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# Matematyka GMAT: 17

## Transkrypcja filmu video

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.