CBSE Class 10 Mathematics · Ch 1: Real Numbers

Topics in this test

Difficulty: 3 easy · 9 medium · 18 hard

Sample questions with answers

The first 5 questions of this free test.

1. Q1. Match each number in Column I with its correct prime factorisation in Column II. Column I: (a) $140$, (b) $156$, (c) $3825$ Column II: (p) $2^{2} \times 3 \times 13$, (q) $3^{2} \times 5^{2} \times 17$, (r) $2^{2} \times 5 \times 7$

   • A. (a)-(r), (b)-(p), (c)-(q)
   • B. (a)-(r), (b)-(q), (c)-(p)
   • C. (a)-(q), (b)-(p), (c)-(r)
   • D. (a)-(p), (b)-(r), (c)-(q)
   Solution

   Factorising gives $140 = 2^{2} \times 5 \times 7$, $156 = 2^{2} \times 3 \times 13$ and $3825 = 3^{2} \times 5^{2} \times 17$, so (a)-(r), (b)-(p), (c)-(q).

2. Q2. Consider the following statements: (i) The sum of a rational number and an irrational number is always irrational. (ii) The product of two irrational numbers is always irrational. (iii) $\sqrt{2}$ is an irrational number. Which of the statements are correct?

   • A. (i) and (ii) only
   • B. (ii) and (iii) only
   • C. (i) and (iii) only
   • D. (i), (ii) and (iii)
   Solution

   Statements (i) and (iii) are true, but (ii) is false since $\sqrt{2} \times \sqrt{2} = 2$ is rational.

3. Q3. The largest number which divides $70$ and $125$, leaving remainders $5$ and $8$ respectively, is:

   • A. $9$
   • B. $13$
   • C. $65$
   • D. $25$
   Solution

   The required number exactly divides $70-5=65$ and $125-8=117$, so it is $\text{HCF}(65,117)=13$.

4. Q4. Three bells toll at intervals of $9$, $12$ and $15$ minutes respectively. If they all toll together at $6{:}00$ a.m., at what time will they next toll together?

   • A. $9{:}30$ a.m.
   • B. $9{:}00$ a.m.
   • C. $7{:}30$ a.m.
   • D. $8{:}00$ a.m.
   Solution

   They toll together after $\text{LCM}(9,12,15)=180$ minutes $=3$ hours, so the next time is $6{:}00$ a.m. $+ 3$ hours $= 9{:}00$ a.m.

5. Q5. For how many natural numbers $n$ can the number $12^{n}$ end with the digit $0$?

   • A. For no value of $n$
   • B. Only for $n = 5$
   • C. For every value of $n$
   • D. For every even value of $n$
   Solution

   A number ends in $0$ only if it has both $2$ and $5$ as factors, but $12^{n} = 2^{2n} \times 3^{n}$ has no factor of $5$, so it can never end with $0$.

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