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General Aptitude

1

The mean and the standarddeviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

A

0

B

1

C

2

D

4

Here mean = $$\overline x $$ = 9

$$ \Rightarrow $$ $$\overline x $$ = $${{\sum {{x_i}} } \over n}$$ = 9

$$ \Rightarrow $$ $${\sum {{x_i}} }$$ = 9 $$ \times $$ 5 = 45

Now, standard deviation = 0

$$\therefore\,\,\,$$ all the five terms are same i.e.; 9

Now for changed observation

$${\overline x _{new}}$$ = $${{36 + {x_5}} \over 5} = 10$$

$$ \Rightarrow $$ x_{5} = 14

$$\therefore\,\,\,$$ $$\sigma $$_{new} = $$\sqrt {{{\sum {{{\left( {{x_i} - {{\overline x }_{new}}} \right)}^2}} } \over n}} $$

= $$\sqrt {{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}} \over 5}} $$ = 2

$$ \Rightarrow $$ $$\overline x $$ = $${{\sum {{x_i}} } \over n}$$ = 9

$$ \Rightarrow $$ $${\sum {{x_i}} }$$ = 9 $$ \times $$ 5 = 45

Now, standard deviation = 0

$$\therefore\,\,\,$$ all the five terms are same i.e.; 9

Now for changed observation

$${\overline x _{new}}$$ = $${{36 + {x_5}} \over 5} = 10$$

$$ \Rightarrow $$ x

$$\therefore\,\,\,$$ $$\sigma $$

= $$\sqrt {{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}} \over 5}} $$ = 2

2

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

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

$$ = 22821 - 22801$$

$$ = 20$$

3

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

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

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

4

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

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