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