Olympiad Maths Prep

Track / Stage 3 / 81 of 260 #81 of 2000

Problem 81

AMC 10/12, early questions
Algebra Difficulty 3.2 Find the answer

Calculate: (1)38÷(2)+4×5(-1)^{3}-8\div (-2)+4\times |-5|.

Official solution

To calculate (1)38÷(2)+4×5(-1)^{3}-8\div (-2)+4\times |-5|, we follow the order of operations, which is parentheses, exponents, multiplication and division (from left to right), and addition and subtraction (from left to right). Let's break it down step by step:

1. Calculate the exponent: (1)3=1(-1)^{3} = -1.
2. Divide: 8÷(2)=48\div (-2) = -4.
3. Calculate the absolute value: 5=5|-5| = 5.
4. Multiply: 4×5=204\times 5 = 20.
5. Add and subtract in order from left to right: 1+(4)+20-1 + (-4) + 20.

Putting it all together:

(1)38÷(2)+4×5=14+20=1+(4)+20=5+20=15. \begin{aligned} (-1)^{3}-8\div (-2)+4\times |-5| &= -1 - 4 + 20 \\ &= -1 + (-4) + 20 \\ &= -5 + 20 \\ &= 15. \end{aligned}

However, it seems there was a mistake in my initial calculation. Let's correct that by following the correct order and operations accurately:

1. Exponent: (1)3=1(-1)^{3} = -1.
2. Division correctly: 8÷(2)=48\div (-2) = -4 (This was incorrect; the correct operation is 8÷(2)=48 \div (-2) = 4).
3. Absolute value: 5=5|-5| = 5.
4. Multiplication: 4×5=204 \times 5 = 20.
5. Adding all together correctly: 1+4+20-1 + 4 + 20.

Correcting the steps:

(1)38÷(2)+4×5=1+4+20=3+20=23. \begin{aligned} (-1)^{3}-8\div (-2)+4\times |-5| &= -1 + 4 + 20 \\ &= 3 + 20 \\ &= 23. \end{aligned}

So, the correct final answer is 23\boxed{23}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.