54 is divided into two parts such that sum of 10 times the first part and 22 times the second part is 780. What is the bigger part?

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Solution

The correct option is A 34
Let the 2 parts of 54 be x and y
$x + y = 54$ ...(i)
and $10 x + 22 y = 780$ ...(ii)
Multiply (i) by 10, we get
$10 x + 10 y = 540$ ...(iii)
Substracting (ii) from (iii)
$- 12 y = - 240$
$y = 20$
Substituting $y = 20$ in $x + y = 54$ ,
$\Rightarrow x + 20 = 54$
$\Rightarrow x = 34$
Hence, $x = 34$ and $y = 20$

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54 is divided into two parts such that sum of 10 times the first part and 22 times the second part is 780. What is the bigger part?

The correct option is A 34Let the 2 parts of 54 be x and y x+y=54...(i) and 10x+22y=780...(ii) Multiply (i) by 10, we get 10x+10y=540...(iii) Substracting (ii) from (iii) −12y=−240 y=20 Substituting y=20 in x+y=54, ⇒x+20=54 ⇒x=34 Hence, x=34 and y=20