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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeIf the number of vertices, edges and faces of a rectangular parallelepiped are denoted by v, E, and f respectively, the value of (v – e + f) is
Correct
Explanation:
In a rectangular parallelepiped, v = 8, e = 12, f = 6Incorrect
Explanation:
In a rectangular parallelepiped, v = 8, e = 12, f = 6 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeThe area of the triangle formed by the graphs of the equations x = 0, 2x + 5y = 10 and x + y = 5
Correct
Explanation:
Area of shaded portion to be found
Area of triangle = (1/2)*base*height = (1/2)*3*5Incorrect
Explanation:
Area of shaded portion to be found
Area of triangle = (1/2)*base*height = (1/2)*3*5 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeIf the side of an equilateral triangle is 16√3 cm, then its area would be?
Correct
Explanation:
Area = (√3/4)*a^2
(√3/4) * 16√3 * 16√3Incorrect
Explanation:
Area = (√3/4)*a^2
(√3/4) * 16√3 * 16√3 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeThe HCF of x^{4} – 1 and x^{4} + x^{3} – x^{2} + x – 2 is
Correct
Explanation:
x^{4} – 1 = (x^{2} – 1) (x^{2} + 1) = (x – 1)(x+1) (x^{2} + 1)
Now x^{4} + x^{3} – x^{2} + x – 2
Putting x = 1 in this equation gives 0, so (x1) is a factor, divide x^{4} + x^{3} – x^{2} + x – 2 by (x1) gives x^{3} – x^{2} + x – 1
Again put x = 1, gives 0, so another factor is (x1) again, divide (x1) gives x^{3} – x^{2} + x – 1 by (x1), gives (x^{2} + 1) which cannot be further divided
So x^{4} + x^{3} – x^{2} + x – 2 = (x^{2} + 1)(x1)(x1)
Now common factors in both expressions are (x^{2} + 1)(x1) which is the HCF thenIncorrect
Explanation:
x^{4} – 1 = (x^{2} – 1) (x^{2} + 1) = (x – 1)(x+1) (x^{2} + 1)
Now x^{4} + x^{3} – x^{2} + x – 2
Putting x = 1 in this equation gives 0, so (x1) is a factor, divide x^{4} + x^{3} – x^{2} + x – 2 by (x1) gives x^{3} – x^{2} + x – 1
Again put x = 1, gives 0, so another factor is (x1) again, divide (x1) gives x^{3} – x^{2} + x – 1 by (x1), gives (x^{2} + 1) which cannot be further divided
So x^{4} + x^{3} – x^{2} + x – 2 = (x^{2} + 1)(x1)(x1)
Now common factors in both expressions are (x^{2} + 1)(x1) which is the HCF then 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeInternal bisectors of angle B and angle C of ∆ABC intersect at O. If angle BOC = 108 degree, then value of angle BAC in degrees is
Correct
Explanation:
Formula is B/2 + C/2 = 180 – BOC = 180108
Solve, B+C = 144
So A = 180 144 = 36Incorrect
Explanation:
Formula is B/2 + C/2 = 180 – BOC = 180108
Solve, B+C = 144
So A = 180 144 = 36 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeIf 2x + 5y = 6 and 8x^{3} + 125y^{3} = 210, then value of xy is
Correct
Explanation:
Use formula x^{3} + y^{3} = (x+y)(x^{2} + y^{2} – xy)
8x^{3} + 125y^{3} = 210
(2x+5y)( (2x)^{2} + (5x)^{2} – 2x*5y) = 210
Use a^{2} + b^{2} – ab = (a+b)^{2} – 3ab
So
6 * ( (2x+5y)^{2} – 3*2x*5y) = 210
So (6)^{2} – 3*2x*5y = 35
So 1 = 3*2x*5y
So xy = 1/30Incorrect
Explanation:
Use formula x^{3} + y^{3} = (x+y)(x^{2} + y^{2} – xy)
8x^{3} + 125y^{3} = 210
(2x+5y)( (2x)^{2} + (5x)^{2} – 2x*5y) = 210
Use a^{2} + b^{2} – ab = (a+b)^{2} – 3ab
So
6 * ( (2x+5y)^{2} – 3*2x*5y) = 210
So (6)^{2} – 3*2x*5y = 35
So 1 = 3*2x*5y
So xy = 1/30 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe minimum value of sin^{2}ɸ + 2 cos^{2}ɸ
Correct
Explanation:
a sin^{2}ɸ + b cos^{2}ɸ
If a>b, maximum value = a, min value = b
If aIncorrect
Explanation:
a sin^{2}ɸ + b cos^{2}ɸ
If a>b, maximum value = a, min value = b
If a 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe difference between successive discounts of 20% followed by 25% and 30% followed by 40% on the marked price of an article is Rs 144. Find the marked price of the article.
Correct
Explanation:
Let MP = x
So x*(10020)/100 * (10025)/100 – x*(10030)/100 * (10040)/100 = 144
x*(80/100)(75/100) – x*(70/100)*(60/100) = 144
Solve, x = 800Incorrect
Explanation:
Let MP = x
So x*(10020)/100 * (10025)/100 – x*(10030)/100 * (10040)/100 = 144
x*(80/100)(75/100) – x*(70/100)*(60/100) = 144
Solve, x = 800 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeA’s 2 days work is equal to B’s 3 days work. If A can complete the work in 8 days then to complete the work B will take how many days?
Correct
Explanation:
A completes 1 work in 8 days, so in 2 days, 2/8 work = 1/4 work
So B in 3 days complete 1/4 work, so 1 work in 3*4 = 12 daysIncorrect
Explanation:
A completes 1 work in 8 days, so in 2 days, 2/8 work = 1/4 work
So B in 3 days complete 1/4 work, so 1 work in 3*4 = 12 days 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeIf x = 2, then the value of x^{3} + 27x^{2} + 243x + 631 is
Correct
Explanation:
Put x = 2 in expression and solve.Incorrect
Explanation:
Put x = 2 in expression and solve.
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