Solve: (27-5×3)÷(6×2) + (15×4)÷3 | Order of Operations Challenge

Complete the following exercise:

275362+1543= \frac{27-5\cdot3}{6\cdot2}+\frac{15\cdot4}{3}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:12 Let's solve this problem together.
00:15 Remember, we multiply and divide before we add and subtract.
00:22 First, calculate the multiplication in the denominator.
00:27 Next, do the last multiplication in the numerator.
00:31 Now, let's solve step by step from left to right.
00:38 Work on sixty divided by three before we add.
00:42 Simplify whatever you can.
00:47 And that's how we find the solution!

Step-by-step written solution

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1

Understand the problem

Complete the following exercise:

275362+1543= \frac{27-5\cdot3}{6\cdot2}+\frac{15\cdot4}{3}=

2

Step-by-step solution

According to the order of arithmetic operations, first we place the multiplication exercises within parentheses:

27(53)(62)+(154)3= \frac{27-(5\cdot3)}{(6\cdot2)}+\frac{(15\cdot4)}{3}=

We then solve the exercises within parentheses:

5×3=15 5\times3=15

6×2=12 6\times2=12

15×4=60 15\times4=60

Now we obtain the exercise:

271512+603= \frac{27-15}{12}+\frac{60}{3}=

We solve the numerator of the fraction:

2715=12 27-15=12

We obtain:

1212+603= \frac{12}{12}+\frac{60}{3}=

We solve the fractions:

1212=1 \frac{12}{12}=1

60:3=20 60:3=20

Finally we obtain the exercise:

1+20=21 1+20=21

3

Final Answer

21

Practice Quiz

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\( 100+5-100+5 \)

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