Four Operations Problems
Example:
Let’s find the number whose one-third of 5 less than 2 times is 15. Let the desired number be \( x \). \[ \frac{2x-5}{3 } = 15 \Rightarrow x = 25 \] is found.
Example:
Of two consecutive odd numbers, one-fifth of the smaller one is 5 more than one-sixth of 7 less than the larger one; based on this, let’s find the smaller of these numbers. If the smaller of two consecutive odd numbers is called \(x \), the larger becomes \( x + 2 \). Accordingly, if the equation for this problem is written as
\[\frac{x}{5} = \frac{(x+2) – 7}{6} +5 \] the smaller number is found to be \(x = 125 \).
Example:
When 1 is subtracted from the numerator of a fraction whose value is \(\large \frac{2}{3} \) and 1 is added to its denominator, the value of the fraction becomes \( \large \frac{3}{5} \). Let’s find how much more the denominator of this fraction is than its numerator.
Since this fraction is in the form of \(\large \frac{2x}{3x} \), let’s find the value of \(3x – 2x = x \). If the equation for the given problem is written as \[ \frac{2x-1}{3x+1 } = \frac{3}{5} \] then \( x = 8 \) is found.
Question 1
When 2 times the multiplicative inverse of a negative number is subtracted from 3 times this number, the result is 1. What is this number?
\[
\text{A)} -3 \quad
\text{B) } -\frac{3}{2} \quad
\text{C) } -2\quad
\text{D) } -\frac{2}{3} \quad
\text{E) } -1
\]
Solution:
If the desired number is called \( x \), the equation for this problem is written as \(3x-2 \cdot \large{ \frac{1}{x}} = 1 \).
If both sides of the equation are multiplied by \( x \) and rearranged, we get:
\[ 3x^2 -x- 2 = 0 \Rightarrow (3x+2 ) (x-1) = 0 \]
\[\Rightarrow 3x+2 = 0 \] or \[ x-1 = 0 \]
\[\Rightarrow x= -\frac{2}{3} \quad \text{or } \quad x= 1 \]
Since the number is negative, the desired number is \[ x = -\frac{2}{3} \]
\(\textbf{Answer: D} \)
Question 2
When the students in a class sit three per desk, 11 students remain standing. When they sit four per desk, there is empty space for 3 people. How many students are there in this class?
\[
\text{A)} 55 \quad
\text{B) } 53 \quad
\text{C) } 50\quad
\text{D) } 48 \quad
\text{E) } 45
\]
Solution:
If the number of desks in the class is called x, the number of students in the class becomes \(3x + 11 \) or \( 4x – 3 \). Accordingly, if the equation for this problem is written as \[ 3x + 11 = 4x – 3 \] then \[ x = 14 \] is found. Therefore, the number of students in the class is:
\[ 3 \cdot 14 + 11 = 53 \]
\(\textbf{Answer: B} \)
Question 3
Sait gives one-fourth of his walnuts to Sinan, and two-thirds of the remaining ones to Özcan. If Sait has 25 walnuts left, how many walnuts did he have at the beginning?
\[
\text{A)} 25 \quad
\text{B) } 40 \quad
\text{C) } 60\quad
\text{D) } 80 \quad
\text{E) } 100
\]
Solution:
Let Sait have \(x \) walnuts. If he gives \(\large \frac{1}{4} \) of his walnuts to Sinan, he has \( \large \frac{3x}{4} \) walnuts left. If he gives two-thirds of the remaining \( \frac{3x}{4} \) walnuts to Özcan, he is finally left with: \[\frac{3x}{4 } \cdot \frac{1}{3 } = 25 \] walnuts. According to this, the number of walnuts Sait had at the beginning is \( x = 100 \).
\(\textbf{Answer: E} \)
Question 4
The number of Uğur’s marbles is 18 more than one-third of Selim’s marbles. The number of Selim’s marbles is 10 more than one-third of Uğur’s marbles. Based on this, how many marbles do Uğur and Selim have in total?
\[
\text{A)} 32 \quad
\text{B) } 36 \quad
\text{C) } 38\quad
\text{D) } 40 \quad
\text{E) } 42
\]
Solution:
If the number of Selim’s marbles is called \( 9x\), the number of Uğur’s marbles becomes \( 3x + 18 \). Accordingly, if the equation belonging to the given problem is written as:
\[9x = \frac{3x+ 18}{3} + 10 \] it simplifies to \[9x = x + 6 + 10 \Rightarrow x= 2 \] is found. The total number of Uğur’s and Selim’s marbles is:
\[9x + (3x +18 ) = 42 \] is found.
\(\textbf{Answer: E} \)
Question 5
In a group of 15 people, some individuals did not pay the bill at the restaurant where they ate because they were guests. For this reason, the non-guest individuals paid 62,500 liras more each, resulting in a bill payment of 312,500 liras each.
Based on this, what is the number of guests in the group?
\[
\text{A)} 7 \quad
\text{B) } 6 \quad
\text{C) } 5\quad
\text{D) } 3 \quad
\text{E) } 2
\]
Solution:
If the number of guests in the group is called \( x\), then \(15 – x \) people paid 312,500 liras each, paying a total bill of \( 312,500 \cdot (15 – x) \) liras. If everyone in the group had paid the bill, they would have paid per person:
\[ 312,500 – 62,500 = 250,000 \]
liras. Accordingly, if the equation for the given problem is written as:
\[15 \cdot 250,000 = (15-x) \cdot 312,500 \] the number of guests in the group is found to be \( x = 3 \).
\(\textbf{Answer: D} \)
Question 6
Uğur buys a pen with one-fifth of the money in his pocket, and a book with two-thirds of his remaining money, leaving him with 40 liras in his pocket.
Accordingly, how much money did Uğur have in his pocket initially?
\[
\text{A)} 100 \quad
\text{B) } 120 \quad
\text{C) } 150 \quad
\text{D) } 200 \quad
\text{E) } 250
\]
Solution:
If the money Uğur initially had in his pocket is called \(x \), since he buys a pen with one-fifth of his money, the remaining money in his pocket after buying the pen is \(x \cdot \large{ \frac{4}{5} } \). Since he buys a book with two-thirds of the remaining money \( (\large \frac{4x}{5} )\), his final remaining money is one-third of this \( \large \frac{4x}{5} \). Therefore, the money left in Uğur’s pocket is \( \large \frac{4x}{5} \cdot \frac{1}{3} \). Since he has 40 liras left, if we set up the equality:
\[40= \frac{4x}{5} \cdot \frac{1}{3} \] the money is found to be \( x= 150 \) liras.
\(\textbf{Answer: C} \)
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