Fundamental Concepts of Algebra — Multiple Choice Questions

The sum of $6x$ and $2y$ is

1. $8xy$

2. $8 (x + y)$

3. $8 + x + y$

4. none of these

Answer

$6x$ and $2y$ are unlike terms, so they cannot be added into a single term.

Sum of $6x$ and $2y = 6x + 2y$, which is not given in options 1, 2 and 3.

Hence, option 4 is the correct option.

$(6 \times a \times a \times 3 \times b \times b \times b \times 2 \times c \times c \times c \times c)$ = ?

1. $36(a^2 + b^3 + c^4)$

2. $36a^2b^3c^4$

3. $11a^2b^3c^4$

4. none of these

Answer

Solving,

$\Rightarrow 6 \times a \times a \times 3 \times b \times b \times b \times 2 \times c \times c \times c \times c\\[1em] \Rightarrow (6 \times 3 \times 2) \times (a \times a) \times (b \times b \times b) \times (c \times c \times c \times c)\\[1em] \Rightarrow 36 \times a^2 \times b^3 \times c^4\\[1em] \Rightarrow 36a^2b^3c^4$

Hence, option 2 is the correct option.

If $x = 3$ and $y = -2$, then the value of $(7x - 9y + 2)$ is

1. 5

2. 41

3. 23

4. none of these

Answer

Substituting $x = 3$ and $y = -2$, we get:

$\Rightarrow 7x - 9y + 2\\[1em] \Rightarrow 7 \times 3 - 9 \times (-2) + 2\\[1em] \Rightarrow 21 + 18 + 2\\[1em] \Rightarrow 41$

Hence, option 2 is the correct option.

If $x = \dfrac{1}{2} \text{ and } y = \dfrac{1}{4}$, then the value of $(16x + 8y - 10)$ is

1. −1

2. 1

3. −2

4. 0

Answer

Substituting $x = \dfrac{1}{2} \text{ and } y = \dfrac{1}{4}$, we get:

$\Rightarrow 16x + 8y - 10\\[1em] \Rightarrow 16 \times \dfrac{1}{2} + 8 \times \dfrac{1}{4} - 10\\[1em] \Rightarrow 8 + 2 - 10\\[1em] \Rightarrow 0$

Hence, option 4 is the correct option.

If $a + 2b = 5$ and $c = 3$, then the value of $a + 2(b - c)$ is

1. 1

2. −1

3. 11

4. −11

Answer

solving,

$\Rightarrow a + 2(b - c)\\[1em] \Rightarrow a + 2b - 2c\\[1em] \Rightarrow (a + 2b) - 2c$

Substituting $a + 2b = 5$ and $c = 3$, we get:

$\Rightarrow 5 - 2 \times 3\\[1em] \Rightarrow 5 - 6\\[1em] \Rightarrow -1$

Hence, option 2 is the correct option.

9 taken away from the product of $x$ and $y$ is

1. $9 - xy$

2. $xy - 9$

3. $9 - \dfrac{x}{y}$

4. $\dfrac{x}{y} - 9$

Answer

Product of $x$ and $y = xy$.

9 taken away from it means 9 is subtracted from $xy$, i.e., $xy - 9$.

Hence, option 2 is the correct option.

In $-\dfrac{7}{8}a^2b$, the coefficient of $b$ is

1. $\dfrac{-7}{8}$

2. $\dfrac{7}{8}a^2$

3. $\dfrac{-7}{8}a^2$

4. $-\dfrac{8}{7}a^2$

Answer

$-\dfrac{7}{8}a^2b = b \times \left(\dfrac{-7}{8}a^2\right)$

So, the coefficient of $b$ is $\dfrac{-7}{8}a^2$.

Hence, option 3 is the correct option.

The degree of $x^2y^2 + 3xy^2 - 5xy$ is

1. 2

2. 3

3. 4

4. 0

Answer

Degrees of the terms are (2 + 2), (1 + 2) and (1 + 1), i.e., 4, 3 and 2.

The highest of these is 4.

Hence, option 3 is the correct option.

$6a + (-3b) + (-2a) + (-4b)$ = ?

1. $4a - 7b$

2. $-10ab$

3. $-3(a + b)$

4. none of these

Answer

solving,

$\Rightarrow 6a + (-3b) + (-2a) + (-4b)\\[1em] \Rightarrow (6a - 2a) + (-3b - 4b)\\[1em] \Rightarrow (6 - 2)a + (-3 - 4)b\\[1em] \Rightarrow 4a - 7b$

Hence, option 1 is the correct option.

Twice a variable added to 4 is added to 3 subtracted from thrice the same variable. This statement can be written as

1. $(2x + 4) - (3x - 3)$

2. $(4 + 2x) + (3x - 3)$

3. $(2x + 4) + (3 - 3x)$

4. none of these

Answer

Let the variable be $x$.

Twice the variable added to $4 = 4 + 2x$.

3 subtracted from thrice the same variable $= 3x - 3$.

Adding the two expressions, we get $(4 + 2x) + (3x - 3)$.

Hence, option 2 is the correct option.