Question

Three consecutive numbers such that thrice the first, 4 times the second and twice the third together make 188. Find the least of the consecutive numbers. * (a)18 (b)21 (c)19 (d)20​

Answer 1

Answer:

le first no be x

second =x+1

third=x+2

atq

3x+4x+4+2x+4=188

9x=188-8

9x=180

x=20

as first is the least no so least no is d 20

Answer 2

Answer:

The least of the three consecutive numbers is 19, and the reply is (c) 19.

Step-by-step explanation:

From the above question,

           Let's count on the first of the three consecutive numbers is x. Then, the subsequent two numbers would be (x+1) and (x+2), respectively.

According to the hassle statement, thrice the first range (3x), four instances the 2d quantity (4(x+1)), and twice the 1/3 wide variety (2(x+2)) collectively make 188. We can write this as an equation:

                   3x + 4(x+1) + 2(x+2) = 188

Simplifying the equation, we get:

                   9x + 14 = 188

Subtracting 14 from each sides, we get:

                   9x = 174

Dividing each facets by means of 9, we get:

                   x = 19.33 (approx)

Since we are searching for the least of the three consecutive numbers, we want to spherical down this fee to the nearest integer, which is 19.

Therefore, the least of the three consecutive numbers is 19, and the reply is (c) 19.

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