Test: Algebra - 1


30 Questions MCQ Test Topic-wise Past Year Questions for CAT | Test: Algebra - 1


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QUESTION: 1

If a1 + a2 + a3 + ... + an = 3(2n+1 – 2), for every n ≥ 1, then a11 equals

(2019)

Solution:

If n = 1, then a1 = 3(21 + 1 – 2) = 6 = 3 × 21
If n =  2, then a1 + a2 = 3(22 + 1 – 2) = 18 ⇒ a2
= 18 – a1 = 18 – 6 = 12 = 3 × 22
If n = 3, then a1 + a2 + a3 = 3(23 + 1 – 2) = 42 ⇒ a3
= 42 – (a1 + a2) = 24 = 3 × 23
Similarly, an = 3 × 2n
Therefore, a11 = 3 × 211 = 6144

QUESTION: 2

The number of solutions to the equation |x|(6x2 + 1) = 5x2 is

(2019)

Solution:

Case I: x = 0.
Clearly, x = 0 satisfy the equation
Case II: x > 0 |x| (6x2 + 1) = 5x2 ⇒ x(6x2 + 1) = 5x2
⇒ x(6x2 + 1 – 5x) = 0
∴ x ≠ 0
∴ 6x2 + 1 – 5x = 0
On solving the quadratic equation, we get

Case III: x < 0
|x| (6x2 + 1) = 5x2 ⇒ –x(6x2 + 1) = 5x2
⇒ –x(6x2 + 1 + 5x) = 0
⇒ x(6x2 + 5x + 1) = 0
∵ x ≠ 0, ∴ 6x2 + 5x + 1 = 0
On solving the quadratic equation, we get

Hence there are 5 solutions.

QUESTION: 3

The product of the distinct roots of |x2 – x – 6| = x + 2 is

(2019)

Solution:

x2 – x – 6 = (x + 2)(x – 3)
Case 1: x2 – x – 6 < 0 ⇒ (x + 2)(x – 3) < 0
⇒ –2 < x < 3 and |x2 – x – 6| = –(x2 – x – 6)
∴ |x2 – x – 6| = x + 2 ⇒ –(x + 2)(x – 3) = x + 2
⇒ (x – 3) = –1 ⇒ x = 2
Case 2: x2 – x – 6 ≥ 0 ⇒ (x + 2) (x – 3) ≥ 0
⇒ x ≤ –2 or x ≥ 3
Checking for boundary conditions:
For x = –2, |x2 – x – 6| = x + 2, therefore,x = -2 is also the root.
But for x = 3, |x2 – x – 6| ≠ x + 2
Hence x = 3 is NOT the root.
And for the interval x < –2 or x > 3 the expression |x2 – x – 6|
= x2 – x – 6
∴ |x2 – x – 6| = x + 2
⇒ (x + 2)(x – 3) = x + 2
⇒ (x – 3) = 1 ⇒ x = 4
Hence, the root are –2, 2, and 4.
∴ Required product = (2)(–2)(4) = –16

QUESTION: 4

The number of the real roots of the equation 2cos (x(x + 1)) = 2x + 2–x is

(2019)

Solution:

Given: 2 cos (x(x + 1)) = 2x + 2–x
Now, by AM – GM inequality, we get


∴ 2 cos (x(x + 1)) ≥ 2
We know, –1 ≤ cosθ ≤ 1
∴ 2 cos(x(x + 1)) = 2
Hence, the expression is valid only if 2x + 2–x = 2, which is true for only one value of x i.e. 0.
Therefore, the expression has only one real solution.

QUESTION: 5

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) – f(m) = 2, then m equals

(2019)

Solution:

Case 1: m is even
Given, 8f(m + 1) – f(m) = 2
⇒ 8(m + 1 + 3) – m(m + 1) = 2
⇒ 8m + 32 – m2 – m = 2
⇒ m2 – 7m – 30 = 0
⇒ (m – 10)(m + 3) = 0
⇒ m = 10 or –3
As m is a positive integer, therefore m = 10.
Case 2: If m is odd, then
8f(m + 1) – f(m) = 2
⇒ 8(m + 1)(m + 2) – (m + 3) = 2
⇒ 8(m2 + 3m +  2) – m – 3 = 2
⇒ 8m2 + 24m + 16 – m – 3 = 2
⇒ 8m2 + 23m + 11 = 0,
Which is not possible and hence no solution.

QUESTION: 6

Consider a function f satisfying f (x + y) = f (x) f (y) where x, y are positive integers, and f (1) = 2. If f (a + 1) + f (a + 2) + ... + f (a + n) = 16(2n – 1) then a is equal to 

(2019)

Solution:

Given,
f(a + 1) + f(a + 2) + .. + f(a + n) = 16(2n – 1)
⇒ f(a) f(1) + f(a) f(2) + ... + f(a) f(n) = 16(2n – 1)
⇒ f(a) (f(1) + f(2) + ... + f(n)) = 16 (2n – 1)
When n = 1, then f(a)f(1) = 16(21 – 1) = 16
⇒ f(a) × 2 = 16 ⇒ f(a) = 8
∴ f(a) (f(1) + f(2) + ... + f(n)) = 16(2n – 1)
⇒ f(1) + f(2) + ... + f(n) = 2(2n – 1)
When n = 2, then f(1) + f(2) = 2(22 – 1) = 6
⇒ f(2) = 6 – f(1) = 6 – 2 = 4
When n = 3, then f(1) + f(2) + f(3) = 2(23 – 1) = 14
⇒ f(3) = 14 – f(1) – f(2) = 14 – 2 – 4 = 8 = f(a)
∴ a = 3

QUESTION: 7

The real root of the equation 26x + 23x+2 – 21 = 0 is

(2019)

Solution:

26x + 23x+2 – 21 = 0
⇒ 26x + 23x × 22 – 21 = 0
Put 23x = y, then y2 + 4y – 21 = 0
⇒ (y – 3) (y + 7) = 0 ⇒ y = 3 or y = –7
⇒ 23x = 3 or 23x = –7 (No solution)
⇒ 3x = log23 ⇒ x = log2 3 / 3

QUESTION: 8

What is the largest positive integer n such that  is also a positive integer?

(2019)

Solution:


The expression is positive integer if  is non-negative integer.
For this (n – 4) must be factor of 8.
For n to be largest, n – 4 = 8
∴ n = 12

QUESTION: 9

If 5x – 3y = 13438 and 5x – 1 + 3y + 1 = 9686, then x + y equals

(2019)

Solution:

Consider the second given equation
5x–1 + 3y+1 = 9686 ...(i)
Last digit of 5x–1 will always be 5 for all positive integral values of (x – 1)
The power cycle of 3 is
34k+1 ≡ 3
34k+2 ≡ 9
34k+3 ≡ 7
34k ≡ 1
Clearly 3y+1 must be in the form of 34k as the unit digit of
R.H.S. = 6 of eq. (i)
We have 34 = 81, and 38 = 6561
Also, 9686 – 81 = 9605 and 9686 – 6561 = 3125
We observe that 3125 = 55
∴ 5x–1 = 55 ⇒ x = 6
and 3y+1 = 38 ⇒ y = 7
x = 6 and y = 7 also satisfies the first given equation 5x – 3y = 13438
∴ x + y = 6 + 7 = 13

QUESTION: 10

The quadratic equation x2 + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b2 + c?

(2019)

Solution:

Sum of roots = 4a + 3a = 7a = –b
∴ b = –7a
Product of roots = 4a × 3a = c
∴ c = 12a2
Now, b2 +  c = (–7a)2 + 12a2 = 61a2
Comparing the options
Option (a) : 61a2 = 3721 ⇒ a2 = 61, clearly a is not an integer.
Option (b) : 61a2 = 361 ⇒ a2 = 361 / 61, clearly a is not an integer.
Option (c) : 61a2 = 549 ⇒ a2 = 9, we can have a = –3 or 3 (an integer)
Option (d) : 61a2 = 427 ⇒ a2 = 7, clearly a is not an integer

QUESTION: 11

Let a, b, x, y be real numbers such that a2 + b2 = 25, x2 + y2 = 169, and ax + by = 65. If k = ay – bx, then

(2019)

Solution:

We can take a = 5, b = 0, x = 13 and y = 0 as these values that satisfies all three equations
∴ k = ay – bx = 5 × 0 – 0 × 13 = 0

QUESTION: 12

If u2 + (u − 2v − 1)2 = −4v(u + v), then what is the value of u + 3v?

(2018)

Solution:

Cost of table of Amal = 1.2p
Cost of table for Asim = 0.8p
Amal sells to Bimal at 1.3 × 1.2p = 1.56p
= cost of table for Bimal = x
Asim sells table to Barun at 0.7 × 0.8p = 0.56p
= cost of table for Barun = y

QUESTION: 13

If f(x) = 5x + 2 / 3x - 5 and g(x) = x2 – 2x – 1, then the value of g (f(f(3))) is

(2017)

Solution:


g (f(f(3))) = (3)2 – 3 × 2 – 1 = 9 – 6 – 1 = 9 – 7 = 2

QUESTION: 14

The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12,and z ≤ 12 is

(2017)

Solution:

x = 25 + y + z
Possible value table below.
27 ≤ x ≤ 40

The no. of solution is 1 + 2 + ... + 12 + 11 + 10 = 99

QUESTION: 15

Suppose, log3 x = log12 y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log6 G is equal to

(2017)

Solution:

log3x = a, x = 3a
log12y = a, y = 12a
Therefore, the geometric mean of x and y equals 

Hence, 6a or log6 G = a

QUESTION: 16

If the sum to infinity of the series 2 + (2 – d) 2/3 + (2 + d) 4/9 + (2 + 3d) 8/27 + .... is 5/2, what is the value of d?

(2016)

Solution:


Subtracting,

QUESTION: 17

If S = 2/10 + 6/102 + 12/103 + 20/104 + 30/105 + 42/106 + ......., find the value of S?

(2016)

Solution:

∵ S = 0.2 + 0.06 + 0.012 + 0.0020 + 0.00030 + 0.000042 + ...
⇒ S = 0.274342 + ...............
Going from answer choices, choice (a) = 0.2666 ....
Choice (b) = 0.268888 .....; choice (c) = 0.27222 ...
Choice (d) = 0.27434 ...
So, choice (d) is answer.

QUESTION: 18

If the roots of the equation (x + 1) (x + 9) + 8 = 0 are a and b, then the roots of the equation (x + a) (x + b) – 8 = 0 are

(2016)

Solution:

Here, (x + 1) (x + 9) + 8 = 0
x2 + 10x + 17 = 0
The roots of the equation are a and b
∴ a + b = – 10
ab = 17
(x + a) (x + b) – 8 = 0
x2 + (a + b) x + ab – 8 = 0
x2 – 10x  +  9 = 0
Therefore, roots of (x + a) (x + b) – 8 = 0 are 1 and 9.

QUESTION: 19

If x2 + (x + 1) (x + 2) (x + 3) (x + 6) = 0, where x is a real number, then one value of x that satisfies this equation is

(2015)

Solution:

x2 + ( x + 1) (x + 2) (x + 3) (x + 6) = 0
(x + 1) (x + 2) (x + 3) (x + 6) = –x2
(x2 + 7x + 6) (x2  + 5x + 6) = –x2


then, (y + 7) (y + 5) = –1
⇒ y2 + 12y + 35 = –1
y2 + 12y + 36 = 0
⇒ (y + 6)2  = 0
⇒ y = – 6

QUESTION: 20

Find the solution set for [x] + [2x] + [3x] = 8, where x is a real number and [x] is the greatest integer less than or equal to x.

(2015)

Solution:

By observing we can find that x > 1 and x < 2.
Else the RHS ≠ 8.
So the combinations are [x] = 1, [2x] = 2 or 3.
[3×] = 4 or 5
The combinations that give RHS = 8 are 1 + 2 + 5 or 1 + 3 + 4.
For any value of x, the case of "1 + 2 + 5" is not possible. Hence it has to be the case of "1 + 3 + 4".
Which will occur

Hence the solution is 

QUESTION: 21

If 3x + y + 4 = 2xy, where x and y are natural numbers, then find the ratio of the sum of all possible values of x to the sum of all possible values of y.

(2015)

Solution:

3x + y + 4 = 2x y
⇒ 3x + 4 = y (2x – 1)

When x = 6 ⇒ y = 2
When x = 1 ⇒ y = 7
These two are the only possible pairs of values of x and y. Where x and y are natural numbers.
∴  Required ratio 

QUESTION: 22

Let x, y, z and t be the positive numbers which satisfy the following conditions:
I. If x > y, then z > t and
II. If x > z, then y < t
Which of the following is necessarily true?

(2015)

Solution:

By assuming the values of x, y, z and t, (a) and (b) can be very easily ruled out.
Checking option (c), if x > y + z. then x > y and x > z (since all numbers are positive).
So, using statements I and II, x > z > t > y.
So, option (c) is correct.

QUESTION: 23

Both the roots of the quadratic equation x2 + rx + s = 0 are real and greater than 1. If  then which of the following is definitely true?

(2015)

Solution:

Take roots as 2, 2
⇒ r = – 4 & s = 4

QUESTION: 24

If x(x – 3) = –1, then the value of x3(x3 – 18) is

(2014)

Solution:

According to question,
x (x – 3) = –1
cubing on both side.
⇒ x3 (x – 3)3 = (–1)3
⇒ x3 (x3– 27 – 9x2 + 27x) = – 1
⇒ x3 (x3 – 18) + x3 (–9 – 9x2 + 27x) = –1
⇒ x3 (x3 – 18) – 9x3 (x2 – 3x + 1) = –1
⇒ x3 (x3 – 18) – 9x3 (–1 + 1) = –1
⇒ x3 (x3 – 18) = –1

QUESTION: 25

 Find the value of P.

(2014)

Solution:

nth term of the series can be written as

QUESTION: 26

What is the sum of the roots of all the quadratic equations that can be formed such that both the roots of the quadratic equation are common with the roots of equation (x – a) (x – b) (x – c) = 0?

(2014)

Solution:

The equations formed by the roots of the equation
(x – a) (x – b)(x – c) can be as follows:
(i)  (x – a)(x – b) ⇒ Roots are a, b
(ii) (x – b)(x – c) ⇒ Roots are b, c
(iii) (x – c)(x – a) ⇒ Roots are c, a
(iv) (x – a)2 ⇒ Roots are a, a
(v) (x – b)2 ⇒ Roots are b, b
(vi) (x – c)2 ⇒ Roots are c, c
Adding all these roots, we get 4(a + b + c).

QUESTION: 27

The sum of the coefficients of the polynomial (x – 1)9 (x – 2)4 (x – 4) is

(2014)

Solution:

We obtain the sum of all the coefficients of a polynomial by equating all the variables to 1. Here by putting x = 1 in the polynomial, the required sum comes out to be zero.

QUESTION: 28

An equation with all positive roots is written as Xn + anXn-1 + an-1Xn-2 + .... a1 = 0. Which of the following in necessarily true?

(2014)

Solution:

Here, (x – 1) (x  – 2) (x – 3) = 0
⇒ (x2 – 3x + 2) (x – 3) = 0
⇒ x3 – 6x2 + 11x – 6 = 0
According to question,
x = 3, an = –6, a1 = –6
Substitute the above values in the option,
Option (a) : (–6)3 ≥ 33 × – 6
⇒ – 216 ≥ – 168
This is incorrect, thus option (a) is incorrect.
Opiton (b) : 33 ≥ (–6)3 × –6                       27 ≥ 64
This is incorrect, thus option (b) is incorrect.
Option (c) : (–6)3 ≥ 33 × – 6
⇒  –216 ≥ – 168
This is incorrect, thus option (c) is incorrect.
Thus, none of the options is necessarily true.

QUESTION: 29

The coefficient of a12b8 in the expansion of (a2 + b)13 is

(2013)

Solution:

General term expansion of
(a2 + b)13 = nCr xn-r yr
To get coefficient of a12  b8
(a2)6 b8 where 6 + 8 ≠ 13.
So the term is not possible

QUESTION: 30

Let R(x) = mx3 – 100x2 + 3n, where m and n are positive integers. For how many ordered pairs (m, n) will (x – 2) be a factor of R(x)?

(2013)

Solution:

(x – 2) is a factor of R(x)
∴ R (2) = 0
⇒ m(2)3 – 100 (2)3 + 3n = 0
⇒ 8m – 400 + 3n = 0

∵ m and n are positive integers, so n must be a multiple of 8. i.e. n = 8, 16, 24, ....., 128, then we get m = 47, 44, 41, ....,2 respectively.
So, the number of ordered pairs (m, n) is 16.

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