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Questions Related to maths

The marks obtained by $20$ students of Class $X$ of a certain school in a English paper consisting of $100$ marks are presented in table below. Find the mean of the marks obtained by the students using step deviation method.

maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $61$

  2. $62$

  3. $63$

  4. $64$

Show answer
Correct Option: C
Explanation:

Answer:- Using Shortcut Method

Class interval width (i) = $50-40 = 10$

Marks  X  F  $d=\cfrac{X-A}{i}$  Fd 
40-50  45  5  -2  -10 
50-60  55   4  -1 -4 
60-70  65 = A  3  0 0 
70-80  75   6  1 6 
80-90  85   2  2 4 
    $\Sigma F = 20$    $\Sigma Fd = -4$ 

Mean ( By step deviation formula) = $A+\cfrac{\Sigma Fd}{\Sigma F} \times i = 65+\left(\cfrac{-4}{20} \times 10 \right) = 65=2 = 63$

C) Mean = $63$

Find the arithmetic mean using step deviation method for the following data shows distance covered by $40$ passengers to perform their work. (Round off your answer to the nearest whole number).

Distance (km) 1-5 5-9 9-13 13-17 17-21 21-25 25-29
Number of passengers 2 4 6 8 10 5 5


maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $15$

  2. $16$

  3. $17$

  4. $19$

Show answer
Correct Option: C
Explanation:
 Distance (km) Mid point (X)  Number of passengers(F)  i=class interval width  A=11 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 1-5 3  2  4  11  -2  -4 
5-9  7  4  4  11  -1  -4 
9-13  11  6  4  11  0  0 
13-17  15  8  4  11  1  8 
17-21  19  10  4  11  2  20 
21-25  23  5  4  11  3  15 
25-29  27  5  4  11  4  20 
    $\Sigma f=40$        $\Sigma f d'=55$ 

The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \frac{\sum fd'}{\sum f} \times i$ 
A = Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \frac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =11 +\frac{55}{40}\times 4$
$= 11 + 5.5$
$= 16.5$ $\approx$ $17$

In a study on a certain population, the following data was given.

Population (X) 2000-2001 2001-2002 2002-2003 2003-2004 2004-2005 2005-2006 2006-2007
Number of people 10 20 30 40 50 60 70


Find the average number of population using step deviation method.

maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $2002$

  2. $2003$

  3. $2004$

  4. $2005$

Show answer
Correct Option: D
Explanation:

 X Midpoint (X)  Frequency (F)  i=class interval width  A=2003.5 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 2000-2001 2000.5  10  1  2003.5  -3  -30 
2001-2002  2001.5  20  1  2003.5  -2  -40 
2002-2003  2002.5  30  1  2003.5  -1  -30 
2003-2004  2003.5=A  40  1  2003.5  0   0
2004-2005  2004.5  50  1  2003.5  1  50 
2005-2006  2005.5  60  1  2003.5  2  120 
2006-2007  2006.5  70  1  2003.5  3  210 
    $\Sigma f=280$        $\Sigma fd'=280$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =2003.5 +\dfrac{280}{280}\times 1$
$= 2003.5 + 1$
$= 2004.5$ $\approx$ $2005$

The frequency distribution of marks in English are given in the table:

Marks 50-60 60-70 70-80 80-90
Number of students 12 24 14 10

Find the mean by step deviation method.

maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $58$

  2. $48$

  3. $69$

  4. $71$

Show answer
Correct Option: C
Explanation:

 X Mid point(x)  Frequency (F)  i=class interval width  A=65 Assumed mean  d'=$\dfrac{x-A}{i} $  fd'
 50-60 55  12  10  65  -1  -12 
 60-70 65=A  24  10  65  0  0 
70 -80  75  14 10  65  1  14 
80-90  85  10  10  65  2  20 
    $\Sigma f=60$        $Sigma fd'=22$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =65 +\dfrac{22}{60}\times 10$
$= 65 + 3.666$
$= 68.666$ $\approx$ $69$ marks

Using step deviation method find the mean.

X 20-40 40-60 60-80 80-100
frequency 4 8 12 16


maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $40$

  2. $50$

  3. $60$

  4. $70$

Show answer
Correct Option: D
Explanation:

 X Mid point (X)  Frequency (F)  i=class interval width  A=50 Assumed Means  d'=$\dfrac{x-A}{i}$  fd' 
 20-40 30  4 20  50  -1  -4 
40-60  50=A  8  20  50  0  0 
60-80  70  12  20  50  1  12 
80-100  90  16  20  50  2  32 
    $\Sigma f=40$        $\Sigma f d'=40$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \frac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =50 +\dfrac{40}{40}\times 20$
$= 50 + 20$
$= 70$

 Marks 0-10  10-20  20-30  30-40  40-50  50-60  60-70  70- 80 80-90  90-100 
 Frequency 3  5  6  7  8   9  10  12  6 4 


Find the mean mark using step deviation method:

maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $54$

  2. $55$

  3. $56$

  4. $57$

Show answer
Correct Option: A
Explanation:

 X Mid Point (X)  Frequency (F)  i=class interval width  A=45 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 0-10 5  3  10  45  -4  -12 
10-20  15  5  10  45  -3  -15
20-30  25  6  10  45  -2  -12 
30-40  35  7  10  45  -1  -7 
40-50  45=A 8  10  45  0  0 
50-60  55  9  10  45  1  9 
60-70  65  10  10  45  2  20 
70-80  75  12  10  45  3  36 
80-90  85  6  10  45  4  24 
90-100   95 4  10  45  5  20 
    $\Sigma f=70$        $\Sigma fd'=63$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =45 +\dfrac{63}{70}\times 10$
$= 45 + 9$
$= 54$

 Marks 0-5  5-10  10-15  15-20  20-25  25-30  30-35  35-40  40-45  45-50 
Frequency  3  5  7  8  10  11  14  19  15  13 

For the following distribution, find the mean using step deviation method. (Round off your answer to the nearest whole number)

maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $29$

  2. $31$

  3. $35$

  4. $37$

Show answer
Correct Option: B
Explanation:

 X Mid point (x)  Frequency (F)  i=class  interval width  A=22.5 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
0-5  2.5  3  5  22.5  -4  -12 
5-10  7.5  5  5  22.5  -3  -15 
10 -15 12.5  7  5  22.5  -2  -14 
15-20  17.5  8  5  22.5  -1  -8 
20-25  22.5=A  10  5  22.5  0  0 
25-30  27.5  11  5  22.5  1  11 
30-35  32.5  14  5  22.5  2  28 
35-40  37.5  19  5  22.5  3  57 
40-45  42.5  15  5  22.5  4  60 
45-50  47.5  13  5  22.5  5  65 
    $\Sigma f=105$        $\Sigma fd'=172$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =22.5 +\dfrac{172}{105}\times 5$
$= 22.5 + 8.19$
$= 30.69$ $\approx$ $31$

Construct a triangle $PQR$, whose perimeter is $22 cm$ and whose sides are in the ratio $2 : 4 : 5$. Measure the sides of the triangle.

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $5, 7, 10$

  2. $4, 8, 10$

  3. $5, 8, 9$

  4. None of these

Show answer
Correct Option: B
Explanation:

Let the sides be $2x,4x$ and $5x$

Perimeter $=22cm$
$2x+4x+5x=22$
$11x=22$
$x=2$
So the sides are 
$2x=2\times2=4cm$
$4x=4\times 2=8cm$
$5x=5\times 2=10cm$
Option $B$ is correct.

Construct a $\triangle ABC$ in which $AB= 5.4\ cm, \angle CAB= 45^{\circ}$ and $AC + BC= 9\ cm.$Then, $m\angle ACB$ is:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $55^o$

  2. $75^o$

  3. $85^o$

  4. None of these

Show answer
Correct Option: B

For constructing a triangle whose perimeter and both base angles are given, the base length is equal to:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. the length of the perimeter

  2. the length of the largest side

  3. the difference between the largest and the shortest side

  4. None of these

Show answer
Correct Option: A
Explanation:

To construct a triangle when the perimeter and both base angles are given, we first draw the base with length equal to the perimeter of the triangle. After that we draw the complete triangle with proper method.