Question

Page No a way that one Divide 3080 into three parts in such that one south of first least one third of the second part and half of third part are equal​

Answer 1

Answer:

Option 4 : C

Step-by-step explanation:

Detailed Solution

Let, the parts are x, y and z

According to the question,

⇒ x/2 = y/3 = z/6

Let, x/2 = y/3 = x/6 = k

∴ x = 2k, y = 3k and z = 6k

According to the question,

⇒ 2k + 3k + 6k = 3740

⇒ 11k = 3740

⇒ k = 3740/11 = 340

∴ First part = 2k = 2 × 340 = 680

∴ Second part = 3k = 3 × 340 = 1020

∴ Third part = 6k = 6 × 340 = 2040