Quantitative Aptitude PYQs

Concept:

Matrix Multiplication:

• Matrix multiplication is defined only when the number of columns in the first matrix is equal to the number of rows in the second matrix.
• For two matrices \( A_{m \times n} \) and \( B_{n \times p} \), their product \( AB \) will result in a matrix \( C_{m \times p} \).
• In general, the product of \( X \) and \( Y \) is defined if and only if the number of columns in \( X \) equals the number of rows in \( Y \).

Calculation:

 

We have the following matrix operations:

\( A_{m \times n} B_{n \times p} = (AB)_{m \times p} \)

Now, consider the matrix multiplications:

\([Z_{3 \times 2} . Y_{2\times3}].X_{3 \times 3}] = [ZYX]_{3 \times 3}\)

\( Y_{2 \times 3} [X_{3 \times 3} Z_{3 \times 2}] = [YXZ]_{2 \times 2} \)

Finally, we observe that:

\( X_{3 \times 3} [Y_{2 \times 3} Z_{3 \times 2}] = X_{3 \times 3} [YZ]_{2 \times 2} \)

Conclusion:

\(\text{No. of columns in X} \neq \text{No. of rows in (YZ)}\)

Hence, X(YZ) is not defined.

∴ the Correct answer is Option 4

Concept:

Infinite and Finite Sets:

• A finite set contains a limited number of elements that can be counted to an end.
• An infinite set contains endlessly many elements; you cannot finish counting them.
• Between any two different real numbers there are infinitely many irrational numbers.
• A set described without a lower bound (for integers) typically includes infinitely many negative members.

 

Calculation:

Given

Statements:

I. All irrational numbers between √12 and √15

II. All odd integers less than 1000

Step wise check:

⇒ For I: The interval between √12 and √15 is a normal real interval.

⇒ For I: Any real interval contains infinitely many irrational numbers.

⇒ So I is true.

⇒ For II: "Less than 1000" gives no lower limit.

⇒ For II: This includes ..., -5, -3, -1, 1, 3, ..., 999.

⇒ For II: We can continue to more negative odd integers without end.

⇒ So II is not finite (it is infinite).

∴ Only statement I is correct.

Concept:

Counting 4-digit Numbers with All Odd Digits:

• Each digit in a 4-digit number can be one of the odd digits: 1, 3, 5, 7, or 9.
• There are 5 choices for each digit since there are 5 odd digits available.
• For a 4-digit number, we need to select a digit for each of the 4 positions independently.
• The total number of 4-digit numbers with all odd digits is the product of choices for each digit.

Calculation:

Given,

Choices for each digit = 5 (as the digits are 1, 3, 5, 7, 9)

Number of digits, n = 4

The total number of 4-digit numbers with all odd digits is,

\(5^4\) =  5 × 5 ×5 × 5 = 625

∴ 625 4-digit numbers have all digits as odd.5×5×5×5 5×5×5×\there

∴ The total number of 4-digit numbers with all odd digits is 625.

Explanation:

(1+ω -ω²)¹⁰⁰+ (1-ω +ω2)100

= (– ω² – ω² )¹⁰⁰ + (–ω – ω)¹⁰⁰  {∵ 1 + ω  + ω² = 0}

= 2¹⁰⁰(ω²⁰⁰ + ω¹⁰⁰)

= 2¹⁰⁰(ω² + ω)     (∵ ω³ = 1)

= –2¹⁰⁰ 

∴ Option (d) is correct

Explanation:

Given AB = O

Let  A = \(\left[\begin{array}{lll} a & b \ c & d \end{array}\right]\)

B = \(\left[\begin{array}{lll} p & q \ r & s \end{array}\right]\)

Now AB = \(\left[\begin{array}{lll} a & b \ c & d \end{array}\right] \left[\begin{array}{lll} p & q \ r & s \end{array}\right] = \left[\begin{array}{lll} 0 & 0 \ 0 & 0 \end{array}\right]\)

\(\left[\begin{array}{lll} ap+br & aq+bs \ cp+dr & cq+ds \end{array}\right]= \left[\begin{array}{lll} 0 & 0 \ 0 & 0 \end{array}\right]\)

Comparing both sides, we get

⇒ ap + br = 0...(1)

⇒ aq + bs = 0...(2)

⇒ cp + dr = 0...(3)

⇒ cq + ds = 0...(4)

From Eqs.

(i) and (iii), we get:

⇒ (ad – bc)p = 0 and (ad – bc)r = 0 

From Eqs.

(ii) and (iv), we get:

⇒ (ad – bc) q = 0 and (ad – bc)s = 0 

Now, if A is non-singular

ad – bc ≠ 0

So p = q = r = s = 0.

Then,  B is the null matrix.

∴ Option (c) is correct.

7Concept:

Binomial Expansion:

• The binomial theorem is used to expand expressions of the form \((1 + x)^n\).
• The general term in the expansion of \((1 + x)^n\) is given by \(T_k = C(n, k) \cdot x^k\), where ⁿC_k is the binomial coefficient.
• To find the coefficient of x^3, we identify the corresponding terms from the expansion and set the coefficient equal to 35.

 

Calculation:

Given the expansion of \((1 + x)^p \cdot (1 + x)^q\), we have:

The coefficient of x³ in  the expansion is 35.

We use the binomial expansion formula for the term x³

⇒ (p+ q)_c3 = 35 = ⁷C₃

⇒ p+q =7

∴ the correct answer is Option C

Explanation:

Let 'a' and 'd' be the first term and common difference of AP

(p + q)^th Term = T_p+q 

= a + (p + q - 1)......(1)

Given

p. T_p = q.T_p

⇒ p.[ a + (p -1) d] = q. [ a + (q - 1) d]

⇒ (p - q) [ a + (p + q -1) d] =0

⇒ [ a + (p + q -1) d] =0           (∵ p ≠ q)

 ⇒Tp+q = 0 ......(From i)

∴ Option (a) is correct

Concept:

Arithmetic Progression (AP):

• The terms are in AP if the difference between consecutive terms is constant.
• The common difference is given by: \(d = q - p = r - q\)

Geometric Progression (GP):

• The terms are in GP if the ratio between consecutive terms is constant.
• The common ratio is given by: \(r = q \times p\)

Calculation:

Given

p = ln(x),

q = ln(x3)  = 3 lnx

and r = ln(x5) = 5 lnx

Clearly, p - q  = r - q = \(2lnx\)

⇒ p, q, r are in AP

Also \(\frac{q}{p} ≠ \frac{r}{p} \because \frac{q}{p} = 3 , \frac{r}{p} = \frac{5}{3}\)

∴ p, q, r can never be in GP

∴ Option (c) is correct.

Explanation:

Given

 \(\rm z=\frac{1}{3}\begin{vmatrix}i&2i&1\\ 2i&3i&2\\ 3&1&3\end{vmatrix}\) 

\(Z = \frac{1}{3}[[ (i9 – 2) – 2i (6i – 6) 1(2i – 9i)]\)

\(Z = \frac{1}{3}[3+3i] =1 + i \)

|Z| = \(√{1^2+1^2} =\) √2 

∴ Option (b) is correct.

Explanation:

Let I =  \(\rm \Sigma_{n=1}^{20}(i^{n-1}+i^n+i^{n+1})\)

⇒ I = \(\rm \Sigma_{n=1}^{20}(i^n (\frac{1}{i} +1+ i)\)

⇒ I = \(\rm \Sigma_{n=1}^{20}(i^n (1)\)

⇒ I = \(i^1+ i^2+i^3+i^4i^5+........+ i^{20}\)

⇒ I = \(5.(i^1+ i^2+i^3+i^4) = 0\) ( {∵  i + i² + i³ + i⁴ = 0} 

∴ Option (b) is correct.

Explanation:

Given,

x, y, z are in GP

Thus, 

⇒ y² =xz

⇒ \(2lny = lnx + lnz\)

lnx, ln y and lnz are in AP

⇒ (1 + lnx), (1 + lny) and (1 + lnz) are in AP

⇒ \(\frac{1}{lnx}, \frac{1}{lny}, \frac{1}{lnz}\) are in HP

Concept:

Determinant and Column Operations:

• The determinant of a matrix is a scalar value representing area, volume, or invertibility.
• Column operations like \( C_1 \rightarrow C_1 + C_2 + C_3 \) can simplify a determinant.
• If one column (or row) becomes a linear combination of others or zero, the determinant becomes 0.
• \( \omega \) is a cube root of unity with identity: \( 1 + \omega + \omega^2 = 0 \).

 

Calculation:

Let \( I = \begin{bmatrix} 1+\omega & 1+\omega^2 & \omega + \omega^2 \ 1 & \omega & \omega^2 \ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \end{bmatrix} \)

⇒ Apply column operation: \( C_1 \rightarrow C_1 + C_2 + C_3 \)

⇒ First column becomes:

\( C_1 = \begin{bmatrix} (1+\omega) + (1+\omega^2) + (\omega + \omega^2) \ 1 + \omega + \omega^2 \ \frac{1}{\omega} + \frac{1}{\omega^2} + 1 \end{bmatrix} \)

⇒ Using identity: \( 1 + \omega + \omega^2 = 0 \)

⇒ First column becomes: \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \)

⇒ New matrix:

\( I = \begin{bmatrix} 0 & 1+\omega^2 & \omega + \omega^2 \ 0 & \omega & \omega^2 \ 0 & \frac{1}{\omega^2} & 1 \end{bmatrix} \)

⇒ Since first column is all zero,

∴ Determinant I = 0

Explanation:

Given

S_n = n(2n + 1), S_n–1 = (n – 1)(2n – 1) 

n^th term = S_n – S_{n – 1}

= n(2n + 1) – (n – 1)(2n–1)

=  4n – 1

∴ Option (a) is correct.

Explanation:

Word 'INDIA' contains three vowels (I, I and A) and two consonants (N and D).

Three vowels (I, I and A) can be arranged at odd position in \(\frac{3!}{2!} = 3 \) ways

2 consonants(N and D) can be arranged in 2! ways = 2 ways. 

Thus, required numbers of ways = (3) (2) = 6.

∴ Option (b) is correct.

Concept:

Permutations and Arrangements:

• A permutation is an arrangement of objects in a specific order. The number of ways to arrange 'n' objects is given by the factorial of 'n' denoted by n!.
• For a set of distinct vowels and consonants, the total number of ways to arrange them can be calculated by multiplying the factorials of the number of vowels and consonants separately.

Calculation:

Given the word 'EQUATION', we have:

• 5 vowels: A, E, I, O, U
• 3 consonants: Q, T, N

⇒ Number of ways = (3!) . (5!)

⇒ Numbers of ways = (3!) . (5!)

⇒ Total required number of ways = (3!) . (5!) + (3!) . (5!)

= 1440

∴ Option (c) is correct.

Calculation:

n is a root of the equation x2 + px + m = 0

So, n² + pn + n = 0

 .....

(1) Also, m is the root of the equation of x² + px + n = 0

So, m² + pm + n = 0 ..... (2) Now Eq.

(i) – Eq. (ii), we get

⇒ \((n^2 – m^2) + p(n – m) – (n – m) = 0 \)

⇒ \((n – m) [(n + m) + p – 1] = 0 \)

⇒ \(n + m + p – 1 = 0\) ..... ( n ≠ m)

⇒ n + m + p = 1

∴ Option (3) is correct.

Explanation:

Given,

Number of courses, n > 4

The student needs to choose (n - 2) courses out of n courses, with 2 courses being compulsory

∴Total courses is n – 2 and total remaining courses that can be chosen from (n – 2) courses is ( n – 4).

∴ Required number of ways ^(n-2) C _(n-4).

⇒ \(\frac{(n-2)(n-3)}{2}\)

∴ Option d is correct.

Concept:

Determinant of a 3×3 Matrix:

• The determinant of a 3×3 matrix can be found by the cofactor expansion method.
• It is expressed as: \( \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \)
• If the matrix has a common factor across rows or columns, it can be factored out to simplify calculations.
• \( \sum_{n=1}^{k} f(n) \) is a summation notation indicating summing function values from n = 1 to n = k.
• Important identities used:
• \( \sum_{n=1}^{k} n = \frac{k(k+1)}{2} \)
• \( \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \)
• \( \sum_{n=1}^{k} n^3 = \left( \frac{k(k+1)}{2} \right)^2 \)

 

Calculation:

Given,

\( D_n = \begin{bmatrix} n & 20 & 30 \ n^2 & 40 & 50 \ n^3 & 60 & 70 \ \end{bmatrix} \)

⇒ Extract common factors 20 and 10:

\( D_n = (20)(10) \begin{bmatrix} n & 1 & 3 \ n^2 & 2 & 5 \ n^3 & 3 & 7 \ \end{bmatrix} \)

⇒ Expand determinant using first column:

\( D_n = 200(-n + 2n^2 - n^3) \)

Now,

\( \sum_{n=1}^{4} D_n = 200 \sum_{n=1}^{4} (-n + 2n^2 - n^3) \)

⇒ \( = 200 \left[ -\frac{4(5)}{2} + 2 \cdot \frac{4(5)(9)}{6} - \left( \frac{4(5)}{2} \right)^2 \right] \)

⇒ \( = 200 (-10 + 60 - 100) \)

∴ The value is -10000.

Concept:

Matrix Multiplication and Properties:

• Matrix multiplication involves the dot product of rows and columns.
• A null matrix is a matrix in which all elements are zero.
• An identity matrix is a square matrix with 1's on the diagonal and 0's elsewhere.
• For matrices P and Q, PQ = QP does not generally hold unless P and Q commute.

Matrix Definitions:

• Null Matrix: A matrix where all elements are zero.
• Identity Matrix: A square matrix with 1's on the main diagonal and 0's elsewhere.

 

Calculation:

\(\rm P=\begin{bmatrix}0&c&-b\\ -c&0&a\\ b&-a&0\end{bmatrix}\ and\ \rm Q=\begin{bmatrix}a^2&ab&ac\\ ab&b^2&bc\\ ac&bc&c^2\end{bmatrix}\)

⇒ PQ = \(=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\ \)

⇒QP = \(=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}\ \)

Then PQ = QP

∴ Option (c) is correct

Explanation:

The determinant of any skew-symmetric matrix with odd order is always zero

∴ Option (b) is correct.