125 practice questions on The World of Numbers, a chapter in the NCERT Class 9 Mathematics textbook. Every question includes a written explanation. Try the samples below - the answer stays hidden until you ask for it.
A square park has each side $3$ units long. What is the exact length of the straight path across it from one corner to the opposite corner?
A$\sqrt{6}$ units
B$6$ units
C$3$ units
D$\sqrt{18}$ units
Show answer and explanation
Correct answer: D - $\sqrt{18}$ units
The diagonal satisfies $d^2=3^2+3^2=18$, so $d=\sqrt{18}$, which is about $4.24$ units. Multiplying the side by $2$ instead of squaring gives $\sqrt{6}$. A diagonal of $6$ units would be as long as two full sides laid end to end, and $3$ units is just the side itself.
Q2
hard
Exactly one of these four numbers is irrational. Which one is it?
A$\frac{22}{7}$
B$\sqrt{3}$
C$0.\overline{18}$
D$\sqrt{2.25}$
Show answer and explanation
Correct answer: B - $\sqrt{3}$
No whole number squares to $3$, so $\sqrt{3}=1.732\ldots$ is irrational. The number $\frac{22}{7}$ is a ratio of integers, and it is only an approximation of $\pi$, not $\pi$ itself. A repeating decimal is rational, since $0.\overline{18}=\frac{18}{99}=\frac{2}{11}$. And $\sqrt{2.25}=1.5$, because $1.5\times 1.5=2.25$.
Q3
medium
Assertion (A): $\frac{7}{40}$ has a terminating decimal expansion. Reason (R): A fraction in its lowest terms terminates only when its denominator has a prime factor other than $2$ and $5$.
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, since $40=2^3\times 5$ and $\frac{7}{40}=0.175$. (R) states the test backwards: a fraction terminates only when the denominator has NO prime factor other than $2$ and $5$. As written, (R) would even predict that $\frac{7}{40}$ does not terminate, so it is false.
Q4
medium
Suppose someone insists that $\frac{5}{0}$ has some value $k$. Which consequence shows that no such $k$ can exist?
AIt would need $k$ to be $0$, and $0$ is already used up
BIt would make $0$ a rational number
CIt would need $k$ to be $5$, which is not allowed
DIt would need $k\times 0=5$, but any number multiplied by $0$ is $0$
Show answer and explanation
Correct answer: D - It would need $k\times 0=5$, but any number multiplied by $0$ is $0$
Division undoes multiplication, so $\frac{5}{0}=k$ means $k\times 0=5$. Every product with $0$ is $0$, never $5$, so no such $k$ exists. Nothing forces $k$ to be $0$ or $5$; the contradiction is with the multiplication rule itself. Zero is already a rational number, as $\frac{0}{1}$.
Q5
medium
Statement I: Every natural number is an integer. Statement II: Every integer is a natural number. Which is correct?
ANeither I nor II is true
BOnly II is true
COnly I is true
DBoth I and II are true
Show answer and explanation
Correct answer: C - Only I is true
The naturals $1,2,3,\ldots$ all appear in the list of integers, so Statement I holds. Statement II fails, because $-5$ and $0$ are integers that are not natural numbers. Exactly one statement is true, so saying both hold, or that neither holds, is wrong.
Q6
easy
In the classical proof, we assume $\sqrt{2}=\frac{p}{q}$ with $p$ and $q$ integers sharing no common factor. What does squaring both sides give?
A$p=2q$
B$p^2=q^2$
C$p^2=2q^2$
D$2p^2=q^2$
Show answer and explanation
Correct answer: C - $p^2=2q^2$
Squaring $\sqrt{2}=\frac{p}{q}$ gives $2=\frac{p^2}{q^2}$, and multiplying both sides by $q^2$ gives $p^2=2q^2$. Losing the factor $2$ gives $p^2=q^2$, which would make $\sqrt{2}=1$. Putting the $2$ on the wrong side gives $2p^2=q^2$, and $p=2q$ comes from not squaring at all.
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