Math Problem Statement

For two positive numbers x and y, the A.M. between x and y is 10, the G.M. between x and y is 6, and x > y, then x - y = ?

Solution

Let's go through the questions from the image step by step.

Question 66:

For two positive numbers xx and yy, the Arithmetic Mean (A.M.) between xx and yy is 10, the Geometric Mean (G.M.) between xx and yy is 6, and x>yx > y. Then xy=?x - y = ?

The A.M. is given by:

A.M.=x+y2=10A.M. = \frac{x + y}{2} = 10

Solving for x+yx + y:

x+y=20x + y = 20

The G.M. is given by:

G.M.=xy=6G.M. = \sqrt{x \cdot y} = 6

Squaring both sides:

xy=36x \cdot y = 36

Now, we have two equations:

  1. x+y=20x + y = 20
  2. xy=36x \cdot y = 36

Using these, we can solve for xx and yy. Let’s solve this system.

From x+y=20x + y = 20, we express yy as:

y=20xy = 20 - x

Substitute this into the second equation:

x(20x)=36x(20 - x) = 36

20xx2=3620x - x^2 = 36

Rearrange into a quadratic equation:

x220x+36=0x^2 - 20x + 36 = 0

Now we solve this quadratic equation using the quadratic formula:

x=(20)±(20)24(1)(36)2(1)x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(36)}}{2(1)}

x=20±4001442x = \frac{20 \pm \sqrt{400 - 144}}{2}

x=20±2562x = \frac{20 \pm \sqrt{256}}{2}

x=20±162x = \frac{20 \pm 16}{2}

So, x=18x = 18 or x=2x = 2.

Since x>yx > y, we take x=18x = 18 and y=2y = 2.

Thus:

xy=182=16x - y = 18 - 2 = 16

Answer: D (16).


Question 67:

For two numbers aa and bb, the Arithmetic Mean (A.M.) is 3, and the Geometric Mean (G.M.) is 222\sqrt{2}. Then a2+b2=?a^2 + b^2 = ?

The A.M. is given by:

A.M.=a+b2=3A.M. = \frac{a + b}{2} = 3

So:

a+b=6a + b = 6

The G.M. is given by:

G.M.=ab=22G.M. = \sqrt{a \cdot b} = 2\sqrt{2}

Squaring both sides:

ab=8a \cdot b = 8

Now, we can use the identity:

a2+b2=(a+b)22aba^2 + b^2 = (a + b)^2 - 2ab

Substituting values:

a2+b2=622(8)=3616=20a^2 + b^2 = 6^2 - 2(8) = 36 - 16 = 20

Answer: B (20).


Question 68:

The last A.M. between -3 and 17 is 13. Then the number of A.M.s is ?

The formula to find the number of Arithmetic Means between two numbers aa and bb is:

A.M.s=bacommon difference1\text{A.M.s} = \frac{b - a}{\text{common difference}} - 1

Here, a=3a = -3, b=17b = 17, and the last A.M. is 13. The common difference dd can be calculated by considering the sequence.

The A.M.s between -3 and 17 will form an arithmetic progression:

-3, A.M. 1, A.M. 2, ..., A.M. nn, 17.

Let's calculate this.


Would you like me to proceed with the next questions in a similar way, or would you like details on these steps so far?

Here are some related questions to consider:

  1. What is the relationship between A.M., G.M., and H.M. for two numbers?
  2. How do you solve quadratic equations like the one in question 66?
  3. Can you derive the formula for a2+b2a^2 + b^2 using the identity used in question 67?
  4. What is the geometric interpretation of A.M. and G.M.?
  5. How do you find the number of terms in an arithmetic progression?

Tip: When solving quadratic equations, always check if the discriminant (inside the square root) is positive, negative, or zero to predict the number of real solutions.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Mean
Geometric Mean
Quadratic Equations

Formulas

Arithmetic Mean: A.M. = (x + y) / 2
Geometric Mean: G.M. = √(x * y)
Quadratic Equation: ax^2 + bx + c = 0
Identity: a^2 + b^2 = (a + b)^2 - 2ab

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-12