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 49.6≈5049.6 \approx 50, 0.198≈0.20.198 \approx 0.2, 2.47≈2.52.47 \approx 2.5, 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:

6×104×5×10−43×10−2=303×10−2=10×102=1000\frac{6 \times 10^4 \times 5 \times 10^{-4}}{3 \times 10^{-2}} = \frac{30}{3 \times 10^{-2}} = 10 \times 10^{2} = 1000

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 xx 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 49.6×0.1982.47\frac{49.6 \times 0.198}{2.47}?

0.04, 0.4, 4, 40, 400

  1. The choices are a factor of 10 apart, so only the size matters. Round: 49.6≈5049.6 \approx 50, 0.198≈0.20.198 \approx 0.2, 2.47≈2.52.47 \approx 2.5.

  2. Estimate: 50×0.22.5=102.5=4\frac{50 \times 0.2}{2.5} = \frac{10}{2.5} = 4.

  3. 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 0.1980.198 (treating it as 0.01980.0198) 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

  1. Test the middle value, 11: 11+121=13211 + 121 = 132. Too small, so the answer is larger; 9 and 10 are out.

  2. Test 12: 12+144=15612 + 144 = 156. It works.

  3. Algebra check: n2+n−156=0n^2 + n - 156 = 0 factorises as (n+13)(n−12)=0(n + 13)(n - 12) = 0, and nn is positive, so n=12n = 12.

Answer

The integer is 12, found with two tests. 13 is tempting if you notice the factor (n+13)(n + 13) and forget its root is −13-13.

Example 3: Picking numbers for algebraic choices

If pp pencils cost cc cents, how many pencils can be bought for dd dollars?

dp100c\frac{dp}{100c}, 100cdp\frac{100c}{dp}, 100dpc\frac{100dp}{c}, 100dpc\frac{100d}{pc}, cdp100\frac{cdp}{100}

  1. Pick clean values: p=2p = 2 pencils for c=10c = 10 cents, and d=1d = 1 dollar. Pencils cost 5 cents each, and 1 dollar is 100 cents, so the target is 1005=20\frac{100}{5} = 20 pencils.

  2. Evaluate every choice: 21000\frac{2}{1000}, 500500, 20010=20\frac{200}{10} = 20, 10020=5\frac{100}{20} = 5, 20100=0.2\frac{20}{100} = 0.2.

  3. Only one choice gives 20, so no second pick is needed.

Answer

100dpc\frac{100dp}{c}. The choice 100dpc\frac{100d}{pc} divides by pp 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 x=1x = 1, the choices xx, x2x^2 and 1x\frac{1}{x} 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 nn is odd, or x>yx > y, 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 (6.02×104)(0.00049)3.1×10−2\frac{(6.02 \times 10^4)(0.00049)}{3.1 \times 10^{-2}}?

  1. A.9.5
  2. B.95
  3. C.950
  4. D.9,500
  5. E.95,000
Show the answer

Answer: C. 950

Working

Write 0.00049=4.9×10−40.00049 = 4.9 \times 10^{-4}. The numerator is 6.02×4.9×100≈29.56.02 \times 4.9 \times 10^{0} \approx 29.5. Dividing by 3.1×10−23.1 \times 10^{-2}: 29.53.1×102≈9.5×100=950\frac{29.5}{3.1} \times 10^{2} \approx 9.5 \times 100 = 950. Rough check: 6×104×5×10−43×10−2=300.03=1000\frac{6 \times 10^4 \times 5 \times 10^{-4}}{3 \times 10^{-2}} = \frac{30}{0.03} = 1000, nearest 950.

Why the other options are wrong

  • A. This divides by 3.1 and drops the 10−210^{-2} in the denominator, making the result 100 times too small.
  • B. This reads 3.1×10−23.1 \times 10^{-2} as 0.31, one decimal place off, so the result is 10 times too small.
  • D. This reads 0.00049 as 4.9×10−34.9 \times 10^{-3}, 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 kk kilometres in hh hours. At the same speed, how many minutes does it take to travel 5 kilometres?

  1. A.5hk\frac{5h}{k}
  2. B.hk300\frac{hk}{300}
  3. C.300hk\frac{300h}{k}
  4. D.12kh\frac{12k}{h}
  5. E.300kh\frac{300k}{h}
Show the answer

Answer: C. 300hk\frac{300h}{k}

Working

Pick k=30k = 30 and h=2h = 2: the speed is 15 km per hour, so 5 km takes 13\frac{1}{3} hour, which is 20 minutes. The choices give 13\frac{1}{3}, 0.20.2, 2020, 180180 and 45004500; only 300hk\frac{300h}{k} gives 20. Algebra: time =5k/h=5hk= \frac{5}{k/h} = \frac{5h}{k} hours =300hk= \frac{300h}{k} minutes. A pick of k=10k = 10, h=2h = 2 would make 300hk\frac{300h}{k} and 12kh\frac{12k}{h} 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 hh and kk where one should divide by the other; with k=30k = 30 and h=2h = 2 it gives 0.2.
  • D. This turns the time fraction upside down, dividing the speed kh\frac{k}{h} 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?

  1. A.15
  2. B.17
  3. C.19
  4. D.21
  5. 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 17×19=32317 \times 19 = 323. It works. Algebra check: with smaller integer nn, n(n+2)=323n(n + 2) = 323, so n2+2n−323=0n^2 + 2n - 323 = 0, which factorises as (n+19)(n−17)=0(n + 19)(n - 17) = 0; n=17n = 17, 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 13×15=19513 \times 15 = 195, which is too small.
  • B. 17×19=32317 \times 19 = 323, but 17 is the smaller integer of the pair, not the larger.
  • D. 19×21=39919 \times 21 = 399, which is too large.
  • E. 21×23=48321 \times 23 = 483, 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.

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