Tag: maths

Questions Related to maths

The Simplified form of $0.35$ is

  1. $\dfrac {7}{20}$

  2. $\dfrac {4}{20}$

  3. $\dfrac {35}{100 }$

  4. $None$


Correct Option: A
Explanation:

$0.35$

$=\dfrac{35}{100}=\dfrac{7}{20}$

Given that  $n$  $AM's$  are inserted between two sets of numbers  $a , 2 b$  and  $2 a , b$  where  $a , b \in R .$  Suppose further that  $mth$  mean between these sets of numbers is same, then the ratio  $a : b$  is equal to

  1. $( n - m + 1 ) : m$

  2. $( n - m + 1 ) : n$

  3. $n : ( n - m + 1 )$

  4. $m : ( n - m + 1 )$


Correct Option: D
Explanation:

Let the common difference be d. As there are $n\;AM's$ between $a$ and $b$ and total number of terms in the sequence is $=n+2$ 

$\Rightarrow nth \;term\;b=a+\left( n-1\right)d$
      $d=\dfrac{\left( b-a\right)}{n+1}$
so, 2nd term that is first term $A\left( 1\right) =a+\left[ \dfrac{\left( b-a\right)}{\left( n+1\right)}\right]$
3rd term that is second mean $A\left(2\right)=a+\left[ 2\times \dfrac{ \left( b-a\right)}{\left( n+1\right)}\right]$
In the way $r^{th}$ mean $=a+\left[ r\times \dfrac{ \left( b-a\right) }{\left(n+1\right)}\right]$
In the first sequence first term is a $n^{th}$ term $=2b$ and $n\;AM's$ between them. 
As such from above concept  -
$\Rightarrow r^{th}$ term is $a+\left[ r\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]$
       $m^{th}$ term is $a+\left[ m\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]$
$ii)$ Similarly for second sequence -
    $m^{th}$ mean $=2a+\left[ m\times \dfrac{\left( b-2a\right)}{\left( n+1\right)}\right]$
$iii).$ Since the $m^{th}$ mean, are equal equation like the above. 
$=a+\left[ m\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]$
$=2a+m\times \dfrac{\left( b-a\right) }{\left( n+1\right)}$
$=\left[ m\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]-\left[ m\times \dfrac{\left( b-2a\right)}{\left( n+1\right)}\right]$
$=2a-a$
$\Rightarrow \dfrac{m}{\left( n+1\right)} \times \left( 2b-a-b+2a\right)=a$
$\dfrac{a+b}{a}=\dfrac{n+1}{m}$
subtracting on both sides 
$\Rightarrow \dfrac{a+b}{a}-1=\dfrac{n+1}{m}-1$
$\Rightarrow \dfrac{a+b-a}{a}=\dfrac{n+1-m}{m}$

$\Rightarrow \dfrac{b}{a}=\dfrac{n+1-m}{m}$

$\Rightarrow \dfrac {a}{b}=\dfrac{m}{n-m+1}$
Hence, the answer is $\dfrac{m}{n-m+1}.$

The lowest form of $3.5$ is 

  1. $\frac{7}{20}$

  2. $\frac{4}{20}$

  3. $\frac{35}{100}$

  4. $None$


Correct Option: A

Express in simpest from

  1. $-247/228$

  2. $-68/119$

  3. $87/116$

  4. $299/161$


Correct Option: B

If $\displaystyle\,5\,\dfrac{7}{x}\,\times\,y\,\dfrac{1}{13}\,=\,12$, where fractions are in their lowest terms, then $x - y$ is equal to 

  1. $2$

  2. $4$

  3. $7$

  4. $9$


Correct Option: C
Explanation:

$\displaystyle 5\,\frac{7}{x}\,\times\,y\,\frac{1}{13}\,=\,12$
By Hit and Trial method. 
Let $x = 9, y = 2$
Where the fractions are in their lowest terms, then x should be maximum possible single digit and $y$ is minimum possible single digit.
Putting this value in equ. (1)
$\displaystyle \,5\,\times\,\frac{7}{9}\,\times\,2\,\times\,\frac{1}{13}\,=\,\frac{52}{9}\,\times\,\frac{27}{13}\,=\,12
$
$\therefore \,x\,-\,y\,=7$
Hence, option 'C' is correct.

Simplest form of the ratio 140 : 24 is__

  1. 1 : 3

  2. 70 :12

  3. 6 : 35

  4. 35 : 6


Correct Option: D
Explanation:

We have, $ 140:24 $
Dividing by $ 4 $ we get $ 35:6 $
As there is no other common factor  to divide with, so $ 35:6 $ is the simplest form.

What is the reciprocal of $-3$?

  1. $-3$

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

  3. $\dfrac {1}{3}$

  4. $3$

  5. Undefined


Correct Option: B
Explanation:

Reciprocal of $ -3 = \dfrac {1}{-3} $ or $ \dfrac {-1}{3} $

The value of $\left[\left(-2\displaystyle\frac{3}{4}\right)-\left(\displaystyle -1\frac{3}{4}\right)\right]+\left[\left(\displaystyle -2\frac{3}{4}\right)-\left(\displaystyle -1\frac{3}{4}\right)\right]+......$ upto $30$ times is:

  1. $-1$

  2. $1$

  3. $30$

  4. $-30$


Correct Option: D
Explanation:
Consider the given expression.

$\left [ \left ( -2\dfrac{3}{4} \right )- \left ( -1\dfrac{3}{4} \right )\right ]+\left [ \left ( -2\dfrac{3}{4} \right )- \left ( -1\dfrac{3}{4} \right )\right ]+.......$ upto $30$ times

Sum $=\left[\left(\dfrac{-11}{4}\right)-\left(\dfrac{-7}{4}\right)\right]+\left[\left(\dfrac{-11}{4}\right)-\left(\dfrac{-7}{4}\right)\right]+......$ upto $30$ times

        $=[-1]+[-1]+......upto\ 30\ times$

        $=-30$

Hence, the value of the expression is $-30$. 

The number $2.525252$ can be written as a fraction, when reduced to the lowest term, the sum of the numerator and denominator is:

  1. $7$

  2. $29$

  3. $141$

  4. $349$


Correct Option: D
Explanation:
Let the given number be $x=2.525252....$
multiplying with $100$ on both sides
$\Rightarrow 100x=252.525252...$
$\Rightarrow 100x=250+2.5252...$
$\Rightarrow 100x=250+x\Rightarrow 99x=250$
$\Rightarrow x=\dfrac{250}{99}$
$\therefore$ Sum of numerator and denominator $=25099=349$

The fraction $\dfrac {a^{2} + b^{2} - c^{2} + 2ab}{a^{2} + c^{2} - b^{2} + 2ac}$ is (with suitable restrictions on the values of $a, b,$ and $c$).

  1. Irreducible

  2. Reducible to $-1$

  3. Reducible to a polynomial of three terms

  4. Reducible to $\dfrac {a - b + c}{a + b - c}$

  5. Reducible to $\dfrac {a + b - c}{a - b + c}$


Correct Option: E
Explanation:

$\dfrac {a^{2} + b^{2} - c^{2} + 2ab}{a^{2} + c^{2} - b^{2} + 2ac} = \dfrac {(a + b)^{2} - c^{2}}{(a + c)^{2} - b^{2}} = \dfrac {(a + b + c)(a + b - c)}{(a + c + b)(a + c - b)}$
$= \dfrac {a + b - c}{a + c - b}$ with $(a + c)^{2} \neq b^{2}$.