Tag: maths

Questions Related to maths

For a certain frequency table which has been partly reproduced here, the arithmetic mean was found to be Rs.28.07

Income (in Rs.) 15 20 25 30 35 40
Number of workers 8 12 ? 16 3 10

If he total number of workers is 75, then the missing frequencies are?

maths average arithmetic mean of ap introduction to averages means

  1. 14, 15

  2. 15, 14

  3. 13, 16

  4. 12, 17

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Correct Option: D

The antithetic mean of the masks is _________.

maths average arithmetic mean of ap introduction to averages means

  1. 50.25

  2. 50.75

  3. 51.25

  4. 53.75

Show answer
Correct Option: A

let a and b be the two different natural numbers whose harmonic mean is 10 then their arithmatic mean is ______.

maths average arithmetic mean of ap introduction to averages means

  1. 12

  2. 15

  3. 16

  4. 18

Show answer
Correct Option: A

If 25 is the arithmetic mean between x and 46, then find x.

maths average arithmetic mean of ap introduction to averages means

  1. 2

  2. 4

  3. 8

  4. 16

Show answer
Correct Option: B
Explanation:

Since $25$ is the arithmetic mean of $X$ and $46$, 

$\therefore 25=\displaystyle \frac{(X+46)}{2}$
$\therefore  X+46=50$
$\therefore  X=4$

The arithmetic mean of first ten natural numbers is

maths average arithmetic mean of ap introduction to averages means

  1. $5.5$

  2. $6$

  3. $7.5$

  4. $10$

Show answer
Correct Option: A
Explanation:

The arithmetic mean of first ten natural numbers is the average of these natural numbers.


$\therefore$ arithmetic mean

   $=\dfrac{1+2+3+4+5+6+7+8+9+10}{10}$

   $=\dfrac{n(\dfrac{n+1}{2})}{10}\,\,\,\,\,\,\,\rightarrow Here \,\boxed{n=10}$

   $=\dfrac{10\times \dfrac{11}{2}}{10}$ 

   $=\dfrac{11}{2}=\boxed{5.5}\rightarrow $ option -A

If AM between $\displaystyle p^{th}$ and $\displaystyle q^{th}$ terms of an AP be equal to the AM between $\displaystyle r^{th}$ and $\displaystyle s^{th}$ term of the AP, then $p + q$ is equal to

maths average arithmetic mean of ap introduction to averages means

  1. $r + s$

  2. $\displaystyle \frac{r-s}{r+s}$

  3. $\displaystyle \frac{r+s}{r-s}$

  4. $r + s + 1$

Show answer
Correct Option: A
Explanation:

We know A.P formula for nth terms with 'a' as the first term and 'd' as the common difference as shown below:


${ t } _{ n }=a+\left( n-1 \right) d$

Also AM is given between two numbers a and b. 
       $A=\dfrac { a+b }{ 2 } $

So arithmetic mean of pth and qth terms of AP is as shown below:

$=\dfrac { a+\left( p-1 \right) d+a+\left( q-1 \right) d }{ 2 } $

Similarly we can have AM of rth term and sth term of AP as shown below:

$=\dfrac { a+\left( r-1 \right) d+a+\left( s-1 \right) d }{ 2 } $

Applying the given conditions we get,

$\dfrac { a+\left( p-1 \right) d+a+\left( q-1 \right) d }{ 2 } =\dfrac { a+\left( r-1 \right) d+a+\left( s-1 \right) d }{ 2 } $

      $\dfrac { a+pd-d+a+qd-d }{ 2 } =\dfrac { a+rd-d+a+sd-d }{ 2 } $

$a+pd-d+a+qd-d=a+rd-d+a+sd-d$

         $2a+d\left( p+q \right) -2d=2a+d\left( r+s \right) -2d$

                           $d\left( p+q \right) =d\left( r+s \right) d$

                                 $p+q=r+s$ 

Hence option A is correct.

If $a+b+c+d+e+f=12$ then the maximum value of $ab+bc+cd+de+ef+fa$ is (a, b, c, d, e, f are non negative real numbers)
maths average arithmetic mean of ap introduction to averages means

  1. $36$

  2. $24$

  3. $30$

  4. none of these

Show answer
Correct Option: B
Explanation:
AM $\ge$ GM
$\cfrac { a+b+c+d+e+f }{ 6 } \ge { (abcdef) }^{ 1/6 }\\ { 2 }^{ 2 }\ge { (abcdef) }^{ 1/3 }\quad \quad \quad (1)\\ \cfrac { ab+bc+cd+de+ef+fa }{ 6 } \ge { (abcdef) }^{ 2/6 }\\ ab+bc+cd+de+ef+fa\ge 6{ (abcdef) }^{ 1/3 } \quad (2)$
Dividing $(2)$ by $(1)$
$ab+bc+cd+de+ef+fa \ge 24$

The arithmetic mean of 1, 2, 3, ..., n, is

maths average arithmetic mean of ap introduction to averages means

  1. $\displaystyle \frac{n-1}{2}$

  2. $\displaystyle \frac{n+1}{2}$

  3. $\displaystyle \frac{n}{2}$

  4. $\displaystyle \frac{n}{2}+1$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$   We have the sequence, $1,2,3.......n$

$\Rightarrow$   This is an AP, with the initial term $a=1$ and the common difference $d=1$.
$\therefore$   The sum of $n$ terms of an AP is given by,
$\Rightarrow$  $S _n=\dfrac{n}{2}[2a+(n-1)d]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[2\times 1+(n-1)\times 1]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[2+(n-1)]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[n+1]$
$\rightarrow$   Arithmetic mean of $n$ numbers $a _1,a _2,a _3,a _4,... a _n$ is given by the formula
$\Rightarrow$  $Arithmetic\,mean=\dfrac{a _1+a _2+a _3+a _4+...+a _n}{n}$

$\Rightarrow$  $Arithmetic\,mean=\dfrac{S _n}{n}$

$\Rightarrow$  $Arithmetic\, mean=\dfrac{\dfrac{n}{2}[n+1]}{n}$

$\therefore$     $Arithmetic\, mean=\dfrac{n+1}{2}$

Find the arithmetic mean of the series $1, 3, 5,...........
(2n - 1)$

maths average arithmetic mean of ap introduction to averages means

  1. $n$

  2. $2n$

  3. $n/2$

  4. $n - 1$

Show answer
Correct Option: A
Explanation:

In the given Arithmetic Progression,
First term $ = a  = 1 $
Common difference $ = 3 - 1 = 2 $

Let $ 2n-1 $ be the $  k $ th term.

Then $ {x} _{n} = a + (n-1)d $
$ => 2n-1 = 1 + (k-1)2  $
$ => 2n -1 = 1 +2k-2 $
$ => k = n $

So,  $ 2n-1 $ is the $ n $ th term.

Now, Sum of the given series upto 'n' terms $ = \frac {n}{2} (2a+(n-1)d) =n^2 $

so mean is = sum of series/total no of terms
$= n^2/n=n$

Find the arithmetic mean of the series: $1,3,5 ........... (2n - 1)$

maths average arithmetic mean of ap introduction to averages means

  1. $n$

  2. $2n$

  3. $\dfrac n2$

  4. $n - 1$

Show answer
Correct Option: A
Explanation:

$1,3,5,..., (2n-1)$

Total number of terms: $n$
Arithmetic mean: $\dfrac{1+3+5+\cdots+(2n-1)}{n}$

Using sum of first n terms of an AP:
$1+3+5+\cdots+(2n-1) = n^2$
$\therefore$ AM = $\dfrac{n^2}n = n$