Quantitative Aptitude PYQs

Calculation:

Given: \( p^x = q^y = r^z = k \) and \( x, y, z \) are in GP.

Taking logs: \( \ln(p^x) = \ln(q^y) = \ln(r^z) = \ln k \Rightarrow x\ln p = y\ln q = z\ln r = A \)

Thus: \( \ln p = \frac{A}{x},\quad \ln q = \frac{A}{y},\quad \ln r = \frac{A}{z} \)

Let \( x, y, z \) be GP: \( x=a,\ y=ar,\ z=ar^2 \)

Then: \( \ln p = \frac{A}{a},\quad \ln q = \frac{A}{ar},\quad \ln r = \frac{A}{ar^2} \)

So \( \ln p,, \ln q,, \ln r \) are in GP (reciprocals of a GP form a GP).

Hence Statement II is true.

For Statement I (whether \( p, q, r \) are in AP):

from \( \ln p = \frac{A}{x},\ \ln q = \frac{A}{y},\ \ln r = \frac{A}{z} \), there is no general condition implying \( q = \frac{p+r}{2} \);

logarithms of reciprocals of a GP do not enforce an AP on the original bases. Hence Statement I is false.

Hence the correct answer is II only.

Calculation:

Statement I:

\( \text{L.H.S.} = A - B = A \cap B^{c} \)

\( \text{R.H.S.} = B^{c} - A^{c} = B^{c} \cap (A^{c})^{c} = B^{c} \cap A \)

\( \therefore ; \text{L.H.S.} = \text{R.H.S.} \)

So, Statement I is correct.

Statement II:

\( \text{L.H.S.} = A - B^{c} = A \cap (B^{c})^{c} = A \cap B \)

\( \text{R.H.S.} = A^{c} - B = A^{c} \cap B^{c} = U - (A \cup B) \)

\( \therefore ; \text{L.H.S.} \ne \text{R.H.S.} \)

So, Statement II is incorrect.

Hence, the correct answer is Option 1.

Calculation:

Given: \(y=x!,; z=(2x)!\) and \(\dfrac{z}{y}=120\)

⇒ \(\dfrac{(2x)!}{x!}=120\)

Check small integer values of \(x\):

For \(x=3\), \(\dfrac{(2\cdot 3)!}{3!}=\dfrac{6!}{3!}=\dfrac{720}{6}=120\)

So, \(x=3\).

Now, \((3x)!=(3\cdot 3)!=9!=362880\)

Hence the correct answer is 362880.

Calculation:

The number of trailing zeros in \( n! \) depends on the number of factors of 5 in its prime factorization.

Number of trailing zeros \( = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots \)

We are given that the number of trailing zeros is exactly 2.

Checking possible values of n :

For \( n = 10 \):

⇒ \( \left\lfloor \frac{10}{5} \right\rfloor = 2,; \left\lfloor \frac{10}{25} \right\rfloor = 0 \Rightarrow 2 \)

For \( n = 11, 12, 13, 14 \):

⇒ \( \left\lfloor \frac{n}{5} \right\rfloor = 2,; \left\lfloor \frac{n}{25} \right\rfloor = 0 \Rightarrow 2 \)

For \( n = 15 \):

\( \left\lfloor \frac{15}{5} \right\rfloor = 3 \Rightarrow \) 3 zeros (exceeds)

Hence, the valid values of \( n \) are:

\( n = 10, 11, 12, 13, 14 \)

Therefore, the number of possible values of n  is 5.

Calculation:

Given that:

\( (10+\log_{10}x),\ (10+\log_{10}y),\ (10+\log_{10}z) \text{ are in A.P.} \)

Using the condition of A.P.:

\( 2(10+\log_{10}y)=(10+\log_{10}x)+(10+\log_{10}z) \)

Simplifying:

\( 20+2\log_{10}y=20+\log_{10}x+\log_{10}z \)

\( 2\log_{10}y=\log_{10}x+\log_{10}z \)

This implies:

\( \log_{10}y=\frac{\log_{10}x+\log_{10}z}{2} \)

Hence, the A.M. of \( \log_{10}x \) and \( \log_{10}z \) is \( \log_{10}y \).

So, statement II is correct.

Now,

\( 2\log_{10}y=\log_{10}x+\log_{10}z \Rightarrow \log_{10}y^2=\log_{10}(xz) \)

\( \Rightarrow y^2=xz \)

Therefore, the G.M. of \(x\) and \(z\) is \(y\), not \(y^2\).

So, statement I is incorrect.

Hence, the correct answer is Option 2.

Calculation:

The sum of the first \( n \) odd natural numbers is given by:

\( 1 + 3 + 5 + \cdots + (2n-1) = n^2 \)

Given that the sum is:

\( n^2 = 12345678987654321 \)

Taking square root on both sides:

\( n = \sqrt{12345678987654321} \)

Now,

\( 111111111^2 = 12345678987654321 \)

\( \Rightarrow n = 111111111 \)

Hence, the correct answer is 111111111.

Calculation:

Given two Arithmetic Progressions (APs):

\(\text{AP}_1: 19, 21, 23, \)...... up to 110 terms

\( a_1 = 19, d_1 = 2, T_n = 19 + 2(n-1) = 2n + 17 \) (up to n = 110,where } T₁₁₀ = 237

\(\text{AP}_2: 19, 22, 25, \)...... up to 75 terms

\( a_2 = 19, d_2 = 3, S_m = 19 + 3(m-1) = 3m + 16 \) up to m = 75 , where  S₇₅ = 241

To find the common terms

\( 2n + 17 = 3m + 16 \Rightarrow 2n - 3m = -1 \Rightarrow 2n \equiv 2 \pmod 3 \Rightarrow n \equiv 1 \pmod 3 \)

\( n = 3k + 1 \), Now, we calculate the common terms

\( T_{3k+1} = 2(3k+1) + 17 = 6k + 19 \)

Now, to ensure the terms are within the limits, we use

\( 6k + 19 \le 237 \Rightarrow 6k \le 218 \Rightarrow k \le 36 \)

Values of k  from 0 to 36 give 37 common terms

Hence, the correct answer is 37.

Calculation:

We are given that: \( \alpha = \frac{-1 + \sqrt{3}}{2} \), which is the cube root of unity.

Using the properties of cube roots of unity:

⇒ \( \alpha^3 = 1 \)

⇒ \( 1 + \alpha + \alpha^2 = 0 \)

⇒ \( \alpha^2 = \alpha - 1 \)

Now, the given expression is:

⇒ \( (1 + \alpha^{19} - \alpha^{35})^{100} - (1 - 3\alpha^{25} + \alpha^{38})^{50} \)

Simplifying \( \alpha^{19} \) and \( \alpha^{35} \)

⇒ \( \alpha^{19} = \alpha \) since  \(19 \mod 3 = 1 \)

⇒ \( \alpha^{35} = \alpha^2\) Since \(35 \mod 3 = 2 \)

Thus, \( 1 + \alpha^{19} - \alpha^{35} = 1 + \alpha - \alpha^2 \)

Using \( \alpha^2 = \alpha - 1 \), we get:

⇒ \( 1 + \alpha - (\alpha - 1) = 1 + \alpha - \alpha + 1 = 2 \)

So, \( (1 + \alpha^{19} - \alpha^{35})^{100} = 2^{100} \)

Simplifying \( \alpha^{25} \) and \( \alpha^{38} \)

⇒ \( \alpha^{25} = \alpha \) since \(25 \mod 3 = 1\)

⇒ \( \alpha^{38} = \alpha^2 \) Since \( 38 \mod 3 = 2 \)

Thus, \( 1 - 3\alpha^{25} + \alpha^{38} = 1 - 3\alpha + \alpha^2 \)

Using \( \alpha^2 = \alpha - 1 \), we get:

⇒ \( 1 - 3\alpha + (\alpha - 1) = 1 - 3\alpha + \alpha - 1 = -2\alpha \)

So, \( (1 - 3\alpha^{25} + \alpha^{38})^{50} = (-2\alpha)^{50} = 2^{50} \alpha^{50} \)

Since \( \alpha^3 = 1 \), we have \( \alpha^{50} = \alpha^2 \).

Thus, \( (1 - 3\alpha^{25} + \alpha^{38})^{50} = 2^{50} \alpha^2 \)

Subtracting the two terms

⇒ \( 2^{100} - 2^{50} \alpha^2 \)

Since \( \alpha^2 = \alpha - 1 \), we get:

⇒ \( 2^{100} - 2^{50} (\alpha - 1) = 2^{100} - 2^{50} \alpha + 2^{50} \)

The value of the expression is \( 0 \)

Hence, the correct answer is Option 3.

Calculation:

We are asked to find the remainder when \( 5^{99} \) is divided by 13.

Let’s find the pattern of powers of 5 modulo 13:

\( 5^1 = 5 \mod 13 = 5 \)

\( 5^2 = 25 \mod 13 = 12 \)

\( 5^3 = 125 \mod 13 = 8 \)

\( 5^4 = 625 \mod 13 = 1 \)

So the powers repeat every 4 steps: \( 5^4 \equiv 1 \mod 13 \)

Now reduce the exponent modulo 4:

\( 99 \mod 4 = 3 \Rightarrow 5^{99} \equiv 5^3 \mod 13 \)

\( 5^3 = 125 \mod 13 = 8 \)

Hence, the remainder when \( 5^{99} \) is divided by 13 is 8.

Calculation:

We are asked to find the value of the determinant of the inverse of the matrix:

\(\begin{bmatrix} -4 & -5 \ 2 & 2 \end{bmatrix}\)

Compute the determinant of the original matrix:

⇒ \(\text{det}(A) = (-4)(2) - (-5)(2) = -8 + 10 = 2\)

Use the property of inverse determinant:

⇒ \(\text{det}(A^{-1}) = \frac{1}{\text{det}(A)} = \frac{1}{2}\)

Hence, the value of the determinant of the inverse is 1/2

Calculation:

Let total students = \(45\)

Students who like cricket = \(34\)

Students who like football = \(26\)

Each student likes at least one of the two games.

Using set theory identity:

\( |C \cup F| = |C| + |F| - |C \cap F| \)

\( 45 = 34 + 26 - |C \cap F| \Rightarrow |C \cap F| = 60 - 45 = 15 \)

Students who like exactly one game:

\( |C| + |F| - 2|C \cap F| = 34 + 26 - 2(15) = 60 - 30 = 30 \)

Hence, the number of students who like exactly one game is \(30\).

Calculation:

Given system of equations:

\( 15y - 10x + 50 = 0 \)

\( 2x - 3y - 5 = 0 \)

Rearranging the first equation:

⇒ \( 15y - 10x + 50 = 0 \Rightarrow -10x + 15y = -50 \)

\( \Rightarrow 2x - 3y = 10 \quad \text{(dividing by } -5) \)

Rearranging the second equation:

⇒ \( 2x - 3y - 5 = 0 \Rightarrow 2x - 3y = 5 \)

Now compare both equations:

⇒ \( 2x - 3y = 10 \) and \( 2x - 3y = 5 \)

This is a contradiction: same left-hand side equals two different right-hand sides.

Hence, the system is inconsistent and has no solution.

Calculation:

Given:

\( \left( \frac{1 - i}{1 + i} \right)^{2m} \left( \frac{1 + i}{1 - i} \right)^{2n} = 1 \)

Simplify the base:

\( \frac{1 - i}{1 + i} = \frac{(1 - i)^2}{(1 + i)(1 - i)} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i \)

Hence:

\( \left( \frac{1 - i}{1 + i} \right)^{2(m - n)} = (-i)^{2(m - n)} = 1 \)

Since \( (-i)^2 = -1 \), we have \( (-i)^{2k} = (-1)^k \). Thus \( (-i)^{2(m - n)} = 1 \) iff \( m - n \) is even.

Smallest positive value: \( m - n = 2 \).

Hence, the correct answer is Option  2.

Calculation:

\( D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & 4 \end{vmatrix} \)

⇒ \( D_2 = \begin{vmatrix} 1 & 7 & 1 \ 1 & 16 & 3 \ 1 & 22 & 4 \end{vmatrix} \)

⇒ \( D_2 = 1 \cdot \begin{vmatrix} 16 & 3 \ 22 & 4 \end{vmatrix} - 7 \cdot \begin{vmatrix} 1 & 3 \ 1 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 16 \ 1 & 22 \end{vmatrix} \)

⇒ \( \begin{vmatrix} 16 & 3 \ 22 & 4 \end{vmatrix} = (16 \cdot 4) - (3 \cdot 22) = 64 - 66 = -2 \)

⇒ \( \begin{vmatrix} 1 & 3 \ 1 & 4 \end{vmatrix} = (1 \cdot 4) - (3 \cdot 1) = 4 - 3 = 1 \)

⇒ \( \begin{vmatrix} 1 & 16 \ 1 & 22 \end{vmatrix} = (1 \cdot 22) - (16 \cdot 1) = 22 - 16 = 6 \)

⇒ \( D_2 = 1 \cdot (-2) - 7 \cdot (1) + 1 \cdot (6) \)

 \( D_2 = -2 - 7 + 6 = -3 \)

Hence the correct answer is Option 2.

Calculation:

Given: Non-singular matrices \( A \) and \( B \)

Statement I: \( (AB)^{-1} = A^{-1}B^{-1} \)

This is incorrect. The correct identity is \( (AB)^{-1} = B^{-1}A^{-1} \)

Statement II: \( (BA)(AB)^{-1} = I \)

This is incorrect. Using \( (AB)^{-1} = B^{-1}A^{-1} \), we get:

\( (BA)(AB)^{-1} = BA \cdot B^{-1}A^{-1} = B A B^{-1} A^{-1} \neq I \)

Statement III: \( (AB)^T = A^T B^T \)

This is incorrect. The transpose of a product reverses the order:

\( (AB)^T = B^T A^T \)

Hence, None of the statements are correct.

Calculation:

Given determinant:

\( \begin{vmatrix} a & b & c \ l & m & n \ p & q & r \end{vmatrix} \)

Let this be denoted as \( D \). Now analyze each option:

Option 1: \( \begin{vmatrix} a & b & c \ p & q & r \ l & m & n \end{vmatrix} \)

This swaps rows 2 and 3 → one transposition → \( = -D \)

Option 2: \( \begin{vmatrix} l & m & n \ a & b & c \ p & q & r \end{vmatrix} \)

This swaps rows 1 and 2 → one transposition → \( = -D \)

Option 3: \( \begin{vmatrix} p & q & r \ a & b & c \ l & m & n \end{vmatrix} \)

This is a 3-cycle permutation of rows → even permutation → \( = D \)

Option 4: \( \begin{vmatrix} a & p & l \ b & q & m \ c & r & n \end{vmatrix} \)

This permutes columns, not rows → determinant value changes → not equal to \( D \)

Hence Option 3 is correct.

Calculation:

Let \( \omega \) be a cube root of unity, so:

\( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \)

Given:

⇒ \( x = a + b \)

⇒ \( y = a\omega + b\omega^2 \)

⇒ \( z = a\omega^2 + b\omega \)

Compute \( y + z \)

⇒ \( y + z = a(\omega + \omega^2) + b(\omega^2 + \omega) = (a + b)(\omega + \omega^2) = -x \)

Use identity:

⇒ \( y^2 + z^2 = (y + z)^2 - 2yz = x^2 - 2yz \)

Compute \( yz \)

⇒ \( yz = (a\omega + b\omega^2)(a\omega^2 + b\omega) \)

\( = a^2\omega^3 + ab(\omega^2\omega^2 + \omega\omega) + b^2\omega^3 \)

\( = a^2 + ab(\omega + \omega^2) + b^2 = a^2 + b^2 - ab \)

Final expression:

⇒ \( x^2 + y^2 + z^2 = x^2 + (x^2 - 2yz) = 2x^2 - 2yz \)

\( = 2(a + b)^2 - 2(a^2 + b^2 - ab) = 2(a^2 + 2ab + b^2) - 2(a^2 + b^2 - ab) = 6ab \)

Hence, the correct answer is Option 1.

Calculation:

We are asked to find how many 4-digit numbers divisible by 4 can be formed using digits \( {1, 2, 3, 4} \) without repetition.

we know A number is divisible by 4 if its last two digits form a number divisible by 4.

List all 2-digit endings from the given digits that are divisible by 4:

\( 12, 24, 32 \) are valid (since they are divisible by 4 and use distinct digits).

For each valid ending, choose the remaining two digits from the unused ones and arrange them in the first two positions:

Ending \( 12 \) → remaining digits: \( {3, 4} \) → 2! = 2 arrangements

Ending \( 24 \) → remaining digits: \( {1, 3} \) → 2! = 2 arrangements

Ending \( 32 \) → remaining digits: \( {1, 4} \) → 2! = 2 arrangements

Total valid numbers = \( 2 + 2 + 2 = 6 \)

Hence, the correct answer is option 2.

Calculation:

Let \( p = a + b + c \) be the perimeter of triangle ABC.

Given determinant:

\( D = \begin{vmatrix} p + c & a & b \ c & p + a & b \ c & a & p + b \end{vmatrix} \)

Apply row operations:

\( R_1 \leftarrow R_1 - R_2 \) and \( R_2 \leftarrow R_2 - R_3 \)

New matrix:

⇒ \( \begin{vmatrix} p & -p & 0 \ 0 & p & -p \ c & a & p + b \end{vmatrix} \)

Factor out \( p \) from first two rows:

⇒ \( D = p^2 \cdot \begin{vmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ c & a & p + b \end{vmatrix} \)

Expand along first row:

\( = 1 \cdot \begin{vmatrix} 1 & -1 \ a & p + b \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & -1 \ c & p + b \end{vmatrix} \)

\( = (p + b + a) + c = 2p \)

⇒ \( D = p^2 \cdot 2p = 2p^3 \)

Hence the correct answer is Option 2.

Calculation:

The general term in the expansion of \( (1 + x)^{100} \) is:

\( T_k = \binom{100}{k} x^k \)

Here, \( \binom{100}{k} \) is the binomial coefficient, which represents the coefficient of \( x^k \).

We are asked to find the value of \( k \) for which \( \binom{100}{k} \) is maximum.

It is known that in the expansion of \( (1 + x)^n \), the binomial coefficients increase until \( k = \left\lfloor \frac{n}{2} \right\rfloor \) and then decrease.

For \( n = 100 \), the maximum occurs at \( k = 50 \).

The greatest coefficient is the coefficient of \( x^{50} \)

Hence, the correct answer is Option 4.