SBI PO Quantitative Aptitude Questions 2019 (Day-16) High Level New Pattern
SBI PO 2019 Notification is about to come and it is the most awaited exam among the aspirants. We all know that new pattern questions are introducing every year in the SBI PO exam. Further, the questions are getting tougher and beyond the level of the candidate’s expectations.
Are You preparing for Bank exams 2019? Start your preparation with Free Mock test Series.
Our IBPS Guide is providing High-Level New Pattern Quantitative Aptitude Questions for SBI PO 2019 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these high-level questions daily to familiarize with the exact exam pattern. We wish that your rigorous preparation leads you to a successful target of becoming SBI PO.
“Be not afraid of growing slowly; be afraid only of standing still”
Time limit: 0
Quiz-summary
0 of 10 questions completed
Questions:
1
2
3
4
5
6
7
8
9
10
Information
New Pattern Quantitative Aptitude Questions For SBI PO (Day-16)
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score
Your score
Categories
Not categorized0%
maximum of 10 points
Pos.
Name
Entered on
Points
Result
Table is loading
No data available
Your result has been entered into leaderboard
Loading
1
2
3
4
5
6
7
8
9
10
Answered
Review
Question 1 of 10
1. Question
Direction (1- 5): Study the following information carefully and answer the given questions
The given table shows the five different boxes contain three different color balls.
Probability of drawing one blue from box E is 1/4 and the probability of drawing one red ball from box E is 2/15. The probability of drawing one yellow ball from box D is 5/12. Find the sum of the probabilities of drawing one yellow ball from box E and one red ball from box D?
Correct
Answer: a)
Number of yellow balls from box E=x
Number of blue balls from box E=y
Total number of balls from box E=40+x+y
Number of Red balls from box D=z
Total number of balls from box D = z+30+12=z+42
y/(40 + x + y) = 1/4
40+x+y=4y
X – 3y + 40=0 ——– (1)
40/(40+x+y) = 2/15
600=80+2x+2y
2x+2y-520=0
x+y-260 =0 ———- (2)
By substituting 1 and 2, we get,
-4y=-300
Y=75
X – (3)*75+40=0
x=185
Total number of balls from box E=185+75+40=300
30/(z+42)=5/12
360 = 5z+210
5Z=150
Z = 30
Total balls from box D=30+42=72
Required sum=185C_{1}/300C_{1 }+ 30C_{1}/72C_{1}
= > (37 + 25)/60 = 62/60 = 31/30
Incorrect
Answer: a)
Number of yellow balls from box E=x
Number of blue balls from box E=y
Total number of balls from box E=40+x+y
Number of Red balls from box D=z
Total number of balls from box D = z+30+12=z+42
y/(40 + x + y) = 1/4
40+x+y=4y
X – 3y + 40=0 ——– (1)
40/(40+x+y) = 2/15
600=80+2x+2y
2x+2y-520=0
x+y-260 =0 ———- (2)
By substituting 1 and 2, we get,
-4y=-300
Y=75
X – (3)*75+40=0
x=185
Total number of balls from box E=185+75+40=300
30/(z+42)=5/12
360 = 5z+210
5Z=150
Z = 30
Total balls from box D=30+42=72
Required sum=185C_{1}/300C_{1 }+ 30C_{1}/72C_{1}
= > (37 + 25)/60 = 62/60 = 31/30
Question 2 of 10
2. Question
Direction (1- 5): Study the following information carefully and answer the given questions
The given table shows the five different boxes contain three different color balls.
The probability of drawing one yellow ball from box A is 2/7. Probability of drawing one red ball from box B is 3/10. Find the sum of the probability of drawing 2 blue balls from box A and the probability of drawing two yellow balls from box B?
Correct
Answer: c)
Number of blue balls from box A=x
Total balls from box A = x+20+14 = x+34
14/(x+34) = 2/7
98 = 2x+68
2x = 30
x=15
Total balls from box A=15+34=49
Number of yellow balls from box B=y
Total balls from box B=y+12+18=y+30
12/(y+30)=3/10
3y+90=120
Y=10
Total balls from box B=30+10=40
Required sum=15C_{2}/49C_{2 }+ 10C_{2}/40C_{2}
= > 5/56 + 3/52 = (260+168) /2912
= > 107/728
Incorrect
Answer: c)
Number of blue balls from box A=x
Total balls from box A = x+20+14 = x+34
14/(x+34) = 2/7
98 = 2x+68
2x = 30
x=15
Total balls from box A=15+34=49
Number of yellow balls from box B=y
Total balls from box B=y+12+18=y+30
12/(y+30)=3/10
3y+90=120
Y=10
Total balls from box B=30+10=40
Required sum=15C_{2}/49C_{2 }+ 10C_{2}/40C_{2}
= > 5/56 + 3/52 = (260+168) /2912
= > 107/728
Question 3 of 10
3. Question
Direction (1- 5): Study the following information carefully and answer the given questions
The given table shows the five different boxes contain three different color balls.
If the probability of one red ball drawing from box A is 1/3 and the probability of drawing either a Yellow or Red ball from box C is 4/5, and then find the difference between the probability of drawing 2 balls from box A such that both balls are either yellow or Red and the probability of drawing three balls from box C such that 2 of them are blue balls and one of them is red ball?
Correct
Answer: a)
Number of blue balls from box A=x
20C_{1}/(x+34)C_{1 }= 1/3
20/(x+34) = 1/3
X+34=60
x=26
Total balls from box A=26+34=60
Number of blue balls from box C=y
24C_{1}/(y+40)C_{1 }+ 16C_{1}/(y+40)C_{1}= 4/5
40/(y+40)=4/5
200=4y+160
y=10
Total balls from box C=10+24+16=50
2 red balls from box A=20C_{2}=20*19/2=190
2 yellow balls from box A=14*13/2=91
Total number of possible=60C_{2}=60*59/2=1770
Required probability=281/1770
From box C = (10C_{2}*24C_{1}) / 50C_{3}
=1080/19600
=27/490
Required difference=281/1770 – 27/490
=89900/867300
=899/8673
Incorrect
Answer: a)
Number of blue balls from box A=x
20C_{1}/(x+34)C_{1 }= 1/3
20/(x+34) = 1/3
X+34=60
x=26
Total balls from box A=26+34=60
Number of blue balls from box C=y
24C_{1}/(y+40)C_{1 }+ 16C_{1}/(y+40)C_{1}= 4/5
40/(y+40)=4/5
200=4y+160
y=10
Total balls from box C=10+24+16=50
2 red balls from box A=20C_{2}=20*19/2=190
2 yellow balls from box A=14*13/2=91
Total number of possible=60C_{2}=60*59/2=1770
Required probability=281/1770
From box C = (10C_{2}*24C_{1}) / 50C_{3}
=1080/19600
=27/490
Required difference=281/1770 – 27/490
=89900/867300
=899/8673
Question 4 of 10
4. Question
Direction (1- 5): Study the following information carefully and answer the given questions
The given table shows the five different boxes contain three different color balls.
The probability of drawing a Yellow ball from box E is 2/15 and the probability of drawing a blue ball from box E is 1/5.
Quantity I: The number of total balls in Box F is 5/6 times the total number of balls in Box E. The number of red balls from box F is 6 less than the number of Yellow balls from box F and the number of Blue balls from box F is 2/3 times the number of blue balls in Box E. Find the number of red balls in Box F?
Quantity II: The number of Red balls in Box G is 9 more than the half of the total number of Red balls in E. The number of Yellow balls from box G is same as the number of yellow balls in Box D. The probability of drawing a blue ball from box G is 11/70. Find the number of blue balls in Box G?
Correct
Answer: a)
Number of yellow balls from box E=x
Number of Blue balls from box E=y
X / (x+y+40) = 2/15
15x=2x+2y+80
13x – 2y=80 ——- (1)
Y / (x+y+40) = 1/5
x+y+40=5y
x – 4y + 40 = 0 ——– (2)
(1)*2 – (2)
25x=200
x=8
-4y=-40-8
y=12
Total number of balls from box E=12+8+40=60
Quantity I: The number of total balls in Box F is 5/6 times the total number of balls in Box E. The number of red balls from box F is 6 less than the number of Yellow balls from box F and the number of Blue balls from box F is 2/3 times the number of blue balls in Box E. Find the number of red balls in Box F?
Total number of balls from box F= (5/6)*60=50
Number of blue balls from box F= (2/3)*12=8
Number of red balls from box F=z
Number of yellow balls from F=z+6
Z+z+6+8=50
2z=36
Z=18 balls
Yellow balls from F=24
The total number of red balls in Box F (z) = 18 balls
Quantity II: The number of Red balls in Box G is 9 more than the half of the total number of Red balls in E. The number of Yellow balls from box G is same as the number of yellow balls in Box D. The probability of drawing a blue ball from box G is 11/70. Find the number of blue balls in Box G?
Red balls from box G = (40/2) + 9 = 29 balls
Yellow balls from G = 30
Number of blue balls from G=b
b / (30+29+b) = 11/70
70b=649 + 11b
59b=649
b=11
The total number of blue balls in Box G (b) = 11 balls
Hence, Quantity I > Quantity II
Incorrect
Answer: a)
Number of yellow balls from box E=x
Number of Blue balls from box E=y
X / (x+y+40) = 2/15
15x=2x+2y+80
13x – 2y=80 ——- (1)
Y / (x+y+40) = 1/5
x+y+40=5y
x – 4y + 40 = 0 ——– (2)
(1)*2 – (2)
25x=200
x=8
-4y=-40-8
y=12
Total number of balls from box E=12+8+40=60
Quantity I: The number of total balls in Box F is 5/6 times the total number of balls in Box E. The number of red balls from box F is 6 less than the number of Yellow balls from box F and the number of Blue balls from box F is 2/3 times the number of blue balls in Box E. Find the number of red balls in Box F?
Total number of balls from box F= (5/6)*60=50
Number of blue balls from box F= (2/3)*12=8
Number of red balls from box F=z
Number of yellow balls from F=z+6
Z+z+6+8=50
2z=36
Z=18 balls
Yellow balls from F=24
The total number of red balls in Box F (z) = 18 balls
Quantity II: The number of Red balls in Box G is 9 more than the half of the total number of Red balls in E. The number of Yellow balls from box G is same as the number of yellow balls in Box D. The probability of drawing a blue ball from box G is 11/70. Find the number of blue balls in Box G?
Red balls from box G = (40/2) + 9 = 29 balls
Yellow balls from G = 30
Number of blue balls from G=b
b / (30+29+b) = 11/70
70b=649 + 11b
59b=649
b=11
The total number of blue balls in Box G (b) = 11 balls
Hence, Quantity I > Quantity II
Question 5 of 10
5. Question
Direction (1- 5): Study the following information carefully and answer the given questions
The given table shows the five different boxes contain three different color balls.
The probability of getting one blue ball from Box A is 3/20 and the probability of getting one red ball from box D is 3/10. Total number of balls in Box A is what percentage more/less than the total number of balls in Box D?
Correct
Answer: e)
Probability of one blue ball from Box A = xC_{1}/(20+14+x)C_{1}= 3/20
x/(34 + x) = 3/20
20x = 102 + 3x
17x = 102
X = 6
Total number of balls in Box A = 34+6 = 40
Probability of one red ball from Box D = yC_{1 }/ (30+12+y)C_{1} = 3/10
y/(42+y)= 3/10
10y = 42*3 + 3y
7y = 42 * 3
Y = 3*6 = 18
Total number of balls in Box D = 42+18 = 60
Required percentage = [(60-40)/60] * 100
= > (20/60)*100 = (1/3)*100 = 33.33 %
Incorrect
Answer: e)
Probability of one blue ball from Box A = xC_{1}/(20+14+x)C_{1}= 3/20
x/(34 + x) = 3/20
20x = 102 + 3x
17x = 102
X = 6
Total number of balls in Box A = 34+6 = 40
Probability of one red ball from Box D = yC_{1 }/ (30+12+y)C_{1} = 3/10
y/(42+y)= 3/10
10y = 42*3 + 3y
7y = 42 * 3
Y = 3*6 = 18
Total number of balls in Box D = 42+18 = 60
Required percentage = [(60-40)/60] * 100
= > (20/60)*100 = (1/3)*100 = 33.33 %
Question 6 of 10
6. Question
Direction (6 – 10): Study the following information carefully and answer the given questions:
The given line graph shows the percentage of time taken by five different cars in the journey.
Total time taken in the journey= 60 hours
The given table shows the distance covered by car.
If the train crosses the car A running in the same direction in 36 seconds and the same train crosses the car E running the opposite direction in 6 seconds. Train runs at the speed of 60 kmph and also crosses the pole in 9 seconds. What is the difference between the distance covered by car A and car E?
Correct
Answer: a)
Length of the train=x
Speed of car A=y
Speed of car E=z
x=60*(5/18)*9
x=150 m
150 = (60 – y)*(5/18)*36
Y=45 kmph
150 = (60+z)*(5/18)*6
90=60+z
Z=30 kmph
Time of car A = (20/100)*60 =12 hours
Distance covered by car A=12*45=540 km
Time taken by car E = (20/100)*60 = 12 hours
Distance covered by car E=12*30=360 km
Required difference=540 – 360 =180 km
Incorrect
Answer: a)
Length of the train=x
Speed of car A=y
Speed of car E=z
x=60*(5/18)*9
x=150 m
150 = (60 – y)*(5/18)*36
Y=45 kmph
150 = (60+z)*(5/18)*6
90=60+z
Z=30 kmph
Time of car A = (20/100)*60 =12 hours
Distance covered by car A=12*45=540 km
Time taken by car E = (20/100)*60 = 12 hours
Distance covered by car E=12*30=360 km
Required difference=540 – 360 =180 km
Question 7 of 10
7. Question
Direction (6 – 10): Study the following information carefully and answer the given questions:
The given line graph shows the percentage of time taken by five different cars in the journey.
Total time taken in the journey= 60 hours
The given table shows the distance covered by car.
A train running at the speed of 72 kmph crosses another train in opposite direction at 54 kmph in 12 seconds. The first train crosses the 260m long bridge in 20 seconds and the second train crosses car C running in the opposite direction in 6 seconds, then find the distance covered by car C?
Correct
Answer: a)
Length of first train=x
Length of second train=y
Speed of car C=s
X+260 = 72*(5/18)*20
x=400-260
x=140 m
140+y = (72+54)*(5/18)*12
140+y = 420
Y=280 m
280=(s+54)*(5/18)*6
S=114 kmph
Time taken by car C = (30/100)*60=18 hours
Distance covered by car C =18*114=2052 km
Incorrect
Answer: a)
Length of first train=x
Length of second train=y
Speed of car C=s
X+260 = 72*(5/18)*20
x=400-260
x=140 m
140+y = (72+54)*(5/18)*12
140+y = 420
Y=280 m
280=(s+54)*(5/18)*6
S=114 kmph
Time taken by car C = (30/100)*60=18 hours
Distance covered by car C =18*114=2052 km
Question 8 of 10
8. Question
Direction (6 – 10): Study the following information carefully and answer the given questions:
The given line graph shows the percentage of time taken by five different cars in the journey.
Total time taken in the journey= 60 hours
The given table shows the distance covered by car.
A train can travel 50% faster than a car A. Both start from Chennai at the same time and reach Bangalore 150 km away from Chennai at the same time. On the way, however, the train lost about 25 minutes while stopping at the stations. If the speed of Car A, Car C and Car E in the ratio of 6:4:3 then find the average of the distance covered by all the five cars together?
Correct
Answer: a)
Speed of car A=x
Speed of train= (150/100)*x=3x/2
150/x – 150/(3x/2) = 25/60
X=120 kmph
Speed of car C = (4/6)*120=80 kmph
Speed of car E= (3/6)*120=60 kmph
Time taken by car A= (20/100)*60=12 hours
Time taken by car C= (30/100)*60=18 hours
Time taken by car E= (20/100)*60=12 hours
Distance covered by car A=12*120=1440 km
Distance covered by car C=18*80=1440 km
Distance covered by car E=60*12=720 km
Average distance = (1440+1440+720+450+180)/5=846 km
Incorrect
Answer: a)
Speed of car A=x
Speed of train= (150/100)*x=3x/2
150/x – 150/(3x/2) = 25/60
X=120 kmph
Speed of car C = (4/6)*120=80 kmph
Speed of car E= (3/6)*120=60 kmph
Time taken by car A= (20/100)*60=12 hours
Time taken by car C= (30/100)*60=18 hours
Time taken by car E= (20/100)*60=12 hours
Distance covered by car A=12*120=1440 km
Distance covered by car C=18*80=1440 km
Distance covered by car E=60*12=720 km
Average distance = (1440+1440+720+450+180)/5=846 km
Question 9 of 10
9. Question
Direction (6 – 10): Study the following information carefully and answer the given questions:
The given line graph shows the percentage of time taken by five different cars in the journey.
Total time taken in the journey= 60 hours
The given table shows the distance covered by car.
The sum of the speed of car B and car D is what percent of the distance covered by car C, if the speed of car C is 40kmph?
Correct
Answer: c)
Time taken by car B= (25/100)*60 =15 hours
Time taken by car D= (5/100)*60 =3 hours
Speed of car B= 450/15=30 kmph
Speed of car D=180/3=60 kmph
Time taken by car C= (30/100)*60=18
Distance covered by car C=40*18=720 km
Required percentage = [(30+60)/720] * 100=12.5%
Incorrect
Answer: c)
Time taken by car B= (25/100)*60 =15 hours
Time taken by car D= (5/100)*60 =3 hours
Speed of car B= 450/15=30 kmph
Speed of car D=180/3=60 kmph
Time taken by car C= (30/100)*60=18
Distance covered by car C=40*18=720 km
Required percentage = [(30+60)/720] * 100=12.5%
Question 10 of 10
10. Question
Direction (6 – 10): Study the following information carefully and answer the given questions:
The given line graph shows the percentage of time taken by five different cars in the journey.
Total time taken in the journey= 60 hours
The given table shows the distance covered by car.
If the total distance covered by all the car is 1350 km and the distance covered by car A, car C and car E in the ratio of 2 : 3 : 1. What is the average speed of all the cars together?
Direction (1- 5): Study the following information carefully and answer the given questions
The given table shows the five different boxes contain three different color balls.
1) Probability of drawing one blue from box E is 1/4 and the probability of drawing one red ball from box E is 2/15. The probability of drawing one yellow ball from box D is 5/12. Find the sum of the probabilities of drawing one yellow ball from box E and one red ball from box D?
a) 31/30
b) 53/62
c) 61/52
d) 73/67
e) None of these
2) The probability of drawing one yellow ball from box A is 2/7. Probability of drawing one red ball from box B is 3/10. Find the sum of the probability of drawing 2 blue balls from box A and the probability of drawing two yellow balls from box B?
a) 137/728
b) 127/728
c) 107/728
d) 117/728
e) None of these
3) If the probability of one red ball drawing from box A is 1/3 and the probability of drawing either a Yellow or Red ball from box C is 4/5, and then find the difference between the probability of drawing 2 balls from box A such that both balls are either yellow or Red and the probability of drawing three balls from box C such that 2 of them are blue balls and one of them is red ball?
a) 899/8673
b) 899/6873
c) 899/8373
d) 879/8763
e) None of these
4) The probability of drawing a Yellow ball from box E is 2/15 and the probability of drawing a blue ball from box E is 1/5.
Quantity I: The number of total balls in Box F is 5/6 times the total number of balls in Box E. The number of red balls from box F is 6 less than the number of Yellow balls from box F and the number of Blue balls from box F is 2/3 times the number of blue balls in Box E. Find the number of red balls in Box F?
Quantity II: The number of Red balls in Box G is 9 more than the half of the total number of Red balls in E. The number of Yellow balls from box G is same as the number of yellow balls in Box D. The probability of drawing a blue ball from box G is 11/70. Find the number of blue balls in Box G?
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≤ Quantity II
d) Quantity I ≥ Quantity II
e) Quantity I = Quantity II (or) Relationship cannot be determined.
5) The probability of getting one blue ball from Box A is 3/20 and the probability of getting one red ball from box D is 3/10. Total number of balls in Box A is what percentage more/less than the total number of balls in Box D?
a) 55.55% less
b) 66.67% more
c) 44.44% less
d) 22.22% more
e) 33.33% less
Direction (6 – 10): Study the following information carefully and answer the given questions:
The given line graph shows the percentage of time taken by five different cars in the journey.
Total time taken in the journey= 60 hours
The given table shows the distance covered by car.
6) If the train crosses the car A running in the same direction in 36 seconds and the same train crosses the car E running the opposite direction in 6 seconds. Train runs at the speed of 60 kmph and also crosses the pole in 9 seconds. What is the difference between the distance covered by car A and car E?
a) 180 km
b) 220 km
c) 200 km
d) 150 km
e) None of these
7) A train running at the speed of 72 kmph crosses another train in opposite direction at 54 kmph in 12 seconds. The first train crosses the 260m long bridge in 20 seconds and the second train crosses car C running in the opposite direction in 6 seconds, then find the distance covered by car C?
a) 2052 km
b) 1082 km
c) 2064 km
d) 2084 km
e) None of these
8) A train can travel 50% faster than a car A. Both start from Chennai at the same time and reach Bangalore 150 km away from Chennai at the same time. On the way, however, the train lost about 25 minutes while stopping at the stations. If the speed of Car A, Car C and Car E in the ratio of 6:4:3 then find the average of the distance covered by all the five cars together?
a) 846 km
b) 746 km
c) 946 km
d) 646 km
e) None of these
9) The sum of the speed of car B and car D is what percent of the distance covered by car C, if the speed of car C is 40kmph?
a) 15.25%
b) 10.75%
c) 12.5%
d) 8%
e) None of these
10) If the total distance covered by all the car is 1350 km and the distance covered by car A, car C and car E in the ratio of 2 : 3 : 1. What is the average speed of all the cars together?
a) 28 kmph
b) 25 kmph
c) 26 kmph
d) 24 kmph
e) None of these
Answers :
Direction (1-5) :
1) Answer: a)
Number of yellow balls from box E=x
Number of blue balls from box E=y
Total number of balls from box E=40+x+y
Number of Red balls from box D=z
Total number of balls from box D = z+30+12=z+42
y/(40 + x + y) = 1/4
40+x+y=4y
X – 3y + 40=0 ——– (1)
40/(40+x+y) = 2/15
600=80+2x+2y
2x+2y-520=0
x+y-260 =0 ———- (2)
By substituting 1 and 2, we get,
-4y=-300
Y=75
X – (3)*75+40=0
x=185
Total number of balls from box E=185+75+40=300
30/(z+42)=5/12
360 = 5z+210
5Z=150
Z = 30
Total balls from box D=30+42=72
Required sum=185C_{1}/300C_{1 }+ 30C_{1}/72C_{1}
= > (37 + 25)/60 = 62/60 = 31/30
2) Answer: c)
Number of blue balls from box A=x
Total balls from box A = x+20+14 = x+34
14/(x+34) = 2/7
98 = 2x+68
2x = 30
x=15
Total balls from box A=15+34=49
Number of yellow balls from box B=y
Total balls from box B=y+12+18=y+30
12/(y+30)=3/10
3y+90=120
Y=10
Total balls from box B=30+10=40
Required sum=15C_{2}/49C_{2 }+ 10C_{2}/40C_{2}
= > 5/56 + 3/52 = (260+168) /2912
= > 107/728
3) Answer: a)
Number of blue balls from box A=x
20C_{1}/(x+34)C_{1 }= 1/3
20/(x+34) = 1/3
X+34=60
x=26
Total balls from box A=26+34=60
Number of blue balls from box C=y
24C_{1}/(y+40)C_{1 }+ 16C_{1}/(y+40)C_{1}= 4/5
40/(y+40)=4/5
200=4y+160
y=10
Total balls from box C=10+24+16=50
2 red balls from box A=20C_{2}=20*19/2=190
2 yellow balls from box A=14*13/2=91
Total number of possible=60C_{2}=60*59/2=1770
Required probability=281/1770
From box C = (10C_{2}*24C_{1}) / 50C_{3}
=1080/19600
=27/490
Required difference=281/1770 – 27/490
=89900/867300
=899/8673
4) Answer: a)
Number of yellow balls from box E=x
Number of Blue balls from box E=y
X / (x+y+40) = 2/15
15x=2x+2y+80
13x – 2y=80 ——- (1)
Y / (x+y+40) = 1/5
x+y+40=5y
x – 4y + 40 = 0 ——– (2)
(1)*2 – (2)
25x=200
x=8
-4y=-40-8
y=12
Total number of balls from box E=12+8+40=60
Quantity I: The number of total balls in Box F is 5/6 times the total number of balls in Box E. The number of red balls from box F is 6 less than the number of Yellow balls from box F and the number of Blue balls from box F is 2/3 times the number of blue balls in Box E. Find the number of red balls in Box F?
Total number of balls from box F= (5/6)*60=50
Number of blue balls from box F= (2/3)*12=8
Number of red balls from box F=z
Number of yellow balls from F=z+6
Z+z+6+8=50
2z=36
Z=18 balls
Yellow balls from F=24
The total number of red balls in Box F (z) = 18 balls
Quantity II: The number of Red balls in Box G is 9 more than the half of the total number of Red balls in E. The number of Yellow balls from box G is same as the number of yellow balls in Box D. The probability of drawing a blue ball from box G is 11/70. Find the number of blue balls in Box G?
Red balls from box G = (40/2) + 9 = 29 balls
Yellow balls from G = 30
Number of blue balls from G=b
b / (30+29+b) = 11/70
70b=649 + 11b
59b=649
b=11
The total number of blue balls in Box G (b) = 11 balls
Hence, Quantity I > Quantity II
5) Answer: e)
Probability of one blue ball from Box A = xC_{1}/(20+14+x)C_{1}= 3/20
x/(34 + x) = 3/20
20x = 102 + 3x
17x = 102
X = 6
Total number of balls in Box A = 34+6 = 40
Probability of one red ball from Box D = yC_{1 }/ (30+12+y)C_{1} = 3/10
y/(42+y)= 3/10
10y = 42*3 + 3y
7y = 42 * 3
Y = 3*6 = 18
Total number of balls in Box D = 42+18 = 60
Required percentage = [(60-40)/60] * 100
= > (20/60)*100 = (1/3)*100 = 33.33 %
Direction (6-10) :
6) Answer: a)
Length of the train=x
Speed of car A=y
Speed of car E=z
x=60*(5/18)*9
x=150 m
150 = (60 – y)*(5/18)*36
Y=45 kmph
150 = (60+z)*(5/18)*6
90=60+z
Z=30 kmph
Time of car A = (20/100)*60 =12 hours
Distance covered by car A=12*45=540 km
Time taken by car E = (20/100)*60 = 12 hours
Distance covered by car E=12*30=360 km
Required difference=540 – 360 =180 km
7) Answer: a)
Length of first train=x
Length of second train=y
Speed of car C=s
X+260 = 72*(5/18)*20
x=400-260
x=140 m
140+y = (72+54)*(5/18)*12
140+y = 420
Y=280 m
280=(s+54)*(5/18)*6
S=114 kmph
Time taken by car C = (30/100)*60=18 hours
Distance covered by car C =18*114=2052 km
8) Answer: a)
Speed of car A=x
Speed of train= (150/100)*x=3x/2
150/x – 150/(3x/2) = 25/60
X=120 kmph
Speed of car C = (4/6)*120=80 kmph
Speed of car E= (3/6)*120=60 kmph
Time taken by car A= (20/100)*60=12 hours
Time taken by car C= (30/100)*60=18 hours
Time taken by car E= (20/100)*60=12 hours
Distance covered by car A=12*120=1440 km
Distance covered by car C=18*80=1440 km
Distance covered by car E=60*12=720 km
Average distance = (1440+1440+720+450+180)/5=846 km