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Tables and Shares - Word Problems on Multiplication and Division

Grade 4CBSE

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

🔑Concepts

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Multiplication as Repeated Addition: Multiplication is a shortcut for adding the same number many times. For example, if you have 4 plates with 3 cookies each, it is 3+3+3+33 + 3 + 3 + 3, which is 3×4=123 \times 4 = 12. Visually, imagine 4 separate circles, each containing 3 small dots.

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Arrays and Grids: An array is an arrangement of objects in rows and columns. A 5×45 \times 4 array has 5 rows and 4 columns, forming a rectangular shape. The total number of items is found by multiplying the number of rows by the number of columns: 5×4=205 \times 4 = 20.

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Division as Equal Sharing: Division is used to split a large group into smaller, equal parts. If you have 20 marbles to share among 5 friends, you give each friend the same amount until the marbles are gone. Visually, this is like drawing 5 large boxes and placing one marble in each box repeatedly until all 20 are distributed.

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Division as Repeated Subtraction: Division can be seen as subtracting the same number over and over again until you reach zero. For 12÷312 \div 3, you subtract 3 from 12 four times (12−3=9,9−3=6,6−3=3,3−3=012 - 3 = 9, 9 - 3 = 6, 6 - 3 = 3, 3 - 3 = 0), so the answer is 4.

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The Division House and Parts: In a division problem like 15á3=515 \div 3 = 5, the number being divided (15) is the Dividend, the number you divide by (3) is the Divisor, and the answer (5) is the Quotient. Visually, the Dividend sits inside the 'division house' bracket, the Divisor sits outside to the left, and the Quotient is written on the roof.

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Fact Families: Multiplication and division are inverse (opposite) operations. If you know 6×3=186 \times 3 = 18, you automatically know 18÷3=618 \div 3 = 6 and 18÷6=318 \div 6 = 3. This is often represented as a 'fact triangle' with the largest number at the top and the two factors at the bottom corners.

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Remainders: Sometimes, a number cannot be shared perfectly into equal groups. The amount left over is called the Remainder (RR). For example, if you share 10 candies between 3 children, each gets 3 candies and 1 candy is left over. This is written as 10á3=3 Remainder 110 \div 3 = 3 \text{ Remainder } 1.

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Multiplying by 10 and 100: When multiplying a number by 10, shift the digits one place to the left and add a zero at the end (25×10=25025 \times 10 = 250). When multiplying by 100, add two zeros (25×100=250025 \times 100 = 2500). This looks like moving the number across a place-value chart.

📐Formulae

Total Value=Number of Groups×Items per Group\text{Total Value} = \text{Number of Groups} \times \text{Items per Group}

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

Items per Group=Total ValueáNumber of Groups\text{Items per Group} = \text{Total Value} \div \text{Number of Groups}

Number of Groups=Total ValueáItems per Group\text{Number of Groups} = \text{Total Value} \div \text{Items per Group}

💡Examples

Problem 1:

A box contains 12 packets of crayons. Each packet has 8 crayons. If a teacher buys 5 such boxes for her class, how many crayons does she have in total?

Solution:

Step 1: Find the number of crayons in one box. 12 packets×8 crayons per packet=96 crayons12 \text{ packets} \times 8 \text{ crayons per packet} = 96 \text{ crayons}. Step 2: Find the total number of crayons in 5 boxes. 96 crayons×5 boxes=480 crayons96 \text{ crayons} \times 5 \text{ boxes} = 480 \text{ crayons}. Total = 480480 crayons.

Explanation:

We first use multiplication to find the contents of one box and then multiply that result by the total number of boxes to find the grand total.

Problem 2:

Rohan has 145 flower seeds. He wants to plant them in rows such that each row has 6 seeds. How many full rows can he plant, and how many seeds will be left over?

Solution:

We need to divide the total seeds by the number of seeds per row. Calculation: 145÷6145 \div 6 144÷6=24144 \div 6 = 24 (since 6×20=1206 \times 20 = 120 and 6×4=246 \times 4 = 24) 145−144=1145 - 144 = 1 Quotient = 2424, Remainder = 11. Full rows = 2424, Leftover seeds = 11.

Explanation:

This is a division problem where the quotient represents the number of complete groups (rows) and the remainder represents the leftover items that don't fit into a full group.