Simple Algebraic Expressions: Word Problems and Step-by-Step Solutions

Suppose $x$ and $y$ respectively represent the price of a pencil and the price of a book. Altogether, the student bought $3$ pencils and $5 + 2(5) = 15$ books. The total cost of the pencils is $3x,$ while the total cost of the books is $15y.$ Therefore, the total expenditure is $3x + 15y.$
(Answer C)

The length of the line segment in option A is equal to $x + 5,$ while the length of the line segment in option B is equal to $2 + 3 + x.$ Furthermore, the area of the figure in option C is equal to $2x + 3x,$ while the area of the figure in option D is equal to $5x + 5x.$
Therefore, the figure that can represent the algebraic expression $2x + 3x$ is the area of the figure in option C.
(Answer C)

Since Monica sent $m$ emails every hour to her clients for $6$ hours, the total number of emails she sent is
$$\underbrace{m + m + \cdots + m}_{6~\text{times}} \\ = 6m.$$Similarly, since Teresa sent $n$ emails every hour to her clients for $5$ hours, the total number of emails she sent is
$$\underbrace{n+ n + \cdots + n}_{5~\text{times}} \\ = 5n.$$Therefore, altogether, a total of $6m + 5n$ emails were sent by them on that day.
(Answer C)

The perimeter ($k$) of a plane figure is obtained by adding all of its side lengths. Therefore, we can write the perimeter of the triangle as follows.
$$\begin{aligned} k & = (2x + 1) + (3x-5) + (x+3) \\ & = (2x+3x+x) + (1+(-5) + 3) \\ & = 6x-1 \end{aligned}$$Therefore, the perimeter of the triangle is $\boxed{(6x-1)~\text{m}}.$
(Answer B)

Notice that in the equation $P = 120-24d,$ $P$ represents the number of remaining mobile phones, while $d$ represents the number of days William has worked during that week. As time passes, the value of $d$ increases, while the value of $P$ continues to decrease with a minimum value of $0.$ From the equation, it can be seen that William repairs $24$ mobile phones every day, so this expression is represented by $24d.$ Based on this fact, it can be concluded that the value $120$ in the equation represents the number of mobile phones William needs to repair each week. In other words, William starts his work with $120$ mobile phones to repair every week.
(Answer B)

Notice that in the equation $t = 8u + 73,$ $t$ represents the height of boys (in cm), while $u$ represents their age (in years). As their age increases, their height also increases by $8$ cm. For illustration, if a boy reaches the age of $3$ years, then $u = 3$ so that $t = 8(3) + 73 = 97$ cm. Therefore, his height is $97$ cm. Meanwhile, the value $73$ in the equation represents the initial height. Thus, the estimated increase in the height of boys each year within the age range of $2$ to $5$ years based on the model is $\boxed{8}$ cm.
(Answer A)

The correct algebraic expression to determine Lili’s current age assuming that she had been born $x+7$ years earlier is
$$(19+x)+(x+7) = 2x+26$$years. Therefore, her age three years from now would be $(2x+26)+3=29+2x$ years.
(Answer D)

Since Mono has $n$ cards and it is known that Mono has $15$ more cards than Fifi, it can be said that Fifi has $n-15$ cards. Furthermore, it is known that Rizal has $3$ times as many cards as Mono, so the number of cards owned by Rizal is $3n.$
Therefore, the total number of cards is $\boxed{n + (n-15) + 3n = 5n-15}.$
(Answer C)

The area of a square ($L$) is obtained by multiplying its side lengths together (or squaring its side length).
$$\begin{aligned} L & = s \cdot s \\ & = (3x-2)(3x-2) \\ & = 3x(3x-2)-2(3x-2) \\ & = 9x^2-6x-6x+4 \\ & = 9x^2-12x+4 \end{aligned}$$Therefore, the area of the square is $\boxed{(9x^2-12x+4)~\text{cm}^2}.$
(Answer D)

The area of a rectangle ($L$) is obtained by multiplying its length and width.
$$\begin{aligned} L & = p \cdot \ell \\ & = (2x-1)(x+3) \\ & = 2x(x+3)-(x+3) \\ & = 2x^2 + 6x-x-3 \\ & = 2x^2+5x-3 \end{aligned}$$Therefore, the area of the rectangle is $\boxed{(2x^2+5x-3)~\text{m}^2}$
(Answer D)

It is known that the width of the box = $y$ cm. The length of the box is three times the width minus $3$ cm, which is $3y-3$ cm. The height of the box is four times the width minus $5$ cm, which is $4y-5$ cm. Therefore, the volume of the box is
$$\begin{aligned} V & = \text{length} \times \text{width} \times \text{height} \\ & = (3y-3)(y)(4y-5) \\ & = (12y^3-27y^2+15y)~\text{cm}^3. \end{aligned}$$(Answer B)

From the figure, it is known that the length of the line segment (the rope length) is equal to $(3x+10)$ cm so that the value of $x$ can be determined as follows.
$$\begin{aligned} 3x+10 & = 43 \\ 3x & = 33 \\ x & = 11 \end{aligned}$$The line segment $CB$ is equal to the total length of the line segment minus the length of the line segment $AC.$
$$\begin{aligned} CB & = AB-AC \\ & = (3x+10)-(2x-5) \\ & = (3x-2x) + (10-(-5)) \\ & = x + 15 \end{aligned}$$Since $x = 11,$ we obtain $CB = 11+15 = 26~\text{cm}.$ Therefore, the length of the rope segment from point $C$ to point $B$ is $\boxed{26~\text{cm}}.$
(Answer C)

Suppose $x$ is the smallest integer among the six consecutive even integers. Thus, the following integers are $(x+2), (x+4),$ $(x+6), (x+8),$ and $(x+10)$ so that we obtain
$$\begin{aligned} x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10) & = c \\ (x+x+x+x+x+x) + (2+4+6+8+10) & = c \\ 6x + 30 & = c \\ 6x & = c-30 \\ x & = \dfrac{c-30}{6}. \end{aligned}$$Therefore, the smallest integer referred to is $\boxed{\dfrac{c-30}{6}}.$
(Answer D)

The total number of gold bars is $1 + 2 + 3 + \cdots + 7 = 28$ bars. Suppose there are $x$ fake gold bars, which means the remaining $(28-x)$ bars are genuine. It is known that the weight of a genuine gold bar is $111$ kg and the weight of a fake gold bar is $99$ kg, while the total weight of all sacks is $3.060$ kg. From here, we can construct an algebraic equation to determine the value of $x.$
$$\begin{aligned} 99x + 111(28-x) & = 3.060 \\ 99x + 3.108-111x & = 3.060 \\ -12x & = -48 \\ x & = 4 \end{aligned}$$Therefore, there are $4$ fake gold bars, so it can be concluded that the fake gold is located in the fourth sack.
(Answer C)

Suppose the numbers of yellow, red, blue, and green balls are respectively represented by $K, M, B,$ and $H.$ Thus, we obtain the following algebraic equations.
$$\begin{cases} M+B+H & = 22 \\ K + B + H & = 23 \\ K + M + H & = 24 \\ K + M + B & = 21 \end{cases}$$Add the four equations above according to their corresponding sides so that we get
$$\begin{aligned} 3K + 3M + 3B + 3H & = 22+23+24+21 \\ 3(K+M+B+H) & = 90 \\ K + M + B + H & = 30 \end{aligned}$$Therefore, the total number of balls in the sack is $\boxed{30}.$
(Answer B)

The equation above can be rewritten as $z = 4xy.$
Suppose $z’$ is the new value of $z$ after changes occur in the values of $x$ and $y.$
$$\begin{aligned} z’ & = 4(150\%x)(75\%y) \\ & = 4 \cdot \dfrac32x \cdot \dfrac34y \\ & = \dfrac92xy \end{aligned}$$Thus, the increase in value from $z$ to $z’$ is $\dfrac92-4 = \dfrac12.$ The percentage increase is $\dfrac{\frac12}{4} \times 100\% = 12,5\%.$
Therefore, the equation will remain satisfied if $z$ increases by $12,5\%.$
(Answer C)

Suppose $x$ represents the total number of spectators. Since $20\% = \dfrac15$ of $x$ are child spectators and $\dfrac13$ of $x$ are adult male spectators, we obtain $1-\dfrac15-\dfrac13 = \dfrac{7}{15}$ of $x$ as adult female spectators.
Since the number of adult female spectators is $200$ more than the number of adult male spectators, we obtain
$$\begin{aligned} \dfrac{7}{15}x & = \dfrac13x+200 \\ \dfrac{2}{15}x & = 200 \\ x & = 1.500. \end{aligned}$$Therefore, the total number of spectators at the art performance is $\boxed{1.500}$ people.
(Answer C)

From the given statement, the algebraic expression representing the year when Susi turns $17$ years old is
$$\begin{aligned} ((2000 + x)-5) + 17 & = 2000 + x + 12 \\ & = 2012 + x. \end{aligned}$$Therefore, Susi’s birthday will fall in the year $\boxed{2012 + x}.$
(Answer C)

Notice that the total weight is the sum of the paint weight and the can weight. Suppose $c$ and $k$ respectively represent the weight of the paint and the weight of the can itself. After three-fifths of the paint has been used, the remaining paint is $1-\dfrac35=\dfrac25.$ The two equations representing this problem are as follows.
$$\begin{cases} x & = c + k && (\cdots 1) \\ y & = \dfrac25c + k && (\cdots 2) \end{cases}$$Add the two equations so that we obtain
$$x + y = \dfrac75c + 2k.$$The first equation is equivalent to $c=x-k$ so that we get
$$\begin{aligned} x + y & = \dfrac75(x-k) + 2k \\ x + y & = \dfrac75x-\dfrac75k + 2k \\ \dfrac35k & = y-\dfrac25x \\ k & = \dfrac13(5y-2x) \end{aligned}$$Therefore, the empty weight of the paint can in terms of $x$ and $y$ is $\boxed{\dfrac13(5y-2x)~\text{kg}}.$
(Answer C)

Suppose the total number of Grade X students is $n$ people and the number of male students is $L.$ Notice that they were assigned to record the number of male students in Grade X (without counting each other).

Based on Ardi’s record, we obtain the equation $\dfrac{5}{17}(n-1) = L$ because the remaining recorded students are $n-1$ people (without counting Lily). On the other hand, based on Lily’s record, we obtain the equation $\dfrac27(n-1) = L-1$ (without counting Ardi), or it can be rewritten as $\dfrac27(n-1)+1=L.$ Thus, we obtain the equation
$$\begin{aligned} \dfrac{5}{17}(n-1) & = \dfrac27(n-1)+1 \\ 35(n-1) & = 34(n-1) + 119 && (\text{Multiply by}~119) \\ n-1 & = 119 \\ n & = 120. \end{aligned}$$This means that the total number of Grade X students is $120$ people. Consequently, $$L = \dfrac{5}{17}(n-1) = \dfrac{5}{17}(120-1) = 35.$$Since the number of male students is $35,$ the rest are female students, namely $120-35=85$ people.
(Answer D)

Suppose the smaller integer is $x$ so that the other integer is $(x+1).$ Since the sum of the two integers is $603,$ we write
$$\begin{aligned} x + (x + 1) & = 603 \\ 2x + 1 & = 603 \\ 2x & = 602 \\ x & = 301. \end{aligned}$$Therefore, the smaller integer is $\boxed{301}.$

Suppose the older sibling’s age is $x$ years so that the younger sibling’s age is $(x-5)$ years.
Five years later, the older sibling’s age becomes $(x+5)$ years, while the younger sibling’s age becomes $x$ years.
Since the sum of their ages 5 years later is known to be $35$ years, we write
$$\begin{aligned} (x+5) + x & = 35 \\ 2x + 5 & = 35 \\ 2x & = 30 \\ x & = 15. \end{aligned}$$Therefore, the older sibling is $15$ years old, while the younger sibling is $10$ years old.

Suppose the third child received $x$ cents so that
$$\begin{aligned} \text{Money received by the second child} & = x+25 \\ \text{Money received by the first child} & = 3(x+25) = 3x+75. \end{aligned}$$Since the total amount of money given was $600$ cents, we write
$$\begin{aligned} x + (x+25) + (3x+75) & = 600 \\ (x+x+3x)+(25+75) & = 600 \\ 5x + 100 & = 600 \\ 5x & = 500 \\ x & = 100. \end{aligned}$$Thus, we conclude that:
$$\begin{aligned} \text{Money received by the third child} & = x = 100~\text{cents} \\ \text{Money received by the second child} & = x+25 = 100+25=125~\text{cents} \\ \text{Money received by the first child} & = 3x+75 = 3(100)+75 = 375~\text{cents}. \end{aligned}$$

Suppose $x$ is the smaller number so that the larger number is $5x.$ Since the difference between the two numbers is $48,$ we write
$$\begin{aligned} 5x-x & = 48 \\ 4x & = 48 \\ x & = 12. \end{aligned}$$Therefore, the two numbers are $12$ and $60.$
Currently, Indonesia has $38$ provinces so that the value of $a = 38.$ Thus, the value of $a$ added to the two numbers is $\boxed{38 + 12 + 60 = 110}.$

Suppose the child’s age is $x$ years so that the mother’s age is $3x$ years. Since the difference between their current ages is 26 years, we write
$$\begin{aligned} \text{Mother’s age}-\text{Child’s age} & = 26 \\ 3x-x & = 26 \\ 2x & = 26 \\ x & = 13. \end{aligned}$$Thus, the mother’s current age is $3x = 3(13) = 39$ years, while the child’s age is $13$ years. The sum of their ages 5 years from now is $$\boxed{(39+5) + (13 + 5) = 62~\text{years}}.$$

Suppose the ages of the twin children are each $x.$ It is known that five years ago, the mother’s age was $30$ years and the total age together with her twin children was $40$ years so that we obtain
$$\begin{aligned} 30 + x + x & = 40 \\ 30 + 2x & = 40 \\ 2x & = 10 \\ x & = 5. \end{aligned}$$The ages of the twin children were $5$ years old five years ago. In other words, their current ages are $\boxed{5+5=10~\text{years}}$

Suppose the smaller number is $x$ so that the larger number is $(3x-3).$ Since the sum of the two numbers is $25,$ we write
$$\begin{aligned} x + (3x-3) & = 25 \\ 4x-3 & = 25 \\ 4x & = 28 \\ x & = 7. \end{aligned}$$Therefore, the smaller number is $x = 7,$ while the larger number is $3x-3 = 3(7)-3 = 18.$

Suppose the price of a pencil is $x$ so that the price of a book becomes $3x.$
It is known that the price of $3$ books and $5$ pencils is Rp42,000.00 so that we write
$$\begin{aligned} 3(3x) + 5x & = 42.000 \\ 9x + 5x & = 42.000 \\ 14x & = 42.000 \\ x & = 3.000. \end{aligned}$$Therefore, the price of a pencil is Rp3,000.00, while the price of a book is Rp9,000.00.

Suppose the length of the pool is $p,$ then the width is $\ell = p-7.$ Since the perimeter of the pool is $86$ m, we write
$$\begin{aligned} k & = 86 \\ 2(p + \ell) & = 86 \\ p + \ell & = 43 \\ p + (p-7) & = 43 \\ 2p-7 & = 43 \\ 2p & = 50 \\ p & = 25. \end{aligned}$$Thus, the length of the pool is $25$ m, while the width is $25-7 = 18$ m.

Suppose the smallest odd number is $x,$ then the other two odd numbers are $(x+2)$ and $(x+4).$ Since the sum of the three consecutive positive odd numbers is $21,$ we write
$$\begin{aligned} x + (x+2) + (x+4) & = 21 \\ 3x + 6 & = 21 \\ 3x & = 15 \\ x & = 5. \end{aligned}$$Therefore, the three numbers are $\boxed{5, 7,~\text{and}~9}.$

Suppose $x$ is the first number, then the second number is $3x$ and the third number is $4x.$ Since the sum of the three numbers is $96,$ we write
$$\begin{aligned} x + 3x + 4x & = 96 \\ 8x & = 96 \\ x & = 12. \end{aligned}$$Therefore, the first number is $12,$ the second number is $36,$ and the third number is $48.$

The perimeter $(k)$ is the sum of the three side lengths of the right triangle.
$$\begin{aligned} k & = (4x +5) + (7x-4) + (2x+2) \\ & = (4x + 7x + 2x) + (5+(-4)+2) \\ & = 13x + 3 \end{aligned}$$Therefore, the perimeter in terms of $x$ is $\boxed{k = (13x+3)~\text{cm}}$
The area $(L)$ is obtained by multiplying the lengths of the two legs, then dividing by $2.$
$$\begin{aligned} L & = \dfrac{(7x-4)(2x+2)}{2} \\ &= \dfrac{(7x-4)\cancel{2}(x+1)}{\cancel{2}} \\ & = (7x-4)(x+1) \\ & = 7x(x+1)-4(x+1) \\ & = 7x^2+7x-4x-4 \\ & = 7x^2+3x-4 \end{aligned}$$Therefore, the area of the triangle in terms of $x$ is $\boxed{(7x^2+3x-4)~\text{cm}^2}.$

For $x = 5,$ we obtain the side lengths of the triangle as follows.
$$\begin{aligned} S_1 & = 2x + 4 = 2(5) + 4 = 14~\text{cm} \\ S_2 & = 4x-6 = 4(5)-6 = 14~\text{cm} \\ S_3 & = 5x-1 = 5(5)-1 = 24~\text{cm} \end{aligned}$$Since there are two sides with equal lengths, namely $S_1 = S_2,$ we conclude that the triangle is an isosceles triangle.

Note: If all three sides had equal lengths, then the triangle would be an isosceles triangle, as well as an equilateral triangle. It should be noted that an equilateral triangle is an isosceles triangle in which the remaining side is also equal in length to the other two sides.

Known:
$$\begin{aligned} p & = (2x-5)~\text{cm} \\ \ell & = (3x+1)~\text{cm} \end{aligned}$$Answer a)
The perimeter of a rectangle ($k$) can be calculated by adding the lengths of all four sides.
$$\begin{aligned} k & = p + p + \ell + \ell \\ & = 2(p + \ell) \\ & = 2((2x-5) + (3x + 1)) \\ & = 2(5x-4) \\ & = 10x-8 \end{aligned}$$Therefore, the perimeter of the rectangle in terms of $x$ is $\boxed{(10x-8)~\text{cm}}.$
Answer b)
It is known that $k = 23~\text{cm}$ so that
$$\begin{aligned} 10x-8 & = 23 \\ 10x & = 31 \\ x & = 3,\!1. \end{aligned}$$The length and width of the rectangle are as follows.
$$\begin{aligned} p & = 2x-5 = 2(3,\!1)-5 = 1,2 \\ \ell & = 3x+1 = 3(3,\!1)+1 = 10,\!3 \end{aligned}$$Therefore, the dimensions of the rectangle are $$\boxed{1,\!2~\text{cm} \times 10,\!3~\text{cm}}$$

Answer a)
Distance ($s$) is obtained by multiplying speed by travel time. Since the unit of time is already appropriate, we can directly multiply to obtain the distance.
$$\begin{aligned} \text{s} & = \text{Distance traveled by motorcycle} + \text{Distance traveled by bus} \\ & = 3(2x-5) + 4(5x+8) \\ & = 6x-15+20x+32 \\ & = 26x+17 \end{aligned}$$Therefore, the distance traveled by Mr. Budi is $\boxed{(26x+17)~\text{km}}.$
Answer b)
Since the total distance traveled is known to be $329$ km, we write
$$\begin{aligned} 26x+17 & = 329 \\ 26x & = 312 \\ x & = 12. \end{aligned}$$Therefore, the value of $x$ that satisfies the equation is $\boxed{12}.$

Known:
$$\begin{aligned} p & = (2x-3)~\text{cm} \\ \ell & = (3x+10)~\text{cm} \\ t & = x~\text{cm} \end{aligned}$$Answer a)
The wire length ($k$) is equal to the total length of all edges forming the cuboid. The length, width, and height edges of the cuboid each appear four times, so we write
$$\begin{aligned} k & = 4(p + \ell + t) \\ & = 4((2x-3) + (3x+10) + x) \\ & = 4((2x+3x+x) + (-3+10)) \\ & = 4(6x + 7) \\ & = 24x + 28. \end{aligned}$$Therefore, the wire length in terms of $x$ is $\boxed{(24x+28)~\text{cm}}.$
Answer b)
It is known that the wire length is $388$ cm, so
$$\begin{aligned} 24x + 28 & = 388 \\ 24x & = 360 \\ x & = 15. \end{aligned}$$Therefore, the value of $x$ is $\boxed{15}.$
Answer c)
For $x = 15,$ we obtain
$$\begin{aligned} p & = 2x-3 = 2(15)-3 = 27 \\ \ell & = 3x+10 = 3(15)+10 = 55 \\ t & = x = 15. \end{aligned}$$Therefore, the dimensions of the cuboid frame are $$\boxed{27~\text{cm} \times 55~\text{cm} \times 15~\text{cm}}.$$