Simple Equations — Mental Maths

Fill in the blanks:

(i) The sum of a number and 6 is 15. The number is ...............

(ii) If $\dfrac{5x}{7} - \dfrac{3x}{2} = \dfrac{11}{14}$, then $x$ = ...............

(iii) If $\dfrac{5x}{6} - \dfrac{2x}{3} = 9$, then $x$ = ...............

(iv) If $5(3 − 2x) − 3(5 − 3x) = 6$, then $x$ = ...............

(v) If $\dfrac{x}{3} - 3 = 3$, then $x$ = ...............

(vi) If 3 more than one-third of a number is 5, then we write it as ...............

Answer

(i) Let the number be $x$.

$\Rightarrow x + 6 = 15 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow x = 15 - 6 \\[1em] \Rightarrow x = 9.$

Hence, the number is 9.

(ii) Solving,

$\Rightarrow \dfrac{5x}{7} - \dfrac{3x}{2} = \dfrac{11}{14} \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \dfrac{10x - 21x}{14} = \dfrac{11}{14} \\[1em] \Rightarrow \dfrac{-11x}{14} = \dfrac{11}{14} \\[1em] \text{ [Equating numerators] } \\[1em] \Rightarrow -11x = 11 \\[1em] \text{ [Dividing each side by −11] } \\[1em] \Rightarrow x = -1.$

Hence, $\bm{x = −1}$.

(iii) Solving,

$\Rightarrow \dfrac{5x}{6} - \dfrac{2x}{3} = 9 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \dfrac{5x - 4x}{6} = 9 \\[1em] \Rightarrow \dfrac{x}{6} = 9 \\[1em] \text{ [Multiplying each side by 6] } \\[1em] \Rightarrow x = 54.$

Hence, $\bm{x = 54}$.

(iv) Solving,

$\Rightarrow 5(3 - 2x) - 3(5 - 3x) = 6 \\[1em] \text{ [Removing brackets] } \\[1em] \Rightarrow 15 - 10x - 15 + 9x = 6 \\[1em] \Rightarrow -x = 6 \\[1em] \text{ [Multiplying each side by −1] } \\[1em] \Rightarrow x = -6.$

Hence, $\bm{x = −6}$.

(v) Solving,

$\Rightarrow \dfrac{x}{3} - 3 = 3 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \dfrac{x}{3} = 3 + 3 \\[1em] \Rightarrow \dfrac{x}{3} = 6 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow x = 18.$

Hence, $\bm{x = 18}$.

(vi) Solving,

Let the number be $x$.

One-third of the number = $\dfrac{x}{3}$

3 more than one-third of the number = $\dfrac{x}{3} + 3$

∴ The statement is written as $\dfrac{x}{3} + 3 = 5$.

Hence, the required equation is $\bm{\dfrac{x}{3} + 3 = 5}$.

Write (T) for true and (F) for false for each of the following statements:

(i) If $\dfrac{4x}{3} + 3 = x - 2$, then $x$ = 15 ...............

(ii) If $\dfrac{3y}{2} - 5 = -7$, then $y = \dfrac{4}{3}$ ...............

(iii) If $2x + 3x + 4x = 5x + 1$, then $x = \dfrac{1}{4}$ ...............

(iv) If $\dfrac{x}{5} + 4 = 5$, then $x$ = −5 ...............

(v) If $−6 + 3x$ = −5, then $x = \dfrac{1}{3}$ ...............

Answer

(i) Solving,

$\Rightarrow \dfrac{4x}{3} + 3 = x - 2 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \dfrac{4x}{3} - x = -2 - 3 \\[1em] \text{ [Taking L.C.M.] } \\[1em] \Rightarrow \dfrac{4x - 3x}{3} = -5 \\[1em] \Rightarrow \dfrac{x}{3} = -5 \\[1em] \text{ [Multiplying each side by 3] } \\[1em] \Rightarrow x = -15.$

Since $x$ = −15 and not 15, the given statement is false.

Hence, the given statement is false (F).

(ii) Solving,

$\Rightarrow \dfrac{3y}{2} - 5 = -7 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \dfrac{3y}{2} = -7 + 5 \\[1em] \Rightarrow \dfrac{3y}{2} = -2 \\[1em] \text{ [Multiplying each side by 2] } \\[1em] \Rightarrow 3y = -4 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow y = \dfrac{-4}{3}.$

Since $y = \dfrac{-4}{3}$ and not $\dfrac{4}{3}$, the given statement is false.

Hence, the given statement is false (F).

(iii) Solving,

$\Rightarrow 2x + 3x + 4x = 5x + 1 \\[1em] \Rightarrow 9x = 5x + 1 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 9x - 5x = 1 \\[1em] \Rightarrow 4x = 1 \\[1em] \text{ [Dividing each side by 4] } \\[1em] \Rightarrow x = \dfrac{1}{4}.$

Since $x = \dfrac{1}{4}$, the given statement is true.

Hence, the given statement is true (T).

(iv) Solving,

$\Rightarrow \dfrac{x}{5} + 4 = 5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow \dfrac{x}{5} = 5 - 4 \\[1em] \Rightarrow \dfrac{x}{5} = 1 \\[1em] \text{ [Multiplying each side by 5] } \\[1em] \Rightarrow x = 5.$

Since $x$ = 5 and not −5, the given statement is false.

Hence, the given statement is false (F).

(v) Solving,

$\Rightarrow -6 + 3x = -5 \\[1em] \text{ [By transposition] } \\[1em] \Rightarrow 3x = -5 + 6 \\[1em] \Rightarrow 3x = 1 \\[1em] \text{ [Dividing each side by 3] } \\[1em] \Rightarrow x = \dfrac{1}{3}.$

Since $x = \dfrac{1}{3}$, the given statement is true.

Hence, the given statement is true (T).