12) Word problems

Question 1

Age Problems

Answer 1

LEVEL 1 WORD PROBLEMS (linear)

Comparing ages in the past, present, and future
> usually set variables equal to PRESENT AGE, then add to get future age and subtract to get past age
> set up equations (usually linear) and solve via substitution or elimination

Watchouts:
** > keep track of time periods when setting up equations (need to adjust ALL PEOPLE’S AGES)

e.g., Forty five years ago, Frank was twenty years older than one-third Carl’s age
(F - 45) = 20 + 1/3*(C-45)

e.g., Six years ago, F was nine times as old as Patrick
F - 6 = 9*(P - 6)

Question 2

Length Problems

Answer 2

LEVEL 1 WORD PROBLEMS (linear)

Rope cut into segments and asked about lengths of one or more pieces
> usually set variables equal to individual pieces (shorter piece gets one variable, longer piece gets another variable)
> sum of lengths of pieces = total length

Question 3

Weight Problems

Answer 3

LEVEL 1 WORD PROBLEMS (linear)

Comparing weights (e.g., Lbs, kgs) of different people or objects

Question 4

Money Problems

Answer 4

Different variants:
> comparing prices (do not need to be integers) –> easier to convert dollars with decimals to CENTS

OR get comfortable saving time by using decimals

Question 5

t y In a certain game, a number of 2-point and 3-point baskets were made. If there were 8 more 2-point baskets than 3-point baskets, how many 3-point baskets were made? (Assume no other types of baskets could be made)

(1) There were a total of 44 2-point and 3-point baskets made
(2) Two-point baskets represent 13/22 of all the baskets made

Answer 5

Word problem –> made a mistake because I FORGOT ABOUT THE CONDITION A = B + 8

Ans D

Question 6

General Business Word Problems

Answer 6

(1) Price per item = Total cost of identical items / number of identical items

(2) Profit and loss = Total Revenue - Total cost
= Total Revenue - Fixed Costs - Variable costs\

(3) Splitting the cost problems (e.g., bill, car rental) among group of people —> Cost per person comparison
> usually an initial number of people plan to split the bill, but at the last moment, some people cannot attend and remaining people must continue to split the bill and therefore pay a higher rate than original number

(4) “Which salary should I choose? // How much will I make?” –> what would make someone INDIFFERENT between two compensation methods
> set two compensation methods equal

** (5) “Which price structure makes the most sense? // How much will I pay?” –> which price structure makes most sense given the constraints // differential pricing structures
> e.g., a car rental company charges d dollars for the first day, but d/4 dollars per day for each subsequent day. How much would a customer pay to rent a car for x days
= d + d/4*(x-1)
> e.g., Anthony must pay the full fare retail price when he spends $500 or less in a given year on tickets, but his employee discount allows him to purchase tickets for 80% off the full fare retail price once he has spent $500. If Anthony spent $3000 on tickets one year, what was the total full fare retail price of the tickets he purchased?

3000 (less than full fare retail price due to discounts) = 500 + 0.2*(x - 500)

NEED TO PAY ATTENTION TO NUMBERS (/50, +5 etc.)
> /50 case is weird –> NEED TO NOT RELY ON DIVISION AND RELY ON LOGIC

Also need to PAY ATTENTION TO WORDING
> “savings at least $2000” means savings between options >= 2000
> “justify switching” means NOT INDIFFERENT (one option must cost less, so need to increase breakeven quantity by 1) e.g., N = 50 to be indifferent, so need 51 units to favor one

Question 7

Equation of rate problems

Answer 7

Asked to EQUATE two different wages using addition and subtraction (transfer of $x)

Wage rate of person 1 = Wage rate of person 2

e.g., how much would A have to give B from his total wage payment so that they have the same hourly wage?

Can be applied to any rates (price per pound, miles per hour, widgets per day)

Question 8

Fraction Word Problems
**

Answer 8

Pay attention to determine whether a fraction is being added or removed FROM THE ORIGINAL WHOLE or FROM SOME PORTION of the whole that remains
> let x = whole

e.g., Person A takes 1/3 of the pizza, and Person B takes 2/3 of what remains. How much pizza is left?
> 1/3 * 2/3
= 2/9 of the pizza remains —–> keep track of what is remaining (saves steps)

Otherwise, just calculate what remains STEP BY STEP (e.g., day by day) and keep organized

Remember: Sum of fractional parts = whole

Hypothetical changes to a fraction problems can also be solved algebraically
e.g., numerator of a certain fraction is two more than three times the numerator. If 7 were added to both the numerator and denominator of the fraction, the new fraction would be equivalent to 3/5. What is the equivalent of the original fraction?
> ans: 2/8 —-> EQUIVALENT is 1/4 (but 2/8 is what satisfies the conditions)

Question 9

Keep in mind when solving word problems…

Answer 9

Any of the word problems discussed in this chapter can be represented by variables instead of constants

Pay attention to wording in order to translate problems effectively
> fraction total vs remaining portion

Question 10

Interest rate problems
**

Answer 10

Simple Interest Rate (no compounding)
Interest = Principal * rate % * time —-> CONSTANT regardless of amount of interest already accumulated
> interest is based on ORIGINAL principal
> make sure units for the rate must match the units of time
e.g., 5% per year * years
e.g., 2% per month * months
> GMAT will often express annual rate interest but for a duration < 1 year —> convert time to year (e.g., 8 months = 8/12 years)
> Total Value = Principal + Principalrate %time

Compound Interest Rate –> interest dollar amount INCREASES as base (principle + interest accumulated) gets larger
> Interest is earned IN EACH COMPOUNDING PERIOD –> then informs base for the NEXT COMPOUNDING PERIOD
> interest rate is expressed as a yearly interest rate, but can be compounded annually, semiannually, quarterly, or monthly
> make sure units for the rate must match the units of time
> (effective) interest rate per period = r / n where r = annual interest rate, n = number of compounding periods PER YEAR
> total number of compounding periods = n periods per year * number of years

Interest = Principal * (1 + rate % per period)^n * t periods - Principal

A = P * (1 + rate % per period)^n periods

where n = number of periods PER YEAR
t = number of years

Be ware of:
> decimals (need to square, cube etc. decimal rates) –> ignore decimal places, then bring them back after multiplication

Question 11

Linear Growth Problems
**

Answer 11

Arithmetic sequence with a constant difference (+c or -c):

Option 1) Arithmetic Formula
An = A1 + (n-1)*d where n >= 1
> need to adjust n-1 depending on whether starting value is A1 or A0

e.g., At = A0 + t*d where t >= 1

Option 2) Growth table showing actual increase during each period
> most effective for growth that occurs over a relatively SMALL number of growth periods

HOWEVER:
> in order to determine the amount of growth, we MUST be presented with the growth driver (relevant for DS)

e.g., 2, 4, 6, 8, __ —> you cannot assume 10 because we don’t know what the growth driver is
> could grow by a constant or NON-CONSTANT amount!! (e.g., stop growing or accelerated or decelerated)

Question 12

Exponential Growth Problems
**

Answer 12

Value grows by a constant FACTOR (c, where c > 1)
> y = a^x = a * a * a … (exponents)
> e.g., a plant doubles in ehight every week
> pay attention if problem is saying “doubling” (2) or “increase by 25%” (*1+0.25)
> related to compounding interest

FORMULA:
vn = v1* (growth factor)^(n-1) —> growth factor g
vn = v0 * (growth factor)^n

Pay attention to what is defined at v1 and adjust n accordingly

e.g., v1 = 5, then four years later we are looking for v5 (not v4)

Common scenarios:
> bacteria growth

HOWEVER:
> in order to determine the amount of growth, we MUST be presented with the growth driver (relevant for DS) (e.g., factor or constant difference)

e.g., 2, 4, 6, 8, __ —> you cannot assume 10 because we don’t know what the growth driver is
> could grow by a constant or NON-CONSTANTamount!!

Question 13

A colony of bacteria grew in number between 7AM and 11AM. If the colony began with 1000 bacteria, how many bacteria were in the colony at 11AM?

(1) At 8AM, there were 1500 bacteria in the colony
(2) At 9AM, there were 2000 bacteria in the colony

Answer 13

E —> could be exponential or linear growth –> different answers

> in order to determine the amount of growth, we MUST be presented with the growth driver (relevant for DS) (e.g., factor or constant difference)

NEED TO KNOW
(1) type of growth (linear or exponential)
(2) growth driver (what is the factor or constant difference)

Question 14

Exponential Decay Problems
**

Answer 14

Value shrinks by a constant FACTOR (*c, where 0 < c < 1)

Vn = V1 * (g)^(n-1)
Vn = V0 * (g)^n

Absolute amount decrease = DIFFERENCE in V’s
e.g., the amount of money by which the investment account decreased during the third week is what fraction of the amount of money remaining in the account at the end of the first week? Investment account decreased by 50% each week —> g = 1/2
V1 = amount of money remaining after first week
V2 = amount of money remaining after second week
V3 = amount of money remaining after third week

Looking for (V2 - V3) / V1

ans 1/4

Things to remember:
> Greatest decrease occurs during the first period of decay

Question 15

Digit Problems

Answer 15

Recall you can turn any number into an equation of its parts based on place values

e.g., 48 = 10*(4) + 8

ab = 10*a + b ——-> expressing the VALUE of numbers using its digits

Other tips:
> you can write out POSSIBLE VALUES of ab in DS
> or you can set up equations
e.g., _ _ = a b (a positive two digit integer)
a = b + 5 (tens digit is five more than its unit digit)

a*b = 24 —> creates quadratic equation with two possible values for b one negative, so Sufficient

Question 16

Word problems involving consecutive integers

Answer 16

Incl: consecutive integers, consecutive even integers, consecutive odd integers, consecutive multiples of integers

e.g., consecutive multiples of 8 –> x, x+8, x+16, x+24

As long as EQUAL CONSTANT DIFFERENCE = arithmetic sequence = linear functions

Helpful to set the FIRST (smallest) integer = x or a

Question 17

Mixture Word Problems**

Answer 17

Mixing 2 or more components to create a final mixture
> deals with WEIGHTED AVERAGE
> either can solve algebraically with table
> or solve 2 data values using visual approach to get weights

Type 1) Dry Mixture
> e.g., combining peanuts and raisins to create a trail mix
(components are pure and composed of only one ingredient)
> usually weighting QUANTITIES to determine $ COST per quantity or average unit
> will be important to keep track of: components of mixture, units of each component, quantity of each component —-> helpful to create a matrix (rows represent components and total, one column for dollar, one column for quantities, and one column for total cost)
—-> each row and each col is an equation that just needs 2 info to be Sufficient
> total quantity = sum of individual quantities
> or set up weighted average formula

PRE-MIXTURE quantities = POST-MIXTURE quantities

PRE-MIXTURE concentration or dollar amounts = POST-MIXTURE concentration or dollar amounts (% * Q or total cost)

HARD Qs involve MULTIPLE COMPONENTS (e.g., two brands of trail mix, each containing some percentage of raisins and peanuts, blended together. Therefore we have TWO EQUATIONS for quantities)
> Pre-mixture raisin quantities = Post-mixture raisin quantities
e.g., 0.6x + 0.2y = 0.5(x+y) —-> weighting percentage of trail mix that is raisin
> Pre-mixture peanut quantities = Post-mixture peanut quantities
> both equations yield the SAME RATIO of OVERALL BLEND quantities

Type 2) Wet mixture
> e.g., combining certain amount of a 40 percent salt solution with a certain amount of a 30 percent salt solution
(combining liquids with a specific concentration to create a mixture with its own concentration)
> usually weighting volume quantities to determine CONCENTRATION AMOUNT
> will be important to keep track of: components of mixture, concentration of each component, and quantity of component
> also helpful to create a matrix
> total volume quantity = sum of individual volume quantities

PRE-MIXTURE quantities = POST-MIXTURE quantities (volumes)

PRE-MIXTURE concentration amounts = POST-MIXTURE concentration amounts (% * volume)

Question 18

World problems with inequalities

**

Answer 18

Involves translating words into inequalities and dealing with both inequalities and equations in the same problem

Strategies:
> Express the inequality in terms of one variable by SUBSTITUTING equation INTO the inequality (as long as equation and inequality have the same variables)
> pay attention to when “=” vs inequality e.g., up to, at least, more than, at most, fewer than
> watch out for INTEGER CONSTRAINTS —> better to think about minimum or maximum possible values and test values (otherwise can fall into pseudo NS traps)

e.g., S > 6 and S is an integer, therefore S >= 7 (smallest value of S is 7)
—-> test around boundary

IN GENERAL —> express question inequality as compared to a VALUE (instead of zero) when dealing with one variable

e.g., instead of asking is “3s - 20 >0”, ask is “s > 6.67”

Question 19

Objects in a line problems

Answer 19

Based on positions
> best approach is the VISUALIZE the positions and create a “chain” = number of people before object + 1 object + number of people after object = total number of objects
> pay attention to DIRECTION (let front of the line be the LEFT most position)

_ person _

or _ person a _ person b _

In general, if you or anyone are the mth person counted from the beginning of the line and the nth person counted from the end of the line, then the number of people waiting in line = m + n - 1
> subtraction to remove double count

*** Variations of Q:
> two or more people involved in the line –> need to consider CASES depending on who is in front of the other person

e.g., A is 10th in line counting from the beginning of the line. B is 15th in line counting from the end of the line. If there are 5 people in between them, how many people could be waiting in line?

Case 1: A is in front of B —> ans 30
Case 2: B is in front of A –> ans 18

Question 20

If Alex pays 40% income tax on all gross (before taxes) income over $25000, then Alex’s gross income in 2012 was approximately what?

Chart shows her after tax income in 2012 was $40k

Answer 20

Concept: Tax is only applied on the amount OVER $25000 (not her entire pre-tax income)

Let x = Alex’s pre-tax income

x - taxes = 40k

x - (40%)*(x - 25k) = 40k

x = 50k

Question 21

One day a certain truck driver dropped off 1/3 of his packages at a stop A in the morning. Later that afternoon, at stop B, the driver dropped off 1/4 of his remaining packages and picked up 20 more. Finally, in the evening, the driver dropped off 10 packages at stop C. He now had 55 packages on his truck. How many packages did the driver originally have on his trucks?

Answer 21

Word problems: Fraction
> be careful of calculating the right fraction of “what’s left”
—> check mentally the long way

Start with x packages

After stop A:
> drop off 1x/3
> keep 2x/3

After stop B:
> drop off 1/4(2x/3)
> keep 3/4(2x/3) = 1x/2
> Gain 20
> Have: 1x/2 + 20

After stop C:
> drop off 10
> Keep: 1x/2 + 10 = 55

x = 90

Question 22

A deli sells pints of two flavors of ice cream: vanilla and chocolate. If the deli purchases each pint of the two ice cream flavors for the same amount of money, did the store, last Monday, make a greater profit on the sale of the vanilla ice cream or the chocolate ice cream?

(1) On Monday, the deli sold 1.5 times as many pints of vanilla ice cream as chocolate ice cream
(2) The selling price per pint of the vanilla ice cream is 1/4 the selling price per pint of the chocolate ice cream

Answer 22

Word problems, inequality with multiple variables
> Strategy: Try to express both sides of the inequality with the SAME PRODUCT OF TWO VARIABLES e.g., AV vs AV, or BC vs BC so we can just compare the constants
> Strategy: alternatively, evaluate to compare SIGNS of both sides of the inequality
> OR THIRDLY: MOVE all terms to one side and evaluate SIGN
> generally NS without both prices and quantities

V = quantity of vanilla ice cream sold
C = quantity of chocolate ice cream sold
A = price of vanilla ice cream
B = price of chocolate ice cream
x = cost to buy a pint

Is: (A - x)V > (B - x)C ?

(1) V = 1.5C —> sub into inequality

Is: (A - x)1.5C > (B - x)C —-> NS without knowing relative prices

(2) NS without quantities sold

(3) V = 1.5C
A = 1B/4

EXPRESS BOTH SIDES in terms of BC: EVEN though we have x variable, it might cancel out or since we know it is always positive, that can help too…

Is…
(B/4 - x)1.5C > (B - x)C? ——> Cancel Cs

Is…
1.5B/4 - 1.5x > B - x?

1.5B/4 - B > 0.5x?

LEFT HAND SIDE IS negative, and Right hand side is ALWAYS positive –> inequality is ALWAYS FALSE

Sufficient
C

Question 23

Paula’s high school has 4 grade levels: ninth, tenth, eleventh, and twelfth. Paula is an eleventh grader and ran against 3 other students (one from each of the other 3 grade levels) for the office of class president. According to the school election rules, students from the same grade level as the candidate cannot cast votes for that candidate. If she received 3/4 of the total votes cast among the students who are eligible to vote for her, how many votes did Paula receive?

(1) Eighty students in the eleventh grade and 120 students in the twelfth grade cast their votes
(2) The nine-grade students accounted for 1/3 of the total voters, and the tenth-grade students accounted for 1/4 of the total voters

Answer 23

Word problems - fractions

Need to figure out the sum of 9th + 10th + 12th grader students who voted

(1) 11th = 80 students who voted
12th = 120 students who voted
NS without totals

(2) 9th = (1/3)(all students who voted)
10th = (1/4)(all students who voted)

RATIO (combined)
9th: 10th: 11th: 12th: total
4x: 3x: ?: ?: 12x

Need to know value of 4x + 3x + 12th

NS

(3) 5x = 11th + 12th = 120+80 based on statement 1
we know x
so we can know the actual quantities of every class

Sufficient
C

Question 24

Assume that x, y, and z are consecutive multiples of three that sum to 945 and that a, b, and c are consecutive multiples of 5 that sum to 2715. If x < y < z and a < b < c, what is the value of ay?

Answer 24

Consecutive integer word problems
> To SAVE TIME, set FIRST INTEGER to be “x” and “a” (instead of 3q and 5p)

(x) + (x+3) + (x+6) = 945
3x = 936
x = 312

and

(a) + (a+5) + (a+10) = 2715
3a = 2700
a = 900

Therefore:
a = 900
y = 315

ay = 900*315
= 283,500

Question 25

Mary produced an organic plant growth formula called “Super Grow.” If Mary had a plant that was initially 3 feet tall and she treated the plant with Super Grow, did the plant grow to a height of at least 5 feet in 4 months?

(1) the plant grew at a constant rate of x inches per month

(2) at the end of 2 months, the plant was 1/2 foot taller than it was at the end of the first month

Answer 25

Word problems: growth
> need to determine if it is linear growth of exponential growth

h0 = 3 ft
Is h4 >= 5 ft?

(1) ht = 3 + (x in ft)*t

No idea what x is
NS

(2) h2 = h1 + 0.5 (NOT factor)
No idea if this is linear growth or exponential growth OR IRREGULAR GROWTH

e.g., linear growth –> h1 = 3.5, h2 = 4, h3 = 4.5, h4 = 5 (Yes)

Irregular growth –> h1 = 3.5, h2 = 3.5, h3 = 3.5, h4 = 3.5 … (no)

(3) we now know the plant grows linearly (constant amount)

Constant difference = 0.5 ft

h4 = 3 + 0.5*4 —> sufficient to know yes

Question 26

Tricky wording used in word problems for comparisons:

A is 1/2 foot taller than B
A is 20 years older than B

Answer 26

Fraction + UNITS + Comparative + Than

A = 1/2 + B —> units is feet

A = 20 + B

Constant difference (“1/2 FOOT” –> units are provided)

Question 27

Tricky wording used in word problems for comparisons:

A is 1/2 taller than B.

After 10 minutes the grass was 3/7 longer than it was after 6 minutes

Answer 27

FOR WORD PROBLEMS WHERE VARIABLES REPRESNT SOME VALUE AND DEALING WITH FRACTIONS

A is bigger than B –> same thing as saying “A is 50% taller than B” (1/2 converted to percent change)

Fraction + fewer/more/taller/comparative + (UNITS) + than

> we are NOT GIVEN UNITS after the fraction (just “larger”, “taller”, “longer”, “shorter”)

A = (1 + 1/2)B
A = (1.5)B

IN GENERAL: NOT treated as a factor (not * times)
“Increase of 1/4” –> increase BY 25%
“Decrease of 1/2” –> Decrease BY 50%

Question 28

Tricky wording used in word problems for comparisons:

A is 5/6 times B

Answer 28

“Times” = multiplication

A = 5/6 * B

Question 29

Tricky wording used in word problems for comparisons:

A is 250 percent of B

Answer 29

“percent OF” = times

A = 2.5*B

Question 30

Tricky wording used in word problems for comparisons:

x is n percent less than y

Answer 30

PERCENT LESS THAN = percent change

x = y*(1-n/100)

Question 31

Tricky wording used in word problems for comparisons:

what percent of x is z?

Answer 31

Percent OF __ denom __

z / x * 100

(? / 100) * x = z

Question 32

A grocery store sells a trail mix consisting of chocolate and nuts. The nuts cost $2.75 per pound and the chocolate cost $1.50 per pound. if a customer prefers to purchase a 20-pound bag of trail mix for $2.25 per pound, how many pounds of chocolate must be mixed in the trail mix?

12
11
10
9
8

Answer 32

Mixture weighted average

> either solve algebraically with table ** preferred
or visual approach (since we have two data values) –> have to be very careful which weights belongs to which values

VISUAL APPROACH;
$1.50 ———- $2.75

Simple average is: 4.25/2 = 2.125
Since WA is 2.25, we know there must be MORE NUTS than chocolate in the mixture

Find weights:
> denom = 2.75 - 1.5 = 1.25
> make sure you use the RIGHT numerator (smaller weight, needs smaller distance between values)

Weight of chocolate / total (Ratio) = smaller distance/1.25
= 0.5/1.25

We know 0.5/1.25 = pounds of chocolate / 20

Pounds of chocolate = 8

Question 33

Tricky wording used in word problems for comparisons:

A worked 1/6 fewer hours than B

Answer 33

NOT SURE IF THIS IS RIGHT***

because another question said “there are 1800 fewer grandstand seats than there are stadium-level seats in the stadium”
G = S - 1800

Fraction + fewer/more/taller/comparative + (UNITS) + than
= treat as % change
> (Units) is optional
> same thing as

Could be reworded to “Francesca’s hours worked was 1/6 fewer than B”

B*(1-1/6)

NOT THIS: B - 1/6 where B represents hours

Question 34

Tricky wording used in word problems for comparisons:

A worked 1/2 the number of hours that B worked

Answer 34

Fraction + “the number” = *
or Fraction + value = *
> hidden “of” = *

A = 1/2 * B

e.g., Frank is 20 years older than 1/3 Carl’s age
F = 20 + 1/3*C

Question 35

Tricky wording used in word problems for comparisons:

The flight time will be reduced by 1/4 of the time taken at the previous power setting every 1 percent increase

Answer 35

HARD ONE

Flight time is REDUCED BY 0.25*previous value = DECREASING by 25%
> Decrease = 25% of previous value

So New flight time = (3/4)*Previous Flight Time

NOT *1/4