Arrange the following in descending order: A. √401-√399 B. √301-√299 C. √101-√99 D. √201-√199 Choose the correct answer from the options given below: Option 1: A, C, B, D Option 2: A, C, D, B Option 3: B, D, A, C Option 4: C, D, B, A

Answer: C, D, B, A. Solution: Each option is of the form √(bigger)-√(smaller), which is painful to compare directly. Rationalise each one by multiplying and dividing by its conjugate, using the identity (A-B)(A+B)=A^2-B^2. For any number x, √(x+1)-√(x-1)=((x+1)-(x-1))/(√(x+1)+√(x-1))=2/(√(x+1)+√(x-1)), Option, After rationalising, Size of denominator, A, 2/(√401+√399), largest, B, 2/(√301+√299), third, C, 2/(√101+√99), smallest, D, 2/(√201+√199), second smallest, Every option now has the same numerator 2, so the smaller the denominator, the bigger the value. Descending order: C>D>B>A

Algebra

Indices

Medium

Arrange the following in descending order:

A. $\sqrt{401}-\sqrt{399}$

B. $\sqrt{301}-\sqrt{299}$

C. $\sqrt{101}-\sqrt{99}$

D. $\sqrt{201}-\sqrt{199}$

Choose the correct answer from the options given below:

A, C, B, D

A, C, D, B

B, D, A, C

C, D, B, A

✅ Correct Option: 4

Each option is of the form $\sqrt{\text{bigger}}-\sqrt{\text{smaller}}$ , which is painful to compare directly. Rationalise each one by multiplying and dividing by its conjugate, using the identity $(A-B)(A+B)=A^2-B^2$ .

For any number $x$ ,

$\sqrt{x+1}-\sqrt{x-1}=\dfrac{(x+1)-(x-1)}{\sqrt{x+1}+\sqrt{x-1}}=\dfrac{2}{\sqrt{x+1}+\sqrt{x-1}}$

Option	After rationalising	Size of denominator
A	$\dfrac{2}{\sqrt{401}+\sqrt{399}}$	largest
B	$\dfrac{2}{\sqrt{301}+\sqrt{299}}$	third
C	$\dfrac{2}{\sqrt{101}+\sqrt{99}}$	smallest
D	$\dfrac{2}{\sqrt{201}+\sqrt{199}}$	second smallest

Every option now has the same numerator 2, so the smaller the denominator, the bigger the value.

Descending order:

$C>D>B>A$

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Choose the correct answer from the options given below :

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Simplify : $\dfrac{\sqrt[4]{0.0625}+\sqrt[3]{0.008}+\sqrt{0.09}-1}{\sqrt[3]{62 \cdot 5 \times \sqrt[5]{32}}}$

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$\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$ is equal to

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$\sqrt{3+\sqrt{5}}=$

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$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}+\frac{\sqrt{11+2 \sqrt{30}}}{\sqrt{6}-\sqrt{5}}$

Arrange the following in descending order: A. 401−399\sqrt{401}-\sqrt{399}401​−399​ B. 301−299\sqrt{301}-\sqrt{299}301​−299​ C. 101−99\sqrt{101}-\sqrt{99}101​−99​ D. 201−199\sqrt{201}-\sqrt{199}201​−199​ Choose the correct answer from the options given below:

Solution

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JIPMAT Past Year Questions (Topic-Wise)

Arrange the following in descending order:

A. $\sqrt{401}-\sqrt{399}$

B. $\sqrt{301}-\sqrt{299}$

C. $\sqrt{101}-\sqrt{99}$

D. $\sqrt{201}-\sqrt{199}$

Choose the correct answer from the options given below: