Sample High School Math Problems with Answers

Want some sample high school math problems with answers? Well then here you go!

High School Math Problems

Problem 1:     A change purse has 100 nickels and dimes. The total value of the coins is $7. How many coins of each type does the purse contain?

If the number of nickels is N and the number of dimes is D, then

5N + 10D = 700

(the 5, 10 and 700 representing the number of cents)

However,

N + D = 100

(the number of nickels plus the number of dimes equals 100)

So, solving for N for both equations, we get as the result

N = – 2D + 140

and

N = 100 – D

Thus,

– 2D + 140 = 100 – D

D = 40

Since the number of dimes equals 40, then the number of nickels must equal 60. Does this prove to be true?

60 (5) + 40 (10) = 300 plus 400 = 700 cents. Correct Answer

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High School Math Problem 2:     How Do You Multiply Fractions?

Example Problem: 1/6 x 3-3/4 [one-sixth times three and three-quarters]

Perhaps the easiest way to multiply fractions is to first convert the numbers entirely into their fraction equivalents. Thus,

1/6 x 3-3/4 becomes 1/6 times 3 converted to fourths (12/4) added to the 3 fourths (for a total of 15/4). Then rewrite the problem as 1/6 x 15/4 = What?

Multiply the two top numbers and the two bottom numbers. 1 x 15 = 15, and 6 x 4 = 24.

Then put the resultant top number on top of the resultant bottom number. That is:

15/24

Now both the top and bottom numbers can be divided by 3. So doing that to the 15/24 gives us 5/8.

1/6 x 3-3/4 = 5/8  Correct Answer

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High School Math Problem 3:     Solve by Factoring: x² – 6x – 27 = 0

Look at the coefficient in front of the x (6) and the pure number (– 27). Clearly a factor of 3 is involved. 27 divides by 9, and 6 = 9 – 3. So making an intelligent first guess,

(x + 3) (x – 9) = 0 Answer

Let’s perform a check to see if this is correct. Multiply the first term in the left expression by the first term in the second expression. That gives x times x = x². Now multiply the first term in the left expression by the second term in the right expression. That gives x times – 9 = – 9x. Next, multiply the second term in the left expression times the first term in the right expression. That gives 3 times x = 3x. Finally, multiply the second term in the left expression times the second term in the right expression. That gives 3 times – 9 = – 27.

Putting these all together gives:

x² – 9x + 3x – 27

That is the same as,

x² – 6x – 27  Correct Answer

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High School Math Problem 4:     The sum of two numbers is 400. If the first number is decreased by 20% and the second number is decreased by 15%, then the sum is 68 less. Find the new numbers (the numbers after decreasing).

The two original numbers (we’ll call them x and y) equal 400 total. Thus,

x + y = 400 [Equation 1]

But if we reduce x by 20 percent and y by 15 percent, and the sum is 68 less than the 400 (or 332), we write,

0.8x + 0.85y = 332

We want to solve these two equations for x. So we multiply the second equation by 1.25, which gives,

x + 1.0625y = 415 [Equation 2]

Subtracting the first very first equation from the second, gives us,

0.0625y = 15

This tells us the first value of y was 240. That means the first x was 160.

The new values are reduced by the cited percentages, and we wind up with,

New Y = 204 and New X = 128 Correct Answer

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High School Math Problem 5:    Solve Two Equations for x and y

Use the following two equations to solve for the variables x and y.

2x + 2y = 6 [Equation 1]

– 3x + 5y = – 33 [Equation 2]

In both instances, we will attempt to “solve” for y. That means we will put y on the left of both equations and move everything else to the right.

Signs change (plus to minus and minus to plus) when you cross from one side of the equals symbol to the other.

Equation 1 becomes:

2y = 6 – 2x

This simplifies (by dividing everything by 2) to

y = 3 – x (New Equation 1)

Equation 2 becomes

5y = – 33 + 3x

Dividing by 5 gives,

y = – 33/5 + (3/5)x (New Equation 2)

Now we can combine or solve the two equations. Since both say y equals something, we can join the equations. If a = b and b = c, then a = c. That’s the principal.

So,

3 – x = – 33/5 + (3/5)x

This becomes (again, by moving across the equals sign)

– x – (3/5)x = – 33/5 – 15/5

(– 8/5)x = – 48/5

– 8x = – 48

x = 6  Correct Answer

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High School Math Problem 6:     Find the slope intercept form of the line passing through the point (– 1, 5) and parallel to the line – 6x – 7y = – 3.

The line given is rewritten (in slope-intercept form, y = mx + b) as

y = – 6/7x + 3/7

Thus the slope is,

m = – 6/7

Now a line is parallel to another line if its slope is the same as the other line. So,

y = – 6/7 x + b is the formula for the new line, with the intercept being b because we have not yet solved for it. We now do so, inserting the value of x and the value of y from the point. We get,

5 = – 6/7 (– 1) + b

b =  29/7

The parallel line is represented by the equation:

y = – 6/7x + 29/7  Correct Answer

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High School Math Problem 7:    Write an Equation of the Line Parallel to the Line 3x ‒ 2y = 8

The general equation for a line is usually written,

y = mx + b

The variable m refers to the tilt or “slope” of the line. The letter b refers to where the line crosses the y-coordinate axis.

To make the problem intelligible, the equation 3x ‒ 2y = 8 should be rewritten in the slope-intercept form. Keeping y on the left and moving x to the right, we obtain,

‒ 2y = ‒ 3x + 8

Multiplying by ‒ 1, we get

2y = 3x ‒ 8

Dividing by 2 yields

y = 3/2 x ‒ 4

We will call this Equation 1. We hold onto this and engage in the same process for the second equation.

2y + 3x = ‒ 4

Moving x to the right, we get

2y = ‒ 3x ‒ 4

Then dividing by two, we get,

y = ‒ 3/2 x ‒ 2

We will call this Equation 2.

An equation is parallel if the tilt or slope, the m value, exactly matches for both equations.

This high school math problem calls for us to use the intercept (b value, Equation 2), which is ‒ 2. It wants us to use the slope or m value of Equation 1, which is 3/2.

Our parallel line equation becomes,

y = 3/2x ‒ 2  Correct Answer

Now admit it. High school math problems can be both fun and entertaining.

Note: You might also enjoy More High School Math

References:

• MIT High School Mathematics

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