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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

If x = [ (m+1)^{1/3} + (m-1)^{1/3} ] / [ (m+1)^{1/3} – (m-1)^{1/3} ], then the value of x^{3} – 3mx^{2} + 3x – m is

Correct

Explanation:
x = [ (m+1)^{1/3} + (m-1)^{1/3} ] / [ (m+1)^{1/3} – (m-1)^{1/3} ]
apply componendo and dividendo

(x+1)/(x-1) = (m+1)^{1/3}/(m-1)^{1/3}
Cube both sides
[x^{3} + 3x^{2} + 3x + 1]/ [x^{3} – 3x^{2} + 3x – 1] = (m+1)/(m-1)
Again apply componendo and dividendo
[2x^{2} + 6x]/[ 6x^{2}+2] = 2m/2
Solve this, we get x^{3} – 3mx^{2} + 3x – m = 0

Incorrect

Explanation:
x = [ (m+1)^{1/3} + (m-1)^{1/3} ] / [ (m+1)^{1/3} – (m-1)^{1/3} ]
apply componendo and dividendo

(x+1)/(x-1) = (m+1)^{1/3}/(m-1)^{1/3}
Cube both sides
[x^{3} + 3x^{2} + 3x + 1]/ [x^{3} – 3x^{2} + 3x – 1] = (m+1)/(m-1)
Again apply componendo and dividendo
[2x^{2} + 6x]/[ 6x^{2}+2] = 2m/2
Solve this, we get x^{3} – 3mx^{2} + 3x – m = 0

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

A person got an article after getting two successive discounts of 15% and 10% at Rs 3060. What is the marked price of this article?

Correct

Explanation:
Single equivalent dance is (15 + 10 – 15*10/100) = 23.5%
So MP is 100/(100-23.5) * 3060

Incorrect

Explanation:
Single equivalent dance is (15 + 10 – 15*10/100) = 23.5%
So MP is 100/(100-23.5) * 3060

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

In a rectangle ABCD, the side AB is divided into 4 equal parts by the points X, Y and Z. Find the ratio of: (area of ∆XYC)/(area of rect. ABCD)

Correct

Explanation:
Let AB = 4x and BC = y
Area of rect ABCD = 4xy
Now XY = 4x/4 = x
So area of ∆XYC = ½ * x * y = xy/2
So required ratio = (xy/2)/(4xy)

Incorrect

Explanation:
Let AB = 4x and BC = y
Area of rect ABCD = 4xy
Now XY = 4x/4 = x
So area of ∆XYC = ½ * x * y = xy/2
So required ratio = (xy/2)/(4xy)

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

If the length of a chord of a circle which makes an angle 30 degrees with the tangent drawn at one end point of the chord is 8cm, then the radius of the circle is

Correct

Explanation:

Angle ABC = 30
So angle ABO = 90 – 30 = 60
BD = 8/2 = 4
Cos OBD = BD/OB
So cos 60 = 4/OB, so OB = 8

Incorrect

Explanation:

Angle ABC = 30
So angle ABO = 90 – 30 = 60
BD = 8/2 = 4
Cos OBD = BD/OB
So cos 60 = 4/OB, so OB = 8

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

If tanɸ + cotɸ = 2, then the value of tan^{6}ɸ + cot^{12}ɸ is

A cube of side 12 feet is to be painted from outside. The cost of paint is Rs 21.50 per kg. If 32 square feet are covered by 1 kg, then what is the cost of painting the cube?

Correct

Explanation:
Surface area of cube = 6a^{2} = 6*12^{2} = 864 sq. feet
So quantity of paint required = (864/32) = 27 kg
So cost of painting = 21.50*27

Incorrect

Explanation:
Surface area of cube = 6a^{2} = 6*12^{2} = 864 sq. feet
So quantity of paint required = (864/32) = 27 kg
So cost of painting = 21.50*27

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

P varies directly with the square of Q when R is a constant and inversely with R when Q is constant. When Q = 12 and R = 9, then P = 48. Find P when Q = 20 and R = 15.

Correct

Explanation:
Let k be constant of proportionality,
If P directly varies with Q^{2}, then P = k Q^{2}
And if P inversely varies with R, then P = k/R
So in all P = k*Q^{2}/R
When Q = 12 and R = 9, then P = 48
So 48 = k*12^{2}/9
Solve, k = 3
Now when Q = 20 and R = 15,
P = 2*20^{2}/15

Incorrect

Explanation:
Let k be constant of proportionality,
If P directly varies with Q^{2}, then P = k Q^{2}
And if P inversely varies with R, then P = k/R
So in all P = k*Q^{2}/R
When Q = 12 and R = 9, then P = 48
So 48 = k*12^{2}/9
Solve, k = 3
Now when Q = 20 and R = 15,
P = 2*20^{2}/15

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

The students of section A and section B have average weights of 48 kg and 51 kg respectively. Find the approximate average weight of both the sections together such that the number of students in the sections A and B is in the ratio 5 : 3.

Correct

Explanation:
Students in A section = 5x, students in B section = 3x
So total weight of students of A section = 48*5x = 240x
And total weight of students of B section = 51*3x = 153x
So average weight of both sections = [(240x+153x)/(5x+3x)]

Incorrect

Explanation:
Students in A section = 5x, students in B section = 3x
So total weight of students of A section = 48*5x = 240x
And total weight of students of B section = 51*3x = 153x
So average weight of both sections = [(240x+153x)/(5x+3x)]

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

A and B can complete work in 20 and 12 days respectively. A starts the work, and after 4 days B joins him. How many days they took to complete the whole work?

Correct

Explanation:
Let the work be completed in x days, so A worked for all x days, and B for (x-4) days
1/20 * x + 1/12 (x – 4) = 1

Incorrect

Explanation:
Let the work be completed in x days, so A worked for all x days, and B for (x-4) days
1/20 * x + 1/12 (x – 4) = 1

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

There are two trains of lengths 300 m and 240 m respectively. Both start at same time. If they travel in same direction they cross each other in 36 hours and if they travel in opposite direction towards each other, they cross each other in 4 hours. What is the speed of the fastest train?

Correct

Explanation:
Let speed of 1st train = s1 km/hr, of 2nd train = s2 km/hr
Then when they travel in same direction, relative speed = s1-s2
So (300+240) = (s1-s2)*36. This gives s1 – s2 = 15
When they travel in opposite direction, relative speed = s1+s2
So (300+240) = (s1+s2)*4. This gives s1 + s2 = 135
Solve both equations, s1 = 75

Incorrect

Explanation:
Let speed of 1st train = s1 km/hr, of 2nd train = s2 km/hr
Then when they travel in same direction, relative speed = s1-s2
So (300+240) = (s1-s2)*36. This gives s1 – s2 = 15
When they travel in opposite direction, relative speed = s1+s2
So (300+240) = (s1+s2)*4. This gives s1 + s2 = 135
Solve both equations, s1 = 75