Question

Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is at most 20.​

Answer 1

ANSWER:

Given:

• 3 consecutive positive integers.
• Sum of one-third of first, one-fourth of second and one-fifth of third is at most 20.

To Find:

• The integers

Solution:

Let the integers be x , x + 1, and x + 2.

So, we are given that,

• one-third of first = (x)/3
• one-fourth of second = (x + 1)/4
• one-fifth of third = (x + 2)/5

We also have that the sum is at most 20. This means, the sum of the numbers is less than or equal to 20.

So,

⇒ x/3 + (x + 1)/4 + (x + 2)/60 ≤ 20

Taking LCM,

⇒ (20x + 15(x + 1) + 12(x + 2))/60 ≤ 20

⇒ (20x + 15x + 15 + 12x + 24)/60 ≤ 20

Transposing 60 to RHS,

⇒ (47x + 39) ≤ 1200

Transposing 39 to RHS,

⇒ 47x ≤ 1200 - 39

⇒ 47x ≤ 1161

So,

⇒ x ≤ 1161/47

⇒ x ≤ 24.7

As, x is an integer, value of x is 24.

So, the numbers are:

• x = 24
• x + 1 = 24 + 1 = 25
• x + 2 = 24 + 2 = 26

Therefore, the value of the 3 consecutive largest positive integers are 24, 25 and 26.

Answer 2

Answer:

The numbers are 24, 25 and 26.

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Hope it helps!!!