Quantitative Aptitude PYQs

Concept:

Product rule of differentiation:

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right)} \right) = {\rm{f'}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{g'}}\left( {\rm{x}} \right)\)

Calculation:

Given: \({\rm{y}} = {{\rm{e}}^{{{\rm{x}}^2}}}\sin 2{\rm{x}}\)

To find: \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\)

\({\rm{y}} = {{\rm{e}}^{{{\rm{x}}^2}}}\sin 2{\rm{x}}\)

Differentiating both sides w.r.t x

\( \Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {{\rm{e}}^{{{\rm{x}}^2}}}\left( {2{\rm{x}}} \right)\sin 2{\rm{x}} + \cos 2{\rm{x}} \times \left( 2 \right) \times {{\rm{e}}^{{{\rm{x}}^2}}}{\rm{\;}}\)

\(\Rightarrow {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)_{{\rm{x}} = {\rm{\pi }}}} = {{\rm{e}}^{{{\rm{\pi }}^2}}}\left( {2{\rm{\pi }}} \right)\sin \left( {2{\rm{\pi }}} \right) + 2{{\rm{e}}^{{{\rm{\pi }}^2}}}\cos \left( {2{\rm{\pi }}} \right)\)

 \(\Rightarrow {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)_{{\rm{x}} = {\rm{\pi }}}} = 2{{\rm{e}}^{{{\rm{\pi }}^2}}}\)

Concept:

Formula used:

• \(\smallint \frac{{{\rm{dx}}}}{{{\rm{a}} + {\rm{bx}}}} = \frac{1}{{\rm{b}}}\ln \left( {{\rm{a}} + {\rm{bx}}} \right) + {\rm{c}}\)

Calculation:

Given: (1 +2x) dy – (1 – 2y) dx = 0

⇒ (1 +2x) dy = (1 – 2y) dx

\(\Rightarrow \frac{{{\rm{dy}}}}{{1 - 2{\rm{y}}}} = \frac{{{\rm{dx}}}}{{1 + 2{\rm{x}}}}\)

On integrating both sides,

\( \Rightarrow \smallint \frac{{{\rm{dy}}}}{{1 - 2{\rm{y}}}} = \smallint \frac{{{\rm{dx}}}}{{1 + 2{\rm{x}}}}{\rm{\;}}\)

\(\Rightarrow - \frac{1}{2}\ln \left( {1 - 2{\rm{y}}} \right) = \frac{1}{2}\ln \left( {1 + 2{\rm{x}}} \right) + {{\rm{c}}_1}\)

⇒ ln (1 + 2x) + ln (1 – 2y) = ln c₂

⇒ (1 + 2x) × (1 – 2y) = c₂

⇒ 1 – 2y + 2x – 4xy = c₂

⇒ 2x – 2y – 4xy = c₂ – 1

⇒ x – y – 2xy = c

Where c₁ is integration constant and c₂ & c are some arbitrary constant

Concept:

• Order: The order of a differential equation is the order of the highest derivative appearing in it.
• Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

For example, if the equation is

\({\left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}} \right)^3} + \frac{{{\rm{dy}}}}{{{\rm{dx}}}} - {{\rm{x}}^2} = 0\)

Here, order of the highest derivative is 2. And the degree is 3.

Calculation:

Given: \({\left( {\frac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}}} \right)^2} = {{\rm{y}}^4} + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^5}\)

To find: Order and degree

So, the highest order is \(\frac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}}\)

∴ Order is 3.

Degree of differential equation is the degree highest order. So, degree is 2.

Concept:

Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P(X = k) = ⁿC_k p^k (1 – p)^{(n – k)}

Where, n is number of observations.

p is the probability of success & (1 – p) is probability of failure.

k = 0, 1, 2, …., n

Properties:

• Mean of the distribution (μ_X) = n × p
• The variance (σ²_x) = n × p × (1 – p)
• Standard deviation (σ_x) = √{np(1 – p)}

Calculation:

Given: Mean (μ_X) = 3 × variance (σ²_x)

n × p = 3 × n × p × (1 – p)

⇒ 1 – p = 1/3

⇒ p = 2/3

Probability of exactly 3 successes = P(X = 3)

= ⁵C₃ p³ (1 – p)^{(5 – 3)}

\(= \frac{{5\; \times \;4\; \times \;3!}}{{2!3!}} \times {\left( {\frac{2}{3}} \right)^3} \times {\left( {\frac{1}{3}} \right)^2}\)

= 10 × 8 × (1/3)⁵

= 80/243

Concept:

Formulas used:

• P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
• Conditional Probability:

\({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

Calculation:

For first statement,

We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

So, P(A̅ ∪ B) = P(A̅) + P(B) – P(A̅ ∩ B)

∴ First statement is correct.

For second statement,

We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

So, P(A ∪ B̅) = P(A) + P(B̅) – P(A ∩ B̅)

⇒ P(A ∪ B̅) = P(A) + 1 – P(B) – {P(A) – P(A ∩ B)

⇒ P(A ∪ B̅) = 1 – P(B) + P(A ∩ B)

∴ Second statement is incorrect.

For third statement,

From conditional probability we know,

\({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

⇒ P(A ∩ B) = P(A|B) × P(B)

∴ Third statement is correct.

Concept:

Formulas used:

\({\rm{r}}\left( {{\rm{x}},{\rm{y}}} \right) = \frac{{{\rm{\;Cov}}\left( {{\rm{x}},{\rm{y}}} \right)}}{{{\rm{\sigma }}\left( {\rm{x}} \right){\rm{\sigma }}\left( {\rm{y}} \right)}}
\)

Calculation:

Given:

Correlation coefficient between x and y, r(x, y) = 0.6

Covariance of x and y, Cov(x, y) = 27

Variance of y, Var(y) = σ²(y) = 25

Variance of x, Var(x) = σ²(x)

We know that,\({\rm{r}}\left( {{\rm{x}},{\rm{y}}} \right) = \frac{{{\rm{\;Cov}}\left( {{\rm{x}},{\rm{y}}} \right)}}{{{\rm{\sigma }}\left( {\rm{x}} \right){\rm{\sigma }}\left( {\rm{y}} \right)}}
\)

\(\Rightarrow {\rm{\sigma }}\left( {\rm{x}} \right) = \frac{{{\rm{Cov}}\left( {{\rm{x}},{\rm{y}}} \right)}}{{{\rm{\sigma }}\left( {\rm{y}} \right)\; \times \;{\rm{r}}\left( {{\rm{x}},{\rm{y}}} \right)}}
\)

\(\Rightarrow {\rm{\sigma }}\left( {\rm{x}} \right) = \frac{{27}}{{5\; \times \;0.6{\rm{\;}}}}\)

⇒ σ(x) = 9

Variance of x, σ²(x) = 81

Concept:

Formulas used:

• P(A) = 1 – P(A̅)
• De Morgan’s law:

\({\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right) = {\rm{P}}(\overline {{\rm{A}} \cup {\rm{B}})}\)

If A and B are two independent events, then:

P(A ∩ B) = P(A) × P(B)

Calculation:

Let,

P(A): probability of solving question A.

P(B): probability of solving question B.

Given: P(A) = 0.4 & P(B) = 0.5

Probability of solving at least one of two questions = P(A ∪ B)

\(= 1 - {\rm{P}}(\overline {{\rm{A}} \cup {\rm{B}})}\)

= 1 – P(A̅ ∩ B̅)      [Using De Morgan’s Law]

= 1 – {P(A̅) × P(B̅)}      [∵ A & B are independent]

= 1 – {[1 – P(A)] × [1 – P(B)]}

= 1 – {(1 – 0.4) × (1 – 0.5)}

= 1 – (0.6 × 0.5)

= 1 – 0.3

= 0.7

Concept:

Mean: It is the average of given observation. Let x₁, x₂, …… , x_n be n observations, then

Mean \(= {\rm{\bar X}} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)

Properties:

• If C is added/subtracted to each observation, then the new mean is also increased/decreased by C.
• If C is multiplied to each observation, then the new mean is also C times.
• If each observation is divided by C, then the new mean is old mean divided by C.

Calculation:

Given: \(\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\bar x}}\)

Let y̅ be the mean of y₁, y₂, y₃, …, y_n

\(\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} ({\rm{a}} + {\rm{c}}{{\rm{y}}_{\rm{i}}}){\rm{\;}}}}{{\rm{n}}} = {\rm{\bar x}}\)

We know that, if C is subtracted to each observation, then the new mean is also decreased by C.

\(\Rightarrow \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {\rm{c}}{{\rm{y}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\bar x}} - {\rm{a}}\)

We know that, if each observation is divided by C, then the new mean is old mean divided by C.

\(\Rightarrow \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{y}}_{\rm{i}}}}}{{\rm{n}}} = \frac{{{\rm{\bar x}} - {\rm{a}}}}{{\rm{c}}}\)

\( \Rightarrow {\rm{\bar y}} = \frac{{{\rm{\bar x}} - {\rm{a}}}}{{\rm{c}}}\)

Concept:

Properties of regression line and correlation coefficient:

• If the correlation coefficient r_xy = 0, then the two lines of regression are perpendicular to each other.
• If the correlation coefficient r_xy = + 1, then the two lines of regression are parallel to each other.

Calculation:

Properties:

• If the correlation coefficient r_xy = 0, then the two lines of regression are perpendicular to each other.
• If the correlation coefficient r_xy = + 1, then the two lines of regression are parallel to each other.

So, statement (1) & statement (2) both are wrong.

Concept:

Regression lines: Let two variables x and y, if y depends on x, then the result comes in form of simple regression. 

Regression line for simple regression:

y = a + bx

Where a and b are regression parameters.

Properties:

• The two regression lines (of y on x and x on y) are known to intersect at a specific point (x̅, y̅).

Calculation:

Given: Regression lines,

4x – 5y + 33 = 0      ----(1)

20x – 9y = 107      ----(2)

Now, we know that, intersection point of regression lines gives a point (x̅, y̅).

Multiplying equation (1) with -5 we get,

-20x + 25y – 165 = 0      ----(3)

On adding (2) & (3) we get,

20x – 9y – 107 – 20x + 25y – 165 = 0

⇒ 16y – 272 = 0

⇒ 16y = 272

⇒ y = 17

Put y = 17 in equation (1) we get,

4x – 5(17) + 33 = 0

⇒ 4x = 85 – 33

⇒ 4x = 52

⇒ x = 13

So, x̅ = 13 and y̅ = 17.

Concept:

Properties of mean:

• If C is added to each observation, then the new mean is also increased by C.
• If C is multiplied to each observation, then the new mean is also C times.
• If C is divided to each observation, then the new mean is also old mean divided by C.

Properties of variance:

• Var(X + C) = Var(X), where C is a constant.
• Var(CX) = C² × Var(X), where C is a constant.

Calculation:

Mean:

Change in scale:

We know that, if C is multiplied to each observation, then new mean is also C times old mean.

So, mean is dependent on change in scale.

Change in origin:

We know that, if C is added/subtracted to each observation, then new mean is also increased/decreased by C.

So, mean is dependent on change in origin.

Variance:

Change in scale:

We know that, Var(CX) = C² × Var(X), where C is a constant.

So, variance is dependent on change in scale.

Change in origin:

We know that, Var(X + C) = Var(X), where C is a constant.

So, variance is independent of change in origin.

Hence, both statement (1) & (2) are incorrect.

Concept:

Properties of mean deviation:

• The sum of deviations from the mean is always zero, i.e., ∑(x_i – x̅) = 0.
• The sum of absolute deviations is minimum when taken around median.

Calculation:

Properties of mean deviation:

• The sum of deviations from the mean is always zero, i.e., ∑(x_i – x̅) = 0.
• The sum of absolute deviations is minimum when taken around median.

So, statement (1) & statement (2) both are correct.

Concept:

Median: Let a data of n observations is given, then for finding median first arrange the observation either in increasing order or decreasing order.

• If n is odd, median \(= {\left( {\frac{{\rm{n}}}{2}} \right)^{{\rm{th}}}}\) term.
• If n is even, median \(= \frac{{{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{term\;}} + {{\left( {\frac{{{\rm{n}} }}{2}+1} \right)}^{{\rm{th}}}}{\rm{term}}}}{2}\)

Calculation:

Arranging the given observation in increasing order,

-4.8, 0, 2.6, 3.5, 3.9, 4.6, 5.2, 6.1, 7.6, 8.2, 9.3, 12.7

Number of observation, n = 12

So, we know that if n is even for median then,

\(Median= \frac{{{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{term\;}} + {{\left( {\frac{{{\rm{n}} }}{2}+1} \right)}^{{\rm{th}}}}{\rm{term}}}}{2}\)

\( = \frac{{4.6\; +\; 5.2}}{2}\)

= 9.8/2

= 4.9

Concept:

Formula used:

• If x% is given as n, then: \({\rm{y\% }} = {\rm{n}} \times \frac{{{\rm{y\% }}}}{{{\rm{x\% }}}}\)

Calculation:

Let the circle be the total number of students given mathematics test.

So, 100% will give 360°.

∴ 20% will be \(360^\circ \times \frac{{20{\rm{\% }}}}{{100{\rm{\% }}}}{\rm{\;}}\)

= 72°

Concept:

Probability: It is a measure of likelihood of an event to occur.

Probability of an event to occur, P(E) \(= \frac{{{\rm{number\;of\;favourable\;outcomes}}}}{{{\rm{number\;of\;total\;outcomes}}}}\)

Formula used:

• If a dice is rolled n times then, number of sample space n(S) = 6ⁿ

Calculation:

Let,

S: Event of rolling 2 dice.

A: Event that the sum of the numbers appeared on both dice to be 7.

Number of total outcomes = n(S)

= 6²

= 36

Now, A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

Number of favourable outcomes = n(A)

= 6

Probability of getting sum of 7, P(A) \(= \frac{{{\rm{number\;of\;favourable\;outcomes}}}}{{{\rm{number\;of\;total\;outcomes}}}}\)

= 6/36

= 1/6

Concept:

Formulas used:

• Conditional probability:

\({\rm{P}}({\rm{A}}|{\rm{B}}) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

Calculation:

Given: 2P(A) = 3P(B) & 0 < P(A) < P(B) < 1

Dividing both sides by P(A ∩B) to both sides we get,

\( \Rightarrow \frac{{2{\rm{P}}\left( {\rm{A}} \right)}}{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}} = \frac{{3{\rm{P}}\left( {\rm{B}} \right)}}{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}\)

Taking reciprocal of both sides,

\(\Rightarrow \frac{1}{2} \times \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{A}} \right)}} = \frac{1}{3} \times \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

⇒ (1/2) × P(B|A) = (1/3) × P(A|B)

⇒ P(B|A) = (2/3) P(A|B)

So, P(B|A) < P(A|B)        ----(1)

From conditional probability,

\({\rm{P}}\left( {{\rm{B|A}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{A}} \right)}}\)

Given, P(A) < 1

So, P(B|A) > P(A ∩ B)        ----(2)

\({\rm{P}}({\rm{A}}|{\rm{B}}) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

Given, P(B) < 1

So, P(A|B) > P(A ∩ B)        ----(3)

From (1), (2) & (3)

P(A ∩ B) < P(B|A) < P(A|B)

Hence, option (2) is correct.

Concept:

Probability: It is a measure of likelihood of an event to occur.

Probability of an event to occur, P(E) \(= \frac{{{\rm{number\;of\;favourable\;outcomes}}}}{{{\rm{number\;of\;total\;outcomes}}}}\)

Note:

• Combination is the number of ways in which we can select a group of objects from a set. For example, selecting r things from set of n things is given as ⁿC_r.

Calculation:

Given: A box has 10 chits numbered {0, 1, 2, …,9}.

Let,

S: Event of drawing two chits one by one.

A: Event of second chit drawn is numbered 9.

So, number of total outcomes = n(S)

= ¹⁰C₁ × ⁹C₁

= 10 × 9

= 90

Number of favourable outcomes = n(A) = (0,9) (1,9).........(8,9)

= 9

Probability of second chit drawn to be 9 = \(\frac{{{\rm{n}}\left( {\rm{A}} \right)}}{{{\rm{n}}\left( {\rm{S}} \right)}}\)

= 9/90 = 1/10

Concept:

Formulas used:

Let A and B be two subsets of X.

• P(A – B) = P(A) – P(B) if B ⊂ A
• P( A alone) = P(A) – P(A ∩ B)
• P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
• If A and B are mutually exclusive, then P(A ∩ B) = 0

Calculation:

P(A occurs but not B) = P(A – B)

= P(A) – P(B)

So, statement (1) is correct.

P(A alone or B alone occurs) = P(A) – P(A ∩ B) + P(B) – P(A ∩ B)

= P(A) + P(B) – 2P(A ∩ B)

So, statement (2) is incorrect.

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Since, A and B are mutually exclusive, P(A ∩ B) = 0

P(A ∪ B) = P(A) + P(B)

So, statement (3) is correct.

Concept:

Formulas used:

• Variance, \({{\rm{\sigma }}^2} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{N}}}\)
• Standard deviation = σ
• Summation of square of first N natural number,  \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} {\rm{x}}_{\rm{i}}^2 = \frac{{{\rm{N}}\left( {{\rm{N}} + 1} \right)\left( {2{\rm{N}} + 1} \right)}}{6}\)
• Summation of first N natural number, \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} {{\rm{x}}_{\rm{i}}} = 1 + 2 + 3 + \ldots + {\rm{N}} = \frac{{{\rm{N}}\left( {{\rm{N}} + 1} \right)}}{2}\)
• \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} 1 = {\rm{N}}\)

Properties of summation:

• \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} \left( {{{\rm{a}}_{\rm{i}}} + {{\rm{b}}_{\rm{i}}}} \right) = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} \left( {{{\rm{a}}_{\rm{i}}}} \right) + \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} \left( {{{\rm{b}}_{\rm{i}}}} \right)\)
• \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} \left( {{\rm{c}}{{\rm{a}}_{\rm{i}}}} \right) = {\rm{c}}\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} \left( {{{\rm{a}}_{\rm{i}}}} \right)\)
• \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {\left( {{{\rm{a}}_{\rm{i}}} + {{\rm{b}}_{\rm{i}}}} \right)^2} = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {\left( {{{\rm{a}}_{\rm{i}}}} \right)^2} + \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {\left( {{{\rm{b}}_{\rm{i}}}} \right)^2}\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} \left( {{{\rm{a}}_{\rm{i}}}{{\rm{b}}_{\rm{i}}}} \right)\)

Calculation:

To find: σ of first N natural number

Let x_i = 1, 2, 3, …,n for i = 1, 2, 3, …,n.

Let x̅ be mean of x_i for i = 1, 2, 3, …,n.

Variance, \({{\rm{\sigma }}^2} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{N}}}\)

\( \Rightarrow {{\rm{\sigma }}^2} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} ({\rm{x}}_{\rm{i}}^2 + {{{\rm{\bar x}}}^2} - 2{{\rm{x}}_{\rm{i}}}{\rm{\bar x}})}}{{\rm{N}}}\)

\(\Rightarrow {{\rm{\sigma }}^2} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {\rm{x}}_{\rm{i}}^2{\rm{\;}} + {\rm{\;}}\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {{{\rm{\bar x}}}^2}{\rm{\;}} - {\rm{\;}}\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} 2{{\rm{x}}_{\rm{i}}}{\rm{\bar x}}}}{{\rm{N}}}\)

\(\Rightarrow {{\rm{\sigma }}^2} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {\rm{x}}_{\rm{i}}^2{\rm{\;}} + {\rm{\;}}{{{\rm{\bar x}}}^2}\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} 1 - {\rm{\;}}2{\rm{\bar x}}\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {{\rm{x}}_{\rm{i}}}}}{{\rm{N}}}\)

We know that,

Summation of square of first N natural number, \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} {\rm{x}}_{\rm{i}}^2 = \frac{{{\rm{N}}\left( {{\rm{N}} + 1} \right)\left( {2{\rm{N}} + 1} \right)}}{6}\)

Summation of first N natural number, \(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} {{\rm{x}}_{\rm{i}}} = 1 + 2 + 3 + \ldots + {\rm{N}} = \frac{{{\rm{N}}\left( {{\rm{N}} + 1} \right)}}{2}\)

Mean of first N natural number, \({\rm{\bar x}} = \frac{{1 + 2 + 3 + \ldots + {\rm{N}}}}{{\rm{N}}}\)

\(\Rightarrow {\rm{\bar x}} = \frac{{{\rm{N}}\left( {{\rm{N}} + 1} \right)}}{{2{\rm{N}}}}\)

\(\Rightarrow {\rm{\bar x}} = \frac{{{\rm{N}} + 1}}{2}\)

\(\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{N}} 1 = {\rm{N}}\)

\({{\rm{\sigma }}^2} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {\rm{x}}_{\rm{i}}^2{\rm{\;}} + {\rm{\;}}{{{\rm{\bar x}}}^2}\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} 1 - {\rm{\;}}2{\rm{\bar x}}\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{N}} {{\rm{x}}_{\rm{i}}}}}{{\rm{N}}}\)

\(\Rightarrow {{\rm{\sigma }}^2} = \frac{{\frac{{{\rm{N}}\left( {{\rm{N}} + 1} \right)\left( {2{\rm{N}} + 1} \right)}}{6}{\rm{\;}} + {\rm{\;}}{{\left( {\frac{{{\rm{N}} + 1}}{2}} \right)}^2} \times {\rm{N\;}} - {\rm{\;}}2 \times \left( {\frac{{{\rm{N}} + 1}}{2}} \right) \times \frac{{{\rm{N}}\left( {{\rm{N}} + 1} \right)}}{2}}}{{\rm{N}}}\)

\(\Rightarrow {{\rm{\sigma }}^2} = \frac{{\left( {{\rm{N}} + 1} \right)\left( {2{\rm{N}} + 1} \right)}}{6} + {\left( {\frac{{{\rm{N}} + 1}}{2}} \right)^2} - {\left( {\frac{{{\rm{N}} + 1}}{4}} \right)^2}\)

\( \Rightarrow {{\rm{\sigma }}^2} = \frac{{\left( {{\rm{N}} + 1} \right)\left( {2{\rm{N}} + 1} \right)}}{6} - {\left( {\frac{{{\rm{N}} + 1}}{4}} \right)^2}\)

\(\Rightarrow {{\rm{\sigma }}^2} = \frac{{\left( {{\rm{N}} + 1} \right)\left\{ {2 \times \left( {2{\rm{N}} + 1} \right) - 3 \times \left( {{\rm{N}} + 1} \right)} \right\}}}{{12}}\)

\(\Rightarrow {{\rm{\sigma }}^2} = \frac{{\left( {{\rm{N}} + 1} \right)\left\{ {4{\rm{N}} + 2 - 3{\rm{N}} - 3} \right\}}}{{12}}\)

\(\Rightarrow {{\rm{\sigma }}^2} = \frac{{\left( {{\rm{N}} + 1} \right)\left( {{\rm{N}} - 1} \right)}}{{12}}\)

\( \Rightarrow {{\rm{\sigma }}^2} = \frac{{{{\rm{N}}^2} - 1}}{{12}}\)

\(\Rightarrow {\rm{\sigma }} = \sqrt {\frac{{{{\rm{N}}^2} - 1}}{{12}}} \)

Concept:

1. If two vector a and b then \({\rm{a}}.{\rm{b}} = \left| {{\rm{\vec a}}} \right|{\rm{\;}}\left| {{\rm{\vec b}}} \right|\cos {\rm{\theta }}\)

 Where θ is angle between vector a and b.

1. Vector A = \({\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\) then |A| = \(\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}} \)

 

Calculation:

Given that,

P (1, -1, 1), Q (3, -2, 2) and R (0, 2, 6)

So from diagram

\(\begin{array}{l} \overrightarrow {{\rm{PR}}} = {\rm{\;}} - 1{\rm{\hat i}} + 3{\rm{\hat j}} + 5\hat k\\ \overrightarrow {{\rm{PQ}}} = {\rm{\;}}2{\rm{\hat i}} - {\rm{\hat j}} + \hat k \end{array}\)

Now, Angle between two vectors are

\(\begin{array}{l} \overrightarrow {{\rm{PR}}} {\rm{\;}}.\overrightarrow {{\rm{PQ}}} = \left| {\overrightarrow {{\rm{PQ}}} } \right|{\rm{\;}}\left| {\overrightarrow {{\rm{PR}}} } \right|\cos {\rm{\theta \;}}\\ \Rightarrow \overrightarrow {{\rm{PR}}} {\rm{\;}}.\overrightarrow {{\rm{PQ}}} = \left( { - 1{\rm{\hat i}} + 3{\rm{\hat j}} + 5{\rm{\hat k}}} \right).\left( {2{\rm{\hat i}} - {\rm{\hat j}} + {\rm{\hat k}}} \right)\\ \Rightarrow \overrightarrow {{\rm{PR}}} {\rm{\;}}.\overrightarrow {{\rm{PQ}}} = {\rm{\;}} - 2 - 3 + 5 = 0 \end{array}\)

So,

\(\left| {\overrightarrow {{\rm{PQ}}} } \right|{\rm{\;}}\left| {\overrightarrow {{\rm{PR}}} } \right|\cos \emptyset = \overrightarrow {{\rm{PR}}} {\rm{\;}}.\overrightarrow {{\rm{PQ}}} = 0\)

∴ cos ∅ = 0

⇒ ∅ = 90°

As we know,

Sum of all internal angle of tringle equal to 180

So from the diagram,

⇒ ∅ + θ + x = 180°

Put ∅ = 90°

⇒ ∠PRQ = x = 90 - θ