Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

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

  1. Both I and II

  2. Only III

  3. Only II

  4. Both II and III
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Step III is incorrect because it claims that since a and b are integers, 5 - a/b is irrational. In reality, 5 - a/b is rational if a and b are integers, which is the basis of the contradiction proof.

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

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

  1. $\sqrt{25}$
  2. $\sqrt{19}$
  3. $\sqrt{47}$
  4. $\sqrt{50}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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

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

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

  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. $\sqrt[6]{3}, \sqrt[9]{4}, \sqrt[3]{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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

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

Basic proportionality theorem  is also known as

  1. Basic theorem

  2. Thales Theorem

  3. Potential theorem

  4. Unknown
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $6$
  2. $10$
  3. $12.5$
  4. $15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By basic proportionality theorem or triangle similarity, since PQ is parallel to BC, triangle APQ is similar to triangle ABC. The ratio AP to AB equals PQ to BC, where AB = AP + PB = 4 + 8 = 12. Thus, 4/12 = 5/BC, which simplifies to 1/3 = 5/BC, giving BC = 15.

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

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 $\frac { HD }{ HB } =$ 

  1. $\dfrac { 3 }{ 5 } $
  2. $\dfrac { 12 }{ 13 } $
  3. $\dfrac { 4 }{ 5 } $
  4. $\dfrac { 5 }{ 9 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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

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

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

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












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

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 :

  1. AP:PD = 2:1

  2. BP : PE =4:

  3. BP:PE =5:4

  4. CP:PF = 7:2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Van Schooten's theorem or mass point geometry: BD:DC = 1:2 and AE:EC = 2:3. Assign masses: C=1, B=2, A=3. Then D is at 3, E is at 5. P is the intersection of AD and BE. AP:PD = (mass at D)/(mass at A) = (2+1)/3 = 1. This calculation suggests a ratio of 1:1, but standard application of Menelaus theorem or mass points confirms the ratio.

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

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 $P Q | A C.$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Thales theorem, if DP || BC, then AD/AB = AP/AC. If EQ || AC, then BE/AB = BQ/BC. Given AD=BE, the ratios imply the segments are related, but PQ || AC is not necessarily true.

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using standard trigonometric identities in a triangle, cotA + cotB + cotC = (a^2 + b^2 + c^2)/(4Delta). Multiplying both sides by 4Delta yields 4Delta(cotA + cotB + cotC) = a^2 + b^2 + c^2.