A merchant has 120 liters of oil of one kind, 180 liters of another kind, and 240

Mathematics

1 answer:

balu736 [363]2 years ago

4 0

The merchant needs to fill 60 litres of all types of oils .

What is HCF ?

HCF is the highest common factor, it is the factor in common in more than two numbers.

It is given in the question that

A merchant has 120 litres of oil of one kind, 180 litres of another kind, and 240 litres of the third kind.

He wants to sell the oil by filling the three kinds of equal capacity.

To find the greatest capacity of such a tin , we have to find the highest common factor of the individual capacity.

120=2×2×2×3×5

180=2×2×3×3×5

240=2×2×2×2×3×5

HCF=2×2×3×5=60

The greatest capacity = 60 litres

So the merchant needs to fill 60 litres of all types of oils .

To know more about HCF

brainly.com/question/17240111

#SPJ1

You might be interested in

Solve for f <br> -f+2+4f=8-3f

Fed [463]

In a step by step fashion you solve the problem like so:

Move terms:
3f+2=8-3f

Combine like terms:
3f+3f=8-2

Divide
6f=6

Solution
f=1

4 0

3 years ago

Read 2 more answers

In a class of 6, there are 4 students who are secretly robots. If the teacher chooses 2 students, what is the probability that n

Phoenix [80]

This problem lends itself to the binomial probability approach.

Focus on students who are not secretly robots.

Then P(student is not a robot) = 2/6, or 1/3. Here n=6 and x=2.

Then the desired probability is binompdf(6,1/3, 2), which, by calculator, comes out to 0.329.

7 0

3 years ago

Read 2 more answers

In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find: The length of the angle bisector of a

Aliun [14]

Check the picture below.

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\quad cos(CAB)=\cfrac{5}{13}\implies \measuredangle CAB=cos^{-1}\left( \frac{5}{13} \right)
\\\\\\
\textit{that means that }\measuredangle hAC=\cfrac{cos^{-1}\left( \frac{5}{13} \right)}{2}\impliedby \textit{one of the halves}$

now, notice, for the angle hAC, the hypotenuse is hA, and the adjacent side is CA, therefore,

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(hAC)=\cfrac{5}{hA}\implies hA=\cfrac{5}{cos(hAC)}
\\\\\\
hA=\cfrac{5}{cos\left[ \frac{cos^{-1}\left( \frac{5}{13} \right)}{2} \right]}$

make sure your calculator is in Degree mode, if you need the angle in degrees.