If x and y are integers greater than 1 and x>y, what is the value of

Target question: What is the value of x?

Given: x and y are integers greater than 1 and x > y

Statement 1: x+y=10
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 9 and y = 1. In this case, the answer to the target question is x = 9
Case b: x = 8 and y = 2. In this case, the answer to the target question is x = 8
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: xy=21
Since x and y are INTEGERS greater than 1 AND x > y, we can see that it MUST be the case that x = 7 and y = 3
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2):
\(y = 10 – x\) and\(xy= x(10-x) = 21\)
\(⇔ -x^2 + 10x = 21\)
\(⇔ x^2 - 10x + 21 = 0\)
\(⇔ (x-3)(x-7) = 0\)
\(⇔ x = 3\) and\(y = 7\), or \(x = 7\) and \(y = 3\).
Since \(x > y\), we must have \(x = 7\) and \(y = 3\).
Thus \(x = 7\), and conditions 1) and 2) are sufficient, when taken together.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1)
There are two possible solutions: \(x = 6\) and \(y = 4\), and \(x = 7\) and \(y = 3\).
Since the solution is not unique, condition 1) is not sufficient.

Condition 2):
Either \(x = 3\) and \(y = 7\), or \(x = 7\) and \(y = 3\).
Since \(x > y, x = 7\) and \(y = 3\).
Thus, we have the unique solution, \(x = 7\).
Therefore, condition 2) is sufficient.

Therefore, B is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

Answer: B

1)
x+y=10
x+y=10

2)
xy=21
xy=21


Ans:

Tip:A lot of DS question go wrong if we don't read the question carefully.So, always read the question carefully in DS.

1)X+Y=10, and given X>Y,then ther are various possibilities as below
X=8 and Y=2
X=7 and Y=3
X=6 and Y=4

So option 1 Does not give any definitive answer.

2)XY=21, and given X,Y are positive. so we cant take negative values,also x,y>1 that means 21,1 and 1,21 is not considered


so we only have 3,7 and 7,3

but as we know X>Y thus x=7 and Y=3

So 2) gives definitive answer and we get B as the correct answer for this question.

1) x+y=10

can be many options as
9+1
8+2
7+3
6+4
(insufficient)

2) xy=21
7*3 is only possible as x>y
(sufficient)
Answer (B)

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