Main content

## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
- GMAT: Math 2
- GMAT: Math 3
- GMAT: Math 4
- GMAT: Math 5
- GMAT: Math 6
- GMAT: Math 7
- GMAT: Math 8
- GMAT: Math 9
- GMAT: Math 10
- GMAT: Math 11
- GMAT: Math 12
- GMAT: Math 13
- GMAT: Math 14
- GMAT: Math 15
- GMAT: Math 16
- GMAT: Math 17
- GMAT: Math 18
- GMAT: Math 19
- GMAT: Math 20
- GMAT: Math 21
- GMAT: Math 22
- GMAT: Math 23
- GMAT: Math 24
- GMAT: Math 25
- GMAT: Math 26
- GMAT: Math 27
- GMAT: Math 28
- GMAT: Math 29
- GMAT: Math 30
- GMAT: Math 31
- GMAT: Math 32
- GMAT: Math 33
- GMAT: Math 34
- GMAT: Math 35
- GMAT: Math 36
- GMAT: Math 37
- GMAT: Math 38
- GMAT: Math 39
- GMAT: Math 40
- GMAT: Math 41
- GMAT: Math 42
- GMAT: Math 43
- GMAT: Math 44
- GMAT: Math 45
- GMAT: Math 46
- GMAT: Math 47
- GMAT: Math 48
- GMAT: Math 49
- GMAT: Math 50
- GMAT: Math 51
- GMAT: Math 52
- GMAT: Math 53
- GMAT: Math 54

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# GMAT: Math 17

92-95, pg. 164. Created by Sal Khan.

## Want to join the conversation?

- Question #95

I find the following method to be an easier way to answer this question. Start by finding out how many years it would take to qualify and then add it to her starting year to determine her year of eligibility.

70 >or= 32+2Y and solve for y and add to 1986.

32 is her starting age, and you get twice the credit for each year she works due to age and tenure.(15 votes)- During the GMAT exam the way you describe would probably be more intuitive and faster, but Sal's method is mathematically more basic, which helps people during the learning stages, but requires several more steps to complete and leaves more chance for error when on a time crunch. I solved initially with your method as well, but am glad to see his as it might be useful on other problem types.(1 vote)

- Thank you, now I undserstand(1 vote)
- Question #95

Can't you just do 70=32+2x and solve for x? x=19 added to 1986 = 2005 being the first yeah of eligibility.(1 vote)- Each year that the employee works for the company, she also becomes a year older. So, she gets 2 "years" credit for each year she is employed, one for the year that she worked and one because her age increased by a year. So, for example, she started in 1986 at 32 years old. In 1987, she would have worked 1 year, and she would turn 33, so her age plus her 1 year of employment would = 34. In 1988, she will turn 34 and she will have been employed for 2 years, so age + years employed would equal 36. The amount that counts towards her retirement goal of 70 is Age + Years of employment, so this amount increments at a rate of 2 per calendar year.(1 vote)

- A really easy and quick way to solve 95)

a+y>or=70 so we essentially solve for a+y=70.

When Age = 32, Years = 0 = 36 which is not equal to 70

Simply add 10 to each side till you hit 70

So at A=42, Y=10 which is equal to 52 which is not equal to 70

At age 52, Y =20 which is 72

So at age 51, Y = 19 = 70

So from 1986, with 19 years of experience, the year would be 2005(0 votes) - what is between 1/3 and 87/100 on a number line?(0 votes)
- Well, I hope you are not still waiting for this answer. The fractional difference between those rational numbers is 161/300. So, if you change the two fractions to make equivalent fractions having a denominator of 300, 1/3 would be 100/300 and 87/100 would be 261/300. Now you can start listing just a
**few**of the numbers between 1/3 and 87/100.

So 101/300, 102/300 (same as 17/50), 103/300, 104/300 (same as 26/75), 105/300 (same as 7/20), 106/300 (AKA 53/150), 107/300, 108/300 (same as 9/25), ... 150/300 (AKA 1/2) ... 200/300 (same as 2/3), ... 225/300 (AKA 3/4)... and I have skipped over 140+ of the numbers in this list. And, if we choose even bigger denominators, we can get even more. Oh, and I forgot π/9, 0.85, 3/5, and 0.3879375, which is 3879375/10000000.

And (√2)/2 and (√3)/3 and (√17)/7 ...

In fact, there are an infinite number of numbers between**any**two points on the number line. If that doesn't give you a headache trying to imagine that, I don't know what will!(2 votes)

## Video transcript

Problem 92. In a certain furniture store,
each week Nancy earns a salary of $240, plus 5% of the amount
of total sales that exceeds $800 for the week. Plus 5% of whatever is greater
than 800 for the week. If Nancy earned a total of $450
in a week, what were her total sales that week? So she earns $450, so what
was her 5% bonus part? So 450 minus 240, she
got $210 bonus. So 210 represents 5% of
her sales above 800. 210 is 5% of what number? So you could say 210 is equal to
0.05 times x, and x will be her sales above 800. So x is equal to 210 divided
by 0.05, so 0.05 goes into 210, add some decimal points. That's the same thing as 5 going
into, put the decimal point there, 21,000. So 5 goes into 21 4 times, 4
times 5 is 20, bring down 1. 5 goes into 10 2 times,
2 times 5 is 10. We're done. So her sales were 4,200, but
that's not her total sales. Remember, she only gets 5%
on her sales above 800. So she sold $4,200 more
than $800 that week. So her total sales are going
to be 4,200 plus 800, which is $5,000. And that's choice E. Next question, 93. They give us two lists. So that's list one, and they
give it 3, 6, 8, and 19, and they give us list two, they say
it's x, 3, 6, 8, and 19. It's actually the same list,
except they have an x here. If the median of the numbers in
list one above is equal to the median of the numbers in
list two above, what is the value of x? So what's the median
of this list? Since we have an even number of
numbers, this would be the average of the middle two
numbers, so what's the middle two numbers right here? The middle two numbers are 6 and
8, the average is 6 plus 8 divided by 2, it's 7. So the median is 7. So they're telling us that the
median of this list is also equal to 7, so where can
I put x so that the median doesn't change. Well what if x is 7? If x is 7, then this list
becomes 3, 6, 7, 8, 19, and now the middle number, since
we have odd numbers, is actually the middle number. We don't have to do any of
this averaging business. So now the median is actually
7, if we place 7 there. So x equals 7. Choice B. 94. In a certain city, 60% of the
registered voters are Democrat, and the rest
are Republican. So 40% of voters. In a mayoral race, if 75% of the
registered voters who are Democrats, and 20% of the
registered voters who are Republicans are expected to
vote for candidate A, what percentage of the registered
voters are expected to vote for candidate A? In a mayoral race, if 75% of
Democrats, so 0.75 Democrats are voting for candidate A,
and 20% of Republicans, so plus 0.2 times the Republicans,
are expected to vote for candidate A, what
percent of the voters are voting for candidate A? So let's just express this in
terms of the total voters. The Democrats are 60% of the
voters, Republicans are 40% of, not the Republicans,
of the voters. So 0.75 times the Democrats, the
Democrats are 60% of the voters, so let's just
substitute. 0.6 voters for D, so we get
0.75 times 0.6 of all the voters, that's the number of
Democrats there are, plus 0.2 times the Republicans,
Republicans are 0.4 times all the voters. And so let's see, what's
75 times 6? Let me just do the whole
decimals, 0.75 times 0.6. 6 times 5 is 30. 6 times 7 is 42 plus 3 is 45. And we have one, two, three
numbers behind the decimal point, so it's 0.45 of the
voters plus, let's see, 2 times 4 is 8, and we have two
numbers behind the decimal, so it's plus 0.08 times
the voters. 0.45 plus 0.08 that's 0.53. You have 45 plus 8, is 0.53, or
53% of the voters are going to vote for candidate
A, that's choice B. Problem 95. A certain company retirement
plan has a rule of 70 provision that allows an
employee to retire when the employee's age plus years of
employment within the company total at least 70. So greater than or
equal to 70. In what year could a female
employee hired in 1986, this is interesting, 1986, on her
32nd birthday, first be eligible to retire under
this provision? So let's say that y is
equal to the year. Her age is going to be the
year minus 1986 plus 32. That's her age, because she
starts at 32, so if the year's 1986, we're just going
to get her age as 32. If the year's 1987, we're going
to get 1987 minus 1986, which is 1 year, plus 32,
her age will be 33. So that's her age. Her years of employment
are going to be the years minus 1986. So a plus e is going to be equal
of this, and so that is equal to what? 2y minus 2 times 1986 plus 32
have to be greater than or equal to 70. Just so I don't have to multiply
2 times 1986, let's divide both sides of this
equation by 2, and I don't have to change the inequality,
because that's positive. So I get y minus 1986
plus 32 is greater than or equal to 70. So y is greater than or equal to
70, let me add 1986 to both sides, plus 1986, minus 32. So what's 1986 minus 32? That's 1954, 8 minus 3
is 5, 6 minus 2 is 4. What's 70 plus 1954? Plus 70, is 4, 5 plus 7 is 12,
1 plus 9 is 10, so 2024 is an acceptable year for her. 2024. And that is not one of
the choices, so I clearly made a mistake. Let me see where I might
have made a mistake. So a certain employee retirement
has a rule of 70 provision that allows an
employee to retire when the employee's age, age plus years
of employment within the company total at least 70. In what year could a female
employee, the fact that she's female shouldn't matter, a
female employee hired in 1986 on her 32nd birthday,
first be eligible to retire under this provision? OK, her age is going to be the
year, whatever year we're talking about, minus
1986 plus 32. And then her years of employment
are going to be whatever year we're talking
about minus 1986, fair enough. So I have a year plus year, and
I have two subtractions of my 1986 plus 32 is greater
than or equal to 70. I divide both sides by 30--
oh, I see my mistake. Here, this stuff right here,
where I divided both sides by 2, I divided only the
left side by 2. That by 2, that by 2-- oh,
I didn't even do that. 2 times 1986 divided by 2 is
1986, 32 divided by 2 is 16, 70 divided by 2 is 35. OK, so let me rewrite it here. That's what happens when you
try to do too much at once. So I have y is greater than or
equal to 35 plus 1986 minus 16, I just added 1986 to
both sides, subtracted 16 from both sides. So this is equal to 1970, and
then we have 35 plus 1970, so y has to be greater than
or equal to 2005, which is choice C. And I'm out of time, sorry for
that careless error, and I'll see you in the next video.