Question

Two-third of a number is greater than three-fifths of the number by 6.Then find the number.​

Answer 1

⛦ Given :

• Two-third of a number is greater than three-fifths of a number by 6.

⛦To find :

• The number.

⛦ Solution :

Let's assume the number as x.

So,

• Two-third of a no. = $\sf\dfrac{2x}{3}$
• Three-fifths of a no. = $\sf\dfrac{3x}{5}$

According to Question,

$\mapsto\sf\dfrac{2x}{3}{-}\dfrac{3x}{5}=6$

Hence, Taking LCM as 15.

$\mapsto\sf\dfrac{10x-9x}{15}=6$

$\mapsto\sf\dfrac{x}{15}=6$

$\mapsto\sf{x=6\times15}$

$\mapsto\sf{x=90}$

${\text{\sf{\pink{Thus,the required no. is 90}}}}$

_______________________

$\sf{\fbox{Check:}}$

As per given :

$\red\bullet\sf\dfrac{2}{3}\sf{ \ of \ 90}$

$\to\sf\dfrac{2}{\cancel{3}^1}\times{\cancel{90}^30}$

$\to\sf{2\times30=60}$

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$\red\bullet\sf\dfrac{3}{5}\sf{ \ of \ 90}$

$\to\sf\dfrac{3}{\cancel{5}^1}\times{\cancel{90}^{18}}$

$\to\sf{3\times18=54}$

So,

$:\implies\tt{60-54=6}$

$\underline{\rm{\blue{Thus, Checked}}}$

Answer 2

Given :

• Two-third of a number is greater than three-fifths of a number by 6.

Find :

• The required number.

Required Solution :

• Let the required number be x.

So, As we know,

• Two-third of a number = $\sf\dfrac{2}{3} x$

• Three-fifths of a number = $\sf\dfrac{3}{5} x$

According to Question,

$\\ \therefore\sf\dfrac{2}{3} x \: {-} \: \dfrac{3}{5} x=6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \:➠ \: \bigg( \frac{2}{3} - \frac{3}{5} \bigg)x = 6 \: \: \: \: \: \: \: (Taking \: LCM \: as \: 15)\: \: \: \: \: \: \\ \\ \sf \: ➠ \: \: \frac{(2 \times 5)x - (3 \times 3)x}{15} = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: ➠ \: \frac{10 x- 9x}{15} = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\\ ➠ \: \sf \: \frac{x}{15} = 6 \\ \\ \sf \therefore \: x = 6 \times 15 \\ \\ \sf \: ➠ \: { \boxed { \underline{\pink{90}}}} \: \star{}$

• $\therefore \: {\text{\sf{\purple{The required number is 90}}}}$