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Surds and Indices questions for IPMAT, with solutions

10 free Surds and Indices practice questions for IPMAT Indore, Rohtak and JIPMAT, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Surds and Indices on Learn.

1. Question 1Type the answerWritten by IPMATwayIndices (Exponents)

   If $2^{x-1}+2^{x+1}=1280$, then the value of $2^{x}+1$ is

   Show the answer and solution

   Answer: 513

   Take out $2^{x-1}$: $2^{x-1}+2^{x+1}=2^{x-1}(1+2^2)=5\cdot2^{x-1}$. So $5\cdot2^{x-1}=1280$ and $2^{x-1}=256=2^8$. Then $x=9$ and $2^x=512$. So $2^x+1=513$.

2. Question 2Multiple choiceWritten by IPMATwaySurds

   If $x=\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}$, then $x^{2}$ equals

   1. A.$12$
   2. B.$4$
   3. C.$16$
   4. D.$2\sqrt{15}$
   Show the answer and solution

   Answer: A. $12$

   Denest each root: $8+2\sqrt{15}=(\sqrt5+\sqrt3)^2$ and $8-2\sqrt{15}=(\sqrt5-\sqrt3)^2$. Both $\sqrt5\pm\sqrt3$ are positive, so $x=(\sqrt5+\sqrt3)-(\sqrt5-\sqrt3)=2\sqrt3$. Then $x^2=4\cdot3=12$. (Or square directly: $x^2=16-2\sqrt{64-60}=16-4=12$.)

3. Question 3Type the answerWritten by IPMATwaySurds

   The value of $\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ is

   Show the answer and solution

   Answer: 3

   Let $y=\sqrt{6+\sqrt{6+\cdots}}$. The expression under the first root contains the whole expression again, so $y=\sqrt{6+y}$. Squaring: $y^2-y-6=0$, i.e. $(y-3)(y+2)=0$. A square root is positive, so $y=3$.

4. Question 4Type the answerWritten by IPMATwaySurds

   The value of $\dfrac{\sqrt7+\sqrt5}{\sqrt7-\sqrt5}+\dfrac{\sqrt7-\sqrt5}{\sqrt7+\sqrt5}$ is

   Show the answer and solution

   Answer: 12

   Put both fractions over the common denominator $(\sqrt7-\sqrt5)(\sqrt7+\sqrt5)=7-5=2$. The numerator is $(\sqrt7+\sqrt5)^2+(\sqrt7-\sqrt5)^2=(12+2\sqrt{35})+(12-2\sqrt{35})=24$. So the sum is $\frac{24}{2}=12$.

5. Question 5Multiple choiceWritten by IPMATwayIndices (Exponents)

   The solution set of $\left(\dfrac12\right)^{x^2-4x}<8$ is

   1. A.$(1,3)$
   2. B.$(1,\infty)$
   3. C.$(-\infty,3)$
   4. D.$(-\infty,1)\cup(3,\infty)$
   Show the answer and solution

   Answer: D. $(-\infty,1)\cup(3,\infty)$

   Write $8=2^3=\left(\frac12\right)^{-3}$. The base $\frac12$ is less than $1$, so comparing exponents reverses the inequality: $x^2-4x>-3$. That is $x^2-4x+3>0$, i.e. $(x-1)(x-3)>0$. So $x<1$ or $x>3$. Check $x=2$: $\left(\frac12\right)^{-4}=16$, which is not less than $8$.

6. Question 6Multiple choiceWritten by IPMATwayIndices (Exponents)

   If $x^{x\sqrt{x}}=\left(x\sqrt{x}\right)^{x}$, where $x>0$ and $x\ne1$, then $x$ equals

   1. A.$\dfrac{9}{4}$
   2. B.$\dfrac{3}{2}$
   3. C.$4$
   4. D.$\dfrac{4}{9}$
   Show the answer and solution

   Answer: A. $\dfrac{9}{4}$

   Write the right side with base $x$: $x\sqrt{x}=x^{\frac32}$, so $\left(x\sqrt x\right)^x=x^{\frac{3x}{2}}$. The left side is $x^{x\sqrt x}=x^{x^{3/2}}$. With $x>0$, $x\ne1$, equal powers need equal exponents: $x^{\frac32}=\frac{3x}{2}$. Divide by $x$: $\sqrt x=\frac32$, so $x=\frac94$.

7. Question 7Multiple choiceWritten by IPMATwayIndices (Exponents)

   The values of $x$ for which $0.6^{x^{2}}>0.6^{3x-2}$ are

   1. A.$1<x<2$
   2. B.$x<1$ or $x>2$
   3. C.all real $x$
   4. D.$x>2$
   Show the answer and solution

   Answer: A. $1<x<2$

   For a base $0<b<1$, larger exponents give smaller values, so $b^u>b^v\iff u<v$. Here that means $x^2<3x-2$, i.e. $x^2-3x+2<0$, i.e. $(x-1)(x-2)<0$. This holds strictly between the roots: $1<x<2$. Check $x=1.5$: exponents $2.25$ and $2.5$, and $0.6^{2.25}>0.6^{2.5}$.

8. Question 8Multiple choiceWritten by IPMATwayIndices (Exponents)

   $5^{48}+1$ is exactly divisible by

   1. A.$5^{16}+1$
   2. B.$5^{24}+1$
   3. C.$5^{8}+1$
   4. D.$5^{12}+1$
   Show the answer and solution

   Answer: A. $5^{16}+1$

   Use $t^n+1$ divisible by $t+1$ exactly when $n$ is odd (put $t=-1$: $(-1)^n+1=0$ only for odd $n$). Option A: $5^{48}+1=(5^{16})^3+1$ with odd power $3$ - divisible by $5^{16}+1$. Option B: $(5^{24})^2+1$, power $2$ even. Option C: $(5^8)^6+1$, power $6$ even. Option D: $(5^{12})^4+1$, power $4$ even. So only $5^{16}+1$ divides it.

9. Question 9Multiple choiceWritten by IPMATwaySurds

   If $x=5+2\sqrt6$, then $\sqrt{x}+\dfrac{1}{\sqrt{x}}$ equals

   1. A.$2\sqrt3$
   2. B.$\sqrt{12}+2$
   3. C.$2\sqrt2$
   4. D.$10$
   Show the answer and solution

   Answer: A. $2\sqrt3$

   Note $5+2\sqrt6=(\sqrt3+\sqrt2)^2$, so $\sqrt x=\sqrt3+\sqrt2$. Its reciprocal is $\frac{1}{\sqrt3+\sqrt2}=\sqrt3-\sqrt2$ (since $3-2=1$). Adding: $\sqrt x+\frac{1}{\sqrt x}=(\sqrt3+\sqrt2)+(\sqrt3-\sqrt2)=2\sqrt3$. Check: its square is $x+\frac1x+2=10+2=12$.

10. Question 10Type the answerWritten by IPMATwaySurds

    The value of $\dfrac{1}{\sqrt1+\sqrt3}+\dfrac{1}{\sqrt3+\sqrt5}+\cdots+\dfrac{1}{\sqrt{23}+\sqrt{25}}$ is

    Show the answer and solution

    Answer: 2

    Rationalise one term: $\frac{1}{\sqrt{k}+\sqrt{k+2}}=\frac{\sqrt{k+2}-\sqrt k}{(k+2)-k}=\frac{\sqrt{k+2}-\sqrt k}{2}$. Summing over $k=1,3,\ldots,23$, the numerators telescope to $\sqrt{25}-\sqrt1=5-1=4$. Dividing by $2$ gives $2$. The trap is forgetting the denominator $2$ and answering $4$.