Dear Readers, Exam Race for the Year 2019 has already started, To enrich your preparation here we are providing new series of Practice Questions on Quantitative Aptitude – Section for CWC/FCI Exam. Aspirants, practice these questions on a regular basis to improve your score in aptitude section. Start your effective preparation from the right beginning to get success in upcoming CWC/FCI Exam.
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Directions (Q. 1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,
a) If x > y
b) If x ≥ y
c) If x < y
d) If x ≤ y
e) If x = y or the relation cannot be established
1) I) 7x – 4y = -2
II) 3x – 5y = -14
Â
2) I) 12x2 + 45x + 27 = 0
II) 7y2 + 36y + 32 = 0
Â
3) I) x2 + 20x + 99 = 0
II) y2 + 13y + 42 = 0
Â
4) I) x + ∛658503 = 422 ÷ 14 – 16
II) y = ∜531441
Â
5) I) 12x2 – 5x – 25 = 0
II) 6y2 + 16y – 56 = 0
Directions (Q. 6 – 10): Study the following information carefully and answer the given questions.
The following bar graph shows the total number of tea stalls in two different cities over the years.
6) Find the difference between the total tea stalls in city A in the year 2013, 2015 and 2017 together to that of total tea stalls in city B in the year 2014, 2016 and 2018 together?
a) 380
b) 450
c) 330
d) 420
e) None of these
7) Find the ratio between the total tea stalls in the year 2013 and 2014 together to that of total tea stalls in the year 2016 and 2017 in both the cities together?
a) 68: 97
b) 35: 44
c) 12: 23
d) 49: 62
e) None of these
8) Find the average number of tea stalls in city A in all the given years together?
a) 360
b) 320
c) 440
d) 400
e) None of these
9) Total number of tea stalls in the year 2015 is approximately what percentage of total number of tea stalls in the year 2018 in both the cities together?
a) 90 %
b) 70 %
c) 105 %
d) 125 %
e) 140 %
10) Find the difference between the total number of tea stalls in city A to that of city B in all the given years together?
a) 450
b) 530
c) 370
d) 280
e) None of these
Answers :
Direction (1-5) :
1) Answer: c)
7x – 4y = -2 —> (1)
3x – 5y = -14 —> (2)
By solving the equation (1) and (2), we get,
X = 2, y = 4
Hence, x < y
2) Answer: e)
I) 12x2 + 45x + 27 = 0
12x2 + 36x + 9x + 27 = 0
12x (x + 3) + 9 (x + 3) = 0
(12x + 9) (x + 3) = 0
X = -9/12, -3 = -0.75, -3
II) 7y2 + 36y + 32 = 0
7y2 + 28y + 8y + 32 = 0
7y (y + 4) + 8 (y + 4) = 0
(7y + 8) (y + 4) = 0
Y = -8/7, -4 = -1.142, -4
Can’t be determined
3) Answer: c)
I) x2 + 20x + 99 = 0
(x + 11) (x + 9) = 0
X = -11, -9
II) y2 + 13y + 42 = 0
(y + 6) (y + 7) = 0
Y = -6, -7
Hence, x < y
4) Answer: c)
I) x + ∛658503 = 422 ÷ 14 – 16
x + 87 = (42*42)/14 – 16
x = 126 – 16 – 87 = 23
II) y = ∜531441 = 27
Hence, x < y
5) Answer: e)
I) 12x2 – 5x – 25 = 0
12x2 – 20x + 15x – 25 = 0
4x (3x – 5) + 5 (3x – 5) = 0
(4x + 5) (3x – 5) = 0
X = -5/4, 5/3 = -1.25, 1.667
II) 6y2 + 16y – 56 = 0
6y2 – 12y + 28y – 56 = 0
6y (y – 2) + 28 (y – 2) = 0
(6y + 28) (y – 2) = 0
Y = -28/6, 2 = -4.667, 2
Can’t be determined
Direction (6-10) :
6) Answer: c)
The total tea stalls in city A in the year 2013, 2015 and 2017 together
= > 280 + 350 + 480 = 1110
The total tea stalls in city B in the year 2014, 2016 and 2018 together
= > 400 + 470 + 570 = 1440
Required difference = 1440 – 1110 = 330
7) Answer: a)
The total tea stalls in the year 2013 and 2014 in both the cities together
= > 280 + 360 + 320 + 400 = 1360
The total tea stalls in the year 2016 and 2017 in both the cities together
= > 440 + 470 + 480 + 550 = 1940
Required ratio = 1360: 1940 = 68: 97
8) Answer: d)
The average number of tea stalls in city A in all the given years together
= > (280 + 320 + 350 + 440 + 480 + 530)/6
= > 2400/6 = 400
9) Answer: b)
Total number of tea stalls in the year 2015 in both the cities together
= > 350 + 420 = 770
Total number of tea stalls in the year 2018 in both the cities together
= > 530 + 570 = 1110
Required % = (770/1110)*100 = 69.369 % = 70 %
10) Answer: c)
The total number of tea stalls in city A in all the given years together
= > 280 + 320 + 350 + 440 + 480 + 530 = 2400
The total number of tea stalls in city B in all the given years together
= > 360 + 400 + 420 + 470 + 550 + 570 = 2770
Required difference = 2770 – 2400 = 370
