109 practice questions covering A Square and A Cube, organised into 2 topics.
109 questions include a written explanation.
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Assertion (A): $3600$ is a perfect square. Reason (R): A number ending in two zeros is always a perfect square.
ABoth A and R are true, and R is the correct explanation of A
BBoth A and R are true, but R is not the correct explanation of A
CA is true, but R is false
DA is false, but R is true
Show answer and explanation
Correct answer: C - A is true, but R is false
A is true because $60 \times 60 = 3600$. R is false: $200$ ends in two zeros but is not a square, since $14^2 = 196$ and $15^2 = 225$. So a true A sits with a false R, which rules out the first two choices and the last one.
Q2
medium
What is $1^3 + 2^3 + 3^3 + 4^3 + 5^3$?
A$3375$
B$55$
C$125$
D$225$
Show answer and explanation
Correct answer: D - $225$
The sum is $1 + 8 + 27 + 64 + 125 = 225$, and this equals $(1 + 2 + 3 + 4 + 5)^2 = 15^2$. The value $125$ is only the last cube, $55$ is the sum of the squares, and $3375$ is $15^3$.
Q3
medium
What is the value of $1^3 + 2^3 + 3^3$?
A$36$
B$216$
C$14$
D$18$
Show answer and explanation
Correct answer: A - $36$
Adding gives $1 + 8 + 27 = 36$, which equals $(1 + 2 + 3)^2 = 6^2$. The value $14$ is the sum of the squares, $216$ is the cube of $6$, and $18$ comes from adding $1 + 8 + 9$ with the last cube mistaken for a square.
Q4
easy
What is $45^2$?
A$1625$
B$2020$
C$2025$
D$1620$
Show answer and explanation
Correct answer: C - $2025$
Multiply the tens digit by the next whole number, $4 \times 5 = 20$, then write $25$ after it to get $2025$. Using $4 \times 4 = 16$ gives the wrong $1625$. Every square of a number ending in $5$ must end in $25$, so $2020$ and $1620$ cannot be right.
Q5
easy
The sum of the first $n$ odd numbers is always equal to which of these?
A$2n$
B$n^2$
C$n + 1$
D$n^3$
Show answer and explanation
Correct answer: B - $n^2$
Adding odd numbers from $1$ builds a square: $1 + 3 = 4$, $1 + 3 + 5 = 9$, so the sum of the first $n$ of them is $n^2$. Doubling gives $2n$, which fails already at $n = 3$ since $2 \times 3 = 6$, not $9$. Neither $n^3$ nor $n + 1$ matches these values.
Q6
easy
From $25$ subtract $1$, then $3$, then $5$, then $7$, then $9$. Nothing is left. What does this show?
AFive subtractions were needed, so $\sqrt{25} = 25$
BFour subtractions were needed, so $\sqrt{25} = 4$
CFive subtractions were needed, so $\sqrt{25} = 5$
DNine subtractions were needed, so $\sqrt{25} = 9$
Show answer and explanation
Correct answer: C - Five subtractions were needed, so $\sqrt{25} = 5$
Exactly five odd numbers were used, and $1 + 3 + 5 + 7 + 9 = 25$, so $\sqrt{25} = 5$. The count of subtractions is five, not nine, and $9$ is merely the last odd number used. Stopping after four would leave $9$ still to subtract.
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