Algebra - Surds & Indices - Previous Year CAT/MBA Questions
You can practice all previous year CAT questions from the topic Algebra - Surds & Indices. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.
If = and = for all non-zero real values of a and b, then the value of x + y is
Answer: 14
Explanation :
Givne, =
⇒ = = =
⇒ 3x - y = - 6
⇒ y = 3x + 6 ...(1)
Also, = =
⇒ = 1
⇒ = 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.
Workspace:
For all possible integers n satisfying 2.25 ≤ 2 + 2n+2 ≤ 202, the number of integer values of 3 + 3n+1 is
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.
Workspace:
In ‘n’ is a positive integer such that ... > 999, then the smallest value of n is
Answer: 6
Explanation :
... > 999
⇒ 101/7 × 102/7 ×… × 10n/7 > 999
⇒ > 999
⇒ > 999
This is possible when n(n + 1)/14 ≥ 3.
∴ Least possible positive integral value of n is 6.
Hence, 6.
Workspace:
The number of real-valued solutions of the equation 2x + 2-x = 2 – (x – 2)² is
- A.
0
- B.
Infinite
- C.
2
- D.
1
Answer: Option A
Explanation :
2x + 2-x is of the form y + 1/y
Minimum value the expression can take is 2.
Hence, 2x + 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 = 2x + 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).
Workspace:
If x = (4096)7+4√3, then which of the following equals 64?
- A.
- B.
- C.
- D.
Answer: Option C
Explanation :
Given,
x = (4096)7+4√3
⇒ x = (642)7+4√3
⇒ x = (64)14+8√3
⇒ = 64
Now, let’s rationalize ,
⇒ = = =
⇒ = = 64
⇒ = 64
⇒ 64 =
Hence, option (c).
Workspace:
If a, b, c are non-zero and 14a = 36b = 84c, then is equal to
Answer: 3
Explanation :
Assuming, k = 14a = 36b = 84c
⇒ 14 = k1/a
⇒ 36 = 62 = k1/b or 6 = k1/2b
⇒ 84 = k1/c
We know, 84 = 6 × 14
⇒ k1/c = k1/2b × k1/a
⇒ k1/c – 1/a = k1/2b
⇒
⇒ = 1
Hence, = 3
Hence, 3.
Workspace:
How many pairs (a, b) of positive integers are there such that a ≤ b and ab = 42017?
- A.
2019
- B.
2017
- C.
2020
- D.
2018
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 = 2x and b = 2y. [x and y are non-negative integers]
Also, a ≤ b, Hence, x ≤ y.
Now, 2x × 2y = 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).
Workspace:
If m and n are integers such that (√2)19 34 42 9m 8n = 3n 16m (64)1/4, then m is
- A.
-16
- B.
-12
- C.
-24
- D.
-20
Answer: Option B
Explanation :
(√2)19 34 42 9m 8n = 3n 16m (64)1/4
∴ 219/2 34 24 32m 23n = 3n 24m 26/4
∴ 219/2 34 24 32m 23n = 3n 24m 23/2
∴ 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).
Workspace:
If 5.55x = 0.555y = 1000, then the value of (1/x) − (1/y) is
- A.
1/3
- B.
2/3
- C.
3
- D.
1
Answer: Option A
Explanation :
Given, 5.55x = 0.555y = 1000 = k say.
⇒ 5.55x = k ⇒ 5.55 = k1/x ...(1)
⇒ 0.555y = k ⇒ 0.555 = k1/y ...(2)
⇒ 1000z = 103 = k ⇒ 10 = k1/3 ...(3)
⇒ 5.55 × 10 = 0.555
∴ k1/x × k1/3 = k1/y
⇒ k1/x + 1/3 = k1/y
⇒ 1/x + 1/3 = 1/y
⇒ 1/x - 1/y = 1/3
Alternately,
5.55x = 0.555y = 1000
Taking log to the base 10, we get;
log(5.55x) = log(0.555y) = 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).
Workspace:
If 5x - 3y = 13438 and 5x-1 + 3y+1 = 9686, then x + y equals
Answer: 13
Explanation :
It is given that 5x - 3y = 13438 and 5x-1 + 3y+1 = 9686
Let 5x-1 = k and 3y = 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, 5x-1 = k
So, x - 1 = 5
x = 6
We know, k + 3m = 9686
3125 + 3m = 9686
3m = 6561
m = 2187
m = 37 = 3y
y = 7
x + y = 6 + 7 = 13
Hence, 13.
Workspace:
Given that x2018y2017 = 1/2 and x2016y2019 = 8, the value of x2 + y3 is
- A.
37/4
- B.
33/4
- C.
35/4
- D.
31/4
Answer: Option B
Explanation :
x2018y2017 = 1/2 ...(1)
x2016y2019 = 8 ...(2)
Dividing the 2 given equations, we get
=
x2 = y2/16
From Eq (1)
x2(1009)y2017 = 1/2
∴ = 1/2
∴ y4035 = 1/2 × 161009
∴ y4035 = 1/2 × 24 × 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).
Workspace:
If N and x are positive integers such that NN = 2160 and N2 + 2N is an integral multiple of 2x, then the largest possible x is
Answer: 10
Explanation :
From observation, NN = 2160 = 3232 or N = 32.
Now, N2 + 2N = 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.
Workspace:
If a and b are integers of opposite signs such that (a + 3)2 : b2 = 9 : 1 and (a - 1)2: (b - 1)2 = 4 : 1, then the ratio a2 : b2 is:
- A.
9 : 4
- B.
81 : 4
- C.
1 : 4
- D.
25 : 4
Answer: Option D
Explanation :
and
⇒ and
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).
Workspace:
If 92x–1 – 81x-1 = 1944, then x is
- A.
3
- B.
9/4
- C.
4/9
- D.
1/3
Answer: Option B
Explanation :
92x–1 – 81x-1 = 1944
⇒ 92x–2 × 9 – 92x-2 = 1944
⇒ 92x–2 (9 – 1) = 8 × 243 = 8 × 35
⇒ 34x–4 × 8 = 8 × 35
⇒ ∴ 4x – 4 = 5
∴ x = 9/4.
Hence, option (b).
Workspace:
If - 22x - 2 = 4x - 32x - 3, then x is
- A.
- B.
- C.
- D.
Answer: Option A
Explanation :
- 22x - 2 = 4x - 32x - 3 ...(1)
Now, = = 32x - 1
Also 4x = (22)x = 22x
From (1), we get
32x - 1 - 22x - 2 = 22x - 32x - 3
32x - 1 + 32x - 3 = 22x + 22x - 2
32x - 3 (32 + 1) = 22x - 2 (22 + 1)
Now this can be written as
Now since the bases of the numerator and the dominator are equal on LHS and RHS, the powers should also be equal
⇒ 32x-3 = 30 and 22x - 2 = 21
In both cases x = 3/2
∴ x =
Hence, option (a).
Workspace:
If x = −0.5, then which of the following has the smallest value?
- A.
- B.
- C.
- D.
- E.
Answer: Option B
Explanation :
Out of the options, only is negative.
All the others are positive.
is the smallest.
Hence, option (b).
Workspace:
Which among 21/2, 31/3, 41/4 , 61/6 and 121/12 is the largest ?
- A.
21/2
- B.
31/3
- C.
41/4
- D.
61/6
- E.
121/12
Answer: Option B
Explanation :
Givne, 21/2, 31/3, 41/4 , 61/6 and 121/12
Raising all numbers to power 12, we get
(21/2)12, (31/3)12, (41/4)12 , (61/6)12 and (121/12)12
(26), (34), (43) , (62) and (121)
64, 81, 64, 36, 12
Since 81 is the higest number here, 31/3 will be the highest of the original numbers.
Hence, option (b).
Workspace:
What are the values of x and y that satisfy both the equations?
20.7x . 3−1. 25y =
40.3x . 90.2y = 8 . (81)1/5
- A.
x = 2, y = 5
- B.
x = 2.5, y = 6
- C.
x = 3, y = 5
- D.
x = 3, y = 4
- E.
x = 5, y = 2
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
20.7x × 3−1.25y = 20.7× 5 × 3−1.25 × 2
= 27/2 × 3−5/2
=
These values of x and y satisfy the second equation also.
Hence, option (e).
Workspace:
What values of x satisfy ?
- A.
–8 ≤ x ≤ 1
- B.
–1 ≤ x ≤ 8
- C.
1 < x < 8
- D.
1 ≤ x ≤ 8
- E.
–8 ≤ x ≤ 8
Answer: Option A
Explanation :
...(1)
Put y = x1/3
Then equation (1) becomes y2 + y – 2 ≤ 0
(y + 2)(y – 1) ≤ 0
–2 ≤ y ≤ 1
–8 ≤ x ≤ 1
Hence, option (a).
Workspace:
Let x = Then x equals
- A.
3
- B.
- C.
- D.
Answer: Option C
Explanation :
x =
∴ x2 - 4 =
∴ (x2 - 4)2 - 4 = -
∴ (x2 - 4)2 - 4 = - x
Now, substituting options, we find that only option (c) satisfies the above equation.
Hence, option (c).
Workspace:
Let
What is the value of y?
- A.
- B.
- C.
- D.
Answer: Option D
Explanation :
y =
y =
y =
∴ 2y2 + 7y = 3 + y
∴ 2y2 + 6y – 3 = 0
∴ y = =
As all the terms in y are positive, y is positive.
∴ y =
Hence, option (d).
Workspace:
If pqr = 1 then is equivalent to
- A.
p + q + r
- B.
- C.
1
- D.
p-1 + q-1 + r-1
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,
Case 2:
Let p = 1, q = 1 and r = 1
Substituting the values of p, q, r, we get,
Hence, option (c).
Workspace:
Which of the following is true?
- A.
= (73)2
- B.
> (73)2
- C.
< (73)2
- D.
None of these
Answer: Option B
Explanation :
= 79 and (73)2 = 76. SInce 79 > 76 ⇒ > (73)2
Workspace:
A, B and C are defined as follows.
A = 2.000004 ÷ [(2.000004)2 + (4.000008)]
B = 3.000003 ÷ [(3.000003)2 + (9.000009)]
C = 4.000002 ÷ [(4.000002)2 + (8.000004)]
Which of the following is true about the values of the above three expressions?
- A.
All of them lie between 0.18 and 0.2
- B.
A is twice of C
- C.
C is the smallest
- D.
B is the smallest
Answer: Option D
Explanation :
=
= (Approximately)
Looking at the answer choices, we can see that the only (d) satisfies the relationship, viz. B is the smallest.
Workspace:
273 – 272 – 271 is the same as
- A.
269
- B.
270
- C.
271
- D.
272
Answer: Option C
Explanation :
273 – 272 – 271 = 271(22 – 21 - 1) = 271(4 - 2 - 1) = 271(1) = 271
Hence, option (c).
Workspace:
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