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Directions (6 – 10): What approximate value should come in the place of question mark (?) in the following questions?
42 % of 2149 + (3/11) of 18149 – 30 % of 1699 =?
Correct
Answer: a)
42 % of 2150 + (3/11) of 18150 – 30 % of 1700 = x
X = (42/100)*2150 + (3/11) * 18150 – (30/100)*1700
X = 903 + 4950 – 510
X = 5343 = 5340
Incorrect
Answer: a)
42 % of 2150 + (3/11) of 18150 – 30 % of 1700 = x
X = (42/100)*2150 + (3/11) * 18150 – (30/100)*1700
X = 903 + 4950 – 510
X = 5343 = 5340
Question 11 of 20
11. Question
The area of a square is 1764 sq cm whose side is 2 times of radius of a circle. The circumference of the circle is equal to the perimeter of rectangle. What is the breadth of rectangle, if the length of the rectangle is 42 cm?
Correct
Answer: d)
Area of square = 1764 Sq cm = a^{2}
Side (a) = 42 cm
Side of square = 2 * Radius of circle
Radius = 42/2 = 21 cm
Circumference of the circle = 2πr = 2*(22/7)*21 = 132 cm
Length = 42 cm
Perimeter of rectangle = 2(42 + b) = 132
42 + b = 66
Breadth = 66 – 42 = 24 cm
Incorrect
Answer: d)
Area of square = 1764 Sq cm = a^{2}
Side (a) = 42 cm
Side of square = 2 * Radius of circle
Radius = 42/2 = 21 cm
Circumference of the circle = 2πr = 2*(22/7)*21 = 132 cm
Length = 42 cm
Perimeter of rectangle = 2(42 + b) = 132
42 + b = 66
Breadth = 66 – 42 = 24 cm
Question 12 of 20
12. Question
P, Q and R started a business by investing Rs. 15000, Rs. 18000 and Rs. 24000 respectively. After 3 months, P invested 40 % of the initial investment more. And after 5 months, R withdraws Rs. 4000. Find the share of Q, if the total profit at the end of the year is Rs. 61370?
4 years ago, the ratio of ages of P and Q is 4: 3. The present age of R is one-fifth of the present age of Q. Sum of the ages of P, Q and R, after 4 years will be 74 years. Find the present age of Q?
Correct
Answer: b)
4 years ago, the ratio of ages of P and Q = 4: 3 (4x, 3x)
Present age of P and Q = 4x + 4, 3x + 4
The present age of R = (1/5)*(3x + 4) = (3x + 4)/5
Sum of the ages of P, Q and R, after 4 years = 74 years
4x + 4 + 3x + 4 + (3x + 4)/5 = 74 – 12
35x + 40 + 3x + 4 = 62*5
38x = 310 – 44
X = 266/38 = 7
The present age of Q = 3x + 4 = 25 years
Incorrect
Answer: b)
4 years ago, the ratio of ages of P and Q = 4: 3 (4x, 3x)
Present age of P and Q = 4x + 4, 3x + 4
The present age of R = (1/5)*(3x + 4) = (3x + 4)/5
Sum of the ages of P, Q and R, after 4 years = 74 years
4x + 4 + 3x + 4 + (3x + 4)/5 = 74 – 12
35x + 40 + 3x + 4 = 62*5
38x = 310 – 44
X = 266/38 = 7
The present age of Q = 3x + 4 = 25 years
Question 14 of 20
14. Question
Two pipes can fill a cistern in 15 and 21 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, 15 minutes extra are taken for the cistern to be filled up. If the cistern is full, in what time would the leak empty it?
Correct
Answer: d)
If both filling pipes are working together, portion of the cistern filled in an hour
= 1/15 + 1/21 = (21 + 15)/(15*21) = 4/35
Hence time taken to fill the cistern = 35/4 hours = 8 3/4 hours = 8 hours 45 minutes.
But due to leak it takes 15 min more.
Hence total time taken = 8 hours 45 minutes + 15 min = 9 hours.
Let the leak take n hours to empty the cistern.
So, 1/15 + 1/21 – 1/n = 1/9
=> 4/35 – 1/n = 1/9
=› 1/n = 4/35 – 1/9 = 1/315
Hence the cistern will be emptied in 315 hours.
Incorrect
Answer: d)
If both filling pipes are working together, portion of the cistern filled in an hour
= 1/15 + 1/21 = (21 + 15)/(15*21) = 4/35
Hence time taken to fill the cistern = 35/4 hours = 8 3/4 hours = 8 hours 45 minutes.
But due to leak it takes 15 min more.
Hence total time taken = 8 hours 45 minutes + 15 min = 9 hours.
Let the leak take n hours to empty the cistern.
So, 1/15 + 1/21 – 1/n = 1/9
=> 4/35 – 1/n = 1/9
=› 1/n = 4/35 – 1/9 = 1/315
Hence the cistern will be emptied in 315 hours.
Question 15 of 20
15. Question
A customer sold an article at 20% discount on the marked price, yet he got 12% profit on the cost price. If the cost price of the article is Rs. 250, then find the marked price
Correct
Answer: c)
MP*(80/100) = 250*(112/100)
MP = 250*(112/100)*(100/80)
MP = Rs. 350
Incorrect
Answer: c)
MP*(80/100) = 250*(112/100)
MP = 250*(112/100)*(100/80)
MP = Rs. 350
Question 16 of 20
16. Question
Directions (16 – 20): Study the following information carefully and answer the questions given below:
The following table represents percentage of marks obtained by six different students in six different subjects and maximum marks of each subject.
Marks obtained by Madan in Hindi is what percent of marks obtained by Krishna in General Knowledge?
Correct
Answer: b)
Marks obtained by Madan in Hindi = 55/100 x 125 = 68.75
Marks obtained by Krishna in General Knowledge = 88/100 x 250 = 220
Required percentage = 68.75/220 x 100 = 31.25%
Incorrect
Answer: b)
Marks obtained by Madan in Hindi = 55/100 x 125 = 68.75
Marks obtained by Krishna in General Knowledge = 88/100 x 250 = 220
Required percentage = 68.75/220 x 100 = 31.25%
Question 17 of 20
17. Question
Directions (16 – 20): Study the following information carefully and answer the questions given below:
The following table represents percentage of marks obtained by six different students in six different subjects and maximum marks of each subject.
Find the total marks obtained by all the students in Science.
Correct
Answer: b)
Total marks obtained by all the students in Science
= (60 + 45 + 86 + 65 + 84 + 90) * 150/100
= 430 x 150/100 = 645
Incorrect
Answer: b)
Total marks obtained by all the students in Science
= (60 + 45 + 86 + 65 + 84 + 90) * 150/100
= 430 x 150/100 = 645
Question 18 of 20
18. Question
Directions (16 – 20): Study the following information carefully and answer the questions given below:
The following table represents percentage of marks obtained by six different students in six different subjects and maximum marks of each subject.
Find the overall percentage of marks obtained by Kunal in all the subjects.
Directions (16 – 20): Study the following information carefully and answer the questions given below:
The following table represents percentage of marks obtained by six different students in six different subjects and maximum marks of each subject.
Find the respective ratio of marks obtained by Nitin in Science, Maths and English together and marks obtained by Mohan in Hindi, Social Science and General Knowledge together.
10) 42 % of 2149 + (3/11) of 18149 – 30 % of 1699 =?
a) 5340
b) 5850
c) 5060
d) 5520
e) 4825
11) The area of a square is 1764 sq cm whose side is 2 times of radius of a circle. The circumference of the circle is equal to the perimeter of rectangle. What is the breadth of rectangle, if the length of the rectangle is 42 cm?
a) 18 cm
b) 25 cm
c) 32 cm
d) 24 cm
e) None of these
12) P, Q and R started a business by investing Rs. 15000, Rs. 18000 and Rs. 24000 respectively. After 3 months, P invested 40 % of the initial investment more. And after 5 months, R withdraws Rs. 4000. Find the share of Q, if the total profit at the end of the year is Rs. 61370?
a) Rs. 18360
b) Rs. 19520
c) Rs. 20480
d) Rs. 17500
e) None of these
13) 4 years ago, the ratio of ages of P and Q is 4: 3. The present age of R is one-fifth of the present age of Q. Sum of the ages of P, Q and R, after 4 years will be 74 years. Find the present age of Q?
a) 22 years
b) 25 years
c) 28 years
d) 20 years
e) None of these
14) Two pipes can fill a cistern in 15 and 21 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, 15 minutes extra are taken for the cistern to be filled up. If the cistern is full, in what time would the leak empty it?
a) 285 hours
b) 260 hours
c) 240 hours
d) 315 hours
e) None of these
15) A customer sold an article at 20% discount on the marked price, yet he got 12% profit on the cost price. If the cost price of the article is Rs. 250, then find the marked price
a) 280
b) 360
c) 350
d) 420
e) 400
Directions (16 – 20): Study the following information carefully and answer the questions given below:
The following table represents percentage of marks obtained by six different students in six different subjects and maximum marks of each subject.
16) Marks obtained by Madan in Hindi is what percent of marks obtained by Krishna in General Knowledge?
a) 35.62%
b) 31.25%
c) 21.22%
d) 16.25%
e) None of these
17) Find the total marks obtained by all the students in Science.
a) 465
b) 645
c) 654
d) 456
e) None of these
18) Find the overall percentage of marks obtained by Kunal in all the subjects.
a) 45%
b) 76%
c) 74%
d) 67%
e) None of these
19) Find the respective ratio of marks obtained by Nitin in Science, Maths and English together and marks obtained by Mohan in Hindi, Social Science and General Knowledge together.
a) 455 : 344
b) 344 : 455
c) 341 : 435
d) 134 : 344
e) None of these
20) Find the difference between total marks obtained by Krishna and Rohan in all the subjects together.
a) 135
b) 154.25
c) 145.5
d) 125.75
e) None of these
Answers:
1) Answer: e)
I. 4x^{2} – (4 + 2√3)x + 2√3 = 0
4x^{2} – 4x – 2√3x + 2√3 = 0
4x (x -1) – 2√3 (x -1) = 0
(4x – 2√3) (x – 1) = 0
X = √3/2, 1 = 0.866, 1
II. 8y^{2} – (24 + 3√3)y + 9√3 = 0
8y^{2} – 24y – 3√3y + 9√3 = 0
8y(y – 3) – 3√3 (y – 3) = 0
(8y – 3√3) (y – 3) = 0
Y = 3√3/8, 3 = 0.64, 3
Can’t be determined
2) Answer: c)
I. x^{2} + 9x – 52 = 0
(x + 13) (x – 4) = 0
X = -13, 4
II. y^{2} – 15y + 54 = 0
(y – 9) (y – 6) = 0
Y = 9, 6
X < y
3) Answer: e)
I. 4x^{2} + 20x + 24 = 0
4x^{2} + 12x + 8x + 24 = 0
4x(x + 3) + 8(x + 3) = 0
(4x + 8) (x + 3) = 0
X = -2, -3
II. 3y^{2 }+ 18y + 24 = 0
3y^{2 }+ 12y + 6y + 24 = 0
3y(y + 4) + 6(y + 4) = 0
(3y + 6) (y + 4) = 0
Y = -2, -4
Can’t be determined
4) Answer: e)
I. 7x^{2} – 6x – 81 = 0
7x^{2} + 21x – 27x – 81 = 0
7x(x + 3) – 27(x + 3) = 0
(7x – 27) (x + 3) = 0
X = 27/7, -3 = 3.85, -3
II. 9y^{2} – 10y – 56 = 0
9y^{2} + 18y – 28y – 56 = 0
9y (y + 2) -28 (y + 2) = 0
(9y – 28) (y + 2) = 0
Y = 28/9, -2 = 3.11, -2
Can’t be determined
5) Answer: e)
2x – 5y = 12 –> (1)
5x – 2y = -12 —> (2)
By substituting (1) and (2), we get,
X = -4, y = -4
X = y
Direction (6-10)
6) Answer: c)
X = (4128 + 574)/25 + 18^{2} + (2/15)*12525 – 24
X = (4702/25) + 324 + 1670 – 24
X = 188 + 324 + 1670 – 24
X = 2158 = 2160
7) Answer: e)
(1400/20) = x + 426 – (312/6)*20
70 – 426 + 1040 = x
X = 684 = 680
8) Answer: b)
(22*5) + 64^{2} ÷ 16 = x – (28/100)*900
110 + (64*64)/16 = x – 252
110 + 256 + 252 = x
X = 618 = 620
9) Answer: d)
(15/43)*(301/120)*20 – 620 + (62/100)*2000 = x
X = 18 – 620 + 1240
X = 638 = 640
10) Answer: a)
42 % of 2150 + (3/11) of 18150 – 30 % of 1700 = x
X = (42/100)*2150 + (3/11) * 18150 – (30/100)*1700
X = 903 + 4950 – 510
X = 5343 = 5340
11) Answer: d)
Area of square = 1764 Sq cm = a^{2}
Side (a) = 42 cm
Side of square = 2 * Radius of circle
Radius = 42/2 = 21 cm
Circumference of the circle = 2πr = 2*(22/7)*21 = 132 cm