Lesson 11 of 35 · Quantitative Reasoning: algebra
Estimation and working from the answer choices
Estimate by hand to the precision the choices need, back-solve from numeric choices, and pick numbers when the choices are expressions.
About 20 minutes
In this lesson
- Decide from the spread of the choices how precise an estimate must be, and round accordingly.
- Back-solve by testing numeric choices, starting from the middle value.
- Pick numbers to test algebraic choices, and avoid numbers that make choices coincide.
The idea
The choices are information
A problem-solving question is multiple choice, and only one choice is right. That is useful: you need enough work to tell the right choice from the others, and sometimes far less than a full solution. GMAT Quantitative Reasoning gives you 45 minutes for 21 questions without a calculator, so a method that replaces two minutes of arithmetic with thirty seconds of reasoning is worth learning on purpose.
This lesson covers three such methods: estimating, testing the choices, and picking numbers. Each is a real mathematical method, not a guess, and each has a way to go wrong.
Estimation: let the spread set the precision
Before estimating, look at how far apart the choices are.
- Choices differ by powers of ten (0.4, 4, 40, 400): you only need the size of the answer. Round hard.
- Choices are close (3.8, 4.0, 4.2): rounding will not separate them. Compute properly, or estimate with care about direction.
Round to numbers that are easy to handle, such as , , , and keep track of which way each rounding moves the result. If you round one factor up, rounding another down partly cancels it; rounding everything the same way piles up the error.
For powers of ten, write each number in scientific notation and handle the powers separately:
Testing the choices (back-solving)
When the question asks for a single number and the choices are numbers, you can put each choice into the conditions of the problem and see which one works. Two habits make it fast:
- Start with the middle value. When the choices are in order, a middle choice that comes out too big or too small tells you which direction to move, so you rarely test more than two or three.
- Test against the original wording, not against an equation you may have set up wrongly.
Back-solving shines when setting up the equation is harder than checking a value: quadratics with integer answers, age problems, problems with conditions such as "consecutive odd integers".
Picking numbers
When the choices are expressions in variables, pick simple values for the variables, compute the answer from the story, then evaluate each choice with the same values. The choice that matches is the answer.
Good picks are small, different from each other, and make the arithmetic clean. Avoid 0 and 1, which make many expressions equal (any power of 1 is 1; anything times 0 is 0), and avoid giving two variables the same value. Respect every condition in the stem: if must be an odd integer, pick 3, not 2.
Test every choice. If two choices match your numbers, the pick was unlucky: pick new numbers and test only those two again.
Picking numbers also handles percent and ratio questions with no actual amounts: start from 100 for a percent question, or from a common multiple of the denominators for a fraction question.
Quick elimination
Even when you plan to compute, a few checks can strike out choices first:
- Sign: must the answer be positive?
- Size: is it more or less than a value in the stem? A discounted price cannot exceed the original.
- Type: must it be an integer, or even, or divisible by 3?
- Units digit: the units digit of a product depends only on the units digits of the factors.
When not to use these methods
If the algebra takes one line, do the algebra. These methods are tools for when the direct route is long or error-prone, not a replacement for it. And an estimate is never safe when two choices sit within the error of your rounding.
Worked examples
Example 1: Estimating when choices differ by powers of ten
Which of the following is closest to ?
0.04, 0.4, 4, 40, 400
The choices are a factor of 10 apart, so only the size matters. Round: , , .
Estimate: .
Direction check: the top was rounded up slightly and the bottom up slightly, so the effects partly cancel. The exact value is about 3.98, comfortably nearest 4.
Answer
About 4. Losing a decimal place in (treating it as ) gives 0.4, a choice placed there for that slip.
Example 2: Back-solving from the middle
The sum of a positive integer and its square is 156. What is the integer?
9, 10, 11, 12, 13
Test the middle value, 11: . Too small, so the answer is larger; 9 and 10 are out.
Test 12: . It works.
Algebra check: factorises as , and is positive, so .
Answer
The integer is 12, found with two tests. 13 is tempting if you notice the factor and forget its root is .
Example 3: Picking numbers for algebraic choices
If pencils cost cents, how many pencils can be bought for dollars?
, , , ,
Pick clean values: pencils for cents, and dollar. Pencils cost 5 cents each, and 1 dollar is 100 cents, so the target is pencils.
Evaluate every choice: , , , , .
Only one choice gives 20, so no second pick is needed.
Answer
. The choice divides by where it should multiply; it gives 5, which only happens to equal the price of one pencil in cents.
Common traps
Estimating when the choices are close
If the choices are 3.8, 4.0 and 4.2, rounding each number by a few percent can move you across a boundary. Close choices are a signal to compute exactly.
Picking 0, 1 or equal values
With , the choices , and all equal 1, so the test cannot tell them apart. Pick small, different values such as 2, 3 and 5.
Stopping at the first matching choice
When picking numbers, a wrong choice can match by accident. Evaluate every choice; if two match, pick new values and retest those two.
Picking numbers that break a condition
If the stem says is odd, or , or the result is an integer, your picks must obey that too. A pick that violates the stem can make the right choice look wrong.
Check yourself
Answer each question before you open its answer.
Question 1: Closest value
Problem Solving: approximation and estimation
Which of the following is closest to ?
- A.9.5
- B.95
- C.950
- D.9,500
- E.95,000
Show the answer
Answer: C. 950
Working
Write . The numerator is . Dividing by : . Rough check: , nearest 950.
Why the other options are wrong
- A. This divides by 3.1 and drops the in the denominator, making the result 100 times too small.
- B. This reads as 0.31, one decimal place off, so the result is 10 times too small.
- D. This reads 0.00049 as , one decimal place off, so the result is 10 times too large.
- E. This reads 0.00049 as 0.049, two places off, so the result is 100 times too large.
What carries over
Handle the powers of ten separately from the digits; with choices a factor of 10 apart, a rough estimate is enough.
Question 2: Minutes for 5 kilometres
Problem Solving: algebra
A car travels kilometres in hours. At the same speed, how many minutes does it take to travel 5 kilometres?
- A.
- B.
- C.
- D.
- E.
Show the answer
Answer: C.
Working
Pick and : the speed is 15 km per hour, so 5 km takes hour, which is 20 minutes. The choices give , , , and ; only gives 20. Algebra: time hours minutes. A pick of , would make and both 60, which is why every choice must be tested.
Why the other options are wrong
- A. This is the time in hours; it was never converted to minutes.
- B. This multiplies and where one should divide by the other; with and it gives 0.2.
- D. This turns the time fraction upside down, dividing the speed by the 5 km instead of the 5 km by the speed, then multiplies by 60; it gives 180 for the test values.
- E. This multiplies by the speed instead of dividing by it; it gives 4,500 for the test values.
What carries over
When choices are expressions, pick values, compute the target from the story, and test every choice.
Question 3: Consecutive odd integers
Problem Solving: algebra
The product of two consecutive positive odd integers is 323. What is the larger of the two integers?
- A.15
- B.17
- C.19
- D.21
- E.23
Show the answer
Answer: C. 19
Working
Test the middle choice as the larger integer: the pair would be 17 and 19, and . It works. Algebra check: with smaller integer , , so , which factorises as ; , and the larger is 19.
Why the other options are wrong
- A. With 15 as the larger integer the pair is 13 and 15, and , which is too small.
- B. , but 17 is the smaller integer of the pair, not the larger.
- D. , which is too large.
- E. , far too large.
What carries over
Back-solve from the middle choice, and reread whether the question wants the smaller or the larger value.
Key points
Let the spread of the choices decide how precise an estimate needs to be.
Back-solve from the middle numeric choice and move in the direction the result points.
Pick small, distinct numbers that obey the stem, and test every choice.
If the direct algebra is one line, do the algebra.
All questions here are original and written for UniExamer; none comes from the test maker.
Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.