Quantitative Aptitude PYQs

Concept: 

i² = - 1, i3 = - i

Calculation:

Given: x = 1 + i    ----(i)

On squaring both side, we get 

⇒ x² = 1 + i² + 2i

⇒ x2 = 1 - 1 + 2i

⇒ x2 = 2i    ----(ii)

Again On squaring both side, we get

⇒ x⁴ = 4i²

⇒ x4 = -4    ----(iii)

Now, we have to find the value of x6 + x4 + x2 + 1

⇒ x⁶ + x⁴ + x² + 1

⇒ (x2)³ + x4 + x2 + 1

Now from (i), (ii) and (iii) we get

⇒ (2i) ³ + (- 4) + (2i) + 1

⇒ -8i - 4 + 2i + 1 = -6i - 3

∴ The value of  x6 + x4 + x2 + 1 is (-6i - 3).

Calculation:

Let, \(y = 2 + \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{2 + \ldots \infty }}}}}}\)

\(\Rightarrow y = 2 + \frac{1}{y}\)

⇒ y² - 2y - 1 = 0

By comparing the equation y² - 2y - 1 = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = 1, b = -2 and c = - 1.

\(\Rightarrow y = \frac{{2 \pm \sqrt {4 + 4;} }}{2} = 1 \pm \sqrt 2 \)

 \(\because y = 2 + \frac{1}{y}\) ⇒ y > 2.

Hence, y = 1 + √2

Concept:

_{\({;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}};and;{;^n}{C_r} = ;\frac{{n!}}{{r! × \left( {n - r} \right)!}};\)}

\(\frac{{{;^n}{P_r}}}{{{;^n}{C_r}}} = r!\)

Calculation:

Given: P (n, r) = 2520 and C (n, r) = 21

As we know that, \(\frac{{{;^n}{P_r}}}{{{;^n}{C_r}}} = r!\)

⇒ P (n, r) / C (n, r) = 2520 / 21 = 120 = r!

⇒ r = 5.

As we know that, \({;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\) 

\(⇒ {;^n}{P_5} = \frac{{n!}}{{\left( {n - 5} \right)!}} = 2520\)

\(⇒ \frac{{n(n-1)(n-2)(n-3)(n-4)(n-5)!}}{{\left( {n - 5} \right)!}} = 2520\)

⇒ n(n - 1)(n - 2)(n - 3)(n - 4) = 7 × 6 × 5 × 4 × 3 

On comparing both side, 

⇒ n = 7

⇒ C (n + 1, r + 1) = C (8, 6)

\(⇒ {;^8}{C_6} = ;\frac{{8!}}{{6! × \left( {8 - 6} \right)!}} = 28\)

Concept:

In the binomial expansion of (a + b) ⁿ, there are total n + 1 terms.

Calculation:

⇒ (1 + 2x + x²) ⁵ + (1 + 4y + 4y²) ⁵ = (1 + x) ^2 × 5 

• (1 + 2y)2 × 5

 = (1 + x) ¹⁰ + (1 + 2y)¹⁰

As we know that, in the binomial expansion of (a + b) ⁿ, there are total n + 1 terms.

⇒ The no. of terms in the binomial expansion of (1 + x) ¹⁰ and (1 + 2y)¹⁰ is 11

∴ The no. of terms in the given binomial expansion is 11 + 11 - 1 = 21

 

Mistake Points

We have subtracted 1 from (11 + 11) just because there is only one term common the expression of the binomial which is 1.

Concept:

The general term in the binomial expansion of (a + b) ⁿ is given by:

T_(r-1) = \(^nC_r \times a^{n-r}\times b^r\)

In the expansion of (a + b) ⁿ :

I. The middle term is ^{\(\left( {\frac{{\rm{n}}}{2} + 1} \right)th\)} term if n is even.

II. If n is odd the two middle terms are: \(\left( {\frac{{{\rm{n}} + 1}}{2}} \right){\rm{th;and}};\left( {\frac{{n + 1}}{2} + 1} \right)th\) terms.

Calculation:

Given:

The middle term in the expansion of ^{\({\left( {{x^2} + \frac{1}{x}} \right)^{2n}}\)} is 184756x¹⁰.

As 2n is even

⇒ The middle term is \(\left( {\frac{{2{\rm{n}}}}{2} + 1} \right) = ;\left( {n + 1} \right);th\) term

As we know that, The general term in the binomial expansion of (a + b) ⁿ is given by: \({T_{r + 1}} = {;^n}{C_r} \times {a^{n - r}} \times {b^r}\)

\(\Rightarrow {T_{n + 1}} = {;^{2n}}{C_n} \times {x^{4n - 2n}} \times {\left( {\frac{1}{x}} \right)^n} \= ;{;^{2n}}{C_n} \times {x^n} = 184756 \times {x^{10}}\)

⇒ n = 10

Concept:

Let A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p. Similarly, the product BA will exist if and only if q = m.

Calculation:

Given: \(A = \left( {\begin{array}{*{20}{c}} 1&2\ 2&3\ 3&4 \end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}} 1&2\ 2&1 \end{array}} \right)\)

So, order of the matrix A is 3 × 2 whereas the order of the matrix A is 2 × 2.

As we know that, if A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p

Here, n = 2 = p ⇒ The product AB exists.

As we know that, if A and B be any two matrices of order m × n and p × q respectively. Then the product BA exists if and only if q = m.

Here, q = 2 and m = 3 ⇒ m ≠ q ⇒ The product BA does not exists.

Concept:

The no. of zeros present in a n! = The highest power of 5 in \(\left[ {\frac{n}{5}} \right] + \left[ {\frac{n}{{25}}} \right] + \left[ {\frac{n}{{125}}} \right]\)

Calculation:

Given: n! has 17 zeros

As we know that, the no. of zeros present in a n! = the highest power of 5 in \(\left[ {\frac{n}{5}} \right] + \left[ {\frac{n}{{25}}} \right] + \left[ {\frac{n}{{125}}} \right]\)

If n = 95, then \(\left[ {\frac{{95}}{5}} \right] + \left[ {\frac{{95}}{{25}}} \right] + \left[ {\frac{{95}}{{125}}} \right] = 23 \ne 17\)

Hence, n ≠ 95

If n = 85, then \(\left[ {\frac{{85}}{5}} \right] + \left[ {\frac{{85}}{{25}}} \right] = 20 \ne 17\)

Hence, n ≠ 85

If n = 80, then \(\left[ {\frac{{80}}{5}} \right] + \left[ {\frac{{80}}{{25}}} \right] = 19 \ne 17\)

Hence, n ≠ 80

Calculation:

Given: A ∪ B = {x|(x - a)(x - b) > 0, where a < b}

Here, if we take any value greater than the b 

Let say (c > b)

Now, if we put x = c in the above inequalities we get 

(c - a) (c - b)

Since, c > b > a 

So, (c - a) (c - b) > 0 

So, x > b satisfy the given inequalities 

Now, if we take a value between a and b 

Let say d: a < d < b 

Now, on putting x = d, we get 

(d - a) (d - b) = -ve × +ve = -ve

According to the condition we get 

(d - a) (d - b) < 0 

So, a < x < b doesn't satisfy the above inequalities 

Now, if we take a value between less than a 

Let say e: e < a

Now, on putting x = e in the inequality we get 

(e - a) (e - b) = -ve × -ve = +ve

So, x < a satisfy the given inequalities 

Hence, The correct relation is A = {x|x < a} and B = {x|x > b}.

Concept:

In the binomial expansion of (a + b)ⁿ, the term which does not involve any variable is said to be an independent term.

The general term in the binomial expansion of (a + b)ⁿ is given by: \({T_{r + 1}} = {;^n}{C_r} \times {a^{n - r}} \times {b^r}\)

Calculation:

Given: The constant term in the expansion of \({\left( {\sqrt x - \frac{k}{{{x^2}}}} \right)^{10}}\) is 405

i.e the independent term of \({\left( {\sqrt x - \frac{k}{{{x^2}}}} \right)^{10}}\) is 405.

Let (r + 1)^th term be the independent term.

\(\Rightarrow {T_{r + 1}} = {;^{10}}{C_r} \times {\left( {\sqrt x } \right)^{10 - r}} \times {\left( {\frac{{ - k}}{{{x^2}}}} \right)^r} = ;{;^{10}}{C_r} \times {\left( { - k} \right)^r} \times {\left( x \right)^{\frac{{10 - 5r}}{2}}};\)

∵ (r + 1)th term is the independent term

\(\Rightarrow \frac{{10 - 5r}}{2} = 0 \Rightarrow r = 2\)

\(\Rightarrow {;^{10}}{C_2} \times {\left( { - k} \right)^2} = 405 \Rightarrow 45{k^2} = 405\)

⇒ k = ± 3

Concept:

C (n, r - 1) + C (n, r) = C (n + 1, r)

Calculation:

⇒ C(47, 4) + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) + C(47, 3)

= {C(47, 4) + C(47, 3)} + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3)

As we know that,

C (n, r - 1) + C (n, r) = C (n + 1, r)

⇒ {C(47, 4) + C(47, 3)} + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3)

= {C (48, 4) + C(48, 3)} + C(51, 3) + C(50, 3) + C(49, 3)

⇒ {C (48, 4) + C(48, 3)} + C(51, 3) + C(50, 3) + C(49, 3)

= { C(49, 4) + C(49, 3)} + C(51, 3) + C(50, 3)

⇒ {C(49, 4) + C(49, 3)} + C(51, 3) + C(50, 3)

= {C(50, 4) + C(50, 3)} + C(51, 3)

⇒ {C(50, 4) + C(50, 3)} + C(51, 3)

= { C(51, 4) + C(51, 3)}

= C(52, 4)

Concept:

1. If a, b and c are in A.P 

• Then ka, kb, kc are in AP
• eg. 2, 4, 6 are in A.P then (2 × 2, 4 × 2, 6 × 2) = (4, 8, 12) are also in A.P

2. If a, b and c are in A.P 

• Then k - a, k - b, k - c are in AP
• eg. 2, 4, 6 are in A.P then (8 - 2, 8 - 4, 8 - 6) = (6, 4, 2) are also in A.P

3. If a, b and c are in A.P 

• Then \(\frac{a}{k},\frac{b}{k},\frac{c}{k}\) are in AP
• eg. 2, 4, 6 are in A.P then (2/2, 4/2, 6/2) = (1, 2, 3) are also in A.P
• Hence, All the options are correct.

Concept:

If a₁, a₂, ….,a_n forms an AP then the nth term of an AP is given by : a_n = a + (n - 1) × d, where a is the first term and d is the common difference.

Calculation:

Two digit numbers which are divisible by 4 are:

12, 16, 20, ..., 96 forms an AP with,

First term (a) = 12,

Common difference (d) = 4 and,

n^th terms (a_n) = 96.

⇒ a_n = a + (n - 1) × d = 96

⇒ 12 + (n - 1) × 4 = 96

∴ n = 22

Concept:

If a and d are the first term and common difference of arithmetic progression. Then sum of first n terms of an arithmetic progression is given by: \({S_n} = \frac{n}{2} \times \left( {2a + \left( {n - 1} \right)d} \right)\)

Calculation:

Given: S_2n = 3n + 14n²

If n = 1, then S₂ = 17

⇒ S₂ = 2a + d = 17     ---(1)

If n = 2, then S₄ = 62

⇒ S₄ = 4a + 6d = 62

⇒ 2a + 3d = 31     ---(2)

Subtracting (1) from (2), we get

⇒ 2d = 14 ⇒ d = 7.

Concept:

If a is the first term and r is the common ratio of a GP then the nth term of GP is given by: a_n = ar^{n - 1}

Calculation:

Given: 3^rd, 8^th and 13^th terms of a GP are p, q and r respectively.

Let a is the first term and R is the common ratio.

⇒ a₃ = aR² = p     ---(1)

⇒ a₈ = aR⁷ = q    ----(2)

⇒ a₁₃ = aR¹² = r     ---(3)

Dividing (2) by (1), we get

⇒ q / p = R⁵          ---(4)

Similarly, by dividing (3) by (2), we get

⇒ r / q = R⁵          ---(5)

From (4) and (5), we get

⇒ q² = p × r

When x ≤ 4, y ≥ 0, it represents the region R₁

When x ≤ -4, y ≤ 0, it represents the region R₂

The solution set is represented by R₁ ∩ R₂ \(\Rightarrow {R_1} \cap ;{R_2};:x \le - ;4;and;y = 0\)

Concept:

\({{\log }_{a}}b=\frac{\log b}{\log a}\)

Calculation:

\({{x}^{{{\log }_{7}}x}}>7\)

By taking log both the sides we get,

\(⇒ \left( {{\log }_{7}}x \right)\times ~\log x>~\log 7\)

As we know that, \({{\log }_{a}}b=\frac{\log b}{\log a}\)

\(⇒ \frac{{{\left( \log x \right)}^{2}}}{\log 7}>\log 7⇒ {{\left( \log x \right)}^{2}}>{{\left( \log 7 \right)}^{2}}\)

\(⇒ {{\left( \log x \right)}^{2}}-{{\left( \log 7 \right)}^{2}}>0\)

\(⇒ \left( \log x+\log 7 \right)\times \left( \log x-\log 7 \right)>0\)

⇒ logx + log 7 < 0, logx - log 7 > 0

⇒ logx < -log 7 , logx > log 7

⇒ logx < log 7^-1 , logx > log 7

⇒ x < 1/7, x > 7

∴ \(x\epsilon \left( 0,\frac{1}{7} \right)\cup \left( 7,~\infty \right)\)

Concept:

If \(f\left( x \right)=\left| x \right|~then~f\left( x \right)=~\left{ \begin{matrix} x,~x\ge 0 \ -x,~x<0 \ \end{matrix} \right.\)

Calculation:

Given: x² + 3|x| + 2 = 0

Case - 1: when x ≥ 0

⇒ x² + 3x + 2 = 0

⇒ x² + x + 2x + 2 = 0

⇒ (x + 1) (x + 2) = 0

⇒ x = - 1 or - 2

∵ x ≥ 0 ⇒ The given quadratic equation has no real roots for x ≥ 0.

Case - 2: when x < 0

⇒ x² - 3x + 2 = 0

⇒ x² - x - 2x + 2 = 0

⇒ (x - 1) (x - 2) = 0

⇒ x = 1 or 2

∵ x < 0 ⇒ The given quadratic equation has no real roots for x < 0.

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² - 4ac then the given quadratic equation has:

1. Distinct and real roots if D > 0.
2. Real and repeated roots, if D = 0.
3. Complex roots and conjugate of each other, D < 0.

Calculation:

Given: 4(x - p)(x - q) - r² = 0

⇒ x² - (p + q) x + pq - r² /4 = 0

By comparing the quadratic equation x2 - (p + q) x + (pq - (r /2)2) = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = 1, b = - (p + q) and c =  pq - r2 /4

⇒ D = b² - 4ac = (p + q)² - 4pq + r² = (p - q)² + r² ≥ 0

Hence statement I is true

The given quadratic equation 4(x - p)(x - q) - r² = 0 will have repeated roots if D = 0.

⇒ D = (p - q)2 + r2 = 0 if and only if p = q and r = 0.

Hence statement II is true.

Concept:

1. Cardinality of a set A is defined as the no. of elements present in the set A, denoted by n(A).
2. Power set: Let A be a non-empty set. Then, \(P\left( A \right)=\left{ B~|~B~\subseteq A \right}\) is called as the power set of A.
3. For any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x.

i.e n(P(A)) = 2^x.

Calculation:

Given: S = {2, 4, 6, 8, ______ 20}

⇒ The no. of elements in set S, n (S) = 10.

As we know that, for any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x.

⇒ The maximum number of subsets of S = 2¹⁰ = 1024.

Calculation:

A binary number is a number expressed in the base-2 numeral system or binary numeral system, a method of mathematical expression that uses only two symbols: typically "0" (zero) and "1" (one).

Here we have,

c > d

Hence, c must be 1 and d must be 0

Hence, (cdccddcccddd) can be written as,

⇒ (cdccddcccddd)₂ = (101100111000)₂

Using concept of binary to decimal conversion,

⇒ (101100111000)₂ = (1 × 2¹¹) + (1 × 2⁹) + (1 × 2⁸) + (1 × 2⁵) + (1 × 2⁴) + (1 × 2³) + 0 = (2872)₁₀