Question 1
|x − 12| = 6
What is the sum of the solutions?
- 6
- 12
- 18
- 24
- 64
Show the worked solution
- The solutions are 6 and 18.
- Sum = 24 (twice the centre).
- Answer: 24.
Absolute value
Absolute value questions are distance questions. The expression |x − a| is the distance from x to a, so |x − a| = b means x sits b units either side of a, giving two solutions that always sum to twice a. The inequality |x − a| < b means x lies strictly between a − b and a + b.
Reading the bars as distance turns most of this topic into mental arithmetic. |x − 4| = 6 asks which numbers are six away from four, which is −2 and 10, with no algebra at all. The same reading handles the inequalities, where the interval is simply centred on a and reaches b in each direction.
Try it against the clock. Absolute Value runs at 90s easy, 105s medium and 120s hard. The clock is stamped and judged on our server, so the limit you see is the deadline that is actually enforced — and it starts when you tap Start, not while the question is loading.
In |x − 4| = 6 the centre is 4 and the distance is 6. Every question of this shape is fully described by those two numbers.
Six units either side of 4 gives −2 and 10. Their sum is 8, which is always twice the centre — a fact that answers sum questions without finding either root.
|x − 2| < 6 means x lies strictly between −4 and 8. The interval is centred on 2 and extends 6 in each direction, so no case-splitting is needed.
Strictly between −4 and 8 the integers run from −3 to 7, which is 11 values. Counting across zero is where this step usually goes wrong.
| The expression | Reads as |
|---|---|
| |x − a| = b | x is b away from a — two solutions |
| |x − a| < b | x is strictly inside (a − b, a + b) |
| |x − a| > b | x is outside the interval, on either side |
| |x − a| = 0 | Exactly one solution, x = a |
| |x − a| = negative | No solutions at all |
Generated by the same engine that mints the Quant Daily daily. Each answer is computed from the numbers printed in the question, and the walkthrough below each one is the engine's own working — not a solution written afterwards. Reload this page's live drill and you get different numbers.
|x − 12| = 6
What is the sum of the solutions?
|x − 6| < 4
How many integer values of x satisfy the inequality?
|x − 4| = 6
What is the sum of the solutions?
The wrong options are not random numbers. Each one is the result of a specific careless error for this topic, so picking one tells you which habit is costing you marks.
|x − 4| = 6 has two answers, not one. Solving x − 4 = 6 and stopping drops the −2 entirely.
From −3 to 7 there are 11 integers, because zero is one of them. Subtracting the endpoints and forgetting to add one is the standard slip.
A distance can never be negative, so an absolute value set equal to a negative number has no solutions. This is a real answer, not a trick.
The specific wrong numbers this topic produces, and what each one tells you about the step you took. If you have just got a question wrong and want to know which habit did it, start here.
It has two: 10 and −2. The bars mean distance, and six units from four reaches in both directions. Solving only x − 4 = 6 finds the upper one and silently drops the lower, and questions asking for the sum or the count are built to catch precisely that.
It is 11. The interval is −4 < x < 8 with both ends strict, so the integers run from −3 to 7. Counting 12 usually means an endpoint was included. Subtracting the bounds gives the width, not the count — remember to exclude both open ends.
It has none. An absolute value is a distance and can never be negative, so no x makes this true. The two-solution rule only applies when the right-hand side is positive; zero gives one solution and a negative gives none at all.
Because the bars measure distance, and there are always two points at a given distance from a centre — one on each side. |x − 4| = 6 is asking which numbers are six units from four, and both −2 and 10 qualify. Only a distance of zero gives a single solution.
It is twice the centre. The two solutions sit symmetrically either side of a, so their average is a and their sum is 2a. For |x − 4| = 6 the sum is 8, and you never need to compute −2 and 10 at all.
Convert it to an interval first. |x − 2| < 6 becomes −4 < x < 8, and then count the integers strictly inside: −3 through 7, which is 11. Converting first removes the case-splitting and leaves an ordinary counting problem.
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