The majority suffered rando miseds dummy about Lorem Ipsum available .

There are many variations of passages about dummy about Lorem Ipsum.

There are many variations of passages about dummy about Lorem Ipsum.

There are many variations of passages about dummy aboutLorem Ipsum available.

There are many variations of passages about dummy aboutLorem Ipsum available.

There are many variations of passages about dummy about Lorem Ipsum available.

There are many variations of passages about dummy aboutLorem Ipsum available but the majority suffered randomiseds.

There are many variations of passages about dummy aboutLorem Ipsum available but the majority suffered randomiseds.

There are many variations of passages about dummy about Lorem Ipsum.

Dummy about Lorem Ipsum There are many variations of passages about .

The majority suffered rando miseds dummy about Lorem Ipsum available .

Welcome to WordPress. This is your first post. Edit or delete it, then start writing!

Welcome to WordPress. This is your first post. Edit or delete it, then start writing!

[WpProQuiz 330]

Previous Post:→ The Hindu Editorial Vocabulary – Set 190

Previous Quant Post:→ Mixed Quantitative Aptitude Questions Set 245

1. A and B can do a work in 12 days. B and C can do the same work in 20 days. A and C can also do the same work in 30 days. If they all work together in how many days will they complete the work ?
   12 days
   8 days
   14 days
   18 days
   17 days
   Option A
   LCM of 12, 20 and 30 = 120
   efficiency of A and B = 120/12 = 10
   efficiency of B and C = 120/20 = 6
   efficiency of A and C = 120/30 = 4
   2(A + B + C) = 20
   A + B + C = 10
   time = 120/10 = 12 days

    

2. Sujit, Rajat and Sanjay started a partnership business rs.8400 , rs.5000 and rs.6000 respectively. If they invested their capital for 5 months, 6 months and 8 months respectively and total profit of the business is rs. 2000, then how much profit Sanjay got at the end of the year ?
   400
   580
   420
   800
   750
   Option D
   Profit sharing ratio of Sujit, Rajat and Sanjay = (8400 * 5) : (5000 * 6) : (6000 * 8) = 7 : 5 : 8
   profit of Sanjay = 2000/20 * 8 = 800

    

3. Total income of A, B and C is rs.9400. If A’s income is 4/5th of B’s income and B’s income is 3/4th of C’s income, then find the income of A.
   1800
   2200
   2800
   2400
   3500
   Option D
   A/B = 4/5
   B/C = 3/4
   by balancing, A/B = 4 *3/5 *3 = 12/15
   B/C = 3 *5/4*5 = 15/20
   ratio of A, B and C = 12 : 15 : 20
   A’s income = 9400 * 12/47 = 2400

    

4. Pipes And B together can fill a tank in 20 hours while pipe A alone can fill the same tank in 30 hours. Find how much time taken by B to fill the same tank ?
   40 hours
   30 hours
   60 hours
   45 hours
   50 hours
   Option C
   LCM of 20 and 30 = 60
   let capacity of tank = 60
   efficiency of pipe A and B together = 60/20 = 3
   efficiency of pipe A = 60/30 = 2
   efficiency of pipe B = 3 – 2 = 1
   time taken by B to fill the tank = 60/1 = 60 hours

    

5. Boat A covers 180 km in downstream in 9 hours. If the speed of boat in still water is 12 km/hr, then find time taken by boat A to cover 96 km in upstream ?
   12 hours
   13 hours
   22 hours
   24 hours
   15 hours
   Option D
   Downstream speed = 180/9 = 20 km/hr
   speed of boat = 12 km/hr
   speed of stream = 20 – 12 = 8 km/hr
   upstream speed = 12 – 8 = 4 km/hr
   time = 96/4 = 24 hours

    

6. I. x^2 – 11x – 80 = 0
   II. y^2 – 20y + 64 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relations.
   Option E
   I. x^2 – 16x + 5x – 80 = 0
   x = 16 , -5
   II. y^2 – 16y – 4y + 64 = 0
   y = 16, 4

    

7. I. x^2 = 196
   II. y^2 – 19y + 70 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relations.
   Option E
   I. x^2 = 196
   x = 14, -14
   II. y^2 – 14y – 5y + 70 = 0
   y = 14, 5

    

8. I. 3x^2 – 13x + 12 = 0
   II. 2y^2 + 14y +12 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relations.
   Option A
   I. 3x^2 – 9x – 4x + 12 = 0
   3x = 9, 4
   x = 3, 1.33
   II. 2y^2 + 14y + 12 = 0
   2y = -12, -2
   y = -6, -1

    

9. I. 2y^2 – 11y + 15 = 0
   II. x^2 + 5x + 4 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relations.
   Option B
   I. 2y^2 – 6y – 5y + 15 = 0
   2y = 6, 5
   y = 3, 2.5
   II. x^2 + 4x + x + 4 = 0
   x = – 4, -1

    

10. I. x^2 – 28 = 36
    II. y^3 = 512
    X > Y
    X < Y
    X ≥ Y
    X ≤ Y
    X = Y or no relations.
    Option D
    I. x^2 = 64
    x = 8, -8
    II. y^3 = 512
    y = 8

     

Previous Post:→ Static GK Quiz for Upcoming Exams Set – 627

Previous Quant Post:→ Mixed Quantitative Aptitude Questions Set 244

1. A vessel has only two types of oil A and B in the ratio of 9 : 7 respectively. When 25 liters of oil is added to the vessel then quantity of oil A becomes 75% of that of oil B. What is the capacity of vessel if it becomes full after addition of 25 liters oil ?
   120 lt
   105 lt
   100 lt
   130 lt
   80 lt
   Option B
   Let quantity of oil in vessel A and B = 9x and 7x
   9x = (7x + 25) 75/100
   36x = 21x + 75
   15x = 75
   x = 5
   capacity of vessel = 9x + 7x + 25 = 105 liters

    

2. The ratio of present age of A and B is 6 : 5 and the present age of C is 5 years less than B. After 6 years the age of A is two times the present age of C. If the present age D is 8 years more than B, then find the present age of D ?
   26 years
   30 years
   28 years
   23 years
   32 years
   Option C
   Let the present age of A and B = 6x and 5x respectively.
   present age of C = 5x – 5
   ATQ, 6x + 6 = (5x – 5) * 2
   6x + 6 = 10x – 10
   x = 4
   present age of D = 5x + 8 = 28 years

    

3. P and Q together can complete a work in 24 days and Q and R together can complete the same work in 30 days. P and R together work for 15 days and after 15 days P and R left the work remaining work done by done B. If the efficiency ratio of P and Q is 2 : 3, then find in how many days total work will be completed ?
   28 days
   60 days
   40 days
   38 days
   25 days
   Option C
   LCM of 24 and 30 = 120
   total work = 120
   efficiency of P and Q = 120/24 = 5
   efficiency of Q and R = 120/30 = 4
   efficiency of P = 5 * 2/5 = 2
   efficiency of Q = 5 – 2 = 3
   efficiency of R = 4 – 3 = 1
   P and R together work for 15 days = ( 2 + 3) * 15 = 45 work
   remaining work = 120 – 45 = 75
   days required by B to complete remaining work = 75/3 = 25 days
   total days = 15 + 25 = 40 days

    

4. A and B entered into a partnership business by investing rs.12000 and rs.14000 respectively. After x months C joined then by investing rs. 8000 and B left the business. If profit earned by C at the end of year is rs. 800 out of total profit rs.7200, then find for how many months did C invest his money ?
   8 months
   4 months
   2 months
   7 months
   6 months
   Option B
   Profit sharing ratio = (12000 * 12) : (14000 * x) : (8000 * 12 – x)
   = 144 : 14x : 96 – 8x
   96 – 8x/144 + 14x = 1/8
   144 + 14x = 768 – 64x
   78x = 624
   x = 8
   C invest his money for 12 – 8 = 4 months

    

5. Rohit calculates his profit percentage on SP while Sumeet calculate his profit percentage on CP. They find that the difference between their profit is rs.280. If SP of both are same and both get 20% profit, then find CP of Rohit ?
   12400
   14000
   15000
   10800
   12000
   Option D
   Let SP of both Rohit and Sumeet = rs.100x
   profit of Rohit = 100x * 20/100 = 20x
   CP of Rohit = 80x
   CP of Sumeet = 100x/120 * 100 = 500x/6
   Rohit of Sumeet = 100x – 500/6 = 50x/3
   20x – 50x/3 = 450
   x = 135
   CP of Rohit = 135 * 80 = 10800

    

6. Directions : What will come in place of questions (?) marks in the following questions given below ? (you are not calculate exact value)
   49.96% of 1740 + 28% of 1200 – √1764.05 + 120 = ?
   1340
   1250
   1336
   1284
   1265
   Option D
   870 + 336 – 42 + 120 = ?
   ? = 1284
   

    

7. 4799.98 – 320% 400 – √1331 + 3700/50 = ?
   3642
   3485
   2394
   5250
   3583
   Option E
   4800 – 1280 – 11 + 74 = ?
   ? = 3583

    

8. (35 * 24) + (40 * 28) – (38)^2 + ? = 800
   384
   360
   538
   284
   264
   Option D
   840 + 1120 – 1444 + ? = 800
   516 + ? = 800
   ? = 284

    

9. 520% of 180 – ( 42 * 25) + 376 = ?
   275
   263
   340
   345
   262
   Option E
   936 – 1050 + 376 = ?
   ? = 262

    

10. 387.04 + 19.99% of 230 – 1540/19.95 + ? = 420
    56
    64
    62
    84
    48
    Option B
    387 + 46 – 77 + ? = 420
    ? = 64

     

1. 8 men and 6 women together can complete a piece of work in 20 days. If the efficiency of a man is 50% more than efficiency of a woman, then find in how many days 12 men and 14 women can complete the same piece of work ?
   14 days
   12.5 days
   90/7 days
   90/8 days
   18 days
   Option D
   Efficiency ratio of a man and a woman = 3 : 2
   total work =( 8 * 3 + 6 * 2) * 20 = 720 work
   days required to complete work by 12 men and 14 women = 720/(12 * 3 + 14 * 2) = 90/8 days

    

2. A bag contains 72 balls out of which green balls are 28 and orange balls are x and remaining balls are pink. If two balls are randomly picked up from the bag then the probability of getting green ball and a orange ball is 56/213. Find the number of pink balls in that bag ?
   14
   22
   18
   25
   20
   Option E
   No. of green balls = 28
   no.of orange balls = x
   no.of pink balls = 72 – (28 + x) = 44 – x
   probability = (28C1) * (xC1)/72C2 = 56/213
   28 * x * 2/72 * 71 = 56/213
   x = 24
   no. of pink balls = 44 – 24 = 20

    

3. Ranju goes from home to office at speed of (x + 25) km/h and returns with a speed of (x + 10) km/h. If average speed of Ranju for the whole journey is 36 km/h, then find time taken by him to cover a distance of 140 km if his speed is (x + 8) km/h ?
   4 hours
   5 hours
   8 hours
   2 hours
   6 hours
   Option B
   2 (x + 25) (x + 10)/x + 25 + x + 10 = 36
   x^2 + 10x + 25x + 250 = 36x + 630
   x^2 – x – 380 = 0
   x = 20, -19
   x = 20
   time = 140/20 + 8 = 5 hours
   

    

4. A and B started a business by investing certain sum in the ratio of 7 : 6 for 5 months and 6 months respectively. If total profit of the business is rs.2840, then find how much amount investment by A in the business ?
   3500
   3200
   4300
   4400
   2800
   Option E
   Let investment amount at A and B is 7x and 6x respectively.
   profit sharing ratio of A and B = (7x * 5 : 6x * 6) = 35x : 36x
   71x = 2840
   x = 40
   amount invested by A = 40 * 7 = 2800

    

5. A person purchased rice at rs. 8 rs. 12 and rs.20 per kg in three consecutive years. If he spent rs.2400 each year on rice, then find the approximate average cost per kg paid by him in the given three consecutive years.
   12
   11
   11 19/31
   12 19/35
   14
   Option C
   Quantity of rice purchased by him in 1st year = 2400/8 = 300kg
   2nd year = 2400/12 = 200 kg
   3rd year = 2400/20 = 120 kg
   total quantity purchased = 300 + 200 + 120 = 620
   total amount paid = 2400 * 3 = 7200
   average = 7200/620 = 11 19/31

    

6. Directions : In each of these questions, two equation (I) and (II) are given. You have both the equations and give answer.
   I. x^2 – x – 380 = 0
   II. y^2 + 29y + 210 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option E
   I. x^2 – 20x + 19x – 380 = 0
   x = 20, -19
   II. y^2 + 15y + 14y + 210 = 0
   y = -15, -14

    

7. I. y^2 – 18y + 72 = 0
   II. x^2 – 31x + 238 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option A
   I. y^2 – 12y – 6y + 72 = 0
   y = 12, 6
   II. x^2 – 17x – 14x + 238 = 0
   x = 17 , 14
   

    

8. I. x^2 – 8√2 + 30 = 0
   II. y^2 – 7√2 + 24 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option E
   I. x^2 – 5√2 – 3√2 + 30 = 0
   x = 5√2, 3√2
   II. y^2 – 4√2- 3√2 + 24 = 0
   y = 4√2, 3√2

    

9. I. x^2 = 1024
   II. y = √1764
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option B
   I. x^2 = 1024
   x = 32, -32
   II. y = √1764
   y = 42

    

10. I. y^3 = 2197
    II. x^2 + 50 = 275
    
    X > Y
    X < Y
    X ≥ Y
    X ≤ Y
    X = Y or no relation.
    Option E
    I. y^3 = 2197 X > Y
    y = 13
    II. x^2 = 275 – 50 = 225
    x = 15, -15
    

     

Previous Post:→ The Hindu Editorial Vocabulary – Set 166

Previous Quant Post: → Mixed Quantitative Aptitude Questions Set 242

1. The discount allowed on an article is rs.960 more than the profit earned on the article. If the selling price of the article is rs.1920 and shopkeeper marked the article 100% above its cost price, then find profit% earned on the article.
   10%
   20%
   12%
   18%
   40%
   Option B
   Let cost price of the article = rs.100x
   MP = 100x *2/1 = 200x
   ATQ, (200x – 1920 ) – (1920 – 100x) = 960
   300x = 4800
   x = 16
   CP = 1600
   Profit % = 1920 – 1600/1600 * 100 = 20%
   

    

2. The height of cylinder is equal to the side of square. If area of square is 196m^2 and ratio and ratio of radius to height of cylinder is 3 : 2, then find total surface area of such cylinder.
   5630
   6450
   3420
   4620
   6430
   Option D
   Side of square = √196 = 14m
   height = 14m
   radius = 14/7 * 6 = 21
   TSA = 2πr ( h + r)
   = 2 * 22/7 * 21(14 + 21)
   = 4620

    

3. A man invested rs. P and (P + 350) each at the rate of 20% p.a on CI and SI respectively for two years. If ratio of total CI to total SI received by man is 44 : 75, then find amount invested by man on CI ?
   1050
   840
   450
   400
   800
   Option D
   Rate of CI for 2 years = 20 + 20 + 20 *20/100 = 44%
   CI received = P * 44/100 = 44P/100
   SI received by man = ( P + 350) 40/100 = 40P + 14000/100
   44P/40P + 14000 = 44/75
   P = 400
   

    

4. If three unbiased coins are tossed simultaneously, then find the probability of getting at most two heads ?
   3/5
   7/8
   2/5
   4/7
   2/17
   Option B
   Total outcome = 2^3 = 8
   favorable outcome = 7
   probability = 7/8
   

    

5. Pipe A and B alone can fill a tank in 20 hours and 24 hours respectively and outlet pipe C can empty same tank in 40 hours. If three pipes are opened simultaneously in how much time required to fill a empty tank ?
   20 hours
   15 hours
   18 hours
   14 hours
   28 hours
   Option B
   LCM of 20, 24 and 40 = 120
   efficiency of A = 120/20 = 6
   efficiency of B = 120/24 = 5
   efficiency of C = 120/40 = 3
   time required = 120/(6 + 5) – 3 = 15 hours
   

    

6. Directions : What will come in place of questions marks (?) in the following number series ?
   85, 104, 189, 293, 482, ?
   875
   775
   456
   565
   235
   Option B
   The series is, (85 + 104), (104 + 289), (189 + 293), (293 + 482)

    

7. 8, ?, 6, 15, 52.5, 236.25
   5
   4
   4.5
   2.5
   6
   Option B
   The series is, (* .5), (* 1.5), (* 2.5), (* 3.5), (* 4.5)

    

8. ?, 2, 3, 8, 35, 204
   2
   4
   5
   6
   8
   Option A
   The series is, (* 2 – 2), (* 3 – 3), (* 4 – 4), (* 5 – 5), (* 6 – 6),

    

9. 27, ?, 159, 324, 555, 852
   60
   86
   78
   34
   80
   Option A
   The series is, difference of difference.

    

10. 68, 92, 66, 94, 64, ?
    84
    90
    102
    96
    86
    Option D
    The series is, (+ 24), (- 26), (+ 28), (- 30), (+ 32)

     

Previous Post:→ Daily Current Affairs 7th December, 2021

Previous Quant Post:→ Mixed Quantitative Aptitude Questions Set 241

1. A bag contains (x + 1), green balls and (x + 4) yellow balls. If one ball is drawn at random from the bag, the probability of getting a green ball is 3/7. Find the total number of balls in the bag ?
   34
   23
   28
   42
   21
   Option E
   Probability = 3/7
   x + 1/x + 1 + x + 4 = 3/7
   x + 1/2x + 5 = 3/7
   x = 8
   total balls in the bag = 9 + 12 = 21 balls
   

    

2. A alone can do a work in 20 days and efficiency of A is 7 1/7% more than B. If efficiency of C is 20% less than A, then find in how many days B and C together can complete the same work ?
   100/9 days
   15 days
   18 days
   200/3 days
   30 days
   Option A
   Fraction value 7 1/7% = 1/14
   efficiency of B = 14
   efficiency of A = 15
   total work = 15 * 20 = 300
   efficiency of C = 15 * 4/5 = 12
   time = 300/15 + 12 = 100/9 days

    

3. The ratio of milk and water in a mixture is 11 : 4. When 30 liters of mixture is taken out and 18 liters of milk is added in the remaining mixture, the ratio of milk to water is 5 : 1, find the initial quantity of milk.
   44 liters
   65 liters
   40 liters
   50 liters
   62 liters
   Option A
   Let quantity of milk = 11x
   and water = 4x
   (11x – 30 * 11/15 + 18)/4x – 30 * 4/15 = 5/1
   11x – 4/4x – 8 = 5/1
   11x – 4 = 20x – 40
   x = 4
   initial quantity of milk = 11 * 4 = 44 liters

    

4. An amount is divided among A, B and C, amount of B is average of amount A and C and when amount of B is reduced by 20% of that of A, it becomes equal to that of C. Find amount of C is what percent of total amount ?
   20%
   30%
   40%
   15%
   25%
   Option E
   Let amount of A and C is rs.x and rs.y respectively.
   amount of B = x + y/2
   x + y/2 – x/5 = y
   3x = 5y
   x/y = 5/3
   amount of A = rs.5a
   amount of B = rs.4a
   amount of C = rs.3a
   % = 3a/12a * 100 = 25%

    

5. Train A running at speed of 108 km/hr crosses a platform having twice the length of train in 24 sec. Train B whose length is 420m crosses same platform in 36 sec, then find the speed of train B ?
   108 km/hr
   45 k/hr
   76 km/hr
   90 km/hr
   72 km/hr
   Option D
   Let length of train A = x
   length of platform = 2x
   x + 2x/24 = 108 * 5/18
   x = 240m
   let speed of train B is y km/hr
   420 + 2 * 240/36 = y * 5/18
   y = 90 km/hr
   

    

6. Directions : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
   I. x^2 – 17x – 84 = 0
   II. y^2 + 4y – 117 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option E
   I. x^2 – 21x + 4x – 84 = 0
   x = 21, -4
   II. y^2 + 13y – 9y – 117 = 0
   y = -13, 9
   

    

7. I. x^2 + 23x + 132 = 0
   II. y^2 + 21y + 110 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option D
   I. x^2 + 12x + 11x + 132 = 0
   x = -12, -11
   II. y^2 + 11y + 10y + 110 = 0
   y = -11, -10

    

8. I. x^2 = 196
   II. y = √196
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option C
   I. x^2 = 196
   x = 14, -14
   II. y = √196
   y = 14, 14

    

9. I. 3x^2 + 13x + 12 = 0
   II. y^2 + 9y + 20 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option A
   I. 3x^2 + 9x 4x + 12 = 0
   3x = -9, -4
   x = -3, -1.33
   II. y^2 + 5y + 4y + 20 = 0
   y = -5, -4

    

10. I. x^3 = 216
    II. 2y^2 – 25y + 78 = 0
    X > Y
    X < Y
    X ≥ Y
    X ≤ Y
    X = Y or no relation.
    Option D
    I. x^3 = 216
    x = 6
    II. 2y^2 – 13y – 12y + 78 = 0
    2y = 13, 12
    y = 6.5, 6

     

1. Directions : What will come in place of questions mark (?) in the following questions. (You are not calculate exact value )
   ∛216.84 – 23 * 3 + (8.99)^3 = ?
   684
   673
   573
   275
   680
   Option B
   13 – 69 + 729 = ?
   ? = 673

    

2. 24.86% of 168 * 5.04 + (12.02)^2 – 19.81 = ?
   284
   224
   342
   334
   128
   Option D
   1/4 * 168 * 5 + 144 – 20 = ?
   ? = 334

    

3. 14.28% of 279.86 * 15 + 21 * 4.05 = ?
   285
   684
   234
   724
   580
   Option B
   40 * 15 + 84 = ?
   ? = 600 + 84
   ? = 684

    

4. 32% of 374.98 + 644.02 ÷ 23 + 185.99 = ?
   334
   234
   134
   448
   528
   Option A
   32% of 375 + 28 + 186 = ?
   ? = 120 + 28 + 186
   ? = 334

    

5. (22)^2 + (28)^2 – (15)^2 = ?
   1245
   1043
   1050
   845
   1440
   Option B
   484 + 784 – 225 = ?
   ? = 1043

    

6. Directions : In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
   I. 5x^2 + 11x + 2 = 0
   II. 4y^2 + 13y + 3 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option E
   I. 5x^2 + 10x + x + 2 = 0
   5x = – 10 , – 1
   x = -2, -.2
   II. 4y^2 + 12y + y + 3 = 0
   4y = -12, – 1
   y = -3, .25

    

7. I. x^2 + 5x + 6 = 0
   II. y^2 + 9y + 14 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option E
   I. x^2 + 3x + 2x + 6 = 0
   x = -3, -2
   II. y^2 + 7y + 2y + 14 = 0
   y = -7, -2

    

8. I. x^2 – 11x + 30 = 0
   II. y^2 – 15y + 56 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option B
   I. x^2 – 6x – 5x + 30 = 0
   x = 6, 5
   II. y^2 – 7y – 8y + 56 = 0
   y = 7, 8

    

9. I. x^2 + 17x + 72 = 0
   II. y^2 + 13y + 42 = 0
   
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option B
   I. x^2 + 9x + 8x + 72 = 0
   x = -9, -8
   II. y^2 + 7y + 6y + 42 = 0
   y = -7, -6

    

10. I. x^2 + 23x + 132 = 0
    II. y^2 + 21y + 110 = 0
    X > Y
    X < Y
    X ≥ Y
    X ≤ Y
    X = Y or no relation
    Option D
    I. x^2 + 12x + 11x + 132 = 0
    x = -12, -11
    II. y^2 + 11y + 10y + 110 = 0
    y = -11, -10
    

     

1. A train , 240 m long crosses a man, walking along the line in opposite direction at the rate of 3 km/hr in 10 seconds. What is the speed of train ?
   80 km/hr
   75 km/hr
   83.4 km/hr
   92.8 km/hr
   68 km/hr
   Option C
   Let speed of train = x m/hr
   240/(3 + x) 5/18 = 10
   x = 83.4 km/hr

    

2. A boat takes a total of 24 hours to over 90 km in downstream and 90 km in upstream. If the ratio of speed of boat in sill water and speed of stream is 4 : 1, then find how much time is required to cover 120 km distance in down stream speed ?
   8 hours
   14 hours
   20 hours
   15 hours
   12 hours
   Option E
   Let speed of boat in still water = 4x
   speed of stream = x
   downstream speed = 4x + x = 5x
   upstream speed = 4x – x = 3x
   90/5x + 90/3x = 24
   x = 2
   downstream speed = 5 * 2 = 10 km/hr
   time = 120/10 = 12 hours

    

3. A equilateral tringle and a square have the same area. If the perimeter of the square is 88cm, then find what is area of the equilateral tringle ?
   196 cm^2
   256 cm^2
   144 cm^2
   484 m^2
   125 cm^2
   Option D
   Perimeter of square = 88cm
   side of square = 88/4 = 22cm
   area of square = 22 * 22 = 484 cm^2
   area of tringle = 484 cm^2

    

4. Pipe A can fill a tank in 15 hours and pipe B can fill the same tank in 60 hours, pipe C can empty in 10 hours. Pipe A and B are kept opened for 10 hours in the beginning and then pipe C is also opened, what is the time taken to empty the tank ?
   60 hours
   40 hours
   30 hours
   25 hours
   50 hours
   Option E
   Capacity of the tank = 15, 60, 10 = 60 units
   efficiency of pipe A = 60/15 = 4 units
   efficiency of pipe B = 60/60 = 1
   efficiency of pipe C = 60/10 = 6 units
   tank filled in 1st 10 hours = (4 + 1) * 10 = 50 units
   now, when all the pipe work together, 1 unit of water will out in every minute from tank.
   50 units of water will be in 50 hours.

    

5. The ratio of age of A and B, 6 years ago was 3 : 4. Sum of the present age of B and C is 80 years. C is 12 years elder to A. Find the difference of age of B and C after 4 years ?
   5 years
   6 years
   8 years
   4 years
   12 years
   Option D
   6 years ago age of A = 3x
   age of B = 4x
   present age of A = 3x + 6
   present age of B = 4x + 6
   present age of C = 3x + 6 + 12 = 3x + 18
   4x + 6 + 3x + 18 = 80
   7x = 56
   x = 8
   age of B after 4 years = 4x + 6 + 4 = 42 years
   age of C after 4 years = 3x + 18 + 4 = 46 years
   difference = 46 – 42 = 4 years
   

    

6. Directions : What will be wrong number in the following wrong number series ?
   120, 83, 113, 93, 105, 99, 101
   113
   83
   105
   120
   99
   Option D
   The series is, (- 6 * 7), (+ 5 * 6), ( – 4 * 5), ( + 3 * 4), ( – 2 * 3), ( + 1 * 2)

    

7. 1898, 1870, 1657, 1608, 1096, 1015, 15
   1096
   1015
   1898
   15
   1870
   Option E
   The series is ( – 5^2 ), (- 6^3), (- 7^2), (- 8^3 ), (- 9^2), (- 10^3)

    

8. 48, 84, 156, 300, 590, 1164
   48
   156
   300
   590
   1164
   Option D
   The series is, ( * 2 – 12), ( * 2 – 12)…

    

9. 2, 6, 12, 24, 30, 42, 56
   24
   12
   30
   42
   2
   Option A
   The series is , (1^2 + 1), (2^2 + 2), (3^2 + 3)…

    

10. 1, 6, 24, 60, 120, 210
    6
    1
    60
    24
    210
    Option B
    The series is, (1^3 – 1), (2^3 – 2), (3^3 – 3), (4^3 – 4)…

     

1. Directions : What will come in place of questions marks (?) in the following questions ? ( You are not to calculate the exact value )
   (20.24% of 749.81) ÷ 2.89 + 85.20 = ?
   125
   135
   120
   130
   145
   Option B
   20% of 750/3 + 85 = ?
   ? = 135

    

2. (12.138 * 4 + 4.98 * 7.99) ÷ ( 9.12 + 1.85) = ?
   5
   6
   8
   12
   15
   Option C
   ( 12 * 4 + 5 * 8) ÷ 9 + 2 = ?
   ? = 88/11
   ? = 8

    

3. √100.12 – √63.87 + √144.01 * 5.32 = ?
   58
   82
   94
   62
   52
   Option D
   10 – 8 + 12 * 5 = ?
   ? = 62

    

4. (34.07)^2 + (15.89)^2 -√576.02 + 45.88 = ?
   1520
   1430
   1434
   1334
   1324
   Option C
   1156 + 256 – 24 + 46 = ?
   ? = 1434

    

5. 449. 75 ÷ 8.82 + 19.95 * 3.22 + √675.82 = ?
   152
   183
   130
   136
   120
   Option D
   450 ÷ 9 + 20 * 3 + √676 = ?
   ? = 50 + 60 + 26
   ? = 136

    

Directions : What will be wrong number in the given wrong number series ?

6. 
   12, 13, 24, 72, 288, 1440
   12
   13
   1440
   24
   72
   Option B
   The series is, * 1, * 2, * 3 …

    

7. 520, 525, 540, 585, 720, 1120
   525
   720
   1120
   720
   585
   Option C
   The series is, + 5, + 15, + 45, + 135, + 405

    

8. 185, 166, 189, 160, 191, 156
   185
   189
   160
   156
   166
   Option D
   The series is, -19, + 23, -29, + 31, -33

    

9. 9, 12, 27, 84, 339, 1544
   339
   84
   27
   9
   1544
   Option E
   The series is, (* 1 + 3), (* 2 + 3), (* 3 + 3)…

    

10. 325, 329, 340, 354, 379, 415
    329
    379
    354
    340
    415
    Option D
    The series is, + 4, + 9, + 16, + 25, + 36

     

1. A starts a business with rs.4000 and B joins him after 3 months with rs. 16000. If total profit of the business is rs.2000, then find profit share of B.
   2000
   1000
   1200
   850
   1500
   Option E
   Profit sharing ratio of A and B = (4000 * 12) : (16000 * 9)
   = 48 : 144 = 1 : 3
   profit share of B = 2000 * 3/4 = 1500

    

2. The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio ?
   5 : 3
   18 : 19
   21 : 22
   3 : 8
   5 : 4
   Option C
   Let number of boys and girls are 7x and 8x respectively
   ratio after increased =
   7x * 6/5 : 8x * 11/10 = 21 : 22

    

3. The ratio between the present ages of P and Q is 5 : 7 respectively. If the difference between Q’s present age and P’s age after 6 years is 2 what is the total P’s and Q’s present age ?
   32 years
   68 years
   50 years
   48 years
   26 years
   Option D
   Let present age of P and Q = 5x and 7x
   7x – (5x + 6) = 2
   x = 4
   total = 12x = 12 * 4 = 48 years

    

4. The ratio of milk to water in total 56 liter of mixture is 5 : 3. If ‘x’ liter of milk and water is added in the mixture so that ratio becomes 25 : 18, then find the value of ‘x’.
   20 liters
   12 liters
   18 liters
   15 liters
   22 liters
   Option D
   Quantity of milk in the mixture = 56 * 5/8 = 35
   water = 56 * 3/8 = 21
   35 + x/21 + x = 25/18
   x = 15 liter

    

5. 4 kg of rice at rs.8 per kg is mixed with 8 kg of wheat at rs.12 per kg. What is the average price of the mixture ?
   12
   10.66
   5
   8
   12
   Option B
   Total cost of 4 kg rice = 4 * 8 = 32
   total cost of 8 kg wheat = 8 * 12 = 96
   average = 128/12 = 10.66

    

6. Directions : What will come in place of questions marks (?) in the following questions given below ?
   15% of 220 + 20 * 15 – √1156 = ?
   320
   300
   284
   299
   340
   Option D
   33 + 300 – 34 = ?
   ? = 299

    

7. 3240 – 2240 + 2050 – 1800 = ?
   2030
   1250
   3140
   1240
   1350
   Option B
   ? = 5290 – 4040
   ? = 1250

    

8. √1296 – 30% of 120 + 40 + ? = 188
   152
   320
   145
   148
   200
   Option D
   36 – 36 + 40 + ? = 188
   ? = 148

    

9. (18)^2 – (12)^2 + 420 – 120 = ?
   320
   126
   440
   480
   520
   Option D
   324 – 144 + 420 – 120 = ?
   ? = 480

    

10. (15)^2 – 30 * 8 + (40)^2 = ?
    1485
    1575
    1340
    1585
    1235
    Option D
    225 – 240 + 1600 = ?
    ? = 1585

     

1. Directions : In the following questions, two equations I and II are given. You have to solve the equations and give answer the questions.
   I. x^2 + 13x + 42 = 0
   II. y^2 + 8y + 12 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option D
   I. x^2 + 7x + 6x + 42 = 0
   x = -7, -6
   II. y^2 + 6y + 2y + 12 = 0
   y = -6, -2

    

2. I. x^3 = 2744
   II. y^2 = 289
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option E
   I. x^3 = 2744
   x = 14
   II. y^2 = 289
   y = 17, -17

    

3. I. x^2 – 2x – 24 = 0
   II. y^2 + 3y – 40 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option E
   I. x^2 – 6x + 4x – 24 = 0
   x = 6, -4
   II. y^2 + 8y – 5y – 40 = 0
   y = -8, 5

    

4. I. 2x^2 + 10x + 12 = 0
   II. y^2 + 10y + 25 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option A
   I. 2x^2 + 6x + 4x + 12 = 0
   x = -3, -2
   II. y^2 + 5y + 5y + 25 = 0
   y = -5, -5

    

5. I. (x – 2)^2 = x – 2
   II. y^2 – 14y + 48 = 0
   
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option B
   I. (x – 2)^2 = x – 2
   x^2 – 4x + 4 = x – 2
   x^2 – 5x + 6 = 0
   x^2 – 3x – 2x + 6 = 0
   x = 3, 2
   II. y^2 – 8y – 6y + 48 = 0
   y = 8, 6
   

    

6. Directions : What will come in place of questions mark (?) in the following questions below ?
   36% of 1800 + 80% 495 – (8)^2 = ?
   970
   880
   440
   980
   1020
   Option D
   648 + 396 – 64 = ?
   ? = 980

    

7. 12 * ? + (10)^2 = (16)^2 – 10 * 6
   5
   8
   2
   14
   12
   Option B
   12 * ? + 100 = 256 – 60
   12 * ? = 196 – 100
   ? = 8

    

8. 32 * 3 + 60% of 300 = (?) + √400
   14
   15
   12
   16
   24
   Option D
   96 + 180 = (?)^2 + 20
   (?)^2 = 256
   ? = 16

    

9. 264 ÷ 24 + 190 ÷ 5 = ? ÷ 5
   245
   315
   255
   320
   240
   Option A
   264/24 + 190/5 = ?/5
   ? = (11 + 38) * 5
   ? = 245

    

10. ?/27 * 60 = 12 * 3 * 5/?
    520
    620
    440
    540
    480
    Option D
    (?)^2 = 12 * 3 * 5 * 27 * 60
    (?)^2 = 36 * 5 * 9 * 3 * 12 * 5
    ? = √(36 * 25 * 9 * 36 )
    ? = 36 * 5 * 3
    ? = 540

     

1. 6 men or 8 women or 3 boys completed certain work in a day is equal. Find in how many days a man and a boy can complete the three-fourth of same work together ?
   2 days
   3 days
   5 days
   6 days
   1.5 days
   Option E
   Let total work = 48
   6 men = 8men = 3 boys
   efficiency of a man = 48/6 = 8
   efficiency of a woman = 48/8 = 6
   efficiency of a boy = 48/3 = 16
   days require = 48/(16 + 8) * 3/4 = 1.5 days
   

    

2. A sum becomes rs.9280 in 6 years and becomes rs.10440 in 8 days at the same rate on simple interest. Find compound interest earned in 2 years on rs.6000 at same rate of interest ?
   1520
   1840
   2540
   1340
   1260
   Option E
   SI for 2 years = 10440 – 9280 = 1160
   SI for 6 years = 1160/2 * 6 = 3480
   sum = 9280 – 3480 = 5800
   rate interest = 3480/(5800 * 6) *100 = 10%
   CI for 2 years = 10 + 10 + 10 * 10/100 = 21%
   CI = 6000 * 21/100 = 1260

    

3. A bat was sold at rs,1260 after giving 30% discount on market price. If discount was not given, then he would have earned 50% profit. Find cost price of bat ?
   1400
   1280
   1800
   1200
   1440
   Option D
   Market price of bat = 1260/70 * 100 = 1800
   CP = 1800/150 * 100 = 1200
   

    

4. Two trains travel from Mumbai to Bangalore and Bangalore to Mumbai in 10 hours and 12 hours respectively. Both trains started simultaneously at 8 am and the train started from Mumbai met with accident at 12 pm. Find percentage reduction in speed of train started from Mumbai, if they meet each other at 2 pm, assume there is no loss of time in accident ?
   30%
   48%
   50%
   42%
   20%
   Option C
   Ratio of speed is inversely proportional to time taken if distance is constant.
   Let speed of Mumbai to Bangalore travelling train = 6x km/hr
   speed of Bangalore to Mumbai travelling train = 5x km/hr
   Let total distance = 60x km
   let speed of train after accident = y km/hr
   60x = 4 (6x + 5x) + 2 (y + 5x)
   y = 3x km/hr
   reduction in speed = 6x – 3x/6x * 100 = 50%

    

5. The radius of a circle is 14 km. What is area of another circle having radius 1.5 times of the actual circle ?
   1540 cm^2
   1386 cm^2
   1440 cm^2
   1450 cm^2
   1240 cm^2
   Option B
   Radius of another circle = 14 * 1.5 = 21 cm
   area = πr^2 = 22/7 * 21 * 21 = 1386 cm^2

    

6. Directions : What will come in place of questions (?) mark in the following missing series ?
   4, 2, 3, 7.5, 26.25, ?
   120.125
   118.5
   114.5
   118.125
   120
   Option D
   The series is, (* .5), (* 1.5), (* 2.5), (* 3.5), (* 4.5)

    

7. 148, 196, 268, 364, 484, ?
   650
   640
   452
   620
   628
   Option E
   The series is, (+ 48), (+ 72), (+ 96), (+ 120), (+ 144),

    

8. 46, 52, ?, 136, 256, 466
   84
   78
   68
   76
   80
   Option D
   The series is, (+ 2^3 – 2), (+ 3^3 – 3), (+ 4^3 – 4), (+ 5^3 – 5), (+ 6^3 – 6),

    

9. 16000, 8000, 24000, 6000, ?, 30000, 5000
   40000
   18000
   12000
   42000
   30000
   Option E
   The series is, (÷ 2), (* 3), (÷ 4), (* 5), (÷ 6)

    

10. 24, 28, 36, 52, 84, ?
    240
    148
    120
    124
    168
    Option B
    The series is, (+ 4), (+ 8), (+ 12), (+ 32), (+ 64),

     

1. 6 men or 8 women or 3 boys completed certain work in a day is equal. Find in how many days a man and a boy can complete the three-fourth of same work together ?
   2 days
   3 days
   5 days
   6 days
   1.5 days
   Option E
   Let total work = 48
   6 men = 8men = 3 boys
   efficiency of a man = 48/6 = 8
   efficiency of a woman = 48/8 = 6
   efficiency of a boy = 48/3 = 16
   days require = 48/(16 + 8) * 3/4 = 1.5 days
   

    

2. A sum becomes rs.9280 in 6 years and becomes rs.10440 in 8 days at the same rate on simple interest. Find compound interest earned in 2 years on rs.6000 at same rate of interest ?
   1520
   1840
   2540
   1340
   1260
   Option E
   SI for 2 years = 10440 – 9280 = 1160
   SI for 6 years = 1160/2 * 6 = 3480
   sum = 9280 – 3480 = 5800
   rate interest = 3480/(5800 * 6) *100 = 10%
   CI for 2 years = 10 + 10 + 10 * 10/100 = 21%
   CI = 6000 * 21/100 = 1260

    

3. A bat was sold at rs,1260 after giving 30% discount on market price. If discount was not given, then he would have earned 50% profit. Find cost price of bat ?
   1400
   1280
   1800
   1200
   1440
   Option D
   Market price of bat = 1260/70 * 100 = 1800
   CP = 1800/150 * 100 = 1200
   

    

4. Two trains travel from Mumbai to Bangalore and Bangalore to Mumbai in 10 hours and 12 hours respectively. Both trains started simultaneously at 8 am and the train started from Mumbai met with accident at 12 pm. Find percentage reduction in speed of train started from Mumbai, if they meet each other at 2 pm, assume there is no loss of time in accident ?
   30%
   48%
   50%
   42%
   20%
   Option C
   Ratio of speed is inversely proportional to time taken if distance is constant.
   Let speed of Mumbai to Bangalore travelling train = 6x km/hr
   speed of Bangalore to Mumbai travelling train = 5x km/hr
   Let total distance = 60x km
   let speed of train after accident = y km/hr
   60x = 4 (6x + 5x) + 2 (y + 5x)
   y = 3x km/hr
   reduction in speed = 6x – 3x/6x * 100 = 50%

    

5. The radius of a circle is 14 km. What is area of another circle having radius 1.5 times of the actual circle ?
   1540 cm^2
   1386 cm^2
   1440 cm^2
   1450 cm^2
   1240 cm^2
   Option B
   Radius of another circle = 14 * 1.5 = 21 cm
   area = πr^2 = 22/7 * 21 * 21 = 1386 cm^2

    

6. Directions : What will come in place of questions (?) mark in the following missing series ?
   4, 2, 3, 7.5, 26.25, ?
   120.125
   118.5
   114.5
   118.125
   120
   Option D
   The series is, (* .5), (* 1.5), (* 2.5), (* 3.5), (* 4.5)

    

7. 148, 196, 268, 364, 484, ?
   650
   640
   452
   620
   628
   Option E
   The series is, (+ 48), (+ 72), (+ 96), (+ 120), (+ 144),

    

8. 46, 52, ?, 136, 256, 466
   84
   78
   68
   76
   80
   Option D
   The series is, (+ 2^3 – 2), (+ 3^3 – 3), (+ 4^3 – 4), (+ 5^3 – 5), (+ 6^3 – 6),

    

9. 16000, 8000, 24000, 6000, ?, 30000, 5000
   40000
   18000
   12000
   42000
   30000
   Option E
   The series is, (÷ 2), (* 3), (÷ 4), (* 5), (÷ 6)

    

10. 24, 28, 36, 52, 84, ?
    240
    148
    120
    124
    168
    Option B
    The series is, (+ 4), (+ 8), (+ 12), (+ 32), (+ 64),

     

1. A container which contains 80 liters of milk and water in the ratio of 3 : 2. How much quantity of mixture should be replaced with water so that in the final mixture, ratio of milk to water is 9 : 11 ?
   25 liters
   32 liters
   28 liters
   20 liters
   42 liters
   Option D
   Quantity of milk = 80 * 3 /5 = 48 liters
   Quantity of water = 80 * 2 /5 = 32 liters
   let ‘x’ liters of mixture is replaced with water.
   (48 – x * 3/5 ) / (32 – x * 2/5 + x) = 9/11
   x = 20 liters

    

2. Ram bought a chair for rs.480 and marked is at rs.720. If he made a profit of 25% on selling the table then find percentage discount allowed by Ram on the chair ?
   15%
   20%
   18%
   45%
   16.67%
   Option E
   SP of chair = 480 * 125/100 = 600
   MP = 720
   discount percentage = 720 – 600/720 * 100 = 16.67%
   

    

3. A alone can do a work in 25 days whereas B and C together can do some work in 20 days. A and B together worked for 8 days then A left the work. If efficiency of C is half of efficiency of A, then in how many days remaining work will be completed b A and C together ?
   5 days
   7 1/3 days
   15 days
   8 days
   14 days
   Option B
   LCM of 25 and 20 = 100
   efficiency of A = 100/25 = 4
   efficiency of B and C = 100/20 = 5
   efficiency of C = 4/2 =2
   efficiency of B = 5 – 2= 2
   A and B worked for 8 days = 8 ( 4 + 3) = 56 work
   remaining work = 100 – 56 = 44
   remaining work will be completed A and C together = 44/4 + 2 = 7 1/3 days

    

4. Train A travelling at 72 km/hr crosses a platform of 160 m in 18 seconds and train B travelling at 90 km/hr crosses the same platform in 15 seconds. Find the length of train B ?
   320m
   240m
   210m
   225m
   215m
   Option E
   Speed of train A = 72 * 5/18 = 20 m/sec
   distance covered by train A = 20 * 18 = 360 m
   length of train A = 360 – 160 = 200m
   speed of train B = 90 * 5/18 = 25 m/sec
   distance covered by train B = 25 * 15 = 375 m
   length of train B = 375 – 160 = 215 m
   

    

5. A is 6 years younger than B and ratio of present age of B to C is 12 : 5. If ratio of present age of A to C is 2 : 1, then find present age of B ?
   32 years
   34 years
   42 years
   36 years
   40 years
   Option D
   Let present age of B = 12x
   age of C = 5x
   age of A = 12x – 6
   12x – 6/5x = 2/1
   12x – 6 = 10x
   2x = 6
   x = 3
   present age of B = 12 * 3 = 36 years
   

    

6. Directions : What will come in place of questions marks ( ? ) in the following questions below ?
   35% of 320 + 440 – √1024 = ?
   540
   320
   340
   620
   520
   Option E
   112 + 440 – 32 = ?
   ? = 520

    

7. 124 * 4 – 320 ÷ 5 + √361 – 18 = ?
   225
   320
   435
   433
   520
   Option D
   496 – 64 + 19 – 18 = ?
   ? = 433

    

8. 495 ÷ 5 + (21)^2 – (20)^2 = ?
   324
   420
   325
   140
   260
   Option D
   99 + 441 – 400 = ?
   ? = 140

    

9. 1240 + 2520 – 3540 + 4250 = ?
   4340
   5470
   4240
   3450
   4470
   Option E
   ? = 8010 – 3540
   ? = 4470

    

10. (12)^2 + (18)^2 – (20)^2 + (14)^2 = ?
    365
    264
    345
    420
    540
    Option B
    144 + 324 + 196 – 400 = ?
    ? = 264

     

1. The ratio of the milk to water in a mixture is 4 : 3. 21 liter of mixture is taken out and 8 liter of milk and 14 liter of water is added in mixture, so that new ratio of milk to water becomes 320 : 17, then find initial quantity of milk in the mixture ?
   84 liters
   35 liters
   63 liters
   80 liters
   72 liters
   Option A
   Let quantity of milk and water in the mixture = 4x and 3x respectively
   (4x – 21 * 4/7 + 8)/(3x – 21 * 3/7 + 14) = 20/17
   4x – 4/3x + 5 = 20/17
   x = 21
   Initial quantity of milk in the mixture = 4x = 4 * 21 = 84 liters

    

2. A alone can do 40% of work in 8 days and B alone can do 30% of work in 12 days. If efficiency of C is 50% more than A, then find in how many days A, B and C together will be completed 90% of work ?
   5 days
   8 days
   12 days
   6 days
   4 days
   Option D
   Total days required to complete the work by A = 8/40 * 100 = 20
   total days required to complete the work by B = 12/30 * 100 = 40
   LCM of 20 and 40 = 40
   efficiency of A = 40/20 = 2
   efficiency of B = 40/40 = 1
   efficiency of C = 2 * 150/100 = 3
   days required = 40/6 * 90/100 = 6 days

    

3. The ratio of present age of A and B is 3 : 4. If the ratio of age of A after 6 years and age of B before 4 years is 6 : 5, then find the age of B after 4 years ?
   42 years
   25 years
   26 years
   28 years
   40 years
   Option D
   Let the present age of A and B = 3x and 4x
   3x + 6/ 4x – 4 = 6/5
   15x + 30 = 24x – 24
   x = 6
   the age of B after 4 years = 4x + 4 = 28 years

    

4. A man goes from a place A to B at a speed of 12 km/hr and returns from B to A at a speed of 18 km/hr. What is the average speed for the whole journey ?
   12 km/hr
   24 km/hr
   14 km/hr
   14 2/5 km/hr
   15 2/5 km/hr
   Option D
   Average speed = 2xy /x + y
   = 2 * 12 * 18/12 + 18 = 14 2/5 km/hr

    

5. A boat covers 12 km upstream and 18 km downstream in 3 hours while it covers 36 km upstream and 24 downstream in 6 1/2 hours, what is the speed of stream ?
   5 km/hr
   4 km/hr
   6 km/hr
   2 km/hr
   8 km/hr
   Option D
   12/x – y + 18/x + y = 3
   36/x – y + 24/x + y = 13/2
   let x – y = a, x + y = b
   12/a + 18/b = 3
   36a + 24/b = 13/2
   By multiplying ‘3’ in equation (i)
   36/a + 54/b = 9
   36/a + 24/b = 13/2
   30/b = 9 – 13/2
   b = 12
   36/a + 9/2 = 9
   36/a = 9/2
   a = 8
   x – y = 8
   x + y = 12
   2x = 20
   x = 10
   y = 2
   

    

6. Directions : What will come in place of questions marks (?) in the following number series ?
   345, 358, 375, ?, 417, 446
   380
   382
   392
   394
   396
   Option D
   The series is, (+ 13), (+ 17), (+ 19), (+ 23), (+ 29),

    

7. 5, 11, 35, 143, 719, ?
   4240
   4319
   4465
   4978
   4565
   Option B
   The series is, ( * 2 + 1), ( * 3 + 2), ( * 4 + 3), ( * 5 + 4), ( * 6 + 5),

    

8. 20, 18, 38, 56, 94, ?
   155
   100
   120
   125
   150
   Option E
   The series is, (20 + 18), (18 + 56), (56 + 94)

    

9. 35, 40, 55, 80, 115, ?
   145
   180
   160
   172
   165
   Option C
   The series is, (+ 5), (+ 15), (+ 25), (+ 35), (+ 45),

    

10. 35, 18, 19, 30, ?, 157.5
    64
    62
    58
    96
    86
    Option B
    The series is, ( * 5 + .5), (* 1 + 1), (* 1.5 + 1.5), (* 2 + 2), (* 2.5 + 2.5)

     

1. I. x^2 – 11x + 18 = 0
   II. y^2 – 16 y + 39 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option E
   I. x^2 – 9x – 2x + 18 = 0
   x = 9, 2
   II. y^2 – 13y – 3y + 39 = 0
   y = 13, 3

    

2. I. 2x^2 + 19x + 35 = 0
   II. 2y^2 + 7y + 5 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option D
   I. 2x^2 + 14x + 5x + 35 = 0
   2x = -14, – 5
   x = -7, -2.5
   II. 2y^2 + 5y + 2y + 5 = 0
   2 = -5, -2
   y = -2.5, -1

    

3. I. x^3 = 512
   II. y^2 – 13y + 40 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option C
   I. x^3 = 512
   x = 8
   II. y^2 – 8y – 5y + 40 = 0
   y = 8, 5

    

4. I. (x – 4)^2 = 0
   II. 2y^2 – 5y + 2 = 0
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option A
   I. x^2 – 4x – 4x + 16 = 0
   x = 4, 4
   II. 2y^2 – 4y – y + 2 = 0
   y = 2, .5
   

    

5. I. x^4 = 1296
   II. y^3 = 216
   X > Y
   X < Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation
   Option E
   I. x^4 = 1296
   x = 6
   II. y^3 = 216
   y = 6

    

6. Directions : What will come in place of questions (?) marks in the following questions given below ?
   21 * 25 – (8)^3 + (16)^2 – ? = (17)^2
   25
   22
   38
   64
   20
   Option E
   525 – 512 + 256 + ? = (17)^2
   269 + ? = 289
   ? = 20

    

7. 672 ÷ 28 + 22 * 19 – (15)^2 = ?
   218
   256
   320
   217
   215
   Option D
   24 + 418 – 225 = ?
   ? = 217

    

8. 55% of 320 + √? = 16 * 8 + 84
   36
   84
   216
   1296
   26
   Option D
   176 + √? = 212
   √? = 36
   ? = 1296
   

    

9. √2025 ÷ 9 * 4 + 25 of 20 ÷ 4 = ?
   120
   132
   138
   145
   136
   Option D
   45 ÷ 9 * 4 + 25 * 20/4 = ?
   ? = 20 + 125
   ? = 145

    

10. (?)^3 ÷ 180 = 1200
    25
    62
    65
    60
    36
    Option D
    (?)^3 = 1200 * 180
    ? = 60

     

1. P is 25% more efficient than Q and he completes a task in 36 days. If R alone complete the same task in 30 days, then find all three together can complete the same task in how many days ?
   15 days
   18 days
   8 days
   14 days
   12 days
   Option E
   Ratio of efficiency of A and B = 5 : 4
   total work = 36 * 5 = 180
   efficiency of C = 180/30 = 6
   required time = 180/5 + 4 + 6 = 12 days

    

2. A shopkeeper marked up an article 60% above cost price and sell it at rs.960 after allowing discount of 25%. If profit of shopkeeper is rs.160, then find the marked price of article ?
   1420
   1200
   1140
   1540
   1280
   Option E
   Let CP of article = 100x
   MP of article = 160x
   CP = 960 – 160 = 800
   100x = 800
   x = 8
   MP of article = 160 * 8 = 1280

    

3. Average of serious of six consecutive even numbers is 25, find the sum of second and fourth number of serious.
   43
   35
   24
   48
   56
   Option D
   Let series be x, x + 2 , x + 4, x + 6, x + 8, x + 10
   6x + 30/6 = 25
   x = 20
   2nd and 4th number = (20 + 2 = 22), ( 20 + 6 = 26)
   sum = 48

    

4. Ratio of investment of A, B and C is 4 : 6 : 9 respectively. At the end of a year, ratio of profit share of A, B and C is 4 : 5 : 6 respectively. If A has invested for a year then find the sum of period of investment of B and C ?
   20 months
   18 months
   14 months
   24 months
   28 months
   Option B
   Let investment of A, B and C = 4x, 6x and 9x respectively
   period of investment of B and C = a and b months respectively.
   ratio of profit share ,
   A : B : C = 4x * 12 : 6x * a : 9x * b
   4x * 12/6x * a = 4/5
   a = 10 months
   4x * 12/9x * b = 4/6
   b = 8 months
   required sum = 10 + 8 = 18 months

    

5. Ratio between income and expenditure of Sujit increased by 20% and 33.33% in two successive years and expenditure remains same, then his saving is increased by rs.18000. What is his new income ?
   45000
   38000
   65000
   32000
   48000
   Option E
   Income and expenditure of Sujit 15x and 6x respectively.
   saving = 15x – 6x = 9x
   new income = 15x * 6/5 * 4/5 = 24x
   24x – 6x = 9x + 18000
   x = 2000
   new income = 24 * 2000 = 48000

    

6. Directions : What will come in place of questions (?) marks in the following number series ?
   36, 18, ?, 27, 54, 135
   12
   19
   20
   18
   22
   Option D
   The series is, (* .5), (* 1), (* 1.5), (* 2), (* 2.5)

    

7. 26, 170, 366, 622, ? 1346
   834
   946
   870
   650
   820
   Option B
   The series is, (+ 12)^2, (+ 14)^2, (+ 16)^2, (+ 18)^2, (+ 20)^2

    

8. 10, 11, 15, 24, ? 65
   45
   30
   38
   40
   45
   Option D
   The series is, (+ 1)^2, (+ 2)^2, (+ 3)^2, (+ 4)^2, (+ 5)^2,

    

9. 8, 16, 43, 168, 511, ?
   1452
   1234
   1020
   1842
   2024
   Option D
   The series is, (+ 2)^3, (+ 3)^3, (+ 5)^3, (+ 7)^3, (+ 11)^3,

    

10. 2, 5, 12, 27, 58, ?
    124
    240
    234
    121
    320
    Option D
    The series is, (* 2 + 1), (* 2 + 2), (* 2 + 3), (* 2 + 4), (* 2 + 5),

     

1. 375 ml of mixture contains milk and water in the ratio of 16 : 9 respectively. If x ml of water is added into the mixture then the ratio of milk to water in the resultant mixture becomes 3 : 2, find the value x.
   30
   25
   35
   40
   45
   Option B
   Quantity of milk = 375 * 16/25 = 240
   quantity of water = 375 * 9/25 = 135
   ATQ,
   240/135 + x = 3/2
   3x + 405 = 480
   x = 25

    

2. A and B alone can complete a piece of work in 24 days and 20 days respectively. If efficiency of C is 60% more than A, then find in how many days C complete the work ?
   15 days
   17 days
   20 days
   25 days
   30 days
   Option A
   LCM of 24 and 20 = 120
   efficiency of A = 120/24 = 5
   efficiency of B = 120/20 = 6
   efficiency of C = 5 * 160/100 = 8
   time required = 120/8 = 15 days

    

3. A shopkeeper marked up an article 25% above its cost price and allowed a discount of 40% on it. If shopkeeper made a loss of rs.128, then find what is the cost price ?
   620
   340
   580
   512
   480
   Option D
   Let CP of article = 100x
   MP of article = 100x * 125/100 = 125x
   SP of article = 125x * 60/100 = 75x
   loss = 100x – 75x = 25x
   CP of article = 128/25x * 100x = 512

    

4. P and Q started a partnership business with investment of rs.15000 and rs.12000 for 8 months and 6 months respectively. If total profit of the business is 6400, then what is profit share of P ?
   3200
   4450
   3800
   4200
   4000
   Option E
   Profit sharing ratio of P and Q = 15000 * 8 : 12000 * 6 = 5 : 3
   profit share of P = 6400 * 5/8 = 4000
   

    

5. A man invested rs. 25000 at the rate of 20% in compound interest for 1 year. How much amount he received if he invested for compounded half yearly.
   42000
   22000
   30250
   28140
   34400
   Option C
   Rate of compound interest for half yearly = 20/2 = 10%
   rate of CI after 1 year = 10 + 10 + 10 * 10/100 = 21%
   CI = 25000 * 21/100 = 5250
   amount = 25000 + 5250 = 30250
   

    

6. Directions : What will come in place of questions mark in the following questions given below ?
   20^2 + 328 – 84 = 28% of 300 + ?
   
   320
   480
   340
   610
   560
   Option E
   400 + 328 – 84 = 84 + ?
   ? = 560

    

7. √1156 + 12^2 – 19 * 5 = ?
   82
   84
   83
   92
   93
   Option C
   34 + 144 – 95 = ?
   ? = 83

    

8. 42% of 800 + 120% of 1400 – 520 = ?
   3200
   1496
   1445
   1296
   2050
   Option B
   336 + 1680 – 520 = ?
   ? = 1496

    

9. 450/3 + 320 – 210 + 150 = ?
   520
   440
   420
   450
   410
   Option E
   150 + 320 + 150 – 210 = ?
   ? = 410

    

10. 350 ÷ 7 of 25 + 340 + (12)^2 = ?
    328
    486
    284
    320
    325
    Option B
    350 /(7 * 25) + 340 + 144 = ?
    ? = 486

     

1. 4 men or 8 women can do a work in 15 days, then in how many days 6 men and 12 women can complete the whole work ?
   4 days
   3 days
   5 days
   6 days
   8 days
   Option C
   Efficiency ratio of men and women = 2 : 1
   total work = 4 * 2 * 15 = 120
   days required = 120/ 6 * 2 + 12 * 1 = 5 days

    

2. If simple interest earned on a certain sum of money at certain rate of interest after 5 years is 2/7 times the amount earned on the same sum of money after 5 years, then what is the rate interest ?
   10%
   6%
   12%
   8%
   15%
   Option D
   Let amount = 7x
   SI = 2x
   P = 7x – 2x = 5x
   rate of interest = 2x * 100/5x * 5 = 8%

    

3. Average of 12 numbers is 60. If average of first six and last four numbers is 70 and 50 respectively and ratio of seventh and eighth number is 3 : 2, then find the 8th number ?
   50
   40
   80
   70
   75
   Option B
   Total 12 numbers = 12 * 60 = 720
   total of first six numbers = 70 * 6 = 420
   total of last four numbers = 50 * 4 = 200
   8th number = (720 – 420 + 200) 2/5 = 100 * 2/5 = 40
   

    

4. Sanjay saves 25% of his monthly income. If his monthly expenditure is increased by 40% while his monthly savings remain same, then find the percentage increase in his monthly income ?
   36%
   40%
   28%
   30%
   40%
   Option D
   Let income of Sanjay = 100x
   savings = 100x * 25/100 = 25x
   expenditure = 100x – 25x = 75x
   increased expenditure = 75x * 140/100 = 105x
   income = 105x + 25x = 130x
   % increased income = 130x – 100x/100 * 100 = 30%

    

5. 490 ml of mixture contains milk and water in the ratio of 4 : 3 respectively. If 140 ml of mixture is taken out and 50 ml of water is added, then what is the ratio of milk to water in final mixture ?
   2 : 3
   4 : 5
   3 : 2
   1 : 1
   2 : 5
   Option D
   Quantity of milk = 490 * 4/7 = 280
   quantity of water = 490 * 3/7 = 210
   140 ml of mixture is taken out,
   (280 – 140 * 4/7)/(210 – 140 * 3/7 + 50) = 200/200 = 1 : 1

    

6. Directions : What will come in place of questions marks (?) in the following given number series ?
   30, 62, 126, 254, 510, ?
   920
   1022
   648
   840
   1220
   Option B
   The series is (*2 + 2), (*2 + 2), (*2 + 2)…

    

7. ?, 9, 25, 61, 125, 225
   5
   4
   6
   8
   2
   Option A
   The series is (+2^2), (+4^2), (+6^2), (+8^2), (+10^2)

    

8. 35, 42, 56, 77, ?, 140
   135
   140
   130
   105
   112
   Option D
   The series is (+7), (+14), (+21), (+28), (+35)

    

9. 8, 16, 48, 192, 960, ?
   4820
   5760
   1920
   3200
   9600
   Option B
   The series is (*2), (*3), (*4), (*5), (*6)

    

10. 320, 330, 350, 385, 430, ?
    320
    520
    640
    510
    650
    Option D
    The series is difference of difference.

     

1. A, B and C entered into a partnership business and A gets profit rs.960 out of total profit rs.5120. C invested rs.14000 for 8 months and B invested rs. 8000. If A invested rs.4000 more than that of B and 4 months less that of C, then find what is the profit share of C ?
   2120
   1345
   1240
   2240
   1050
   Option D
   Investment of A = 8000 + 4000 = 12000
   time period for investment of A = 8 – 4 = 4months
   profit sharing ratio of A, B and C = (12000 * 4) : (8000 * 12) : (14000 * 8) = 3 : 6 : 7
   profit share of C = 5120 * 7/16 = 2240

    

2. Amit invested rs.22500 in scheme X offering 10% compound interest for 2 years. If he invested 40% of the money received by him from scheme X in another scheme Y offering 20% simple interest for 4 years, then how much amount received by him from scheme Y ?
   12340
   38450
   19602
   20505
   23420
   Option C
   CI received after 2 years = 22500 * 10/100 * 10/100 = 4725
   total amount = 22500 + 4725 = 27225
   40 % of 27225 = 10890
   SI received after 4 years = 27225 * 40/100 * 4 * 20/100 = 8712
   total amount = 10890 + 8712 = 19602

    

3. The monthly income of A is rs.28000 and he saves 28 4/7% of his monthly income. If the monthly expenditure of B is rs.4000 more than A and also expenditure of B is 20% less than his income, then find saving of B is how much percent more or less than savings of A ?
   20%
   30%
   15%
   25%
   22%
   Option D
   Income of A = 28000
   savings of A = 28000 * 2/7 = 8000
   expenditure of A = 28000 – 8000 = 20000
   expenditure of B = 20000 + 4000 = 24000
   income of B = 24000/4 * 5 = 30000
   saving of B = 30000 – 24000 = 6000
   % less = 8000 – 6000/8000 * 100 = 25%

    

4. The ratio of male to female in a company is 5 : 3 and the number of male in the company is 1250. If the number of female and male is increased by 10% and 8% respectively, then find what percent change in total population ?
   5.2%
   6%
   8%
   7.25%
   8.75%
   Option E
   The number of male = 1250
   the number of female = 1250/5 * 3 = 750
   the number of male after increased = 1250 * 108/100 = 1350
   the number of female after increased = 750 * 110/100 = 825
   total population = 1250 + 750 = 2000
   total population after increased = 1350 + 825 = 2175
   change in population = 2175 – 2000/2000 = 8.75%

    

5. A man travels 150 km to a place in 3 hours 20 minutes and he returns to the starting point in 4 hours 10 minutes. Let x be the rate for the entire trip. Then in how many kilometers per hour more or less than x ?
   4 km/hr
   2 km/hr
   4.5 km/hr
   5 km/hr
   8 km/hr
   Option D
   Time taken to cover 150 km in going trip
   3 hours 20 minutes = 3 20/60 = 3 1/3 hours
   speed in going trip = 150 *3/10 = 45 km/hr
   time taken to cover 150 km in return trip = 4 hours 10 minutes = 4 1/6 hours
   speed in return trip = 150 * 6/25 = 36 km/hr
   average speed = 2 * 45 * 36/ 45 + 36 = 40 km/hr
   required difference = 45 – 40 = 5 km/hr

    

6. Direction : What will come in place of question(?) mark in the following questions given below ?
   60% of 1800 + 24% of 3500 – 32% of 150 = ?
   1992
   1872
   2092
   1532
   1240
   Option B
   1080 + 840 – 48 = ?
   ? = 1872

    

7. 3/5 * 4/7 * 5/12 * 1015 = ?
   120
   145
   167
   89
   52
   Option B
   ? = 3/5 * 4/7 * 5/12 * 1015
   ? = 145

    

8. 4545 + 4320 – 2540 + 320 = ?
   3642
   2720
   4520
   6645
   3435
   Option D
   4545 + 4320 + 320 – 2540 = ?
   ? = 6645

    

9. 330 – 345 + 421 – 22% of 1500 = ?
   76
   32
   45
   52
   65
   Option A
   330 + 421 – 345 – 330 = ?
   ? = 76

    

10. 552 ÷ 24 * 5 + 113 – 85 = ?
    144
    143
    320
    245
    145
    Option B
    23 * 5 + 113 – 85 = ?
    ? = 143

     

1. A shopkeeper X sold an article for rs.850 at a profit of 25%. Another shopkeeper Y sold the another article and earned 30% more profit than the profit earned by X. If the selling price of article Y is three-fifth of selling price of article X, then find the profit percent of article Y ?
   64 2/17%
   76 8/17%
   52 3/34%
   52 3/8%
   78%
   Option B
   SP of article X = 850
   CP of article X = 850 * 100/125 = 680
   profit = 170
   profit of article Y = 170*130/100 = 221
   SP of article Y = 850 * 3/5 = 510
   CP = 510 – 221 = 289
   profit percent = 221/289* 100 = 76 8/17%

    

2. Mixture A contains milk and water in the ratio of 3 : 2 respectively, while mixture B contains water and milk in the ratio of 4 : 3 respectively. If the quantity of water in mixture B is twice the quantity of water in mixture A, then find the quantity of milk in mixture A is what percent quantity of water in mixture B ?
   68%
   60%
   82%
   75%
   56%
   Option D
   Quantity of milk and water in mixture A = 3x and 2x
   according to questions,
   quantity of water in mixture B = 2*2x = 4x
   quantity of milk in mixture B = 4x/4 *3 = 3x
   required percentage = 3x/4x*100 = 75%

    

3. 340 liters of mixture of milk and water is contained in a container having milk and water mixed in the ratio of 3 : 1 respectively. 76 liters of this mixture is taken out from the container and replaced with (x + 7) liters of milk so that the ratio of water to milk is 6 : 11. What is the value of ‘x’ ?
   42
   48
   38
   35
   52
   Option D
   Quantity of milk in the container = 340 * 3/4 = 255
   quantity of water in the container = 340*1/4 = 85
   76 liters of mixture is taken out and (x + 7) liter of water is added.
   (255 – 57) / 85 – 19 + (x + 7) = 11/6
   198/(73 + x) = 11/6
   11x = 385
   x = 35
   

    

4. A can do a piece of work in 30days while B and C together can complete the work in 20 days and A and C together can complete the work in 24 days. If B is worked in two-fifth of his efficiency, then in how many days A and B together complete half of the work ?
   8 days
   6 days
   8 1/3 days
   5 days
   10 days
   Option E
   LCM of 30 , 20 and 24 = 120
   efficiency of A = 120/30 = 4
   efficiency of B and C = 120/20 = 6
   efficiency of A and C together = 120/24 = 5
   efficiency of C = 5- 4 = 1
   efficiency of B 6- 1 = 5
   two-fifth of efficiency B = 5 * 2/5 = 2
   time required to complete half of the work = 60/2 + 4 = 10 days

    

5. Rohan and Sivan started a business together with initial investments of rs. 1500 and rs.1800 respectively. If after 2.5 years profit share of Rohan and Sivan are rs(x + 80) and (x + 120) respectively, then find the difference between profit of Rohan and Sivan ?
   180
   120
   320
   130
   340
   Option B
   Profit sharing ratio of Rohan and Sivan = (1500 * 2.5) : (1800 * 2.5) = 5 : 6
   x + 80/x + 120 = 5/6
   6x + 480 = 5x + 600
   x = 120

    

6. Directions : In each of these questions given below two equations. You have to solve both the equations and give answer.
   I. x^2 – 7√3 + 36 = 0
   II. y^2 – 6√2 + 16 = 0
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. x^2 – 4√3 – 3√3 + 36 = 0
   x = 4√3 , 3√3
   II. y^2 – 4√2 – 2√2 + 16 = 0
   y = 4√2, 2√2

    

7. I. x^2 + √2x – 12 = 0
   II. y^2 – 8√3 + 48 = 0
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option B
   I. x^2 + 3√2 – 2√2 – 12 = 0
   x = -3√2, 2√2
   II. y^2 – 4√3 – 4√3 + 48 = 0
   y = 4√3, 4√3

    

8. I. x^2 – 36 = 0
   II. y^2 – 15y + 36 = 0
   
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. x^2 = 36
   x = 6, -6
   II. y^2 – 12y – 3y + 36 = 0
   y = 12, 3

    

9. I. y^2 = 144
   II. 2x^2 – 12x + 16 = 0
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. y^2 = 144
   y = 12, -12
   II. 2x^2 – 8x – 4x + 16 = 0
   2x = 8, 4
   x = 4, 2

    

10. I. x^2 – 2x – 195 = 0
    II. y^2 – 24y + 128 = 0
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option E
    I. x^2 – 15x + 13x – 195 = 0
    x = 15, -13
    II. y^2 – 16y – 8y + 128 = 0
    y = 16, 8

     

1. What is the monthly income of Jitu ?
   I. Jitu spends 85% of his income on various expenses and remaining amount he saved.
   II. Monthly saving of Jitu are rs.7500.
   III. Out of total money spent by Jitu in a month, one-fifth is spent on rent and remaining amount of rs.34000 on other items.
   Only I and II sufficient
   Only II and III sufficient
   Only I and II sufficient
   Neither I nor II and III is sufficient
   Any of the two statements is sufficient
   Option E
   Let income of Jitu = rs.x
   from I and II,
   15% of x = 7500
   x = 7500 * 100/15 = 50000
   From I and III,
   x * 85/100 * 4/5 = 34000
   x = 50000
   from I and III
   4/5 th of expenditure = 34000
   expenditure = 42500
   income = 42500 + 7500 = 50000
   thus, answer can be find out by any of two given statement

    

2. How much time will require the train to reach from point X to point Y ?
   I. The train will pass the other train of equal length of 400m running opposite in direction in 16 secs
   II. Distance between point X and Y is 252 km.
   III. The 400m long train crosses a signal pole in 20sec.
   Only I and II is sufficient
   Only II and III is sufficient
   Neither of any statement is sufficient
   All statements are sufficient
   Only I and III is sufficient
   Option B
   Statement I is not required to get answer
   from statement III,
   Speed of train = 400/20 = 20 m/sec
   speed( in km) = 20 * 18/5 = 72 km/hr
   from statement II,
   time = 252/72 = 3.5 hr
   statement II and III required to get answer.
   

    

3. What is the length of a running train A crossing another running train B ?
   I. A and B two train take 18secs to cross each other, while running opposite direction.
   II. The length of train B is 180m
   Only I and II is sufficient
   Only II is sufficient
   Either I or II statement
   Neither I nor II is sufficient
   Only I sufficient
   Option D
   Length of train A = l meters
   from I, time taken by train to cross each other = 18secs
   let speed of train A and B = x and y respectively
   relative speed of A and B = (x + y) m/s
   from II, length of train B = 180 meters
   180 + l/x + y = 18
   thus, we can not calculate answer by using these two statements.

    

4. What will be respective ratio of saving of A and B.
   I. Income of A is 4% less than that of C and also expenditure of A is 12.5% less than that of C. B spend 3/5 th of his income.
   II. C save rs. 7000 and A save rs.7400. Income of B is rs.1000 more than that of C.
   Only I is sufficient
   Only II is sufficient
   Either I or II statement
   Neither I nor II is sufficient
   Only I and II is sufficient
   Option E
   Let income of C = 25x
   income of A = 25x * 96/100 = 24x
   let expenditure of A = 7y
   expenditure of C = 8y
   B spend 3/5 th of his his income
   from II
   saving of C = 7000
   saving of A = 7400
   Income of B is 1000 that of C
   from I and II,
   expenditure of A = 24x – 7y = 7400
   expenditure of C = 25x – 8y = 7000
   by solving two equations,
   x = 600 and y = 1000
   income of B = 25 * 600 + 1000 = 16000
   saving B = 16000 * 2/5 = 6400
   Ratio = 7400 : 6400 = 37 : 32
   statement I and II is required

    

5. What is the CI on a sum at the end of 3 years ?
   I. CI at the end of two years is rs.110
   II. Difference between CI and SI at the end of two year is rs.100 and rate of percent is 10%.
   Only I is sufficient
   Only II is sufficient
   Either I or II statement
   Neither I nor II is sufficient
   Only I and II is sufficient
   Option B
   From I,
   Sum can not be find out as rate is not given
   From II,
   Difference PR^2/100^2
   100 = P *100/10000
   P = 10000
   only statement II is sufficient.

    

6. I. x^2v- 41x + 348 = 0
   II. y^2 – 20y + 99 = 0
   
   X < Y
   X > Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option B
   I. x^2 – 29x – 12x + 99 = 0
   x = 29, 12
   II. y^2 – 11y – 9y + 99 = 0
   y = 11, 9
   

    

7. I. 2x^2 – 2x _ 24 = 0
   II. 3y^2 – 8y + 4 = 0
   
   X < Y
   X > Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option E
   I. 2x^2 – 8x + 6x – 24 = 0
   x = 8, – 6
   x = 4, -3
   II. 3y^2 – 6y – 2y + 4 = 0
   3y = 6, 2
   y = 2, .67

    

8. I. 5x^2 – 28x + 15 = 0
   II. 3y^2 – 29 y + 68 = 0
   
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. 5x^2 – 25x – 3x + 15 = 0
   5x = 25, 3
   x = 5, .6
   II. 3y^2 – 17y – 12y + 68 = 0
   y3y = 17, 12
   y = 5.66, 4

    

9. I. x^3 = 4913
   II. y^2 = 225
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option B
   I. x^3 = 4913
   x = 17
   II. y^2 = 225
   y = 15, -15

    

10. I. 2x^2 – 20x + 48 = 0
    II. y^2 – 15y + 56 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option A
    I. 2x^2 – 12x – 8x + 48 = 0
    2x = 12, 8
    x = 6, 4
    II. y^2 – 8y – 7y + 56 = 0
    y = 8, 7

     

1. What is the monthly income of Jitu ?
   I. Jitu spends 85% of his income on various expenses and remaining amount he saved.
   II. Monthly saving of Jitu are rs.7500.
   III. Out of total money spent by Jitu in a month, one-fifth is spent on rent and remaining amount of rs.34000 on other items.
   Only I and II sufficient
   Only II and III sufficient
   Only I and II sufficient
   Neither I nor II and III is sufficient
   Any of the two statements is sufficient
   Option E
   Let income of Jitu = rs.x
   from I and II,
   15% of x = 7500
   x = 7500 * 100/15 = 50000
   From I and III,
   x * 85/100 * 4/5 = 34000
   x = 50000
   from I and III
   4/5 th of expenditure = 34000
   expenditure = 42500
   income = 42500 + 7500 = 50000
   thus, answer can be find out by any of two given statement

    

2. How much time will require the train to reach from point X to point Y ?
   I. The train will pass the other train of equal length of 400m running opposite in direction in 16 secs
   II. Distance between point X and Y is 252 km.
   III. The 400m long train crosses a signal pole in 20sec.
   Only I and II is sufficient
   Only II and III is sufficient
   Neither of any statement is sufficient
   All statements are sufficient
   Only I and III is sufficient
   Option B
   Statement I is not required to get answer
   from statement III,
   Speed of train = 400/20 = 20 m/sec
   speed( in km) = 20 * 18/5 = 72 km/hr
   from statement II,
   time = 252/72 = 3.5 hr
   statement II and III required to get answer.
   

    

3. What is the length of a running train A crossing another running train B ?
   I. A and B two train take 18secs to cross each other, while running opposite direction.
   II. The length of train B is 180m
   Only I and II is sufficient
   Only II is sufficient
   Either I or II statement
   Neither I nor II is sufficient
   Only I sufficient
   Option D
   Length of train A = l meters
   from I, time taken by train to cross each other = 18secs
   let speed of train A and B = x and y respectively
   relative speed of A and B = (x + y) m/s
   from II, length of train B = 180 meters
   180 + l/x + y = 18
   thus, we can not calculate answer by using these two statements.

    

4. What will be respective ratio of saving of A and B.
   I. Income of A is 4% less than that of C and also expenditure of A is 12.5% less than that of C. B spend 3/5 th of his income.
   II. C save rs. 7000 and A save rs.7400. Income of B is rs.1000 more than that of C.
   Only I is sufficient
   Only II is sufficient
   Either I or II statement
   Neither I nor II is sufficient
   Only I and II is sufficient
   Option E
   Let income of C = 25x
   income of A = 25x * 96/100 = 24x
   let expenditure of A = 7y
   expenditure of C = 8y
   B spend 3/5 th of his his income
   from II
   saving of C = 7000
   saving of A = 7400
   Income of B is 1000 that of C
   from I and II,
   expenditure of A = 24x – 7y = 7400
   expenditure of C = 25x – 8y = 7000
   by solving two equations,
   x = 600 and y = 1000
   income of B = 25 * 600 + 1000 = 16000
   saving B = 16000 * 2/5 = 6400
   Ratio = 7400 : 6400 = 37 : 32
   statement I and II is required

    

5. What is the CI on a sum at the end of 3 years ?
   I. CI at the end of two years is rs.110
   II. Difference between CI and SI at the end of two year is rs.100 and rate of percent is 10%.
   Only I is sufficient
   Only II is sufficient
   Either I or II statement
   Neither I nor II is sufficient
   Only I and II is sufficient
   Option B
   From I,
   Sum can not be find out as rate is not given
   From II,
   Difference PR^2/100^2
   100 = P *100/10000
   P = 10000
   only statement II is sufficient.

    

6. I. x^2v- 41x + 348 = 0
   II. y^2 – 20y + 99 = 0
   
   X < Y
   X > Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option B
   I. x^2 – 29x – 12x + 99 = 0
   x = 29, 12
   II. y^2 – 11y – 9y + 99 = 0
   y = 11, 9
   

    

7. I. 2x^2 – 2x _ 24 = 0
   II. 3y^2 – 8y + 4 = 0
   
   X < Y
   X > Y
   X ≥ Y
   X ≤ Y
   X = Y or no relation.
   Option E
   I. 2x^2 – 8x + 6x – 24 = 0
   x = 8, – 6
   x = 4, -3
   II. 3y^2 – 6y – 2y + 4 = 0
   3y = 6, 2
   y = 2, .67

    

8. I. 5x^2 – 28x + 15 = 0
   II. 3y^2 – 29 y + 68 = 0
   
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. 5x^2 – 25x – 3x + 15 = 0
   5x = 25, 3
   x = 5, .6
   II. 3y^2 – 17y – 12y + 68 = 0
   y3y = 17, 12
   y = 5.66, 4

    

9. I. x^3 = 4913
   II. y^2 = 225
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option B
   I. x^3 = 4913
   x = 17
   II. y^2 = 225
   y = 15, -15

    

10. I. 2x^2 – 20x + 48 = 0
    II. y^2 – 15y + 56 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option A
    I. 2x^2 – 12x – 8x + 48 = 0
    2x = 12, 8
    x = 6, 4
    II. y^2 – 8y – 7y + 56 = 0
    y = 8, 7

     

1. Train A leaves kolkata at 8am and is travelling at a speed of 70 km/hr, 4 hours later another train B leaves kolkata and is travelling in same direction as train A in how many time (in minutes) train B will be 25 km ahead train A, if speed of train B is 100 km/hr ?
   475 minutes
   610 minutes
   340 minutes
   620 minutes
   368 minutes
   Option B
   Distance covered by train A in 4 hours = 70 * 4 = 280 km
   total distance should be covered by train B from kolkata for ahead 25 km to train A
   required time = 280 + 25/100-70 = 305/30 * 60 = 610 minutes
   

    

2. In a box there are 3 red balls, 7 white balls and 5 orange balls. If one bottle is drawn at random from the cartoon what is the probability that it is either red balls or white balls ?
   2/7
   8/9
   3/4
   5/7
   2/3
   Option E
   probability = (3C1) + (7C1)/15C1 = (3 + 7)/15 = 2/3

    

3. Two persons A and B alone can do a piece of work in 10 days and 12 days respectively. A and B together worked for 4 days and remaining worked by C alone. If efficiency of C is 20% less than the efficiency of B then find in how many days total work will be completed ?
   5 days
   8 days
   7 days
   12 days
   6 days
   Option B
   LCM of 10 and 12 = 60
   efficiency of A = 60/10 = 6
   efficiency of B = 60/12 = 5
   work completed by A and B together in 4 days = (6 + 5)* 4 = 44 work
   remaining work = 60 – 44 = 16 work
   efficiency of C = 5 * 80/100 = 4
   time required to complete the work = 16/4 = 4 work
   total time to complete the work = 4 + 4 = 8 days

    

4. a and B invested rs.2500 and rs.3100 in a business respectively. After 6 months A left the business and C joined with a capital of rs.2200. At the end of 1.5 years the profit received by A and B together is how much percent more or less than that by C ?
   125 3/8%
   140 10/11%
   48 3/5%
   152 3/7%
   215 3/8%
   Option B
   Profit sharing ratio of A , B and C = (2500 * 1.5) : (3100 * .5) : (2200 * 1) = 75 : 31 : 44
   more % = (75 + 31) – 44/44 * 100 = 140 10/11%

    

5. Sudha is travelling from point A to B which is 560 km apart with a certain speed. While returning he covers half the distance with 30% less speed than the usual speed and cover the remaining distance with a speed of 140 km/hr. If time taken in both cases is equal, then find the speed of Sudha while going from point A to B ?
   75 km/hr
   80 km/hr
   72 km/hr
   84 km/hr
   108 km/hr
   Option B
   Let speed of Sudha while travelling from point A to point B = 10x km/hr
   time taken by Sudha while travelling from A to B = 560/10x
   time taken by Sudha while returning = 280/7x + 280/140
   560/10x = 280/7x + 2
   56/x – 40/x = 2
   x = 8
   speed of Sudha = 10 * 8 = 80 km/hr

    

6. Directions : What will come in place of questions (?) marks in the following questions below ?
   (15)^2 – (18)^2 + 312 = ? + 47
   319
   120
   184
   133
   166
   Option E
   225 – 324 + 312 = ? + 47
   – 99 + 312 = ? + 47
   ? = 166
   

    

7. ∛1331 + 38% of 1200 – ? = 98.20
   
   421.50
   213.40
   368.80
   254.20
   281.20
   Option C
   11 + 456 – ? = 98.20
   ? = 368.80

    

8. 1/14 + 2/7 – 1/42 = ?
   2/7
   1/3
   4/5
   3/5
   4/7
   Option B
   (3 + 12 – 1) /42 = ?
   ? = 1/3

    

9. 42.30 – (11)^2 + 520 – ? = 218.40
   240.80
   312.90
   422.90
   222.90
   128.46
   Option D
   562.30 – 121 – ? = 218.40
   ? = 222.90

    

10. 3/4 of 420 + 320 – 4/5 of 215 = ? + 85
    420
    422
    540
    325
    378
    Option E
    315 + 320 – 172 = ? + 85
    ? = 378

     

1. While we add 4 to the numerator of the fraction and increases denominator by 50% then both become equal. When the numerator is increased by 5 and denominator is doubled then it becomes equal to the original fraction. What will be two times of that fraction ?
   2/5
   5/3
   2/7
   4/5
   4/3
   Option B
   Let the numerator and denominator of a fraction be x and y respectively.
   ATQ, x + 4 = 3y/2
   x + 5/2y = x/y
   by solving two equations,
   x = 5, y = 6
   original fraction = 5/6
   required fraction = 5/6 * 2 = 5/3

    

2. Two dices are rolled together, then find probability of getting a number of one dice greater than the number on other dice ?
   2/3
   7/6
   5/6
   2/5
   4/7
   Option C
   total number of cases = 6 * 6 = 36
   total cases of getting same number = (1,1), (2,2), 3,3), (4,4), (5,5), (6,6) = 6
   probability = 1 – 6/36 = 5/6

    

3. P, Q, R are three inlet pipes. Time taken by P and Q together to fill half of the tank is same as time taken by pipe R alone to fill one-sixth of the tank. If P, Q, R together can fill the tank in 9 hours, then find time taken by pipe R alone to fill the tank ?
   34 hours
   42 hours
   24 hours
   36 yours
   40 yours
   Option D
   ATQ, 2(P+Q) = 6R
   P + Q = 3R
   P + Q + R = 1/9
   4R = 1/9
   R = 36 hours

    

4. A man cuts a square field of maximum possible area from his circular field. Find the area of the remaining circular field, if the cost of grass cutting of the square field at rs. 4 per m^2 is rs.882 ?
   154 m^2
   126 m^2
   200 m^2
   348 m^2
   180 m^2
   Option B
   Let side of square be = a meters
   radius of circle = r meters
   area of the square field = 220.5 m^2
   diagonal of the square field = 21 m
   radius of the circular field = 10.5 m
   area of the circular field = 346.5 m^2
   area of the remaining circular field = 126 m^2
   

    

5. Rohit invested rs.1600 on CI for two years at the rate of R% p.a and gets amount of rs.2304. If man invested same sum on SI for same period of time at the rate of ( R – 8)%, then find interest he will get ?
   450
   384
   420
   640
   260
   Option B
   ATQ,
   2304 = 1600 ( 1 + R/100)^2
   48/40 = ( 1 + R/100)
   R = 20%
   New rate = (20 – 8) = 12%
   required interest = 1600 * 12 * 2/100 = 384

    

6. directions : What will be come in place of questions marks (?) in the following questions given below ?
   428, 434, 446, 466, 496 ?
   620
   538
   580
   650
   940
   Option B
   The series is, +(2 * 3), +(3 * 4), +(4 * 5), +(5 * 6), +(6 * 7)

    

7. 187, 193, 205, ?, 277, 373
   198
   229
   240
   350
   270
   Option B
   The series is (+6), (+12), (+24), (+48), (+96)

    

8. 144, 72, 108, 270, ?, 4252.5
   320
   620
   825
   945
   480
   Option D
   The series is, (* .5), (* 1.5), (* 2.5), (* 3.5), (* 4.5)

    

9. 81, ? 76, 92, 67, 103
   73
   85
   65
   76
   84
   Option B
   The series is, (+ 2^2), (- 3^2), (+ 4^2), (- 5^2), (+ 6^2)

    

10. 11, 13, 22, ?, 115, 241
    48
    56
    75
    50
    82
    Option D
    the series is, +(1^3 + 1), +(2^3 + 1), +(3^3 + 1), +(4^3 + 1), + (5^3 + 1)

     

1. The number of boys in a class are x and the number of girls are 4 less than the number of boys. The sum of weight of boys is 630 and the average weight of boys is 45 kg. If 2 students are selected for a exam then what will be the probability that the number of boys the numbers of girls are equal ?
   4/9
   35/69
   82/265
   36/129
   1/5
   Option B
   Total number of boys = x
   number of girls = x – 4
   the number of boys = 630/45 = 14
   the number of girls = 14 – 4 = 10
   probability = 14C1 * 10C1 / 24C2
   = 14 * 10 * 2/24 * 23 = 35/69

    

2. Difference between CI received in first 1.5 years at 20% per annum compounded annually and CI received in last 1.5 years at same rate of interest in compounded half yearly on the same sum is rs.495, then find the sum ?
   32000
   48000
   30000
   52000
   45000
   Option E
   Let sum = 100x
   total rate of in 1st 1.5 years = 20 + 10 + 20 * 10 /100 = 32%
   CI received = 100x * 32/100 = 32x
   CI received in last 1.5 years = 100x * 1.1 * 1.1 * 1.1 – 100x = 33.1x
   difference = 33.1x – 32x = 495
   1.1x = 495
   x = 450
   sum = 100x = 100* 450 = 45000

    

3. X and Z alone can do a piece of work in 25 days and 30 days respectively, while Y takes as half time as X and Z take together. If they start working alternatively starting by Y, followed X and Z respectively, then find in how many days work will be completed ?
   
   5 8/17 days
   12 9/11 days
   8 2/5 days
   6 3/5 days
   26 3/8 days
   Option B
   LCM of 25 and 30 = 150
   efficiency of X = 150/25 = 6
   efficiency of Z = 150/30 = 5
   efficiency of Y = 22
   when all three works alternatively
   3 days work = 22 + 6 + 5 == 33 work
   in total 12 days = 12/3 * 33 = 132 work
   remaining work = 150 – 132 = 18
   remaining work completed by Y = 18/22 = 9/11 day
   total days = 12 9/11 days

    

4. A boat takes total 10 hours to cover the distance of 84 km in upstream and 84 km in downstream. If the speed of boat is increased by 12km/hr, then the new upstream speed is doubled of its usual speed. Find the time taken by boat to cover 140 km in downstream.
   5
   8
   2
   6.5
   2.5
   Option A
   Let speed of boat = x km/hr
   speed of stream = y km/hr
   downstream speed = x + y
   upstream speed = x – y
   new speed of boat = x + 12
   upstream speed = x + 12 – y
   2 (x – y ) = x + 12 – y
   x – y = 12
   upstream speed = 12
   time taken by boat to cover 84 km in upstream 84/12 = 7
   time taken by boat in downstream = 10 – 7 = 3
   downstream speed = 84/3 = 28 km
   time taken by boat to cover 140 km in downstream = 140/28 = 5 hours

    

5. A and B started a partnership business with investment (x + 500) and (x – 1000) respectively. After 6 months a withdrew 40% of his amount. If profit received by A at the end of the year is 2800 out of the total profit rs.4800, then what is the value of ‘x’ ?
   4000
   2000
   1500
   3000
   5600
   Option D
   Investment of A = (x + 500)* 6 + ( x + 500)* .6 * 6 = 6(1.6x + 800)
   investment of B = (x – 1000) * 12
   profit of A = 2800
   profit B = 4800 – 2800 = 2000
   6( 1.6x + 800)/(x – 1000)12 = 2800/2000
   8x + 4000 = 14x – 14000
   x = 3000

    

6. I. 2x^2 – 13x + 15 = 0
   II. y^2 – 6y + 8 = 0
   
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. 2x^2 – 10x – 3x + 15 = 0
   2x = 10, 3
   x = 5 , 1.5
   II. y^2 – 4y – 2y + 8 = 0
   y = 4, 2
   

    

7. I. x^2 – 27x + 92 = 0
   II. y^2 – 17y + 60 = 0
   
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. x^2 – 23x – 4x + 92 = 0
   x = 23, 4
   II. y^2 – 12y – 5y + 60 = 0
   y = 12 , 5
   

    

8. I. x^2 = 196
   II. y^3 = 4096
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option B
   I. x^2 = 196
   x = 14, -14
   II. y^3 = 4096
   y = 16

    

9. I. 5x^2 – 7x + 2 = 0
   II. y^2 + 12y + 32 = 0
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option A
   I. 5x^2 – 5x – 2x + 2 = 0
   5x = 5, 2
   x = 1, .4
   II. y^2 + 8y + 4y + 32 = 0
   y = -8, -4

    

10. I. 3x^2 – 19x + 20 = 0
    II. 4y^2 – 6y + 2 = 0
    
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option A
    I. 3x^2 – 15x – 4x + 20 = 0
    3x = 15, 4
    x = 5, 1.33
    II. 4y^2 – 4y – 2y + 2 = 0
    4y = 4, 2
    y = 1. .5

     

1. A shopkeeper X sold an article for rs.850 at a profit of 25%. Another shopkeeper Y sold the another article and earned 30% more profit than the profit earned by X. If the selling price of article Y is three-fifth of selling price of article X, then find the profit percent of article Y ?
   64 2/17%
   76 8/17%
   52 3/34%
   52 3/8%
   78%
   Option B
   SP of article X = 850
   CP of article X = 850 * 100/125 = 680
   profit = 170
   profit of article Y = 170*130/100 = 221
   SP of article Y = 850 * 3/5 = 510
   CP = 510 – 221 = 289
   profit percent = 221/289* 100 = 76 8/17%

    

2. Mixture A contains milk and water in the ratio of 3 : 2 respectively, while mixture B contains water and milk in the ratio of 4 : 3 respectively. If the quantity of water in mixture B is twice the quantity of water in mixture A, then find the quantity of milk in mixture A is what percent quantity of water in mixture B ?
   68%
   60%
   82%
   75%
   56%
   Option D
   Quantity of milk and water in mixture A = 3x and 2x
   according to questions,
   quantity of water in mixture B = 2*2x = 4x
   quantity of milk in mixture B = 4x/4 *3 = 3x
   required percentage = 3x/4x*100 = 75%

    

3. 340 liters of mixture of milk and water is contained in a container having milk and water mixed in the ratio of 3 : 1 respectively. 76 liters of this mixture is taken out from the container and replaced with (x + 7) liters of milk so that the ratio of water to milk is 6 : 11. What is the value of ‘x’ ?
   42
   48
   38
   35
   52
   Option D
   Quantity of milk in the container = 340 * 3/4 = 255
   quantity of water in the container = 340*1/4 = 85
   76 liters of mixture is taken out and (x + 7) liter of water is added.
   (255 – 57) / 85 – 19 + (x + 7) = 11/6
   198/(73 + x) = 11/6
   11x = 385
   x = 35
   

    

4. A can do a piece of work in 30days while B and C together can complete the work in 20 days and A and C together can complete the work in 24 days. If B is worked in two-fifth of his efficiency, then in how many days A and B together complete half of the work ?
   8 days
   6 days
   8 1/3 days
   5 days
   10 days
   Option E
   LCM of 30 , 20 and 24 = 120
   efficiency of A = 120/30 = 4
   efficiency of B and C = 120/20 = 6
   efficiency of A and C together = 120/24 = 5
   efficiency of C = 5- 4 = 1
   efficiency of B 6- 1 = 5
   two-fifth of efficiency B = 5 * 2/5 = 2
   time required to complete half of the work = 60/2 + 4 = 10 days

    

5. Rohan and Sivan started a business together with initial investments of rs. 1500 and rs.1800 respectively. If after 2.5 years profit share of Rohan and Sivan are rs(x + 80) and (x + 120) respectively, then find the difference between profit of Rohan and Sivan ?
   180
   120
   320
   130
   340
   Option B
   Profit sharing ratio of Rohan and Sivan = (1500 * 2.5) : (1800 * 2.5) = 5 : 6
   x + 80/x + 120 = 5/6
   6x + 480 = 5x + 600
   x = 120

    

6. Directions : In each of these questions given below two equations. You have to solve both the equations and give answer.
   I. x^2 – 7√3 + 36 = 0
   II. y^2 – 6√2 + 16 = 0
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. x^2 – 4√3 – 3√3 + 36 = 0
   x = 4√3 , 3√3
   II. y^2 – 4√2 – 2√2 + 16 = 0
   y = 4√2, 2√2

    

7. I. x^2 + √2x – 12 = 0
   II. y^2 – 8√3 + 48 = 0
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option B
   I. x^2 + 3√2 – 2√2 – 12 = 0
   x = -3√2, 2√2
   II. y^2 – 4√3 – 4√3 + 48 = 0
   y = 4√3, 4√3

    

8. I. x^2 – 36 = 0
   II. y^2 – 15y + 36 = 0
   
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. x^2 = 36
   x = 6, -6
   II. y^2 – 12y – 3y + 36 = 0
   y = 12, 3

    

9. I. y^2 = 144
   II. 2x^2 – 12x + 16 = 0
   X > Y
   X < Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation.
   Option E
   I. y^2 = 144
   y = 12, -12
   II. 2x^2 – 8x – 4x + 16 = 0
   2x = 8, 4
   x = 4, 2

    

10. I. x^2 – 2x – 195 = 0
    II. y^2 – 24y + 128 = 0
    X > Y
    X < Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation.
    Option E
    I. x^2 – 15x + 13x – 195 = 0
    x = 15, -13
    II. y^2 – 16y – 8y + 128 = 0
    y = 16, 8

     

1. A man deposited a total amount of rs. 36000 partially in scheme A and scheme B. Scheme A offers simple interest at the rate of 14% p.a while scheme B offers compound interest at the rate of 20% p.a compounded annually. If the interest earned from scheme A after 2 years is rs.2160 less than the interest earned from scheme B after 2 years, then what is the amount deposited in scheme B ?
   15000
   17000
   23000
   14000
   17500
   Option B
   Let amount deposited in scheme B = rs. x
   amount deposited in scheme A = rs.(36000 – x)
   Rate of compound interest after 2 years = 20 + 20 + 20 * 20/100 = 44%
   x * 44/100 – (36000 – x) * 14/100 * 2 = 2160
   (44x – 108000 + 28x)/100 = 2160
   72x = 1224000
   x = 17000
   amount deposited in scheme B is rs.17000

    

2. The income of A and B are in the ratio 4 : 5 respectively. If the expenditure of A and B are rs.14000 and rs.18000 and ratio of saving of A and B is 6 : 7, then what is the saving of B ?
   8000
   9200
   7000
   6800
   8200
   Option C
   Let income of A and B = 4x and 5x respectively
   4x – 14000/5x – 18000 = 6/7
   28x – 98000 = 30x – 108000
   x = 5000
   income of B = 5 * 5000 = 25000
   saving of B = 25000 – 18000 = 7000

    

3. A takes 5 days less than B to complete a work alone. A and B together can complete the work in 13 7/11 days. Find the time taken by B to complete the whole work ?
   25 days
   30 days
   45 days
   43 days
   40days
   Option B
   Let B alone can complete the whole work = x days
   A alone can complete the whole work = (x – 5) days
   1/x – 5 + 1/x = 11/150
   2x – 5/x^2 – 5x = 11/150
   11x^2 – 55x = 300x – 750
   11x^2 – 355x + 750 = 0
   11x^2 – 330x – 25x + 750 = 0
   x = 30
   B alone can complete the whole work in 30 days

    

4. Three containers having capacity in the ratio of 4 : 5 : 3 contain mixture of milk and water in the ratio of 2 : 3, 4 : 1 and 3 : 1 respectively. If all three containers poured in a big container, then find the ratio of water to milk in final mixture ?
   28 : 137
   85 : 152
   83 : 157
   134 : 49
   5 : 2
   Option C
   Let total mixture in three container are 400, 500 and 300 respectively
   total quantity of milk in big container = 400 * 2/5 + 500 * 4/5 + 300 * 3/4 = 785
   total quantity of water in big container = 400 * 3/5 + 500 * 1/5 + 300 * 1/4 = 415
   required ratio = 415 : 785 = 83 : 157
   

    

5. B and C started a partnership business with the capital of rs.(x + 4000) and rs.(x + 2000) respectively. After one year A joined the business with capital of rs.(x + 12000) while B left the business. If at the end of three years the profit sharing ratio of A , B and C is 20 : 6 : 15, then find the difference between amount invested by A and B in the business ?
   15000
   13000
   14000
   8000
   12000
   Option D
   A : B : C = (x + 12000) * 2 : (x + 4000) * 1 : (x + 2000) * 3
   = 2x + 24000 : x + 4000 : 3x + 6000
   2x + 24000/x + 4000 = 20/6
   20x + 80000 = 12x + 144000
   8x = 64000
   x = 8000
   difference = (8000 + 12000) – (8000 + 4000) = 8000

    

6. Directions : In these questions two equations is given, solve the equations and answer the questions given below.
   I. x^2 – 42x + 392 = 0
   II. y^2 – 27y + 180 = 0
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option E
   I. x^2 – 28x – 14x + 392 = 0
   x = 28 , 14
   II. y^2 – 15y – 12y + 180 = 0
   y = 15, 12

    

7. I. x^2 – 8x – 65 = 0
   II. 2y^2 – 27y + 91 = 0
   
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option E
   I. x^2 – 13x + 5x – 65 = 0
   x = 13, -5
   II. 2y^2 – 14y – 13y + 91 = 0
   2y = 14, 13
   y = 7, 6.5
   

    

8. I. x^3 = 216
   II. 5y^2 + 2y – 185 = 0
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option B
   I. x^3 = 216
   x = 6, 6
   II. 5y^2 + 15y – 13y – 37 = 0
   5y = – 15, 13
   y = -3 , 2.6

    

9. I. x^2 + 17x + 60 = 0
   II. y^2 – 2y – 15 = 0
   X < Y
   X > Y
   X ≤ Y
   X ≥ Y
   X = Y or no relation
   Option A
   I. x^2 + 12x + 5x + 60 = 0
   x = -12, -5
   II. y^2 – 5y + 3y – 15 = 0
   y = 5, -3

    

10. I. x^2 – 15x + 36 = 0
    II. y^2 – 19y + 60 = 0
    X < Y
    X > Y
    X ≤ Y
    X ≥ Y
    X = Y or no relation
    Option E
    I. x^2 – 12x – 3x + 36 = 0
    x = 12 , 3
    II. y^2 – 15y – 4y + 60 = 0
    y = 15, 4

     

1. Ratio of number of male of company X to company Y is 8 : 5. The ratio of female to male of company X and company Y is 3 : 4 and 7 : 5 respectively. The difference between number of female of company X and company Y is 150, then find total number of male and female in company Y ?
   2700
   2200
   1800
   1500
   1200
   Option C
   Let the number of male of company X and Y = 8a and 5a
   number of male and female of company X = 4a and 3a
   number of male and female of company Y = 5a and 7a
   The number of male and female of company X = 8a and 6a
   difference = 7a – 6a = 150
   a = 150
   sum of number of male and female of company Y = 5a + 7a = 12a
   = 12 * 15 = 1800
   

    

2. A and B entered into a partnership business with initial investments of rs.4000 and rs.5000 respectively. After 6 months A and B withdrew rs.1000 and rs.2000 respectively, while C joined with them capital of 4x. If after 14 months, C received rs.4000 as profit share out of total profit of rs.12500, then find how much amount invested by C ?
   4000
   4500
   4800
   6000
   3200
   Option D
   Profit sharing ratio of A, B and C = (4000 * 6 + 3000 * 8) : (5000 * 6 + 3000 * 8) : 4x * 8
   = 48000 : 54000 : 32x
   profit of A and B = 12500 – 4000 = 8500
   32x/102000 = 4000/8500
   x = 1500
   C invested = 4 * 1500 = 6000

    

3. Rohit invested rs.100x in a scheme offering compound interest at the rate of 20% p.a and after two years he again invested rs.80x in the same scheme. He received rs. 1243.2 as total interest after 3 years. If he rs.120x is invested at 10% p.a simple interest for 4 years the find how much interest he received ?
   1128
   672
   845
   862
   920
   Option B
   Compound interest after 2 years = 100x * 120/100 * 120/100 – 100x = 44x
   now total principal = 100x + 44x + 80x = 224x
   CI received after one year = 224x * 20/100 = 44.8x
   total CI after 3 years = 88.8x
   88.8x = 1243.2
   x = 14
   If he received 120x at 10% p.a in SI for 4 years
   120 * 14 * 40/100 = 672

    

4. The ratio of milk and water in Jar A is 5 : 3 and total mixture of container A is 480 liters. Jar B contains mixture of water to milk in the ratio of 7 : 9. Total mixture of jar A and B poured into another jar C such that quantity of milk in jar C is 60% of total mixture. Find the quantity of mixture in jar B ?
   440 lt
   480 lt
   380 lt
   240 lt
   320 lt
   Option E
   Quantity of milk in jar A = 480 * 5/8 = 300
   quantity of water in jar B = 480 * 3/8 = 180
   let the quantity of milk and water in jar B = 9x and 7x
   300 + 9x = (480 + 16x) * 60/100
   1500 + 45x = 48x + 1440
   3x = 60
   x = 20
   total mixture in jar B = (9x + 7x) * 20 = 320 liters
   

    

5. A man deposited rs.8000 in a bank offering compound interest of 20% p.a compound annually for first 1.5 years and after 1.5 years he offering simple interest 50% p.a for rest of the time. If at the end of 1.5 years, he withdrew rs.3560, then how much amount received by him at the end of 3 years ?
   12400
   10500
   18000
   12250
   15240
   Option D
   Rate of compound interest after 1.5 years = 20 + 10 + 20 * 10/100 = 32%
   total amount after 1.5 years = 8000 * 132/100 = 10560
   remaining amount after withdrawal by him = 10560 – 3560 = 7000
   simple interest received for rest of time = 7000 * 1.5 * 50/100 = 5250
   total amount received = 7000 + 5250 = 12250

    

6. Directions : What will come in place of questions (?) marks in the following questions given below ?
   30% of 168 + √1024 + 42 = ?
   84.2
   94.8
   124.4
   65
   84
   Option C
   50.4 + 32 + 42 = ?
   ? = 124.4

    

7. (25^2) + (38^2) – 1220 + ? = 1000
   151
   162
   170
   175
   320
   Option A
   625 + 1444 – 1220 + ? = 1000
   ? = 1000 – 849
   ? = 151

    

8. (32 * 8 + 440 + 15 * 12) * 6 = ?
   4340
   3240
   2015
   5256
   8450
   Option D
   (256 + 440 + 180) * 6 = ?
   ? = 5256
   

    

9. 5^2 + 387 = 180/? + 376
   6
   8
   4
   5
   7
   Option D
   25 + 387 = 180/? + 376
   412 – 376 = 180/?
   ? = 5

    

10. 75 * 8 + √1156 = ?^3 + 122
    4
    32
    14
    8
    4
    Option D
    600 + 34 = ?^3 + 122
    ?^3 = 512
    ? = 8

     

Direction: What is the approximate value in the place of questions mark (?) in the following questions ?

1. 887.97÷12.06*√16.05-36=?
   268
   260
   280
   300
   240
   Option B
   888/12*4-36=?
   ?=296-36=260

    

2. ?^2+56.01-24*5.99=168.01
   16
   15
   13
   20
   24
   Option A
   168+144-56=?^2
   ?=16^2

    

3. 45.01% of 4500÷15.05+139.88-220.01=?
   60
   65
   55
   70
   85
   Option C
   2025/15+140-220=?
   135+140-220=55

    

4. (7.99)^2+(10.01)^2-(18.05)^2+(12.99)^2=?
   15
   13
   9
   12
   28
   Option C
   8^2+10^2-18^2+13^2=?
   64+100-324+169=?
   ?=9

    

5. 539.98/?=?/60.01
   200
   160
   120
   180
   200
   Option D
   ?^2=540*60
   ?^2=540*60
   ?=180

    

Direction: What will come in the of questions mark(?) in the following questions.

6. 36% of 1000+42% of 3000=?-180+1050
   750
   650
   850
   950
   550
   Option A
   360+1260+180-1050=750

    

7. 5/11 of 1331+4/9 of 270=30% of 400+30+?
   680
   575
   675
   550
   740
   Option B
   605+120-120-30=?
   ?=575

    

8. 2748+1323-2238-1235+3248=?
   3646
   3948
   3748
   3846
   2896
   Option D
   7319-3473=3846

    

9. 4515÷15+324-25% of 1200=?
   325
   425
   320
   440
   345
   Option A
   301+324-300=325

    

10. 48/3*45/80*1/9+135-40=?
    118
    105
    78
    96
    120
    Option D
    48/3*45/80*1/9+135-40=?
    1+135-40=96