# IBPS RRB PO Prelims Quantitative Aptitude Questions 2019 (Day-08)

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Directions (1 – 5): Following question contains two equations as I and II You have to solve both equations and determine the relationship between them and give answer as,

a) If x> y

b) If x ≥ y

c) If x = y or relationship can’t be determined.

d) If x< y

e) If x ≤ y

1) I) x2 + 31x + 240 = 0

II) y2 –9y–90 = 0

2) I) x2 – 0.58x + 0.084 = 0

II) y2–0.48y + 0.056 = 0

3) I) x2 – x –72 = 0

II) y2–19y + 90 = 0

4) I) x = 89%

II) y2 – y + 0.0979 = 0

5) I) x3 = 216

II)y2 – 13y + 42 = 0

Directions (6 – 10): Read the following information carefully and answer the following question. The table below shows number of student in different section of classes: 6) The average number of students in class V is 88 and the number of students in section B is 20% more than the number of students in section D then find the ratio between the number of students in Section B of class V and section D of class X?

a) 20:3

b) 40:9

c) 50:69

d) 30:37

e) 69:50

7) If the ratio between number of students in section B and section C of class VI becomes 4:5 when 20 more students joins section C then find the number of students initially in section C?

a) 20

b) 40

c) 55

d) 30

e) 60

8) If the average weight of students in section A of class IX is ‘x’ while that of section C is ‘x-10’ kg and the average weight of all the students in these sections is 50 kg, then find the new average weight of section C if the teacher is also included whose weight is 44 kg?

a) 38

b) 40

c) 52

d) 44

e) 60

9) If the total number of students in class VII  and class VIII is 620 then find the average number of students in section D of class VII and VIII?

a) 137

b) 140

c) 105

d) 130

e) 160

10) If the ratio between the number of students in section C of class VIII and X is 5:4 and the number of students in section A of class X is 50% more than the section C of the same class then find the average number of students in class X?

a) 20

b) 40

c) 10

d) 30

e) 60

Direction (1-5) :

x2 + 31x + 240 = 0

x2 + 15x + 16x + 240 = 0

x(x + 15) + 16(x + 15) = 0

(x + 16)(x + 15) = 0

x = -16, -15

y2 – 9y – 90 = 0

y2 – 15y + 6y – 90 = 0

y(y – 15) + 6(y – 15) = 0

(y + 6)(y – 15) = 0

y = -6, 15

Hence x<y

x2 – 0.58x + 0.084 = 0

x2 – 0.28x – 0.30x + 0.084 = 0

x(x – 0.28) – 0.30(x – 0.28) = 0

(x – 0.30)(x – 0.28) = 0

x = 0.30, 0.28

y2 – 0.48y + 0.056 = 0

y2 – 0.28y – 0.20y + 0.056 = 0

y(y – 0.28) – 0.20(y – 0.28) = 0

(y – 0.28)(y – 0.20) = 0

y = 0.28, 0.20

Hence x ≥ y

x2 – x – 72 = 0

x2 – 9x + 8x – 72 = 0

x(x – 9) + 8(x – 9) = 0

(x + 8)(x – 9) = 0

x = -8, 9

y2 – 19y + 90 = 0

y2 – 10y – 9y + 90 = 0

y(y – 10) – 9(y – 10) = 0

(y – 9)(y – 10) = 0

y = 9, 10

Hence x ≤ y

x = 89%

x = 89/100 = 0.89

y2 – y + 0.0979 = 0

y2 – 0.89y – 0.11y + 0.0979 = 0

y(y – 0.89) – 0.11(y – 0.89) = 0

(y – 0.11)(y – 0.89) = 0

y = 0.11, 0.89

Hence x ≥ y

x3 = 216

x = 6

y2 – 13y + 42 = 0

y2 – 6y – 7y + 42 = 0

y(y – 6) – 7(y – 6) = 0

(y – 7)(y – 6) = 0

y = 7, 6

Hence x ≤ y

Direction (6-10) :

Total number of students in class V = 88*4= 352

Let the number of students in section D be x

Section B= x*120/100= 6x/5

So,

x+6x/5+ 60+39= 352

11x/5= 352-99

11x/5= 253

x= 253*5/11

x= 115

Number of students in section B= 6*115/5= 138

Required ratio= 138:100= 69:50

Let the let the number of students initially in section C be x

Now,

Number of students in section C = x+20

60/x+20 = 4/5

300= 4x+80

220= 4x

x= 55

Total weight of all students= 50*160= 8000 kg

So,

96*x + 64(x-10) = 8000

96x + 64x – 640= 8000

160x= 8000+640

x= 8640/160

x= 54 kg

Average weight of section C = 54 -10= 44 kg

Total weight if we include weight of the teacher

= (44*64) + 44 = 2816 + 44 = 2860 kg

Required average weight= 2860/65= 44 kg

Let the number of students in section D of class VII and VIII be x and y respectively

So,

(80+90+16+x) + (85+30+45+y) = 620

186+x+160+y = 620

x + y = 620- 346

x + y = 274

Required average = 274/2 = 137 