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[WpProQuiz 6655]
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
Â
2) I) x2 – 0.58x + 0.084 = 0
II) y2–0.48y + 0.056 = 0
Â
3) I) x2 – x –72 = 0
II) y2–19y + 90 = 0
Â
4) I) x = 89%
II) y2 – y + 0.0979 = 0
Â
5) I) x3 = 216
II)y2 – 13y + 42 = 0
Â
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 VIIÂ 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
Answers :
Direction (1-5) :
1) Answer: d)
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
2) Answer: b)
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
3) Answer: e)
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
4) Answer: b)
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
5) Answer: e)
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) :
6) Answer: e)
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
7) Answer: c)
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
8) Answer: d)
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
9) Answer: a)
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
10) Answer: e)
Number of students in section C of class X= 45*4/5= 36
Number of students in section A of class X= 36*150/100= 54
Required average= (54+50+36+100)/4 = 240/4 = 60 students
