# Quantitative Aptitude: Inequalities Set 2 (New Pattern)

Directions: Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.

1. The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.
Quantity I: Cost Price of Item X.
Quantity II:  Selling Price of item Y after two successive discount of 10% and 15% is provided on the Marked Price of Rs 10,000.
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option A
Solution:
I : Mp =Rs 10,000; SP after 4% discount = Rs 9600; CP = Rs 8000
II : SP= 10,000*90/100 * 85/100 = Rs 7650
Hence I > II
2. Quantity I: Simple Interest on a sum of Rs 6200 at the rate of 6% per annum for 5 years
Quantity II:  Compound Interest on a sum of Rs 6400 at rate of 10% compounded semi-annually for one year.
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option A
Solution:
I: SI = 6200*6*5/100 = Rs 1860
II: CI = 6400 [1 – 5/100]<sup>2</sup> – 6400 = Rs 1376
Hence I > II
3. The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.
Quantity I: Cost Price of an article if it is sold making a loss of 10% given that if the cost price was 20% less, a profit of Rs 12 could be made.
Quantity II:  Labeled Price of an article if a discount of 30% is given on it and then sold for  Rs 105 at a loss of 20%
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option E
Solution:
I: Let CP = Rs 100, then at 10% loss, SP = Rs 90
Now if CP is 20% less, means CP = Rs 80, then profit is = 90 – 80 = Rs 10
but given profit is Rs 12
So if profit is Rs 10, then CP = Rs100
If profit is Rs 12, then CP is 100/10 * 12 = Rs 120
II: Use formula MP = (100-loss%)/(100-dicount%) * CPS
So MP = (100-20)/(100-30) * 105 = Rs 120
Hence I = II
4. Given that x < 0 and y > 0
Quantity I: 14x3y2
Quantity II:  28x2y3
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option C
Solution:
I: Since x < 0, so I will always be negative
II: It will be always positive
Hence I < II
5. Quantity I: Area of rectangle whose sides are in the ratio 3 : 1 and perimeter twice the circumference of a semi-circle with area 77 cm<sup>2</sup>
Quantity II:  Curved surface area of a cylinder of radius 12 cm and height equal to the side of square of area 49cm<sup>2</sup>
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option C
Solution:
I: ½ πr<sup>2</sup> = 77 [area of semicircle]
So r = 7 cm
So circumference of semicircle  = πr = 22 cm
So perimeter of rectangle = 2*22 = 44 cm
So 3x+x = 44 cm, x = 11,
So area of rectangle = 3x<sup>2<sup> = 363cm<sup>2</sup>
II: height of cylinder = √49 = 7 cm
So CSA = 2πrh = 528cm<sup>2</sup>
Hence I < II
6. Quantity I: Time taken by A to complete 1/5th of work if B takes 6 days to complete 3/5th of work and together they take 5 days to complete 3/4th of work
Quantity II:  Time taken by a train to cross a platform of length 60 km given that it crosses a pole in two and a half hours running at 60 km/hr
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option A
Solution:
I: A and B completes 3/4th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days
B complete 3/5 work in 6 days, so complete work in 5/3 * 6  = 10 days
So in 1 day A completes = 3/20 – 1/10 = 1/20
So to complete 1/5th work = 1/5 * 20 = 4 days
II: length of train = 2.5 * 60 = 150 km
So time taken to cross platform of length 60 km with speed 60 km/hr = (150+60)/60 = 3.5 hrs
Hence I > II
7. Quantity I: Age of A five years ago if 6 years hence ratio of age of A to B will be 14 : 11, and 1 year ago the ratio was 7 : 5.
Quantity II:  Average age of 2 students included in a group of 5 students with average age 18 years given that addition of these 2 students in group makes average increase by 1.
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option C
Solution:
I: (A+6)/(B+6) = 14/11
And (A-1)/(B-1) = 7/5
Solve the equations, A = 22, so 5 years ago = 22 – 5 = 17 years
II: Total age of 2 students added to group = 18*2 + 7*1 = 43
So their average age = 43/2 = 21.5 years
Hence I < II
8. Quantity I: x, such that 6x2 + 22x + 15 = 0
Quantity II:  y, such that 6y2 – 7y – 10 = 0
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option D
Solution:
I: x = -3, -5/6
II: y= -5/6, 2
Hence I ≤ II
9. A shopkeeper gets a discount of 20% on an article and sells it at 10% profit.
Quantity I: Labeled price as a percent of cost price.
Quantity II:  Profit percent if discount given is 60%
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option C
Solution:
I: MP = (100+10)/(100-20) * CP = 11/8 of CP
So required % = MP/CP * 100 = (11CP/8)/CP * 100 = 137.5%
II: % profit = (100+10) [(100-10)/(100-60)] – 100 = 590/4 = 147.5%
Hence I < II
10. Quantity I: Speed of boat in still water f it took 24 hours to cover a distance of 64 km going downstream and back. Given the speed of current to be 2 km/hr
Quantity II:  Average speed of a boy to go from home to school if his average speed is 2 km/hr from school to home and he covers a total of 12 km in 5 hours.
A) Quantity I > Quantity II
B) Quantity I ≥  Quantity II
C) Quantity II > Quantity I
D) Quantity II ≥  Quantity I
E) Quantity I = Quantity II or Relation cannot be established
Option A
Solution:
I: Let speed of boat = x km/hr
So 64/(x+2) + 64/(x-2) = 24
Solve, x = 6 km/hr
II : Distance from home to school is 12/2 = 6 km
Let speed from home to school is x km/hr
So using distance = total time * average speed
6 = 5 *[2*x/(2+x)]
Solve, x = 3 km/hr
Hence I > II

## 4 Thoughts to “Quantitative Aptitude: Inequalities Set 2 (New Pattern)”

1. ^^^Jaga^^^....

no 10 ka quantity ii without formula kese hoga?

1. Shubhra

Total time 5 hrs hai jisse break nhi kar sakte so use T = d/s
So
5 = 6/2 + 6/x
X = 3

1. ^^^Jaga^^^....

thank u mam