Class 10 Real Numbers Mock Test (CBSE)

Topics in Real Numbers

• Fundamental theorem of arithmetic
• HCF and LCM using prime factorisation
• Revisiting irrational numbers
• Proofs of irrationality

Real Numbers practice questions with answers

1. Q1Multiple choice·1 mark
   Which of the following numbers is an irrational number?

   1. (a)$5 - \sqrt{16}$
   2. (b)$2 + \sqrt{3}$
   3. (c)$\frac{\sqrt{50}}{\sqrt{2}}$
   4. (d)$\sqrt{4} + \sqrt{9}$
   Show answer and step marks

   1. Correct option: (b) $2 + \sqrt{3}$+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   The HCF of $2^3 \times 3^2 \times 5$ and $2 \times 3^3 \times 7$ is:

   1. (a)$2$
   2. (b)$18$
   3. (c)$54$
   4. (d)$7560$
   Show answer and step marks

   1. Correct option: (b) $18$, since HCF takes the smallest power of each common prime: $2^1 \times 3^2 = 18$+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   The exponent of $2$ in the prime factorisation of $144$ is:

   1. (a)2
   2. (b)4
   3. (c)6
   4. (d)3
   Show answer and step marks

   1. Correct option: (b) $4$ (since prime factorisation of $144$ is $2^4 \times 3^2$)+1

   Answer: (b)
4. Q4Short answer·2 marks
   Let $p = 2^4 \times 3^2 \times 7$ and $q = 2^4 \times 3 \times 11$. (a) Name the prime that is a factor of $p$ but not a factor of $q$. [1] (b) Is $q$ divisible by $7$? Give a reason. [1]
   Show answer and step marks

   1. (a) The primes in $p$ are $2, 3, 7$ and the primes in $q$ are $2, 3, 11$, so the prime in $p$ but not in $q$ is $7$.+1
   2. (b) By the fundamental theorem of arithmetic, the prime factorisation of $q$ is unique: $q = 2^4 \times 3 \times 11$ has no prime factor $7$, so $7$ does not divide $q$. Hence $q$ is not divisible by $7$.+1

   Answer: (a) $7$; (b) No, $q = 2^4 \times 3 \times 11$ has no prime factor $7$, so it is not divisible by $7$.
5. Q5Short answer·2 marks
   A square floor tile has an area of $10$ cm$^2$. The side length of the tile is $\sqrt{10}$ cm. (a) Between which two consecutive whole numbers does the side length of the tile lie? [1] (b) Is the side length of the tile a rational number? Give a reason. [1]
   Show answer and step marks

   1. (a) Since $9 < 10 < 16$, we get $\sqrt{9} < \sqrt{10} < \sqrt{16}$, i.e. $3 < \sqrt{10} < 4$, so the side lies between 3 and 4.+1
   2. (b) Prime factorisation of $10$ is $2 \times 5$, which has no repeated pair of primes, so $10$ is not a perfect square and $\sqrt{10}$ is irrational; the side is not a rational number.+1

   Answer: (a) Between 3 and 4. (b) No, $\sqrt{10}$ is irrational because $10 = 2 \times 5$ is not a perfect square.
6. Q6Long answer·5 marks
   Priya wants to locate $\sqrt{2}$ on a number line without using a calculator. She traps $\sqrt{2}$ between two decimals by squaring them. (a) Verify, by squaring, that $\sqrt{2}$ lies between $1.414$ and $1.415$. [2] (b) Write a terminating decimal with four decimal places that lies strictly between $1.414$ and $\sqrt{2}$. [2] (c) Using the bounds in (a), write $\sqrt{2}$ correct to two decimal places. [1]
   Show answer and step marks

   1. (a) $1.414^2 = 1.999396 < 2$, so $1.414 < \sqrt{2}$.+1
   2. (a) $1.415^2 = 2.002225 > 2$, so $\sqrt{2} < 1.415$. Hence $1.414 < \sqrt{2} < 1.415$.+1
   3. (b) $1.4142^2 = 1.99996164 < 2$, so $1.4142 < \sqrt{2}$.+1
   4. (b) Since $1.414 < 1.4142 < \sqrt{2}$, the decimal $1.4142$ lies strictly between $1.414$ and $\sqrt{2}$ (any decimal from $1.4141$ to $1.4142$ also works).+1
   5. (c) Since $1.414 < \sqrt{2} < 1.415$, $\sqrt{2}$ is less than $1.415$, so it rounds down to $1.41$ correct to two decimal places.+1

   Answer: (a) $1.414 < \sqrt{2} < 1.415$ is verified. (b) $1.4142$ (or $1.4141$). (c) $\sqrt{2} \approx 1.41$

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