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Bihar Board Class 8 Mathematics · Ch 1: A Square and a Cube

Bihar Board Class 8 Mathematics · Ch 1: A Square and a Cube is an online chapter test in the Bihar Board Class 8 pattern, covering A Square and a Cube. Attempt it on SeatBeat with a timer like the real exam, then see a detailed solution for every question and your score and weak-topic analysis.

30
Questions
20
Minutes
30
Marks
None
Negative marking
Attempt this test free

Topics in this test

TopicQuestions
A Square and a Cube30

Difficulty: 3 easy · 9 medium · 18 hard

Sample questions with answers

The first 5 questions of this free test.

  1. Q1. To find the square root of $81$ by repeatedly subtracting consecutive odd numbers ($1, 3, 5, \dots$) until the result is $0$, how many subtractions are needed?

    • A. $40$
    • B. $8$
    • C. $81$
    • D. $9$
    Solution

    Subtracting consecutive odd numbers from a perfect square reaches $0$ after $n$ steps, where $n$ is the square root. Since $81=9^2$, it takes $9$ subtractions.

  2. Q2. The triangular numbers are $1, 3, 6, 10, 15, \dots$. Adding the two consecutive triangular numbers $10$ and $15$ gives a number that is the square of which number?

    • A. $5$
    • B. $25$
    • C. $4$
    • D. $6$
    Solution

    The sum of two consecutive triangular numbers is always a square number. Here $10+15=25=5^2$, so it is the square of $5$.

  3. Q3. Consider these statements about square numbers. (i) A perfect square ends in an even number of zeros. (ii) The square of an even number is always even. (iii) A perfect square can end in the digit $8$. Which statements are correct?

    • A. Only (i) and (ii)
    • B. Only (i)
    • C. Only (ii) and (iii)
    • D. All of (i), (ii) and (iii)
    Solution

    Statements (i) and (ii) are true. Statement (iii) is false because no square ends in $8$.

  4. Q4. Read the two statements. Assertion (A): $1+3+5+7+9+11+13+15+17+19 = 100$. Reason (R): The sum of the first $n$ odd numbers is always $n^2$. Choose the correct option.

    • A. A is true, but R is false
    • B. A is false, but R is true
    • C. Both A and R are true, but R is not the correct explanation of A
    • D. Both A and R are true, and R is the correct explanation of A
    Solution

    There are $10$ odd numbers, so their sum is $10^2=100$, making A true. R correctly states the rule that gives this result, so R explains A.

  5. Q5. The sum $31+33+35+37+39+41$ is equal to the cube of which number?

    • A. $36$
    • B. $5$
    • C. $7$
    • D. $6$
    Solution

    A cube $n^3$ can be written as the sum of $n$ consecutive odd numbers. This sum has $6$ terms and equals $216=6^3$, so the number is $6$.

Frequently asked questions

How many questions are in the Bihar Board Class 8 Mathematics · Ch 1: A Square and a Cube?

30 questions for 30 marks, to be answered in 20 minutes.

Is there negative marking in the Bihar Board Class 8 Mathematics · Ch 1: A Square and a Cube?

No — wrong answers are not penalised in this test.

Which topics does the Bihar Board Class 8 Mathematics · Ch 1: A Square and a Cube cover?

A Square and a Cube (30 questions).

Is the Bihar Board Class 8 Mathematics · Ch 1: A Square and a Cube free?

Yes — this test is free. Create a free SeatBeat account and attempt it online; you get the solutions and your score analysis after submitting.

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Bihar Board Class 8 Mathematics · Ch 1: A Square and a Cube — Free Online Mock Test | SeatBeat