Tag: maths

Questions Related to maths

$\sqrt { 7 }$ is a

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. an integer

  2. an irrational number

  3. a rational number

  4. none of these

Show answer
Correct Option: B
Explanation:

$\sqrt 7 $ is a rational number


State True or False:

$4\, - \,5\sqrt 2 $ is irrational if $\sqrt 2 $ is irrational.

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

We know that difference of a rational and irrational number is also irrational, 


So if $\sqrt{2}$ is irrational 4-5$\sqrt{2}$ is also irrational. 

GCF of $99$ and $100$ is __________

maths be my multiple, i'll be your factor proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A
Explanation:

when we divide $100$ with $99$ we will get remainder $1$ and hence next we divide 99 with $1$ which gives us remainder $0$.


therefore $1$ is the HCF or GCF of $99 $ and $100$

The area of two similar triangles are $200$ and $128$, then the ratio of their corresponding altitude is __________

maths triangles areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $25:16$

  2. $5:4$

  3. $4:5$

  4. $16:25$

Show answer
Correct Option: B
Explanation:

Since we know that ratio of areas of two similar triangles is equal to the square of the ratio of their altitude
therefore
Ratio of their altitude=$\sqrt {\dfrac{{200}}{{128}}} $
$ = \sqrt {\dfrac{{100}}{{64}}} $
$ = \dfrac{{10}}{8}$
$ = \dfrac{5}{4}$
$ = 5:4$

The value of $3 - 1 + \frac{1}{3} - \frac{1}{9} +  \ldots $ is equal to

sum to infinite terms of a gp sequence, progression and series maths

  1. $\dfrac{{20}}{9}$

  2. $\dfrac{{9}}{20}$

  3. $\dfrac{{9}}{4}$

  4. $\dfrac{{4}}{9}$

Show answer
Correct Option: C

Let $P = 3^{1/3} . 3^{2/9} . 3^{3/27} ...\infty$, then $P^{1/3}$ is equal to

sum to infinite terms of a gp sequence, progression and series maths

  1. $3^{2/3}$

  2. $\sqrt {3}$

  3. $3^{1/3}$

  4. $3^{1/4}$

Show answer
Correct Option: D
Explanation:

Above in an infinite A.G.S. with $a = 1, d = 1$ for A.P., $b = \dfrac {1}{3}, r = \dfrac {1}{3}$ for G.P.
$\therefore S _{\infty} = \dfrac {ab}{1 - r} + \dfrac {dbr}{(1 - r)^{2}} = \dfrac {\dfrac {1}{3}}{1 - \dfrac {1}{3}} + \dfrac {1 . \dfrac {1}{3} . \dfrac {1}{3}}{\left (1 - \dfrac {1}{3}\right )^{2}} = \dfrac {1}{2} + \dfrac {1}{4} = \dfrac {3}{4}$
$\therefore P = 3^{S} = 3^{3/4} \therefore P^{1/3} = 3^{1/4}$.

The first term of a $G.P.$ whose second term is $2$ and sum to infinity is $8$ will be

sum to infinite terms of a gp sequence, progression and series maths

  1. $6$

  2. $3$

  3. $4$

  4. $1$

Show answer
Correct Option: C
Explanation:

$ar = 2, \dfrac {a}{1 - r} = \dfrac {1}{8}$. Eliminate $r$ and we get
$a^{2} - 8a + 16 = 0 \therefore (a - 4)^{2} = 0$ or $a = 4$

The value of $9^{1/3}\times 9^{1/9} \times 9^{1/27} \times .....\infty$ is

sum to infinite terms of a gp sequence, progression and series maths

  1. $9$

  2. $1$

  3. $3$

  4. None of these

Show answer
Correct Option: C
Explanation:

$(9)^{S _{\infty}} = 9^{1/2} = 3\because S _{\infty} = \dfrac {1/3}{1 - (1/3)} = \dfrac {1}{2}$.

The value of $9^\cfrac{1}{3}.9^\cfrac{1}{9}.9^\cfrac{1}{27}...........$ upto $\infty$, is

sum to infinite terms of a gp sequence, progression and series maths

  1. $1$

  2. $3$

  3. $9$

  4. $None\ of\ these$

Show answer
Correct Option: B
Explanation:
${9^{\cfrac{1}{3}}}{.9^{\cfrac{1}{9}}}{.9^{\cfrac{1}{{27}}}} -  -  -  - upto\,\,\infty $
$ = {9^{\left( {\cfrac{1}{3} + \cfrac{1}{9} + \cfrac{1}{{27}} +  -  -  - } \right)}}$
$ = {9^{\left( {\cfrac{{\cfrac{1}{3}}}{{1 - \cfrac{1}{3}}}} \right)}}$
$ = {9^{\left( {\cfrac{{\cfrac{1}{3}}}{{\cfrac{2}{3}}}} \right)}}$
$ = {9^{\cfrac{1}{2}}}$
$ = 3$

If $x=1+a+{ a }^{ 2 }+{ a }^{ 3 }+....$ to $\infty \left( \left| a \right| <1 \right) $ and 
$y=1+b+{ b }^{ 2 }+{ b }^{ 3 }+...$ to $\infty \left( \left| b \right| <1 \right) $ then
$1+ab+{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 3 }{ b }^{ 3 }+...$ to $\infty =\cfrac { xy }{ x+y-1 } $

sum to infinite terms of a gp sequence, progression and series maths

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

${ x }=\dfrac { 1 }{ 1-a } $ ${ y }=\dfrac { 1 }{ 1-b } $ [summing infinite $G.P's$].
$\therefore a=\dfrac { x-1 }{ x } $, $b=\dfrac { y-1 }{ y } $
$\therefore 1+ab+{ a }^{ 2 }{ b }^{ 2 }+...\infty $
$=\dfrac { 1 }{ 1-ab } =\dfrac { 1 }{ 1-\dfrac { (x-1)(y-1) }{ xy }  } =\dfrac { xy }{ x+y-1 }. $