If a, b, c, and d, are positive numbers, is a/b < c/d?

Hi catty,

We're looking for whether a/b < c/d. Fortunately, we're told a useful bit of info in the question stem. All four terms are positive. That's very important with inequalities, because it means that we can multiply and divide without having to worry about the direction of the inequality signs. In this case, we could rephrase the question to whether ad < bc by cross-multiplying. This will be useful laters.

Statement 1) is not useful, however. (c-a) and (d-b) could both be positive or negative; that means that when me multiply to get rid of a term, we might or might not have to flip the terms. Since any of the variables could be greater or less than any of the other variables, this statement is insufficient.

Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get ad/bc < 1, and can cross-multiply to get ad < bc. That answers our question with a definite yes, so it's sufficient and the answer is (B)

We know that a,b,c and d are positive numbers.
This is a Yes/No DS question type - is a/b < c/d.
Since we are certain that we have no negative values, we can manipulate the inequality question to - is ad < bc? It's much easier to look at.

(1) (c-a)/(d-b) - a positive fraction or whole number

Say c=d=5 and a=2 and b=1 for 3/4, then ad < bc is false
Say c=d=5 and a=1 and b=2 for 4/3, then ad < bc is true
thus (1) is INSUFFICIENT

(2) Thus, YES! SUFFICIENT. See attachment.

Answer: B

Answer: Option B

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Hi could you please explain the part on cross multiplication? I am getting \(a/b\) > \(b/c\).

\((\frac{ad}{bc})^2 < \frac{ad}{bc}\) --> reduce by ad/bc: \(\frac{ad}{bc} <1\) --> multiply by bc: \(ad<bc\).

Hope it's clear.
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If \(ad<bc\), then \(\frac{a}{b}\) < \((\frac{c}{d})\)?

For positive values, yes.
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Bunuel,

2) This quite clearly sufficient
1) However, it was not very clear to me how this one is insufficient. Is there any way we know this choice is not sufficient w/o resorting to number picking?

Thanks!

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a, b, c, and d, are positive numbers, is a/b < c/d?

(1) 0 < (c-a) / (d-b)

(2) (ad/bc)^2 < (ad)/(bc)

If we modify the question, the sign of the inequality does not change as a,b,c,d are positive integers. Hence, we want to know whether a/b < c/d?, or ad<bc.
From condition 1, 0 < (c-a) / (d-b), and if we multiply (d-b)^2 on both sides, 0<(c-a)(d-b). We cannot know whether ad<bc, so this is insufficient.
From condition 2, we can divide both sides by (ad)/(bc), which gives us (ad/bc)^2 < (ad)/(bc), or (ad/bc)<1, and when we multiply bc on both sides, (ad/bc)<1, or ad<bc. This answers the question 'yes' so this is sufficient, and the answer becomes (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

I know your question is to Bunuel, but I am just sharing how I eliminated the first statement:

Statement 1 tells us that \(\frac{c-a}{d-b}>0\), i.e. \(c-a\) and \(d-b\) have same signs (either both are +ve or both are -ve). However that doesn't tells us anything about their individual values, which makes this statement insufficient.

Hope it helps.

hey,

for statement 1:

can't we just multiply the equation by d-b ?


1. 0 < c-a/d-b --> multiply by d-b --> 0 < c-a --> a < c so a < c but we don't know whether a/b < c/d


Is that correct ?

We cannot multiply the equation by d-b because we do not know if d-b is less or greater than zero. For instance, d=3 and b=4, then 3-4 = -1. Rule of thumb: we cannot manipulate inequalities without knowing the signs. (and as I mentioned before we don't know the sign of d-b). Hope it helps

This is perhaps off topic but it caught my attention. How can a square of a number x be strictly less than x? I can't seem to find it logical.

x^2 < x is true for 0 < x < 1.

For example, (1/2)^2 < 1/2.

Check the links below for more:
Inequality tips

Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach

Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

Hope it helps.
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Plug in Approach

Statement 1: a and b and also c and d can each be different numbers for the same value of a/b and c/d resp. Hence will give different answers and not sufficient

Statement 2: At least one of a/b or d/c is less than 1 and the other has a ceiling on its value. For convenience we will round values to one decimal.

If d/c =0.9, a/b should be less than 1.1. Since d/c=0.9, c/d=1.1. Hence c/d >a/b
If a/b=0.9, d/c should be less than 1.1, and c/d>0.9. Hence c/d>a/b
If both are less than 1, then c/d>a/b
Hence c/d in all the cases gretaer than a/b. Sufficient

Hence B.
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Hello everyone,

I see many great explanations regarding this question but I still cannot understand how to know quickly that 1 is not sufficient. I understand that to satisfy the condition 0<c-a/d-b, either both must be negative or both must be positive. But from there I'm lost... I know that c-a can be positive or negative, d-b can also be positive or negative. But so what? How does this help to determine if this statement is sufficient? Can someone please explain? Thank you!

kamrim322 Genoa2000
Let's solve once again
(1) \(0 < \frac{(c-a)}{(d-b)}\)
Either both numerator and denominator are positive or both are negative.
Case I: Let's take them positive. So, c > a and d > b
But if we divide
\(\frac{c}{d} > \frac{a}{b}\) (\(\frac{3}{4} > \frac{1}{2}\)) \(\implies\) bc > ad; YES

Case II: Both are negative. So, c < a and d < b
\(\frac{c}{d} < \frac{a}{b}\) (\(\frac{1}{3} < \frac{2}{4}\)) \(\implies\) bc < ad; NO

(Note that taking either both numerator and denominator are positive or both are negative, we are making an assumption which leads to other possibilities thus no unique solution)
Also, beware of dividing or multiplying by variable with negative sign. I have taken only positive values and still reached insufficiency of statement 1.

INSUFFICIENT.

(2) \((\frac{ad}{bc})^2 < \frac{(ad)}{(bc)}\)
Here you can solve it for a unique solution:
Dividing the statement by \(\frac{(ad)}{(bc)}\)
\((\frac{ad}{bc}) < 1\)
ad < bc; YES

or

\((\frac{ad}{bc})^2 - \frac{(ad)}{(bc)} < 0\)
\(\frac{ad}{bc}(\frac{ad}{bc} - \frac{(ad)}{(bc)}) < 0\)
So, either \((\frac{ad}{bc}) < 0\) OR \((\frac{ad}{bc}) < 1\)
The former is not true as a, b, c and d all are positive.

Thus ad < bc.

SUFFICIENT.

Answer B.

HTH
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Solution:

We are asked if a/b < c/d => Is ad <b c ?
​
St(1):- 0 < (c-a) / (d-b)

0 < (c-a)(d-b) (Multiply (d-b)^2 on both sides)

Hence either both (c-a) and (d-b) are positive or both are negative.

When both are positive-

c-a>0 = > c>a & d-b>0=>d>b

If a=1,b=2,c=3,d=4 (here c>a and d>b)then ad =4 < b c=6 –(Yes)

If a=3,b=2,c=4,d=5(here c>a and d>b) then ad = 15 < b c = 8 (No)

The insufficiency is proved and hence you do not need to check for both negative.
Hence insufficient with contradictory answers.

St(2):- (ad/bc )^2 < (ad/bc)

=>(ad/bc) < 1 (Divide by ad/bc on both sides)

=> ad<bc (Sufficient) Option(b)
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