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Rajasthan Board Class 10 (RBSE) Mathematics — practice by topic

Browse 502 questions across 37 Mathematics topics. Choose what you want to revise, then open a question to check its answer and worked solution.

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Mathematics topics

Sequences (74)Binomial Theorem (72)Geometry (72)Probability & Statistics (72)Sequences and Series (72)Vectors & 3D Geometry (72)Probability (58)Correlation and Regression (55)Linear Programming (55)Mean (55)Median (55)Permcomb (55)Permutations and Combinations (55)Quadratic Equations (48)Algebra (46)Logarithms (46)Number Theory (37)Areas Related to Circles (36)Circles (34)Surface Areas and Volumes (34)Triangles (34)Arithmetic Progressions (32)Polynomials (32)Coordinate Geometry (31)Statistics (31)Polygons (30)Real Numbers (30)HCF and LCM (29)Probability Density Function (28)Probability Distribution (28)Some Applications of Trigonometry (20)Intro Trigonometry (19)Linear Equations (18)Pair of Linear Equations in Two Variables (18)Introduction to Trigonometry (13)Pair Of Linear Equations (13)Section Formula (2)

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For any integer $n$, the expression $n^2 - 1$ is always divisible by 8 when $n$ isRead the answer and worked solution →The product of any three consecutive positive integers is always divisible byRead the answer and worked solution →The number of trailing zeros at the end of $50!$ isRead the answer and worked solution →If $\text{HCF}(a, b) = 12$ and $\text{LCM}(a, b) = 180$, then the number of possible ordered pairs $(a, b)$ isRead the answer and worked solution →

Worked practice questions

For any integer $n$, the expression $n^2 - 1$ is always divisible by 8 when $n$ is

  1. any integer
  2. an even integer
  3. a prime number
  4. an odd integer
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Correct answer: D. an odd integer

If $n$ is odd, write $n = 2k+1$. Then $n^2 - 1 = 4k(k+1)$. Since $k(k+1)$ is a product of two consecutive integers it is even so $4k(k+1)$ is divisible by 8. For even $n$, $n^2-1$ is odd, so not divisible by 8.

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The product of any three consecutive positive integers is always divisible by

  1. 4
  2. 12
  3. 9
  4. 6
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Correct answer: D. 6

Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3. Hence the product is divisible by $2 \times 3 = 6$. It need not be divisible by 4 (e.g. $1\cdot2\cdot3 = 6$).

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The number of trailing zeros at the end of $50!$ is

  1. 12
  2. 13
  3. 11
  4. 10
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Correct answer: A. 12

Trailing zeros are governed by the power of 5 in $50!$: $\lfloor 50/5 \rfloor + \lfloor 50/25 \rfloor = 10 + 2 = 12$. (Powers of 2 are more plentiful, so 5 is the limiting factor.)

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