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

Answer: 4

Explanation :

Given, (x² - $10{)}^{x^2-3x-10}$

For this to be equal to 1

Case 1: x² - 10 = 1

⇒ x = $\sqrt{11}$
Rejected as x is not an integer.

Case 2: x² – 3x – 10 = 0
⇒ (x – 5)(x + 2) = 0
⇒ x = 5 or -2 i.e., 2 integral values of x.

Case 3: x² – 10 = -1 and x² – 3x – 10 = even
If x² – 10 = -1 ⇒ x = ±3
For both x = +3 and -3 x² – 3x – 10 is even, hence, 2 integral values of x.

⇒ Total 4 integral values of x are possible.

Hence, 4.

Answer: 14

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

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

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

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

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

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: 24

Explanation :

Given, $\frac{2\times 4\times 8\times 16}{\left({{\log }_{2}4)}^2\right({{\log }_{4}8)}^3({{\log }_{8}16)}^4}$

= $\frac{2\times 2^2\times 2^3\times 2^4}{\left({{\log }_2{2}^2)}^2\right({{\log }_{2^2}{2}^3)}^3({{\log }_{2^3}{2}^4)}^4}$

= $\frac{2^{10}}{{\left(2\right)}^2{\left(\frac{3}{2}\right)}^3{\left(\frac{4}{3}\right)}^4}$

= $\frac{2^{10}\times 2^3\times 3^4}{2^2\times 3^3\times 2^8}$

= 2³ × 3 = 24 

Hence, 24.

Answer: Option D

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

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 2.

Answer: Option A

Explanation :

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 1.

Answer: 13

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

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 2.

Answer: 10

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

Explanation :

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

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

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 2.

Answer: Option A

Explanation :

$9^{x-\frac{1}{2}}$ - 2^{2x - 2} = 4^x - 3^{2x - 3}

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

Also 4^x = (2²)^x = 2^2x

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

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

Explanation :

2^1/2 = 2^6/12 = (26)^1/12 = 64^1/12

Similarly, 3^1/3 = 81^1/12 , 4^1/4 = 64^1/12, 6^1/6 = 36^1/12

Now, all the powers are equal. Thus the option with the largest base is the largest.

3^1/3 is the largest.

Hence, option (b).

Answer: Option E

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

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).

Answer: Option C

Explanation :

x = $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-....to infinity}}}}$

∴ x² - 4 = $\sqrt{4-\sqrt{4+\sqrt{4-.... to infinity}}}$

∴ (x² - 4)² - 4 = - $\sqrt{4+\sqrt{4-...to infinity}}$

∴ (x² - 4)² - 4 = -x

Now, substituting options, we find that only option 3 satisfies the above equation.

Hence, option (c).

Answer: Option D

Explanation :

y = $\frac{1}{2+{\displaystyle \frac{1}{3+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{3+....}}}}}}}$

y = $\frac{1}{2+{\displaystyle \frac{1}{3+y}}}$

y = $\frac{3+y}{2y+7}$

∴ 2y² + 7y = 3 + y

∴ 2y² + 6y – 3 = 0

∴ y = $\frac{-6\pm \sqrt{36-4(-3)\left(2\right)}}{4}$ = $\frac{\pm \sqrt{15}-3}{2}$

As all the terms in y are positive, y is positive.

∴ y = $\frac{\sqrt{15}-3}{2}$

Hence, option (d).

Answer: Option C

Explanation :

∵ pqr = 1

Case 1:

Let p = 2/3, q = 3/2 and r = 1

Substituting the values of p, q, r, we get,

$\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=\frac{3}{7}+\frac{2}{7}+\frac{2}{7}=1$

Case 2:

Let p = 1, q = 1 and r = 1

Substituting the values of p, q, r, we get,

$\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$

Hence, option 3.

Answer: Option B

Explanation :

$7^{3^2}$ = 7⁹ and (7³)² = 7⁶. SInce 7⁹ > 7⁶ ⇒ $7^{3^2}$ > (7³)²