How to Solve Word Problems by Working Backwards Part 3

In part 1 and part 2 of this series, we have learned how to solve number age problems by working backward. In this post, we are going to learn how to solve backward using inverse operations. Recall that multiplication and division are inverse operations and addition and subtraction are inverse operations.

Example 5

A number is multiplied by 4 and then, 3 is added to the product. The result is 31. What is the number?

Solution

The key phrases in this problem are (1) multiplied by 4 and (2) added to (3) the result is 31. Since we are working backward, we start with 31, and then find the inverse of “added to 3” which is “subtract 3.” So, 31 – 3 = 28.

Next, we find the inverse of “multiplied by 4,” which is “divided by 4.” So, 28/4 = 7.

So, the answer to this problem is 7.

Check: 7(4) + 3 = 31

Example 6

Think of a number. Divide it by 8. Then subtract 4 from the quotient. The result is 5. What is the number?

Solution

The key phrases in this problem are (1) divided by 8 (2) subtract 4 and (3) the result is (3) the result is 5.

We start with the result which is 5 and find the inverse of “subtract 4” which is “add 4.” So, 5 + 4 = 9. Next, we find the inverse of “divide by 8” which is “multiply by 8.” So, 9(8) = 72.

So, the correct answer is 72.

Check: 72/8 – 4 = 9 – 4 = 5.

In the next post, we will discuss more about solving math word problems by working backward.

How to Solve Problems by Working Backwards Part 2

In the previous post, we have learned how to solve number problems by working backward.  In this post, we discuss age problems. We already had 2 examples in the previous post in this series, so we start with the third example.

Example 3

Arvin is 5 years older than Michael.  The sum of their ages is 37.  What are their ages?

Solution

This is very similar to the problems in the previous post in this series.  Arvin is 5 years older than Michael, so if we subtract 5 from Arvin’s age, their ages will be equal.  But if we subtract 5 from Arvin’s age, we also have to subtract 5 from the sum of their ages. That is,

37 – 5 = 32.

Now, their ages are equal, so we can divide the sum by 2.  That is 32 ÷ 2 = 16.  This means that the younger person is 16 since we subtracted 5 from Arvin’s age. Therefore, Michael is 16 and Arvin is 16 + 5 = 21.

Check

16 + 21 = 37.

Example 4

Mia is 3 years older than Pia.  In 4 years, the sum of their ages is 35. What are their ages?

Solution

There are two of them and we added 4 to both ages, so subtracting 8 from 35 (the sum) will determine the sum of their present age.  That is 35 – 8 = 27 is the sum of their present age.

Next, Mia is 3 years older than Pia, so if we subtract 3 from the sum of their present age, their ages will be equal. So, 27 – 3 = 24.

We can now divide the sum of their ages by 2.  That is 24/2 = 12.

This means that 12 is the age of the younger person because we subtracted 3 from the age of the older person.

So, Pia is 12 and Mia is 15.

Check 12 + 15 = 22.

In the next post, we will discuss more problems that can be solved by working backward.

How to Solve Word Problems by Working Backwards Part 1

Most of us would always take a pen and solve for x if we see word problems. But did you know that you can solve them by working backward or even mentally? In this post, I am going to teach you some techniques on solving problems by working backward.

Example 1: One number is three more than the other. Their sum is 45. What are the numbers?

Solution

In the given, one number is 3 more than the other. This means that if you subtract 3 from the larger number they will be equal. Note that if we subtract 3 from one of the numbers, then we should also subtract 3 from their sum. Therefore, their sum will be 45 – 3 = 42. Since the numbers are equal, we now divide the sum by 2. That is, 42/2 = 21.

So, the smaller number is 21 and the larger is 21 + 3 = 24.

Check: 21 + 24 = 45

Example 2: One number is 5 less than the other. Their sum is 43. What is the smaller number?

Solution

This is very similar to Example 1. Here, one number is 5 less than the other; so, if we add 5 to the smaller number, they will be equal. If we add 5 to the smaller number, we should also add 5 to their sum. Therefore, their sum will be 43 + 5 = 48. Since the two numbers are equal, we can divide the sum by 2. That is 48/2 = 24. Since we added 5, it means that 24 is the larger number. So, the smaller number is 24 – 5 = 19.

Check: 19 + 24 = 43

In the next post, we are going to discuss more examples.

How to Solve Quadratic Word Problems Part 3

In the previous two posts (Part 1 and Part 2), we have discussed two problems involving quadratic equations. The first one is

A rectangular flower garden with dimensions 3m by 7m is surrounded by a walk of uniform width. If the area of the walk is 11 square meters, what is the width of the walk in meters?

Solution

If we let $x$ be the width of the walk, then the length of the walk becomes $x + 7 + x = 7 + 2x$ and the width becomes $x + 3 + x = 3 + 2x$ as shown in the figure below.

We know that the area of the garden (the inner rectangle) is equal $7 \times 3 = 21$ square meters. The area of the larger rectangle is equal to

$(3 + 2x)(7 + 2x) = 21 + 14x + 6x + 4x^2 = 21 + 20x + 4x^2$.

Now, to get the area of the walk, we have to subtract the area of the garden from the area of the large rectangle. That is,

area of the larger rectangle – area of the smaller rectangle = 11

$4x^2 + 20x + 21 - 21 = 11$

$4x^2 + 20x = 11$

Subtracting 11 from both sides, we have

$4x^2 + 20x -11 = 0$.

Now, the quadratic equation is not factorable, so we use the quadratic formula in order to find the value of x. In using the quadratic formula, we want to identify the values of a, b, and c in the equation $ax^2 + bx + c = 0$ and substitute in the quadratic formula

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

From the equation above, $a = 4$, $b = 20$ and $c = -11$. Substituting these values in the quadratic formula, we have

$x = \dfrac{-20 \pm \sqrt{(-20)^2 - 4(4)(-11)}}{2(4)}$

$= \dfrac{-20 \pm \sqrt{400 + 176}}{8}$

$= \dfrac{20 \pm 24}{8}$

That means that we have two roots

$x_1 = \dfrac{-20 + 24}{8} = \dfrac{1}{2}$
and

$x_2 = \dfrac{-20 - 24}{8} = \dfrac{-44}{8} = \dfrac{-11}{2}$.

As we can see, we $x_2$ is not a possible root because it is negative. Therefore, the acceptable answer to the problem is

$x_1 = \dfrac{1}{2}$.

This means that the width of the walk is ½ meters.

Check: If the width of the walk is ½ meters, then the length of the larger rectangle is $1/2 + 7 + 1/2 = 8$ and the width is $1/2 + 3 + 1/2 = 4$. The area is $(8)(4) = 32$ sq. m.

Now the area of the walk is $21$ meters and $32 - 21 = 11$ sq. m. which is the area of the walk indicated in the problem. Therefore, we are correct.

You can also watch the Youtube video below regarding the discussion above.

Enjoy!

How to Solve Quadratic Problems Part 2

In the previous post, we have used quadratic equations to solve a word problem involving consecutive numbers. In this post, we discuss more quadratic problems. This is the second problem in the series.

Problem 2

Miel is 12 years older than Nina. The product of their ages is 540.

Solution

Let x = age of Nina
x + 12 = age of Miel

The product of their ages is 540, so we can multiply the expressions above and equate the product to 540. That is,

x(x + 12) = 540.

Multiplying the expressions, we have

$x^2 + 12x = 540$.

Subtracting 540 from both sides, we obtain

$x^2 + 12x - 540 = 0$.

We want to find two numbers whose product is -540 and whose sum is 12. Those numbers are -18 and 30.

This means that the factors are

(x – 18)(x + 30) = 0.

Equating each expression to 0, we have

x – 18 = 0, x = 18
x + 30 = 0, x = – 30.

Since we are talking about age, we take the positive answer x = 18.

This means that Nina is 18 years old. Therefore, Miel is 18 + 12 = 30 years old.

Area of Circles: Worked Examples

In the previous posts, we have solved problems on how to calculate the area of square and rectangle. We continue this series by solving problems involving area of circles.

The formula for the area of a circle is

A = pi r2

Where A is the Area, r is the radius, and pi is the irrational number which is approximately equal to 3.1416. In solving area of circles, the approximate value of pi is usually given.

Problem 1

Find the area of a circle whose radius is 12 cm. Use pi = 3.14.

Solution

Substituting to the values we have

A = pi r2
A = (3.14)(122)
A = (3.14)(144)
A = 452.16.

So, the area of the circle is equal to 452.16 sq.cm.

Problem 2

The area of a circular garden is 50.24 sq. m. Find its diameter. Use pi = 3.14.

Solution

In this problem, area is given. We are looking for the diameter which is twice the radius.

Substituting the values to the formula, we have

A = \pi r2

50.24 = (3.14) ( r2)

Dividing both sides by 3.14, we have
(50.24)/(3.14)= (3.14)( r2)/(3.14)

16 = r2

Getting the square root of both sides, we have

4 = r

Therefore, the radius of the circle is 4cm. Since the diameter of a circle is twice its radius, the answer is (4 cm)(2) = 8 cm.

Problem 3

The diameter of a circle is 15 cm. Find its area. Use pi = 3.14 and round to the nearest tenths.

Solution

We are given the diameter, but we need the radius which is half the diameter. Therefore, the radius is 7.5 cm.

Substituting to the formula, we have

A = pi r2
A = (3.14)(7.52)
A = (3.14)(56.25)
A = 176.625

Rounding to the nearest tenths, we have 176.6.

Therefore, the area of the circle is equal to 176.6 sq. cm.

Area of Rectangle: Worked Examples

In the previous post, we have learned some problems on how to solve area of squares. In this post, we are going to solve problems involving area of rectangles. The area of rectangle is one of the most common questions in the Geometry part of the Civil Service Examination. In this post, we are going to learn how to solve problems involving area of rectangles. The formula for the Area (A) can be calculated by multiplying the length (l) and the width (w). That is,

$A = l \times w$.

Problem 1

Find the area of a rectangle with length 8cm and width 5 cm.

Solution

$A = l \times w$

$A = 8 \times 5$

$A = 40$

So, the area of the rectangle is $40 cm^2$

Problem 2

The area of a rectangle is $12 cm^2$ and its width is $3$ cm. What is its length?

Solution

$A = l \times w$

Substituting the given, we have

$12 = l \times 3$.

To get $l$, we have to divide both sides by 3.

$\dfrac{12}{3} = \dfrac{l \times 3}{3}$

$4 = l$

Therefore, the length is equal to 3.

Problem 3

The length of a rectangle is twice its width. Its area is 32 square units. What are the dimensions of the rectangle?

Solution

Let $x$ = width of the rectangle

Since the length is twice, it is $2x$.

So,

$w = x$
$l = 2x$

$A = l \times w$

$A = 2w \times w$

$32 = 2w^2$

Dividing both sides by 2, we have

$16 = w^2$.

Getting the square root of both sides, we have

$4 = w$

So, $l = 2w = 2(4) = 8$.

So, the dimension of the rectangle is 4 by 8.

Problem 4

The length of a rectangle is two more than its width. Its area is 48 square units. What are the dimensions of the rectangle?

Solution

Let length be equals x and width be equals x + 2. That is

$w = x$
$l = x + 2$

$A = lw$
$48 = x(x + 2)$
$48 = x^2 + 2x$

Subtracting 48 from both sides, we have
$x^2 + 2x - 48 = 0$.

Factoring, we have

$(x + 8)(x - 6) = 0$

$x = -8, x = 6$

So, the width is 6 units and the length is 6 + 2 = 8 units.

Check:
Is the length two more than $6$?
Is $6 \times 8$ equals $48$?

Area of a Square: Worked Examples

One of the word problems involved in Geometry in the Civil Service Examinations is about area. In this series of posts, we are going to discuss sample problems which are typically used in the exams. We start with the easiest problems where we substitute the given to the formulas and then discuss more complicated problems at the end. In this post, we will discuss how to solve problems involving area of a square.

The formula area of a square (A) is the product of the length the sides (s).

$A = s \times s$ or $A = s^2$.

Example 1

The side length of a square is 6 units. What is its area?

Solution

$A = s \times s$

$A = 6 \times 6$

$A = 36$

Therefore, the area of a square is $36$ square units.

Example 2

The area of a square is 144 square units. What is the side length?

Solution

$A = s \times s$

Substituting the given, we have

$144 = s \times s$.

Since $s \times s = s^2$, to get the value of $s$, we have to get the square root of both sides.

$\sqrt{144} = \sqrt{s^2}$

$12 = s$

Therefore, the side length of the square is $12$ units.

Example 3

If the side length of a square is increased by 4, its area increases 80. What is the side length of the square?

Solution

Let $x$ be the side length of the original square. When it is increased by $4$, then its new length is $x + 4$. With this information, we can get the following:

Original area increased by 80: $x^2 + 80$
Area of the new square: $(x + 4)^2$

Now, the original area above increased by 80 is equal to the area of the new square, so we can equate the two expressions above. That is,

$(x + 4)^2 = x^2 + 80$
$x^2 + 8x + 16 = x^2 + 80$.

Subtracting $x^2$ from both sides, we have

$8x + 16 = 80$
$8x = 64$
$x = 8$.

Therefore, the side length of the square is 8 units.

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