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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirection (15): In the following questions, two equations are given. You have to solve both the equations and give answer –
4X^2+12x+9 = 0, 2Y^2+11y+14 = 0
Correct
Explanation :
4X^2+12x+9 = 0
(2x+3)(2x+3) = 0
X = 3/2, 3/2
(y+2)(2y+7) = 0
Y = 2, 7/2
3/2, 3/2, 2, 7/2
x>yIncorrect
Explanation :
4X^2+12x+9 = 0
(2x+3)(2x+3) = 0
X = 3/2, 3/2
(y+2)(2y+7) = 0
Y = 2, 7/2
3/2, 3/2, 2, 7/2
x>y 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirection (15): In the following questions, two equations are given. You have to solve both the equations and give answer –
X^4 21^2 = 855, Y^2+677 = 713
Correct
Explanation :
X^4 21^2 = 855,
X^4 = 855 + 441 = 1296
X = 6
Y^2+677 = 713
Y^2 = 713 – 677 = 36
Y = 6Incorrect
Explanation :
X^4 21^2 = 855,
X^4 = 855 + 441 = 1296
X = 6
Y^2+677 = 713
Y^2 = 713 – 677 = 36
Y = 6 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirection (15): In the following questions, two equations are given. You have to solve both the equations and give answer –
√676 x – √6084 = 0, (900)^1/2 Y = 30
Correct
Explanation :
26X – 6084 = 0
X = 6084/26 = 234
30y = 30
Y = 1Incorrect
Explanation :
26X – 6084 = 0
X = 6084/26 = 234
30y = 30
Y = 1 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirection (15): In the following questions, two equations are given. You have to solve both the equations and give answer –
X^2/3 × 45 = 1215÷ X^1/3 , Y^3 × 17 = Y^2 × 459
Correct
Explanation :
x^2/3+1/3 = 1215/45 = 27
x = 27
Y= 459/17 = 27Incorrect
Explanation :
x^2/3+1/3 = 1215/45 = 27
x = 27
Y= 459/17 = 27 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirection (15): In the following questions, two equations are given. You have to solve both the equations and give answer –
6x^2x 2 = 0, 1/ 2y^2+3y+28 = 1/30
Correct
Explanation :
(2x+1)(3x2) = 0
X = 1/2, 2/3
y^23y28 = 30 => 2y^2+3y2 = 0
2y^2+4yy+2 = 0
2y(y2)(y2) = 0
(2y1)(y2) = 0
Y = ½, y = 2
2, 2/3,1/2, 1/2Incorrect
Explanation :
(2x+1)(3x2) = 0
X = 1/2, 2/3
y^23y28 = 30 => 2y^2+3y2 = 0
2y^2+4yy+2 = 0
2y(y2)(y2) = 0
(2y1)(y2) = 0
Y = ½, y = 2
2, 2/3,1/2, 1/2 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeThe ratio of amount for two years under CI per annum and for one year under SI is 4/3, when the rate of interest is same, Find the rate of Interest ?
Correct
Explanation :
Note : SI for 1 year = CI for 1 year
P(1+r/100)2 /P(1+r/100) = 4/3
1+ r/100 = 4/3
r/100 = 4/3 1 = 1/3
r = 100/3 = 33 1/3Incorrect
Explanation :
Note : SI for 1 year = CI for 1 year
P(1+r/100)2 /P(1+r/100) = 4/3
1+ r/100 = 4/3
r/100 = 4/3 1 = 1/3
r = 100/3 = 33 1/3 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeIf a river flowing at 3kmph, a boat travels at 40km upstream and then returns downstream to the starting point. If its speed in still water is 5kmph, Find the total journey time ?
Correct
Explanation :
D = t(x2y2)/2x
40 = t(52 – 32)/2×5
400 = 16t
T = 25Incorrect
Explanation :
D = t(x2y2)/2x
40 = t(52 – 32)/2×5
400 = 16t
T = 25 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe sum of ages of 4 members of a family 5 years ago was 80year. Now the daughter has been married off and replaced by a daughterinlaw, the sum of their ages is 94 years. What is the difference between the ages of daughter and daughterinlaw ?
Correct
Explanation :
Present Sum of their ages with daughter = 80+20 = 100
Present Sum of their ages with daughterinlaw = 92
Difference = 100 94 = 6 yrIncorrect
Explanation :
Present Sum of their ages with daughter = 80+20 = 100
Present Sum of their ages with daughterinlaw = 92
Difference = 100 94 = 6 yr 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThe average weight of 32 students in a class was calculated as 27kg. It was later found that the weight of two students in a class was wrongly added. The actual weight of one boy was 30kg, but it was calculated as 28kg and the weight of another boy in the class was 21 whereas it was calculated as 25kg. Find the difference between original average weight and calculated average weight ?
Correct
Explanation :
Original Avg weight = 32*27+24 / 32 = 862/32 = 26.9
Given = 27kg
Difference = 27 – 26.9 = 0.1kgIncorrect
Explanation :
Original Avg weight = 32*27+24 / 32 = 862/32 = 26.9
Given = 27kg
Difference = 27 – 26.9 = 0.1kg 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThree pipes A , B and C can fill a tank in 8 hours , After working it at together for 2 hours , C is closed and A and B can fill the remaining part in 8 hours . How much time will C take alone to fill the tank?
Correct
Explanation :
In 2 hour the part filled by all the three pipes = 2 X 1/8 = 1/4
Remaining part = 1 – 1/4 = 3/4, which is filled by ( A + B ) in 8 hours.
1 hour the part filled by ( A + B ) = 3/(4 x 8) = 3/32
1 hour the part filled by C alone = 1/8 – 3/32= 43/32 = 1/32Incorrect
Explanation :
In 2 hour the part filled by all the three pipes = 2 X 1/8 = 1/4
Remaining part = 1 – 1/4 = 3/4, which is filled by ( A + B ) in 8 hours.
1 hour the part filled by ( A + B ) = 3/(4 x 8) = 3/32
1 hour the part filled by C alone = 1/8 – 3/32= 43/32 = 1/32
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