Surds & Indices Questions and Answers

Q: If \(pqr \neq 0\) and \(p^{-x} = \frac{1}{q}, q^{-y} = \frac{1}{r}, r^{-z} = \frac{1}{p}\), what is the value of the product \(xyz\)?

The product is \(xyz = 1\).Taking reciprocals, \(p^{-x} = \frac{1}{q}\) gives \(p^{x} = q\); likewise \(q^{y} = r\) and \(r^{z} = p\). Substitute each into the next:\[ \begin{aligned} r &= q^{y} = (p^{x})^{y} = p^{xy} \\ p &= r^{z} = p^{xyz} \end{aligned} \]So \(p^{xyz} = p^{1}\), and \(xyz = 1\) for a general base \(p\). The answer \(-1\) comes from forgetting that the negative exponents were cleared by the reciprocals, and \(\frac{1}{pqr}\) and \(pqr\) mix the bases into the exponents, which the equations never do.

Q: If (1.001)1259 = 3.52 and (1.001)2062 = 7.85, then (1.001)3321 =

(1.001)3321 = 27.64.The exponents add up: 1259 + 2062 = 3321. Powers of the same base multiply when their exponents add, so:\[ \begin{aligned} (1.001)^{3321} &= (1.001)^{1259} \times (1.001)^{2062} \\ &= 3.52 \times 7.85 = 27.632 \end{aligned} \]which agrees with 27.64 within the rounding of the two given values (3.52 and 7.85 are themselves rounded). 11.37 is 3.52 + 7.85, adding the values instead of multiplying them. 4.33 and 2.23 are a difference and a ratio of the two values, which would answer a subtraction of exponents, not an addition.

Q: If \((1.001)^{1259} = 3.52\) and \((1.001)^{2062} = 7.85\), then \((1.001)^{3321} =\)

The value is about 27.64.The exponents add: \(1259 + 2062 = 3321\). So by the law of indices\[ \begin{aligned} (1.001)^{3321} &= (1.001)^{1259}\times(1.001)^{2062} \\ &= 3.52 \times 7.85 = 27.632 \end{aligned} \]which rounds to the option 27.64. 11.37 adds the two values instead of multiplying them, 4.33 is their difference, and 2.23 is their ratio; none follows from the rule \(a^{m}a^{n} = a^{m+n}\).

Q: If a = \(b^{x}\), b = \(c^{y}\) and c = \(a^{z}\), then the value of xyz is equal to :

The value of xyz is 1.Substitute each relation into the one before it so that everything is expressed as a power of a.\[ \begin{aligned} a &= b^{x} = (c^{y})^{x} = c^{xy} \\ &= (a^{z})^{xy} = a^{xyz} \end{aligned} \]Since \(a = a^{1}\), the exponents must match, so xyz = 1 (for a base other than 0 or 1, the usual assumption in such questions).−1 would give \(a = 1/a\), and 0 would give a = 1 for every a. abc is a product of the bases, not of the exponents, so it is not what the relations determine.

Q: Which number amongst \(3^{31}\), \(4^{25}\), \(5^{21}\), \(9^{16}\) is the largest?

The largest is \(9^{16}\).Compare the numbers in a common form. \(9^{16} = (3^2)^{16} = 3^{32}\), which is bigger than \(3^{31}\) by a factor of 3.Take logarithms (base 10) for the other two:\[ \begin{aligned} \log 3^{32} &= 32 \times 0.4771 \approx 15.27 \\ \log 4^{25} &= 25 \times 0.6021 \approx 15.05 \\ \log 5^{21} &= 21 \times 0.6990 \approx 14.68 \end{aligned} \]So \(9^{16} \gt 4^{25} \gt 5^{21}\), and \(3^{31}\) is a third of \(9^{16}\). The trap is to assume the largest exponent wins; here the base matters as much as the power.

Q: If \(9\sqrt{x} = \sqrt{12} + \sqrt{147}\) then x =?

The value of x is 3.Simplify each root by taking out square factors:\[ \begin{aligned} \sqrt{12} + \sqrt{147} &= 2\sqrt{3} + 7\sqrt{3} \\ &= 9\sqrt{3} \end{aligned} \]So \(9\sqrt{x} = 9\sqrt{3}\) and \(x = 3\). Adding the numbers under the roots (12 + 147 = 159) is the common error and gives none of the options.

Q: If \(\sqrt{15} = 3.8729\), evaluate \((\sqrt{5}+\sqrt{3})/(\sqrt{5}-\sqrt{3})\)

The value is 7.8729.Rationalise by multiplying the top and bottom by \(\sqrt{5}+\sqrt{3}\):\[ \begin{aligned} \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} &= \frac{(\sqrt{5}+\sqrt{3})^{2}}{5 - 3} \\ &= \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15} \\ &= 4 + 3.8729 = 7.8729 \end{aligned} \]6.8729 and 8.8729 are off by one in the whole number (3 + \(\sqrt{15}\) and 5 + \(\sqrt{15}\)), and 4.7289 scrambles the digits.

Q: Value of \(\sqrt{3.\sqrt{3.\sqrt{3.\sqrt{3....}}}}\)

The value is 3.Let the infinite expression equal \(x\). Because it repeats without end, the part under the first root sign is again \(x\):\[ \begin{aligned} x &= \sqrt{3x} \\ x^{2} &= 3x \Rightarrow x = 3 \end{aligned} \](\(x = 0\) is rejected because the expression is clearly positive.) Another way: the expression is \(3^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots} = 3^{1} = 3\). It does not grow to infinity, and \(\sqrt{3}\) is only the value of the innermost root.

Q: If \((\frac{3}{8})^{-5} \times (\frac{16}{21})^{-5} = (\frac{2}{7})^{x}\), then the value of x3 is

The value of \(x^3\) is −125.Both factors have the same power, so they can be multiplied inside one bracket:\[ \begin{aligned} \left(\frac{3}{8} \times \frac{16}{21}\right)^{-5} &= \left(\frac{2}{7}\right)^{x} \\ \frac{3 \times 16}{8 \times 21} &= \frac{48}{168} = \frac{2}{7} \\ \left(\frac{2}{7}\right)^{-5} &= \left(\frac{2}{7}\right)^{x} \end{aligned} \]So \(x = -5\) and \(x^3 = (-5)^3 = -125\). The option 125 forgets that the cube of a negative number is negative, and ±27 would need \(x = \pm 3\).

Q: If \(\left(\frac{60}{7}\right)^{x-2} = \left(\frac{7}{60}\right)^{x-6}\), then x is equal to

x is equal to 4.Turn the right-hand side into the same base by flipping the fraction, which changes the sign of the power:\[ \begin{aligned} \left(\frac{7}{60}\right)^{x-6} &= \left(\frac{60}{7}\right)^{6-x} \\ x - 2 &= 6 - x \\ 2x &= 8,\ x = 4 \end{aligned} \]Check: with \(x = 4\) the left side is \(\left(\frac{60}{7}\right)^{2}\) and the right side is \(\left(\frac{7}{60}\right)^{-2}\), which is the same number. Equating \(x - 2\) with \(x - 6\) directly, without flipping, has no solution, which is the trap.

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11.

If \(pqr \neq 0\) and \(p^{-x} = \frac{1}{q}, q^{-y} = \frac{1}{r}, r^{-z} = \frac{1}{p}\), what is the value of the product \(xyz\)?

A.−1
B.\(\frac{1}{pqr}\)
C.1
D.\(pqr\)

Exam Year(s)

GATE CS - 2018,

Answer: C

The product is \(xyz = 1\).

Taking reciprocals, \(p^{-x} = \frac{1}{q}\) gives \(p^{x} = q\); likewise \(q^{y} = r\) and \(r^{z} = p\). Substitute each into the next:

\[ \begin{aligned} r &= q^{y} = (p^{x})^{y} = p^{xy} \\ p &= r^{z} = p^{xyz} \end{aligned} \]

So \(p^{xyz} = p^{1}\), and \(xyz = 1\) for a general base \(p\). The answer \(-1\) comes from forgetting that the negative exponents were cleared by the reciprocals, and \(\frac{1}{pqr}\) and \(pqr\) mix the bases into the exponents, which the equations never do.

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12.

If (1.001)¹²⁵⁹ = 3.52 and (1.001)²⁰⁶² = 7.85, then (1.001)³³²¹ =

A.2.23
B.4.33
C.11.37
D.27.64

Exam Year(s)

GATE Chemistry - 2012,

Answer: D

(1.001)³³²¹ = 27.64.

The exponents add up: 1259 + 2062 = 3321. Powers of the same base multiply when their exponents add, so:

\[ \begin{aligned} (1.001)^{3321} &= (1.001)^{1259} \times (1.001)^{2062} \\ &= 3.52 \times 7.85 = 27.632 \end{aligned} \]

which agrees with 27.64 within the rounding of the two given values (3.52 and 7.85 are themselves rounded). 11.37 is 3.52 + 7.85, adding the values instead of multiplying them. 4.33 and 2.23 are a difference and a ratio of the two values, which would answer a subtraction of exponents, not an addition.

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13.

If \((1.001)^{1259} = 3.52\) and \((1.001)^{2062} = 7.85\), then \((1.001)^{3321} =\)

A.2.23
B.4.33
C.11.37
D.27.64

Exam Year(s)

GATE Mathematics - 2012, GATE Life Sciences - 2012, GATE XE - 2012,

Answer: D

The value is about 27.64.

The exponents add: \(1259 + 2062 = 3321\). So by the law of indices

\[ \begin{aligned} (1.001)^{3321} &= (1.001)^{1259}\times(1.001)^{2062} \\ &= 3.52 \times 7.85 = 27.632 \end{aligned} \]

which rounds to the option 27.64. 11.37 adds the two values instead of multiplying them, 4.33 is their difference, and 2.23 is their ratio; none follows from the rule \(a^{m}a^{n} = a^{m+n}\).

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14.

If a = \(b^{x}\), b = \(c^{y}\) and c = \(a^{z}\), then the value of xyz is equal to :

A.- 1
B.0
C.1
D.abc

Exam Year(s)

HPPSC HPAS - 2021,

Answer: C

The value of xyz is 1.

Substitute each relation into the one before it so that everything is expressed as a power of a.

\[ \begin{aligned} a &= b^{x} = (c^{y})^{x} = c^{xy} \\ &= (a^{z})^{xy} = a^{xyz} \end{aligned} \]

Since \(a = a^{1}\), the exponents must match, so xyz = 1 (for a base other than 0 or 1, the usual assumption in such questions).

−1 would give \(a = 1/a\), and 0 would give a = 1 for every a. abc is a product of the bases, not of the exponents, so it is not what the relations determine.

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15.

Which number amongst \(3^{31}\), \(4^{25}\), \(5^{21}\), \(9^{16}\) is the largest?

A.\(3^{31}\)
B.\(4^{25}\)
C.\(5^{21}\)
D.\(9^{16}\)

Exam Year(s)

Haryana Civil Services - 2026, HPSC - 2026,

Answer: D

The largest is \(9^{16}\).

Compare the numbers in a common form. \(9^{16} = (3^2)^{16} = 3^{32}\), which is bigger than \(3^{31}\) by a factor of 3.

Take logarithms (base 10) for the other two:

\[ \begin{aligned} \log 3^{32} &= 32 \times 0.4771 \approx 15.27 \\ \log 4^{25} &= 25 \times 0.6021 \approx 15.05 \\ \log 5^{21} &= 21 \times 0.6990 \approx 14.68 \end{aligned} \]

So \(9^{16} \gt 4^{25} \gt 5^{21}\), and \(3^{31}\) is a third of \(9^{16}\). The trap is to assume the largest exponent wins; here the base matters as much as the power.

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16.

If \(9\sqrt{x} = \sqrt{12} + \sqrt{147}\) then x =?

Exam Year(s)

ISRO Assistant - 2023,

Answer: B

The value of x is 3.

Simplify each root by taking out square factors:

\[ \begin{aligned} \sqrt{12} + \sqrt{147} &= 2\sqrt{3} + 7\sqrt{3} \\ &= 9\sqrt{3} \end{aligned} \]

So \(9\sqrt{x} = 9\sqrt{3}\) and \(x = 3\). Adding the numbers under the roots (12 + 147 = 159) is the common error and gives none of the options.

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17.

If \(\sqrt{15} = 3.8729\), evaluate \((\sqrt{5}+\sqrt{3})/(\sqrt{5}-\sqrt{3})\)

A.4.7289
B.6.8729
C.7.8729
D.8.8729

Exam Year(s)

Northern Coalfields Limited - 2020, NCL Technician - 2020,

Answer: C

The value is 7.8729.

Rationalise by multiplying the top and bottom by \(\sqrt{5}+\sqrt{3}\):

\[ \begin{aligned} \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} &= \frac{(\sqrt{5}+\sqrt{3})^{2}}{5 - 3} \\ &= \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15} \\ &= 4 + 3.8729 = 7.8729 \end{aligned} \]

6.8729 and 8.8729 are off by one in the whole number (3 + \(\sqrt{15}\) and 5 + \(\sqrt{15}\)), and 4.7289 scrambles the digits.

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18.

Value of \(\sqrt{3.\sqrt{3.\sqrt{3.\sqrt{3....}}}}\)

A.3
B.infinity
C.0
D.\(\sqrt{3}\)

Exam Year(s)

TNPSC Group 1 - 2017, TNPSC - 2017,

Answer: A

The value is 3.

Let the infinite expression equal \(x\). Because it repeats without end, the part under the first root sign is again \(x\):

\[ \begin{aligned} x &= \sqrt{3x} \\ x^{2} &= 3x \Rightarrow x = 3 \end{aligned} \]

(\(x = 0\) is rejected because the expression is clearly positive.) Another way: the expression is \(3^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots} = 3^{1} = 3\). It does not grow to infinity, and \(\sqrt{3}\) is only the value of the innermost root.

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19.

If \((\frac{3}{8})^{-5} \times (\frac{16}{21})^{-5} = (\frac{2}{7})^{x}\), then the value of x³ is

A.125
B.−27
C.27
D.−125

Exam Year(s)

Tripura Public Service Commission - 2021,

Answer: D

The value of \(x^3\) is −125.

Both factors have the same power, so they can be multiplied inside one bracket:

\[ \begin{aligned} \left(\frac{3}{8} \times \frac{16}{21}\right)^{-5} &= \left(\frac{2}{7}\right)^{x} \\ \frac{3 \times 16}{8 \times 21} &= \frac{48}{168} = \frac{2}{7} \\ \left(\frac{2}{7}\right)^{-5} &= \left(\frac{2}{7}\right)^{x} \end{aligned} \]

So \(x = -5\) and \(x^3 = (-5)^3 = -125\). The option 125 forgets that the cube of a negative number is negative, and ±27 would need \(x = \pm 3\).

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20.

If \(\left(\frac{60}{7}\right)^{x-2} = \left(\frac{7}{60}\right)^{x-6}\), then x is equal to

Exam Year(s)

Tripura Public Service Commission - 2021,

Answer: B

x is equal to 4.

Turn the right-hand side into the same base by flipping the fraction, which changes the sign of the power:

\[ \begin{aligned} \left(\frac{7}{60}\right)^{x-6} &= \left(\frac{60}{7}\right)^{6-x} \\ x - 2 &= 6 - x \\ 2x &= 8,\ x = 4 \end{aligned} \]

Check: with \(x = 4\) the left side is \(\left(\frac{60}{7}\right)^{2}\) and the right side is \(\left(\frac{7}{60}\right)^{-2}\), which is the same number. Equating \(x - 2\) with \(x - 6\) directly, without flipping, has no solution, which is the trap.

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