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Number - Whole-Number Operations and Multi-Step Problems

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Understanding the place value system for large whole numbers up to billions is essential for performing accurate operations.

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The Order of Operations (BODMAS/PEMDAS) must be followed for multi-step problems: Brackets, Orders (indices), Division and Multiplication (left to right), and Addition and Subtraction (left to right).

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The Commutative Property states that the order of numbers does not change the result for addition (a+b=b+aa + b = b + a) and multiplication (a×b=b×aa \times b = b \times a).

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The Associative Property allows for regrouping of numbers in addition (a+b)+c=a+(b+c)(a + b) + c = a + (b + c) and multiplication (a×b)×c=a×(b×c)(a \times b) \times c = a \times (b \times c).

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The Distributive Property relates multiplication and addition: a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c).

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Estimation and rounding are used to check the reasonableness of an answer before and after performing complex calculations.

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Inverse operations are used to check accuracy; subtraction is the inverse of addition, and division is the inverse of multiplication.

📐Formulae

a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c)

Dividend=(Divisor×Quotient)+Remainder\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}

B→O→D/M→A/S\text{B} \rightarrow \text{O} \rightarrow \text{D/M} \rightarrow \text{A/S}

💡Examples

Problem 1:

A factory produces 2,4502,450 widgets every day. If the factory operates for 2525 days in a month, how many widgets are produced in total? If 1,2501,250 widgets are found to be defective and discarded, how many widgets remain?

Solution:

2,450×25=61,2502,450 \times 25 = 61,250 61,250−1,250=60,00061,250 - 1,250 = 60,000

Explanation:

First, we multiply the daily production by the number of days to find the total production. Then, we subtract the defective widgets to find the final remaining quantity.

Problem 2:

Solve the following vertical subtraction: 85,000,000−34,567,89285,000,000 - 34,567,892

Solution:

85000000−3456789250432108\begin{array}{r} 85000000 \\ -34567892 \\ \hline 50432108 \end{array}

Explanation:

Align the numbers by place value and subtract column by column starting from the ones place, regrouping (borrowing) as necessary.

Problem 3:

Evaluate the expression: 120+(15×4)÷2−10120 + (15 \times 4) \div 2 - 10

Solution:

120+60÷2−10120 + 60 \div 2 - 10 120+30−10120 + 30 - 10 150−10=140150 - 10 = 140

Explanation:

Using the order of operations (BODMAS): 1. Calculate the brackets (15×4=6015 \times 4 = 60). 2. Perform division (60÷2=3060 \div 2 = 30). 3. Perform addition (120+30=150120 + 30 = 150). 4. Perform subtraction (150−10=140150 - 10 = 140).

Problem 4:

A school has 1,5601,560 students. They want to organize a field trip using buses that can carry 4848 students each. How many buses are needed, and how many empty seats will be on the last bus?

Solution:

1,560÷48=32 remainder 241,560 \div 48 = 32 \text{ remainder } 24 Buses needed =32+1=33= 32 + 1 = 33 Empty seats =48−24=24= 48 - 24 = 24

Explanation:

Divide the total students by the bus capacity. Since there is a remainder of 2424, we need an additional bus (total 3333 buses). The number of empty seats is the capacity minus the students remaining for the last bus.