IBPS RRB PO Mains Quantitative Aptitude Questions 2019 (Day-04)

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IBPS RRB PO Mains Quantitative Aptitude Questions 2019 (Day-04)

maximum of 10 points
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Directions (1 – 5): Study the following graph carefully and answer the given questions

The bar graph shows the no. of Red, Blue and Pink colour toys in a box A, B, C, D and E

The line graph shows the probability of a green colour toy taken from each boxes

1) If 25% of red toys taken from box A and added to box E, then find the probability of taken two toys both are pink colour in box E

a) 98/1378

b) 105/1353

c) 105/1378

d) 102/1378

e) 115/1378

2) If the total no of red colour toys and pink colour toys from all the boxes (A, B, C, D and E) are taken out and filled in a new box P, then 2 toys taken out from box P find the probability of different colours

a) 3854/7875

b) 3869/7875

c) 3824/7875

d) 3857/7875

e) 3814/7875

3) Four toys taken from box C, then find the probability of different colour

a) 116/2303

b) 226/2303

c) 256/2303

d) 216/2303

e) 96/2303

4) Total number of balls in box B and D together is approximately what percentage of the total number of balls in box C and E together?

a) 60

b) 50

c) 40

d) 80

e) None of these

5) Two balls taken randomly from box B. What is the probability of at least one pink colour ball?

a) 42/65

b) 41/65

c) 36/65

d) 38/65

e) 48/65

Directions (6 – 10): Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

6) Quantity I: 5, 8, 21, 80, 395, ?

Quantity II: 15, 40, 100, 230, 500, ?

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

7) Quantity I: 6x2+11x-35=0

Quantity II: 2x2 – 40x + 200= 0

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

8) Quantity I: Time taken by P to complete 1/5th of work if Q takes 6 days to complete 3/5th of work and together they take 5 days to complete 3/4th of work

Quantity II:  Time taken by a train to cross a platform of length 60 km given that it crosses a pole in two and a half hours running at 60 km/hr

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

9) Quantity I: Age of M five years ago if 6 years hence ratio of age of M to N will be 14: 11, and 1 year ago the ratio was 7: 5.

Quantity II:  Average age of 2 students included in a group of 5 students with average age 18 years given that addition of these 2 students in group makes average increase by 1.

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

10) A shopkeeper gets a discount of 20% on an article and sells it at 10% profit.

Quantity I: Labeled price as a percent of cost price.

Quantity II:  Profit percent if discount given is 60%

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

Answers :

Directions (1 – 5):

Box A:

Let us take the no. of green colour toys be x

Probability = xc1/(32+x)c1

9/25 = x/(32+x)

32*9 +9x = 25x

16x = 32*9=> x= 18 green colour toys

Box B:

Let us take the no. of green colour toys be x

Probability = xc1/(34+x)c1

3/20 = x/(34+x)

3*34+ 3x = 20x

=>3*34= 17x

X= 6 green colour toys

Box C:

Let us take the no. of green colour toys be x

Probability = xc1/(40+x)c1

1/5 = x/(40+x)

=>40+x= 5x

=>4x=40 =>x= 10 green colour toys

Box D:

Let us take the no. of green colour toys be x

Probability = xc1/(26+x)c1

2/15 = x/(26+x)

52+2x= 15x

=>13x= 52

=> x= 4 green colour toys

Box E:

Let us take the no. of green colour toys be x

Probability = xc1/(40+x)c1

1/5 = x/(40+x)

=>40+x= 5x

=>4x=40 =>x= 10 green colour toys

1) Answer: c)

Total number of balls in box E = [(12*25/100)+15] +10+15+10

= 18+10+15+10

= 53

Required probability = 15c2/53c2 = (15*14)/(53*52)

= 105/1378

2) Answer: b)

Total number of red colour boys in box P = (12+8+10+8+15) = 53

Total number of pink colour balls in box P = (12+16+18+12+15) = 73

Required probability = (53*73)/(53+73)c2

= 53*73/(126*125/1*2) = 3869/7875

3) Answer: d)

Required probability = (10*12*18*10)/50c4

= (10*12*18*10)/(50*49*48*47/1*2*3*4)

= (24*9)/(49*47)

= 216/2303

4) Answer: e)

Required percentage

= [(8+10+16+6+8+6+12+4)/(10+12+18+10+15+10+15+10)]*100

= 70/100 * 100 = 70%

5) Answer: a)

Required probability = 1- 24c2/40c2

= 1 – (24*23/40*39)

= 1- 23/65

= 42/65

Direction (6-10) :

6) Answer: a)

Quantity I:

5*2-2 = 8

8*3-3 = 21

21*4-4 = 80

80*5-5 = 395

395*6-6 = 2364

Quantity II:

15*2+10 = 40

40*2+20 = 100

100*2+30 =230

230*2+40 =500

500*2+50 =1050

Quantity I > Quantity II

7) Answer: c)

Quantity I: 6x2 + 11x – 35 = 0

6x2 – 21x + 10x – 35 = 0

3x (2x – 7) + 5(2x – 7) = 0

(2x – 7) + (3x + 5) = 0

X = 7/2, -5/3

Quantity II: 2x2 – 40x + 200= 0

x2– 20x + 100 = 0

x (x – 10) – 10 (x – 10) = 0

(x – 10) (x – 10) = 0

X = 10, 10

Quantity I < Quantity II

8) Answer: a)

Quantity I: P and Q completes 3/4th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days

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

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

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

Quantity II: length of train = 2.5 * 60 = 150 km

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

= (150+60)/60 = 3.5hrs

Hence, Quantity I >Quantity II

9) Answer: c)

Quantity I: (M+6)/(N+6) = 14/11

And (M-1)/(N-1) = 7/5

Solve the equations, M = 22, so 5 years ago = 22 – 5 = 17 years

Quantity II: Total age of 2 students added to group = 18*2 + 7*1 = 43

So their average age = 43/2 = 21.5 years

Hence Quantity I < Quantity II

10) Answer: c)

Quantity I: MP = (100+10)/(100-20) * CP = 11/8 of CP

So required % = MP/CP * 100 = (11CP/8)/CP * 100 = 137.5%

Quantity II: % profit = (100+10) [(100-10)/(100-60)] – 100 = 590/4 = 147.5%

Hence Quantity I < Quantity II

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