The sum of 2 positive numbers is 151

The sum of 2 positive numbers is 151. The lesser number is 19 more than the square root of the greater number. What is the value of the greater number, minus the lesser number?
F. 19
G. 66
H. 85

J. 91
Κ. 121

The correct answer and explanation is :

We are tasked with finding the difference between two positive numbers where:

• The sum of the two numbers is 151.
• The lesser number is 19 more than the square root of the greater number.

Let’s define the two numbers:

• Let $x$ represent the greater number.
• Let $y$ represent the lesser number.

From the problem, we have two key pieces of information:

1. The sum of the two numbers is 151, so we can write the equation: $$
   x + y = 151
   $$
2. The lesser number $y$ is 19 more than the square root of the greater number $x$, which gives us the equation: $$
   y = \sqrt{x} + 19
   $$

Step 1: Substitute the second equation into the first

We already know that $x + y = 151$. Substituting $y = \sqrt{x} + 19$ into this equation, we get:

$$
x + (\sqrt{x} + 19) = 151
$$

Simplifying the equation:

$$
x + \sqrt{x} + 19 = 151
$$

$$
x + \sqrt{x} = 132
$$

Step 2: Solve the equation

Let’s solve for $x$. To eliminate the square root, we’ll first make a substitution. Let $z = \sqrt{x}$. Therefore, $x = z^2$, and the equation becomes:

$$
z^2 + z = 132
$$

Rearranging the equation:

$$
z^2 + z – 132 = 0
$$

This is a quadratic equation, and we can solve it using the quadratic formula:

$$
z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
$$

where $a = 1$, $b = 1$, and $c = -132$. Substituting these values into the quadratic formula:

$$
z = \frac{-1 \pm \sqrt{1^2 – 4(1)(-132)}}{2(1)}
$$

$$
z = \frac{-1 \pm \sqrt{1 + 528}}{2}
$$

$$
z = \frac{-1 \pm \sqrt{529}}{2}
$$

$$
z = \frac{-1 \pm 23}{2}
$$

Thus, we have two possible solutions for $z$:

$$
z = \frac{-1 + 23}{2} = \frac{22}{2} = 11
$$

or

$$
z = \frac{-1 – 23}{2} = \frac{-24}{2} = -12
$$

Since $z = \sqrt{x}$ and square roots are non-negative, we discard $z = -12$ and keep $z = 11$.

Step 3: Find $x$ and $y$

Now that we know $z = 11$, we can find $x$:

$$
x = z^2 = 11^2 = 121
$$

Substitute $x = 121$ into the equation for $y$:

$$
y = \sqrt{x} + 19 = 11 + 19 = 30
$$

Step 4: Find the difference between $x$ and $y$

We are asked to find the value of the greater number minus the lesser number:

$$
x – y = 121 – 30 = 91
$$

Thus, the correct answer is J. 91.