Quantitative Aptitude PYQs

CONCEPT:

Formulas and Properties of Logarithms:

1. \({\log _a}a = 1\)
2. \({\log _a}\left( {x.y} \right) = \;{\log _a}x + {\log _a}y\)
3. \({\log _a}\left( {\frac{x}{y}} \right) = \;{\log _a}x - {\log _a}y\)
4. \({\log _a}(\frac{1}{x}) = - {\log _a}x\)
5. \({\log _a}{x^p} = p{\log _a}x\)

CALCULATION:

Given:

\({\rm{x}} + {\rm{\;lo}}{{\rm{g}}_{10}}\left( {1 + {2^{\rm{x}}}} \right) = {\rm{xlo}}{{\rm{g}}_{10}}5 + {\rm{lo}}{{\rm{g}}_{10}}6\)

\(\begin{array}{l} \Rightarrow {\rm{x}} - {\rm{xlog}}5 = {\rm{log}}6 - \log \left( {1 + {2^{\rm{x}}}} \right)\\ \Rightarrow {\rm{x}}\left( {1 - {\rm{log}}5} \right) = {\rm{log}}6 - \log \left( {1 + {2^{\rm{x}}}} \right)\\ \Rightarrow {\rm{x}}\left( {{\rm{log}}10 - {\rm{log}}5} \right) = {\rm{log}}6 - \log \left( {1 + {2^{\rm{x}}}} \right){\rm{\;}}\left( {{\rm{\;log}}_{\rm{a}}^{\rm{a}} = 1} \right) \end{array}\)

\(\Rightarrow {\rm{xlog}}\left( {\frac{{10}}{5}} \right) = \log \left( {\frac{6}{{1 + {2^{\rm{x}}}}}} \right){\rm{\;}}\left( {{\rm{lo}}{{\rm{g}}_{\rm{a}}}\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right) = {\rm{\;lo}}{{\rm{g}}_{\rm{a}}}{\rm{x}} - {\rm{\;lo}}{{\rm{g}}_{\rm{a}}}{\rm{y}}} \right)\)

\(\Rightarrow {\rm{xlog}}\left( 2 \right) = \log \left( {\frac{6}{{1 + {2^{\rm{x}}}}}} \right)\)

\(\Rightarrow \log \left( {{2^{\rm{x}}}} \right) = \log \left( {\frac{6}{{1 + {2^{\rm{x}}}}}} \right){\rm{\;}}\left( {{\rm{lo}}{{\rm{g}}_{\rm{a}}}{{\rm{x}}^{\rm{p}}} = {\rm{plo}}{{\rm{g}}_{\rm{a}}}{\rm{x}}} \right)\)

Taking “ANTILOG” on both sides

\(\Rightarrow {2^{\rm{x}}} = \frac{6}{{1 + {2^{\rm{x}}}}}\)

Now simplify this equation

⇒ 2^x (1 + 2^x) = 6

Let’s take 2^x = u

⇒ u(1 + u) = 6

⇒ u² + u = 6

⇒ u² + u - 6 = 0

⇒ u² + 3u - 2u - 6 = 0

⇒ u(u + 3) - 2(u + 3) = 0

⇒ (u + 3) (u - 2) = 0

⇒ u = -3, 2

Now replace “u” with “2^x”

⇒ 2^x = 2, 2^x = -3

Observe both equations.

Clearly, 2^x = -3 is impossible

So, 2^x = 2

⇒ x = 1

CONCEPT:

BINARY TO DECIMAL CONVERSION:

For Example, to convert 10011011 into decimal the following process must be followed.

DECIMAL TO BINARY CONVERSION:

To convert 57 from decimal to binary, the following two methods can be used.

CALCULATION:

The easiest way to solve this problem is by converting the given binary numbers into decimal numbers.

(101110)₂ = (1 × 2⁵) + (0 × 2⁴) + (1 × 2³) + (1 × 2²) + (1 × 2¹) + (1 × 2⁰)

= 32 + 0 + 8 + 4 + 2 + 0

= (46)₁₀

(110)₂ = (1 × 2²) + (1 × 2¹) + (0 × 2⁰)

= 4 + 2 + 0

= (6)₁₀

Now divide the decimal numbers,

46/6 ⇒ Quotient =7, Remainder = 4

Now again convert 7, 4 into binary numbers

⇒ (7)₁₀ = (1 × 2²) + (1 × 2¹) + (1 × 2⁰)

= (111)₂

(4)₁₀ = (1 × 2²) + (0 × 2¹) + (1 × 2⁰)

= (100)₂

Hence, Remainder = (100)₂ and Quotient = (111)₂

CONCEPT:

If A, B are two matrices, then matrix AB exist when “No. of columns of Matrix A = No. of rows of Matrix B “

Similarly, matrix BA exist when “No. of columns of Matrix B = No. of rows of Matrix A “

CALCULATION:

For Matrix AB to exist, No. of columns of Matrix A = No. of rows of Matrix B)

⇒ x + 5 = y

For Matrix BA to exist, No. of columns of Matrix B = No. of rows of Matrix A

⇒ 11 - y = x

Solving both the equations, we get: x = 3, y = 8.

CONCEPT:

Arithmetic Progression:

The standard form of arithmetic progression (AP) is given by: a, a + d, a + 2d, a + 3d……. so on.

The nth term of an AP series is T_n = a + (n-1) d,

Sum of n terms of an A.P is S_n = n/2 [2a + (n − 1) d] 

where ‘a’= first term and ‘d’= common difference.

CALCULATION:

Given, \({{\rm{S}}_{\rm{n}}} = {\rm{nP}} + \frac{{{\rm{n}}\left( {{\rm{n}} - 1} \right){\rm{Q}}}}{2}\)

Comparing the given equation with the formula of S_n of an A.P

\(\Rightarrow {\rm{nP}} + \frac{{{\rm{n}}\left( {{\rm{n}} - 1} \right){\rm{Q}}}}{2} = \frac{{\rm{n}}}{2}\left[ {2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right]\)

RHS equation can be written as,

\(\Rightarrow \frac{{\rm{n}}}{2}\left[ {2{\rm{P}} + \left( {{\rm{n}} - 1} \right){\rm{Q}}\left] { = \frac{{\rm{n}}}{2}} \right[2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right]\)

By comparing LHS and RHS, we get: a = P and d = Q

So, common difference d = Q.

CONCEPT:

Quadratic Equation:

The standard quadratic equation is given by: f(x) = ax² + bx + c = 0

The roots of a quadratic equation can be calculated by making f(x) = 0

Thus, the roots of a quadratic function are given by: \(x = \;\frac{{ - b\; \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

CALCULATION:

Given: (q - r)x² + (r - p)x + (p - q) = 0

By Comparing the given equation with the standard quadratic equation, ax² + bx + c = 0 , we get: a = (q − r), b = (r − p) and c = (p − q)

So, Discriminant b²- 4ac = (r - p)² - {4 × (p - q) × (q - r)}

⇒ b² - 4ac = r² +p² - 2rp - 4[pq - rp - q² + rq]

⇒ b² - 4ac = r² +p² - 2rp - 4pq + 4rp - 4q² - 4rq

⇒ b² - 4ac = r² +p² - 4pq + 2rp - 4q² - 4rq

⇒ b² - 4ac = (r + p - 2q)²

Roots of the given Quadratic Equation is given by: \({\rm{x}} = \frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right){\rm{\;}} \pm \sqrt {{{\left( {{\rm{r}} + {\rm{p}} - 2{\rm{q\;}}} \right)}^2}} }}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}\)

\(\Rightarrow {\rm{x}} = \frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) + \sqrt {{{\left( {{\rm{r}} + {\rm{p}} - 2{\rm{q\;}}} \right)}^2}} }}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}{\rm{\;}},{\rm{\;x}} = {\rm{\;}}\frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) - \sqrt {{{\left( {{\rm{r}} + {\rm{p}} - 2{\rm{q\;}}} \right)}^2}} }}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}\)

\(\Rightarrow {\rm{x}} = \frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) + \left( {{\rm{r}} + {\rm{p}} - 2{\rm{q}}} \right)}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}{\rm{\;}},{\rm{\;x}} = {\rm{\;}}\frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) - \left( {{\rm{r}} + {\rm{p}} - 2{\rm{q}}} \right)}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}\)

\(\Rightarrow {\rm{x}} = \frac{{2{\rm{p}} - 2{\rm{q}}}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}{\rm{\;}},{\rm{\;x}} = {\rm{\;}}\frac{{2{\rm{q}} - 2{\rm{r}}}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}\)

Hence, \({\rm{x}} = \frac{{{\rm{p}} - {\rm{q}}}}{{{\rm{q}} - {\rm{r}}}},{\rm{\;}}1\)

CONCEPT:

If there are two sets P and Q, then

1. n(P U Q) represents the number of elements present in both the sets P or Q.
2. n(P ⋂ Q) represents the number of elements which are common in both the sets P & Q.
3. n(P U Q) = n(P) + (n(Q) - n (P ⋂ Q)

 

If U is a Universal set, A is a subset, then

1. U - A = A’ (A’ is the complement of A).
2. U - A’ = A

 

CALCULATION:

Given that, E is a universal set and A is a set.

⇒ Here, E - A = A’ called as (A complement)

⇒ Similarly, E - A’ = A

⇒ Hence, E-(E-(E-(E-(E-A)))) = E-(E-(E-(E-A’))) (∵ E - A = A’)

= E-(E-(E-A)) (∵ E - A’ = A)

= E-(E-A’)

= E-A

= A’

= (B∪ C)’ = B’∩ C’ (∵ we were given that A = B∪ C)

CONCEPT:

The intersection of two or more sets is the set of elements that are common to each of the sets.

i.e If A and B are any two sets, then A ∩ B = {x : x ∈ A and B}

CALCULATION:

Given,

A = {x : x is a multiple of 2},

B = {x : x is a multiple of 5},

C = {x : x is a multiple of 10}

From the given information, it is clear that multiples of 2 include multiples of 10. So, C is subset of A.

Similarly, multiples of 5 include multiples of 10. So, C is subset of B.

Also, C = A ∩ B

But, B = A ∩ C, B ∩ c = B

Hence, (A ∩ B ∩ C) = A ∩ B = C

CONCEPT:

Quadratic Equation Roots:

The Standard Quadratic Equation is given by: f(x) = ax² + bx + c

The roots of a Quadratic Equation f(x) = 0, Satisfies the Given Quadratic Equation.

If α, β are roots of a Quadratic Equation f(x) = 0, then f(α) = 0 and f(β) = 0.

Sum of the roots \({\rm{\alpha }} + {\rm{\beta }} = \frac{{ - {\rm{b}}}}{{\rm{a}}}\), Product of the roots \(\left( {{\rm{\alpha }} \times {\rm{\beta }}} \right) = \frac{{\rm{c}}}{{\rm{a}}}\)

Matrix Multiplication:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{b}}\\ {\rm{c}}&{\rm{d}} \end{array}} \right],{\rm{\;B}} = \left[ {\begin{array}{*{20}{c}} {\rm{e}}&{\rm{f}}\\ {\rm{g}}&{\rm{h}} \end{array}} \right],\) then AB is given by:

 \({\rm{AB}} = \left[ {\begin{array}{*{20}{c}} {ae + bg}&{af + bh}\\ {ce + dg}&{cf + dh} \end{array}} \right]\)

CALCULATION:

Given that α , β are the roots of the Quadratic Equation 1+ x + x² = 0.

Hence, 1 + α + α² = 0 and 1+ β + β² = 0

⇒ Sum of the roots \({\rm{\alpha }} + {\rm{\beta }} = \frac{{ - {\rm{b}}}}{{\rm{a}}} = {\rm{\;}} - 1\)

⇒ Product of the roots \(\left( {{\rm{\alpha }} \times {\rm{\beta }}} \right) = \frac{{\rm{c}}}{{\rm{a}}} = 1{\rm{\;}}\)

\(\left[ {\begin{array}{*{20}{c}} 1&{\rm{\beta }}\\ {\rm{\alpha }}&{\rm{\alpha }} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{\alpha }}&{\rm{\beta }}\\ 1&{\rm{\beta }} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{\rm{\alpha }} + {\rm{\beta }}}&{{\rm{\beta }} + {{\rm{\beta }}^2}}\\ {{\rm{\alpha }} + {{\rm{\alpha }}^2}}&{{\rm{\alpha \beta }} + {\rm{\alpha \beta }}} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1}\\ { - 1}&2 \end{array}} \right]\)

CONCEPT:

PROPERTIES OF MODULUS FUNCTION:

1. |ab| = |a||b|
2. |a + b| ≤ |a| + |b|
3. |a - b| ≥ ||a| - |b||

 

EXPLANATION:

Given that,

Statement 1: |ab| = |a||b|

⇒ From the properties of Modulus Function, it is clear that Statement 1 is true.

Statement 2: |a + b|

⇒ From the properties of Modulus Function, it is clear that Statement 2 is true.

Statement 3: |a - b|

⇒ From the properties of Modulus Function, it is clear that Statement 3 is true.

Hence, All the given statements are true.

CONCEPT:

PERMUTATION:

Number of Permutations of ‘n’ things taken ‘r’ at a time:

\({\rm{p}}\left( {{\rm{n}},{\rm{\;r}}} \right) = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!}}\)

Number of Permutations of ‘n’ objects where there are n₁ repeated items, n₂ repeated items, ….. n_k repeated items taken ‘r’ at a time:

\({\rm{p}}\left( {{\rm{n}},{\rm{\;r}}} \right) = \frac{{{\rm{n}}!}}{{{{\rm{n}}_1}!{{\rm{n}}_2}!{{\rm{n}}_3}! \ldots .{{\rm{n}}_{\rm{k}}}!}}\)

CALCULATION:

For the given word ‘PERMUTATION’

The number of letters in the given word is 11.

The letter ‘t’ is repeated 2 times and there is no repetition of any other letter.

So, different permutations that can be made out of the letters of the word ‘PERMUTATION’ is: \(\frac{{11!}}{{2!}} \)

CONCEPT:

Scalar multiplication:

If A = (a_ij) is a matrix and k is any scalar then the matrix kA is obtained by multiplying every element of matrix A by a scalar k.

CALCULATION:

Given: \(A = \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\\ {14i}&{6 + 4i} \end{array}} \right]\) and \(k = \frac{1}{{2i}}\)

\(\Rightarrow kA = \frac{1}{{2i}} \times \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\\ {14i}&{6 + 4i} \end{array}} \right]\)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{2i}} \times \left( {4i - 6} \right)}&{\frac{1}{{2i}} \times \left( {10i} \right)}\\ {\frac{1}{{2i}} \times \left( {14i} \right)}&{\frac{1}{{2i}} \times \left( {6 + 4i} \right)} \end{array}} \right]\)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {2 - \frac{3}{i}}&5\\ 7&{\frac{3}{i} + 2} \end{array}} \right]\)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {2 - \frac{3}{i}\left( {\frac{i}{i}} \right)}&5\\ 7&{\frac{3}{i}\left( {\frac{i}{i}} \right) + 2} \end{array}} \right]\)     …(by multiplying and dividing by “i”)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\\ 7&{2 - 3i} \end{array}} \right]\)

Hence, \(kA = \left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\\ 7&{2 - 3i} \end{array}} \right]\)

CONCEPT:

Modulus function:

Modulus of a real number x, denoted by |x| is the non-negative value of x without regard to its sign. Namely, |x| = x if x is positive, and |x| = -x if x is negative.

The modulus function is defined as: \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right| = \left\{ {\begin{array}{*{20}{c}} { - x,\;\;x < 0}\\ {x,\;\;x \ge 0} \end{array}} \right.\)

CALCULATION:

Given equation is |x – 3|² + |x – 3| - 2 = 0

Let us assume x – 3 = t     …equation (1)

So, |t|² + |t| - 2 = 0

We know that*,* \(\left| {\rm{t}} \right| = \left\{ {\begin{array}{*{20}{c}} { - t,\;\;t < 0}\\ {t,\;\;t \ge 0} \end{array}} \right.\)

Case 1: (t ≥ 0)

t² + t - 2 = 0

⇒ t² + 2t - t - 2 = 0

⇒ t(t + 2) - 1(t + 2) = 0

⇒ (t - 1) (t + 2) = 0

⇒ t = 1, -2

As we have taken in the above case that (t > 0) so, t = 1 is the possible solution.

Now substitute t = 1 in equation (1)

⇒ x – 3 = 1

⇒ x = 4.

CASE 2: (t < 0)

(-t)² + (-t) - 2 = 0

⇒ t² - t - 2 = 0

⇒ t² – 2t + t - 2 = 0

⇒ t (t +1) - 2(t + 1) = 0

⇒ (t – 2) (t +1) = 0

⇒ t = -1, 2

As we have taken in the above case that (t < 0) so, t = -1 is the possible solution.

Now substitute t = -1 in equation (1)

⇒ x – 3 = -1

⇒ x = 2.

Hence, the sum of the real roots of the given equation is 6.

CONCEPT:

Nature of roots of a quadratic equation:

The roots of a quadratic function are given by, \(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

We call the term b² – 4ac as discriminant (Δ).

Case 1: If Δ = 0, then the roots are real and equal.

Case 2: If Δ > 0, then the roots are real and unequal.

Case 3: If Δ < 0, then the roots are imaginary.

CALCULATION:

Given that roots of the equation x² – 4x – log₃ P = 0 are real.

Comparing the given equation with the standard form of quadratic equation: f(x) = ax² + bx + c = 0

⇒ a = 1

⇒ b = -4

⇒ c = -log­₃ P

For roots to be real, Δ ≥ 0

⇒ Δ = b² – 4ac = [(- 4)]² – (4× 1× (- log₃ P)

⇒ Δ = 16 + (4 × log₃ P)

⇒ 16 + (4 × log₃ P) ≥ 0

⇒ 4log₃ P ≥ -16

⇒log₃ P ≥ -4

Applying “ANTILOG” on both sides

⇒ P ≥ 3^-4

⇒ P ≥ 1/81

The minimum value of P for roots to be real is 1/81.

CONCEPT:

Adjoint Matrix:

If B_{n× n} is a cofactor matrix of matrix A_{n× n} then the adjoint matrix of A_{n× n} is denoted by adj(A) and is defined as B^T. So, adj(A) = B^T.

Property of adjoint matrix:

If A is a square matrix of order ‘n’ then adj(A^T) = (adj(A))^T

CALCULATION:

As we know that  adj (A^T) = (adj (A))^T

Hence, adj (A^T) - (adj (A))^T = adj(A^T) – adj(A^T) = O (Null Matrix whose order is same as the order of A)

CONCEPT:

Infinite Geometric series:

Standard infinite geometric series is given by: a₁ + a₁r + a₁r² + a₁r³ + …… +a₁r^n-1 + ….

Sum of infinite geometric series is given by \({{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}\) as long as -1 < r < 1

CALCULATION:

Given that: \({6^{\frac{1}{2}}} \times {6^{\frac{1}{4}}} \times {6^{\frac{1}{8}}} \times {6^{\frac{1}{{16}}}} \times \cdots\) upto infinite terms

⇒ \({6^{\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots } \right)}}\)     ...(∵ a^m × aⁿ = a^m+n)

The series: \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\)is in form of infinite geometric series.

Comparing it with standard infinite G.P series, we get a = 1/2 and r = 1/2

Sum of infinite terms of a G.P, \({s_n} = \frac{a}{{1 - r}}\)

\( \Rightarrow {{\rm{s}}_{\rm{n}}} = \left( {\frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}}} \right)\)

\(\Rightarrow {{\rm{s}}_{\rm{n}}} = \left( {\frac{{\frac{1}{2}}}{{\frac{1}{2}}}} \right) = 1\)

\(\Rightarrow {6^{\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots } \right)}} = {6^1} = 6\)

Hence, the value of product of \({6^{\frac{1}{2}}} \times {6^{\frac{1}{4}}} \times {6^{\frac{1}{8}}} \times {6^{\frac{1}{{16}}}} \times \cdots\) upto infinite terms = 6

CONCEPT:

Determinant of a square matrix of order two:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) is a square matrix of order 2 then determinant is given by:

\(\det \left( A \right) = \left| A \right| = {\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right| = {a_{11}}{a_{22}} - \;{a_{12}}{a_{21}}\)

CALCULATION:

Given that: \(\left| {\begin{array}{*{20}{c}} {{{\cos }^2}\frac{\theta }{2}}&{{{\sin }^2}\frac{\theta }{2}}\\ {{{\sin }^2}\frac{\theta }{2}}&{{{\cos }^2}\frac{\theta }{2}} \end{array}} \right|\)

Determinant of the given matrix is given by: \({\rm{\Delta }} = co{s^4}\left( {\frac{\theta }{2}} \right) - si{n^4}\left( {\frac{\theta }{2}} \right)\)

\(\Rightarrow {\rm{\Delta }} = \left[ {co{s^2}\left( {\frac{\theta }{2}} \right) - si{n^2}\left( {\frac{\theta }{2}} \right)} \right]\left[ {co{s^2}\left( {\frac{\theta }{2}} \right) + si{n^2}\left( {\frac{\theta }{2}} \right)} \right]\)     …{∵ A⁴ – B⁴ = (A² – B²)(A² + B²)}

\(\Rightarrow {\rm{\Delta }} = \left[ {co{s^2}\left( {\frac{\theta }{2}} \right) - \;si{n^2}\left( {\frac{\theta }{2}} \right)} \right]\left[ 1 \right]\)     …{∵ cos²A – sin²A = cos(2A)}

\(\Rightarrow {\rm{\Delta }} = [\cos \left( {\frac{{\theta \times 2}}{2}} \right)]\left[ 1 \right]\)

⇒ Δ = cos θ

CONCEPT:

Binomial Theorem:

According to the theorem, it is possible to expand any non-negative power of x + y into a sum of the form:

(x + y)ⁿ = ⁿC₀ xⁿy⁰ + ⁿC₁x^n-1y­_­¹+ ⁿC₂x^n-2y­_­²+ ⁿC₃x^n-3y­_­³+ …………. + ⁿC_nx⁰y­_­ⁿ where n≥ 0 is an integer.

CALCULATION:

(x + a)¹⁰⁰ = ¹⁰⁰C₀ x¹⁰⁰ + ¹⁰⁰C₁x⁹⁹a­_­¹ + ¹⁰⁰C₂x⁹⁸a­_­² + ¹⁰⁰C₃x⁹⁸y­_­² +     ………… + ¹⁰⁰C₁₀₀a­_­¹⁰⁰     …equation (1)

(x – a)¹⁰⁰ = ¹⁰⁰C₀ x¹⁰⁰ - ¹⁰⁰C₁x⁹⁹a­_­¹+ ¹⁰⁰C₂x⁹⁸a­_­² - ¹⁰⁰C₃x⁹⁸y­_­²+     ………… + ¹⁰⁰C₁₀₀a­_­¹⁰⁰     …equation (2)

The above two equations are written with the help of “BINOMIAL THEOREM”

By adding equation (1) and equation (2), we get

(x + a)¹⁰⁰ = 2× [¹⁰⁰C₀ x¹⁰⁰ + ¹⁰⁰C₂x⁹⁸a­_­² + ………… + ¹⁰⁰C₁₀₀a­_­¹⁰⁰]     …equation (3)

Total no. of terms in equation (1) = 101 (odd terms = 50, even terms = 51)

By adding equation (1) and equation (2), odd terms will get cancelled

Hence, the total no. of terms in the expansion (x + a)¹⁰⁰ + (x - a)¹⁰⁰ = 51

Formula used:

(1 + x)n  = [nC0 + nC1 x + nC2 x2 + … +nCn xn]

• C₀ + C₁ + C₂ + … + C_n = 2ⁿ
• C₀ + C₂ + C₄ + … =  2n-1
• C₁ + C₃ + C₅ + … = 2^n-1

 

Calculation:

(1 + x)⁵⁰  = [⁵⁰C0 + ⁵⁰C1 x + ⁵⁰C2 x2 + … +⁵⁰Cn x⁵⁰]    ----(1)

Here, n = 50

Using the above formula, the sum of odd terms of the coefficient is

S = (50C1 + 50C3­ + 50C5 + ……. + 50C49)

⇒ S = 2^50-1 = 2⁴⁹

∴ Sum of odd terms of the coefficient = 249

CONCEPT:

Inverse of a matrix: The Inverse of a n × n matrix is given by:

\({A^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}\) where adj(A) is called adjoint matrix.

Adjoint Matrix: If B_{n× n} is a cofactor matrix of matrix A_{n× n} then the adjoint matrix of A_{n× n} is denoted by adj(A) and is defined as B^T. So, adj(A) = B^T.

CALCULATION:

Given: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} a&0&0\\ 0&b&0\\ 0&0&c \end{array}} \right]\)

We know that, \({{\rm{A}}^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}\)

⇒ |A| = [a (bc - 0) - 0(0) - 0(0)]

⇒ |A| = abc

\(\Rightarrow {\rm{adj}}\left( {\rm{A}} \right) = {\left[ {\begin{array}{*{20}{c}} {bc}&0&0\\ 0&{ac}&0\\ 0&0&{ab} \end{array}} \right]^T}\)

\(\Rightarrow {\rm{adj}}\left( {\rm{A}} \right) = \left[ {\begin{array}{*{20}{c}} {bc}&0&0\\ 0&{ac}&0\\ 0&0&{ab} \end{array}} \right]\)

\(\Rightarrow {A^{ - 1}} = \frac{{\left[ {\begin{array}{*{20}{c}} {bc}&0&0\\ 0&{ac}&0\\ 0&0&{ab} \end{array}} \right]}}{{abc}}\)

\(\Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {{a^{ - 1}}}&0&0\\ 0&{{b^{ - 1}}}&0\\ 0&0&{{c^{ - 1}}} \end{array}} \right]\)

CONCEPT:

Arithmetic progression:

The Standard form of an arithmetic progression (AP) is given by: a, a+ d, a + 2d, a + 3d ……. so on.

⇒ The nth term of an AP series is T_n = a+ (n - 1) d,

⇒ Sum of n terms of an A.P is S_n = n/2 [2a + (n − 1) d]

‘a’= first term and

‘d’= common difference.

CALCULATION:

Given that: a₁ = a₂ = a₃ = ……. = a₁₀ = 150 and a₁₀, a₁₁, a₁₂, … are in AP.

So, the person counts 1500 notes in 10 minutes. The remaining notes which he need to count are

⇒ 4500 – 1500 = 3000

It is given that a₁₁, a₁₂, a₁₃ are in AP with a common difference of -2.

Hence, a₁₀ = 150, a₁₁ = 148, a₁₂ = 146, a₁₃ = 144 … and so on.

So, the sum of all the notes must be 3000.

Hence, s_n = 3000, a = 148 (as he already counted 1500 notes in 10 minutes, so we need to start from

a₁₁ = 148, d = -2

\( \Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{\rm{n}}}{2}[\left( {2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right]\)

\(\Rightarrow 3000 = \frac{{\rm{n}}}{2}\left[ {\left( {2 \times 148} \right) + \left( {{\rm{n}} - 1} \right)\left( { - 2} \right)} \right]\)

\(\Rightarrow 3000 = \frac{{\rm{n}}}{2}\left[ {296 + 2 - 2{\rm{n}}} \right]\)

⇒ 3000 = n (149 – n)

⇒ 3000 = 149n – n²

⇒ n² – 149n + 3000 = 0

⇒ n² – 125n – 24n + 3000 = 0

⇒ n (n – 125) – 24(n – 125) = 0

⇒ (n – 24) (n – 125) = 0

⇒ n = 24, 125

So, the time taken by the person to count 3000 notes is 24 minutes.

Now, the count will take place until a₃₅ term because the count of 3000 notes started from a₁₁ but not a₁.

Hence, the total time required to count 4500 notes = 10 + 24 = 34 minutes