Practice Questions for Algebra - Surds & Indices

1. CRE 1 - Indices | Algebra - Surds & Indices

$3^{\frac{1}{2}} \times 3^{\frac{3}{2}} \times 3^{\frac{5}{2}} \times 3^{\frac{7}{2}} \times 3^{\frac{9}{2}} \times 3^{\frac{11}{2}}$ = 27^x. Find x.

(a)
6
(b)
4
(c)
8
(d)
10

Answer: Option A

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Explanation :

Given, $3^{\frac{1}{2}} \times 3^{\frac{3}{2}} \times 3^{\frac{5}{2}} \times 3^{\frac{7}{2}} \times 3^{\frac{9}{2}} \times 3^{\frac{11}{2}}$ = 27^x

⇒ $3^{\frac{1}{2} + \frac{3}{2} + \frac{5}{2} + \frac{7}{2} + \frac{9}{2} + \frac{11}{2}}$ = 27^x

⇒ $3^{\frac{1 + 3 + 5 + 7 + 9 + 11}{2}}$ = (3³)^x

⇒ 3^36/2= 3^3x

⇒ 3¹⁸= 3^3x

⇒ 3x = 18

⇒ x = 6

Hence, option (a).

2. CRE 1 - Indices | Algebra - Surds & Indices

Find x, if $3^{x^{x}}$ = 81

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Answer: 2

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Explanation :

Given, $3^{x^{x}}$ = 81

⇒ $3^{x^{x}}$ = 3⁴

⇒ x^x = 4 = 2²

∴ x = 2.

Hence, 2.

3. CRE 1 - Indices | Algebra - Surds & Indices

Simplify: ${[{(\frac{2}{3})}^{2}]}^{3} \times {(\frac{1}{3})}^{- 4} \times 3^{- 1} \times \frac{1}{6}$

(a)
27/48
(b)
32/81
(c)
9/16
(d)
16/27

Answer: Option B

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Explanation :

Given, ${[{(\frac{2}{3})}^{2}]}^{3} \times {(\frac{1}{3})}^{- 4} \times 3^{- 1} \times \frac{1}{6}$

= ${(\frac{2}{3})}^{2 \times 3} \times 3^{4} \times \frac{1}{3} \times \frac{1}{6}$

= $\frac{2^{6}}{3^{6}} \times 3^{4} \times \frac{1}{3} \times \frac{1}{2 \times 3}$

= $\frac{2^{5}}{3^{4}}$ = $\frac{32}{81}$

Hence, option (b).

4. CRE 1 - Indices | Algebra - Surds & Indices

Given that 10^0.5 = x, 10^0.7 = y and x^z = y², then the value of z is close to:

(a)
1.45
(b)
1.88
(c)
2.8
(d)
3.7

5. CRE 1 - Indices | Algebra - Surds & Indices

Find the value of x, 2^5x-1 = $2^{2^{4^{1^{5^{2^{9^{2}}}}}}}$

(a)
-7/5
(b)
2
(c)
5
(d)
17/5

Answer: Option D

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Explanation :

2^5x-1 = $2^{2^{4^{1^{5^{2^{9^{2}}}}}}}$ = $2^{2^{4^{1}}}$ = $2^{2^{4}}$ = 2¹⁶

⇒ 5x - 1 = 16

⇒ x = 17/5

Hence, option (d).

6. CRE 1 - Indices | Algebra - Surds & Indices

Find x: 15^3x+3 = 225^x-1

(a)
-7
(b)
5
(c)
-3
(d)
-5

7. CRE 1 - Indices | Algebra - Surds & Indices

If 2^a3^b = 864. Find a + b if a and b are integers.

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8. CRE 1 - Indices | Algebra - Surds & Indices

Solve for integral value x, if ${(5 \sqrt{7})}^{5 x - 4}$ = ${(35)}^{3} {(25)}^{3 / 2}$

(a)
2
(b)
7/2
(c)
5/4
(d)
3

Answer: Option A

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Explanation :

Given, ${(5 \sqrt{7})}^{5 x - 4}$ = ${(35)}^{3} {(25)}^{3 / 2}$

⇒ $5^{5 x - 4} \times {(7^{\frac{1}{2}})}^{5 x - 4}$ = $5^{3} \times 7^{3} \times {(5^{2})}^{3 / 2}$

⇒ $5^{5 x - 4} \times 7^{\frac{5 x - 4}{2}}$ = $5^{6} \times 7^{3}$

Since 5 and 7 are prime numbers, power of 5 should be same in LHS and RHS. (Same can be solved w.r.t. power of 7)

⇒ 5x – 4 = 6

⇒ x = 2

Hence, option (a).

9. CRE 1 - Indices | Algebra - Surds & Indices

Simplify: ${(\frac{a^{x + y}}{a^{y}})}^{\frac{x}{y z}} \times {(\frac{a^{y + z}}{a^{z}})}^{\frac{y}{z x}} \times {(\frac{a^{z + x}}{a^{x}})}^{\frac{z}{y x}}$ if x + y + z = 0.

(a)
1
(b)
a³
(c)
0
(d)
a

Answer: Option B

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Explanation :

Given: ${(\frac{a^{x + y}}{a^{y}})}^{\frac{x}{y z}} \times {(\frac{a^{y + z}}{a^{z}})}^{\frac{y}{z x}} \times {(\frac{a^{z + x}}{a^{x}})}^{\frac{z}{y x}}$

Considering ${(\frac{a^{x + y}}{a^{y}})}^{\frac{x}{y z}}$

⇒ ${(\frac{a^{x + y}}{a^{y}})}^{\frac{x}{y z}} = {(a^{x + y - y})}^{\frac{x}{y z}}$ = ${(a^{x})}^{\frac{x}{y z}} = a^{\frac{x^{2}}{y z}}$

Similarly:

⇒ ${(\frac{a^{y + z}}{a^{z}})}^{\frac{y}{z x}} = a^{\frac{y^{2}}{z x}}$

⇒ ${(\frac{a^{z + x}}{a^{x}})}^{\frac{z}{y x}} = a^{\frac{z^{2}}{x y}}$

Hence,

${(\frac{a^{x + y}}{a^{y}})}^{\frac{x}{y z}} \times {(\frac{a^{y + z}}{a^{z}})}^{\frac{y}{z x}} \times {(\frac{a^{z + x}}{a^{x}})}^{\frac{z}{y x}}$

= $(a^{\frac{x^{2}}{y z}}) \times (a^{\frac{y^{2}}{z x}}) \times (a^{\frac{z^{2}}{x z}})$

= $a^{\frac{x^{2}}{y z} + \frac{y^{2}}{z x} + \frac{z^{2}}{x z}}$ = $a^{\frac{x^{3} + y^{3} + z^{3}}{x y z}}$ = $a^{\frac{3 x y z}{x y z}}$ = $a^{3}$

Note: a³ + b³ + c³ = 3abc when a + b + c = 0.

Hence, option (b).

10. CRE 1 - Indices | Algebra - Surds & Indices

Which of the following is correct?

(a)
2³⁰⁰ > 3²⁰⁰
(b)
2³⁰⁰ < 3²⁰⁰
(c)
2³⁰⁰ = 3²⁰⁰
(d)
More than 1 of the above

11. CRE 1 - Indices | Algebra - Surds & Indices

Arrange in decreasing order: 3^1/2, 4^1/5, 5^1/4

(a)
3^1/2 > 4^1/5 > 5^1/4
(b)
5^1/4 > 3^1/2 > 4^1/5
(c)
4^1/5 > 5^1/4 > 3^1/2
(d)
3^1/2 > 5^1/4 > 4^1/5

Answer: Option D

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Explanation :

Given, 3^1/2, 4^1/5, 5^1/4

LCM (2, 5, 4) = 20

Raising all the numbers to power 20, we get:

⇒ ${(3^{\frac{1}{2}})}^{20}, {(4^{\frac{1}{5}})}^{20}, {(5^{\frac{1}{4}})}^{20}$

⇒ 3¹⁰, 4⁴, 5⁵

⇒ 9⁵, 256, 3125

⇒ 81 × 729, 256, 3125

Since, 81 × 729 > 3125 > 256

⇒ 3^1/2 > 5^1/4 > 4^1/5

Hence, option (d).

12. CRE 1 - Indices | Algebra - Surds & Indices

If $\sqrt{x \sqrt{x \sqrt{x \dots \dots \infty}}} = 11^{x}$ , what is the value of $\sqrt[x]{x}$ ?

(a)
11
(b)
1
(c)
√11
(d)
11^1/x

Answer: Option A

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Explanation :

Given, $\sqrt{x \sqrt{x \sqrt{x \dots \dots \infty}}} = 11^{x}$

Let’s assume √ $\sqrt{x \sqrt{x \sqrt{x \dots \dots \infty}}}$ = y

⇒ √xy = y

⇒ xy = y²

⇒ x = y

Hence,

$\sqrt{x \sqrt{x \sqrt{x \dots \dots \infty}}}$ = x = 11^x

⇒ x = 11^1/x

⇒ $\sqrt[x]{x}$ = 11

Hence, option (a).

13. CRE 1 - Indices | Algebra - Surds & Indices

Find the value of $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots \dots \infty}}}$

(a)
2
(b)
-1
(c)
3
(d)
More than one of the above

Answer: Option A

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Explanation :

Let y = $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots \dots \infty}}}$

⇒ y = $\sqrt{2 + y}$

⇒ y² = 2 + y

⇒ y² – y – 2 = 0

⇒ y² – 2y + y – 2 = 0

⇒ (y - 2)(y + 1) = 0

Hence, y = 2 or -1

Since $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots \dots \infty}}}$ cannot be negative.

Hence, $\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots \dots \infty}}}$ = 2

Hence, option (a).

14. CRE 1 - Indices | Algebra - Surds & Indices

If 5^x= (0.125)^y= 10^3z, then what is the relation between x, y and z, given that they are non-zero real numbers?

(a)
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$
(b)
$\frac{3}{x} = \frac{1}{y} + \frac{1}{z}$
(c)
$\frac{1}{z} - \frac{1}{y} - \frac{1}{x} = 1$
(d)
xyz = 1

Answer: Option B

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Explanation :

Let 5^x = (0.125)^y = 10^3z = k

Hence, 5 = k^1/x

0.125 = k^1/y and

10 = k^1/3z

We know that 0.125 = (0.5)³ = (5/10)³

Hence, $k^{\frac{1}{y}} = {(\frac{k^{\frac{1}{x}}}{k^{\frac{1}{3 z}}})}^{3} = k^{\frac{3}{x} - \frac{1}{z}}$

⇒ $\frac{1}{y} = \frac{3}{x} - \frac{1}{z}$

⇒ $\frac{3}{x} = \frac{1}{y} + \frac{1}{z}$

Hence, option (b).

15. CRE 1 - Indices | Algebra - Surds & Indices

What is the value of z in terms of x and y, if 3^x= 2^y= 6^z, and x, y and z are non-zero?

(a)
$z = \frac{x + y}{4 xy}$
(b)
$z = \frac{xy}{x + y}$
(c)
$z = \frac{x + y}{2}$
(d)
$z = \frac{x^{2} + y^{2}}{xy}$

Answer: Option B

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Explanation :

Let 3^x = 2^y = 6^z = k

Now, 3x = k ⇒ 3 = k^1/x …(1)

Similarly,

2y = k ⇒ 2 = k^1/y …(2)

6y = k ⇒ 6 = k^1/z …(3)

We know, 3 × 2 = 6

∴ (1) × (2) = (3)

⇒ k^1/x × k^1/y = k^1/z

⇒ $k^{\frac{1}{x} + \frac{1}{y}} = k^{\frac{1}{z}}$

⇒ $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$

⇒ $z = \frac{x y}{x + y}$

Hence, option (b).

16. CRE 1 - Indices | Algebra - Surds & Indices

If p²+ q²+ r²= 1 what is the value of $(\sqrt[- pq]{a^{r}}) \times (\sqrt[- qr]{a^{p}}) \times (\sqrt[- pr]{a^{q}}) ?$ ?

(a)
a^pqr
(b)
a^1/pqr
(c)
1/a^pqr
(d)
a^-1/pqr

Answer: Option D

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Explanation :

Given, $(\sqrt[- pq]{a^{r}}) \times (\sqrt[- qr]{a^{p}}) \times (\sqrt[- pr]{a^{q}})$

Let’s consider $\sqrt[- pq]{a^{r}}$ .

⇒ $\sqrt[- pq]{a^{r}}$ = ${(a^{r})}^{- \frac{1}{p q}}$ = $a^{- \frac{r}{pq}}$ …(1)

Similarly,

$\sqrt[- qr]{a^{p}}$ = $a^{- \frac{p}{qr}}$ …(2)

$\sqrt[- rp]{a^{q}}$ = $a^{- \frac{q}{rp}}$ …(3)

Multiplying all three equation,

$(\sqrt[- pq]{a^{r}}) \times (\sqrt[- qr]{a^{p}}) \times (\sqrt[- pr]{a^{q}})$ = $a^{- \frac{r}{pq}} \times a^{- \frac{p}{qr}} \times a^{- \frac{q}{rp}}$

= $a^{- \frac{r}{pq}} \times a^{- \frac{p}{qr}} \times a^{- \frac{q}{rp}}$

= $a^{- (\frac{r^{2} + p^{2} + q^{2}}{pqr})}$

= $a^{- (\frac{1}{pqr})}$ [p² + q² + r²= 1]

Hence, option (d).

CRE 1 - Indices | Algebra - Surds & Indices

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