Tag: maths

Questions Related to maths

State True or False.

The five rational numbers between $\dfrac{3}{5}$ and $\dfrac{4}{5}$ are $ \displaystyle \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}$.

maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Since we want five  numbers we write $\frac{3}{5}$ and $\frac{4}{5}$
So multiply in numerator and denominator by 5+1 = 6 we get
$\Rightarrow \frac{3}{5}=\frac{3\times 6}{5\times 6}=\frac{18}{30}$
$\Rightarrow \frac{4}{5}=\frac{4\times 6}{5\times 6}=\frac{24}{30}$
We know that $18<19<20<21<22<23<24$
$\Rightarrow \frac{18}{30}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{24}{30}$
Hence 5 rational number between $\frac{3}{5} and  \frac{4}{5}$are
$\frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30},$

There are 50 numbers Each numbers is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5 The mean of the given numbers is 

maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

  1. $48.9$

  2. $49.5$

  3. $52.5$

  4. $56.5$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$   Total observation is $50$

$\Rightarrow$  Let sum of $50$ number be $x$
$\therefore$    $\dfrac{x-(50\times 53)}{50}=-3.5$
$\therefore$    $x-2650=-3.5\times 50$
$\therefore$    $x-2650=-175$
$\therefore$    $x=-175+2650$
$\therefore$    $x=2475$
$\Rightarrow$   Original mean = $\dfrac{2475}{50}=49.5$

Write any $10$ rational numbers between $0\;and\;2$.

maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

  1. $\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{88}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$

  2. $\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{21}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$

  3. $\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{35}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$

  4. $\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$

Show answer
Correct Option: D
Explanation:

Let us write $0$ as $\displaystyle\frac{0}{10}\;and\;2$ as $\displaystyle\frac{20}{10}$.

The rational numbers between these are


$\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10},\,\displaystyle\frac{11}{10},\,\displaystyle\frac{12}{10},\,\displaystyle\frac{13}{10},\,\displaystyle\frac{14}{10},\,\displaystyle\frac{15}{10},\,\displaystyle\frac{16}{10},\,\displaystyle\frac{17}{10},\,\displaystyle\frac{18}{10},\,\displaystyle\frac{19}{10}$

Choose the rational number which does not lie between rational numbers $ \displaystyle \frac{3}{5} $ and $ \displaystyle \frac{2}{3} $ :

maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

  1. $ \displaystyle \frac{46}{75} $

  2. $ \displaystyle \frac{47}{75} $

  3. $ \displaystyle \frac{49}{75} $

  4. $ \displaystyle \frac{50}{75} $

Show answer
Correct Option: D
Explanation:

For $\dfrac{3}{5}$ multiply numerator and denominator by $15$ to make denominator $75$ that comes into $\dfrac{45}{75}.$
Similarly doing for second then we have $\dfrac{50}{75}.$
Now question is asking about rational lying between them.

So, we need to check the numerator only that lies in between $45$ and $50$ or not.
Clearly $D$ is correct.

Find five rational numbers between $1$ and $2.$

maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

  1. $\dfrac {1}{10}, \dfrac {2}{10}, \dfrac {3}{10}, \dfrac {4}{10}, \dfrac {5}{10}$

  2. $\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}, \dfrac {5}{5}$

  3. $\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}, \dfrac {1}{5}, \dfrac {1}{6}$

  4. $\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$

Show answer
Correct Option: D
Explanation:

The given rational numbers are $1$ and $2$.

Let us multiply both the numbers by $\dfrac {7}{7}$.
$1\times \dfrac {7}{7} = \dfrac {7}{7}$ and $2\times \dfrac {7}{7} = \dfrac {14}{7}$.
Thus, five rational numbers between $\dfrac {7}{7} = 1$ and $\dfrac {14}{7} = 2$ are $\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$.

Choose the rational number which does not lie between rational numbers $-\cfrac {2}{5}$ and $-\cfrac {1}{5}$

maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

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

  2. $-\dfrac {3}{10}$

  3. $\dfrac {3}{10}$

  4. $-\dfrac {7}{20}$

Show answer
Correct Option: C
Explanation:

Since the given rational numbers $-\dfrac {2}{5}$ and $-\dfrac {1}{5}$ are negative rational numbers, therefore, none of the positive rational number can lie between them.


Hence, the rational number $\dfrac {3}{10}$ does not lie between the rational numbers $-\dfrac {2}{5}$ and $-\dfrac {1}{5}$ 

Identity the rational number that does not lie between  $ \cfrac{3}{5}$ and $ \cfrac{2}{3}$.

maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

  1. $ \cfrac{46}{75}$

  2. $ \cfrac{47}{75}$

  3. $ \cfrac{49}{75}$

  4. $ \cfrac{50}{75}$

Show answer
Correct Option: D
Explanation:

Changing the fractions to denominator $=75$


$\dfrac{3}{5} = \dfrac{45}{75}$

$\dfrac{2}{3} = \dfrac{50}{75}$

$\therefore \dfrac{50}{75}$ doesn't lie in between $\dfrac{3}{5}$ and $\dfrac{2}{3}$

Arvind has Piggybank. It is full of one-rupee and fifty paise coins. It contains $3$ times as many fifty paise coins as one rupee coins. The total amount of money in the bank is $Rs\ 35$. How many one-rupee and fifty paise coins are there in the bank ?(respectively)

maths equations and rearranging formulae further equations solving linear equations with variable on both sides equations and simple functions

  1. $14,\ 42$

  2. $15,\ 42$

  3. $14,\ 52$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let no. of fifty paise coins be $x$ and no.of one Rupees coins be $y$

then $x=3y$

According to question 
$0.50 \times x+1 \times y=35\$
$\Rightarrow 0.5x+y=35\$
$\Rightarrow 0.5 \times 3y+y=35\$
$1.5y+y=35\$
$\Rightarrow 2.5y=35\$
$\Rightarrow y=\dfrac{350}{25}\$
$\Rightarrow y=14\$
So,
    $x=3y\$
$=3 \times 14\$
$=42$

Therefore no.of fifty paise coins is $42$ ans no.of one Ruppee coin is $14$.

Sum of the ages of three friends $x$ years ago was $y$ years. Then what will be the sum of their ages now?

maths equations and rearranging formulae further equations solving linear equations with variable on both sides equations and simple functions

  1. $3x+y$

  2. $x+3y$

  3. $3x-y$

  4. $x-3y$

Show answer
Correct Option: A
Explanation:

Let their present age be $a,b,c$


Sum of their present age $=a+b+c$

Their ages $x$ years ago $a-x,b-x,c-x$

Sum $=a+b+c-3x$

Given $a+b+c-3x=y$

$a+b+c=y+3x=3x+y$


So option $A$ is correct.

$\sum{n^3}=$

maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

  1. $(\sum{n})^3$

  2. $(\sum{n})^2$

  3. $(\sum{n})^3+(\sum{n})^2$

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

Show answer
Correct Option: B
Explanation:
We have, $ \sum { { n }^{ 3 } } = { 1 }^{ 3 }+{ 2 }^{ 3 }+....+{ n }^{ 3 } $
The formula to find the sum of cubes of natural numbers is $ ={( \dfrac {n(n+1)}{2} )}^2 $

But sum of first $ n $ natural numbers is$  \sum { { n } } = \dfrac {n(n+1)}{2} $ 

So, $ \sum { { n }^{ 3 } } = { (\sum { { n } })}^2 $