Solved Questions

 

Question 1:

$$ \frac{2x-5}{y+4} = \frac{1}{5} $$

 

Solution:

The value that makes the fraction \[ \frac{x}{y-1} \] undefined is $y = 1$. Accordingly, calculating the corresponding value of $x$ for $y = 1$ using the given equation:

\[ \frac{2x-5}{y+4} = \frac{1}{5} \Rightarrow \frac{2x-5}{1+4} = \frac{1}{5} \]

\[ \Rightarrow 2x – 5= 1 \Rightarrow x = 3 \]

\(\textbf{Answer: C} \)

 

Solution:

 

What is the result of this operation?

\[ \text{A)} \frac{11}{6} \quad \text{B) } \frac{13}{6} \quad \text{C) } \frac{19}{6} \quad \text{D) } \frac{5}{3} \quad \text{E) } \frac{7}{3} \]

Solution:

We can group the first $20$ terms into pairs:

\[
\underbrace{
\left( \frac{3}{2} – \frac{4}{3} \right) + \left( \frac{3}{2} – \frac{4}{3} \right) + \dots + \left( \frac{3}{2} – \frac{4}{3} \right)
}_{20 \, \text{terms}} + \frac{3}{2}
\]

Evaluating each pair yields $\frac{3}{2} – \frac{4}{3} = \frac{9-8}{6} = \frac{1}{6}$. Since there are $10$ such pairs:

\[
= \underbrace{\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \dots + \frac{1}{6}}_{10 \, \text{times}}
+ \frac{3}{2} = 10 \cdot \frac{1}{6} + \frac{3}{2} = \frac{5}{3} + \frac{3}{2} = \frac{10+9}{6} = \frac{19}{6}
\]

\(\textbf{Answer: C} \)

 

Question 4:

The expressions $m – n$ and $p + n$ are coprime numbers. Given that $$ \frac{m-n}{p+n}= \frac{12}{18} $$ what is the value of the sum $m + p$?

\[ \text{A)} 1 \quad \text{B) } 2 \quad \text{C) } 3 \quad \text{D) } 4 \quad \text{E) } 5 \]

Solution:

 

Question 5:

\[
\left[ \left( \frac{1}{3} – \frac{1}{5} \right) – \left( \frac{2}{9} – \frac{1}{5} + \frac{1}{3} \right) \right] : \frac{2}{9}
\]

 

Solution:

\[
\text{Initial expression:} \quad \left[ \left( \frac{1}{3} – \frac{1}{5} \right) – \left( \frac{2}{9} – \frac{1}{5} + \frac{1}{3} \right) \right] : \frac{2}{9}
\]
Expanding the brackets by distributing the negative sign:
\[
= \left[ \frac{1}{3} – \frac{1}{5} – \frac{2}{9} + \frac{1}{5} – \frac{1}{3} \right] : \frac{2}{9}
\]
\[
\text{Step 1:} \quad \text{Group like terms together:}
\]
\[
= \left[ \left(\frac{1}{3} – \frac{1}{3}\right) + \left(-\frac{1}{5} + \frac{1}{5}\right) – \frac{2}{9} \right] : \frac{2}{9}
\]
\[
\text{Step 2:} \quad \text{Simplify by canceling out terms:}
\]
\[
= \left[ -\frac{2}{9} \right] : \frac{2}{9}
\]
\[
\text{Step 3:} \quad \text{Convert the division into multiplication by the reciprocal:}
\]
\[
= -\frac{2}{9} \cdot \frac{9}{2}
\]
\[
\text{Step 4:} \quad \text{Perform the multiplication:}
\]
\[
= -1
\]

\(\textbf{Answer: A} \)

 

Question 6:

$$ \frac{n+1}{n} = 1.05 $$ what is the value of $n$?

\[ \text{A)} 17 \quad \text{B) } 18 \quad \text{C) } 19 \quad \text{D) } 20 \quad \text{E) } 21 \]

Solution:

We can rewrite the fraction and the decimal as follows:
\[
\frac{n+1}{n} = 1.05 \Rightarrow 1 + \frac{1}{n} = 1 + \frac{5}{100}
\]
\[
\Rightarrow 1 + \frac{1}{n} = 1 + \frac{1}{20} \Rightarrow \frac{1}{n} = \frac{1}{20} \Rightarrow n = 20
\]

\(\textbf{Answer: D} \)

 

Question 7:

Given $$ x =-\frac{15}{14}- \frac{16}{15} – \frac{17}{16} $$ what is the value of the expression \(\frac{1}{14} + \frac{1}{15} + \frac{1}{16}\) in terms of $x$?

\[ \text{A)} 3-x \quad \text{B) } x-3 \quad \text{C) } -x-3 \quad \text{D) } x+3 \quad \text{E) } 3x \]

Solution:

Let’s break down each improper fraction into a whole number and a proper fraction:
\[
x = -\frac{15}{14} – \frac{16}{15} – \frac{17}{16}
\]
\[
\Rightarrow x = -\left(1 + \frac{1}{14}\right) – \left(1 + \frac{1}{15}\right) – \left(1 + \frac{1}{16}\right)
\]
\[
\Rightarrow x = -1 – \frac{1}{14} – 1 – \frac{1}{15} – 1 – \frac{1}{16}
\]
\[
\Rightarrow x = -3 – \left( \frac{1}{14} + \frac{1}{15} + \frac{1}{16} \right)
\]
Isolating the target expression:
\[
\Rightarrow \frac{1}{14} + \frac{1}{15} + \frac{1}{16} = -x – 3
\]

\(\textbf{Answer: C} \)

 

Question 8:

What is the value of the following nested fraction?

\[
\Large
2 – \frac{1}{2 + \frac{1}{2 – \frac{1}{2}}}
\]

\[ \text{A)} \frac{7}{8} \quad \text{B) } \frac{9}{8} \quad \text{C) } \frac{11}{8} \quad \text{D) } \frac{13}{8} \quad \text{E)} \frac{15}{8} \]

Solution:

We evaluate the fraction step-by-step, starting from the bottom:

\[
\text{Given expression:} \quad 2 – \frac{1}{2 + \frac{1}{2 – \frac{1}{2}}}
\]

\[\textbf{Step 1: Simplify the innermost denominator:} \quad 2 – \frac{1}{2} = \frac{3}{2}
\]
\[
\Rightarrow 2 – \frac{1}{2 + \frac{1}{\frac{3}{2}}}
\]

\[\textbf{Step 2: Take the reciprocal of that innermost fraction:} \quad \frac{1}{\frac{3}{2}} = \frac{2}{3}
\]
\[
\Rightarrow 2 – \frac{1}{2 + \frac{2}{3}}
\]

\[\textbf{Step 3: Evaluate the next denominator down:} \quad 2 + \frac{2}{3} = \frac{8}{3}
\]
\[
\Rightarrow 2 – \frac{1}{\frac{8}{3}}
\]

\[\textbf{Step 4: Take the reciprocal again:} \quad \frac{1}{\frac{8}{3}} = \frac{3}{8}
\]
\[
\Rightarrow 2 – \frac{3}{8}
\]

\[\textbf{Step 5: Perform the final subtraction:} \quad \frac{16}{8} – \frac{3}{8} = \frac{13}{8}
\]

\(\textbf{Answer: D} \)

 

Question 9:

What is the result of the expression \[ \left( 1 – 0.2: 0.4 + \frac{3}{5} \right) \cdot \frac{1}{0.1} \]?

\[ \text{A)} 26 \quad \text{B) } 11 \quad \text{C) } 2 \quad \text{D) } 1.1 \quad \text{E)} 0.11 \]

Solution:

$$(1-0.2: 0.4 +\frac{3}{5} ) \cdot \frac{1}{0.1} = \left(1- \frac{0.2}{0.4} + \frac{3}{5}\right) \cdot 10$$
$$=\left(1 -\frac{1}{2} + \frac{3}{5}\right) \cdot 10 $$
$$= 10 – \left(\frac{1}{2} \cdot 10\right) + \left(\frac{3}{5} \cdot 10\right) = 10 – 5 + 6 = 11 $$

\(\textbf{Answer: B} \)

 

Question 10:

What is the result of the expression $$ \frac{3}{0.5} – \frac{2}{0.25} + \frac{1}{0.125} $$?

\[ \text{A)} 6 \quad \text{B) } 8 \quad \text{C) } 10 \quad \text{D) } 0.2 \quad \text{E)} 0.04 \]

Solution:

\[
\textbf{Method 1 (Using Fractions):}
\]
\[
\frac{3}{0.5} – \frac{2}{0.25} + \frac{1}{0.125}
= \frac{3}{\frac{1}{2}} – \frac{2}{\frac{1}{4}} + \frac{1}{\frac{1}{8}}
\]
\[
= (3 \cdot 2) – (2 \cdot 4) + (1 \cdot 8) = 6 – 8 + 8 = 6
\]

\[
\textbf{Method 2 (Scaling Decimals):}
\]
\[
\frac{3}{0.5} – \frac{2}{0.25} + \frac{1}{0.125}
= \frac{30}{5} – \frac{200}{25} + \frac{1000}{125}
\]
\[
= 6 – 8 + 8 = 6
\]

\(\textbf{Answer: A} \)

Question 11:

\[ \text{If} \quad 1 + \cfrac{1+\cfrac{1+\cfrac{x}{2} }{2} } {2}= 2 \quad \text{then what is the value of } x? \]

\[ \text{A)} 0 \quad \text{B) } 1 \quad \text{C) } 2 \quad \text{D) } \frac{6}{5} \quad \text{E)} \frac{7}{2} \]

Solution:

Working outwards from the main structure:
\[
1 + \underbrace{\cfrac{1+\cfrac{1+\cfrac{x}{2}}{2}}{2}}_{1} = 2
\]
This implies the numerator of that main fraction must equal its denominator:
\[ 1+\cfrac{1+\cfrac{x}{2}}{2} = 2 \Rightarrow \cfrac{1+\cfrac{x}{2}}{2} = 1 \]
\[ \Rightarrow 1+\cfrac{x}{2} = 2 \Rightarrow \frac{x}{2} = 1 \Rightarrow x=2 \]

\(\textbf{Answer: C} \)

Question 12:

Given that \(\large \frac{1}{2a} = \frac{2}{3b} = \frac{3}{4c} \) and $a, b, c$ are negative numbers, which of the following is the correct ordering of $a, b,$ and $c$?

\[ \text{A)} a<b

Solution:

Method 1:

Multiply all parts of the equality \(\Large \frac{1}{2a} = \frac{2}{3b} = \frac{3}{4c} \) by the LCM of the numerators, which is $\text{LCM}(1, 2, 3) = 6$:
\[ \frac{6}{2a} = \frac{12}{3b} = \frac{18}{4c} \Rightarrow \frac{3}{a} = \frac{4}{b} = \frac{4.5}{c} \]
Taking the reciprocals gives:
\[ \frac{a}{3} = \frac{b}{4} = \frac{c}{4.5} \]
Since $a, b,$ and $c$ are negative numbers, their absolute values scale with their denominators: $|a| < |b| < |c|$. Reintroducing the negative signs reverses this order:
$$ c < b < a $$

Method 2:

\(\textbf{Answer: B} \)

Question 13:

 

The value \(\large 1.\overline{3}\) represents a repeating decimal.

If $$ x-1.\overline{3} = \frac{x}{3}$$, what is the value of $x$?

\[ \text{A)} 1 \quad \text{B) } -1 \quad \text{C) } 2 \quad \text{D) } -2 \quad \text{E)} 3 \]

Solution:

Converting the repeating decimal to a rational fraction:
$$ 1.\overline{3}= \frac{13-1}{9} =\frac{12}{9} = \frac{4}{3} $$
Substitute this back into the equation and rearrange terms to solve for $x$:
$$ x – \frac{x}{3} = 1.\overline{3} \Rightarrow \frac{2x}{3} = \frac{4}{3} \Rightarrow 2x = 4 \Rightarrow x=2 $$

\(\textbf{Answer: C} \)

Question 15:

$$\large \text{What is the result of the operation } \frac{\frac{3}{5} }{6}- \frac{3}{\frac{5}{6} }? \quad\quad\quad\quad\quad \quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad $$

\[ \text{A)} \frac{3}{2} \quad \text{B) } \frac{5}{2} \quad \text{C) } -\frac{5}{2} \quad \text{D) } \frac{7}{2} \quad \text{E)} -\frac{7}{2} \]

Solution:

Carefully handle the primary fraction bars:
$$ \frac{\frac{3}{5} }{6}- \frac{3}{\frac{5}{6}} = \left(\frac{3}{5} \cdot \frac{1}{6}\right) – \left(3 \cdot \frac{6}{5}\right) = \frac{3}{30} – \frac{18}{5} = \frac{1}{10} – \frac{36}{10} $$
$$ = -\frac{35}{10} = -\frac{7}{2} $$

\(\textbf{Answer: E} \)

Question 16:

$$ \text{What is the result of the operation } \frac{0.000099+ 0.0099+ 0.99}{0.000011}? \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$

\[ \text{A)} 111 \quad \text{B) } 10101 \quad \text{C) } 9099 \quad \text{D) } 90909 \quad \text{E)} 999 \]

Solution:

First, accurately align and add the decimal terms in the numerator:
$$ 0.99 + 0.0099 + 0.000099 = 0.999999 $$
Substitute this back into the expression:
$$ \frac{0.999999}{0.000011} = \frac{999999}{11} $$
Performing the division:
$$ 999999 \div 11 = 90909 $$

\(\textbf{Answer: D} \)

Question 17:

 

Question 22:

\[
\text{If} \quad 1 – \cfrac{3}{1 + \cfrac{1}{1 + \cfrac{1}{x}}} = \frac{2}{3} \quad \text{then what is } x?
\]

\[ \text{A)} -\frac{1}{7} \quad \text{B) } -\frac{2}{7} \quad \text{C) } -\frac{8}{7} \quad \text{D) } -\frac{1}{8} \quad \text{E)} -\frac{8}{3} \]

Solution:

Working step-by-step from the outer shell inwards:
\[
1 – \frac{2}{3} = \cfrac{3}{1 + \cfrac{1}{1 + \cfrac{1}{x}}} \Rightarrow \frac{1}{3} = \cfrac{3}{1 + \cfrac{1}{1 + \cfrac{1}{x}}}
\]
Cross-multiplying reveals that the denominator must equal $9$:
\[
1 + \cfrac{1}{1 + \cfrac{1}{x}} = 9 \Rightarrow \cfrac{1}{1 + \cfrac{1}{x}} = 8
\]
Taking the reciprocal:
\[
1 + \frac{1}{x} = \frac{1}{8} \Rightarrow \frac{1}{x} = \frac{1}{8} – 1 = -\frac{7}{8} \Rightarrow x = -\frac{8}{7}
\]

\(\textbf{Answer: C} \)

Question 23:

\[
\text{What is the absolute value of the infinite continued fraction: } 2 + \cfrac{3}{2 + \cfrac{3}{2 + \cfrac{3}{2 + \cfrac{3}{\ddots}}}} \,?
\]

\[ \text{A)} \frac{5}{2} \quad \text{B) } 3 \quad \text{C) } \frac{7}{2} \quad \text{D) } 4 \quad \text{E)} \frac{9}{2} \]

Solution:

Let the entire value of this repeating infinite structure be equal to $x$:
\[
x = 2 + \cfrac{3}{2 + \cfrac{3}{2 + \cfrac{3}{2 + \cfrac{3}{\ddots}}}}
\]
Notice that the nested denominator is structurally identical to the complete expression $x$:
\[
x = 2 + \frac{3}{x}
\]
Multiply the entire equation by $x$ to clear the fraction format:
\[
x^2 = 2x + 3 \Rightarrow x^2 – 2x – 3 = 0
\]
Factoring this quadratic equation yields:
\[
(x – 3)(x + 1) = 0
\]
This gives two possible mathematical roots: $x = 3$ or $x = -1$. Because all components within the fraction are positive, the result must be positive, confirming $x = 3$.

\(\textbf{Answer: B} \)

Question 24:

 

Question 28:

\[ \text{If} \quad 1 – \frac{2}{3 – \frac{1}{2 + \large \frac{1}{x}}} = -1 \quad \text{then find } x. \]

\[ \text{A)} \frac{1}{2} \quad \text{B) } \frac{2}{3} \quad \text{C) } -\frac{1}{2} \quad \text{D) }- \frac{2}{3} \quad \text{E)} -2 \]

Solution:

Isolating the fraction block step-by-step:
\[ 1 – \cfrac{2}{3 – \cfrac{1}{2 + \cfrac{1}{x}}} = -1 \Rightarrow \cfrac{2}{3 – \cfrac{1}{2 + \cfrac{1}{x}}} = 2 \]
This means the denominator must equal $1$:
\[ 3 – \frac{1}{2 + \frac{1}{x}} = 1 \Rightarrow \frac{1}{2 + \frac{1}{x}} = 2 \]
Taking the reciprocal:
\[ 2 + \frac{1}{x} = \frac{1}{2} \Rightarrow \frac{1}{x} = \frac{1}{2} – 2 = -\frac{3}{2} \Rightarrow x = -\frac{2}{3} \]

\(\textbf{Answer: D} \)

Question 29:

What is the result of the operation $$( \frac{0.001}{0.04} + \frac{0.002}{0.08} ) : \frac{0.003}{0.06} $$?

\[ \text{A)} \frac{1}{10} \quad \text{B) } \frac{1}{20} \quad \text{C) } 10 \quad \text{D) }20 \quad \text{E)} 1 \]

Solution:

Simplify each individual fraction component first by shifting the decimal points:
– $\frac{0.001}{0.04} = \frac{1}{40}$
– $\frac{0.002}{0.08} = \frac{2}{80} = \frac{1}{40}$
– $\frac{0.003}{0.06} = \frac{3}{60} = \frac{1}{20}$

Substitute these simplified values back into the expression:
$$ \left( \frac{1}{40} + \frac{1}{40} \right) : \frac{1}{20} = \frac{2}{40} : \frac{1}{20} = \frac{1}{20} : \frac{1}{20} = 1 $$

\(\textbf{Answer: E} \)

Question 30: