# Quantitative Aptitude: Inequalities Set 9 (Important for Upcoming Exams)

Directions(1-10): Establish the correct relation between these two quantities and choose a correct option.

1. Quantity I: A vessel contains a mixture of milk and water mixed in the ratio 9:5 resp. 168 litres of the mixture is taken out of the vessel and is replaced with 48 litres of milk such that the ratio of the milk to water in the vessel becomes 15:7 resp. Find the initial quantity of water in the vessel.
Quantity II: Four pipes are attached to a tank, and they together can fill the tank in 2250 seconds. If each pipe can supply water at rate 20 litres per 15 min. Find the capacity of the tank.

Quantity I < Quantity II
Quantity I >= Quantity II
Quantity II > Quantity I
Quantity I > Quantity II
No relation.
Option E
From I: (9x – 108 +48)/(5x – 60) = 15/7
=> x = 40
Initial quantity of water in the vessel = 40*5 = 200 litres
From II: Supplying capacity of 4 pipes = 20*4 = 80 litres/15 min.
Capacity of tank = 80/(15*60)*2250 = 200 litres
No relation.

2. Quantity I: A, B and C started a business with initial investments of Rs.1200, Rs.1500 and Rs. 1800 resp. A is the working partner so 16% of the profit is given to A for handling the business and rest of the profit is shared among A, B and C in the ratio of their investments. Find the amount received by A in a profit of Rs.2500.
Quantity II: A and B started a business with initial investments of Rs.4000 and Rs.3000 resp. After six months, C joined them, with an investment of Rs.2000. After one year, a profit of Rs.2520 is generated. Find the profit share of B.

Quantity I > Quantity II
Quantity II > Quantity I
No relation.
Quantity I >= Quantity II
Quantity I < Quantity II
Option A
From I: Profit share of A:B:C = 4:5:6
Profit share of A as partner = 16% of 2500 = Rs.400
Profit share of A for his investment = 4/5*2100 = Rs.560
Total profit of A = 400+560 = Rs.960
From II: Ratio of profit share A:B:C = 4000*12:3000*12:2000*6 = 4:3:1
Profit share of B = 3/8*2520 = Rs.945
Quantity I > Quantity II

3. Quantity I: The length of a rectangular field twice the side of a square field. The cost of watering the square field at Rs.8 per m^2 is Rs.X. The cost of watering the rectangular field at Rs. 6 per m^2 is Rs. 6000.
Quantity II: Seeta and Geeta start a business together with ratio of investment as 5:4. The profit of Geeta out of total profit of Rs.3600 is Rs.Y and 25% of the profit is given to Geeta for taking care of all the marketing work.

Quantity II > Quantity I
No relation.
Quantity I < Quantity II
Quantity I > Quantity II
Quantity I >= Quantity II
Option D
From I: Area of the rectangular field = 6000/6
= 1000 m^2
Let length be x and breadth be (x-15) m.
Now, x(x-15) = 1000
=> x = 40
side of square field = 40/2 = 20 m
20*20 * 8 = x
=> x = Rs.3200
From II: Profit received by Geeta = 0.25*3600 = Rs.900
Remaining profit = 3600 – 900 = Rs.2700
Profit share of Geeta in remaining profit = 4/9*2700 = Rs.1200
Total profit share of Geeta = 1200+900 = Rs.2100
Quantity I > Quantity II

4. Quantity I: A shopkeeper sold two books X and Y of same marked price but discounts given on both the books are different. On book X, the shopkeeper gives two successive discounts of 10% and 15% and on book Y, he gives a discount of 28%. If the difference between the selling prices of both the books is Rs.27. Find the marked price of each book.
Quantity II: A shopkeeper sold an article at Rs.490 at a loss of 12.5%. If the shopkeeper marked the article at 7(1/7)% above the cost price. Find the marked price of the article.

Quantity I > Quantity II
Quantity I < Quantity II
No relation.
Quantity I >= Quantity II
Quantity II > Quantity I
Option C
From I: Let MP be Rs.x.
Total discount on book X = 10+15-{(10*15)/100} = 23.5%
SP of book X = 0.765x
Discount on book Y = 28%
SP on book Y = 0.72x
Now, 0.765x – 0.72x = 27
=> x = Rs.600
From II: SP of the article = Rs.490
Loss% = 12.5%
CP = 490/0.875 = Rs.560
MP of the article = 560*(7.5/7) = Rs.600
No relation.

5. Quantity I: A rectangular path of width 3 m is made outside around a rectangular park of perimeter 90 m. The ratio of length and breadth of rectangle is 32:13. Find the area of path.
Quantity II: Find the area of rectangle of perimeter of 72 m having ratio of length and breadth as 11:7.

Quantity I < Quantity II
No relation.
Quantity I > Quantity II
Quantity II > Quantity I
Quantity I >= Quantity II
Option D
From I: 2(32x+13x) = 90
=> x = 1
length = 32
Area of required path = (32+3*2) (13+2*3) – 32*13 = 306 m^2
From II: 2(11x+7x) = 72
=> x = 2 length = 22 m
Breadth = 14 m
Area of the rectangle = 22*14 = 308 m^2
Quantity II > Quantity I

6. Quantity I: A, B and C can do a certain piece of work in 24, 20 and 36 days resp. They started working together but after 5 days left the work and A left the work before 7 days of the completion of the work. If the work is completed in x days. Find the value of x.
Quantity II: 6 years ago, the ratio of the age of the Harish to that of Hari was 4:3 resp. After 15 years, the ratio of the age of Hari to that of Harish will become 13:15 resp. If the present age of Sahil is ‘y’ years and he is 8 years younger than Hari. Find the value of y.

No relation.
Quantity I > Quantity II
Quantity II > Quantity I
Quantity I < Quantity II
Quantity I >= Quantity II
Option D
From I: Total work = 360 units
Work done by A in a day = 15 units
Work done by B in a day = 18 units
Work done by C in a day = 10 units
Total work in first five days = 43*5 = 215 units
Total work by C in last 7 days = 10*7 = 70 units
Remaining work = 360 – (215 + 70) = 75 units
75 units of work done by A and C = 75/25 = 3 days
Total time to complete the work = 5+7+3 = 15 days
From II: (4x+21)/(3x+21) = 15/13
=> x = 6
Present age of Harish = 4x+6 = 30 years
y = 24 – 8 = 16
Quantity I < Quantity II

7. Quantity I: Time taken by a boat to travel a distance of 120 km downstream is 48/9 hours less than the time taken by boat to travel the same distance upstream and the ratio of the speed of the boat upstream is 5:3 resp. Find the speed of the stream.
Quantity II: 2.5 km/hr.

Quantity I < Quantity II
Quantity II > Quantity I
Quantity I > Quantity II
No relation.
Quantity I >= Quantity II
Option C
From I: 120/3x – 120/5x = 48/9
=>x = 3
Speed of stream = (15-9)/2 = 3 km/hr.
Quantity I > Quantity II

8. Quantity I: A man invested Rs.600 in a scheme and received Rs.30 as simple interest after two years. Find the rate of interest.
Quantity II: Kamal bought a cycle for Rs.2500 and he spent Rs.500 on its maintenance. If he marked it up by x% and sold it for Rs.3050 after offering a discount of Rs.250. Find the value of x.

Quantity I < Quantity II
No relation.
Quantity I > Quantity II
Quantity I >= Quantity II
Quantity II > Quantity I
Option E
From I: Let the rate of the interest be r% pa.
Now, 600*r%*2 = 30
=> r = 2.5%
From II: Total CP = 2500+500 = Rs.3000 3000*{(100+x)/100} = 3050+250
=> x = 10%
Quantity II > Quantity I

9. Two numbers are in the ratio 3:5, and the difference between the square of the given numbers is equal to the square of 36.
Quantity I: Find the difference between the digits of the smaller number.
Quantity II: Find the tens place digit of the larger number.

Quantity I < Quantity II
Quantity I > Quantity II
Quantity II > Quantity I
Quantity I >= Quantity II
No relation.
Option B
(5x)^2 – (3x)^2 = 36^
=> x = 9 27 and 45 are the two numbers.
From I: 7-2 = 5
From II: 4
Quantity I > Quantity II

10. Quantity I: x^2 -15 x + 54 = 0
Quantity II: 3y^2 – 25y + 42 = 0

Quantity I < Quantity II
Quantity II > Quantity I
Quantity I >= Quantity II
Quantity I > Quantity II
No relation.
Option C
From I: x^2 -15 x + 54 = 0
=> x^2 – 9x – 6x + 54 = 0
=> (x-9)(x-6) = 0
=> x = 9,6
From II: 3y^2 – 25y + 42 = 0
=> 3y^2 -18y – 7y + 42 = 0
=> (y -6)(3y- 7) = 0
=>y = 6,7/3
Quantity I >= Quantity II