Test: Linear Equations In Two Variables - 1

In the graph of y = x², the point (0, 0) is called the vertex. The vertex is the minimum point in a parabola that opens upward. In a parabola that opens downward, the vertex is the maximum point. We can graph a parabola with a different vertex.

Solution : 

The correct option is Option B.

n² + p² = (q + n)²

n² + p² = q² - 2qn + n²

Eliminating n² from both sides,

p² = q² - 2qn

n = q² - p² / 2q

let the unit place digit be x and tens place digit be y

then the two-digit number will be  10y + x

and the number formed by interchanging the unit place and tens place digits will be 10x + y

according to the first condition given in the qs i.e, the sum of two numbers is 110 that  is

10y + x + 10x + y = 110

=> 11x + 11y = 110

divide the above equation by 11 we get

x + y = 10

x = 10 - y ....(i)

now according to the second equation,

if 10 is subtracted from the first number i.e, the new number is  10y + x - 10

given that the new number is 4 more than 5 time the sum of its digits in the first number i.e

the sum of its digits in the first number is  x + y,  now 5 times of its, 5(x + y), and now 4 more that is, 4 + 5(x + y)

therefore new number = 4 + 5(x + y)

10y + x - 10 = 4 +5(x + y)

10y - 5y + x = 4 +10 +5x

5y = 14 + 4x.....(ii)

substitute the value of x from eq(i) to eq (ii)

we  get , 5y = 14 + 4(10 - y)

5y = 14 + 40 - 4y

y = 6

and from eq(i)

x = 4

then the first number 10y + x = 10x6 + 4 = 64

first number is 64.