Question

The sum of three numbers is 84. The second number is twice the first, and the third is 4 more than the second. Find the numbers.

Answer 1

a+b+c=84. (1)

b=2a ==> a=b/2

c=4b

b=b

substitute these values in equation(1)

a+b+c=84

b/2+b+4b=84

b+2b+8b=2×84

11b=168

b=168/11 =15.2

a=b/2=168×2/11=326/11 =29.6

c=4b=4×168/11=652/11 =59.2

Answer 2

$\bf{\red{\underline{\bf{Given}}}}$

• Sum of three numbers = 84

• Second Number = 2 × First number

• Third Number = Second Number + 4

$\bf{\red{\underline{\bf{ToFind}}}}$

• Numbers

$\bf{\red{\underline{\bf{Solution}}}}$

➥ Let the first no be x

Then,

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ Qᴜᴇsᴛɪᴏɴ :-

➥ Second Number = 2x

➥ Third Number = (2x + 4)

Now,

➠ Sum of three numbers = 84

So,

➮ x + 2x + (2x + 4) = 84

➮ 5x + 4 = 84

➮ 5x = 84 - 4

➮ x = 80/5

➮ x = 16

Therefore,

➤ First number = x = 16

➤ Second Number = 2x = 2×16 = 32

➤ Third Number = (2x + 4) = 2×16 + 4 = 32 + 4 = 36