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Updated on: 31 Aug 2026, 01:41
We ran a live Quant workshop on inequalities, hosted by hr1212 Harsh Rumalwala. One method carried the entire session, and then five questions to apply it to.
Full solutions are behind spoilers so you can attempt them first.
THE METHOD: NUMBER LINE, NOT SUBSTITUTION
Shalmali Sriram named it early when asked how she approaches inequalities, and the rest of the session was built on it.
The instinct most people have is to plug in random numbers and check. Harsh's argument against that is simple: substitution costs you time and never tells you when to stop, because you are guessing which numbers to try and then guessing whether you have tried enough of them.
The number line method removes the guessing.
The procedure
The warm-up we used
Critical points are 2, 3 and 5. Mark them, then take any number in each gap and read the sign of the whole expression:
Pick x = 5.5, above 5: all three factors positive, so the result is positive. Step left of 5 and the denominator flips negative, so the whole thing goes negative. Step left of 3 and another factor flips, so it returns to positive. The signs alternate as you cross each critical point.
We want the expression to be less than or equal to zero, so we take the negative intervals.
Please note the bracket on 5. The inequality allows equality, so 2 and 3 are included. But 5 makes the denominator zero, so it is excluded no matter what the inequality sign says. That distinction between a bracket and a parenthesis is where a lot of otherwise correct work gets thrown away.
Once you see the pattern, it gets faster. The signs simply alternate across the critical points, so you can often place one sign and fill in the rest without testing every interval.
QUESTION 1 - RATIONAL INEQUALITY
What is the range of x such that
The answer is (A), 1 < x < 2.5.
There is x on both sides, so the first move is to get everything on one side. Do not be tempted to cross-multiply by (x − 1), because you do not know its sign, and multiplying an inequality by a negative flips it.
The 4 is positive, so dividing it out changes nothing. Now the critical points are 1 and 2.5.
We want strictly less than zero, so the middle interval: 1 < x < 2.5.
Notice how much the algebra simplified. A quadratic over a linear collapsed into a linear over a linear, because the x2 terms cancelled. That cancellation is the whole design of the question.
QUESTION 2 - RECIPROCALS
If t denotes a negative number such that 1/81 < t2 < 1/9, then 1/t must be
The answer is (B), between −9 and −3.
Take the square root of all three parts. The square root of t2 is |t|, not t, and this is the step where most of the errors happen.
Writing the modulus explicitly gives two branches:
The question says t is negative, so the second branch survives.
Now reciprocate. Both bounds have the same sign, both negative, so the inequality signs flip.
That is (B).
The rule worth memorising: when both sides of an inequality carry the same sign, reciprocating flips the direction. When the two sides straddle zero, the rule does not apply and you have to split into cases.
Harsh's advice here was to physically write the modulus line out rather than doing it in your head. It is one extra line and it stops you from silently dropping the negative branch, which is exactly the trap this question is built around.
Try these to check the reflex
QUESTION 3 - LATTICE POINT COUNTING
The number of ordered pairs of integers (x, y) satisfying x2 + y2 < 100 and |x| > |y| is
The answer is (D), 138.
x2 + y2 < 100 confines both variables to −9 through 9. The counting decision that matters is which variable to organise by. Harsh's suggestion was to fix y and count x, because |x| > |y| means y is the smaller one, so working upward from y = 0 keeps the ranges shrinking in a predictable way.
For each y, x has to satisfy two things at once: |x| > |y|, and x2 < 100 − y2.
The first ×2 is for the sign of x, the second is for the sign of y. At y = 0 there is only one y, so there is no second doubling.
Two places this goes wrong.
The upper limits tighten as |y| grows and it is easy to miss. At y = ±5 you need x2 < 75, so |x| can only reach 8, not 9. At y = ±6 you need x2 < 64, so |x| = 7 is the only value that also satisfies |x| > 6. At y = ±7 the two conditions contradict each other and nothing survives, which is why the table stops.
The other trap is the word ordered. (3, 1) and (−3, 1) and (3, −1) and (−3, −1) are four different pairs. Forgetting a sign factor is the fastest route to 120 or 126, both of which are sitting there as answer choices.
Running total as you go: 18, 50, 78, 102, 122, 134, 138.
QUESTION 4 - ABSOLUTE VALUES
If |8x − 1| < |3x + 9|, which of the following CANNOT be the product of two possible values of x?
The answer is (A), −3/2.
When both sides of an inequality are absolute values, square both sides. Squaring is safe here because both sides are non-negative, and it removes both moduli in one step rather than forcing you through four sign cases.
That is a difference of two squares, so factor it rather than expanding:
Critical points −8/11 and 2, and we want the product negative, so:
Now the actual question. We need the product of two values drawn from that range. The most negative product comes from pairing the two extremes, one near each end:
And that bound is not even attainable, since both endpoints are excluded. So every achievable product is greater than −16/11.
−3/2 is −1.5, which is below −16/11. It cannot be reached. Every other option sits inside the achievable range.
The trap: −3/2 and −16/11 differ by less than 0.05. If you approximate early or round −8/11 to −0.7, you will not be able to separate them.
QUESTION 5 - LINEAR INEQUALITIES
If 2x + 3y < 15 and 3x − y ≥ −5, which of the following is true?
The answer is (B), y < 5.
You can see it from a graph, but here is the algebra so you are not relying on a sketch.
Rearrange the second inequality to bound x from below, and the first to bound x from above.
For any solution to exist at all, the lower bound has to sit below the upper bound:
So y < 5 is forced by the two constraints together.
Why (A) fails. Set y = 0. The constraints become 2x < 15 and 3x ≥ −5, so x can be anywhere from −5/3 up to 7.5. x = −1 satisfies both. x > 0 is not required, which also kills (D) and (E).
On (C). y ≤ 5 is not false, but it is a weaker statement than the bound we actually derived, and y = 5 is never attainable. (B) is the precise answer.
THREE RULES THAT CAME OUT OF THE SESSION
And underneath all three, the point Harsh kept returning to: you do not need to memorise sign rules if you can put the critical points on a number line and test one value per interval. The number line reconstructs the rule for you in about five seconds, which is faster than trying to remember which case you are in.
Thanks to Shalmali Sriram for naming the method, and to Nitesh Motwani, Praveen Nedunuri, Sai Sandeep, Akhil T and Sayi Shharadha for working through these out loud. Hosted by @hr1212 Harsh Rumalwala.
Post your approach below, particularly on question 3. There are at least two sensible ways to organise that count and the one you choose changes how error-prone it is.
Full solutions are behind spoilers so you can attempt them first.
THE METHOD: NUMBER LINE, NOT SUBSTITUTION
Shalmali Sriram named it early when asked how she approaches inequalities, and the rest of the session was built on it.
The instinct most people have is to plug in random numbers and check. Harsh's argument against that is simple: substitution costs you time and never tells you when to stop, because you are guessing which numbers to try and then guessing whether you have tried enough of them.
The number line method removes the guessing.
The procedure
- Rearrange so one side is zero. Everything on the left, zero on the right.
- Find the points where each factor equals zero. These are your critical points.
- Mark them on a number line. They cut it into intervals.
- Determine the sign in each interval, then read off the intervals that satisfy your inequality.
The warm-up we used
Code:
(x − 2)(x − 3) / (x − 5) ≤ 0
Critical points are 2, 3 and 5. Mark them, then take any number in each gap and read the sign of the whole expression:
Code:
− − − − − 2 + + + + + 3 − − − − − 5 + + + + +
Pick x = 5.5, above 5: all three factors positive, so the result is positive. Step left of 5 and the denominator flips negative, so the whole thing goes negative. Step left of 3 and another factor flips, so it returns to positive. The signs alternate as you cross each critical point.
We want the expression to be less than or equal to zero, so we take the negative intervals.
Code:
Answer: x ≤ 2 or 3 ≤ x < 5
(−∞, 2] ∪ [3, 5)
(−∞, 2] ∪ [3, 5)
Please note the bracket on 5. The inequality allows equality, so 2 and 3 are included. But 5 makes the denominator zero, so it is excluded no matter what the inequality sign says. That distinction between a bracket and a parenthesis is where a lot of otherwise correct work gets thrown away.
Once you see the pattern, it gets faster. The signs simply alternate across the critical points, so you can often place one sign and fill in the rest without testing every interval.
QUESTION 1 - RATIONAL INEQUALITY
What is the range of x such that
Code:
(x − 2)(x + 5) / (x − 1) < x
- (A) 1 < x < 2.5
- (B) −5 < x < 1
- (C) −2 < x < 5
- (D) x < −5 or x > 1
- (E) −5 < x < 5
Show SpoilerSolution
The answer is (A), 1 < x < 2.5.
There is x on both sides, so the first move is to get everything on one side. Do not be tempted to cross-multiply by (x − 1), because you do not know its sign, and multiplying an inequality by a negative flips it.
Code:
((x − 2)(x + 5)) / (x − 1) − x < 0
(x2 + 3x − 10) / (x − 1) − x < 0
((x2 + 3x − 10) − (x2 − x)) / (x − 1) < 0
(4x − 10) / (x − 1) < 0
4(x − 2.5) / (x − 1) < 0
(x − 2.5) / (x − 1) < 0
(x2 + 3x − 10) / (x − 1) − x < 0
((x2 + 3x − 10) − (x2 − x)) / (x − 1) < 0
(4x − 10) / (x − 1) < 0
4(x − 2.5) / (x − 1) < 0
(x − 2.5) / (x − 1) < 0
The 4 is positive, so dividing it out changes nothing. Now the critical points are 1 and 2.5.
Code:
+ + + + + 1 − − − − − 2.5 + + + + +
We want strictly less than zero, so the middle interval: 1 < x < 2.5.
Notice how much the algebra simplified. A quadratic over a linear collapsed into a linear over a linear, because the x2 terms cancelled. That cancellation is the whole design of the question.
QUESTION 2 - RECIPROCALS
If t denotes a negative number such that 1/81 < t2 < 1/9, then 1/t must be
- (A) Less than −9
- (B) Between −9 and −3
- (C) Between −3 and 0
- (D) Between 0 and 3
- (E) Greater than 9
Show SpoilerSolution
The answer is (B), between −9 and −3.
Take the square root of all three parts. The square root of t2 is |t|, not t, and this is the step where most of the errors happen.
Code:
1/81 < t2 < 1/9
1/9 < |t| < 1/3
1/9 < |t| < 1/3
Writing the modulus explicitly gives two branches:
Code:
1/9 < t < 1/3 or −1/3 < t < −1/9
The question says t is negative, so the second branch survives.
Code:
−1/3 < t < −1/9
Now reciprocate. Both bounds have the same sign, both negative, so the inequality signs flip.
Code:
−3 > 1/t > −9
which is −9 < 1/t < −3
which is −9 < 1/t < −3
That is (B).
The rule worth memorising: when both sides of an inequality carry the same sign, reciprocating flips the direction. When the two sides straddle zero, the rule does not apply and you have to split into cases.
Harsh's advice here was to physically write the modulus line out rather than doing it in your head. It is one extra line and it stops you from silently dropping the negative branch, which is exactly the trap this question is built around.
Try these to check the reflex
Code:
1/4 < y < 1/2 → 2 < 1/y < 4 (both positive, flips)
−2 < x < −0.1 → reciprocate carefully
−2 < x < 2 → straddles zero, split it
−2 < x < −0.1 → reciprocate carefully
−2 < x < 2 → straddles zero, split it
QUESTION 3 - LATTICE POINT COUNTING
The number of ordered pairs of integers (x, y) satisfying x2 + y2 < 100 and |x| > |y| is
- (A) 120
- (B) 126
- (C) 132
- (D) 138
- (E) 144
Show SpoilerSolution
The answer is (D), 138.
x2 + y2 < 100 confines both variables to −9 through 9. The counting decision that matters is which variable to organise by. Harsh's suggestion was to fix y and count x, because |x| > |y| means y is the smaller one, so working upward from y = 0 keeps the ranges shrinking in a predictable way.
For each y, x has to satisfy two things at once: |x| > |y|, and x2 < 100 − y2.
Code:
y permitted |x| count
0 1 to 9 9 × 2 = 18
±1 2 to 9 8 × 2 × 2 = 32
±2 3 to 9 7 × 2 × 2 = 28
±3 4 to 9 6 × 2 × 2 = 24
±4 5 to 9 5 × 2 × 2 = 20
±5 6 to 8 3 × 2 × 2 = 12
±6 7 only 1 × 2 × 2 = 4
Total = 138
0 1 to 9 9 × 2 = 18
±1 2 to 9 8 × 2 × 2 = 32
±2 3 to 9 7 × 2 × 2 = 28
±3 4 to 9 6 × 2 × 2 = 24
±4 5 to 9 5 × 2 × 2 = 20
±5 6 to 8 3 × 2 × 2 = 12
±6 7 only 1 × 2 × 2 = 4
Total = 138
The first ×2 is for the sign of x, the second is for the sign of y. At y = 0 there is only one y, so there is no second doubling.
Two places this goes wrong.
The upper limits tighten as |y| grows and it is easy to miss. At y = ±5 you need x2 < 75, so |x| can only reach 8, not 9. At y = ±6 you need x2 < 64, so |x| = 7 is the only value that also satisfies |x| > 6. At y = ±7 the two conditions contradict each other and nothing survives, which is why the table stops.
The other trap is the word ordered. (3, 1) and (−3, 1) and (3, −1) and (−3, −1) are four different pairs. Forgetting a sign factor is the fastest route to 120 or 126, both of which are sitting there as answer choices.
Running total as you go: 18, 50, 78, 102, 122, 134, 138.
QUESTION 4 - ABSOLUTE VALUES
If |8x − 1| < |3x + 9|, which of the following CANNOT be the product of two possible values of x?
- (A) −3/2
- (B) −1
- (C) 1/4
- (D) 1
- (E) 3
Show SpoilerSolution
The answer is (A), −3/2.
When both sides of an inequality are absolute values, square both sides. Squaring is safe here because both sides are non-negative, and it removes both moduli in one step rather than forcing you through four sign cases.
Code:
(8x − 1)2 < (3x + 9)2
(8x − 1)2 − (3x + 9)2 < 0
(8x − 1)2 − (3x + 9)2 < 0
That is a difference of two squares, so factor it rather than expanding:
Code:
(8x − 1 − 3x − 9)(8x − 1 + 3x + 9) < 0
(5x − 10)(11x + 8) < 0
(x − 2)(x + 8/11) < 0
(5x − 10)(11x + 8) < 0
(x − 2)(x + 8/11) < 0
Critical points −8/11 and 2, and we want the product negative, so:
Code:
−8/11 < x < 2
Now the actual question. We need the product of two values drawn from that range. The most negative product comes from pairing the two extremes, one near each end:
Code:
−8/11 × 2 = −16/11 ≈ −1.4545
And that bound is not even attainable, since both endpoints are excluded. So every achievable product is greater than −16/11.
−3/2 is −1.5, which is below −16/11. It cannot be reached. Every other option sits inside the achievable range.
The trap: −3/2 and −16/11 differ by less than 0.05. If you approximate early or round −8/11 to −0.7, you will not be able to separate them.
QUESTION 5 - LINEAR INEQUALITIES
If 2x + 3y < 15 and 3x − y ≥ −5, which of the following is true?
- (A) x > 0
- (B) y < 5
- (C) y ≤ 5
- (D) A and B
- (E) A and C
Show SpoilerSolution
The answer is (B), y < 5.
You can see it from a graph, but here is the algebra so you are not relying on a sketch.
Rearrange the second inequality to bound x from below, and the first to bound x from above.
Code:
3x − y ≥ −5 → x ≥ (y − 5)/3
2x + 3y < 15 → x < (15 − 3y)/2
2x + 3y < 15 → x < (15 − 3y)/2
For any solution to exist at all, the lower bound has to sit below the upper bound:
Code:
(y − 5)/3 < (15 − 3y)/2
2(y − 5) < 3(15 − 3y)
2y − 10 < 45 − 9y
11y < 55
y < 5
2(y − 5) < 3(15 − 3y)
2y − 10 < 45 − 9y
11y < 55
y < 5
So y < 5 is forced by the two constraints together.
Why (A) fails. Set y = 0. The constraints become 2x < 15 and 3x ≥ −5, so x can be anywhere from −5/3 up to 7.5. x = −1 satisfies both. x > 0 is not required, which also kills (D) and (E).
On (C). y ≤ 5 is not false, but it is a weaker statement than the bound we actually derived, and y = 5 is never attainable. (B) is the precise answer.
THREE RULES THAT CAME OUT OF THE SESSION
- Never multiply or divide an inequality by an expression whose sign you do not know. On question 1, cross-multiplying by (x − 1) is the obvious move and it is wrong, because x − 1 changes sign inside the region you care about. Move everything to one side instead.
- Square roots produce a modulus, and a modulus has two branches. Write the branch line out explicitly. On question 2 the negative branch is the entire answer, and it is the branch people drop.
- Reciprocating flips the inequality when both sides share a sign. If the range straddles zero, split it before reciprocating.
And underneath all three, the point Harsh kept returning to: you do not need to memorise sign rules if you can put the critical points on a number line and test one value per interval. The number line reconstructs the rule for you in about five seconds, which is faster than trying to remember which case you are in.
Thanks to Shalmali Sriram for naming the method, and to Nitesh Motwani, Praveen Nedunuri, Sai Sandeep, Akhil T and Sayi Shharadha for working through these out loud. Hosted by @hr1212 Harsh Rumalwala.
Post your approach below, particularly on question 3. There are at least two sensible ways to organise that count and the one you choose changes how error-prone it is.
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