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Joined: 02 Jan 2020
Status:GRE Quant Tutor
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Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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How to Solve: Positive and Negative Numbers
[#permalink]
08 Jun 2024, 04:33
08 Jun 2024, 04:33
1
Attachment:
How to Solve: Positive and Negative Numbers
Hi All,
I have posted a video on YouTube to discuss Positive and Negative Numbers
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
- ⁍ Types of Numbers
⁍ Zero
⁍ Properties of Positive & Negative Numbers
⁍ Solved Problems
Types of Numbers
Three types of numbers are there in a number line
⁍ Positive Numbers – All numbers to the right of 0 on the number line
⁍ Negative Numbers – All numbers to the left of 0 on the number line
⁍ Zero (neither +ve nor –ve)
Attachment:
img-1.jpg [ 23.04 KiB | Viewed 889 times ]
Zero
⁍ Zero is neither positive nor negative
⁍ Zero is an even number
⁍ For all x ≠ 0, \(x^0\) = 1
⁍ Division by zero is not defined
Properties of Positive and Negative Numbers
⁍ Addition
(P is Positive and N is Negative)
P + P = P, Ex: 2 + 3 = 5
P + N = P or N or 0, Ex1: 3 + (-2) = 1, Ex2: 2 + (-3) = -1, Ex3: 3 + (-3) = 0
N + P = P or N or 0, Ex: Same as above
N + N = N, Ex: -2 + (-3) = = -5
⁍ Subtration
P – P = P or N or 0, Ex1: 3 - 2 = 1, Ex2: 2 - 3 = -1, Ex3: 3 - 3 = 0
P – N = P, Ex: 3 - (-2) = 5
N – P = N, Ex: -3 - 2 = -5
N – N = P or N or 0, Ex1: -3 - (-4) = 1, Ex2: -3 - (-2) = -1, Ex3: -3 - (-3) = 0
⁍ Division
P / P = P, Ex: 6/2 = 3
P / N = N, Ex: 6/-2 = -3
N / P = N, Ex: -6/2 = -3
N / N = P, Ex: -6/-2 = 3
⁍ Multiplication
P * P = P, Ex: 2 * 3 = 6
P * N = N, Ex: 2 * -3 = -6
N * P = N, Ex: -2 * 3 = -6
N * N = P, Ex: -2 * -3 = 6
⁍ IF we are multiplying ODD number of Negative numbers then we will get a NEGATIVE number (Assuming they are not getting multiplied with zero)
⁍ IF we are multiplying EVEN number of Negative numbers then we will get a POSITIVE number (Assuming they are not getting multiplied with zero)
⁍ If Product/Ratio of two numbers is positive, then both the numbers will have the SAME sign.
Example : xy > 0 or x/y > 0
=> x and y have the SAME sign
=> Either x > 0 and y > 0
Or x < 0 and y < 0
⁍ If Product/Ratio of two numbers is negative, then both the numbers will have DIFFERENT signs.
Example : xy < 0 or x/y < 0
=> x and y have DIFFERENT signs
=> Either x > 0 and y < 0
Or x < 0 and y > 0
Solved Problems
Q1. Which of the following cannot be the value of x if y/(x-2) = 5 ?
A. 1
B. 2
C. 3
D. 4
E. 5
Show: ::
Sol: Denominator cannot be equal to 0
=> x - 2 ≠ 0
=> x ≠ 2
So, Answer will be B
=> x - 2 ≠ 0
=> x ≠ 2
So, Answer will be B
Q2. Given that a and b are positive, and c and d are negative, which of the following will be positive for sure? (select all possible)
A. ab + cd
B. ab + acd
C. acd + bcd
D. abc + d
E. c + cd
Show: ::
Sol:
A. ab + cd = P*P + N*N = P + P = P = TRUE
B. ab + acd = P*P + P*N*N = P + P = P = TRUE
C. acd + bcd = P*N*N + P*N*N = P + P = P = TRUE
D. abc + d = P*P*N + N = N + N = N = FALSE
E. c + cd = N + N*N = N + P = N or P or 0 = FALSE
So, Answer will be A, B, C
A. ab + cd = P*P + N*N = P + P = P = TRUE
B. ab + acd = P*P + P*N*N = P + P = P = TRUE
C. acd + bcd = P*N*N + P*N*N = P + P = P = TRUE
D. abc + d = P*P*N + N = N + N = N = FALSE
E. c + cd = N + N*N = N + P = N or P or 0 = FALSE
So, Answer will be A, B, C
Q3. Is bd>0 ?
A. ab > 0
B. cd > 0
Show: ::
Sol:
Stat A : ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: cd > 0
There are two cases
c>0 and d>0
c<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT sufficient
Combining both the statements we will have four cases
(1) a>0 b>0 c> 0 d>0
(2) a>0 b>0 c< 0 d<0
(3) a<0 b<0 c>0 d>0
(4) a<0 b<0 c<0 d<0
In case 1 and 4 bd > 0 and in case 2 and 3 bd < 0
So, Together also NOT sufficient.
So, Answer will be E
Stat A : ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: cd > 0
There are two cases
c>0 and d>0
c<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT sufficient
Combining both the statements we will have four cases
(1) a>0 b>0 c> 0 d>0
(2) a>0 b>0 c< 0 d<0
(3) a<0 b<0 c>0 d>0
(4) a<0 b<0 c<0 d<0
In case 1 and 4 bd > 0 and in case 2 and 3 bd < 0
So, Together also NOT sufficient.
So, Answer will be E
Q4. Is bd>0 ?
A. ab > 0
B. ad > 0
Show: ::
Sol:
Stat A : ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: ad > 0
There are two cases
a>0 and d>0
a<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT sufficient
Combining both the statements we will have two cases
(Since we have a common variable “a” in both the statements so we will combine the two statements based on the sign of the common variable)
First case of STAT A will be combined with the first case of Stat B and
Second case of STAT A will be combined with the second case of Stat B
(1) a>0 b>0 d>0 (2) a<0 b<0 d<0
In both the cases bd > 0
So, Together the two statements are SUFFICIENT.
So, Answer will be C
Stat A : ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: ad > 0
There are two cases
a>0 and d>0
a<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT sufficient
Combining both the statements we will have two cases
(Since we have a common variable “a” in both the statements so we will combine the two statements based on the sign of the common variable)
First case of STAT A will be combined with the first case of Stat B and
Second case of STAT A will be combined with the second case of Stat B
(1) a>0 b>0 d>0 (2) a<0 b<0 d<0
In both the cases bd > 0
So, Together the two statements are SUFFICIENT.
So, Answer will be C
Hope it helps!
Re: How to Solve: Positive and Negative Numbers
[#permalink]
08 Jun 2024, 05:11
08 Jun 2024, 05:11
Expert Reply
Added to the GRE Math Essentials - A most comprehensive handout!! [COMPLETED]
https://gre.myprepclub.com/forum/gre-ma ... tml#p95352
https://gre.myprepclub.com/forum/gre-ma ... tml#p95352
