Night Class: Quant Quiz


Directions (Q. 1 – 5) : In the following questions two equations numbered (I) and (II) are given. You have to solve both the equations and give answer:
1) If x > y 
2) If x >= y 
3) If x < y 
4) If x <= y
5) If x = y or the relationship can’t be established.

1. 
I. 14x^2 + 17x – 6 = 0 
II. 6y^2 – 13y + 5 = 0
2. 
I. x = √7 
II. 6y^2 – 7y – 20 = 0
3. 
I. 3x^2 + 8x – 35 = 0 
II. y^2 – 2y – 48 = 0
4. 
I. x^2 – 23x + 132 = 0 
II. y = 3√1331
5. 
I. 7x – 5y = 64 
II. 4x + 3y = 19
Directions (Q. 6 – 10): Following table shows the sale of cars of six different companies during the period of 2008 to 2013. (Sale of cars is given in thousand):
6. What is the average number of cars sold by company ‘F’ during this whole period. (Answer in thousands)
1) 22.5 
2) 23 
3) 23.5 
4) 24 
5) 24.5
7. Average number of cars sold by all companies in year 2012 is approximately what percentage of total number of cars sold by company A in year 2013?
1) 96% 
2) 92% 
3) 88% 
4) 84% 
5) 76%
8. Total number of cars sold by all companies in year 2008 is approximately what percentage of total number of cars sold by company C in this whole period ?
1) 81% 
2) 85% 
3) 87% 
4) 91% 
5) 94%
9. What is the difference between average number of cars sold in year 2013 and average number of cars sold in 2010 ?
1) 2.5 thousand 
2) 3 thousand 
3) 3.5 thousand 
4) 4 thousand 
5) 4.5 thousand
10. Total number of cars sold by company D in all six years is approximately what percentage of total number of cars sold in year 2009 and 2011 together ?
1) 36% 
2) 42% 
3) 48% 
4) 51% 
5) 54%
Answers:
1. 3
2. 1
3. 5
4. 2
5. 1
6. 4
Avg = (25+19+27+22+30+21)/6 = 24 thousand
7. 3
Avg(2012) = (16+23+27+19+17+30)/6 = 132/6 = 22 thousand
Reqd% = 22/25*100 = 88%
8. 2
Total(2008) = 199 thousand 
C(Total) = 140 thousand 
Req% = 119/140 * 100 = 85%
9. 4
Avg(2013) = 141/6 = 23.5 thousand 
Avg(2010)117/6 = 19.5 thousand 
Difference = 4 thousand 
100. 3
D(Total) = 119 thousand 
Total(2009+2011) 119 + 129 = 248 thousand 
Req% = (119*100)/248 = 48%

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