# New Pattern Inequality Quantity Based Shortcut Tricks Tips & Concepts : How to Solve Inequality Question Short Trick ?

## New Pattern Inequality Quantitative Aptitude Short Tricks & Tips

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Sometimes, Quantitative Comparison (Quant Comp) and Algebra may turn out to be a deadly combination on the Exams. Not so if you are good at handling inequalities! After all Quant Comp questions are nothing but inequality questions.

Quant Comp questions ask you whether Quantity A is always >Quantity B, or whether Quantity B is always > Quantity A, or whether Quantity A is always = Quantity B. If the answer to any of these questions is yes, you have got your answer – (A), (B), and (C) respectively! If the answer to any of the above questions is “may be or may not be”, even then you have got your answer! In that case, your answer is (D)!

Let’s discuss just two simple inequality related properties and see how that can help us in solving Quant Comp questions involving Algebraic expressions.

Inequality Property 1: In any inequality you can add or subtract same quantity on both the sides and the inequality still holds good.

For example,

If x > y, where x, y, and a are real numbers,

x +a > y+ a and

x − a > y – a

It does not matter whether x, y, and a are positive or negative.

Inequality Property 2: In any inequality you can multiply or divide both the sides by a positive number and the inequality still holds good. However, if you multiply or divide both the sides by a negative number the inequality gets reversed.

If x > y, where x, y, and a are real numbers,

ax > ay and x/a > y/a  if a is positive, but

ax < ay and x/a< y/a  if a is negative.

It doesn’t matter whether x and y are positive or negative.

Now, let’s see how knowing these properties can help us solve some GRE Quant Comp questions.

Example 1 A:

x is a real number

Quantity A                                                              Quantity B

4x − 7                                                                          2x+3

As per Inequality Property 1 above, in any inequality, you can add or subtract same quantity on both the sides and the inequality still holds good.

Let us subtract 2x from and add 7 to both the quantities.

We get

Quantity A                                                                 Quantity B

2x                                                                                     10

As per Inequality Property 2, in any inequality you can multiply or divide both the sides by a positive number and the inequality still holds good.

Thus by dividing both sides by 2, we get

Quantity A                                                                 Quantity B

x                                                                                        5

Now, let’s look at a small variation.

Example 1B:

x is a real number greater than 5

Quantity A                                                                   Quantity B

4x-7                                                                                 2x+3

The same way, this will lead us to

Quantity A                                                                Quantity B

x                                                                                       5

Since x is a real number greater than 5, the answer becomes (A)

Here’s another variation.

Example 1C:

x is a real number greater than 5

Quantity A                                                                   Quantity B

4x-8                                                                                    2x+3

This time, we subtract 2x from and add 8 to both the quantities and divide both the quantities by 2.

We get

Quantity A                                                                     Quantity B

x                                                                                            5.5

We are told that x is greater than 5. Obviously, x is not necessarily an integer. Therefore, x may be less than, equal to, or greater than 5.5, and therefore the answer is (D).

Example 2A:

x is a real number greater than 5

Quantity A                       Quantity B

1/4x-7                                1/2x+3

Even if we want to plug in numbers for x, it may not be very convenient to do so with the question in its present form. A good idea is to multiply both sides by (4x – 7) (2x + 3)

We get

x is a real number greater than 5

Quantity A                       Quantity B

2x+3                                  4x-7

Use the same process as used in Examples 1A, 1B, and 1C, and you will get the answer (B).

The most important point to note here is that we are able to do this multiplication because we know that (4x – 7)(2x + 3) is always positive as both (4x – 7) and (2x + 3) are always positive.

Let’s look at a variation of this question.

Example 2B:

x is a real number greater than 1

Quantity A                       Quantity B

1/4x-7                                1/2x+3

In this case, it will be erroneous to multiply both sides by (4x – 7) (2x + 3) as, although (2x + 3) is always positive, (4x – 7) can be either positive or negative, and, therefore, (4x – 7) (2x + 3) can be either positive or negative!

However, there’s still a way to simplify this to some extent. We can multiply both sides by (2x+3), and we get

x is a real number greater than 1

Quantity A                       Quantity B

2x+3/4x-7                        1

Now it may be a good idea to plug in values for x.

Quantity A                       Quantity B

2x+3/4x-7                        1

Put x = 1.1    5.2/-2.6

Quantity B is greater; cancel out (A) and (C).

Quantity A                       Quantity B

Put x = 2     7/1                 1

Let’s look at another variation of this question.

Example 2C:

x is a real number less than − 5

Quantity A                       Quantity B

1/4x-7                                1/2x+3

Now, we note that both (4x – 7) and (2x + 3) are always negative and, therefore, (4x – 7)(2x + 3) is always positive. We can just multiply both sides by (4x – 7)(2x + 3) and follow the same process.

We get

x is a real number less than − 5

Quantity A                       Quantity B

2x+3                                 4x-7

Use the same process as used in Examples 1A, 1B, and 1C.

This time, however, your answer will be (A) as x is a real number less than − 5

Here is one more variation — this time with a twist!

Example 2D:

x is a real number less than 7 but greater than −3

Quantity A                       Quantity B

1/x-7                                  1/x+3

Hey! Don’t just start putting values for x or start manipulating the expressions algebraically. This time you really don’t have to do anything. You just have to figure out that (x–7) is always negative and (x+3) is always positive. Consequently, your answer is (B). This one is really an easy question. You don’t have to do anything more.

As we say at Manya-The Princeton Review, “Work smarter, not harder.”

Problems On Quantity :-

The ratio of the present age of Bala to that of Arnav is 3 : 11. Arnav is 12 years younger than Rahim. Rahim’s age after 7 years will be 85 years.
Quantity I: The present age of Bala’s father, who is 25 years older than Bala
Quantity II: Rahim’s present age
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I < Quantity II

11x = 85 – 7 – 12
x = 6
Present age of Bala = 18
Present age of Bala’s father = 18 + 25 = 43; Rahim’s present age = 78

  Mr. Ramesh bought two watches which together cost him Rs.440. He sold one of the watches at a loss of 20% and the other one at a gain of 40%. The selling price of both watches are same.
Quantity I: SP and CP one of the watches sold at a loss of 20%
Quantity II: SP and CP one of the watches sold at a profit of 40%
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I ≤ Quantity II

80/100 * x = 140/100 * y
x = 7/4y
x + y = 440
7/4 y + y = 440
y = 160 ; x = 280

  Ravi, Hari and Sanjay are three typists, who working simultaneously, can type 228 pages in four hours. In one hour, Sanjay can type as many pages more than Hari as Hari can type more than Ravi. During a period of five hours, Sanjay can type as many passages as Ravi can, during seven hours.
Quantity I: Number of pages typed by Ravi
Quantity II: Number of pages typed by Hari
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I < Quantity II

Let Ravi, Hari and Sanjay can type x, y, and z pages respectively in 1 h.
Therefore, they together can type 4(x + y + z) pages in 4 h
∴ 4(x + y + z) = 228
⇒ x + y + z = 57 …..(i)
Also, z – y = y – x
i.e., 2y = x + z ……(ii)
5z = 7x ……(iii)
From Eqs. (i) and (ii), we get
3y = 57
⇒ y = 19
From Eq. (ii), x + z = 38
x = 16 and z = 22

  The length of a rectangle wall is 3/2 times of its height. The area of the wall is 600m².
Quantity I: Height of the wall
Quantity II: Length of the wall

A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I > Quantity II

length = 3x
height = 2x
Area of the wall = 3x * 2x = 6x² = 600
Length = 30 & Height = 20

  Quantity I: x² – 26x + 168 = 0
Quantity II: y² – 29y + 210 = 0
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I ≤ Quantity II

x² – 26x + 168 = 0
x = 12, 14
y² – 29y + 210 = 0
y = 14, 15

  Quantity I: x² – 21x + 110 = 0
Quantity II: y² – 18x + 80 = 0

A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I ≥ Quantity II

x² – 21x + 110 = 0
x = 10 11
y² – 18y + 80 = 0
y = 10 8

  A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is plugged.
Quantity I: X = Inlet Pipe Efficiency
Quantity II: Y = Outlet Pipe Efficiency
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I > Quantity II

Inlet pipe Efficiency = 100/(8/6) = 75%
Outlet pipe Efficiency = 100/(6) = 16.66%

  Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job.
Quantity I: Probability of selecting no woman
Quantity II: Probability of selecting at least one woman

A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I < Quantity II

Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting at least one woman = 1 – 14/91 = 77/91

  A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at random,
Quantity I: Probability that at least one is Black.
Quantity II: Probability that all is Black.

A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I > Quantity II

Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91

  Two pipes A and B can fill a tank in 12 hours and 18 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom of the tank it took 48 minutes excess time to fill the cistern.
Quantity I: Due to leakage, time taken to fill the tank
Quantity II: Time taken to empty the full cistern

A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established

1. Quantity I < Quantity II

Work done by the two pipes in 1 hour = (1/12)+(1/18) = (15/108).
Time taken by these pipes to fill the tank = (108/15)hrs = 7 hours 12 min.
Due to leakage, time taken to fill the tank = 7 hours 12 min + 48 min = 8 hours
Work done by two pipes and leak in 1 hour = 1/8.
Work done by the leak in 1 hour =(15/108)-(1/8)=(1/72).
Leak will empty the full cistern in 72 hours.

Quantity I: The age of teacher, if the average age of 36 students is 14. When teacher’s age is included the average increases by 1.
Quantity II: The age of teacher, if the average age of 19 students is 35. When teacher’s age is included the average increases by 0.5.
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=a +or – nd
14+ (37*1) =14+37=51yrs.
II 35 + (20*0.5) =35+ 10=45yrs.

  Quantity I: Profit Percentage , if Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6.
Quantity II: Profit Percentage, if 100 toys are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen.
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 If the no of article bought is LCM of 6 and 5 is 30
CP of 30 articles =5/6*30=Rs25
SP of 30 articles =6/5*30=Rs36
Profit 36-25=11
Profit %ge =11/25 *100=44%
II CP of 1 toy =350/100=3.50
SP of 1toy = 48/12 =4.
Profit=4-3.5=0.5.
Profit %ge (0.5/3.5) *100=14 2/7%

  Quantity I: On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
Quantity II: A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. The selling price is:
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. C.P. of 12 balls = S.P. of 17 balls = Rs.720.
CP of 1 ball = 720/12=Rs60.
II. SP =85% of 1400
=85/100*1400
=Rs1190.

  Quantity I: A and B together can do a piece of work in 4 days. If A alone can do the same work in 6 days, then B alone can do the same work in?
Quantity II: A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?
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. B work= 1/4 – 1/6 = 2/24 ==> 12 days
II.  A’s 1hr work 1/4 .
(B+C’s) 1 hr work 1/3.
(A+C’s) 1hr work 1/2.
A+B+C 1hr work = 1/4+1/3=7/12.
B’s work =7/12 – 1/2=1/12
12hours.

  Quantity I: A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed of the tour is:
Quantity II: A went from P to Q with the speed of 60km/hr. and return back with the speed of 90km/hr. Find the average speed.
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. Total time taken = (160/64 + 160/80) =9/2hrs
Then avg speed =320/(9/2)
=320*2/9= 71.11km/hr.
II. (2*60*90)/150=72km/hr.

  Quantity I: The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
Quantity II: Find the speed of a train which passes a tree in 12 seconds. The length of the train is 264m.
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 the speed of two trains be 7x and 8x.
400/4=100
8x=100==>x=12.5.
Then speed of first train =7*12.5 =87.5km/hr.
II. Length of the train = 264m.
Time taken to pass the tree = 12 seconds.
Speed of the train = 264/12 m/sec = 22 m/sec = 22 * 18/5 km/hr = 79.2 km/hr.

  Quantity I: A and B started a business by investing Rs. 20000 and Rs. 35000 respectively. Find the share of B out of an annual profit of Rs. 3520.
Quantity II: X and Y invested in a business. Their profit ratio is 2:3. If X invested Rs. 4000. Find the amount invested by Y?
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 Ratio 20_35=4:7
11 == 3520
7 ?==>Rs2240.
II 4000/y = 2/3
y = 6000.

  Quantity I: The age of P is twelve times that of her daughter Q. If the age of Q is 3 years, what is the age of P?
Quantity II: The ratio between the present ages of A and B is 2:3. 4 years ago the ratio between their ages was 5:8. What will be A’s age after 7 years?
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
Ratio P:Q 12:1
1…….== 3
12 ? == 12*3 =36years.
II
(5x+4)/(8x+4) =2/3
15x+12 = 16x+8
x=4.
A’s age 4 yrs ago 5*4=20
Then A’s age after 7yrs is 20+4+7=31yrs.

  Quantity I: The difference between SI and CI compounded annually on a certain sum of money for 2 years at 8% per annum is Rs. 12.80. Find the principal.
Quantity II: A sum fetched a total simple interest of Rs. 800 at the rate of 8 %per annum in 5 years. What is the sum?
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