Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

Simplify : $\frac{(-18\frac{1}{3}\times 2\frac{8}{11})}{|\frac{3}{5}+(\frac{-9}{10})| + |-(\frac{-3}{5})|}$

  1. 63$\frac{4}{81}$
  2. -23$\frac{7}{9}$
  3. -67$\frac{7}{9}$
  4. 12$\frac{6}{17}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\cfrac {\left(-18\cfrac{1}{3}\times 2\cfrac {8}{11}\right)}{\left|\cfrac {3}{5}+\left(\cfrac {-9}{10}\right)\right|+\left|-\left(\cfrac{-3}{5}\right)\right|}$

$=\cfrac{-\cfrac{55}{3}\times \cfrac{30}{11}}{\left|\cfrac {6-9}{10}\right|+\cfrac {3}{5}}$
$=\cfrac {\cfrac {-55\times 30}{33}}{\cfrac {3}{10}+\cfrac {3}{5}}$
$=\cfrac {\cfrac {-55\times 30}{33}}{\cfrac{9}{50}}$
$=\cfrac {-55\times 30\times 50}{33\times 9}$
$=\cfrac {-5\times 10\times 50}{9}$
$=-67\cfrac {7}{9}$

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

Which of the following options is arranged in descending order?

  1. $\frac{1}{4},\frac{6}{4},\frac{16}{9},\frac{25}{4}$
  2. $\frac{-3}{6},\frac{-4}{3},\frac{-9}{4},\frac{-13}{4}$
  3. $\frac{-5}{8},\frac{-3}{8},\frac{0}{8},\frac{1}{8}$
  4. $\frac{-7}{4},\frac{-3}{4},\frac{5}{4},\frac{8}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To check the order, first make the denominator same of all the fractions, then compare the numerator.

Note- To make the denominator same, multiply both numerator and denominator by HCF of denominator values.
$\cfrac {-3}{6}, \cfrac {-4}{3},\cfrac {-9}{4}, \cfrac {-13}{4}$
Multiply by $12$ both numerator and denominator.
$\cfrac {-36}{12}, \cfrac {-43}{12},\cfrac {-108}{12}, \cfrac {-156}{12}$

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

Compare the given fractions and specify the correct operator

$\dfrac{9}{16}$ ___ $\dfrac{13}{5}$

  1. $\displaystyle = $
  2. $\displaystyle > $
  3. $\displaystyle < $

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To compare fractions, make the denominators equal

LCM of denominators $16$ and $5$ is $80$
$\therefore \dfrac{9}{16} \times \dfrac{5}{5}$ $=\dfrac{45}{80}$ and $\dfrac{13}{5} \times  \dfrac {16}{16}$ $=\dfrac{208}{80}$


By making denominators equal, we find that $208$ is greater than $45$ 
$\therefore \dfrac{9}{16} < \dfrac{13}{5} $.

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

If a, b, c, are positive $\displaystyle \frac{a+c}{b+c}$ is 

  1. always smaller than $\displaystyle \frac{a}{b}$
  2. always greater than $\displaystyle \frac{a}{b}$
  3. greater than $\displaystyle \frac{a}{b}$ only if a > b
  4. greater than $\displaystyle \frac{a}{b}$ only if a < b
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since we need to compare the fraction $\displaystyle \frac{a+c}{b+c}$ with $\displaystyle \frac{a}{b}$, we cross multiply the terms and check since $a,b,c$ are all given to be positive.
We thus get $b(a + c)$ on the L.H.S. and $a(b + c)$ on R.H.S.
Thus, simplifying we are left with $ab + bc$ on the L.H.S. and $ab + ac$ on the R.H.S.
Now, which side is greater depends on $ac$ and $bc$, which in turn depends upon $a$ & $b.$
L.H.S. is greater if $b > a$, which implies $\frac{a + c}{b + c}$ is greater when $b > a.$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The positive root of ${x}^{2}-98.8=0$ after first approximation by Newton Raphson method assuming initial approximation to the root is $14$ is

  1. $9.821$
  2. $9.814$
  3. $9.715$
  4. $9.915$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here ${x} _{0}=14, f(x)={x}^{2}-78.8$
and $f'(x)=2x$
$\therefore$ ${x} _{1}={x} _{0}-\cfrac{f({x} _{0}}{f'({x} _{0})}$
$=14-\cfrac { { \left( 14 \right)  }^{ 2 }-\left( 78.8 \right)  }{ 2\times 14 } =9.814$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

State the following statement is True or False

According to the Newton-Raphson's method the approximate root of the equation $f(x) = 0$ is $x _{n}$ then be $x _{n} = x _{n + 1} - \dfrac {f(x)}{f'(x _{n})}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If $x _n$ is the approximate root of $f(x)=0$ then the value of $x _n$ given by Iterative eqn. for Newton-Raphson method is 
    ${ x } _{ n+1 }= { x } _{ n } - \dfrac { f({ x } _{ n }) }{ f' ({ x } _{ n }) } $, where $f'(x _n)$ is derivative of $f$ at $x _n$

$\therefore $ The given statement is FALSE.