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Indices — PYPs

3 solved Indices previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:jipmat 2025QA › IndicesHardQA · MCQ

\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

A1
B2
C4
D $\sqrt{6}$

Pick an option to attempt

The Setup: This looks like a terrifying infinite radical, but it's actually a satisfying chain reaction. We work from right to left, using the difference of squares identity:

(a+b)(a-b)=a^2-b^2

. Step 1: Multiply the last two terms together. They are identical except for the sign in the middle. Let

X=\sqrt{2+\sqrt{3}}

. The last two terms are

\sqrt{2+\sqrt{X}}

and

\sqrt{2-\sqrt{X}}

\sqrt{2+\sqrt{X}} \times \sqrt{2-\sqrt{X}}=\sqrt{(2)^2-(\sqrt{X})^2}=\sqrt{4-X}

Substitute

X

back in:

\sqrt{4-(2+\sqrt{3})}=\sqrt{2-\sqrt{3}}

Step 2: Now our expression is shorter. Multiply this new result by the second term from the original equation. Let's drop the visual noise.

\sqrt{2+\sqrt{3}} \times (\sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2-\sqrt{3}})

Apply the difference of squares again! Let

Y=\sqrt{3}

\sqrt{2+\sqrt{Y}} \times \sqrt{2-\sqrt{Y}}=\sqrt{4-Y}=\sqrt{4-(2+\sqrt{3})}=\sqrt{2-\sqrt{3}}

Step 3: Final stage. Multiply this with the very first term.

\sqrt{2+\sqrt{3}} \times \sqrt{2-\sqrt{3}}

\sqrt{(2)^2-(\sqrt{3})^2}=\sqrt{4-3}=\sqrt{1}=1

Final Answer: 1

Q2:ipmat indore 2024QA › IndicesEasyMCQ · MCQ

The greatest number among

2^{300}

3^{200}

4^{100}

2^{100} + 3^{100}

A $2^{300}$
B $3^{200}$
C $2^{100} + 3^{100}$
D $4^{100}$

Pick an option to attempt

The Setup: We need to identify the greatest number among the expressions

2^{300}

3^{200}

4^{100}

, and

2^{100}+3^{100}

. Step 1: Normalize the exponents to a common power for direct comparison. We can rewrite each single-term expression by factoring out a power of

100

in the exponent:

2^{300} = (2^3)^{100} = 8^{100}

3^{200} = (3^2)^{100} = 9^{100}

4^{100} = 4^{100}

Step 2: Evaluate the additive term against the largest single base. Compare the addition term

2^{100} + 3^{100}

to the largest normalized term

9^{100}

: Since

2^{100} < 3^{100}

, their sum satisfies

2^{100} + 3^{100} < 3^{100} + 3^{100} = 2 \times 3^{100}

. Clearly,

2 \times 3^{100}

is vastly smaller than

9^{100}

(which is

3^{200}

). Step 3: Conclude the greatest term. Among the expressions

8^{100}

9^{100}

, and

4^{100}

, the term with the largest base is

9^{100}

, which corresponds back to

3^{200}

. Final Answer:

3^{200}

Q3:ipmat indore 2019QA › IndicesMediumMCQ · MCQ

Determine the greatest number among the following four numbers:

A $2^{300}$
B $3^{200}$
C $2^{100} + 3^{100}$
D $4^{100}$

Pick an option to attempt

The Setup: A power-scaling comparison. Computing these directly is hopeless, so rewrite every term with the same exponent - then only the bases need comparing. Step 1: Normalise to the exponent 100. Each exponent is a multiple of 100, so pull that out:

2^{300}=\left(2^3\right)^{100}=8^{100}, \qquad 3^{200}=\left(3^2\right)^{100}=9^{100}, \qquad 4^{100}=4^{100}

Step 2: Compare the three pure powers. With a common positive exponent,

t\mapsto t^{100}

is increasing on positive bases, so the ordering of the bases carries straight over:

9^{100}>8^{100}>4^{100} \implies 3^{200}>2^{300}>4^{100}

Step 3: Dispose of the sum. The remaining candidate

2^{100}+3^{100}

is not a pure power, so bound it. Since

2^{100}<3^{100}

2^{100}+3^{100}<3^{100}+3^{100}=2\cdot 3^{100}

Compare that generous overestimate against the leader:

\frac{3^{200}}{3^{100}}=3^{100} \quad\text{which is vastly larger than }2

2\cdot 3^{100}<3^{100}\cdot 3^{100}=3^{200}

. The sum loses even after being inflated, and the gap is not close -

3^{200}

exceeds it by a factor of roughly

\frac{3^{100}}{2}

. Step 4: Conclude. The full ordering is

3^{200}>2^{300}>4^{100}>2^{100}+3^{100}

, and the greatest is

3^{200}

. The lesson: adding two large powers is worth far less than multiplying the exponent.

2^{100}+3^{100}

is barely bigger than its larger half, while squaring

3^{100}

multiplies it by itself. Final Answer:

3^{200}

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