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JIPMAT 2026 PYQsQuantitative Ability. Free, no login required.

Algebra

past year questions, JIPMAT 2026

Indices

past year questions, Algebra, JIPMAT 2026

Medium

The value of 3+13+13+3+13−33+\frac{1}{\sqrt3}+\frac{1}{3+\sqrt3}+\frac{1}{\sqrt3-3} is

Solution

✅ Correct Option: 2

Add the last two fractions first, since those are the messy ones.

13+3+13−3=(3−3)+(3+3)(3+3)(3−3)\dfrac{1}{3+\sqrt3}+\dfrac{1}{\sqrt3-3}=\dfrac{(\sqrt3-3)+(3+\sqrt3)}{(3+\sqrt3)(\sqrt3-3)}

Numerator=23\text{Numerator}=2\sqrt3

Using the identity (A+B)(A−B)=A2−B2(A+B)(A-B)=A^2-B^2 on the denominator:

Denominator=(3)2−32=3−9=−6\text{Denominator}=(\sqrt3)^2-3^2=3-9=-6

So their sum is

23−6=−33=−13\dfrac{2\sqrt3}{-6}=-\dfrac{\sqrt3}{3}=-\dfrac{1}{\sqrt3}

Put this back into the original expression:

3+13−13=33+\dfrac{1}{\sqrt3}-\dfrac{1}{\sqrt3}=3

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B. 301−299\sqrt{301}-\sqrt{299}

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D. 201−199\sqrt{201}-\sqrt{199}

Choose the correct answer from the options given below:

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If 2x=3y=6−z2^{x} = 3^{y} = 6^{-z}, then (1x+1y+1z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) is equal to

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

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5+38−215+11+2306−5\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}+\frac{\sqrt{11+2 \sqrt{30}}}{\sqrt{6}-\sqrt{5}}

JIPMAT 2024

Simplify : 0.06254+0.0083+0.09−162⋅5×3253\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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