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Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

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Fluid Mechanics and Hydraulic Machines

General Aptitude

1

If the mean deviation of the numbers 1, 1 + d, ..., 1 +100d from their mean is 255, then a value of d is :

A

10.1

B

20.2

C

10

D

5.05

Given numbers are,

1, 1 + d, 1 + 2d . . . . . 1 + 100d

$$ \therefore $$ Total 101 number are present.

$$ \therefore $$ n = 101

$$ \therefore $$ mean $$\left( {\overline x } \right)$$ = $${{1 + \left( {1 + d} \right) + ......\left( {1 + 100d} \right)} \over {101}}$$

= $${1 \over {101}} \times {{101} \over 2}$$ [1 + (1 + 100d)]

= 1 + 50d

$$ \therefore $$ Mean deviation from mean

= $${1 \over {101}}$$ $$\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right|} \right.$$

$$\left. { + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right]$$

= $${{2\left| d \right|} \over {101}}$$ ( 1 + 2 + 3 + . . . . . + 50)

= $${{2\left| d \right|} \over {101}} \times {{50 \times 51} \over 2}$$

= $${{2550} \over {101}}\left| d \right|$$

From question,

$${{2550} \over {101}}\left| d \right|$$ = 255

$$ \Rightarrow $$ $$\left| d \right|$$ = 10.1

1, 1 + d, 1 + 2d . . . . . 1 + 100d

$$ \therefore $$ Total 101 number are present.

$$ \therefore $$ n = 101

$$ \therefore $$ mean $$\left( {\overline x } \right)$$ = $${{1 + \left( {1 + d} \right) + ......\left( {1 + 100d} \right)} \over {101}}$$

= $${1 \over {101}} \times {{101} \over 2}$$ [1 + (1 + 100d)]

= 1 + 50d

$$ \therefore $$ Mean deviation from mean

= $${1 \over {101}}$$ $$\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right|} \right.$$

$$\left. { + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right]$$

= $${{2\left| d \right|} \over {101}}$$ ( 1 + 2 + 3 + . . . . . + 50)

= $${{2\left| d \right|} \over {101}} \times {{50 \times 51} \over 2}$$

= $${{2550} \over {101}}\left| d \right|$$

From question,

$${{2550} \over {101}}\left| d \right|$$ = 255

$$ \Rightarrow $$ $$\left| d \right|$$ = 10.1

2

The mean of 5 observations is 5 and their variance is 124. If three of the observations
are 1, 2 and 6 ; then the mean deviation from the mean of the data is :

A

2.4

B

2.8

C

2.5

D

2.6

Let 5 observations are x_{1}, x_{2}, x_{3}, x_{4}, x_{5}

given, x_{1} = 1, x_{2} = 2, x_{3} = 6

Mean = 5

$$ \therefore $$ Mean$$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$$ = 5

$$ \Rightarrow $$ 1 + 2 + 6 + x_{4} + x_{5} = 25

$$ \therefore $$ x_{4} + x_{5} = 16

$$ \Rightarrow $$ (x_{4} $$-$$ 5) + (x_{5} $$-$$ 5) + 10 = 16

$$ \Rightarrow $$ (x_{4} $$-$$ 5) + (x_{5} $$-$$ 5) = 6

$$ \therefore $$ Mean deviation about mean,

= $${{\sum {\left| {{x_i} - \overline x } \right|} } \over n}$$

= $${{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}$$

= $${{4 + 3 + 1 + 6} \over 5}$$

= $${{14} \over 5}$$

= 2.8

given, x

Mean = 5

$$ \therefore $$ Mean$$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$$ = 5

$$ \Rightarrow $$ 1 + 2 + 6 + x

$$ \therefore $$ x

$$ \Rightarrow $$ (x

$$ \Rightarrow $$ (x

$$ \therefore $$ Mean deviation about mean,

= $${{\sum {\left| {{x_i} - \overline x } \right|} } \over n}$$

= $${{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}$$

= $${{4 + 3 + 1 + 6} \over 5}$$

= $${{14} \over 5}$$

= 2.8

3

The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :

A

25

B

30

C

35

D

40

Mrean $$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2}..... + {x_n}} \over n}$$ = $${{\sum x } \over n}$$

Here mean = 40 of 25 teachers

$$\therefore\,\,\,$$ 40 = $${{\sum x } \over {25}}$$

$$ \Rightarrow $$$$\,\,\,$$ $$\sum x $$ = 40 $$ \times $$ 25 = 1000

After retireing of a 60 year old teacher, total age of 24 teachers,

x_{1} + x_{2} + . . . . . .x_{24} = 1000 $$-$$ 60 = 940

Now a new teacher of age A year is appointed.

$$\therefore\,\,\,$$ Now total age of this 25 teachers

x_{1} + x_{2} + x_{3} + . . . . . + x_{25} = 940 + A

$$\therefore\,\,\,$$ mean age = $${{940 + A} \over {25}}$$

According to question,

$${{940 + A} \over {25}}$$ = 39

$$ \Rightarrow $$$$\,\,\,$$ A = 35

Here mean = 40 of 25 teachers

$$\therefore\,\,\,$$ 40 = $${{\sum x } \over {25}}$$

$$ \Rightarrow $$$$\,\,\,$$ $$\sum x $$ = 40 $$ \times $$ 25 = 1000

After retireing of a 60 year old teacher, total age of 24 teachers,

x

Now a new teacher of age A year is appointed.

$$\therefore\,\,\,$$ Now total age of this 25 teachers

x

$$\therefore\,\,\,$$ mean age = $${{940 + A} \over {25}}$$

According to question,

$${{940 + A} \over {25}}$$ = 39

$$ \Rightarrow $$$$\,\,\,$$ A = 35

4

The sum of 100 observations and the sum of their squares are 400 and 2475,
respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If
the incorrect observations are omitted, then the variance of the remaining observations
is :

A

8.25

B

8.50

C

8.00

D

9.00

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