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Introduction to averages - class-V

Description: introduction to averages
Number of Questions: 95
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Tags: average binomial theorem, sequence and series sequence, progression and series sequences and series maths arithimetic progression
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The $A.M.$ of the observations $1.3.5,3.5,7,9,.,(2n-1)(2n+1) (2n+3)$ is $(\forall\ n\ \in\ N)$

  1. $2n^{3}+6n^{2}+7n-2$

  2. $n^{3}+8n^{2}+7n-2$

  3. $2n^{3}+5n^{2}+6n-2$

  4. $2n^{3}+8n^{2}+7n-2$


Correct Option: D

The mean of ${1^{2,}}{2^2},{3^2},{4^2},{5^2},{6^2},{7^2}$ is:

  1. 10

  2. 20

  3. 30

  4. None of these


Correct Option: B

The A.M. of a + 2, a, 2-a is

  1. $a$

  2. $\cfrac { a+4 }{ 3 } $

  3. $\cfrac { a-4 }{ 3 } $

  4. $\cfrac { a }{ 2 } $


Correct Option: B
Explanation:

Given values are : $a+2, a, 2-a$


Arithmetic Mean, $AM = \dfrac{a+2+a+2-a}{3}$

$AM = \dfrac{a+4}{3}$

Hence, option B is corret

Find the mean of $ 43,54,64,53,36$

  1. $50$

  2. $ 40$

  3. $60$

  4. $ 30$


Correct Option: A
Explanation:

Given data $43,54,64,53,36$


Sum of observations $43+54+64+53+36=250$


No . of observations $5$

Mean of data $=\dfrac{250}{5}=50$

The arithmetic mean of $1 + \sqrt { 2 }$ and $7 + 5 \sqrt { 2 }$ is $\sqrt { a } + \sqrt { b }$ . Then $a - b =$

  1. -1

  2. 1

  3. 2

  4. -2


Correct Option: D
Explanation:

The arithmetic mean of $1+\sqrt{2}$ and $7+5\sqrt{2}$ is $\dfrac{1+\sqrt{2}+7+5\sqrt{2}}{2}=\dfrac{8+6\sqrt{2}}{2}=4+3\sqrt{2}=\sqrt{16}+\sqrt{18}$

$\implies \sqrt{a}+\sqrt{b}=\sqrt{16}+\sqrt{18}$
$\implies a=16,b=18$
$a-b=16-18=-2$

The airthmatic mean of $1 + \sqrt { 2 }$ and $7 + 5 \sqrt { 2 }$ is $\sqrt { a } + \sqrt { b }$ . Then a $- b =$

  1. -1

  2. 1

  3. 2

  4. -2


Correct Option: D
Explanation:

The arithmetic mean of $1+\sqrt{2}$ and $7+5\sqrt{2}$ is $\dfrac{1+\sqrt{2}+7+5\sqrt{2}}{2}=\dfrac{8+6\sqrt{2}}{2}=4+3\sqrt{2}=\sqrt{16}+\sqrt{18}$

$\implies \sqrt{a}+\sqrt{b}=\sqrt{16}+\sqrt{18}$
$\implies a=16,b=18$
$a-b=16-18=-2$

Mean of the first $n$ terms of the A.P. $a, (a + d), (a + 2d), ........$ is

  1. $\displaystyle a + \frac{nd}{2}$

  2. $\displaystyle a + \frac{(n - 1)d}{2}$

  3. $a + (n - 1) d$

  4. $a + nd$


Correct Option: B
Explanation:

Required mean $= \displaystyle \frac{a + (a + d) + (a + 2d) + ....... + { a + (n - 1) d }}{n}$
$\displaystyle = \frac{\displaystyle \frac{n}{2} [a + a + (n - 1) d]}{n} = a + \frac{(n - 1)d}{2}$

If the arithmetic mean of $6, 8, 5, 7, x$ and $4$ is $7,$ then $x$ is

  1. $12$

  2. $6$

  3. $8$

  4. $4$


Correct Option: A
Explanation:

By definition,
$\displaystyle AM=\frac{6+8+5+7+x+4}{6}=7$
or $x=12.$

Find the Arithmetic mean between 24 and 36 

  1. 26

  2. 28

  3. 30

  4. 32


Correct Option: C
Explanation:

The arithmetic mean(AM) between 2 numbers is the simple average of the two numbers.

So, AM = $\qquad \dfrac { 24\quad +\quad 36 }{ 2 } \ =\quad 30$.

AM = 30.

The arithmetic mean of $12$ and $20$ is :

  1. $12$

  2. $14$

  3. $16$

  4. $18$


Correct Option: C
Explanation:

The AM of two numbers is their average.
So, the AM of $12$ and $20$ is $ \dfrac { 12+ 20 }{ 2 }= 16$.

Find the AM between $20$ and $26$.

  1. $23$

  2. $22$

  3. $21$

  4. $24$


Correct Option: A
Explanation:

$\displaystyle AM=\frac { 20+26 }{ 2 } =\frac { 46 }{ 2 } =23$

The arithmetic mean of 5, 6, 8, 9, 12, 13, 17 is

  1. $20$

  2. $15$

  3. $10$

  4. $25$


Correct Option: C
Explanation:

Arithmetic mean $=\dfrac {\text{Sum of Terms}}{\text{No. of Terms}}=\dfrac {5+6+8+9+12+13+17}{7}$


Arithmetic mean $=\dfrac {70}{7} = 10$

If each observation is multiplied by $\displaystyle \frac{1}{3}$ then the mean of the new data will de

  1. $\displaystyle \frac{1}{3}$ times

  2. 3 times

  3. $\displaystyle \frac{1}{\sqrt{3}}$ times

  4. $\displaystyle \frac{2}{3}$ times


Correct Option: A
Explanation:

Let set of data is 9,10,11,17 and 13

Then mean =$\frac{9+10+11+17+13}{5}=\frac{60}{5}$
If Each observation is multiply by $\frac{9+10+11+17+13}{5}=\frac{60}{5}=12$
Then data are $\frac{9}{3},\frac{10}{3},\frac{11}{3},\frac{17}{3}and \frac{13}{3}$
Then sum of data=$\frac{9}{3}+\frac{10}{3}+\frac{11}{3}+\frac{17}{3}+ \frac{13}{3}=\frac{60}{3}=20$
Then new mean of data=$\frac{20}{5}=4$
Then new data mean =$\frac{4}{12}=\frac{1}{3}$ times of old data

The mean of $x, y, z$ is $y$, then $x + z = .............$

  1. $y$

  2. $3y$

  3. $2y$

  4. $4y$


Correct Option: C
Explanation:

Mean of 3 numbers $=\dfrac {\mbox {Sum of three numbers}}{3}$

$\Rightarrow y = \dfrac {x+y+z}{3}$

$\Rightarrow 3y=x+y+z$

$\Rightarrow x+z = 2y$

The arithmetic mean of first five natural number is

  1. $2$

  2. $3$

  3. $4$

  4. $8$


Correct Option: B
Explanation:

$Arithmetic\ mean=\cfrac{1+2+3+4+5}{5}$
$=3$

The arithmetic mean between $2+\sqrt {(2)}$ and $2-\sqrt {(2)}$ is

  1. $2$

  2. $\sqrt {(2)}$

  3. $0$

  4. $4$


Correct Option: A
Explanation:

Since, the arithmetic mean between $a$ and $b$ is $\displaystyle  \frac {a+b}{2}$
$\therefore $the arithmetic mean between $2+\sqrt 2$ and $2-\sqrt 2$  $=\displaystyle \frac {2+\sqrt 2+2-\sqrt 2}{2}$

$=\dfrac {4}{2}$

$=2$
Option A is correct.

Find the arithmetic mean of the progression $2, 4, 6, 8, 10.$

  1. $10$

  2. $20$

  3. $30$

  4. $6$


Correct Option: D
Explanation:

Using the formula for Required arithmetic mean $=\dfrac{\text{sum of the terms}}{\text{number of terms}}$


After substituting the values we get$=\dfrac{2+4+6+8+10}{5}=\dfrac{30}{5}=6$

What is the arithmetic mean of the progression $11, 22, 33, 44, 55, 66, 77?$

  1. $44$

  2. $208$

  3. $308$

  4. $48$


Correct Option: A
Explanation:

Using the formula for required arithmetic mean $=\dfrac{\text{sum of the terms}}{\text{number of terms}}$


After substituing the values we get: $=\dfrac{11+22+33+44+55+66+77}{7}$

                                                $\quad \quad \quad =\dfrac{11(1+2+3+4+5+6+7)}{5}=\dfrac{11\cdot 28}{7}=11\cdot 4=44$

Find the arithmetic mean of first $10$ natural numbers.

  1. $55$

  2. $550$

  3. $5.5$

  4. None of the above


Correct Option: C
Explanation:

Arithmetic mean of first 10 natural numbers is:

$=\dfrac{1+2+3+4+5+6+7+8+9+10}{10}$
$=\dfrac{(\dfrac{n.(n+1)}{2})}{10}$,  where $n=10$. 
$=\dfrac{(\dfrac{10.11}{2})}{10}$
$=5.5$                                             

Find  AM  of  first $250$  natural numbers.

  1. $115$

  2. $225$

  3. $125$

  4. $125.5$


Correct Option: D
Explanation:

Sum of first 250 natural numbers:

$S=\dfrac{n(n+1)}{2}=\dfrac{250\times 251}{2}$
$\therefore A.M.=\dfrac{S}{n}=\dfrac{250\times 251}{250\times 2}=125.5$

If $n^{th}$ term of AP is $4n+1$, then AM of $11^{th}$ to $ 20^{ th}$ terms is 

  1. $61.5$

  2. $63$

  3. $63.5$

  4. $62$


Correct Option: B
Explanation:

Given: $t _{n}=4n+1$

$A.M.=\cfrac{t _{11}+t _{20}}{2}$
           $=\cfrac{4\times 11+1+4\times 20+1}{2}$
           $=63$

If  $n^{th}$ term of AP is $t _n=4n+1$. Find mean of first $10$ terms.   

  1. $85$

  2. $95$

  3. $23$

  4. $7.5$


Correct Option: C
Explanation:
Given:
$t _{n}=4n+1$
$\therefore A.M.=\dfrac{\sum _{n=1}^{10}4n+1}{n}$
               $=\dfrac{2n(n+1)+n}{n}$
               $=2n+3$
               $=2\times 10+3$
               $=23$

Find AM of multiple of $3$  from natural numbers $1$ to $100$.

  1. $48$

  2. $51$

  3. $36$

  4. $57$


Correct Option: B
Explanation:

First term $=3$

Last term $=99$
$\therefore A.M.=\dfrac{3+99}{2}=51$

The  arithmatic mean of $4,6,8$ is

  1. $4$

  2. $6$

  3. $8$

  4. $4.5$


Correct Option: B
Explanation:

$\Rightarrow$  First adding numbers $4,\, 6$ and $8$ = $4+6+8=18$

$\Rightarrow$   We have number of terms 3
$\Rightarrow$   $Arithemetic\, mean=\dfrac{S}{N}=\dfrac{18}{3}=6$

Find AM of  $ 3$  digit even numbers between $1$  to $500$.

  1. $200$

  2. $400$

  3. $300$

  4. $150$


Correct Option: C
Explanation:

First term $=100$

Last term $=500$
$\therefore A.M.=\cfrac{100+500}{2}=300$

Find AM of divisors of $100$.

  1. $24$

  2. $25.5$

  3. $24.11$

  4. $21.9$


Correct Option: C
Explanation:

Divisors of 100 are:

$1, 2, 4, 5, 10, 20, 25, 50, 100$
$\therefore n=9$
$\therefore S=1+2+4+5+10+20+25+50+100=217$
$\therefore A.M.=\dfrac{S}{n}=\dfrac{217}{9}=24.11$

If the nth term of AP is $2n+5$. Then find the  AM of first $38$  terms.

  1. $99$

  2. $98$

  3. $100$

  4. $44$


Correct Option: D
Explanation:
Given:
$t _{n}=2n+5$
First term $=2\times 1+5=7$
Last term $=2\times 38+5=81$
$\therefore A.M.=\dfrac{7+81}{2}=44$

The AM of  multiple of $5$ from numbers $1$ to $500$ is

  1. $250$

  2. $\dfrac{500}{2}$

  3. $\dfrac{505}{2}$

  4. $252.5$


Correct Option: D
Explanation:

First term $=5$

Last term $=500$
$\therefore A.M.=\dfrac{5+500}{2}=252.5$

The  Sum of three numbers in AP is $75$, and product of extremities is $609$. The numbrs and AM of 1st two numbers is 

  1. ${21,25,29}$, AM $= 23$

  2. ${13,17,21}$, AM $= 22$

  3. ${21,25,29}$, AM $= 25$

  4. ${21,22,29}$, AM $= 23$


Correct Option: A
Explanation:

Let the numbers in A.P.  be (a-d), a, (a+d)

$\therefore (a-d)+a+(a+d)=75$
$\Rightarrow a=25$
Also, $(a-d)(a+d)=609$
$\Rightarrow a^{2}-d^{2}=609$
$\Rightarrow 25^{2}-d^{2}=609$
$\therefore d=\pm4$
The numbers are {21,25,29} or {29,25,21}
According to option, we take {21,25,29}
A.M. of 1st two numbers $=\dfrac{21+25}{2}=23$

If the Arithmetic mean of $8, 6, 4, x, 3, 6, 0$ is $4$; then the value of $x =$

  1. $7$

  2. $6$

  3. $1$

  4. $4$


Correct Option: C
Explanation:

Arithmetic mean $= \cfrac{\text{sum of all observations}}{\text{no. of observations}}$

$\Rightarrow 4=\cfrac { 8+6+4+x+3+6+0 }{ 7 } \ \Rightarrow 28=27+x\ \Rightarrow x=1$

Arithmetic mean of $2$ and $8$ is

  1. $5$

  2. $10$

  3. $16$

  4. $3.2$


Correct Option: A
Explanation:

Arithmetic mean of $a$ and $b$ is $\dfrac{a+b}{2}$

$\therefore$ arithmetic mean of 2 and 8 is $\dfrac{2+8}{2}=\dfrac{10}{2}=5$
Hence, option $A$ is correct.

State the following statement is True or False
Arithmetic mean of first five natural numbers is $3$.

  1. True

  2. False


Correct Option: A
Explanation:

First five natural numbers are $1 ,2 , 3 ,4, 5$

arithmetic mean = $\dfrac { sum\quad of\quad numbers }{ count\quad of\quad numbers } $
$ = \dfrac{1 + 2 + 3 + 4  + 5 }{ 5} $
$=\dfrac { 15 }{ 5 } =3$
Arithmetic mean of first five natural number is 3.
Hence the given statement is true.

The arithmetic mean of the squares of the first $n$ natural numbers is

  1. $\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $

  2. $\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 2 } $

  3. $\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 } $

  4. $\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 3 } $


Correct Option: C
Explanation:

Sum of squares of the first $n$ natural numbers $=\dfrac { n(n+1)(2n+1) }{ 6 } $


Their arithmetic mean is $\dfrac { Sum }{ n } =\dfrac { n(n+1)(2n+1) }{ 6n } =\dfrac { (n+1)(2n+1) }{ 6 } $
Hence, C is correct.

The middle terms , if four different numbers are in proportion are called ______ .

  1. Antecedents

  2. Means

  3. Extremes

  4. Consequents


Correct Option: B
Explanation:

$\Rightarrow$  The middle terms  of four different numbers are in proportion are called $Means.$

$\Rightarrow$  A proportion is simply a statement that two ratios are equal. It can be written in two ways:  $\dfrac{a}{b} = \dfrac{c}{d}$ or  $a:b = c:d.$
$\Rightarrow$  Here, outer terms $a$ and $d$ called extremes.
$\Rightarrow$  Middle terms $b$ and $c$ are called means.

The arithmetic mean between $\cfrac { x+a }{ x } $ and $\cfrac { x-a }{ x } $ when $x\ne 0$, is (the symbol $\ne$ means "not equal to"):

  1. $2$, if $a\ne 0$

  2. $1$

  3. $1$, only if $a=0$

  4. $\dfrac {a}{x}$

  5. $x$


Correct Option: B

The arithmetic mean (average) of a set of $50$ numbers is $38$. If two numbers, namely, $45$ and $55$, are discarded, the mean of the remaining set of numbers is :

  1. $36.5$

  2. $37$

  3. $37.2$

  4. $37.5$

  5. $37.52$


Correct Option: D
Explanation:

$Arithmetic\quad sum=\cfrac { sum\quad of\quad numbers }{ number\quad of\quad numbers } $
$\therefore S=50\times 38=1900;\quad \therefore x=\cfrac { 1900-45-55 }{ 48 } =37.5$

If $A _1,A _2$ be two arithmetic means between $\dfrac{1}{3}$ and $\dfrac{1}{24}$, then their value are 

  1. $\dfrac{7}{72},\dfrac{5}{36}$

  2. $\dfrac{17}{72},\dfrac{5}{36}$

  3. $\dfrac{7}{36},\dfrac{5}{72}$

  4. $\dfrac{5}{72},\dfrac{17}{72}$


Correct Option: B

Sum of $4$ numbers in GP is $60$. And the AM of first and last no. is $18$ find the first term and common difference of the GP

  1. $a=4, r=2$

  2. $a=32, r=\dfrac {1}{2}$

  3. $a=3, r=1$

  4. $a=6, r=3$


Correct Option: A,B
Explanation:

Let 4 numbers in G.P. be $a,ar,{ ar }^{ 2 }{ ,ar }^{ 3 }\ a(1+r+{ r }^{ 2 }+{ r }^{ 3 })=60\ \cfrac { a+{ ar }^{ 3 } }{ 2 } =18=>a+{ ar }^{ 3 }=36\ a(1+r)(1+{ r }^{ 2 })=60\ a(1+r)(1+{ r }^{ 2 }-r)=36\ \cfrac { 1+{ r }^{ 2 }-r }{ 1+{ r }^{ 2 } } =\cfrac { 36 }{ 60 } \ 5+5{ r }^{ 2 }-5r=3+3{ r }^{ 2 }\ 2{ r }^{ 2 }-5r+2=0\ r=2,\cfrac { 1 }{ 2 } \ if\quad r=2,a(1+8)=36=>a=4\ if\quad r=\cfrac { 1 }{ 2 } ,a(1+\cfrac { 1 }{ 8 } )=36=>a=32\ a=4,r=2\quad (or)\quad a=32,r=\cfrac { 1 }{ 2 } $

The A.M. of the observations $1.3.5, 3.5.7, 5.7.9,...,(2n-1)(2n+1)(2n+3)$ is $(\forall n\in N)$

  1. $2n^3+6n^2+7n-2$

  2. $n^3+8n^2+7n-2$

  3. $2n^3+5n^2+6n-1$

  4. $2n^3+8n^2+7n-2$


Correct Option: A
Explanation:
$1.3.5, 3.5.7, 5.7.9, ........, (2n-1)(2n+1)(2n+3) $
 $x _1$       $x _2$       $x _3$                                   $x _n=n \,  terms$

$AM=\displaystyle\sum _{i=1}^{n}\dfrac{x _{i}}{n}$

$\displaystyle\sum _{i=1}^{n}x _{i}=\sum _{r=1}^{n}(2r-1)(2r+1)(2r+3)$

$=\displaystyle\sum _{r=1}^{n} 8r^{3}-2r+12r^{2}-3$

$=\displaystyle 8\sum _{r=1}^{n}+12\sum _{r=1}^{n}r^{2}-2\sum _{r=1}^{n}-3\sum _{r=1}^{n}(1) ....... (1)$

$1^{2}+2^{2}+.........+x^{2}=\dfrac{n(2n+1)(n+1)}{6}=\displaystyle\sum _{r=1}^{n}r^{2}$

$\displaystyle\sum _{r=1}^{n}r^{3}=1^{3}+2^{3}+......+ n^{3}=\left(\dfrac{n(n+1)}{2}\right)^{2}$

$\displaystyle\sum _{r=1}^{n}r=1+2+3+.......+ n=\dfrac{n(n+1)}{2}$

from $(1)$

$8\left[\dfrac{n(n+1)}{2}\right]^{2}+12\left[\dfrac{n(2n+1)(n+1)}{6}\right]-\dfrac{2n(n+1)}{2}-3n$

$=2(n(n+1))^{2}+2n(2n+1)(n+1)-n(n+1)-3n$

$=n(n+1)[2n^{2}+2n+4n+2-1]-3n$

$\displaystyle\sum =n(n+1)[2x^{2}+6x+1]-3n$

$A.M=\dfrac{n(n+1)(2n^{2}+6n+1)-3n}{n}$

$=(n+1)(2n^{2}6n+1)-3$

$=2n^{3}+6n^{2}+n-2n^{2}+6n+1-3=2n^{3}+8n^{2}+7n-2$

If $n\ AM's$ are inserted between $1$ and $31$ and ratio of ${7}^{th}$ and $(n-1)^{th}$ $A.M.$ is $5:9$ then $n$ equals ?

  1. $12$

  2. $13$

  3. $14$

  4. $None$


Correct Option: C

In a Maths test the average score of the $10$ girls in a class is $15$ and the average score of the $15$ boys is $10$. The average score of the class in the test is 

  1. $12$

  2. $12.5$

  3. $13$

  4. $12.75$`


Correct Option: A

Mean deviation of first $7$ natural no. about their A.M. is?

  1. $2$

  2. $\sqrt{2}$

  3. $\dfrac{12}{7}$

  4. $0$


Correct Option: A

The ratio of sum of n arithmetic means between two given numbers to that of single arithmetic mean between them id

  1. n : 1

  2. n$^2$ : 1

  3. 1 : 1

  4. $\sqrt{n}$ : 1


Correct Option: A

If $\cfrac {a^n+b^n}{a^{n-1}+b^{n-1}}$ is the AM between a and b, then the value of n is 

  1. 0

  2. 1

  3. -1

  4. none of these


Correct Option: A
Explanation:

The A.M. of a and b is $\dfrac{a+b}{2}$.


Therefore, $\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}$ will be the A.M. of a and b, if 

$\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\dfrac{a+b}{2}$

$\Rightarrow 2(a^{n+1}+b^{n+1})=(a^n+b^n)(a+b)$

$\Rightarrow 2a^{n+1}+2b^{n+1}=a^{n+1}+a^nb+b^na+b^{n+1}$

$\Rightarrow a^{n+1}+b^{n+1}=a^nb+b^na$

$\Rightarrow a^n(a-b)=b^n(a-b)$

$\Rightarrow a^n=b^n$

$\Rightarrow \dfrac{a^n}{b^n}=1$

$\Rightarrow (\dfrac{a}{b})^n=1$

$\Rightarrow (\dfrac{a}{b})^n=(\dfrac{a}{b})^0$

$\Rightarrow n=0$

Find the average of the following set of scores $253,124,255,534,836,375,101,443,760$

  1. $427$

  2. $413$

  3. $141$

  4. $409$


Correct Option: A

The average age of $30$ girls. is $13\ yr$. The average of first $18$ girls is $15\ yr$. Find out the average age of remaining $12$ girls? 

  1. $12\ yr$

  2. $10\ yr$

  3. $16\ yr$

  4. $10.5\ yr$


Correct Option: A

A student bought $4$ books for $Rs.120$ from one book shop and $6$ books for $Rs.150$ from another. The average price (in  rupees), he paid per book was:

  1. $27$

  2. $27.50$

  3. $135$

  4. $138$


Correct Option: A

Let $(1-2x+3x^{2})^{10}=a _{0}+a _{1}x+a _{2}x^{2}+....+a _{n}x^{n},a _{n}\neq 0$, then the arithmetic mean of $a _{0},a _{1},a _{2},....a _{n}$ is

  1. $\dfrac{1024}{11}$

  2. $\dfrac{512}{7}$

  3. $\dfrac{512}{11}$

  4. $\dfrac{1024}{21}$


Correct Option: A

If $a,b,c,d,e,f$ are $A.M.s$ between $2$ and $12$ then $a+b+c+d+e+f$ is equal to

  1. $14$

  2. $42$

  3. $84$

  4. $None\ of\ these$


Correct Option: A

The average height of 25 boys is 1.4 m. When 5 boys leave the group, then the average height increases by 0.15 m. What is the average height of the 5 boys who leave?

  1. 0.8 m

  2. 0.9 m

  3. 0.95 m

  4. 1.05 m


Correct Option: A

The average age of 36 student in a group is 14 years. When teacher's age is included to it, the average increases by one. What is the teacher's age in years?

  1. 31

  2. 36

  3. 51

  4. Cannot be determined

  5. None of these


Correct Option: A

The sum of first $n$ natural numbers is given by the expression $(2n^{2}+3n)$. The mean of the given numbers is 

  1. $\left(\dfrac{2}{n}+3\right)$

  2. $\left(2+\dfrac{3}{n}\right)$

  3. $(2+3n)$

  4. $(2n+3)$


Correct Option: A

The average weight of 45 students in a class is 25 kg. Five of them whose average weight is 48 kg leave the class and other 5 students whose average weight is 54 kg join the class. What is the new average weight (in kg) of the class?

  1. $52\frac{1}{3}$

  2. $52\frac{1}{2}$

  3. $52\frac{2}{3}$

  4. None of these


Correct Option: A

If n A.M.'s are inserted between 3 and 17 such that the ratio of the last mean to the first mean is 3:1, then the value of n is

  1. 4

  2. 6

  3. 8

  4. 9


Correct Option: A

Four numbers are in proportion. The sum of the square of the four numbers is $50$ and the sum of the means is $5$. The ratio of first two terms is $1 : 3$. What is the average of the four numbers ?

  1. $2$

  2. $3$

  3. $5$

  4. $6$


Correct Option: A

Average of first ten prime numbers is :

  1. 12.6

  2. 12.9

  3. 13.9

  4. 14.9


Correct Option: A

A line is such that the algebraic sum of the perpendicular on it from a number of point is zero. The line always passes through a fixed point that is 

  1. A. M of the given points

  2. GM of the given points

  3. HM of the given points

  4. None of these


Correct Option: A

Coefficient of variance of a distribution is 60% and the standard deviation is 25. The arithmetic mean of the distribution is

  1. $\cfrac {25}{3}$

  2. 35

  3. $\cfrac {125}{3}$

  4. $\cfrac {25}{6}$


Correct Option: A

If the arithmetic mean of the numbers $x _{1}, x _{2}, x _{3}.......,x _{3}$ is $\overline{X}$, then the arithmetic mean of numbers $ax _{1}+b, ax _{2}+b, ax _{3}+b,........, ax _{n}+b$, where  $a, b$ are two constants would be 

  1. $\overline{X}$

  2. $na\overline{X}+nb$

  3. $a\overline{X}$

  4. $a\overline{X}+b$


Correct Option: A

The average score of a cricketer for 10 matches is 42, find the average for last four matches :

  1. $34\frac { 1 }{ 4 } $

  2. 35

  3. $36\frac { 1 }{ 4 } $

  4. $35\frac { 1 }{ 4 } $


Correct Option: A

If the total incomes of $M,N,O,P$ are in the rate a $2:3:4:5$ and the total income of $M$ is $Rs\ 8000$ , then find approximate average salary of all four?

  1. $Rs\ 14000$

  2. $Rs\ 7000$

  3. $Rs\ 9950$

  4. $Rs\ 4875$


Correct Option: A

The artimetic mean of $2 sin 2^o, 4 sin 4^o, 6sin6^o,...,180sin 180^o$ is equal to 

  1. $cosec1^o$

  2. $sec1^o$

  3. $cot1^o$

  4. None of these


Correct Option: A

Arithmetic Mean is ______ affected by extreme values.

  1. Not

  2. Highly

  3. Less

  4. None of these


Correct Option: A

Suppose a population $A$ has $100$ observations $101,102...200$ and other population $B$ has $100$ observations $151,152...250$. 

Find the difference in their means

  1. $49$

  2. $50$

  3. $51$

  4. $52$


Correct Option: B
Explanation:

The observations of $A$ are $101,102,......,200$

The sum is $\dfrac{100}{2}[2(101)+(100-1)1]\50(202+99)\50(301)=15050$
Mean is $\dfrac{15050}{100}\150.5$
The observations of $B$ are $151,152,......,250$
The sum is $\dfrac{100}{2}[2(151)+(100-1)1]\50(302+99)\50(401)=20050$
Mean is $\dfrac{20050}{100}=200.50$
The difference is $200.5-150.5=50$

The mean of 2, 7, 6 and x is 15 and the mean of 18, 1, 6, x and y is 10. What is the value of y ?

  1. $-5$

  2. $-10$

  3. $-20$

  4. $-30$


Correct Option: C
Explanation:

$\dfrac{2+7+6+x}{4}=15$


$15+x=60$

$x=45$

Now,
$\dfrac{18+1+6+45+y}{5}=10$

$70+y=50$

$y=-20$

Sum of $50$ A.M. between $20$ and $30$ is :

  1. $1255$

  2. $1205$

  3. $1250$

  4. $1225$


Correct Option: A

$11\,AM's$ are inserted between $28$ and $10$ then ${6}^{th}\,AM$ is 

  1. $19$

  2. $\displaystyle 17\frac{1}{2}$

  3. $\displaystyle 20\frac{1}{2}$

  4. $22$


Correct Option: A

The mean of $3$ observations is $12$ and mean of $5$ observations is $4$ the combined mean is 

  1. 7

  2. 8

  3. 9

  4. 10


Correct Option: A
Explanation:

The no .of  observations is $3+5=8$

The sum of observations is $3(12)+5(4)=36+20=56$
The mean is $\dfrac{56}{8}=7$

The sum of $9$ numbers is $246$. If the average of three of them is $24$, what is the average of the remaining numbers?

  1. $30$

  2. $29$

  3. $31$

  4. $25$


Correct Option: A

A boy draws n squares with sides $1,2,3,4,5....$ in inches.The average area covered by these n squares  will be:  

  1. $\left(\dfrac{n+1}{2}\right)$

  2. $\left(\dfrac{n+1}{2}\right)\left(\dfrac{2n+1}{3}\right)$

  3. $\left(\dfrac{n+1}{2}\right)\left(\dfrac{2n+1}{3}\right)^{-1}$

  4. $\left(\dfrac{n+1}{2}\right)-1 \left(\dfrac{2n+1}{3}\right)$


Correct Option: A

The mean marks of $120$ students is $20$. It was later discovered that two marks were wrongly taken as $50$ and $80$ instead of $15$ and $18$. The correct mean of mark is

  1. $19.19$

  2. $19.17$

  3. $19.21$

  4. $19.14$


Correct Option: A

The Arithmetic mean of first $5$ whole numbers.

  1. $1$

  2. $2$

  3. $3$

  4. $5$


Correct Option: A

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?

  1. 14, 15

  2. 15, 14

  3. 13, 16

  4. 12, 17


Correct Option: D

The antithetic mean of the masks is _________.

  1. 50.25

  2. 50.75

  3. 51.25

  4. 53.75


Correct Option: A

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

  1. 12

  2. 15

  3. 16

  4. 18


Correct Option: A

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

  1. 2

  2. 4

  3. 8

  4. 16


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

  1. $5.5$

  2. $6$

  3. $7.5$

  4. $10$


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

  1. $r + s$

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

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

  4. $r + s + 1$


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)
  1. $36$

  2. $24$

  3. $30$

  4. none of these


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

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

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

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

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


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

  1. $n$

  2. $2n$

  3. $n/2$

  4. $n - 1$


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

  1. $n$

  2. $2n$

  3. $\dfrac n2$

  4. $n - 1$


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 = $\dfrac{n^2}n = n$

What is the average of the first $300$ terms of the given sequence?
$1, -2, 3, -4, 5, -6, ....., n.(-1)^{n + 1}$

  1. $-1$

  2. $0.5$

  3. $0$

  4. $-0.5$


Correct Option: D
Explanation:
Avg $=\cfrac{[1+3+5+7+...+(2n-1)]-[2+4+6+...+2n]}{n}$
Here $n=300$
No. of even terms $=150$
No. of odd terms $=150$
Now,
$1+3+5+7+...+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n1}=\cfrac{150}{2} [2\times 1 +149\times 2]=22500$
and $2+4+6+8+....+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n2}=\cfrac{150}{2}[2\times 2+149\times 2]=22650$
Average $=\cfrac{S _{n1}-S _{n2}}{300}=\cfrac{22500-22650}{300} \\ =-0.5$

The A.M. of 'n' observations is M. If the sum of $(n - 4)$ observation is 'a', what is the mean of remaining $4$ observations?

  1. $nM + a$

  2. $\dfrac {nM - a}{2}$

  3. $\dfrac {nM + a}{2}$

  4. $\dfrac {nM - a}{4}$


Correct Option: D
Explanation:
Given $\cfrac { { a } _{ 1 }+{ a } _{ 2 }+{ a } _{ 3 }+{ a } _{ 4 }+....+{ a } _{ n } }{ n } =M\quad \quad \quad (1)\\ \left( { a } _{ 1 }+{ a } _{ 2 }+{ a } _{ 3 }+{ a } _{ 4 }+....+{ a } _{ n-4 } \right) =a\quad \quad \quad (2)$

From $(1)$ and $(2)$
$a+{ a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n }=nM\\ { a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n }=(nM-a)\\ Mean=\cfrac { { a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n } }{ 4 } =\cfrac { nM-a }{ 4 } \\ Mean=\cfrac { (nM-a) }{ 4 } $

The mean  of five numbers in AP is $89$. The product of first and last terms is $7885$. The AM of first, third and fifth term is

  1. $83$

  2. $86$

  3. $89$

  4. $90$


Correct Option: C
Explanation:
Let the numbers in A.P. be  (a-2d), (a-d), a, (a+d), (a+2d)
$\therefore \dfrac{(a-2d)+(a-d)+a+(a+d)+(a+2d)}{5}=89$
$\Rightarrow \dfrac{5a}{5}=89$
$\therefore a=89$
$\therefore$ The AM of 1st, 3rd & 5th term is:
$=\dfrac{(a-2d)+a+(a+2d)}{3}$
$=a=89$                           

The sum of four numbers in AP is $176$. The product of 1st and last is $1855$.  The mean of middle two is 

  1. $42$

  2. $41$

  3. $44$

  4. $53$


Correct Option: C
Explanation:
Let the numbers in A.P. be  (a-3d), (a-d), (a+d), (a+3d)
$\therefore (a-3d)+(a-d)+(a+d)+(a+3d)=176$
$\Rightarrow 4a=176$
$\therefore a=44$
$\therefore$ The mean of middle two is:
$=\dfrac{(a-d)+(a+d)}{2}$
$=a=44$                           

The arithmetic mean of 1, 8, 27, 64, ....... up to n terms is given by

  1. $\dfrac{n(n+1)}{2}$

  2. $\dfrac{n(n+1)^2}{2}$

  3. $\dfrac{n(n+1)^2}{4}$

  4. $\dfrac{n^2(n+1)^2}{4}$


Correct Option: C
Explanation:

Solution:

Given:
$1,8,27,64,...........$ upto $n$ terms
or,$1,2^3,3^3,4^3,..............$ upto $n$ terms
$\therefore AM=\cfrac{1+2^3+3^3+4^3............+n^3}{n}=\cfrac{\left[\cfrac{n(n+1)}2\right]^2}{n}$
$=\cfrac{n^2(n+1)^2}{4n}=\cfrac{n(n+1)^2}4$
Hence, C is the correct option.

Say true or false.
A.M. of any $n$ positive numbers $a _1, a _2, a _3, ...., a _n$ is $A$ then $A=\dfrac {a _1+a _2+....+a _n}{n}$

  1. True

  2. False


Correct Option: A
Explanation:

Arithmetic mean is the average of all the numbers.
Hence for n positive numbers:
$a _1$,$a _2$,$a _3$,....$a _n$, the arithmetic mean is $\dfrac{a _1+a _2+a _3+...+a _n}{n}$.

Say true or false.

The A.M. between $(a-b)^2$ and $(a+b)^2$  is $a^2+b^2 $.

  1. True

  2. False


Correct Option: A
Explanation:

If the two terms are $x$ and $y$, then their AM is $\dfrac{x+y}{2}$
AM between $(a-b)^2$ and $(a+b)^2$
$=$ $\dfrac{(a-b)^2 + (a+b)^2}{2}$
$=$ $\dfrac{(a^2 +b^2 - 2ab + a^2+b^2 + 2ab)}{2}$
$=$ $a^2 + b^2$

If the arithmetic mean of n numbers of a series is $\overline{x}$ and the sum of the first (n-1) numbers is k, then the nth number is

  1. n+k

  2. $n\overline{x}+k$

  3. $n\overline{x}-k$

  4. n-k


Correct Option: C
Explanation:

$Sum$ $of$ $all$ $the$ $n$ $numbers$ $= Mean \times no. of numbers = n \times \overline x$
$Sum$ $of$ $n-1$ $numbers$ $= k$
$Thus$ $nth$ $number$ $= n \times \overline x - k$

The arithmetic mean (average) of the first $n$ positive integers is

  1. $\dfrac {n}{2}$

  2. $\dfrac {n^{2}}{2}$

  3. $n$

  4. $\dfrac {n - 1}{2}$

  5. $\dfrac {n + 1}{2}$


Correct Option: E
Explanation:

The sum of the first $n$ positive integers is $s = \dfrac {n(n + 1)}{2}$. Since $A.M. = \dfrac {s}{n}$, the correct choice is (e).

The mean of the cubes of the first $n$ natural numbers is :

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

  2. $n^2$

  3. $\displaystyle \frac{n(n+1)(n+2)}{8}$

  4. $(n^2+n+1)$


Correct Option: A
Explanation:

Sum of the cubes of first $n$ natural numbers
$=\displaystyle \left [ \frac{n(n+1)}{2}\right]^2 = \frac{n^2 (n+1)^2}{4}$
$\therefore $ Mean $= \displaystyle \frac{n(n+1)^2}{4}$

If the arithmetic mean of $n$ numbers of a series is $\bar{x}$ and sum of the first $(n - 1)$ numbers is $k$, then which one of the following is the nth number of the series ?

  1. $\bar{x} - nk$

  2. $n\bar{x} - k$

  3. $k\bar{x} - n$

  4. $nk\bar{x}$


Correct Option: B
Explanation:

Mean of n terms $=\bar{x}$
Sum of n terms  $=n\bar{x}$
Sum of $(n-1)$  terms $=k$

$n^{th}$ term $=$ Sum of n terms $-$ Sum of $(n-1)$ terms $=n\bar{x} - k$

The mean marks got by $300$ students in the subject of statistics was $45$. The mean of the top $100$ of them was found to be $70$ and the mean of the last $100$ was known to be $20$, then the mean of the remaining $100$ students is 

  1. $45$

  2. $58$

  3. $68$

  4. $88$


Correct Option: A
Explanation:
Let mean marks of remaining student be $M$
then Mean $=\dfrac{\displaystyle\sum (fx)}{\displaystyle\sum f}=45$        $\displaystyle\sum f=300$
$\therefore \dfrac{\displaystyle\sum (fx)}{\displaystyle\sum f}=100\times 70+100\times M+100\times 20$
$\dfrac{\displaystyle\sum (f)(x)}{\displaystyle\sum f}=\dfrac{100(90+M)}{\displaystyle\sum f}$
$45=\dfrac{100(90+M)}{300}$
$135=90+M$
$\Rightarrow M=135-90$
$=45$

If $a _{1}=0$ and $a _{1}, a _{2}, a _{3}, ...., a _{n}$ are real numbers such that $|a _{i}|=|a _{i-1}+1|$ for all $i$ then the Arithmetic mean of the numbers $a _{1}, a _{2}, ..., a _{n}$ has value $x$ where

  1. $x<-1$

  2. $x<-\dfrac{1}{2}$

  3. $x>-\dfrac{1}{2}$

  4. $x=-\dfrac{1}{2}$


Correct Option: C
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