Question

Tank a has a capacity of 9.5 gallons. 6 1/3 gallons of the tank's water are poured out. how many gallons of water are left in the tank?

Answer 1

$3\frac{1}{6}$ gallons are left in the tank.

• The total capacity of the tank = 9.5 gallons
• The total amount of water poured out in gallons = $6\frac{1}{3}$

Further Explanation

To determine the total amount left in the tank = Total capacity of the tank – Total amount of water poured out in gallons

= 9.5 (also expressed as 95/10) – $6\frac{1}{3}$ (otherwise expressed as 19/3)

= 95/10 – 19/3

= (95 x 3) – (19 x 10) / 30

= (285) – (190) / 30

= 285 – 190 / 30

= 95/30

= 19/6

= $3\frac{1}{6}$ gallons

The above solution to the given mathematical problem was derived as a result of following the rules of BODMAS.

The math problem has an element of fractions in it.

9.5 is the given value of the tank, and was expressed as fraction i.e 95/10 because the volume of water poured out of the tank was given as a fraction and whole number ( $6\frac{1}{3}$ ), which was then factored to get a fractional value (19/3).

To determine the remaining gallons left in the tank, the Lowest Common Multiples of the denominators (30) was factored out, and is divided by the denominators and then multiplied to by the numerators of both fractions.

The derived values were subtracted from each other, and broken down to their lowest factors to finally derive the total amount of gallons of water left in the tank.

LEARN MORE:

• fraction brainly.com/question/6201432
• mixed numbers brainly.com/question/12853142
• Lowest common denominator  brainly.com/question/10962736

KEYWORDS:

• bodmas
• lowest common denominator
• mixed numbers
• fractions
• gallons of water

Answer 2

Answer:

Water left in tank is $3\frac{1}{6}$ gallons.

Step-by-step explanation:

Given :

Capacity of tank = 9.5 gallons  $=\frac{95}{10}$

Water taken out of tank =$6\frac{1}{3}=\frac{19}{3}$ gallons

We have to find the water left in the tank.

Water left in the tank = Total capacity of tank - water taken out of tank

                                    =$\frac{95}{10}-\frac{19}{3}$

Taking LCM of (3,10)= 30

$\Rightarrow \frac{95 \times 3-19 \times10}{30}$

Solving further, we get,

$\Rightarrow \frac{285-190}{30}$

$\Rightarrow \frac{95}{30}=\frac{19}{6}=3\frac{1}{6}$

Thus, Water left in tank is $3\frac{1}{6}$ gallons..