Mathematics - MCQ Practice Questions

Understanding:

Count integers from $1$ to $300$ divisible by neither $3$ nor $5$ using the Inclusion–Exclusion Principle.

Formula:

Step 1: Count multiples of $3$

Step 2: Count multiples of $5$

Step 3: Count multiples of $15$ (both $3$ and $5$)

Step 4: Apply Inclusion–Exclusion

Answer:

The number of integers divisible by neither $3$ nor $5$ is $160$.

Quick Tip:

Alternatively, use Euler's product: $300\times \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=300\times \frac{2}{3}\times \frac{4}{5}=160$.

Understanding:

Find prime numbers $p$ and $q$ such that $p^2-2q^2=1$.

Formula:

Rearrange and test small primes:

Step 1: Test small prime values of $q$

Step 2: Check parity for $q>2$ (odd primes)

If $q$ is odd, $q^2$ is odd, so $2q^2$ is even, $2q^2+1$ is odd. $p^2$ is odd, so $p$ is odd — consistent. But checking shows no further prime solutions for small $q$.

For large $q$, $p=\sqrt{2q^2+1}$ is between $q\sqrt{2}$ and $q\sqrt{2}+1$, making integer solutions rare (Pell equation). The only small prime solution is $p=3,q=2$.

Step 3: Compute the sum

Answer:

The value of $p+q$ is $5$.

Quick Tip:

Whenever a Diophantine equation involves primes, always test $2$ first since it is the only even prime — it often yields the only solution.

Understanding:

The number of trailing zeros in $100!$ equals the highest power of $10$ dividing $100!$, which equals the highest power of $5$ (since there are always more factors of $2$ than $5$).

Formula:

Legendre's formula for the largest power of a prime $p$ dividing $n!$:

Step 1: Apply with $p=5$, $n=100$

Step 2: Sum the terms

Step 3: Confirm factor of $2$ is not limiting

So the number of trailing zeros $=24$.

Answer:

The number of trailing zeros in $100!$ is $24$.

Quick Tip:

Multiples of $25$ contribute two factors of $5$ each, which is why the second term $\lfloor 100/25\rfloor =4$ adds $4$ extra zeros beyond the naive count of $20$.

Understanding:

Find the sum of all two-digit integers $n$ satisfying $n\equiv 3(\mathrm{mod}7)$.

Formula:

Sum of an arithmetic progression with first term $a$, last term $l$, and $k$ terms:

Step 1: Identify the first term

The sequence $3,10,17,24,\dots$ consists of numbers $\equiv 3(\mathrm{mod}7)$.

Verify: $10=7\times 1+3$, so the first two-digit term is $a=10$.

Step 2: Identify the last term

Check downward from $99$:

So the last term is $l=94$.

Step 3: Count the number of terms

Step 4: Compute the sum

Answer:

The sum of all two-digit positive integers leaving remainder $3$ on division by $7$ is $676$.

Quick Tip:

The common difference in such a sequence is always equal to the divisor ($7$ here). Identifying first and last terms in range is the only careful step needed.

Understanding:

Find $(13^{99}+7^{99})(\mathrm{mod}20)$.

Formula:

Use the algebraic identity: for odd exponent $n$,

and compute directly mod $20$.

Step 1: Apply the sum-of-powers divisibility rule

Since $13+7=20$ and the exponent $99$ is odd:

This follows from the factorisation $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots +b^{n-1})$ when $n$ is odd.

Step 2: Verify by computing mod $20$

So both $13$ and $7$ have order $4$ mod $20$.

Answer:

The remainder when $13^{99}+7^{99}$ is divided by $20$ is $0$.

Quick Tip:

Whenever you see $a^n+b^n$ with $a+b$ being the divisor and $n$ odd, the remainder is immediately $0$. This is one of the most useful number theory shortcuts in competitive exams.

Understanding:

We are given $\sin \theta +\cos \theta =\sqrt{2}\cos \theta$ and must find $\tan \theta$.

Formula:

Step 1: Isolate $\sin \theta$.

Step 2: Divide both sides by $\cos \theta$.

Answer:

The value of $\tan \theta$ is $\sqrt{2}-1$.

Quick Tip:

A common trap is to transpose $\cos \theta$ to the other side incorrectly and get $\sqrt{2}+1$. Always isolate $\sin \theta$ first, then divide.

Understanding:

We need to evaluate $\sin 20°\cdot \sin 40°\cdot \sin 80°$.

Formula:

For any angle $\theta$, the following product identity holds:

Step 1: Recognise the pattern with $\theta =20°$.

Step 2: Substitute $\sin 60°=\frac{\sqrt{3}}{2}$.

Answer:

The value of $\sin 20°\cdot \sin 40°\cdot \sin 80°$ is $\frac{\sqrt{3}}{8}$.

Quick Tip:

The identity $\sin \theta \cdot \sin (60°-\theta )\cdot \sin (60°+\theta )=\frac{1}{4}\sin 3\theta$ is extremely useful in competitive exams and worth memorising.

Understanding:

We need to evaluate ${\cos }^{2}10°-\cos 10°\cos 50°+{\cos }^{2}50°$.

Formula:

We use the identity $a^2-ab+b^2=(a-b{)}^2+ab$ and the product-to-sum formula:

Step 1: Use the substitution. Let $a=\cos 10°$, $b=\cos 50°$.

Step 2: Compute $a^2+b^2={\cos }^{2}10°+{\cos }^{2}50°$.

Step 3: Apply sum-to-product: $\cos 20°+\cos 100°=2\cos 60°\cos 40°=2\cdot \frac{1}{2}\cdot \cos 40°=\cos 40°$.

Step 4: Compute $ab=\cos 10°\cos 50°$.

Step 5: Combine.

Answer:

The value of the expression is $\frac{3}{4}$.

Quick Tip:

For expressions of the form ${\cos }^{2}A-\cos A\cos (A+40°)+{\cos }^2(A+40°)$, the result is always $\frac{3}{4}$ — a useful pattern in competitive exams.

Understanding:

We are given $\tan A=\frac{1}{2}$ and $\tan B=\frac{1}{3}$ and need to find $A+B$.

Formula:

Step 1: Substitute the values.

Step 2: Simplify the numerator and denominator.

Step 3: Compute $\tan (A+B)$.

Step 4: Since $A$ and $B$ are acute, $A+B\in (0°,180°)$ and $\tan (A+B)=1$ gives:

Answer:

The value of $A+B$ is $45°$.

Understanding:

We need the general solution of $\sin 2\theta =\cos \theta$.

Formula:

Step 1: Rewrite the equation.

Step 2: Set each factor to zero.

Step 3: Solve $\cos \theta =0$.

Step 4: Solve $\sin \theta =\frac{1}{2}$.

Answer:

The general solution is $\theta =(2n+1)\frac{\pi }{2}$ or $\theta =n\pi +(-1{)}^n\frac{\pi }{6}$.

Quick Tip:

Always factorise trigonometric equations rather than dividing by a trig function — dividing by $\cos \theta$ would lose the solutions where $\cos \theta =0$.

Understanding:

We need to simplify $\frac{\sin 5A-\sin 3A}{\cos 3A+\cos 5A}$.

Formula:

Sum-to-product identities:

Step 1: Apply to the numerator with $C=5A$, $D=3A$.

Step 2: Apply to the denominator with $C=3A$, $D=5A$.

Step 3: Divide.

Answer:

The expression simplifies to $\tan A$.

Quick Tip:

Whenever you see a difference of sines over a sum of cosines (or vice versa) with arguments in arithmetic progression, sum-to-product identities reduce the expression to a single trig ratio in one step.

Understanding:

Given $3\sin \theta +4\cos \theta =5$, we must find $3\cos \theta -4\sin \theta$.

Formula:

For any expression of the form $a\sin \theta +b\cos \theta$, the maximum value is $\sqrt{a^2+b^2}$.

Step 1: Square both expressions and add them.

Step 2: Substitute the known value.

Answer:

The value of $3\cos \theta -4\sin \theta$ is $0$.

Quick Tip:

The identity $(a\sin \theta +b\cos \theta {)}^2+(a\cos \theta -b\sin \theta {)}^2=a^2+b^2$ is a powerful tool. When the first expression attains the maximum $\sqrt{a^2+b^2}$, the second must equal $0$.

Understanding:

We need to evaluate the product $\cos 1°\cdot \cos 2°\cdot \cos 3°\cdots \cos 90°$.

Formula:

A product is zero if any one of its factors equals zero.

Step 1: Identify whether any factor in the product is zero.

Step 2: Since $\cos 90°$ is one of the factors in the product, the entire product equals zero regardless of the other factors.

Answer:

The value of the product is $0$.

Quick Tip:

Before computing any complex product, always check whether any single factor equals zero — that immediately makes the entire product zero.

Understanding:

We need to find ${\alpha }^3+{\beta }^3$ where $\alpha$ and $\beta$ are roots of $x^2-5x+6=0$.

Formula:

Using the identity:

Step 1: Identify sum and product of roots.

By Vieta's formulas:

Step 2: Substitute into the identity.

Answer:

The value of ${\alpha }^3+{\beta }^3$ is $35$.

Quick Tip:

A common error is computing $(\alpha +\beta {)}^3$ alone without subtracting $3\alpha \beta (\alpha +\beta )$. Always use the full identity.

Understanding:

Given $x+\frac{1}{x}=3$, we need to find $x^4+\frac{1}{x^4}$.

Formula:

Repeatedly use the squaring identity:

Step 1: Find $x^2+\frac{1}{x^2}$.

Step 2: Find $x^4+\frac{1}{x^4}$.

Answer:

The value of $x^4+\frac{1}{x^4}$ is $47$.

Quick Tip:

Square the given expression step by step rather than expanding directly. Each squaring increases the power by a factor of $2$.

Understanding:

We need to count real solutions of $x^2-5|x|+6=0$.

Since $x^2=|x{|}^2$, substituting $t=|x|\ge 0$ simplifies the equation.

Formula:

Let $t=|x|$. Then:

Step 1: Solve the quadratic in $t$.

Both values are non-negative, so both are valid.

Step 2: Recover $x$ from each value of $t$.

Step 3: Count total real solutions.

There are $4$ distinct real values: $\{-3,-2,2,3\}$.

Answer:

The equation has $4$ real solutions.

Quick Tip:

Whenever $|x|$ appears, substitute $t=|x|\ge 0$. Each positive value of $t$ gives two real values of $x$; $t=0$ gives only one.

Understanding:

For the quadratic $2x^2-3x+k=0$ to have equal roots, the discriminant must be zero.

Formula:

For equal roots, the discriminant condition is:

Step 1: Apply the condition.

Verification:

Substitute back: $\Delta =9-8\cdot \frac{9}{8}=9-9=0$. ✓

Answer:

The value of $k$ for equal roots is $\frac{9}{8}$.

Quick Tip:

Equal roots $\iff$ discriminant $=0$. Never confuse this with real roots ($\Delta \ge 0$) or distinct roots ($\Delta >0$).

Understanding:

We solve ${\log }_{2}x+{\log }_{4}x+{\log }_{8}x=11$.

Convert all logarithms to base $2$.

Formula:

Change-of-base rule:

Step 1: Express each term in base $2$.

Step 2: Substitute and simplify. Let $u={\log }_{2}x$.

Step 3: Recover $x$.

Answer:

The value of $x$ is $2^6=64$.

Quick Tip:

Convert all log terms to a single base before combining. The LCD of the denominators saves time when adding the fractions.

Understanding:

The roots of $x^2-3x+2=0$ are $\alpha$ and $\beta$. We need the equation with roots $2\alpha$ and $2\beta$.

Formula:

If the new roots are $2\alpha$ and $2\beta$, the new equation is:

Step 1: Find sum and product of original roots.

Step 2: Compute new sum and product.

Step 3: Write the new equation.

Verification:

Original roots: $x^2-3x+2=0\Rightarrow (x-1)(x-2)=0$, so $\alpha =1,\beta =2$.

New roots: $2,4$. Check: $(x-2)(x-4)=x^2-6x+8=0$. ✓

Answer:

The required quadratic equation is $x^2-6x+8=0$.

Quick Tip:

To obtain an equation with roots $k$ times those of $f(x)=0$, replace $x$ with $\frac{x}{k}$ and simplify. Here, replace $x$ with $\frac{x}{2}$: $\frac{x^2}{4}-\frac{3x}{2}+2=0\Rightarrow x^2-6x+8=0$.

Understanding:

Given $a+b+c=0$, evaluate $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}$.

Formula:

Combine fractions over a common denominator $abc$:

Step 1: Use the identity when $a+b+c=0$.

There is a known algebraic identity:

Step 2: Substitute.

Answer:

The value of the expression is $3$.

Quick Tip:

The identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ immediately gives $a^3+b^3+c^3=3abc$ when $a+b+c=0$. Memorise it — it appears very frequently.