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.
The number of integral solutions of the equation = 1 is
Answer: 4
Explanation :
Given, (x2 -
For this to be equal to 1
Case 1: x2 - 10 = 1
⇒ x =
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.
Workspace:
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:
equals
Answer: 24
Explanation :
Given,
=
=
=
= 23 × 3 = 24
Hence, 24.
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 2.
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 :
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 1.
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 2.
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
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 2.
Workspace:
If - 22x - 2 = 4x - 32x - 3, then x is
- A.
- B.
- C.
- D.
Answer: Option A
Explanation :
- 22x - 2 = 4x - 32x - 3
= = 32x - 1
Also 4x = (22)x = 22x
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 :
21/2 = 26/12 = (26)1/12 = 641/12
Similarly, 31/3 = 811/12 , 41/4 = 641/12, 61/6 = 361/12
Now, all the powers are equal. Thus the option with the largest base is the largest.
31/3 is the largest.
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 3 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 3.
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:
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