IBPS PO MAINS 2016 - Quant Quiz-3 (Quadratic Equations)

1. 15x² – 14x – 32 = 0
    6y² – y – 5 = 0

A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

2. 2x² + 11x + 14 = 0
    4y² + 12y + 9 = 0

A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

3. 5x² – 18x + 9 = 0
    3y² + 5y – 2 = 0

A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

4. x² – 30x + 104 = 0
    y² – 53y + 702 = 0

A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

5. x² – 82x – 255 = 0
    y² – 29y + 208 = 0

A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established

11. A certain number of tennis balls were purchased for Rs 450. Five more balls could have been purchased for the same amount if each ball was cheaper by Rs 15. Find the number of balls purchased.

A. 10

B. 16

C. 25

D. 18

E.. 45

12. The sum of the squares of two numbers is 170. The difference between thrice the smaller number and the larger number is 10. Find the numbers.   

A. 7 and 12

B. 7 and 10

C. 8 and 11

D.  7 and 11

E. None of these

13. The length of a rectangular filed is seven times its breadth. If the area of the filed be 252 sq.m, find its length.   

A. 40m

B. 45m

C. 42m

D. 43m

E. 46m

14. A packet food costs Rs. 360. If the quantity was 6 kg more and per kg costs Rs 3 less, the total cost would remain unchanged. How much the packet contains?  

A. 21

B. 31

C. 29

D. 25

E. 24

15. A cyclist covers a distance of 24 km in a certain time with a certain fixed uniform speed. If he increases his speed by 2 km/hour, he takes 2 hr less to cover the same distance. Find his original speed.  

A. 8

B. 5.5

C. 4

D. 5

E. 6

                                                                                ANSWERS

1. E

    15x² – 14x – 32 = 0
    x = 2, -1.06
    

    6y² – y – 5 = 0
    y = 1, -0.8

2. B

    2x² + 11x + 14 = 0
    x = -3.5, -2
    

    4y² + 12y + 9 = 0
    y = -1.5, -1.5

3. A

    5x² – 18x + 9 = 0
    x = 0.6, 3
    

    3y² + 5y – 2 = 0
    y = 0.33, -2

4. D

    x² – 30x + 104 = 0
    x = 26, 4

    y² – 53y + 702 = 0
    y = 26, 27

5. E

    x² – 82x – 255 = 0
    x = 85, -3

    y² – 29y + 208 = 0
    y = 13, 16

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