Question

Can you find the number which is greater than the aggregate of its third, tenth and the twelfth parts by 58?

Answer 1

Answer:120

Step-by-step explanation:

Let the number be X

X=X/3 +X/10 +X/12 +58

X(1-1/3 -1/10 -1/12)=58

X(29/60)=58

X=58*60÷29=120

Answer 2

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$\bold{\underline{Question-}}$Can you find the number which is greater than the aggregate of its third, tenth and the twelfth parts by 58?




$\bold{\underline{Answer-}}$Let the number be x

now,

$x = \frac{x}{3} + \frac{x}{10} + \frac{x}{12} + 58 \\ \\ \\ \\ \\ 58 = x(1 - \frac{1}{3} - \frac{1}{10} - \frac{1}{12})$



$58 = x(\frac{29}{60} )$


$x = \frac{58 \times 60}{29}$


$x = 120$