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17 Jul 2019, 04:47
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:36) correct
37%
(02:15)
wrong
based on 276
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Product of three distinct positive integers p, q and r is 120. Is p + q > r?
(1) 4p + 5r = 50
(2) p < r
(1) 4p + 5r = 50
(2) p < r
18 Jul 2019, 03:45
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This is a great question on equations. It tests whether you are able to break the stereotype of “Number of equations should be equal to number of variables” and solve it using logic.
For you to be able to do it though, you need to extract every bit of data that you possibly can from the statements and the question. In this question, we are lucky that the question mentions explicitly that the integers p,q and r are positive integers. And distinct too. This is a BIG clue.
Also, p*q*r = 120. For example, we could take p = 10, q = 2 and r = 6; or p = 2, q = 6 and r = 10. Depending on the relative values of p, q and r, we need to figure out if p+q > r.
From statement I, 4p + 5r = 50. Looking at the equation, if you thought that this equation cannot be solved because there are more variables than equations, you cannot be more wrong.
In 4p + 5r = 50, you need to understand that 4p will always be even, irrespective of the value of p. 50 is also even. Therefore, 5r HAS to be even since only EVEN + EVEN = EVEN.
5r can be even only when r is even.
So, the only values we need to try for r are the multiples of 120 which are even numbers and which satisfy the equation. This means, r = 2 or 4 or 6 or 8 or 10. The other even factors of 120 do not satisfy the equation given.
For r = 2, p = 10. Therefore, q = 6. In this case, p + q > r.
For r = 4, p = \(\frac{30}{4}\) which is not an integer. Ruled out.
For r = 6, p = 5. Therefore, q = 4. In this case also, p + q > r.
For r = 8, p = \(\frac{10}{4}\), not an integer. Ruled out.
For r = 10, p = 0, which is not a positive integer. Ruled out.
Although we have two cases, in both the cases, we are able to say p+q is always more than r. Statement I alone is sufficient. Possible answer options are A or D. Answer options B, C and E can be ruled out.
From statement II, p<r.
p = 3 and r = 4 is one case. Here, q = 10 and so p+q>r.
p = 3 and r = 10 is another case. Here, q = 4 and so p+q<r.
Statement II alone is insufficient. Answer option D can be eliminated.
The correct answer option is A.
When there is ONE linear equation with two variables, do not conclude straight away that it cannot be solved. Depending on what additional conditions are given, you may just be able to solve it, even uniquely sometimes.
So, the next time you are faced with an equation like 4p + 5r = 50, look for additional constraints.
Hope this helps!
For you to be able to do it though, you need to extract every bit of data that you possibly can from the statements and the question. In this question, we are lucky that the question mentions explicitly that the integers p,q and r are positive integers. And distinct too. This is a BIG clue.
Also, p*q*r = 120. For example, we could take p = 10, q = 2 and r = 6; or p = 2, q = 6 and r = 10. Depending on the relative values of p, q and r, we need to figure out if p+q > r.
From statement I, 4p + 5r = 50. Looking at the equation, if you thought that this equation cannot be solved because there are more variables than equations, you cannot be more wrong.
In 4p + 5r = 50, you need to understand that 4p will always be even, irrespective of the value of p. 50 is also even. Therefore, 5r HAS to be even since only EVEN + EVEN = EVEN.
5r can be even only when r is even.
So, the only values we need to try for r are the multiples of 120 which are even numbers and which satisfy the equation. This means, r = 2 or 4 or 6 or 8 or 10. The other even factors of 120 do not satisfy the equation given.
For r = 2, p = 10. Therefore, q = 6. In this case, p + q > r.
For r = 4, p = \(\frac{30}{4}\) which is not an integer. Ruled out.
For r = 6, p = 5. Therefore, q = 4. In this case also, p + q > r.
For r = 8, p = \(\frac{10}{4}\), not an integer. Ruled out.
For r = 10, p = 0, which is not a positive integer. Ruled out.
Although we have two cases, in both the cases, we are able to say p+q is always more than r. Statement I alone is sufficient. Possible answer options are A or D. Answer options B, C and E can be ruled out.
From statement II, p<r.
p = 3 and r = 4 is one case. Here, q = 10 and so p+q>r.
p = 3 and r = 10 is another case. Here, q = 4 and so p+q<r.
Statement II alone is insufficient. Answer option D can be eliminated.
The correct answer option is A.
When there is ONE linear equation with two variables, do not conclude straight away that it cannot be solved. Depending on what additional conditions are given, you may just be able to solve it, even uniquely sometimes.
So, the next time you are faced with an equation like 4p + 5r = 50, look for additional constraints.
Hope this helps!
18 Jul 2019, 11:51
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Dillesh4096
The source JAMBOREE says, "Keep it simple".
So why not
Given- positive integers, pqr=120
We have an equation from st 1
possible set of (p,r)
(5,6)(10,2)
In each case p+q>r
A
