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General Aptitude

1

5 students of a class have an average height 150 cm and variance 18 cm^{2}. A new student, whose height is 156 cm, joined them. The variance (in cm^{2}) of the height of these six students is :

A

16

B

22

C

20

D

18

Average height of 5 students,

$$\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150$$

$$ \Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750$$

We know,

Variance $$\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$$

given that,

$${{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18$$

$$ \Rightarrow \,\,\,\sum {x_i^2} = 112590$$

Height of new student, x_{6} $$=$$ 156 cm

New average height $$\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151$$

New variance $$ = {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}$$

$$ = {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}$$

$$ = 22821 - 22801$$

$$ = 20$$

$$\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150$$

$$ \Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750$$

We know,

Variance $$\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$$

given that,

$${{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18$$

$$ \Rightarrow \,\,\,\sum {x_i^2} = 112590$$

Height of new student, x

New average height $$\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151$$

New variance $$ = {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}$$

$$ = {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}$$

$$ = 22821 - 22801$$

$$ = 20$$

2

A data consists of n observations : x_{1}, x_{2}, . . . . . . ., x_{n}.

If $$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$$ and

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$$

then the standard deviation of this data is :

If $$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$$ and

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$$

then the standard deviation of this data is :

A

2

B

$$\sqrt 5 $$

C

5

D

$$\sqrt 7 $$

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n $$

$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)$$

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n $$

$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)$$

Performing (1) + (2), we get

$$2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n$$

$$\sum\limits_{i = 1}^n {x_i^2} = 6n$$

Performing (1) $$-$$ (2), we get

$$ \Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n$$

$$ \Rightarrow $$$$ \Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n$$

S.D($$\sigma $$)$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}} $$

$$\sigma $$ $$ = \sqrt {{{6n} \over n} - \left( 1 \right)} $$

$$\sigma $$ $$ = \sqrt 5 $$

$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)$$

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n $$

$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)$$

Performing (1) + (2), we get

$$2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n$$

$$\sum\limits_{i = 1}^n {x_i^2} = 6n$$

Performing (1) $$-$$ (2), we get

$$ \Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n$$

$$ \Rightarrow $$$$ \Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n$$

S.D($$\sigma $$)$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}} $$

$$\sigma $$ $$ = \sqrt {{{6n} \over n} - \left( 1 \right)} $$

$$\sigma $$ $$ = \sqrt 5 $$

3

In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is

A

42

B

102

C

1

D

38

Let n(A) = number of students opted mathematic = 70,

n(B) = number of studens opted Physics = 46,

n(C) = number of students opted Chemistry = 28,

n(A $$ \cap $$ B) = 23,

n(B $$ \cap $$ C) = 9,

n(A $$ \cap $$ C) = 14,

n(A $$ \cap $$ B $$ \cap $$ C) = 4,

Now n(A $$ \cup $$ B $$ \cup $$ C)

= n(A) + n(B) + n(C) $$-$$ n(A $$ \cap $$ B) $$-$$ n(B $$ \cap $$ C)

$$-$$ n(A $$ \cap $$ C) + n(A $$ \cap $$ B $$ \cap $$ C)

= 70 + 46 + 28 $$-$$ 23 $$-$$ 9 $$-$$ 14 + 4 = 102

So number of students not opted for any course

Total $$-$$ n(A $$ \cup $$ B $$ \cup $$ C)

= 140 $$-$$ 102 = 38

n(B) = number of studens opted Physics = 46,

n(C) = number of students opted Chemistry = 28,

n(A $$ \cap $$ B) = 23,

n(B $$ \cap $$ C) = 9,

n(A $$ \cap $$ C) = 14,

n(A $$ \cap $$ B $$ \cap $$ C) = 4,

Now n(A $$ \cup $$ B $$ \cup $$ C)

= n(A) + n(B) + n(C) $$-$$ n(A $$ \cap $$ B) $$-$$ n(B $$ \cap $$ C)

$$-$$ n(A $$ \cap $$ C) + n(A $$ \cap $$ B $$ \cap $$ C)

= 70 + 46 + 28 $$-$$ 23 $$-$$ 9 $$-$$ 14 + 4 = 102

So number of students not opted for any course

Total $$-$$ n(A $$ \cup $$ B $$ \cup $$ C)

= 140 $$-$$ 102 = 38

4

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -

A

6 : 7

B

10 : 3

C

4 : 9

D

5 : 8

Let two observations are x_{1} & x_{2}

mean = $${{\sum {{x_i}} } \over 5} = 5 $$

$$\Rightarrow 1 + 3 + 8 + {x_1} + {x_2} = 25$$

$$ \Rightarrow {x_1} + {x_2} = 13$$ . . . . (1)

variance $$\left( {{\sigma ^2}} \right)$$ = $${{\sum {x_i^2} } \over 5} - 25 = 9.20$$

$$ \Rightarrow $$ $${\sum {x_i^2 = 171} }$$

$$ \Rightarrow $$ $$x_1^2 + x_2^2 = 97$$ . . . . . (2)

$$ \Rightarrow $$(x_{1} + x_{2})^{2} $$-$$ 2x_{1}x_{2} = 97

$$ \Rightarrow $$ 169 - 2x_{1}x_{2} = 97

or x_{1}x_{2} = 36

$$ \therefore $$ x_{1} : x_{2} = 4 : 9

mean = $${{\sum {{x_i}} } \over 5} = 5 $$

$$\Rightarrow 1 + 3 + 8 + {x_1} + {x_2} = 25$$

$$ \Rightarrow {x_1} + {x_2} = 13$$ . . . . (1)

variance $$\left( {{\sigma ^2}} \right)$$ = $${{\sum {x_i^2} } \over 5} - 25 = 9.20$$

$$ \Rightarrow $$ $${\sum {x_i^2 = 171} }$$

$$ \Rightarrow $$ $$x_1^2 + x_2^2 = 97$$ . . . . . (2)

$$ \Rightarrow $$(x

$$ \Rightarrow $$ 169 - 2x

or x

$$ \therefore $$ x

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