New Pattern Quant Inequalities Questions for SBI PO Mains 2018
New Pattern Inequalities Questions. Quant Inequalities Questions for SBI PO Mains 2018. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for SBI PO/Clerk, BOB PO, IBPS and other competitive exams, you will come across a section on the Quantitative Aptitude. Here we are providing you a quiz on “New Pattern Quant Inequalities Questions” based on the latest pattern for your daily practice.
The pattern of the Inequalities Questions asked was changed nowadays. So we are providing you “New Pattern Quant Inequalities Questions” for banking exams such as IBPS PO, IBPS Clerk, SBI Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, IBPS SO, SBI SO and other competitive exams.
New Pattern Inequalities Questions for SBI PO Mains  Set – 12
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Question 1 of 10
1. Question
Direction (14) Given below are two quantities named A and B. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose between the possible answers.
A committee of five members is to be formed out of 5 students, 3 teachers and 2 coaches.
Quantity A: The number of ways, the committee can be formed which consists of 2 students, 2 teachers and 1 coach.
Quantity B: The number of ways, the committee can be formed which consists of any five people.
Correct
A. Quantity B > Quantity A
Quantity A:
Number of ways to choose 2 students from 5 students = ^{5}C_{2} = 10
Number of ways to choose 2 teachers from 3 teachers = ^{3}C_{2} = 3
Number of ways to choose 1 coach from 2 coaches = ^{2}C_{1} = 2
∴ Total number of ways = 10 × 3 × 2 = 60
Now,
Quantity B:
Total number of people = 5 + 3 + 2 = 10
Number of ways to choose 5 people from 10 people = ^{10}C_{5} = 252
Quantity B > Quantity A
Incorrect
A. Quantity B > Quantity A
Quantity A:
Number of ways to choose 2 students from 5 students = ^{5}C_{2} = 10
Number of ways to choose 2 teachers from 3 teachers = ^{3}C_{2} = 3
Number of ways to choose 1 coach from 2 coaches = ^{2}C_{1} = 2
∴ Total number of ways = 10 × 3 × 2 = 60
Now,
Quantity B:
Total number of people = 5 + 3 + 2 = 10
Number of ways to choose 5 people from 10 people = ^{10}C_{5} = 252
Quantity B > Quantity A

Question 2 of 10
2. Question
A box contains 3 black and 5 white balls. A ball is drawn at random from the box.
Quantity A: Probability that the ball is black
Quantity B: Probability that the ball is white
Correct
B. Quantity A < Quantity B
Total number of balls in the box = 3 + 5 = 8
∵ A ball is drawn, Sample space = n(S) = ^{8}C_{1} = 8
First we will find Quantity A,
Quantity A: Probability that the ball is black.
Quantity A = n(A)/n(S) = ^{3}C_{1}/8 = 3/8
Now,
Quantity B: Probability that the ball is white.
Quantity B = n(B)/n(S) = ^{5}C_{1}/8 = 5/8
∴ Quantity B > Quantity A
Incorrect
B. Quantity A < Quantity B
Total number of balls in the box = 3 + 5 = 8
∵ A ball is drawn, Sample space = n(S) = ^{8}C_{1} = 8
First we will find Quantity A,
Quantity A: Probability that the ball is black.
Quantity A = n(A)/n(S) = ^{3}C_{1}/8 = 3/8
Now,
Quantity B: Probability that the ball is white.
Quantity B = n(B)/n(S) = ^{5}C_{1}/8 = 5/8
∴ Quantity B > Quantity A

Question 3 of 10
3. Question
Quantity A: Profit percent when an article of cost price Rs. 140 is sold for Rs. 168.
Quantity B: Percentage discount availed by customer when he buys an article with marked price Rs. 2000 for Rs. 1800.
Correct
B. Quantity B < Quantity A
Quantity A:
Here, Cost price (CP) = 140 and
Selling Price (SP) = 168
% Profit = (SP – CP)/CP × 100
⇒ % Profit = (168 – 140)/140 × 100 = 20
Now,
Quantity B:
Here, Marked Price (MP) = 2000
Selling Price (SP) = 1800
% Discount = (MP – SP)/MP × 100
⇒ % Discount = (2000 – 1800)/2000 × 100 = 10
Clearly, Quantity B < Quantity A
Incorrect
B. Quantity B < Quantity A
Quantity A:
Here, Cost price (CP) = 140 and
Selling Price (SP) = 168
% Profit = (SP – CP)/CP × 100
⇒ % Profit = (168 – 140)/140 × 100 = 20
Now,
Quantity B:
Here, Marked Price (MP) = 2000
Selling Price (SP) = 1800
% Discount = (MP – SP)/MP × 100
⇒ % Discount = (2000 – 1800)/2000 × 100 = 10
Clearly, Quantity B < Quantity A

Question 4 of 10
4. Question
Quantity A:
The curved surface area of the cone whose slant height is 16 cm and diameter of the same is 14 cm
Quantity B:
The curved surface area of the hemisphere whose diameter is 28 cm
Correct
B. Quantity A < Quantity B
For Quantity A,
Curve surface area of cone = πrl
Where, r = radius of cone = 14/2 = 7 cm
l = slant height of cone = 16 cm
Area =227×7×16=352=227×7×16=352 cm^{2}
For Quantity B,
Surface area of hemisphere = 2πr^{2}
Where, r = radius of hemisphere = 28/2 = 14 cm
Area = 2 × 22/7 × 14 × 14 = 1232 cm^{2}
∴ Quantity A < Quantity B
Incorrect
B. Quantity A < Quantity B
For Quantity A,
Curve surface area of cone = πrl
Where, r = radius of cone = 14/2 = 7 cm
l = slant height of cone = 16 cm
Area =227×7×16=352=227×7×16=352 cm^{2}
For Quantity B,
Surface area of hemisphere = 2πr^{2}
Where, r = radius of hemisphere = 28/2 = 14 cm
Area = 2 × 22/7 × 14 × 14 = 1232 cm^{2}
∴ Quantity A < Quantity B

Question 5 of 10
5. Question
Quantity I: A man deposited Rs. 24800 on scheme 1 which offers simple interest at the rate of 7%. The simple interest obtained from scheme 1 is deposited in scheme 2 at the rate of 8% simple interest. If the time period for deposition on scheme 1 and scheme 2 was 3 years and 4 years respectively, then find the total simple interest earned by the man.
Quantity II: Find the interest earned by a man after 3 years, if he deposits Rs. 21000 on compound interest at the rate of 10% per annum.
Correct
B. QuantityI < QuantityII
Quantity I:
Simple interest earned from scheme 1 = (24800 × 7 × 3)/100 = Rs. 5208
Simple interest earned from scheme 2 = (5208 × 8 × 4)/100 = Rs. 1666.56
Total simple interest earned = 5208 + 1666.56 = Rs. 6874.56
Quantity II:
Compound interest earned = 21000 × (1.1^{3} – 1)
= 21000 × 0.331 = Rs. 6951
Incorrect
B. QuantityI < QuantityII
Quantity I:
Simple interest earned from scheme 1 = (24800 × 7 × 3)/100 = Rs. 5208
Simple interest earned from scheme 2 = (5208 × 8 × 4)/100 = Rs. 1666.56
Total simple interest earned = 5208 + 1666.56 = Rs. 6874.56
Quantity II:
Compound interest earned = 21000 × (1.1^{3} – 1)
= 21000 × 0.331 = Rs. 6951

Question 6 of 10
6. Question
Quantity I: The difference between the curved surface area and total surface area of the cylinder of height 31 cm is 1734 cm^{2}. If radius of a cone is 5 cm less than the radius of cylinder and the height of the cone is 15 cm less than the height of the cylinder, then find the curved surface area of the cone. [Use π = 3]
Quantity II: A semicircle is drawn on each side of the square of side 18 cm such that the side of the square is the diameter of the semicircle. Find the area of the entire figure. [Use π = 3]
Correct
C. QuantityI < QuantityII
Quantity I:
According to question,
Total surface area of cylinder – curved surface area of cylinder = 1734 cm^{2}
2 × π × r × h + 2 × π × r^{2} – 2 × π × r × h = 1734
2 × π × r^{2} = 1734
6r^{2} = 1734
r^{2} = 289
r = 17 cm (Since ‘r’ cannot be negative)
So, radius of cone = 17 – 5 = 12 cm
Height of cone = 31 – 15 = 16 cm
Slant height of cone = (12^{2} + 16^{2})^{0.5}
= (144 + 256) = 400^{0.5 }= 20 cm
Curved surface area of the cone = π × r × l
= 3 × 12 × 20 = 720 cm^{2}
Quantity II:
Side of square, a = 18 cm
Radius of semicircle, r = 9 cm
Area of the entire figure = a^{2} + 4 × 0.5 × π × r^{2}
= 18^{2} + 2 × 3 × 9^{2}
= 324 + 486 = 810 cm^{2}
Incorrect
C. QuantityI < QuantityII
Quantity I:
According to question,
Total surface area of cylinder – curved surface area of cylinder = 1734 cm^{2}
2 × π × r × h + 2 × π × r^{2} – 2 × π × r × h = 1734
2 × π × r^{2} = 1734
6r^{2} = 1734
r^{2} = 289
r = 17 cm (Since ‘r’ cannot be negative)
So, radius of cone = 17 – 5 = 12 cm
Height of cone = 31 – 15 = 16 cm
Slant height of cone = (12^{2} + 16^{2})^{0.5}
= (144 + 256) = 400^{0.5 }= 20 cm
Curved surface area of the cone = π × r × l
= 3 × 12 × 20 = 720 cm^{2}
Quantity II:
Side of square, a = 18 cm
Radius of semicircle, r = 9 cm
Area of the entire figure = a^{2} + 4 × 0.5 × π × r^{2}
= 18^{2} + 2 × 3 × 9^{2}
= 324 + 486 = 810 cm^{2}

Question 7 of 10
7. Question
Quantity I: The present ages of Geeta and Sita are in the ratio 3:4 respectively. The ages of Radha and Sita, 4 years hence would be in the ratio of 5: 4 respectively. Mohan is 8 years elder to Radha. The average of present ages of Geeta, Sita, Radha and Mohan is 28 years. Find Mohan’s present age.
Quantity II: The present age of Gaurav is ‘3x’ years and Manish’s age after 4 years will be ‘2x’ years. Difference between the ages of Aditya and Gaurav is 3 years. Ashwani’s age 5 years ago was ‘x – 1’ years. If the average of present ages of Gaurav, Manish and Ashwani is 24 years then find the present age of Aditya.
Correct
E. QuantityI ≥ QuantityII
Quantity I:
Let the present ages of Geeta and Sita be 3x years and 4x years.
Sita’s age after 4 years = 4x + 4
Radha’s age after 4 years = [(4x + 4) / 4] × 5 = 5x + 5
Radha’s present age = 5x + 5 – 4 = 5x + 1
Mohan’s present age = 5x + 9
So according to question: 3x + 4x + 5x + 1 + 5x + 9 = 28 × 4
17x + 10 = 112
17x = 102, x = 6
So, Mohan’s present age = 5 × 6 + 9 = 39 years
Quantity II:
Gaurav’s present age = 3x
Manish’s present age = 2x – 4
Ashwani’s present age = x – 1 + 5 = x + 4
So according to question: 3x + 2x – 4 + x + 4 = 3 × 24
6x = 72, x = 12
So, Gaurav’s present age = 3 × 12 = 36 years
Aditya’s present age = either 36 + 3 = 39 years or 36 – 3 = 33 years
So, Quantity I ≥ Quantity II
Incorrect
E. QuantityI ≥ QuantityII
Quantity I:
Let the present ages of Geeta and Sita be 3x years and 4x years.
Sita’s age after 4 years = 4x + 4
Radha’s age after 4 years = [(4x + 4) / 4] × 5 = 5x + 5
Radha’s present age = 5x + 5 – 4 = 5x + 1
Mohan’s present age = 5x + 9
So according to question: 3x + 4x + 5x + 1 + 5x + 9 = 28 × 4
17x + 10 = 112
17x = 102, x = 6
So, Mohan’s present age = 5 × 6 + 9 = 39 years
Quantity II:
Gaurav’s present age = 3x
Manish’s present age = 2x – 4
Ashwani’s present age = x – 1 + 5 = x + 4
So according to question: 3x + 2x – 4 + x + 4 = 3 × 24
6x = 72, x = 12
So, Gaurav’s present age = 3 × 12 = 36 years
Aditya’s present age = either 36 + 3 = 39 years or 36 – 3 = 33 years
So, Quantity I ≥ Quantity II

Question 8 of 10
8. Question
Quantity I: A bag contains ‘x’ blue, ‘x + 3’ pink and 11 black balls. Two balls are randomly drawn from the bag and the probability that the both the balls are of same color is (121/378). Find the value of x^{2} – 2x.
Quantity II: A bag contains ‘x’ watches out of which 7 are analog watches and rest are digital watches. Two watches are randomly drawn from the bag and the probability that both the watches are digital watches is (54/85). Find the value of x.
Correct
D. QuantityI = QuantityII or No relation
Quantity I:
Probability that both the balls are of same color = P(both balls are blue) + P(both balls are pink) + P(both balls are black)
[^{x}C_{2} / ^{2x+4}C_{2}] + [^{x+3}C_{2} / ^{2x+4}C_{2}] + [^{11}C_{2} / ^{2x+4}C_{2}] = 121 / 378
[x(x1) + (x+3)(x+2) + 110] / [(2x+14)(2x+13)] = 121/378
[x^{2} – x + x^{2} + 5x + 6 + 110] / [4x^{2} + 54x + 182] = 121/378
[2x^{2} + 4x + 116] / [4x^{2} + 54x + 182] = 121/378
[x^{2} + 2x + 58] / [2x^{2} + 27x + 91] = 121/378
378x^{2} + 756x + 21924 = 242x^{2} + 3267x + 11011
136x^{2} – 2511x + 10913 = 0
136x^{2} – 952x – 1559x + 10913 = 0
136x(x – 7) – 1559(x – 7) = 0
(136x – 1559)(x – 7) = 0
x = 7, (1559/136)
Number of balls can’t be in fraction. So the value of x = 7
So the value of x^{2} – 2x = 49 – 14 = 35
Quantity II:
Total number of digital watches = x – 7
The probability that both the watches are digital = [^{x7}C_{2} / ^{x}C_{2}] = 54/85
[(x – 7)(x – 8)] / [(x)(x – 1)] = 54/85
[x^{2} – 15x + 56] / [x^{2} – x] = 54/85
85x^{2} – 1275x + 4760 = 54x^{2} – 54x
31x^{2} – 1221 + 4760 = 0
31x^{2} – 1085x – 136x + 4760 = 0
31x(x – 35) – 136(x – 35) = 0
(31x – 136)(x – 35) = 0
x = 35, (136/31)
Number of watches can’t be in fraction.
So the value of x = 35
So Quantity I = Quantity II
Incorrect
D. QuantityI = QuantityII or No relation
Quantity I:
Probability that both the balls are of same color = P(both balls are blue) + P(both balls are pink) + P(both balls are black)
[^{x}C_{2} / ^{2x+4}C_{2}] + [^{x+3}C_{2} / ^{2x+4}C_{2}] + [^{11}C_{2} / ^{2x+4}C_{2}] = 121 / 378
[x(x1) + (x+3)(x+2) + 110] / [(2x+14)(2x+13)] = 121/378
[x^{2} – x + x^{2} + 5x + 6 + 110] / [4x^{2} + 54x + 182] = 121/378
[2x^{2} + 4x + 116] / [4x^{2} + 54x + 182] = 121/378
[x^{2} + 2x + 58] / [2x^{2} + 27x + 91] = 121/378
378x^{2} + 756x + 21924 = 242x^{2} + 3267x + 11011
136x^{2} – 2511x + 10913 = 0
136x^{2} – 952x – 1559x + 10913 = 0
136x(x – 7) – 1559(x – 7) = 0
(136x – 1559)(x – 7) = 0
x = 7, (1559/136)
Number of balls can’t be in fraction. So the value of x = 7
So the value of x^{2} – 2x = 49 – 14 = 35
Quantity II:
Total number of digital watches = x – 7
The probability that both the watches are digital = [^{x7}C_{2} / ^{x}C_{2}] = 54/85
[(x – 7)(x – 8)] / [(x)(x – 1)] = 54/85
[x^{2} – 15x + 56] / [x^{2} – x] = 54/85
85x^{2} – 1275x + 4760 = 54x^{2} – 54x
31x^{2} – 1221 + 4760 = 0
31x^{2} – 1085x – 136x + 4760 = 0
31x(x – 35) – 136(x – 35) = 0
(31x – 136)(x – 35) = 0
x = 35, (136/31)
Number of watches can’t be in fraction.
So the value of x = 35
So Quantity I = Quantity II

Question 9 of 10
9. Question
Quantity I: Ram and Shyam together start a business with investments of Rs. 12400 and Rs. 14800, respectively. After a year, Ram increases his investment by 15% while Shyam decreases his investment by 20%. After one more year, Shyam increases his investment by 35% while Ram decreases his investment by 10%. Find the difference between the total investment of Ram and Shyam in 3 years.
Quantity II: Akhil and Nikhil together start a business with investments of Rs. 8500 and Rs. 9200, respectively. After a year, Akhil increases his investment by 22% while Nikhil decreases his investment by 10%. After one more year, Nikhil increases his investment by 40% while Akhil decreases his investment by 25%. Find the difference between the total investment of Akhil and Nikhil in 3 years
Correct
A. QuantityI > QuantityII
Total investment of Ram in 3 years = 12400 + 115% of 12400 + 90% of 115% of 12400
= 12400 + 14260 + 12834
= Rs. 39494
Total investment of Shyam in 3 years = 14800 + 80% of 14800 + 135% of 14800
= 14800 + 11840 + 15984
= Rs. 42624
Required difference = 42624 – 39494 = Rs. 3130
Quantity II:
Total investment of Akhil in 3 years = 8500 + 122% of 8500 + 75% of 122% of 8500
= 8500 + 10370 + 7777.5
= Rs. 26647.5
Total investment of Nikhil in 3 years = 9200 + 90% of 9200 + 140% of 90% of 9200
= 9200 + 8280 + 11592
= Rs. 29072
Required difference = 29072 – 26647.5 = Rs. 2424.5
Incorrect
A. QuantityI > QuantityII
Total investment of Ram in 3 years = 12400 + 115% of 12400 + 90% of 115% of 12400
= 12400 + 14260 + 12834
= Rs. 39494
Total investment of Shyam in 3 years = 14800 + 80% of 14800 + 135% of 14800
= 14800 + 11840 + 15984
= Rs. 42624
Required difference = 42624 – 39494 = Rs. 3130
Quantity II:
Total investment of Akhil in 3 years = 8500 + 122% of 8500 + 75% of 122% of 8500
= 8500 + 10370 + 7777.5
= Rs. 26647.5
Total investment of Nikhil in 3 years = 9200 + 90% of 9200 + 140% of 90% of 9200
= 9200 + 8280 + 11592
= Rs. 29072
Required difference = 29072 – 26647.5 = Rs. 2424.5

Question 10 of 10
10. Question
Quantity I: Gagan express and Pawan express start from station A and station B, respectively towards each other at the same time. The ratio of the speed of Gagan express to that of Pawan express is 4:5, respectively. Total distance travelled by Gagan express to meet Pawan express is 280 km in 3.5 hours. Find the total time taken by Pawan express to reach station A.
Quantity II: Rajdhani express starts from station P and after one hour, Garib Rath starts from station Q towards P with speed 80 km/h. Garib Rath meets Rajdhani express after 3 hours of its travel. If Garib Rath takes 9 hours to reach station P then find the time taken by Rajdhani express to reach station Q.
Correct
A. QuantityI > QuantityII
Quantity I:
Speed of Gagan express = 280/3.5 = 80 km/h
Therefore, speed of Pawan express = (5/4) * 80 = 100 km/h
Distance travelled by Pawan express in 3.5 hours = 100 * 3.5 = 350 km
Total distance between station A and station B = 350 + 280 = 630 km
Required time = 630/100 = 6.3 hours
Quantity II:
Let, speed of Rajdhani express = ‘x’ km/h
So, x + 3x + 3 * 80 = 9 * 80
4x + 240 = 720
4x = 480
x = 480/4
x = 120 km/h
Required time = 720/120 = 6 hours
Incorrect
A. QuantityI > QuantityII
Quantity I:
Speed of Gagan express = 280/3.5 = 80 km/h
Therefore, speed of Pawan express = (5/4) * 80 = 100 km/h
Distance travelled by Pawan express in 3.5 hours = 100 * 3.5 = 350 km
Total distance between station A and station B = 350 + 280 = 630 km
Required time = 630/100 = 6.3 hours
Quantity II:
Let, speed of Rajdhani express = ‘x’ km/h
So, x + 3x + 3 * 80 = 9 * 80
4x + 240 = 720
4x = 480
x = 480/4
x = 120 km/h
Required time = 720/120 = 6 hours
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