Tag: maths

Questions Related to maths

Which is the wrong step that shows $\displaystyle 5-\sqrt{3}$ is irrational?
(I) Contradiction : Assume that $\displaystyle 5-\sqrt{3}$ is rational
(II) Find coprime a & b $\displaystyle \left ( b\neq 0 \right )$ such that $\displaystyle 5-\sqrt{3}=\frac{a}{b},\therefore 5-\frac{a}{b}=\sqrt{3}$
Rearranging above equation $\displaystyle \sqrt{3}=5-\frac{a}{b}=\frac{5b-a}{b}$
(III) Since a & b are integers we get $\displaystyle 5-\frac{a}{b}$ is irrational and so $\displaystyle \sqrt{3}$ is irrational
(IV) But this contradicts the fact that $\displaystyle \sqrt{3}$ is irrational Hence $\displaystyle 5-\sqrt{3}$ is irrational

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. Both I and II

  2. Only III

  3. Only II

  4. Both II and III

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Correct Option: B

Which of the following irrational numbers lie between $4$ and $7$?

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\sqrt{25}$

  2. $\sqrt{19}$

  3. $\sqrt{47}$

  4. $\sqrt{50}$

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Correct Option: B,C
Explanation:

$4^{2} = 16$

$5^{2} = 25$
$6^{2} = 36$
$7^{2} = 49$


$\Rightarrow \sqrt19$ and $\sqrt47$ are irrational numbers which lie between $4$ and $7$


$\sqrt25 = 5$ which is a rational number

The ascending order of the surds $\sqrt[3]{2}, \sqrt[6]{3}, \sqrt[9]{4}$ is 

maths number system representation of irrational numbers on number line existence of irrational numbers irrational numbers

  1. $\sqrt[9]{4}, \sqrt[6]{3}, \sqrt[3]{2}$

  2. $\sqrt[9]{4}, \sqrt[3]{2}, \sqrt[6]{3}$

  3. $\sqrt[3]{2}, \sqrt[6]{3}, \sqrt[9]{4}$

  4. $\sqrt[6]{3}, \sqrt[9]{4}, \sqrt[3]{2}$

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Correct Option: A
Explanation:

Lets keep in mind the following common log values:

$log(2)=0.30$
$log(3)=0.47$
Lets compare these three surds by their log values.
Reason: Since $log(x) $ is an increasing function  when $x>1$ so we can compare these values by comparing their log values.

$log(\sqrt[3]{2})=log(2^{\frac{1}3})=\left(\dfrac{1}3\right)log(2)=\left(\dfrac{1}3\right) \times 0.3=0.100$

$log(\sqrt[6]{3})=log(3^{\frac{1}6})=\left(\dfrac{1}6\right)log(3)=\left(\dfrac{1}6\right) \times 0.47=0.078$

$log(\sqrt[9]{4})=log(\sqrt[9]{2^2})=log(2^{\frac{2}9})=\left(\dfrac{2}9\right)log(2)=\left(\dfrac{2}9\right) \times 0.3=0.060$

By looking at the log values,it is clear that,

$\sqrt[9]{4} < \sqrt[6]{3} < \sqrt[3]{2}$

Basic proportionality theorem  is also known as

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. Basic theorem

  2. Thales Theorem

  3. Potential theorem

  4. Unknown

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Correct Option: B
Explanation:

Basic proportionality theorem is also known as Thales Theorem. Thales was a famous Greek mathematician who gave an important truth relating two equiangular triangles.
Therefore, C is the correct answer.

In $\triangle ABC,A-P-B, A-Q-C$ and $\overline {PQ} \parallel \overline {BC}$. If $PQ=5, AP=4$ and $PB=8$, then $BC=$.....

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. $6$

  2. $10$

  3. $12.5$

  4. $15$

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Correct Option: A

ABC is a triangle with AB = $13$ cm, BC =$14$ cm and CA=$15$ cm. AD and BE are the altitudes from A to B to BC and AC respectively. H is the point of intersection of the AD and BE. Then the ratio of $\frac { HD }{ HB } =$ 

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. $\dfrac { 3 }{ 5 } $

  2. $\dfrac { 12 }{ 13 } $

  3. $\dfrac { 4 }{ 5 } $

  4. $\dfrac { 5 }{ 9 } $

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Correct Option: A
Explanation:
According to the question,
Triangle $BEC$ and triangle $BDH$ are similar, because they have the same angles this means that the sides  of these two triangles are in the same ratio.

So,

$\dfrac{{HD}}{{BD}} = \dfrac{{CE}}{{BC}}$

Note,However that $\displaystyle \frac{{CE}}{{BC}}$=$cosC$, 

Hence$\displaystyle \frac{{HD}}{{BD}}$=$cosC$, so we proceed to find $cosC$ using the cosine rule,

${c^2} = {a^2} + {b^2} - 2ab\cos C$

${13^2} = {14^2} + {15^2} - 2(14)(15)cosC$

$\cos C = \dfrac{{{{13}^2} - {{14}^2} - {{15}^2}}}{{ - 2 \times 14 \times 15}} = \dfrac{3}{5}$

$so\, \, \dfrac{{HD}}{{HB}} = \dfrac{3}{5}$












In a triangle ABC, D and E are the point on the line segment BC and AC respectively, such that 2 BD = DC and 3 AE = 2 EC. The lines AD and BE meet at P,the line CP and AB F, then :

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. AP:PD = 2:1

  2. BP : PE =4:

  3. BP:PE =5:4

  4. CP:PF = 7:2

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Correct Option: A

Let  $ABC$  be a triangle and  $D$  and  $E$  be two points on side  $AB$  such that  $AD = BE$.  If  $D P | B C$  and  $E Q | A C,$ then $P Q | A C.$

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. True

  2. False

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Correct Option: B

If the sides a, b, c, of a triangle are such that a: b: c: :1:$\sqrt{3}$: 2, then the A:B:C is -

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. 3 : 2 : 1

  2. 3 : 1 : 2

  3. 1 : 3 : 2

  4. 1 : 2 : 3

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Correct Option: A

In any $\Delta$ABC , $4\Delta(cotA+cotB+cotC)$ is equal to 

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. $3(a^2+b^2+c^2)$

  2. $2(a^2+b^2+c^2)$

  3. $(a^2+b^2+c^2)$

  4. none of these

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Correct Option: A