CAT4MBA

Question 2: Given a³ = 150b, and a and b are positive integers. Find the least value of b.

(a) 180

(b) 90

(c) 270

(d) None of these

Solution:

150 = 2 x 3 x 5²

Now, for a³ = 150b to be a perfect cube, the powers of each of the prime factors of a³ has to be a multiple of 3 and to obtain the least such value of b, the powers of the prime factors have to be the least multiple of 3 and that is 3 itself and so the value of b has to contain 2², 3² and 5 as its factors.

Hence, the minimum value of b = 2² x 3² x 5 = 180

Thank You.

Ravi Raja

Question 3: If x, y, z are non-zero real numbers such that:

(x + y – z)/ z = (x – y + z)/y = (– x + y + z)/x = 1 and a = [ (x + y) (y + z) (z + x)] / xyz , find the value of a?

(a) 1

(b) 9

(c) 8

(d) 1/9

Solution:

(x + y – z)/ z = 1

or, x + y – z = z

or, x + y = 2z ------------- (1)

Similarly, from the other two relations: (x – y + z)/y = 1 and (– x + y + z)/x = 1, we get:

y + z = 2x ------------- (2)

z + x = 2y ------------- (3)

Substituting these values of (x + y), (y + z) and (z + x) in a = [ (x + y) (y + z) (z + x)] / xyz , we get:

a = [ (x + y) (y + z) (z + x)] / xyz

or, a = [(2x)(2y)(2z)]/(xyz) = [8(xyz)]/(xyz) = 8

Thank You.

Ravi Raja

Question 4: How many solutions does the equation | x + 2 | = 2x have?

(a) 0

(b) 1

(c) 2

(d) 3

Solution:

Case I: When | x + 2 | = 0

Then we have: x + 2 = 0

x = – 2

Note that the left hand side equals zero but the right hand side equals – 4, which is not possible. Hence | x + 2 | cannot be equal to zero.

Case II: When | x + 2 | > 0

Then we can replace | x + 2 | by (x + 2) and get the following equation:

x + 2 = 2x

or, x = 2

Note that when x = 2, the left hand side and the right hand side both equals 4. Hence, x = 2 is a solution to the given equation.

Case III: When | x + 2 | < 0

Then we can replace | x + 2 | by – (x + 2) and get the following equation:

– (x + 2) = 2x

or, – x – 2 = 2x

or, 3x = – 2

or, x = – (2/3)

Note that when x = –  (2/3), the left hand side equals 4/3 but the right hand side equals – 4/3, which is not possible. Hence | x + 2 | cannot be less than zero.

Hence, for the given equation, we have only one solution and that is x = 2.

Thank You.

Ravi Raja

Question 5: What is the value of n, if 2 log a + 2 log a² + 2 log a³ + … + 2 log aⁿ = 56 log a?

(a) 7

(b) 8

(c) 6

(d) None of these

Solution:

2 log a + 2 log a² + 2 log a³ + … + 2 log aⁿ = 56 log a

or, 2 log a + 4 log a + 6 log a + … + 2n log a = 56 log a             [Since log aⁿ = n log a]

or, (2 + 4 + 6 + … to n terms) log a = 56 log a

or, 2 + 4 + 6 + … to n terms = 56

or, n (n + 1) = 56

or, n² + n – 56 = 0

or, (n – 7)(n + 8) = 0

or, n = – 8, 7

but n cannot be negative.

Hence n = 7.

Thank You.

Ravi Raja

Question 7: If the roots of the equation 2mx² + 8x + 32m  = 0 are equal and real, what is the value of m?

(a) 2/3                         

(b) ½

(c) 2

(d) 5/2

Solution:

If the roots of the equation 2mx² + 8x + 32m  = 0 are equal and real, then the discriminant of the given equation should be equal to 0.

That is, b² = 4ac

In the given problem, a = 2m, b = 8 and c = 32m

Substituting the values in the above expression, we get:

b² = 4ac

or, (8)² = 4(2m)(32m)

or, 4(64m²) = 64

or, m² = ¼

or, m = – ½ , ½

Hence, if we check the options, we have m = ½

Thank You.

Ravi Raja

Q8. Two lines are placed symmetrically around the line y = 4. One line is defined by y  = 3x + 1. Which equation yields the other line?

(a) y = 1/3x + 1

(b) y = – 1/3x + 7

(c) y = – 1/3x + 6

(d) y = – 3x + 7

Solution:

Since the two lines are placed symmetrically about the line y = 4 and the equation of one of the lines is y = 3x + 1, we can say that the other line will be a reflection of this line about the line y = 4.

Now, we can check that the lines y = 4 and y = 3x + 1 are not parallel lines and so they will intersect at a point and that point of intersection can be calculated by solving the two equations and we obtain the point as (1, 4). So, the required line should also pass through this point and note that only option (d) satisfies this condition.

Thank You. 

Ravi Raja

Question 7: If the roots of the equation 2^mx² + 8x + 32^m  = 0 are equal and real, what is the value of m?

(a) 2/3                         

(b) ½

(c) 2

(d) 5/2

Solution:

If the roots of the equation 2^mx² + 8x + 32^m  = 0 are equal and real, then the discriminant of the given equation should be equal to 0.

That is, b² = 4ac

In the given problem, a = 2^m, b = 8 and c = 32^m = 2^5m

Substituting the values in the above expression, we get:

b² = 4ac

or, (8)² = 4(2^m)(32^m)

or, 4(2^6m) = 64

or, 2^6m = 16

or, 2^6m = 2⁴

or, 6m = 4

or, m = 4/6 = 2/3

Hence, we have m = 2/3

Thank You.

Ravi Raja

as per my knowledge the answer of question 1 is (b) 83 .....if u got differnt answer plz let me know

Hello Praveen,

You are most welcome. Will try to post the solutions of the remaining problems as soon as possible.

Thank You.

Ravi Raja