Tag: maths

Questions Related to maths

If $x=\sqrt{2}+1, y=\sqrt{17}-\sqrt{2}$, then .............

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $x< y$

  2. $x>y$

  3. $x=y$

  4. $x\ge y$

Show answer
Correct Option: A
Explanation:

$\sqrt{2} =1.414$

$\sqrt{17} =4.123$

Hence,

$x = \sqrt{2}+1 =1.414+1 = 2.414$

$ y= \sqrt{17} -\sqrt{2} =4.123- 1.414 = 2.709$

Hence , $x<y$

$\sqrt{11}-\sqrt{10}$ $\Box$ $ \sqrt{12}-\sqrt{11}$

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $<$

  2. $>$

  3. $=$

  4. cannot be determined

Show answer
Correct Option: B
Explanation:

$\sqrt{11} =3.316$

$\sqrt{10} =3.162$

$\sqrt{12} =3.464$

Hence,

$\sqrt{11} -\sqrt{10} =3.316-3.162 = 0.154$

$\sqrt{12} -\sqrt{11} =3.464-3.316 = 0.148$

Hence , $\sqrt{11} -\sqrt{10} >\sqrt{12} -\sqrt{11}$

Which among the following numbers is the greatest?
$\displaystyle \sqrt[3]{4},\sqrt{2},\sqrt[6]{13},\sqrt[4]{5}$

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\displaystyle \sqrt[3]{4}$ is the greatest

  2. $\sqrt{2}$ is the greatest

  3. $\sqrt[6]{13}$ is the greatest

  4. $\sqrt[4]{5}$ is the greatest

Show answer
Correct Option: A
Explanation:

LCM of $3, 6, 4 = 12$

So, raising each given number to power $12$.
$\Rightarrow \sqrt[3]{4}=(4)^{1/3}=(4^{1/3})^{12}=4^{4}=256$
$\Rightarrow \sqrt{2}=(2)^{1/2}=(2^{1/2})^{12}=2^{6}=64$
$\Rightarrow \sqrt[6]{13}=(13)^{1/6}=(13^{1/6})^{12}=13^{2}=169$
$\Rightarrow \sqrt[4]{5}=(5)^{1/4}=(15^{1/4})^{12}=5^{3}=125$

$\therefore \sqrt[3]{4}$ is the greatest.

If $x=\sqrt{7}-\sqrt{5}, y=\sqrt{13}-\sqrt{11}$, then:

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $x=y$

  2. $x> y$

  3. $x< y$

  4. $x\ge y$

Show answer
Correct Option: B
Explanation:

$x = \sqrt{7}-\sqrt{5}$

$= 2.64 – 2.23= 0.41$

$y=  \sqrt{13}-\sqrt{11}$

$= 3.60- 3.31 = 0.29$

$Hence, x> y$

If $A=\sqrt{7}-\sqrt{6}$ and $B=\sqrt{6}-\sqrt{5}$, then identify the true statement.

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $A> B$

  2. $A=B$

  3. $A< B$

  4. $A\ge B$

Show answer
Correct Option: C
Explanation:
$A=\sqrt{7}-\sqrt{6}\Rightarrow \dfrac{1}{A}=\dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}$
$\Rightarrow \boxed{\dfrac{1}{A}=\sqrt{7}+\sqrt{6}}$
$\boxed{\dfrac{1}{B}=\sqrt{6}+\sqrt{5}}$
As $\sqrt{7} > \sqrt{5}\Rightarrow \dfrac{1}{A} > \dfrac{1}{B}$
$\Rightarrow \boxed{B > A}$

The smallest between $\sqrt{17} - \sqrt{12}$ and $\sqrt{11} - \sqrt{6}$ is _________.

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\sqrt{17} - \sqrt{12}$

  2. $\sqrt{11} - \sqrt{6}$

  3. Both are equal

  4. Can't be determined

Show answer
Correct Option: A
Explanation:

$\sqrt{17} \approx 4.123$

$\sqrt{12} \approx 3.464$
$\sqrt{11} \approx 3.316$
$\sqrt{6} \approx 2.449$

$\Rightarrow \sqrt{17} - \sqrt{12} = 0.659$
$\Rightarrow \sqrt{11} - \sqrt{6} = 0.867$

Hence, $\sqrt{17}-\sqrt{12}$ is smaller.

The smallest of $\sqrt [ 3 ]{ 4 } , \sqrt [ 4 ]{ 5 } , \sqrt [ 4 ]{ 6 } , \sqrt [ 3 ]{ 8 } $ is:

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\sqrt [ 3 ]{ 8 } $

  2. $\sqrt [ 4 ]{ 5 } $

  3. $\sqrt [ 3 ]{ 4 } $

  4. $\sqrt [ 4 ]{ 6 } $

Show answer
Correct Option: B
Explanation:
$\sqrt[3]{4}=\sqrt[12]{44}=\sqrt[12]{256}$
$\sqrt[4]{6}=\sqrt[12]{5^3}=\sqrt[12]{125}$
$\sqrt[4]{6}=\sqrt[12]{6^{3}}=\sqrt[12]{216}$
$\sqrt[3]{8}=\sqrt[12]{8^{4}}=\sqrt[12]{64^{2}}$
As $'125'$ is smallest
$\therefore \boxed{4\sqrt{5}}$ is smallest

Let x and y be rational and irrational numbers, respectively, then x + y necessarily an irrational number.


State True or False.

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Yes.

Let x $= 21, y =\sqrt{2}$ be a rational number
Now $x+y=21 +\sqrt{2}=21+1.4142....=22.4142....$ , which is non-terminating and non-recurring. Hence x+y is irrational.

If $A=\sqrt [ 3 ]{ 3 } , B=\sqrt [ 4 ]{ 5 } $, then which of the following is true?

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $A=\cfrac{3}{5}B$

  2. $A< B$

  3. $A> B$

  4. $A={B}^{4/5}$

Show answer
Correct Option: B
Explanation:
$A=3^{\dfrac{1}{3}}=\sqrt[12]{34}=\sqrt[12]{81}$
$B=5^{\dfrac{1}{4}}=\sqrt[12]{5^{3}}=\sqrt[12]{125}$
As $125 > 81$
$\Rightarrow \boxed{B > A}$

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

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

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

Show answer
Correct Option: C
Explanation:

$\sqrt[3]{2} \approx 1.26$

$\sqrt[6]{3} \approx 1.201$
$\sqrt[9]{4} \approx 1.166$

$\therefore$ Ascending order is $\sqrt[9]{4} < \sqrt[6]{3} < \sqrt[3]{2}$