I. x2 – x – 42 = 0, II. y2 – 17y + 72 = 0 to solve both the equations to find the values of x and y?

I. x2 – x – 42 = 0,
II. y2 – 17y + 72 = 0 to solve both the equations to find the values of x and y?

A. If x < y
B. If x > y
C. If x ≤ y
D. If x ≥ y
E. If x = y or the relationship between x and y cannot be established.
Explanation:
I. x2 – 7x + 6x – 42 = 0
=> (x – 7)(x + 6) = 0 => x = 7, -6
II. y2 – 8y – 9y + 72 = 0
=> (y – 8)(y – 9) = 0 => y = 8, 9
=> x < y

I. x2 + 5x + 6 = 0, II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?

I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?

A. If x < y
B. If x > y
C. If x ≤ y
D. If x = y or the relationship between x and y cannot be established.
Explanation:
I. x2 + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.

(i). a2 – 9a + 20 = 0, (ii). 2b2 – 5b – 12 = 0 to solve both the equations to find the values of a and b?

(i). a2 – 9a + 20 = 0,
(ii). 2b2 – 5b – 12 = 0 to solve both the equations to find the values of a and b?

A. If a < b
B. If a ≤ b
C. If the relationship between a and b cannot be established
D. If a ≥ b
Explanation:
I. (a – 5)(a – 4) = 0
=> a = 5, 4
II. (2b + 3)(b – 4) = 0
=> b = 4, -3/2 => a ≥ b

(i). a2 – 7a + 12 = 0, (ii). b2 – 3b + 2 = 0 to solve both the equations to find the values of a and b?

(i). a2 – 7a + 12 = 0,
(ii). b2 – 3b + 2 = 0 to solve both the equations to find the values of a and b?

A. if a < b
B. if a ≤ b
C. if the relationship between a and b cannot be established.
D. if a > b
Explanation:
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b

A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?

A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?

A. 10
B. 8
C. 15
D. 7.50
Explanation:
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10

Find the quadratic equations whose roots are the reciprocals of the roots of 2×2 + 5x + 3 = 0?

Find the quadratic equations whose roots are the reciprocals of the roots of 2×2 + 5x + 3 = 0?

A. 3×2 + 5x – 2 = 0
B. 3×2 + 5x + 2 = 0
C. 3×2 – 5x + 2 = 0
D. 3×2 – 5x – 2 = 0

The quadratic equation whose roots are reciprocal of 2×2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3×2 + 5x + 2 = 0