moving towards each other at the same time. After some time when 1 has travelled 360 km he meets 2. What is the ratio of the speed of 1 to that of 2?
   8 : 11
   7 : 5
   8 : 9
   4 : 5
   8 : 5
   Option E
   Distance travelled by 1 when he met = 360 km

   Distance travelled by 2 when he met = 585 – 360 = 225 km

   The ratio of the speed of 1 to that of 2

   = > 360 : 225

   = > 8 : 5

    

5. Juice and water in a jar contains 32 liters Juice and 10 liters water. X liter of Juice and X liter of water is added to a new mixture. If 40 % of new mixture is 24 liters, then find the value of X (In liters)?
   5
   6
   7
   8
   9
   Option E
   40/100) * New mixture = 24

   New mixture = 24 * (5/2) = 60 liters

   Given,

   32 + X + 10 + X = 60

   42 + 2X = 60

   2X = 18

   X = 9

    

Directions (6-10): Following Radar graph shows the total number of Tulips in 5 different boxes and the blue Tulips in them.



Blue line—–> total tulips
Red line——> blue tulips

6. Find the difference between the average number of Tulips in the box A, B and C to the average number of Tulips other than the blue colour in the box B, D and E.
   25
   26
   27
   28
   29
   Option A
   Number of Tulips other than the blue colour in the box,
   B = 40 – 10 = 30
   D = 75 – 45 = 30
   E = 30 – 15 = 15
   The average = (30 + 30 + 15)/3 = 25
   Average number of Tulips in the box A, B and C =(60 + 40 + 50)/3 = 50
   Required difference = 50 – 25 = 25
   

    

7. Find the ratio of sum of Red color Tulips in A and D together to the pink color Tulips in same boxes, if the boxes have only three color Tulips and the ratio of pink and red color Tulips in the boxes A and D is 13: 7 and 4: 11 respectively.
   8: 17
   18: 37
   10: 17
   18: 10
   18: 17
   Option E
   Box A:
   Red + Pink = 60 – 20 = 40
   Red = (40/20) * 7 = 14
   Pink = (40/20) * 13 = 26
   Box D:
   Red + Pink = 75 – 45 = 30
   Red = (30/15) * 11 = 22
   Pink = (30/15) * 4 = 8
   Required ratio = (14 + 22): (26 + 8)
   = 18: 17
   

    

8. The boxes have only three color Tulips. The red color Tulips in the boxes B and C is what percentage of number of blue Tulips in the box D, if the grey and red color Tulips in the box B and C is in the ratio 3: 2 and 4: 3 respectively?
   20
   40
   60
   80
   85
   Option C
   Number of Tulips other than the blue colour in the box,
   B = 40 – 10 = 30
   Red Tulips in B = (30/5) * 2 = 12
   C = 50 – 15 = 35
   Red Tulips in C = (35/7) * 3 = 15
   Required percentage = [(12 + 15)/45] * 100 = 60%
   

    

9. The boxes have only three color Tulips. If the ratio of yellow color Tulips and pink color Tulips in the boxes is 3: 2, find the average difference between them.
   6
   7
   8
   9
   4
   Option A
   Number of Tulips other than the blue colour in the box,
   A = 60 – 20 = 40
   B = 40 – 10 = 30
   C = 50 – 15 = 35
   D = 75 – 45 = 30
   E = 30 – 15 = 15
   The sum = 40 + 30 + 35 + 30 + 15 = 150
   The ratio of yellow color Tulips and pink color Tulips in the boxes = 3: 2
   Yellow color Tulips = (150/5) * 3 = 90
   Pink color Tulips = (150/5) * 2 = 60
   Required difference = (90/5)- (60/5) = 30/5 = 6
   

    

10. Total number of blue Tulips in all the boxes together is approximately what percentage of total Tulips in all the boxes together?
    40%
    30%
    45%
    20%
    25%
    Option A
    Required % = [(20 + 10 + 15 + 45 + 15)/(60 + 40 + 50 + 75 + 30)] * 100
    = (105/255) * 100
    = 40%
    

     

(1-5):The following information gives the percentage distribution of total number of Tablets sold by Samsung in 6 different months in the year 2019.
Jan – 15%, Feb – 18%, Mar – 23%, Apr – 20%, May – 14%, Jun-10%
Total number of tablets= 15000

1. Find the difference between the total number of Tablets sold by Samsung in Jan and may together to that of total number of Tablets sold by Nokia in Feb and Mar together, if the ratio between the total Tablets sold by Samsung to that of Nokia is 3 : 4 and the percentage distribution of total Tablets sold by Nokia is same as Samsung.
   
   3450
   3850
   4450
   4560
   4455
   Option B
   The total number of Tablets sold by Samsung in Jan and may together
   = >15000 * (29/100) = 4350
   The total number of Tablets sold by Nokia in Feb and Mar together
   = >[15000 * 4/ 3] * (41/100) = 8200
   Required difference = 8200 – 4350 = 3850
   

    

2. Find the difference between the average number of Tablets sold by Samsung in Feb and Mar together to that of the average number of Tablets sold by Samsung in may and June together.
   1275
   1335
   1445
   1555
   1455
   Option A
   The average number of Tablets sold by Samsung in Feb and Mar together
   = > 15000 * (41/100) * (1/2) = 3075
   The average number of Tablets sold by Samsung in May and June together
   = > 15000 * (24/100) * (1/2) = 1800
   Required difference = 3075 – 1800 = 1275
   

    

3. If the total number of Tablets sold by Samsung in Aug 2019 is 45% more than the total number of Tablets sold by the same company in April 2019 and the total number of Tablets sold by Samsung in Sep 2019 is 60% more than the total number of Tablets sold by the same company in May 2019, then find the total number of Tablets sold by the Samsung in Aug and Sep together.
   5590
   7890
   7710
   3450
   7800
   Option C
   The total number of Tablets sold by the Samsung in Aug and Sep together
   = >15000 * (20/100) * (145/100) + 15000 * (14/100) * (160/100)
   = > 4350 + 3360 = 7710
   

    

4. Find the ratio between the total number of Tablets sold by Samsung in Feb and Apr together to that of total number of Tablets sold by Samsung in Jan and June together.
   15:12
   15:90
   38:25
   56:23
   45:23
   Option C
   The total number of Tablets sold by Samsung in Feb and Apr together
   = > 15000 * (38/100)
   The total number of Tablets sold by Samsung in Jan and June together
   = > 15000 * (25/100)
   Required ratio = [15000 * (38/100)]: [15000 * (25/100)]
   =38: 25
   

    

5. What is the average number of Tablets sold by Samsung in Feb?
   4350
   3470
   6590
   3340
   6690
   Option A
   The total number of Tablets sold by Samsung in Jan and may together
   = >15000 * (29/100) = 4350

    

(6-10): Study the following information carefully and answer the questions given below.



Note: For each correct answer the student is given mark and for every incorrect answer 0.25 marks are deducted. No marks are deducted for un – attempted questions.

6. Number of incorrect answers marked by Naren is what percentage More/less than the number of incorrect answers marked by Arun?
   
   20
   10
   15
   25
   30
   Option A
   Arun:
   Answered=85/100 * 200 =170
   Correct=170 * 12/17 = 120
   Wrong=170-120=50
   score= 120-0.25*50 = 107.5
   Akil:
   Answered=65/100 * 200 =130
   Correct=130 * 11/13 = 110
   Wrong=130-110=20
   score= 110-0.25*20 = 105
   Vinoth:
   Answered=50/100 * 200 =100
   Correct=100 * 22/25 = 88
   Wrong=100-88=12
   score= 88-0.25*12 = 85
   Naren:
   Answered=70/100 * 200 =140
   Correct=140 * 31/35 = 124
   Wrong=140-124=16
   score= 124-0.25*16 = 120

   Number of incorrect answers marked by Naren= 16
   the number of incorrect answers marked by Arun= 20
   Required percentage = (20-16) / 20 * 100
   = 20 % Less
   

    

7. What is the difference between the marks scored by Vinoth and Arun?
   5
   10
   15
   20
   25
   Option D
   Marks scored by Vinoth = 85
   Marks scored by Arun = 105
   Difference = 20
   

    

8. What is the ratio of correct answers marked by Arun to the wrong answers marked by Naren?
   10:8
   14:8
   15:2
   16:5
   11:9
   Option C
   Correct answers marked by Arun = 120
   Wrong answers marked by Naren = 16
   Ratio = 120/16 = 15: 2
   

    

9. Who scored the 2nd highest marks?
   Arun
   Akil
   Vinoth
   Naren
   Arun and akil
   Option A
   Mark of,
   Arun = 107.5
   Akil = 105
   Vinoth = 85
   Naren= 120
   Arun has 2nd highest mark.
   

    

10. What is the score obtained by Arun?
    145.9
    122.8
    125.6
    107.5
    156.9
    Option D
    Arun’s score = 120 – (0.25 * 50) = 107.50

     

1. An admission of a school has increased in every academic year by 6%. If the number of students in the previous academic year is 2500, then what is the number of students in the next year?
   5690
   2916
   1456
   3456
   1146
   Option B
   8% of 2500 = 8/100 x2500 = 200
   Total no., of students in the current year = 2500 + 200 = 2700
   8% of 2700= 216
   So number of students in the next year = 2700 + 216 = 2916
   

    

2. 5 1/8 – 2 1/16 – 1 1/ 16 – 1 1/ 8 = x + 2 1/8.Find the value of x?
   -11/67
   -12/89
   -34/56
   -66/90
   -29/16
   Option E
   Take 1/8 as common ,
   Then 1/8 ( 5 – 2/2 – ½ -1) = x + 2 1/8
   1/8 ( 5 -1 -1/2 -1) = x + 2 1/8
   1/8 ( 5-2-1/2) = x + 17 /8
   1/8 ( 5 /2) = x + 17 /8
   5/16 = x + 17/8
    X = – 29/16
   

    

3. A shoots the balloons 6 times to B is 4 but A blasts only once in 2 shots while B blasts once in 3 shots. When B has missed 28 times, A has blasted :
   56
   67
   78
   63
   61
   Option D
   Let the no of shots be x. Then,
   Shots fired by A =(6/10)x
   Shots fired by B = (4/10)x
   blasting shots by A = 1/2 of (6/10)x = (6/20)x
   Shots missed by B = 1/3 of (4/10)x = (4/30)x
   (4/30)x = 28 ⇒ x = 210
   Balloons blasts by A = (6/20)x = 6 / 20 x 210 =63
   

    

4. In a group of lions and rabbits, the number of legs are 30 more than twice the number of heads. What is the number of lions in the group ?
   10
   13
   14
   15
   17
   Option D
   Let the number of lions be x and number of rabbits be y.
   Then 4x + 2y = 2(x + y) + 30
   => 2 x = 30
   => x = 15

    

5. A tank is 3/5 full. If 28 litres of water is added to the tank, it becomes 5/6 full. The capacity of the tank is:
   110
   120
   130
   140
   150
   Option B
   Let the capacity of the tank be x litres.
   Then, (5x/6 – 3x/5) = 28
   => 25x – 18x / 30
   = > 7x / 30 = 28
   => 7x = 840
   => x = 120
   

    

6. 13.928 % of 210 + 43.0851 % of 350 + 71.428% of 140 = ?
   340
   450
   280
   600
   320
   Option C
   1/7 * 210 + 3/7 * 350 + 5/7 * 140 = ?

   30 + 150 + 100 = ?

   280 = ?

    

7. 23.002 * ? + 117.002 = 672.123– 232.23
   170
   180
   190
   200
   210
   Option B
   23 * ? + 117 = 672 – 232

   23 * ? + 117 = 4489 – 232

   ? * 23 = 4140

   ? = 180

    

8. (21.23 ÷ 797.90) * (49 ÷ 735) * ? = 10
   4700
   5700
   6700
   8700
   9700
   Option B
   (21 ÷ 798) * (49 ÷ 735) * ? = 10

   1/38 * 1/15 * ? = 10

   ? = 5700

    

9. 300.078 = 23.003 × 256.189 ÷ 8.11 – ? – 159.75
   117
   456
   276
   231
   876
   Option C

   300 = 23 * (256/8) – x – 160

   x = 736 – 160 – 300 = 276

    

10. (4.99/8.92) of 180.181 × 60.12 ÷ 14.93 – ? = 122
    167
    178
    267
    278
    567
    Option D
    (5/9) * 180 * (60/15) – 122 = x

    x = 400 – 122 = 278

     

1. Find the wrong term:-
   534…, 266…, 132…, 64…31.5…, 14.75
   65
   132
   534
   31.5
   14.75
   Option A
   The pattern is, ÷ 2 – 1

    

2. 7…, 4…, 5…, 10…, 20…, 52.5…, 160.5
   7
   4
   5
   10
   20
   Option D
   The pattern is, *0.5 + 0.5, *1 + 1, *1.5 + 1.5, *2 + 2, *2.5 + 2.5, *3 + 3,…

    

3. 1924…, 1790…, 1610…, 1370….1064…, 684
   1924
   1790
   1610
   1370
   1064
   Option B
   The pattern is, -(11^2 + 11), -(13^2 + 13), -(15^2 + 15), -(17^2+17), -(19^2 + 19),..

    

4. 172…, 200…, 263…, 388…, 604…, 947…, 1459
   172
   200
   263
   388
   604
   Option B
   The pattern is, +3^3, +4^3, +5^3, +6^3, +7^3, +8^3,..

    

5. 9…, 16…, 45…, 175…, 875…, 5244
   9
   16
   45
   175
   875
   Option D
   The pattern is, *2 – 2, *3 – 3, *4 – 4, *5 – 5, *6 – 6,..

    

Study the following and answer the questions that follow:
There are N number of guest went for marriage and likes veg and non-veg items. Each guest have like at least one items of veg and non-veg. 62 % of the guest like veg items. 15 % of the guest like both veg and non-veg items. The number of guest like s only veg items is 36 more than that of guest like only non-veg items.

6. Find the total number of guest like only veg items?
   
   180
   188
   190
   198
   200
   Option B
   The percentage of guest like both veg and non-veg items = 15 %
   The percentage of guest like only Veg items = 62 – 15 = 47 %
   The percentage of guest like only Non-veg items = 100 – 15 – 47 = 38 %
   According to the question,
   (47 % – 38 %) of total guest went for marriage = 36
   9 % of total guest went for marriage = 36
   Total guest went for marriage = 36 * (100 / 9) = 400
   Total number of guest like only Veg items = 400 * (47/100) = 188
   Total number of guest like only Non-veg items = 400 * (38/100) = 152
   Total number of guest like both Veg and Non-veg items
   = > 400 * (15/100) = 60
   Total number of guest like only Veg items
   = > 400 * (47/100) = 188
   

    

7. Find 175 % of total number of guest like both veg and non-veg items?
   100
   105
   110
   115
   120
   Option B
   Total number of guest like both Veg and Non-veg items
   = > 400 * (15/100) = 60
   175 % of total number of guest like both veg and non-veg items
   = > 60 * (175/100) = 105
   

    

8. Find the difference between the total number of guest like only veg items to that of non-veg items?
   36
   46
   56
   76
   86
   Option A
   The total number of guest like only veg items = 188
   The total number of guest like only non-veg items = 152
   Required difference = 188 – 152 = 36
   

    

9. Find the ratio between the total number of guest like veg items to that of non-veg items?
   62: 13
   2: 53
   10: 53
   62: 53
   62: 14
   Option D
   The total number of guest like veg items = 62 %
   The total number of guest like non-veg items = 53 %
   Required ratio = 62: 53
   

    

10. Total number of guest like only veg items is appronimately what percentage of total number of guest like only non-veg items?
    120%
    125%
    130%
    135%
    140%
    Option B
    Total number of guest like only veg items = 188
    Total number of guest like only non-veg items = 152
    Required % = (188/152) * 100 = 123.68 % = 125 %
    

     

Directions (1-5): The following table shows the different state’s tax collection in lakhs in different month.

1. What is the maximum difference growth tax collection in the month April between the states Karnataka and Tamil nadu?
   0.34
   0.67
   0.24
   0.56
   0.17
   Option C
   Growth collection in the month April in Karnataka = 966 – 674/ 674 = 0.43
   Growth collection in the month April in Tamilnadu = 6765 – 5677/ 5677 = 0.19
   Difference = 0.43 – 0.19= 0.24
   

    

2. What is the ratio of growth in the month and Aug in Goa?
   2.1 : 0.07
   2.3 : 0.5
   2 : 0.07
   2.3 : 0.15
   2.3 : 0.07
   Option E
   Growth in tax collection in Feb = 668 – 45 /45 = 13.8
   Growth in tax collection in Aug = 95 – 66/66 = 0.44
   Ratio = 13.8 : 0.44= 2.3 : 0.07
   

    

3. Among UP and Maharashtra which one the highest tax collection state in all 8 months and Find its Average?
   3459 lakhs.
   3249 lakhs.
   2459 lakhs.
   32459 lakhs.
   32959 lakhs.
   Option D
   Collection of Tax in UP = 5644 + 1230 + 4567 + 4562 + 975 + 345 + 2345 + 3543 = 23211 lakhs
   Collection of Tax in Maharashtra = 3223 + 5456+ 8654 + 4578 + 785 + 5679 + 8654 + 4678 = 41707 lakhs.
   So Maharshtra has collected maximum Tax.
   Average = 23211 + 41707/2 = 32459 lakhs.
   

    

4. Find the percentage of contribution of Haryana tax collection on all the states tax collection in the month Jan?
   18.8%.
   1.8%.
   28.8%.
   18%.
   10.8%.
   Option A
   All the states collection in month Jan = 123 +7747 +5644 + 446 +235+4234+45+3223+43+5454+644+345+686 = 28869 lakhs.
   Haryana state collection on Jan = 5454 lakhs
   Percentage = 5454/28869 *100 = 18.8%.
   

    

5. If 10% of Tax collection in the month march in Odissa was allocate for cyclone disaster. Also, 2000 crore was released by the central government for odissa restoration. Find the amount of compensation .
   796.5 cr
   279.5 cr
   2796.5 cr
   296.5 cr
   279.5 cr
   Option C
   10% of 7965 = 796.5 lakhs was allocated by odissa government.
   Total compensation amount = 2000 cr + 796.5 lakhs=2796.5 cr
   

    

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

Raymond, a Jean seller bought 720 Jeans of 3 different types (J1, J2 and J3) in the ratio 4: 5: 3. They sell the Jeans based on market demand. On the 1st month he sell 10% of the total volume. On the 2nd month they sell (40/3) % of the total volume. On the 3rd month they sell (35/3) % of the total volume. On the 4th month they sell 20% of the total volume. On the 5th month they sell 33(1/3) % of the total volume. On the 6th month they sell the same volume on the 3rd month.
The cost of each J1, J2 and J3 Jeans is Rs. 450, Rs. 540 and Rs. 600 respectively. On each month they sell the Jeans of each variety in the same ratio as they bought.

6. If on the 3rd month they sell J3 Jean for a price which is more than 20% than their cost price whereas the other two Jeans are sold at their cost price, then find the total amount received by Raymond by selling the Jeans on the 3rd month?
   23440
   23678
   46620
   45780
   67908
   Option C
   Total Jeans = 720
   Ratio of J1, J2 and J3 is 4: 5: 3
   J1 type Jeans = 720 * 4/12 = 240
   J2 type Jeans = 720 * 5/12 = 300
   J3 type Jeans = 720 * 3/12 = 180
   On the 1st month:
   Total Jeans sold = 720 * 10/100 = 72
   Ratio of J1, J2 and J3 is 4: 5: 3
   J1 type Jeans = 72 * 4/12 = 24
   J2 type Jeans = 72 * 5/12 = 30
   J3 type Jeans = 72 * 3/12 = 18
   On the 2nd month:
   Total Jeans sold = 720 * 40/300 = 96
   Ratio of J1, J2 and J3 is 4: 5: 3
   J1 type Jeans = 96 * 4/12 = 32
   J2 type Jeans = 96 * 5/12 = 40
   J3 type Jeans = 96 * 3/12 = 24
   On the 3rd month:
   Total Jeans sold = 720 * 35/300 = 84
   Ratio of J1, J2 and J3 is 4: 5: 3
   J1 type Jeans = 84 * 4/12 = 28
   J2 type Jeans = 84 * 5/12 = 35
   J3 type Jeans = 84 * 3/12 = 21
   On the 4th month:
   Total Jeans sold = 720 * 20/100 = 144
   Ratio of J1, J2 and J3 is 4: 5: 3
   J1 type Jeans = 144 * 4/12 = 48
   J2 type Jeans = 144 * 5/12 = 60
   J3 type Jeans = 144 * 3/12 = 36
   On the 5th month:
   Total Jeans sold = 720 * 100/300 = 240
   Ratio of J1, J2 and J3 is 4: 5: 3
   J1 type Jeans = 240 * 4/12 = 80
   J2 type Jeans = 240 * 5/12 = 100
   J3 type Jeans = 240 * 3/12 = 60
   On the 6th month:
   Total Jeans sold = 720 * 35/300 = 84
   Ratio of J1, J2 and J3 is 4: 5: 3
   J1 type Jeans = 84 * 4/12 = 28
   J2 type Jeans = 84 * 5/12 = 35
   J3 type Jeans = 84 * 3/12 = 21

   6) Answer: c)
   The cost of each J1, J2 and J3 Jean is Rs.450, Rs. 540 and Rs.600 respectively
   The selling price of each J1, J2 and J3 Jean is,
   SP of J1 = CP of J1 = 450
   SP of J2 = CP of J2 = 540
   SP of J3 = 120% of CP of J3 = (120/100)*600 = 720
   Total amount received by Raymond = (28 *450) + (35*540) + (21*720)
   = 12600 + 18900 + 15120
   = Rs. 46620
   

    

7. If on the 4th month they marks up 60% for all the three type of Jeans and allows 20%, 15% and 18% discount on J1, J2 and J3 type Jeans respectively. Find the total amount received by Raymond by selling the Jeans on the 4th month?
   Rs. 2111.2
   Rs. 3051.2
   Rs. 1000.2
   Rs. 200051.2
   Rs. 10051.2
   Option
   MP of J1 type Jeans = 450 * 160/100 = 720
   SP of J1 type Jeans = 720 * 80/100 = 576
   MP of J2 type Jeans = 540 * 160/100 = 864
   SP of J2 type Jeans = 864 * 85/100 = 734.4
   MP of J3 type Jeans = 600 * 160/100 = 960
   SP of J3 type Jeans = 960 * 82/100 = 787.2
   Total amount received by Raymond
   = (48*576) + (60*734.4) + (36*787.2)
   = 27648 + 44064 + 28339.2
   = Rs. 100051.2
   

    

8. Total J1 type Jeans sold on the 2nd, 3rd and 4th month is what percentage more/less than the total J3 type Jeans sold on the 1st, 5th and 6th month?
   9.09
   11
   12
   10.8
   14
   Option A
   Total J1 type Jeans sold on 2nd, 3rd and 4th month
   = 32 + 28 + 48 = 108
   Total J3 type Jeans sold on the 1st, 5th and 6th month
   = 18 + 60 + 21 = 99
   Required percentage = [(108 – 99)/99] *100 = 9.09%
   

    

9. In which of the following month the amount received by Raymond is maximum if all the Jeans sold the same as cost price?
   1
   2
   3
   4
   5
   Option E
   On the 1st month,
   Total amount = (24*450) + (30*540) + (18*600)
   = 10800 + 16200 + 10800
   = 37800
   On the 2nd month,
   Total amount = (32*450) + (40*540) + (24*600)
   = 14400 + 21600 + 14400
   = 50400
   On the 3rd month,
   Total amount = (28*450) + (35*540) + (21*600)
   = 12600 + 18900 + 12600
   = 44100
   On the 4th month,
   Total amount = (48*450) + (60*540) + (36*600)
   = 21600 + 32400 + 21600
   = 75600
   On the 5th month,
   Total amount = (80*450) + (100*540) + (60*600)
   = 36000 + 54000 + 36000
   = 126000
   On the 6th month,
   Total amount = (28*450) + (35*540) + (21*600)
   = 12600 + 18900 + 12600
   = 44100
   On 5th month, Raymond received the maximum amount.
   

    

10. It is known that 25% of the Jeans were slightly damaged in each month. If Raymond managed to sell the Jeans which were slightly damaged at 20% loss, then find the amount made by Raymond for selling the damaged Jeans on the 2nd, 4th and 5th month. (Consider the ratio of the damaged Jeans in the three categories is same as the ratio which they bought)?
    Rs. 50455
    Rs. 50000
    Rs. 70400
    Rs. 20400
    Rs. 50400
    Option E
    Total damaged Jeans = 720 * 25/100 = 180
    SP of J1 type Jeans = 450 *80/100 = 360
    SP of J2 type Jeans = 540 * 80/100 = 432
    SP of J3 type Jeans = 600 * 80/100 = 480
    On the 2nd month:
    Total damaged Jeans sold = 96 * 25/100 = 24
    Ratio of J1, J2 and J3 is 4: 5: 3
    J1 type Jeans = 24 * 4/12 = 8
    J2 type Jeans = 24 * 5/12 = 10
    J3 type Jeans = 24 * 3/12 = 6
    Total amount received by Raymond
    = (8*360) + (10*432) + (6*480)
    = 2880 + 4320 + 2880
    = Rs. 10080
    On the 4th month:
    Total damaged Jeans sold = 144 * 25/100 = 36
    Ratio of J1, J2 and J3 is 4: 5: 3
    J1 type Jeans = 36 * 4/12 = 12
    J2 type Jeans = 36 * 5/12 = 15
    J3 type Jeans = 36 * 3/12 = 9
    Total amount received by Raymond
    = (12*360) + (15*432) + (9*480)
    = 4320 + 6480 + 4320
    = Rs. 15120
    On the 5th month:
    Total damaged Jeans sold = 240 * 25/100 = 60
    Ratio of J1, J2 and J3 is 4: 5: 3
    J1 type Jeans = 60 * 4/12 = 20
    J2 type Jeans = 60 * 5/12 = 25
    J3 type Jeans = 60 * 3/12 = 15
    Total amount received by Raymond
    = (20*360) + (25*432) + (15*480)
    = 7200 + 10800 + 7200
    = 25200
    Total amount = 10080 + 15120 + 25200 = Rs. 50400

     

1. The present age of Aisha is 12 years more the present age of Rahuman. After 5 years the ratio of ages of Rahuman and Aisha will be 3: 5, then find the age of Rahuman, after 2 years?
   10
   15
   20
   25
   30
   Option B
   Present age of Aisha = 12 + Present age of Rahul

   After 10 years, the ratio of ages of Rahuman and Aisha = 3: 5 (3x, 5x)

   5x – 3x = 12

   2x = 12

   x = 6

   The age of Rahuman, after 2 years = 3x – 5 + 2 = 15 years

    

2. The present age of N is twice the present age of P. The ratio between the present ages of Q and P is 7: 4. If the age of N after 6 years will be 30 years, then find the age of Q before 6 years?
   15
   20
   25
   30
   25
   Option A
   The ratio of present age of N and P = 2 : 1

   The ratio of present age of Q and P = 7 : 4

   The present age of N, P and Q = 8 : 4 : 7

   The age of N, after 6 years = 30 years

   The present age of N = 24 years

   8’s = 24

   1’s = 3

   The age of Q before 6 years = 7x – 6 = 15 years

    

3. Akil started the business investing Rs. 19000 and after 6 months Anush joined him with Rs. 28000. At the end of one year the total profit is Rs.16500.Akil’s profit share?
   4500
   5500
   7500
   8500
   9500
   Option E
   Akil’s share

   = > 19/33*16500

   = > Rs.9500

    

4. Akil started the business investing Rs. 19000 and after 6 months Anush joined him with Rs. 28000. At the end of one year the total profit is Rs.16500.Anush’s profit share?
   7000
   2000
   5000
   3000
   6000
   Option A
   = > 14/33 * 16500

   = > 7000

    

5. Two expresses running the same direction at the speed of 60kmph and 80kmph respectively. If the length of the expresses is 200m and 100m respectively, then find the time taken by first express crosses the second one?
   51
   52
   53
   54
   55
   Option D
   300 = (80-60) * (5/18) * x

   300 = 20 * (5/18) * x

   x = 54 seconds

    

6. The train 1 crosses the pole in 18 seconds at the speed of 60kmph and the train 1 crosses the 700m train 2 which is running opposite direction of train 1 at the speed of 20kmph, then find the time taken by train 1 crosses the another train?
   40
   45
   30
   35
   20
   Option B
   x = 60 * 5/18 * 18

   x = 300m

   300+700 = 80 * 5/18 * x

   x = 45 seconds

    

7. Time taken by A to complete 1/5th of work, if B takes 6 days to complete 3/5th of work and together they take 5 days to complete 3/4th of work?
   3
   4
   5
   6
   7
   Option B
   A and B completes 3/4th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days

   B complete 3/5 work in 6 days, so complete work in 5/3 * 6 = 10 days

   So in 1 day A completes = 3/20 – 1/10 = 1/20

   So to complete 1/5th work = 1/5 * 20 = 4 days

    

8. Time taken by a Sadhapthi express to cross a plat form of length 60 km given that it crosses a pole in two and a half hours running at 60 km/hr?
   3
   3.5
   4
   4.5
   5
   Option B
   Length of train = 2.5 * 60 = 150 km

   So time taken to cross platform of length 60 km with speed 60 km/hr

   = (150+60)/60

   = 3.5 hrs

    

9. A bag contains 4 Red flowers, 3 white flowers, 5 orange flowers and 6 pink flowers. If two flowerss are drawn out randomly, then find the probability that both are either orange or pink?
   25 / 153
   15 / 153
   25 / 154
   12/ 153
   25 / 143
   Option A
   Total number of flowers = 4 + 3 + 5 + 6 = 18 flowers

   n(S) = 18C2

   Probability that both Orange or pink

   n(E) = 5C2 or 6C2

   P(E) = n(E) / n(S) = [5C2 or 6C2] / 18C2

   P(E) = 25 / 153

    

10. A bag contains 4 Red marbles, 3 white marbles, 5 orange marbles and 6 blue marbles. If four marbles are drawn out randomly, then find the probability that all the marbles are of different colours?
    1/13
    2/17
    3/14
    4/15
    4/18
    Option B
    n(S) = 18C4

    n(E) = 4C1 and 3C1 and 5C1 and 6C1

    P(E) = n(E) / n(S) = [4C1 or 3C1 or 5C1 or 6C1] / 18C4

    P(E) = (4 * 3 * 5 * 6) / [(18 * 17 * 16 * 15) / (1 * 2 * 3 * 4)]

    P(E) = 2 / 17

     

1. Number of ways of drawing 3 flowers from a basket containing 5 pink and 7 yellow flowers if at least one pink flower is to be included in the draw.
   125
   145
   175
   185
   195
   Option D
   Total no. of ways of drawing 3 flowers
   12C3= 12*11*10/3*2=220
   No. of ways of drawing 3 flowers if no pink ball is drawn
   7C3=7*6*5/3*2=35
   Hence req. no. of ways = 220-35=185
   

    

2. There are 5 MEN and 3 WOMEN in a family. They are photographed in groups of 2 MEN and one WOMEN.
   10
   20
   30
   40
   50
   Option C
   No. of different photographs = 5C2 * 3C1
   (5*4/2)*3
   =30
   

    

3. In majority voting between three parties A, B and C. A got 30%, B got 48% vote and C got rest of the votes polled. If B got 720 more votes than A then find how many votes were polled?
   3000
   4000
   5000
   6000
   7000
   Option B
   The difference in the percentage of votes received by A and B= 48-30= 18%
   Total number of votes polled= 720*100/18= 4000
   

    

4. In Parliament election between two Parties A and B, 10% of the votes were invalid. The ratio of votes received by A and B is 7:5. If the number of votes polled in favor of A was 630, then find the number of votes polled?
   1100
   1200
   1300
   1400
   1500
   Option B
   Total number of votes received by A and B= 630*12/7= 1080
   Total number of votes polled= 1080*100/90= 1200 votes.
   

    

5. Persons X, Y, Z jogging around a circular field and complete a single round in 18, 22, 30 seconds respectively. In how much time they will meet again at the starting point?
   16minutes 30 seconds
   6minutes 30 seconds
   16minutes 10 seconds
   16minutes 20 seconds
   5minutes 30 seconds
   Option A

   Required time = LCM of 18, 22, 30
   = 990 seconds
   =990/60 minutes=16minutes 30 seconds
   

    

6. The respective ratio between present age of Kalim’s Aunty, Uncle and cousin is 16:18:3. Sum of ages of aunty and uncle is 68. Difference between the age of cousin after 6 yrs and aunty’s age before 18 yrs.
   5
   4
   3
   2
   1
   Option D
   Ratio of present age = 16:18:3
   Sum of ages 16x+18x=34x=68
   Therefore we can find uncles’s age=36, aunty’s age=32 and cousin’s age=6.
   cousin’s age after 6 yrs= 12 yrs
   aunty’s age before 18 yrs=14 yrs
   Difference=2 yrs
   

    

7. Present age of Arun and Anil are in the ratio 4:5. Five years hence ratio of their ages will be 5:6. Arun’s present age is.
   10
   20
   30
   40
   50
   Option B
   4x+5/5x+5 = 5/6
   24x+30 = 25x+25
   X= 30-25=5 years
   Arun’s present age is 20 years.
   

    

8. In how many different ways can the letters of the word MONICOT can be arranged so that vowels occupy odd positions?
   220
   350
   470
   576
   668
   Option D
   Required no. of ways =4P3* 4p4
   = 576 ways
   

    

9. In how many different ways can the letters of the word HEADING can be arranged so that vowels always come together?
   230
   450
   540
   660
   720
   Option E
   HEADING = 7 letter
   3vowels (EAI) supposed to form one letter
   Now remaining letter arranged in 5!
   Vowels arranged in 3!
   Therefore 5!*3! = 720 ways

    

10. The ratio of men and women in a office is 3: 5. If 10% of men and 15% of women speak Tamil, the difference between men and women who did not speak Tamil is 155, find the total number of employee in the office.
    800
    300
    700
    500
    1000
    Option

    Men: Women = 3: 5
    Unspeak Tamil = 3 * 90/100 : 5 * 85/100
    = 2.7 : 4.25
    Total = 155/1.55 * 8 = 800