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 10x + 22y = 780 ...(ii)

Multiply (i) by 10 , we get

10 x + 10 y = 540 ...(iii)

Subtracting (ii) from (iii)

- 12y = - 240

y = 20

Subsituting y = 20 in x + y = 54 ,
$⟹$ x + 20 = 54
$⟹$ 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 34 Let the 2 parts of 54 be x and y x+y = 54 ....(i) and 10x + 22y = 780 ...(ii) Multiply (i) by 10 , we get 10 x + 10 y = 540 ...(iii) Subtracting (ii) from (iii) - 12y = - 240 y = 20 Subsituting y = 20 in x + y = 54 , ⟹ x + 20 = 54 ⟹x = 34 Hence, x = 34 and y = 20