If P is at \(\sqrt{256}-\sqrt{121}\) on the number line, what is the coordinate of P?
Answer and explanation
Correct answer: 5
The governing concept is evaluation of exact square roots before performing subtraction. Since 256 = 16², √256 = 16; similarly, 121 = 11², so √121 = 11. Therefore the coordinate is √256 − √121 = 16 − 11 = 5. Hence option A is correct. Option B results from adding 16 and 11 rather than subtracting them. Option C does not follow from either the square roots or their difference, while option D is only the first square root and ignores the second term. Because both radicands are perfect squares, no approximation is needed, and the exact coordinate is 5.
Frequently asked questions
What is the correct answer to this question?
5
Why is this the correct answer?
The governing concept is evaluation of exact square roots before performing subtraction. Since 256 = 16², √256 = 16; similarly, 121 = 11², so √121 = 11. Therefore the coordinate is √256 − √121 = 16 − 11 = 5. Hence option A is correct. Option B results from adding 16 and 11 rather than subtracting them. Option C does not follow from either the square roots or their difference, while option D is only the first square root and ignores the second term. Because both radicands are perfect squares, no approximation is needed, and the exact coordinate is 5.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Representing real numbers on the number line.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
