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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeTwo medians AD and BE of ∆ABC intersect at O at right angles. If AD = 18 cm and BE = 12 cm, then the length of BD in cm is
Correct
Explanation:
GD = (1/3)*AD = (1/3)*18 = 6 cm
BG = (2/3)*BE = (2/3)*12 = 8 cm
In rt angled ∆BOD
So BD = √(6^2 + 8^2) = 10Incorrect
Explanation:
GD = (1/3)*AD = (1/3)*18 = 6 cm
BG = (2/3)*BE = (2/3)*12 = 8 cm
In rt angled ∆BOD
So BD = √(6^2 + 8^2) = 10 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeThe incomes of A and B are in the ratio 3 : 5 and their expenditures are in the ratio 1 : 2. If A saves Rs 26,000 and B saves Rs 40,000, what is the income of B?
Correct
Explanation:
3x and 5x
y and 2y
so 3x – y = 26000 and 5x – 2y = 40000
solve these equations, find x = 12000, so B’s income = 5x = 60,000Incorrect
Explanation:
3x and 5x
y and 2y
so 3x – y = 26000 and 5x – 2y = 40000
solve these equations, find x = 12000, so B’s income = 5x = 60,000 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeA shopkeeper marks an article at Rs 120 and offers 20% discount. He also gives an additional article worth Rs 6 with it for free. If he still makes a profit of 25%, what is the cost price of the article?
Correct
Explanation:
MP = 120, so SP = (80/100)*120 = Rs 96
After giving gift, SP = 96 – 6 = Rs 90
At 25% profit, CP = (100/125)*90Incorrect
Explanation:
MP = 120, so SP = (80/100)*120 = Rs 96
After giving gift, SP = 96 – 6 = Rs 90
At 25% profit, CP = (100/125)*90 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeThe time taken by A and B to complete 2/7 of work is same as time taken by B and C to complete 6/7 of work. The ratio of work done by A, B, and C is
Correct
Explanation:
Let A, B and C completes the whole work. So the work done by A+B+C is 1
Work done by A and B = 2/7
So work done by C = 1 – 2/7 = 5/7
Work done by B and C is 6/7, so work done by A = 1 – 6/7 = 1/7
And so work done by B is 1 – (5/7 + 1/7) = 1 – 6/7 = 1/7
So ratio of A : B : C = 1/7 : 1/7 : 5/7 = 1 : 1 : 5Incorrect
Explanation:
Let A, B and C completes the whole work. So the work done by A+B+C is 1
Work done by A and B = 2/7
So work done by C = 1 – 2/7 = 5/7
Work done by B and C is 6/7, so work done by A = 1 – 6/7 = 1/7
And so work done by B is 1 – (5/7 + 1/7) = 1 – 6/7 = 1/7
So ratio of A : B : C = 1/7 : 1/7 : 5/7 = 1 : 1 : 5 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeThe length of each side of an equilateral triangle is 12√3 cm. What is the approximate area of the incircle?
Correct
Explanation:
BD = DC = 6√3 cm
In rt. angled ∆ADB
AD = √[(AB)^2 – (BD)^2] = √[(12√3)^2 – (6√3)^2]
AD = = √[(12√3 – 6√3) (12√3 + 6√3)] = √[(6√3) (18√3)] = 18
Radius of circle = OD = (1/3)*18 = 6 cm
So area = (22/7)*6*6Incorrect
Explanation:
BD = DC = 6√3 cm
In rt. angled ∆ADB
AD = √[(AB)^2 – (BD)^2] = √[(12√3)^2 – (6√3)^2]
AD = = √[(12√3 – 6√3) (12√3 + 6√3)] = √[(6√3) (18√3)] = 18
Radius of circle = OD = (1/3)*18 = 6 cm
So area = (22/7)*6*6 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeIf x^{2} + y^{2} + 1/x^{2} + 1/y^{2} = 4, then 2x^{2} + 3y^{2} equals
Correct
Explanation:
x^{2} + y^{2} + 1/x^{2} + 1/y^{2} = 4
so [x^{2} + 1/x^{2} 2] + [y^{2} + 1/y^{2} 2] = 0
(x – 1/x)^{2} + (y – 1/y)^{2} = 0
So x – 1/x)^{2} = 0,
Solve x= 1
Similarly, y = 1
So 2x^{2} + 3y^{2} = 2*1 + 3*1 = 5Incorrect
Explanation:
x^{2} + y^{2} + 1/x^{2} + 1/y^{2} = 4
so [x^{2} + 1/x^{2} 2] + [y^{2} + 1/y^{2} 2] = 0
(x – 1/x)^{2} + (y – 1/y)^{2} = 0
So x – 1/x)^{2} = 0,
Solve x= 1
Similarly, y = 1
So 2x^{2} + 3y^{2} = 2*1 + 3*1 = 5 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeMean of 10 numbers is 40. But it was observed that numbers 12 and 16 are wrongly taken as 21 and 61. The correct mean is
Correct
Explanation:
Difference = 12+16 – 21 – 61 = 54
So correct average = 40 – 54/10 = 34.6Incorrect
Explanation:
Difference = 12+16 – 21 – 61 = 54
So correct average = 40 – 54/10 = 34.6 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections (810): The histogram shows the marks obtained by 45 students of a class. Answer the questions that follow:
How many students have obtained 60 and above marks?
Correct
Explanation:
7 + 5Incorrect
Explanation:
7 + 5 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections (810): The histogram shows the marks obtained by 45 students of a class. Answer the questions that follow:
If the pass marks are 40, what is the percentage of successful students?
Correct
Explanation:
Students Above 40 marks are 8+4+7+5 = 24
Total students are 2+6+10+3+8+4+7+5 = 45
So required % = (24/45)*100Incorrect
Explanation:
Students Above 40 marks are 8+4+7+5 = 24
Total students are 2+6+10+3+8+4+7+5 = 45
So required % = (24/45)*100 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections (810): The histogram shows the marks obtained by 45 students of a class. Answer the questions that follow:
How many students have obtained 40 or more marks but less than 60?
Correct
Explanation:
8+4Incorrect
Explanation:
8+4
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