SBI Clerk/ IDBI and South Indian Bank Mains Quantitative Aptitude Questions 2019 (Day-04)
Dear Aspirants, Our IBPS Guide team is providing a new series of Quantitative Aptitude Questions for SBI Clerk/IDBI and South Indian Bank Mains 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 new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
Are You preparing for Bank exams 2019? Start your preparation with Free Mock test Series.
SBI Clerk/ IDBI and South Indian Bank Mains Quantitative Aptitude Questions 2019 (Day-04)
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
I) 5x + 4y â€“ 20xy = 1
II) 8 + 6xy â€“ 6x â€“ 8y = 0
Correct
Answer: c)
I) 5x + 4y â€“ 20xy â€“ 1 = 0
Divide by 20,
(5x/20) + (4y/20) â€“ (20xy/20) â€“ (1/20) = 0
(x/4) + (y/5) â€“ xy â€“ (1/20) = 0
xy â€“ (x/4) â€“ (y/5) + (1/20) = 0
x (x â€“ Â¼) â€“ 1/5 (y â€“ 1/4) = 0
(x â€“ 1/5) (y â€“ Â¼) = 0
x = 1/5, y = Â¼
II) 8 + 6xy â€“ 6x â€“ 8y = 0
Divide by 6,
(8/6) + (6xy/6) â€“ (6x/6) â€“ (8y/6) = 0
(4/3) + xy â€“ x â€“ (4y/3) = 0
X (y â€“ 1) â€“ (4/3) (y â€“ 1) = 0
(x â€“ 4/3) (y â€“ 1) = 0
x = 4/3, y = 1
Canâ€™t be determined
Incorrect
Answer: c)
I) 5x + 4y â€“ 20xy â€“ 1 = 0
Divide by 20,
(5x/20) + (4y/20) â€“ (20xy/20) â€“ (1/20) = 0
(x/4) + (y/5) â€“ xy â€“ (1/20) = 0
xy â€“ (x/4) â€“ (y/5) + (1/20) = 0
x (x â€“ Â¼) â€“ 1/5 (y â€“ 1/4) = 0
(x â€“ 1/5) (y â€“ Â¼) = 0
x = 1/5, y = Â¼
II) 8 + 6xy â€“ 6x â€“ 8y = 0
Divide by 6,
(8/6) + (6xy/6) â€“ (6x/6) â€“ (8y/6) = 0
(4/3) + xy â€“ x â€“ (4y/3) = 0
X (y â€“ 1) â€“ (4/3) (y â€“ 1) = 0
(x â€“ 4/3) (y â€“ 1) = 0
x = 4/3, y = 1
Canâ€™t be determined
Question 2 of 10
2. Question
I) 5x^{2} â€“ 20x + 15 = 0
II) y + âˆ›125 = 0.9
Correct
Answer: a)
I) 5x^{2} â€“ 20x + 15 = 0
5x^{2} â€“ 5x â€“ 15x + 15 = 0
5x (x â€“ 1) â€“ 15 (x â€“ 1) = 0
(5x â€“ 15) (x â€“ 1) = 0
x = 3, 1
II) y + âˆ›125 = 0.9
y + 0.5 = 0.9
y = 0.9 â€“ 0.5 = 0.4
x > y
Incorrect
Answer: a)
I) 5x^{2} â€“ 20x + 15 = 0
5x^{2} â€“ 5x â€“ 15x + 15 = 0
5x (x â€“ 1) â€“ 15 (x â€“ 1) = 0
(5x â€“ 15) (x â€“ 1) = 0
x = 3, 1
II) y + âˆ›125 = 0.9
y + 0.5 = 0.9
y = 0.9 â€“ 0.5 = 0.4
x > y
Question 3 of 10
3. Question
I) x^{2} + 7^{3/2} = (7^{1/2} + 7) x
II) y^{13} â€“ (343^{9/2} / âˆšy) = 0
Correct
Answer: e)
I) x^{2} + 7^{3/2} = (7^{1/2} + 7) x
x^{2} â€“ (âˆš7 + 7) x + 7âˆš7 = 0
x = 7, âˆš7
II) y^{13} â€“ (343^{9/2} / âˆšy) = 0
[(y^{13} * y^{1/2}) â€“ (7^{3})^{9/2}] / âˆšy = 0
y^{27/2} â€“ 7^{27/2} = 0
y^{27/2} = 7^{27/2}
y = 7
x â‰¤ y
Incorrect
Answer: e)
I) x^{2} + 7^{3/2} = (7^{1/2} + 7) x
x^{2} â€“ (âˆš7 + 7) x + 7âˆš7 = 0
x = 7, âˆš7
II) y^{13} â€“ (343^{9/2} / âˆšy) = 0
[(y^{13} * y^{1/2}) â€“ (7^{3})^{9/2}] / âˆšy = 0
y^{27/2} â€“ 7^{27/2} = 0
y^{27/2} = 7^{27/2}
y = 7
x â‰¤ y
Question 4 of 10
4. Question
I) 13x^{3} + 142x^{2} + 216x = 5x^{3} + 228x^{2}
II) 5y^{2} – 21y + 18 = 0
Correct
Answer: c)
I) 13x^{3} + 142x^{2} + 216x = 5x^{3} + 228x^{2}
8x^{3} – 86x^{2} + 216x = 0
x (8x^{2} – 86x + 216) = 0
x [8x^{2} – 32x – 54x + 216] = 0
x [8x (x – 4) – 54 (x – 4)] = 0
x [(8x – 54) (x – 4)] = 0
x = 0, 54/8, 4
x = 0, 6.75, 4
II) 5y^{2} – 21y + 18 = 0
5y^{2} – 15y â€“ 6y â€“ 12 = 0
5y (y – 3) â€“ 6 (y – 3) = 0
(5y â€“ 6) (y – 3) = 0
y = 6/5, 3 = 1.2, 3
Canâ€™t be determined
Incorrect
Answer: c)
I) 13x^{3} + 142x^{2} + 216x = 5x^{3} + 228x^{2}
8x^{3} – 86x^{2} + 216x = 0
x (8x^{2} – 86x + 216) = 0
x [8x^{2} – 32x – 54x + 216] = 0
x [8x (x – 4) – 54 (x – 4)] = 0
x [(8x – 54) (x – 4)] = 0
x = 0, 54/8, 4
x = 0, 6.75, 4
II) 5y^{2} – 21y + 18 = 0
5y^{2} – 15y â€“ 6y â€“ 12 = 0
5y (y – 3) â€“ 6 (y – 3) = 0
(5y â€“ 6) (y – 3) = 0
y = 6/5, 3 = 1.2, 3
Canâ€™t be determined
Question 5 of 10
5. Question
I) 8x^{2} â€“ (8 + 4âˆš3) x + 4âˆš3 = 0
II) 12y^{2} â€“ (27 + 4âˆš3) y + 9âˆš3 = 0
Correct
Answer: c)
I) 8x^{2} â€“ (8 + 4âˆš3) x + 2âˆš3 = 0
x = 8/8, 4âˆš3/8
x = 0.866, 1
II) 12y^{2} â€“ (27 + 4âˆš3) y + 9âˆš3 = 0
y = 27/12, 4âˆš3/12
y = 2.25, 0.577
Canâ€™t be determined
Incorrect
Answer: c)
I) 8x^{2} â€“ (8 + 4âˆš3) x + 2âˆš3 = 0
x = 8/8, 4âˆš3/8
x = 0.866, 1
II) 12y^{2} â€“ (27 + 4âˆš3) y + 9âˆš3 = 0
y = 27/12, 4âˆš3/12
y = 2.25, 0.577
Canâ€™t be determined
Question 6 of 10
6. Question
If the time taken by Q is twice the time taken by P to cover the given distance and the ratio between the distance covered by P to that of Q isÂ Â Â Â Â Â Â 5 : 6, then find the speed of Q?
Correct
Answer: a)
The distance covered by P = 105 * 8 = 840 Km
The time taken by Q = 2 * the time taken by P = 2 * 8 = 16 hr
The distance covered by Q = (840/5) * 6 = 1008 Km
Speed of Q = Distance/Time = 1008/16 = 63 Km/hr
Incorrect
Answer: a)
The distance covered by P = 105 * 8 = 840 Km
The time taken by Q = 2 * the time taken by P = 2 * 8 = 16 hr
The distance covered by Q = (840/5) * 6 = 1008 Km
Speed of Q = Distance/Time = 1008/16 = 63 Km/hr
Question 7 of 10
7. Question
With the help of question 6, find the average speed of P and Q?
Correct
Answer: c)
The speed of P = 105 Km/hr
The distance covered by P = 105 * 8 = 840 Km
The time taken by Q = 2 * the time taken by P = 2 * 8 = 16 hr
The distance covered by Q = (840/5) * 6 = 1008 Km
Speed of Q = Distance/Time = 1008/16 = 63 Km/hr
Average Speed = Total distance/Total time
= > (840 + 1008) / (8 + 16)
= > 1848/24 = 77 Km/hr
Incorrect
Answer: c)
The speed of P = 105 Km/hr
The distance covered by P = 105 * 8 = 840 Km
The time taken by Q = 2 * the time taken by P = 2 * 8 = 16 hr
The distance covered by Q = (840/5) * 6 = 1008 Km
Speed of Q = Distance/Time = 1008/16 = 63 Km/hr
Average Speed = Total distance/Total time
= > (840 + 1008) / (8 + 16)
= > 1848/24 = 77 Km/hr
Question 8 of 10
8. Question
If the distance covered by S is 912 km and the time taken by Q to cover the given distance is 12 hours, then find the ratio between the speeds of Q to that of R?
Correct
Answer: d)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
Total degree for circle = 360^{0}
X^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The distance covered by S = 912 km = y^{0}
Y^{0} = (912/40)*3 = 68.4^{0}
The distance covered by Q = x^{0} = 144 â€“ 68.4^{0} = 75.6^{0}
The distance covered by Q = 75.6*(40/3) = 1008 Km
The time taken by Q to cover the given distance = 12 hours
The speed of Q = 1008/12 = 84 Km
The distance covered by R = 72^{0} = 72*(40/3) = 960 Km
The speed of R = (960/12) = 80 Km/hr
Required ratio = 84 : 80 = 21 : 20
Incorrect
Answer: d)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
Total degree for circle = 360^{0}
X^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The distance covered by S = 912 km = y^{0}
Y^{0} = (912/40)*3 = 68.4^{0}
The distance covered by Q = x^{0} = 144 â€“ 68.4^{0} = 75.6^{0}
The distance covered by Q = 75.6*(40/3) = 1008 Km
The time taken by Q to cover the given distance = 12 hours
The speed of Q = 1008/12 = 84 Km
The distance covered by R = 72^{0} = 72*(40/3) = 960 Km
The speed of R = (960/12) = 80 Km/hr
Required ratio = 84 : 80 = 21 : 20
Question 9 of 10
9. Question
Find the total time taken by Q and T to cover the given distance, if the speed of Q is 65 Km/hr and the distance covered by S is 880 Km?
Correct
Answer: b)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
Total degree for circle = 360^{0}
x^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The distance covered by S = 880 km = y^{0}
y^{0} = (880/40) * 3 = 66^{0}
The distance covered by Q = x^{0} = 144 â€“ 66^{0} = 78^{0} = 78*(40/3) = 1040 Km
The speed of Q = 65 Km/hr
The time taken by Q to cover the distance
= > (1040/65) = 16 hr
The distance covered by T = 81^{0} = 81*(40/3) = 1080 Km
The time taken by T to cover the distance
= > (1080/50) = 21.6 hr (or) 21 hr 36 min
Required time = 16 + 21 hr 36 min = 37 hr 36 min
Incorrect
Answer: b)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
Total degree for circle = 360^{0}
x^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The distance covered by S = 880 km = y^{0}
y^{0} = (880/40) * 3 = 66^{0}
The distance covered by Q = x^{0} = 144 â€“ 66^{0} = 78^{0} = 78*(40/3) = 1040 Km
The speed of Q = 65 Km/hr
The time taken by Q to cover the distance
= > (1040/65) = 16 hr
The distance covered by T = 81^{0} = 81*(40/3) = 1080 Km
The time taken by T to cover the distance
= > (1080/50) = 21.6 hr (or) 21 hr 36 min
Required time = 16 + 21 hr 36 min = 37 hr 36 min
Question 10 of 10
10. Question
Find the sum of the average distance covered by P, R and T together and the average distance covered by Q and S together?
Correct
Answer: a)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
The total distance covered by P, R and T together
= > 216 * (40/3) = 2880 Km
The average distance covered by P, R and T together
= > 2880/3 = 960 Km
Total degree for circle = 360^{0}
x^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The total distance covered by Q and S together = 144*(40/3) = 1920 Km
The average distance covered by Q and S together
= > 1920/2 = 960 Km
Required difference = 960 + 960 = 1920 Km
Incorrect
Answer: a)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
The total distance covered by P, R and T together
= > 216 * (40/3) = 2880 Km
The average distance covered by P, R and T together
= > 2880/3 = 960 Km
Total degree for circle = 360^{0}
x^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The total distance covered by Q and S together = 144*(40/3) = 1920 Km
The average distance covered by Q and S together
= > 1920/2 = 960 Km
Required difference = 960 + 960 = 1920 Km
Directions (Q. 1 – 5): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give answer as,
Directions (Q. 6 â€“ 10): Study the following information carefully and answer the questions given below.
The following pie chart shows the distance covered by 5 different persons. (In degrees)
The table below shows the speed and time taken by the person to cover the given distance. Some values are missing here.Â
6) If the time taken by Q is twice the time taken by P to cover the given distance and the ratio between the distance covered by P to that of Q isÂ Â Â Â Â Â Â 5 : 6, then find the speed of Q?
a) 63 Km/hr
b) 57 Km/hr
c) 68 Km/hr
d) 52 Km/hr
e) None of these
7) With the help of question 6, find the average speed of P and Q?
a) 93 Km/hr
b) 65 Km/hr
c) 77 Km/hr
d) 84 Km/hr
e) None of these
8) If the distance covered by S is 912 km and the time taken by Q to cover the given distance is 12 hours, then find the ratio between the speeds of Q to that of R?
a) 29 : 15
b) 15 : 13
c) 23 : 19
d) 21 : 20
e) None of these
9) Find the total time taken by Q and T to cover the given distance, if the speed of Q is 65 Km/hr and the distance covered by S is 880 Km?
a) 42 hr 28 min
b) 37 hr 36 min
c) 35 hr 24 min
d) 40 hr 40 min
e) None of these
10) Find the sum of the average distance covered by P, R and T together and the average distance covered by Q and S together?
a) 1920 Km
b) 1840 Km
c) 1720 Km
d) 2160 Km
e) None of these
Answers :
Direction (1-5) :
1) Answer: c)
I) 5x + 4y â€“ 20xy â€“ 1 = 0
Divide by 20,
(5x/20) + (4y/20) â€“ (20xy/20) â€“ (1/20) = 0
(x/4) + (y/5) â€“ xy â€“ (1/20) = 0
xy â€“ (x/4) â€“ (y/5) + (1/20) = 0
x (x â€“ Â¼) â€“ 1/5 (y â€“ 1/4) = 0
(x â€“ 1/5) (y â€“ Â¼) = 0
x = 1/5, y = Â¼
II) 8 + 6xy â€“ 6x â€“ 8y = 0
Divide by 6,
(8/6) + (6xy/6) â€“ (6x/6) â€“ (8y/6) = 0
(4/3) + xy â€“ x â€“ (4y/3) = 0
X (y â€“ 1) â€“ (4/3) (y â€“ 1) = 0
(x â€“ 4/3) (y â€“ 1) = 0
x = 4/3, y = 1
Canâ€™t be determined
2) Answer: a)
I) 5x^{2} â€“ 20x + 15 = 0
5x^{2} â€“ 5x â€“ 15x + 15 = 0
5x (x â€“ 1) â€“ 15 (x â€“ 1) = 0
(5x â€“ 15) (x â€“ 1) = 0
x = 3, 1
II) y + âˆ›125 = 0.9
y + 0.5 = 0.9
y = 0.9 â€“ 0.5 = 0.4
x > y
3) Answer: e)
I) x^{2} + 7^{3/2} = (7^{1/2} + 7) x
x^{2} â€“ (âˆš7 + 7) x + 7âˆš7 = 0
x = 7, âˆš7
II) y^{13} â€“ (343^{9/2} / âˆšy) = 0
[(y^{13} * y^{1/2}) â€“ (7^{3})^{9/2}] / âˆšy = 0
y^{27/2} â€“ 7^{27/2} = 0
y^{27/2} = 7^{27/2}
y = 7
x â‰¤ y
4) Answer: c)
I) 13x^{3} + 142x^{2} + 216x = 5x^{3} + 228x^{2}
8x^{3} – 86x^{2} + 216x = 0
x (8x^{2} – 86x + 216) = 0
x [8x^{2} – 32x – 54x + 216] = 0
x [8x (x – 4) – 54 (x – 4)] = 0
x [(8x – 54) (x – 4)] = 0
x = 0, 54/8, 4
x = 0, 6.75, 4
II) 5y^{2} – 21y + 18 = 0
5y^{2} – 15y â€“ 6y â€“ 12 = 0
5y (y – 3) â€“ 6 (y – 3) = 0
(5y â€“ 6) (y – 3) = 0
y = 6/5, 3 = 1.2, 3
Canâ€™t be determined
5) Answer: c)
I) 8x^{2} â€“ (8 + 4âˆš3) x + 2âˆš3 = 0
x = 8/8, 4âˆš3/8
x = 0.866, 1
II) 12y^{2} â€“ (27 + 4âˆš3) y + 9âˆš3 = 0
y = 27/12, 4âˆš3/12
y = 2.25, 0.577
Canâ€™t be determined
Direction (6-10) :
6) Answer: a)
The distance covered by P = 105 * 8 = 840 Km
The time taken by Q = 2 * the time taken by P = 2 * 8 = 16 hr
The distance covered by Q = (840/5) * 6 = 1008 Km
Speed of Q = Distance/Time = 1008/16 = 63 Km/hr
7) Answer: c)
The speed of P = 105 Km/hr
The distance covered by P = 105 * 8 = 840 Km
The time taken by Q = 2 * the time taken by P = 2 * 8 = 16 hr
The distance covered by Q = (840/5) * 6 = 1008 Km
Speed of Q = Distance/Time = 1008/16 = 63 Km/hr
Average Speed = Total distance/Total time
= > (840 + 1008) / (8 + 16)
= > 1848/24 = 77 Km/hr
8) Answer: d)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
Total degree for circle = 360^{0}
X^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The distance covered by S = 912 km = y^{0}
Y^{0} = (912/40)*3 = 68.4^{0}
The distance covered by Q = x^{0} = 144 â€“ 68.4^{0} = 75.6^{0}
The distance covered by Q = 75.6*(40/3) = 1008 Km
The time taken by Q to cover the given distance = 12 hours
The speed of Q = 1008/12 = 84 Km
The distance covered by R = 72^{0} = 72*(40/3) = 960 Km
The speed of R = (960/12) = 80 Km/hr
Required ratio = 84 : 80 = 21 : 20
9) Answer: b)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
Total degree for circle = 360^{0}
x^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The distance covered by S = 880 km = y^{0}
y^{0} = (880/40) * 3 = 66^{0}
The distance covered by Q = x^{0} = 144 â€“ 66^{0} = 78^{0} = 78*(40/3) = 1040 Km
The speed of Q = 65 Km/hr
The time taken by Q to cover the distance
= > (1040/65) = 16 hr
The distance covered by T = 81^{0} = 81*(40/3) = 1080 Km
The time taken by T to cover the distance
= > (1080/50) = 21.6 hr (or) 21 hr 36 min
Required time = 16 + 21 hr 36 min = 37 hr 36 min
10) Answer: a)
The distance covered by P = 105*8 = 840 Km
63^{0} = 840 km
3^{0} = 40 km
1^{0} = (40/3) Km
The total distance covered by P, R and T together
= > 216 * (40/3) = 2880 Km
The average distance covered by P, R and T together
= > 2880/3 = 960 Km
Total degree for circle = 360^{0}
x^{0} + y^{0} = 360 â€“ (63 + 72 + 81) = 144^{0}
The total distance covered by Q and S together = 144*(40/3) = 1920 Km
The average distance covered by Q and S together
= > 1920/2 = 960 Km
Required difference = 960 + 960 = 1920 Km
IBPSGuide Recommends Affairs Cloud Current affairs PDF