Equations in One Variable

A letter symbol that means an element of a set is called—

By convention, which letters are usually taken as variables?

By convention, which letters are usually taken as constants?

In the equation $x + a = 5$ with no further direction, $x$ is treated as—

In the equation $x + a = 5$, while solving for $x$ the symbol $a$ is treated as—

If we want to determine the value of $a$ from $x + a = 5$, then $x$ is treated as—

In the set $S = \{x : x \in \mathbb{R},\ 1 \le x \le 10\}$, the symbol $x$ is—

In $x + 3 = 5$, what kind of equation is it?

Which of the following is NOT a linear equation in one variable?

The degree of an equation is—

The degree of $x + 1 = 5$ is—

The degree of $2x^2 + 5x + 6 = 0$ is—

The degree of $y^2 - y = 12$ is—

The degree of $2x^3 - x^2 - 4x + 4 = 0$ is—

An equation of degree 2 is called—

The number of roots of an equation equals—

The roots of $x^2 - 5x + 6 = 0$ are—

The equation $(x - 3)^2 = 0$ has roots—

The roots of an equation always satisfy—

If two roots of a quadratic are equal, then—

How many real roots does $x^2 + 1 = 0$ have?

In an equation, the polynomials on the two sides—

In an identity, the polynomials on the two sides—

The equality $(x + 1)^2 - (x - 1)^2 = 4x$ is—

Which statement is correct?

An identity is satisfied by—

The proper sign used for an identity is—

$(a + b)^2 = a^2 + 2ab + b^2$ is—

$a^2 - b^2 = (a + b)(a - b)$ is an example of—

The equation $x^2 - 5x + 6 = 0$ is satisfied by—

Which of the following is an identity?

If the same number is added to both sides of an equation, the two sides—

If both sides of an equation are multiplied by the same nonzero number, the two sides—

If $a = b + c$, then by transposition—

If $a + c = b$, then by transposition $a$ equals—

The equation $ax = b$ with $a \ne 0$ has the solution—

To clear fractions in $\dfrac{x}{7} - \dfrac{4}{5} = \dfrac{6}{7} - \dfrac{2}{5}$ we may multiply both sides by—

Which operation is NOT a valid step while solving an equation?

Solve $\dfrac{x}{7} - \dfrac{4}{5} = \dfrac{6}{7} - \dfrac{2}{5}$:

Solve $(y - 1)(y + 2) = (y + 4)(y - 2)$:

Solve $\dfrac{6x + 1}{15} - \dfrac{2x - 4}{7x - 1} = \dfrac{2x - 1}{5}$:

Solution set of the equation in question 41 is—

Solve $\dfrac{1}{x - 3} + \dfrac{1}{x - 4} = \dfrac{1}{x - 2} + \dfrac{1}{x - 5}$:

Why does the equation in question 43 reduce to setting only the numerator to zero?

If $(\sqrt{5} + 1)x + 4 = 4\sqrt{5}$, then $x$ equals—

Solve $\dfrac{a}{b}x + \dfrac{b}{a}x = \dfrac{a^2 - b^2}{ab}$:

Solve $(2x + 1)(x - 2) = (x - 4)(2x + 2)$:

Solve $\dfrac{1}{x + 1} - \dfrac{1}{x + 4} = \dfrac{1}{x + 2} - \dfrac{1}{x + 3}$:

Solution set of $2x + \sqrt{2} = 3x - 4 - 3\sqrt{2}$ is—

Solution set of $\dfrac{x - 2}{x - 1} - \dfrac{1}{x - 1} = 2$ is—

To solve a real-life problem by an equation, we first—

A two-digit number with tens digit $t$ and units digit $u$ equals—

If the digits of a two-digit number are interchanged, the new number with original tens digit $t$ and units digit $u$ is—

If a class has $x$ students and $4$ students sit on each bench leaving $3$ benches empty, the number of benches is—

If $3$ students sit on each bench and $6$ students remain standing in a class of $x$ students, the number of benches is—

Profit on Tk. $P$ in $1$ year at $r\%$ per annum (simple) is—

If profit on Tk. $x$ at $12\%$ in one year is $T$, then $T$ equals—

The units digit of a two-digit number is $2$ more than the tens digit. If the digits are interchanged, the new number is less by $6$ than twice the original. The original number is—

In the previous question, with tens digit $x$, the original number is—

If $4$ students sit per bench, $3$ benches stay empty; if $3$ students sit per bench, $6$ students remain standing. Total number of students is—

Mr. Kabir invested some part of Tk. $56000$ at $12\%$ profit and the rest at $10\%$. Total profit after one year is Tk. $6400$. Money invested at $12\%$ is—

In the previous question, money invested at $10\%$ is—

If a number plus its square equals nine times the next natural number, the number is—

In a circle of radius $10$ cm, a chord is drawn so that the perpendicular from the centre to the chord is $2$ cm shorter than the half-chord. The length of the chord is—

If $120$ coins of Tk. $1$ and Tk. $2$ together amount to Tk. $180$, the number of Tk. $2$ coins is—

If a number $N$ is added to both numerator and denominator of $\dfrac{1}{2}$, the fraction becomes $\dfrac{2}{3}$. Then $N$ equals—

The difference of the squares of two consecutive natural numbers is $151$. The smaller number is—

Solve $\dfrac{a}{b}x + \dfrac{b}{a}x = \dfrac{a^2 - b^2}{ab}$:

The equation $(2x + 1)(x - 2) = (x - 4)(2x + 2)$ becomes after simplification—

Solve $(3 + \sqrt{3})x + 2 = 5 + 3\sqrt{3}$:

Sum of two numbers is $98$ and one is $\dfrac{2}{5}$ of the other. The numbers are—

A proper fraction has numerator $1$ less than denominator. Subtracting $2$ from numerator and adding $2$ to denominator gives $\dfrac{1}{6}$. The fraction is—

Sum of digits of a two-digit number is $9$; the number got by interchanging digits is less by $45$ than the original. The number is—

The tens digit of a two-digit number is twice its units digit. If sum of digits is $S$, the number is—

A merchant invested Tk. $5600$, getting $5\%$ on part and $4\%$ on the rest. Total profit is Tk. $256$. Money at $5\%$ is—

In a girls' school, $6$ students per bench leaves $2$ empty; $5$ per bench leaves $6$ standing. Number of benches is—

Number of students in question 76 is—

A launch carries $47$ passengers. Cabin fare is twice the deck fare; deck fare is Tk. $30$. Total collection is Tk. $1680$. Number of cabin passengers is—

$120$ coins of $25$ paisa and $50$ paisa together total Tk. $35$. Number of $25$-paisa coins is—

In question 79, number of $50$-paisa coins is—

A car covers part of $240$ km at $60$ km/h and the rest at $40$ km/h, total time $5$ hours. Distance at $60$ km/h is—

Distance Dhaka NM–Gabtoli is $12$ km. Sajal goes by rickshaw at $6$ km/h, rests $30$ min at Gabtoli, returns at the same speed. Kajal walks at $4$ km/h from NM. They meet at distance from NM—

A steamer carries $376$ passengers. Cabin fare is thrice deck fare; deck fare is Tk. $60$. Total fare is Tk. $33840$. Number of deck passengers is—

In question 83, cabin passengers number—

In question 83, fare per cabin passenger is Tk.—

Solution set of $\dfrac{x - 2}{x - 1} - \dfrac{1}{x - 1} = 2$ is—

Solution set of $\dfrac{a}{x} + \dfrac{a}{x + 1} = \dfrac{2a}{x - 1}$ leads to a quadratic in $x$ whose constant term sign is—

Solution set of $\dfrac{1}{x + 2} - \dfrac{1}{x + 5} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$ is—

Equations of the form $ax^2 + bx + c = 0$, $a \ne 0$ are called—

The general form of a quadratic equation in one variable requires—

The highest degree of the variable in a quadratic equation is—

The equation $x^2 - x - 12 = 0$ comes from a rectangle of area $12$ sq. cm. with breadth $(x - 1)$ cm and length—

Number of roots of a quadratic equation is—

If a product of two real quantities is zero, then—

From $(x + 2)(x - 3) = 0$ we get—

The principle "$ab = 0 \Rightarrow a = 0$ or $b = 0$" is used in the method of—

Solve $(x + 2)(x - 3) = 0$:

Solution set of $y^2 = \sqrt{3}\, y$ is—

Solution set of $x - 4 = \dfrac{x - 4}{x}$ is—

From $\left(\dfrac{x + a}{x - a}\right)^2 - 5\left(\dfrac{x + a}{x - a}\right) + 6 = 0$, with $y = \dfrac{x + a}{x - a}$, the values of $y$ are—

In question 100, the corresponding values of $x$ are—

Comparing $x^2 - 1 = 0$ with $ax^2 + bx + c = 0$, we get—

The degree of $(x - 1)^2$ as a polynomial is—

Equation $(x - 1)^2 = 0$ has roots—

The denominator of a proper fraction is $4$ more than its numerator. When the fraction is squared, the new denominator exceeds the new numerator by $40$. The fraction is—

With numerator $x$ in question 105, equation reduces (after simplification) to—