Tag: history of mathematics

Questions Related to history of mathematics

State the following statement is True or False
The digit 1,2,3,4... which we call as English digits are developed form of Indian numerals

mathematicians and their contribution history of mathematics maths

  1. True

  2. False

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Correct Option: A
Explanation:

10 digit number system(1,2,3,4,..) was first developed in india.

The last two digits in $X=\displaystyle \sum _{k=1}^{100} k$ are 

mathematicians and their contribution history of mathematics maths

  1. $10$

  2. $11$

  3. $12$

  4. $13$

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Correct Option: A

State true or false.
$0\div 4=0$

mathematicians and their contribution history of mathematics maths

  1. True

  2. False

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Correct Option: A
Explanation:

The given statement is true.

Since,
$0\div 4$
$=\dfrac{0}{4}$
$=0$.

State true or false.
$0\div \left(-7\right)=0$

mathematicians and their contribution history of mathematics maths

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Given statement is true.

Since 
$0\div (-7)$

$=\dfrac{0}{-7}$

$=0$ 

State true or false.
$\left(-8\right)\div \left(-1\right)=-8 $

mathematicians and their contribution history of mathematics maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The statement is not correct.

Since 
$(-8)\div (-1)$
$=\dfrac{-8}{-1}$
$=8$.
So $(-8)\div (-1)\ne (-8)$.

Which of the following statements are true ?
$0 \div 0=0 $

mathematicians and their contribution history of mathematics maths

  1. True

  2. False

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Correct Option: B
Explanation:

The given statement is false.

Since $\dfrac{0}{0}$ is indeterminate form.

The digit in the ten's place , in the smallest of such numbers , is 

mathematicians and their contribution history of mathematics maths

  1. 9

  2. 8

  3. 7

  4. 6

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Correct Option: A

State true or false.
$-8 \div \left(-2\right)=4 $

mathematicians and their contribution history of mathematics maths

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Now,

$\dfrac{-8}{-2}$
$=\dfrac{4\times (-2)}{-2}$
$=4$.
So the given statement is true.

If $A, {A} _{1}, {A} _{2}, {A} _{3}$ be the area of the in-circle and ex-circles, then $\dfrac {1}{\sqrt {{A} _{1}}}+\dfrac {1}{\sqrt {{A} _{2}}}+\dfrac {1}{\sqrt {{A} _{3}}}$ is equal to

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $\dfrac {1}{\sqrt {{A}}}$

  2. $\dfrac {2}{\sqrt {{A}}}$

  3. $\dfrac {3}{\sqrt {{A}}}$

  4. $None$

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Correct Option: A
Explanation:

$A _1={\pi}{r _1}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-a)^{2}}$

$A _2={\pi}{r _2}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-b)^{2}}$
$A _3={\pi}{r _3}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-c)^{2}}$
$A={\pi}{r}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s)^{2}}$
$\dfrac{1}{\sqrt{A _1}}+\dfrac{1}{\sqrt{A _2}}+\dfrac{1}{\sqrt{A _3}}=\dfrac{1}{\sqrt{\pi}}\bigg[\dfrac{s-a}{\Delta}+\dfrac{s-b}{\Delta}+\dfrac{s-c}{\Delta}\bigg]=\dfrac{1}{\sqrt{\pi}\Delta}[3{s}-(a+b+c)]=\dfrac{s}{\sqrt{\pi}\Delta}=\dfrac{1}{\sqrt{A}}$

(?)$-19657-33994=9999$

addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

  1. $63650$

  2. $53760$

  3. $59640$

  4. $61560$

  5. None of these

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Correct Option: A
Explanation:

Let $x-53651=9999$
Then, $x=9999+53651=63650$