Question

There are $3$ sharpeners and $7$ erasers. If $10–7=3$ is one of the equations from the fact family of the numbers $3,7,$ and $10,$ what are the remaining three equations that can be formed?

• A. $3–7=10,3+10=7,10+3=7$
• B. 10 – 3 = 7, 3 + 7 = 10, 7 + 3 = 10
• C. $10–3=7,10+7=3,7+10=3$
• D. $10–8=2,2+8=10,8+2=10$

Answer 1

The correct option is B 10 – 3 = 7, 3 + 7 = 10, 7 + 3 = 10

Say: Let us read the question and observe the image.

Ask:
$\bullet$ Just by looking at the image, can you tell me what are the $3$ numbers of the fact family that can be used to frame the equations?
Ans: Yes, $3,7,$ and $10$
$\bullet$ Why do you say so?
Ans: Because in the given number frame, there are $3$ counters that are not crossed and 7 are crossed. There are a total of $10$ counters in the number frame.
$\bullet$ What is the given equation?
Ans: $10–7=3$
$\bullet$ How many more equations can we find?
Ans: 3 more
$\bullet$ How many addition and subtraction equations can be framed?
Ans: $2$ addition and $1$ subtraction equations.
$\bullet$ What will be the other equations?
Ans: $10–3=7,3+7=10,$ and $7+3=10$

Hence, option D is the correct answer.

WB: Use the whiteboard to draw the number frame and write the equations.