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Question 1 of 10
1. Question
1 pointsCategory: Quantitative Aptitudey^{2} – x^{2} = √4096, y – x = 4
Correct
Explanation:
y^{2} – x^{2} = √4096; y^{2} – x^{2} = 64
(y + x)(y – x) = 64
From equation2, y – x = 4
4(y + x) = 64;
y + x = 16 –(A)
y – x = 4 —(B)
Solving A & B,
2y = 20 =>y = 10
10 – x = 4
x = 6Incorrect
Explanation:
y^{2} – x^{2} = √4096; y^{2} – x^{2} = 64
(y + x)(y – x) = 64
From equation2, y – x = 4
4(y + x) = 64;
y + x = 16 –(A)
y – x = 4 —(B)
Solving A & B,
2y = 20 =>y = 10
10 – x = 4
x = 6 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative Aptitudex^{2}– 3x – √5x + 3√5 = 0
y^{2} – √3y – √2y + √6 = 0Correct
Explanation:
x^{2} – 3x – √5x + 3√5 = 0
x (x3) – √5(x3) = 0
(x3) (x√5) = 0
x = 3, 2.23
y^{2} – √3y – √2y + √6 = 0
y (y√3) – √2(y√3) = 0
(y√2) (y√3) = 0
y = 1.414, 1.732
Put on number line 1.414, 1.732, 2, 2.23Incorrect
Explanation:
x^{2} – 3x – √5x + 3√5 = 0
x (x3) – √5(x3) = 0
(x3) (x√5) = 0
x = 3, 2.23
y^{2} – √3y – √2y + √6 = 0
y (y√3) – √2(y√3) = 0
(y√2) (y√3) = 0
y = 1.414, 1.732
Put on number line 1.414, 1.732, 2, 2.23 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative Aptitude√1521 x + √5184 = 0, (256)^{1/4}y + (1728)^{1/3}=0
Correct
Explanation:
√1521 x + √5184 = 0;
39x + 72 = 0; x = 2
(256)^{1/4}y + (1728)^{1/3}=0; 4y + 12 = 0; y = 3
Put on number line 3 2Incorrect
Explanation:
√1521 x + √5184 = 0;
39x + 72 = 0; x = 2
(256)^{1/4}y + (1728)^{1/3}=0; 4y + 12 = 0; y = 3
Put on number line 3 2 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative Aptitude4/√x + 7/√x =√x, y^{2} – (11^{5/2} /√y) = 0
Correct
Explanation:
4/√x + 7/√x =√x; 4+7/√x =√x ; x = 11
y^{2}– (11^{5/2} /√y )= 0; y^{5/2 }– 11^{5/2 }y =0
y = 11Incorrect
Explanation:
4/√x + 7/√x =√x; 4+7/√x =√x ; x = 11
y^{2}– (11^{5/2} /√y )= 0; y^{5/2 }– 11^{5/2 }y =0
y = 11 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative Aptitude3x^{2}– 4x – 4 = 0, 4y^{2} + 23y + 15 = 0
Correct
Explanation:
3x^{2} – 4x – 4 = 0; 3x^{2} + 2x – 6x – 4 = 0
Gives x = 2/3 , 2
4y^{2} + 23y + 15 = 0; 4y^{2} + 20y + 3y + 15 = 0
Gives y = 5, 3/4
Put on number line 5 3/4 2/3 2Incorrect
Explanation:
3x^{2} – 4x – 4 = 0; 3x^{2} + 2x – 6x – 4 = 0
Gives x = 2/3 , 2
4y^{2} + 23y + 15 = 0; 4y^{2} + 20y + 3y + 15 = 0
Gives y = 5, 3/4
Put on number line 5 3/4 2/3 2 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirection: Q(610) Study the given chart carefully and answer the following questions.
TRAIN A
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationChennai Departing 5:00 pm – 400 Arakonnam 6:30 pm 6:35 pm 100 100 Ambur 8:50 pm 9:00 pm 250 90 Krishnajapuram 4:00 am 4:10 am 800 300 Bangarapet 7:30 am 7:45 am 1050 150 Bengaluru 10:20 am Arrival 1280 – TRAIN B
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationBengaluru Departing 6:00 pm – 300 Bangarapet 7:40 pm 7:45 pm 230 150 Krishnajapuram 9:30 pm 9:35 pm 480 270 Ambur 5:40 am 5:55 am 1030 50 Arakonnam 9:00 am 9:10 am 1180 100 Chennai 12:20 noon Arrival 1280 – What is the ratio of Train A’s distance from origin of the station of Krishnajapuram to the Train B’s distance from origin of the station of Chennai ?
Correct
EXPLANATION:
Train A dustance from Krishnajapuram = 800
Train B distance from Chennai = 1280
Required ratio = 800 : 1280 = 5 : 8Incorrect
EXPLANATION:
Train A dustance from Krishnajapuram = 800
Train B distance from Chennai = 1280
Required ratio = 800 : 1280 = 5 : 8 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirection: Q(610) Study the given chart carefully and answer the following questions.
TRAIN A
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationChennai Departing 5:00 pm – 400 Arakonnam 6:30 pm 6:35 pm 100 100 Ambur 8:50 pm 9:00 pm 250 90 Krishnajapuram 4:00 am 4:10 am 800 300 Bangarapet 7:30 am 7:45 am 1050 150 Bengaluru 10:20 am Arrival 1280 – TRAIN B
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationBengaluru Departing 6:00 pm – 300 Bangarapet 7:40 pm 7:45 pm 230 150 Krishnajapuram 9:30 pm 9:35 pm 480 270 Ambur 5:40 am 5:55 am 1030 50 Arakonnam 9:00 am 9:10 am 1180 100 Chennai 12:20 noon Arrival 1280 – What is the difference between the speed of Train A and that of Train B?
Correct
EXPLANATION:
Speed of Train A = 1280/17hrs 20 min
= 1280*3/52 = 73.84 kmph
Speed of train B = 1280*3/55hrs = 69.81kmph
So, difference between the speed of train A and train B = 73.84 ⎯69.81 = 4.03 kmphIncorrect
EXPLANATION:
Speed of Train A = 1280/17hrs 20 min
= 1280*3/52 = 73.84 kmph
Speed of train B = 1280*3/55hrs = 69.81kmph
So, difference between the speed of train A and train B = 73.84 ⎯69.81 = 4.03 kmph 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirection: Q(610) Study the given chart carefully and answer the following questions.
TRAIN A
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationChennai Departing 5:00 pm – 400 Arakonnam 6:30 pm 6:35 pm 100 100 Ambur 8:50 pm 9:00 pm 250 90 Krishnajapuram 4:00 am 4:10 am 800 300 Bangarapet 7:30 am 7:45 am 1050 150 Bengaluru 10:20 am Arrival 1280 – TRAIN B
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationBengaluru Departing 6:00 pm – 300 Bangarapet 7:40 pm 7:45 pm 230 150 Krishnajapuram 9:30 pm 9:35 pm 480 270 Ambur 5:40 am 5:55 am 1030 50 Arakonnam 9:00 am 9:10 am 1180 100 Chennai 12:20 noon Arrival 1280 –
What is the ratio of the total passengers of Train A to that of Train B?Correct
EXPLANATION:
Total passengers in train A = 400 + 100 + 90 + 300 + 150 = 1040
Total passengers in train B = 300 + 150 + 270 + 50 + 100 = 870
So, required ratio = 1040 : 870 = 104 : 87Incorrect
EXPLANATION:
Total passengers in train A = 400 + 100 + 90 + 300 + 150 = 1040
Total passengers in train B = 300 + 150 + 270 + 50 + 100 = 870
So, required ratio = 1040 : 870 = 104 : 87 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirection: Q(610) Study the given chart carefully and answer the following questions.
TRAIN A
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationChennai Departing 5:00 pm – 400 Arakonnam 6:30 pm 6:35 pm 100 100 Ambur 8:50 pm 9:00 pm 250 90 Krishnajapuram 4:00 am 4:10 am 800 300 Bangarapet 7:30 am 7:45 am 1050 150 Bengaluru 10:20 am Arrival 1280 – TRAIN B
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationBengaluru Departing 6:00 pm – 300 Bangarapet 7:40 pm 7:45 pm 230 150 Krishnajapuram 9:30 pm 9:35 pm 480 270 Ambur 5:40 am 5:55 am 1030 50 Arakonnam 9:00 am 9:10 am 1180 100 Chennai 12:20 noon Arrival 1280 – The number of passengers boarding Train A at Arakonnamis what percent of the number of passengers boarding Train B at Krishnajapuram?(approx.)
Correct
EXPLANATION:
Required percentage = 100× 100/270 = 37%Incorrect
EXPLANATION:
Required percentage = 100× 100/270 = 37% 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirection: Q(610) Study the given chart carefully and answer the following questions.
TRAIN A
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationChennai Departing 5:00 pm – 400 Arakonnam 6:30 pm 6:35 pm 100 100 Ambur 8:50 pm 9:00 pm 250 90 Krishnajapuram 4:00 am 4:10 am 800 300 Bangarapet 7:30 am 7:45 am 1050 150 Bengaluru 10:20 am Arrival 1280 – TRAIN B
Station Arrival time Departure
timeDistance
from origin
(in km)Number of
passengers
boarding at
each stationBengaluru Departing 6:00 pm – 300 Bangarapet 7:40 pm 7:45 pm 230 150 Krishnajapuram 9:30 pm 9:35 pm 480 270 Ambur 5:40 am 5:55 am 1030 50 Arakonnam 9:00 am 9:10 am 1180 100 Chennai 12:20 noon Arrival 1280 – If the average speed of Train A increases by 10% then when will it reach to its destination?
Correct
EXPLANATION:
If the average speed of train A increases by 10%
then its new speed = 73.84 ×110/100 = 81.22 kmph
Time taken by train A during the journey = 1280/81.22 = 15.75 hours = 15 hours 45 minutes
The time when the train will reach its destination = 5 pm + 15 hours 45 minutes = 8:45 amIncorrect
EXPLANATION:
If the average speed of train A increases by 10%
then its new speed = 73.84 ×110/100 = 81.22 kmph
Time taken by train A during the journey = 1280/81.22 = 15.75 hours = 15 hours 45 minutes
The time when the train will reach its destination = 5 pm + 15 hours 45 minutes = 8:45 am
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