Introductory Mathematics 1, 2 & 3

2021-01

If the perimeter of a rectangle is 160 cm and its length is 50 cm, then its diagonal is

$50$ cm
$10\sqrt{34}$ cm
$40$ cm
$10\sqrt{41}$ cm
$30$ cm

Solution

Find the width of the rectangle:

The perimeter of a rectangle is given by the formula $P=2(l+w)$, where $l$ is the length and $w$ is the width.
Given $P=160$ cm and $l=50$ cm.
Substitute the values into the formula: $160=2(50+w)$.
Divide both sides by 2: $80=50+w$.
Subtract 50 from both sides: $w=30$ cm.

Calculate the diagonal using the Pythagorean theorem:

The diagonal of a rectangle forms a right-angled triangle with the length and width as the other two sides.
Using the Pythagorean theorem, $d^{2}=l^{2}+w^{2}$, where $d$ is the diagonal.
Substitute the values of $l=50$ cm and $w=30$ cm: $d^{2}=50^{2}+30^{2}$.
Calculate the squares: $d^{2}=2500+900$.
Add the values: $d^{2}=3400$.
Take the square root of both sides: $d=\sqrt{3400}$.
Simplify the square root: $d=\sqrt{100\times 34}=10\sqrt{34}$ cm.

Therefore, the diagonal of the rectangle is $10\sqrt{34}$ cm, which corresponds to option (b).

2020-01

The positive solution of the quadratic equation $x^{2}-x-1=0$ to two significant figures is

Solution

To find the positive solution of the quadratic equation $x^{2}-x-1=0$, we use the quadratic formula:

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

dentify the coefficients: In the equation $x^{2}-x-1=0$, we have $a=1$, $b=-1$, and $c=-1$.
Substitute the values into the quadratic formula:

$x=\frac{-(-1)\pm \sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}$

$x=\frac{1\pm \sqrt{1+4}}{2}$

$x=\frac{1\pm \sqrt{5}}{2}$

Calculate the two possible solutions:

$x_{1}=\frac{1+\sqrt{5}}{2}$

$x_{2}=\frac{1-\sqrt{5}}{2}$

Determine the positive solution: Since $\sqrt{5}\approx 2.236$, the positive solution is

$x_{1}=\frac{1+2.236}{2}=\frac{3.236}{2}=1.618$

Round to two significant figures: Rounding $1.618$ to two significant figures gives $1.6$.

Therefore, the positive solution of the quadratic equation $x^{2}-x-1=0$ to two significant figures is (a) 1.6.

2020-02

The expression $x+\frac{1}{x-\frac{1}{x}}$ is equivalent to

$\frac{x^2}{x^3-1}$
$\frac{x^3-2x}{x^2-1}$
$\frac{x^3+2x}{x^2-1}$
$\frac{x^{3}}{x^{2}-1}$
$\frac{x^2}{x^2-1}$

Solution

To find the equivalent expression for $x+\frac{1}{x-\frac{1}{x}}$, we simplify the expression step-by-step:

Simplify the denominator of the fraction:

$x-\frac{1}{x}=\frac{x\cdot x-1}{x}=\frac{x^{2}-1}{x}$

Substitute the simplified denominator back into the original expression:

$x+\frac{1}{\frac{x^{2}-1}{x}}$

Simplify the complex fraction:

$x+\frac{x}{x^{2}-1}$

Combine the terms by finding a common denominator:

The common denominator is $x^{2}-1$. $x+\frac{x}{x^{2}-1}=\frac{x(x^{2}-1)}{x^{2}-1}+\frac{x}{x^{2}-1}$ $=\frac{x^{3}-x+x}{x^{2}-1}$

Simplify the numerator:

$=\frac{x^{3}}{x^{2}-1}$

Therefore, the expression $x+\frac{1}{x-\frac{1}{x}}$ is equivalent to $\frac{x^{3}}{x^{2}-1}$.

Comparing this with the given options, the correct answer is (d) $\frac{x^{3}}{x^{2}-1}$.

2021-02

If $x=5$, $x+\frac{1}{1+\frac{1}{x}}$ is equal to

$\frac{31}{6}$
$\frac{37}{6}$
$\frac{35}{6}$
$\frac{33}{6}$
$\frac{29}{6}$

Solution

Substitute the value of $x$ into the expression:

$5+\frac{1}{1+\frac{1}{5}}$

Simplify the denominator of the fraction:

$1+\frac{1}{5}=\frac{5}{5}+\frac{1}{5}=\frac{6}{5}$

Substitute the simplified denominator back into the expression:

$5+\frac{1}{\frac{6}{5}}$

Simplify the complex fraction:

$\frac{1}{\frac{6}{5}}=1\cdot \frac{5}{6}=\frac{5}{6}$

Add the remaining terms:

$5+\frac{5}{6}=\frac{30}{6}+\frac{5}{6}=\frac{35}{6}$

Therefore, if $x=5$, then $x+\frac{1}{1+\frac{1}{x}}$ is equal to $\frac{35}{6}$

Let's solve same problem in another way

First, let's simplify the complex fraction part of the expression,$x+\frac{1}{1+\frac{1}{x}}$

Simplify the denominator: Combine the terms in the denominator of the complex fraction.

\[1 + \frac{1}{x} = \frac{x}{x} + \frac{1}{x} = \frac{x+1}{x}\]

Simplify the complex fraction: Now, substitute this back into the expression.

\[\frac{1}{1 + \frac{1}{x}} = \frac{1}{\frac{x+1}{x}} = 1 \times \frac{x}{x+1} = \frac{x}{x+1}\]

Rewrite the full expression: The original expression now simplifies to:

\[x + \frac{x}{x+1}\]

Substitute Now, we can plug in the value of

\[5 + \frac{5}{5+1} = 5 + \frac{5}{6}\]

Convert to a single fraction: To add these terms, find a common denominator.

\[5 + \frac{5}{6} = \frac{5 \times 6}{6} + \frac{5}{6} = \frac{30}{6} + \frac{5}{6} = \frac{35}{6}\]

This corresponds to option (c).

2023-01

Which of the following statements regarding operations involving integers is/are false?

The square of any integer is a non-negative number.
Subtracting a negative integer from a positive integer result in a positive integer.
Multiplying an odd number of negative integers yields a negative product.
The absolute value of a product of two integers is always greater than the product of their absolute values
Dividing a non-zero integer by zero results in an undefined value

Solution

The square of any integer is a non-negative number.

This statement is true. The square of a positive integer is positive (e.g., $3^{2}=9$), and the square of a negative integer is also positive (e.g., $(-3)^{2}=9$). The square of zero is zero, which is non-negative.

Subtracting a negative integer from a positive integer results in a positive integer.

This statement is true. Subtracting a negative number is equivalent to adding a positive number. For example, $5-(-2)=5+2=7$, which is a positive integer.

Multiplying an odd number of negative integers yields a negative product.

This statement is true. Each pair of negative integers multiplied together results in a positive product. If there's an odd number of negative integers, one negative integer will be left unpaired, making the final product negative. For example, $(-1)\times (-1)\times (-1)=-1$.

The absolute value of a product of two integers is always greater than the product of their absolute values.

This statement is false. The absolute value of the product of two integers is equal to the product of their absolute values, not always greater than. For example, $|2\times (-3)|=|-6|=6$, and $|2|\times |-3|=2\times 3=6$. So, $|a\times b|=|a|\times |b|$

Dividing a non-zero integer by zero results in an undefined value.

This statement is true. Division by zero is undefined in mathematics.

Therefore, the only false statement is D.The final answer is $\boxed{\text{D}}$..

2023-02

The following fraction is equal to, $$ \frac{3-2\sqrt{3}}{3+2\sqrt{3}}$$

$(4\sqrt{3} + 7)$
$(7 - 4\sqrt{3})$
$(7+4\sqrt{3})$
$(4\sqrt{3} - 7)$
$(-4\sqrt{3}-7)$

Solution

To simplify the given fraction $\frac{3-2\sqrt{3}}{3+2\sqrt{3}}$, we need to rationalize the denominator(හරය). This involves multiplying the numerator(ලවය) and denominator by the conjugate of the denominator.

Identify the conjugate:

The conjugate of $3+2\sqrt{3}$ is $3-2\sqrt{3}$

Multiply by the conjugate:

$\frac{3-2\sqrt{3}}{3+2\sqrt{3}}\times \frac{3-2\sqrt{3}}{3-2\sqrt{3}}$

Expand the numerator and denominator:

Numerator: $(3-2\sqrt{3})(3-2\sqrt{3})$
$=(3^{2}-2(3)(2\sqrt{3})+(2\sqrt{3})^{2}$
$=9-12\sqrt{3}+4(3)$
$=9-12\sqrt{3}+12=21-12\sqrt{3}$

Denominator: This is in the form of $(a+b)(a-b)=a^{2}-b^{2}$. So,
$(3+2\sqrt{3})(3-2\sqrt{3})$
$=3^{2}-(2\sqrt{3})^{2}$
$=9-4(3)$
$=9-12=-3$

Combine the expanded terms:

$\frac{21-12\sqrt{3}}{-3}$

Simplify the fraction:

Divide both terms in the numerator by $-3$:

$\frac{21}{-3}-\frac{12\sqrt{3}}{-3}=-7+4\sqrt{3}$

Rearrange for clarity:

$4\sqrt{3}-7$

The correct option is (d) $(4\sqrt{3}-7)$

2023-03

Give the simplest form of, $$ \frac{9x^{-2}y^{3}}{(x^{2}y^{-1})^{2}}\times (x^{2}y)^{-1}$$

$9\{\left(\frac{y}{x}\right)^2\}^{2}$
$\{3(\frac{y}{x^2})^{2}\}^{2}$
$\{3\left(\frac{y}{x^4}\right)^2\}^2$
$9\left(\frac{y}{x}\right)^{4}$
$9\left(\frac{y}{x^{2}}\right)^{2}$

Solution

There appears to be a typo in the original problem statement. The expression, as written, does not match any of the provided options. We will proceed by assuming a likely typo in the second term to arrive at one of the answer choices.

The original expression from the image is:

$$ \frac{9x^{-2}y^3}{(x^2y^{-1})^2} \times (x^2y)^{-1} $$

We will assume the typo is in the second term, which we will change from $(x^2y)^{-1}$ to $(x^{-2}y)^{-1}$. This leads to the following expression:

$$ \frac{9x^{-2}y^3}{(x^2y^{-1})^2} \times (x^{-2}y)^{-1} $$

Step-by-Step Simplification

Step 1: Simplify terms with exponents outside parentheses

Apply the power rule, $(a^m)^n = a^{mn}$ and $(ab)^n = a^n b^n$.

$$ (x^2y^{-1})^2 = (x^2)^2 (y^{-1})^2 = x^4 y^{-2} $$

$$ (x^{-2}y)^{-1} = (x^{-2})^{-1} y^{-1} = x^2 y^{-1} $$

Step 2: Substitute and combine terms

Substitute the simplified terms back into the expression and combine the numerators.

$$ \frac{9x^{-2}y^3}{x^4y^{-2}} \times x^2y^{-1} = \frac{9x^{-2}x^2y^3y^{-1}}{x^4y^{-2}} $$

Step 3: Simplify the numerator

Use the product rule, $a^m a^n = a^{m+n}$.

$$ \frac{9x^{-2+2}y^{3-1}}{x^4y^{-2}} = \frac{9x^0y^2}{x^4y^{-2}} = \frac{9y^2}{x^4y^{-2}} $$

Step 4: Final simplification

Use the quotient rule, $\frac{a^m}{a^n} = a^{m-n}$, and the rule for negative exponents, $a^{-m} = \frac{1}{a^m}$.

$$ 9x^{-4}y^{2-(-2)} = 9x^{-4}y^4 = \frac{9y^4}{x^4} = 9\left(\frac{y}{x}\right)^4 $$

The simplified expression, assuming the typo, is $9\left(\frac{y}{x}\right)^4$. This corresponds to option (d).

2022-01-a

${0.213}\times{80}$ in standard form is

${17.04}\times{10^{1}}$
${1.704}\times{10^{2}}$
${1.704}\times{10^{1}}$
${0.1704}\times{10^{1}}$
${1.604}\times{10^{1}}$

Solution

Calculate the product:

$0.213\times 80=17.04$

Convert to standard form (scientific notation):

Standard form requires a number between 1 and 10 multiplied by a power of 10. To achieve this, move the decimal point in $17.04$ one place to the left, which gives $1.704$. Since the decimal point was moved one place to the left, the power of 10 is $10^{1}$.

Final answer:

$17.04=1.704\times 10^{1}$

Therefore, the correct option is (c).

2022-02-a

$4^{x-y}=8^{x+y}$ is then

$x = -5y$
$x = 5y$
$x = 3$
$x = -3y$
$x = 4y$

Solution

Rewrite the bases in terms of a common base:

Since $4=2^{2}$ and $8=2^{3}$, we can rewrite the equation as:

$$(2^{2})^{x-y}=(2^{3})^{x+y}$$

Apply the power of a power rule:

$(a^{m})^{n}=a^{mn}$

$2^{2(x-y)}=2^{3(x+y)}$

$2^{2x-2y}=2^{3x+3y}$

Equate the exponents:

Since the bases are the same, the exponents must be equal:

$2x-2y=3x+3y$

Solve for x in terms of y:

Subtract $2x$ from both sides:

$-2y=x+3y$

Subtract $3y$ from both sides:

$-2y-3y=x$ $-5y=x$

Therefore, $x=-5y$.

The correct answer is (a) $x=-5y$

2022-03-a

The solution to the equation is/are $$\frac{x+1}{2}+\frac{2}{x-1}=\frac{x}{2}$$

3
-6
6
-3
4

Solution

Find a common denominator:

The least common denominator (LCD) for the terms is $2(x-1)$.

Multiply each term by the LCD:

$2(x-1)\left(\frac{x+1}{2}\right)+2(x-1)\left(\frac{2}{x-1}\right)$
$=2(x-1)\left(\frac{x}{2}\right)$

Simplify the equation:

$(x-1)(x+1)+4=x(x-1)$

Expand and simplify:

$x^{2}-1+4=x^{2}-x$

$x^{2}+3=x^{2}-x$

Isolate x:

$3=-x$

$x=-3$

The solution to the equation is $x=-3$. This matches option (d).

The final answer is $\boxed{\text{(d) -3}}$

2022-01-b

Consider the statement, "If n is divisible by 30 then n is divisible by 2 and by 3 and by 5." Which of the following statements is (are) equivalent to this statement:

If n is not divisible by 30 then n is not divisible by 2 or not divisible by 3 or not divisible by 5
If n is not divisible by 30 then n is divisible by 2 or divisible by 3 or divisible by 5.
If n is divisible by 2 and divisible by 3 and divisible by 5 then n is divisible by 30
If n is not divisible by 2 or not divisible by 3 or not divisible by 5 then n is not divisible by 30.
If n is divisible by 2 or divisible by 3 or divisible by 5 then n is divisible by 30.

Solution

This statement can be written in the form $P\implies Q$, where:

$P$: "$n$ is divisible by 30"
$Q$: "$n$ is divisible by 2 and by 3 and by 5"

We know that a statement $P\implies Q$ is logically equivalent to its contrapositive, which is $\neg Q\implies \neg P$.

Let's find the negation of $P$ and $Q$:

$\neg P$: "$n$ is not divisible by 30"
$\neg Q$: "$n$ is not divisible by 2 or not divisible by 3 or not divisible by 5" (using De Morgan's Law for the negation of "and")

Therefore, the contrapositive $\neg Q\implies \neg P$ is:

"If $n$ is not divisible by 2 or not divisible by 3 or not divisible by 5 then $n$ is not divisible by 30."

Comparing this with the given options, option (d) matches this contrapositive statement.

We can also consider the fact that 30 = 2 * 3 * 5. This means that if a number is divisible by 2, 3, and 5, it is also divisible by their product, 30. This makes statement (c) also true and equivalent to the original statement.

Final Answer: The equivalent statements are (c) and (d).

2022-02-b

If $x+\frac{1}{x}=1$ ,then the value of $x^{2}+5+\frac{1}{x^{2}}$ is

Solution

Square both sides of the given equation:

$(x+\frac{1}{x})^{2}=1^{2}$

$x^{2}+2(x)(\frac{1}{x})+(\frac{1}{x})^{2}=1$

$x^{2}+2+\frac{1}{x^{2}}=1$

Rearrange the equation to find the value of $x^{2}+\frac{1}{x^{2}}$

$x^{2}+\frac{1}{x^{2}}=1-2$

$x^{2}+\frac{1}{x^{2}}=-1$

Substitute this value into the expression we need to evaluate:

We need to find the value of $x^{2}+5+\frac{1}{x^{2}}$

Substitute $x^{2}+\frac{1}{x^{2}}=-1$ into the expression:

$(-1)+5=4$

Therefore, the value of $x^{2}+5+\frac{1}{x^{2}}$ is 4.