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Quizzes (518) Questions (9324)

Questions Related to maths

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

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}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$\frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30},$

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

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 

  1. $48.9$
  2. $49.5$
  3. $52.5$
  4. $56.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 = $\dfrac{2475}{50}=49.5$

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

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

  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. $\displaystyle\frac{1}{10},\,\displaystyle\frac{2}{10},\,\displaystyle\frac{3}{10},\,\displaystyle\frac{4}{10},\,\displaystyle\frac{5}{10},\,\displaystyle\frac{6}{10},\,\displaystyle\frac{7}{10},\,\displaystyle\frac{8}{10},\,\displaystyle\frac{9}{10},\,\displaystyle\frac{10}{10}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

The rational numbers between these are


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

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

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

  1. $ \displaystyle \frac{46}{75} $
  2. $ \displaystyle \frac{47}{75} $
  3. $ \displaystyle \frac{49}{75} $
  4. $ \displaystyle \frac{50}{75} $
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  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. $\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$.

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

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

  1. $-\dfrac {1}{4}$
  2. $-\dfrac {3}{10}$
  3. $\dfrac {3}{10}$
  4. $-\dfrac {7}{20}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $-\dfrac {1}{5}$ 

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

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

  1. $ \cfrac{46}{75}$
  2. $ \cfrac{47}{75}$
  3. $ \cfrac{49}{75}$
  4. $ \cfrac{50}{75}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac{2}{3}$

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

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)

  1. $14,\ 42$
  2. $15,\ 42$
  3. $14,\ 52$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $14$.

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

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

  1. $3x+y$
  2. $x+3y$
  3. $3x-y$
  4. $x-3y$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

$\sum{n^3}=$

  1. $(\sum{n})^3$
  2. $(\sum{n})^2$
  3. $(\sum{n})^3+(\sum{n})^2$
  4. $\sum{(n+n^2)}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $