Algebra - Surds & Indices - Previous Year CAT/MBA Questions

Answer: 19

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Text Explanation :

Number of micro-organisms on day 1 = 2

Number of micro-organisms on day 2 = 2 × 2 + 3
                                                                            = 2² + 3
Number of micro-organisms on day 3 = 2 × (2² + 3) + 3
                                                                            = 2³ + 3 × (2 + 1)
Number of micro-organisms on day 4 = 2 × (2³ + 3 × (2 + 1)) + 3
                                                                            = 2⁴ + 3 × (2² + 2 + 1)
Number of micro-organisms on day 5 = 2 × (2⁴ + 3 × (2² + 2 + 1)) + 3
                                                                            = 2⁵ + 3 × (2³ + 2² + 2 + 1)

∴ Number of micro-organisms at the end of day n = 2ⁿ + 3 × (2^n-2 + … + 2² + 2 + 1)
                                                                            = 2 × 2^n-1 + 3 × (2^n-1 - 1)
                                                                            = 5 × 2^n-1 - 3

Now, 5 × 2^n-1 - 3 ≥ 10,00,000
⇒ 2^n-1 ≥ 2,00,000 + 3/5

The least value of (n - 1) satisfying above inequality is 18.
⇒ n - 1 = 18
⇒ n = 19

Hence, 19.

Answer: Option A

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Text Explanation :

Given, $\sqrt{5x+9}$ + $\sqrt{5x-9}$ = 6 + 3√2

⇒ $\sqrt{5x+9}$ + $\sqrt{5x-9}$ = √36 + √18

∴ $\sqrt{5x+9}$= √36 and $\sqrt{5x-9}$= √18

⇒ 5x + 9 = 36 and 5x - 9 = 18

⇒ 5x = 27

⇒ 10x = 54

⇒ 10x + 9 = 63

⇒ $\sqrt{10x+9}$ = √63

⇒ $\sqrt{10x+9}$ = 3√7

Hence, option (a).

Answer: Option B

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Text Explanation :

Given, $\frac{2}{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}}}$ = $\frac{1}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{z}}}$ + $\frac{1}{\sqrt{\mathrm{x}}+\sqrt{y}}$

⇒ $\frac{2}{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}}}$ = $\frac{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}}+\sqrt{\mathrm{z}}}{(\sqrt{\mathrm{x}}+\sqrt{\mathrm{z}})(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})}$

⇒ 2(√x + √z)(√x + √y) = (2√x + √y + √z)(√y + √z)

⇒ 2x + 2√xy + 2√zx + 2√zy = 2√xy + 2√xz + y + √yz + zy + z

⇒ 2x = y + z

∴ y, x and z are in Arithmetic Progression

Hence, option (b).

Concept:

Arithmetic Progression

Answer: Option D

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Text Explanation :

Given, ${2^{4x}}^2$ - $2^{2x^2+x+16}$ + $2^{2x+30}$ = 0

⇒ ${\left({2^{2x}}^2\right)}^2$ - $2\times 2^{2x^2}\times 2^{x+15}$ + ${\left(2^{x+15}\right)}^2$ = 0

⇒ ${\left({2^{2x}}^2-2^{x+15}\right)}^2$ = 0

⇒ ${2^{2x}}^2$ = $2^{x+15}$

⇒ 2x² = x + 15

⇒ 2x² - x - 15 = 0

⇒ (2x + 5)(x - 3) = 0

⇒ x = -5/2 or 3.

Sum of all possible values of x = 3 + (-5/2) = 1/2

Hence, option (d).

Concept:

Indices

Algebraic Identities

Answer: Option A

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Text Explanation :

a^mbⁿ = 144¹⁴⁵ 

⇒ a^mbⁿ = (12²)¹⁴⁵ 

⇒ a^mbⁿ = (2⁴ × 3²)¹⁴⁵ 

⇒ a^mbⁿ = 2⁵⁸⁰ × 3²⁹⁰  

To maximise n - m, we need to maximise n and minimise m.
∴ Let 3²⁹⁰ = X

⇒ a^mbⁿ = X × 2⁵⁸⁰  

∴ a = X, m = 1, b = 2 and n = 580

⇒ n - m = 580 - 1 = 579

Hence, option (a). 

Concept:

Indices

Answer: 5

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Text Explanation :

Given, 5^n-1 < 3ⁿ⁺¹ 

Putting values of n = 1, 2 and so on we see that the above inequality is true for n = 1, 2, 3, 4 and 5 only.

Now, 3ⁿ⁺¹ < 2^n+m is true of all values of n.

Taking n = 5, we get 
3⁶ < 2^5+m 
⇒ 729 < 2^5+m 

The least power of 2 greater than 729 is 1024 (2¹⁰)

∴ 2¹⁰  =  2^5+m 
⇒ 5 + m = 10
⇒ m = 5
∴ Least value of m = 5.

If we check for other values of n, we may get smaller value of m, but those values will not suffice when n = 5.

Hence, 5.

Answer: 14

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Text Explanation :

Givne, ${\left(\sqrt{\frac{7}{5}}\right)}^{3x-y}$ = $\frac{875}{2401}$ 

⇒ ${\left(\frac{7}{5}\right)}^{\frac{3x-y}{2}}$ = $\frac{125\times 7}{7\times 7\times 7\times 7}$ = ${\left(\frac{5}{7}\right)}^3$ = ${\left(\frac{7}{5}\right)}^{-3}$

⇒ 3x - y = - 6   

⇒ y = 3x + 6   ...(1)

Also, ${\left(\frac{4a}{b}\right)}^{6x-y}$ = ${\left(\frac{2a}{b}\right)}^{y-6x}$ = ${\left(\frac{b}{2a}\right)}^{6x-y}$

⇒ ${\left(\frac{4\mathrm{a}}{\mathrm{b}}\times \frac{2\mathrm{a}}{\mathrm{b}}\right)}^{6\mathrm{x}-\mathrm{y}}$ = 1

⇒ ${\left(\frac{8{\mathrm{a}}^2}{{\mathrm{b}}^2}\right)}^{6\mathrm{x}-\mathrm{y}}$ = 1

This is true for all values of a and b. This is only possible when 6x - y = 0
⇒ y = 6x   ...(2)

Solving (1) and (2), we get x = 2 and y = 12.

⇒ x + y = 2 + 12 = 14.

Hence, 14.

Answer: 7

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Text Explanation :

Given, 2.25 ≤ 2 + 2ⁿ⁺² ≤ 202

⇒ 0.25 ≤ 2ⁿ⁺² ≤ 200

Case 1: 0.25 ≤ 2ⁿ⁺²
∴ n + 2 ≥ -2
⇒ n ≥ -4

Case 2: 2ⁿ⁺² ≤ 200
This is true when n ≤ 5

∴ -4 ≤ n ≤ 5

Now, 3 + 3ⁿ⁺¹ should be an integer

This is possible when n = -1, 0, 1, 2, 3, 4 and 5

∴ n can take 7 values.

Hence, 7.

Answer: 6

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Text Explanation :

$\left(\sqrt[7]{10}\right)$${\left(\sqrt[7]{10}\right)}^2$...${\left(\sqrt[7]{10}\right)}^n$ > 999

⇒ 101/7 × 102/7 ×… × 10n/7 > 999

⇒ ${10}^{\frac{1+2+3+...+n}{7}}$ > 999

⇒ ${10}^{\frac{n(n+1)}{14}}$ > 999

This is possible when n(n + 1)/14 ≥ 3.

∴ Least possible positive integral value of n is 6.

Hence, 6.

Answer: Option A

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Text Explanation :

2^x + 2^-x is of the form y + 1/y
Minimum value the expression can take is 2.
Hence, 2^x + 2^-x ≥ 2

Now, 2 - (x - 2)²
We know, (x - 2)² ≥ 0
∴ 2 - (x - 2)² ≤ 2 (From 2 we are subtracting a non-negative number)
Maximum value this expression can have is 2.

The only possibility is both sides are = 2
LHS = 2^x + 2^-x = 0
This is possible only when x = 0.

When x = 0, RHS = 2 - (x - 2)² = 2 - 2² = - 2
Hence, at x = 0, LHS ≠ RHS.
∴ There is no solution possible.
Hence, option (a).

Answer: Option C

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Text Explanation :

Given,

x = (4096)^7+4√3

⇒ x = (64²)^7+4√3

⇒ x = (64)^14+8√3

⇒ $x^{\frac{1}{14+8\sqrt{3}}}$ = 64

Now, let’s rationalize $\frac{1}{14+8\sqrt{3}}$,

⇒ $\frac{1}{14+8\sqrt{3}}$ = $\frac{1}{14+8\sqrt{3}}\times \frac{14-8\sqrt{3}}{14-8\sqrt{3}}$ = $\frac{14-8\sqrt{3}}{196-192}$ = $\frac{7-4\sqrt{3}}{2}$

⇒ $x^{\frac{1}{14+8\sqrt{3}}}$ = $x^{\frac{7-4\sqrt{3}}{2}}$= 64

⇒ $x^{\frac{7}{2}}\times x^{\frac{-4\sqrt{3}}{2}}$ = 64

⇒ 64 = $\frac{x^{\frac{7}{2}}}{x^{2\sqrt{3}}}$

Hence, option (c).

Answer: 3

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Text Explanation :

Assuming, k = 14^a = 36^b = 84^c

⇒ 14 = k^1/a

⇒ 36 = 6² = k^1/b or 6 = k^1/2b 

⇒ 84 = k^1/c

We know, 84 = 6 × 14

⇒ k^1/c = k^1/2b × k^1/a

⇒ k^{1/c – 1/a} = k^1/2b

⇒ $\frac{1}{c}-\frac{1}{a}=\frac{1}{2b}$

⇒ $2\mathrm{b}\left(\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$ = 1

Hence, $6\mathrm{b}\left(\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$ = 3

Hence, 3.

Answer: Option D

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Text Explanation :

Given, a × b = 4²⁰¹⁷ = 2⁴⁰³⁴

Here, a and b and integers and will be some power of 2 since RHS is 2⁴⁰³⁴.

Let’s say a = 2^x and b = 2^y. [x and y are non-negative integers]

Also, a ≤ b, Hence, x ≤ y.

Now, 2^x × 2^y = 2⁴⁰³⁴

Since x ≤ y, x ≤ 4034/2 = 2017

The values x can take is 0, 1, 2, …. till 2017.

Hence, x can take a total of 2018 values.

Hence, option (d).

Answer: Option B

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Text Explanation :

(√2)¹⁹ 3⁴ 4² 9^m 8ⁿ = 3ⁿ 16^m (64)^1/4

∴ 2^19/2 3⁴ 2⁴ 3^2m 2³ⁿ = 3ⁿ 2^4m 2^6/4

∴ 2^19/2 3⁴ 2⁴ 3^2m 2³ⁿ = 3ⁿ 2^4m 2^3/2

∴ 2^{[(19/2) + 4 + 3n]} 3^{(4 + 2m)} = 2^{[4m + (3/2)]} 3ⁿ

Equating the exponents of 2 and 3 on both sides, we get;

Powers of 2 : (19/2) + 4 + 3n = 4m + (3/2) 

⇒ 6n + 24 = 8m ...(1)

Powers of 3 : 4 + 2m = n  ...(2)

Solving (1) and (2), we get;

m = −12.

Hence, option (b).

Answer: Option A

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Text Explanation :

Given, 5.55^x = 0.555^y = 1000 = k say.

⇒ 5.55^x = k ⇒ 5.55 = k^1/x   ...(1)
⇒ 0.555^y = k ⇒ 0.555 = k^1/y   ...(2)
⇒ 1000^z = 10³ = k ⇒ 10 = k^1/3   ...(3)

⇒ 5.55 × 10 = 0.555
∴ k^1/x × k^1/3 = k^1/y
⇒ k^{1/x + 1/3} = k^1/y
⇒ 1/x + 1/3 = 1/y
⇒ 1/x - 1/y = 1/3

Alternately,
5.55^x = 0.555^y = 1000  

Taking log to the base 10, we get; 

log(5.55^x) = log(0.555^y) = log(1000) = log 10³ = 3 

∴ x log 5.55 = y log 0.555 = 3 

So, (1/x) = (log 5.55)/3 

(1/y) = (log 0.555)/3  

Log 0.555 = log 5.55 × 10⁻¹ = log 5.55 + log 10⁻¹ =  (log 5.55) − 1. 

So, (1/y) = [(log 5.55) − 1]/3 

∴  (1/x) − (1/y) = [(log 5.55)/3] − [{(log 5.55) − 1}/3] = 1/3. 

Hence, option (a).

Answer: 13

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Text Explanation :

It is given that 5^x - 3^y = 13438 and 5^x-1 + 3^y+1 = 9686

Let 5^x-1 = k and 3^y = m

So, 5k - m = 13538        …(1)

k + 3m = 9686        …(2)

Multiply (1) with 3,

15k - 3m = 40314 ----(3)

Adding (2) and (3) we get,

16k = 50000

k = 625 × 5

k = 5⁵

We know, 5^x-1 = k

So, x - 1 = 5

x = 6

We know, k + 3m = 9686

3125 + 3m = 9686

3m = 6561

m = 2187

m = 3⁷ = 3^y

y = 7

x + y = 6 + 7 = 13

Hence, 13.

Answer: Option B

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Text Explanation :

x²⁰¹⁸y²⁰¹⁷ = 1/2 ...(1)

x²⁰¹⁶y²⁰¹⁹ = 8    ...(2)

Dividing the 2 given equations, we get

$\frac{x^{2018}{y}^{2017}}{x^{2016}{y}^{2019}}$ = $\frac{{\displaystyle \frac{1}{2}}}{4}=\frac{1}{16}$

x² = y²/16

From Eq (1)

x^2(1009)y²⁰¹⁷ = 1/2

∴ ${\left(\frac{y^2}{16}\right)}^{1009}{y}^{2017}$ = 1/2

∴ y⁴⁰³⁵ = 1/2 × 16¹⁰⁰⁹

∴ y⁴⁰³⁵ = 1/2 × 2^{4 × 1009} = 2⁴⁰³⁵

∴ y = 2

Again, from equation 1

x²⁰¹⁸y²⁰¹⁷ = 1/2

∴ x²⁰¹⁸2²⁰¹⁷ = 1/2

∴ x²⁰¹⁸ = 1/2²⁰¹⁸

∴ x = ± 1/2

Now, x² + y³ = 1/4 + 8 = 33/4

Hence, option (b).

Answer: 10

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Text Explanation :

From observation, N^N = 2¹⁶⁰ = 32³² or N = 32.

Now, N² + 2^N = 32² + 2³² = (2⁵ )² + 2³²

= 2¹⁰ + 2³² = 2¹⁰ (1 + 2²² ).

Here, (1 + 2²² ) is an odd number, hence the higest power of 2 is 10.

Therefore the largest value of x = 10.

Hence, 10.

Answer: Option D

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Text Explanation :

${\left(\frac{a+3}{b}\right)}^2=9$ and ${\left(\frac{a-1}{b-1}\right)}^2=4$

⇒ $\frac{\mathrm{a}+3}{\mathrm{b}}=\pm 3$ and $\frac{\mathrm{a}-1}{\mathrm{b}-1}=\pm 2$

We get following cases:

Case 1(a): a + 3 = 3b and a - 1 = 2b - 2
⇒ a = 3 and b = 2 (Rejected)

Case 1(b): a + 3 = 3b and a - 1 = -2b + 2
⇒ b is not an integer (Rejected)

Case 2(a): a + 3 = -3b and a - 1 = 2b - 2
⇒ b is not an integer (Rejected)

Case 2(b): a + 3 = -3b and a - 1 = -2b + 2
⇒ a = 15 and b = -6
∴ (a/b)² = (15/-6)² = 25/4

Hence, option (d).

Answer: Option B

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Text Explanation :

9^2x–1 – 81^x-1 = 1944

⇒ 9^2x–2 × 9 – 9^2x-2 = 1944

⇒ 9^2x–2 (9 – 1) = 8 × 243 = 8 × 3⁵

⇒ 3^4x–4 × 8 = 8 × 3⁵

⇒ ∴ 4x – 4 = 5

∴ x = 9/4.

Hence, option (b).

Answer: Option A

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Text Explanation :

$9^{\mathrm{x}-\frac{1}{2}}$ - 2^{2x - 2} = 4^x - 3^{2x - 3}   ...(1)

Now, $9^{\mathrm{x}-\frac{1}{2}}$ = ${\left(3^2\right)}^{x-\frac{1}{2}}$ = 3^{2x - 1}

Also 4^x = (2²)^x = 2^2x

From (1), we get

3^{2x - 1} - 2^{2x - 2} = 2^2x - 3^{2x - 3}

3^{2x - 1} + 3^{2x - 3} = 2^2x + 2^{2x - 2}

3^{2x - 3} (3² + 1) = 2^{2x - 2} (2² + 1)

$\frac{3^{2\mathrm{x}-3}}{2^{2\mathrm{x}-2}}=\frac{5}{10} \mathrm{or} \frac{1}{2}$

Now this can be written as

$\frac{3^{2x-3}}{2^{2x-2}}=\frac{3^0}{2^1}$

Now since the bases of the numerator and the dominator are equal on LHS and RHS, the powers should also be equal 

⇒ 3^2x-3 = 3⁰ and 2^{2x - 2} = 2¹

In both cases x = 3/2

∴ x = $\frac{3}{2}$

Hence, option (a).

Answer: Option B

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Text Explanation :

Out of the options, only $\frac{1}{x}$ is negative.

All the others are positive.

$\therefore \frac{1}{x}$ is the smallest.

Hence, option (b).

Answer: Option B

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Text Explanation :

Givne, 2^1/2, 3^1/3, 4^1/4 , 6^1/6  and 12^1/12

Raising all numbers to power 12, we get

(2^1/2)¹², (3^1/3)¹², (4^1/4)¹² , (6^1/6)¹²  and (12^1/12)¹²

(2⁶), (3⁴), (4³) , (6²)  and (12¹)

64, 81, 64, 36, 12

Since 81 is the higest number here, 3^1/3 will be the highest of the original numbers.

Hence, option (b).

Answer: Option E

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Text Explanation :

The fastest way to solve this sum is by substituting the values of x and y from the options.

For x = 5 and y = 2, the first equation becomes

2^0.7x × 3^−1.25y  = 2^{0.7× 5} × 3^{−1.25 × 2}

              = 2^7/2 × 3^−5/2

             = $\frac{8\sqrt{6}}{27}$

These values of x and y satisfy the second equation also.

Hence, option (e).

Answer: Option A

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Text Explanation :

$x^{\frac{2}{3}}+x^{\frac{1}{3}}-2\le 0$   ...(1)

Put y = x^1/3

Then equation (1) becomes y² + y – 2 ≤ 0

(y + 2)(y – 1) ≤ 0

–2 ≤ y ≤ 1

$-2\le x^{\frac{1}{3}}\le 1$

–8 ≤ x ≤ 1

Hence, option (a).