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20 May 2019, 01:56
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B
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Difficulty:
35%
(medium)
Question Stats:
72% (02:05) correct
28%
(02:15)
wrong
based on 171
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If px+qy > rx+sy , and x<y where p,q,r,s,x and y are all positive real numbers, then which of the following must be true?
1. p-q>r-s
2. p+q>r+s
3. p>s
A. only 1
B. only 2
C. only 3
D. Both 1 and 2
E. Both 1 and 3
1. p-q>r-s
2. p+q>r+s
3. p>s
A. only 1
B. only 2
C. only 3
D. Both 1 and 2
E. Both 1 and 3
24 May 2019, 00:48
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Given : x<y
p,q,r,s,x,y are all positive real numbers.
px+qy > rx+sy
==> x(p-r) > y(s-q)
Since given that x< y and all numbers are positive , the above equality will hold only if
p-r > s-q . Solving this we get:
p+q > r+s.
As for equations 1 and 3, they are not necessarily true, this can be proved by plugging in positive numbers in above equation.
Thus answer is B
p,q,r,s,x,y are all positive real numbers.
px+qy > rx+sy
==> x(p-r) > y(s-q)
Since given that x< y and all numbers are positive , the above equality will hold only if
p-r > s-q . Solving this we get:
p+q > r+s.
As for equations 1 and 3, they are not necessarily true, this can be proved by plugging in positive numbers in above equation.
Thus answer is B
24 May 2019, 10:31
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Since this is a ‘Must be’ type of a question, it is a good idea to take simple cases and see if we can make the statements False. This is especially a good approach in this problem because, we know that all the numbers we are dealing with are positive numbers.
Since x < y, let’s take x = 2 and y = 5. Also, px + qy > rx + sy. Let p = 5, q = 4, r = 4, s =3.
Then, 5*2 + 4*5 > 4*2 + 3*5 which is (10 + 20) > (8 + 15) which is true.
But in this case, (p-q) = (r-s). Therefore, the first statement is not true in all cases (we got one false case just noww). So, options containing statement 1 can be eliminated. Option A, option D and option E are ruled out.
For the same set of values of x and y, if we take p = 5, q = 8, r= 8 and s = 5, we have,
5*2 + 8*5 > 8*2+5*5 which is (10 + 40) > (16 + 25) which is true.
But in this case, p = s. So, statement 3 is not always true and the options containing option 3 can be eliminated, which means option C can be eliminated.
The only option left is option B, which has to be the right answer.
In ‘must be’ true kind of questions, the best approach is to always take simple values and eliminate as many options as you can, before you analyse (if you have to, that is). So, this question is actually a moderately difficult question, which should take you about 2 minutes to solve.
Hope this helps!
Since x < y, let’s take x = 2 and y = 5. Also, px + qy > rx + sy. Let p = 5, q = 4, r = 4, s =3.
Then, 5*2 + 4*5 > 4*2 + 3*5 which is (10 + 20) > (8 + 15) which is true.
But in this case, (p-q) = (r-s). Therefore, the first statement is not true in all cases (we got one false case just noww). So, options containing statement 1 can be eliminated. Option A, option D and option E are ruled out.
For the same set of values of x and y, if we take p = 5, q = 8, r= 8 and s = 5, we have,
5*2 + 8*5 > 8*2+5*5 which is (10 + 40) > (16 + 25) which is true.
But in this case, p = s. So, statement 3 is not always true and the options containing option 3 can be eliminated, which means option C can be eliminated.
The only option left is option B, which has to be the right answer.
In ‘must be’ true kind of questions, the best approach is to always take simple values and eliminate as many options as you can, before you analyse (if you have to, that is). So, this question is actually a moderately difficult question, which should take you about 2 minutes to solve.
Hope this helps!
