Magoosh Review Problems

Pataasin ang iyong marka sa homework at exams ngayon gamit ang Quizwiz!

Peter went to the store to buy paint. Small cans cost $30 and larger cans cost $80. How many small cans of paint did he buy? (1) Peter spent $220 on paint. (2) Peter bought four cans of paint in total.

30s +80L S = ? Exact dollar amounts - very specific 1)30s +80L=220 Test out numbers all 30s --> impossible

w, x and y are positive integers such that w ≤ x ≤ y. If the average (arithmetic mean) of w, x and y is 20, is w > 15 ? (1) y = 28 (2) One of the three numbers is 17

Sum = 20*3 = 60 1) y=28 (largest number) 60-28 = 32 w can be less or greater than 15 (NS) 2) 60-17 = 43 Could be any combination - NS Both: know 28 and 17 so the last number we know and we can answer w>15 question Remaining number must be 15 Sufficient

The average (arithmetic mean) of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?

Sum = number of items * average Sum = 4*75 = 300 a+b+c+d = 300 a+b+c+90 = 300 Max --> a+88+89+90 = 300 (numbers are all different) a = 33

∣a∣=b Range of n numbers is even If diagonals are equal, share can be Scale factor for similar triangles

a=b and −a=b number of items in set is odd and avg can be found Square or rectangle Area - scale factor is k^2; if similar, angles are same and lengths are proportional

​159,919/(409)(17)​​​​=

160000-81 / (400+9)(17) Difference of square (400-9)(400+9) / (409)(17) 400-9/17 Difference of square again (20-3)(20+3)/17 = 23

Which data set has the greater standard deviation, data set K or data set L? (1) Every element in data set K is a multiple of 3. (2) Every element in data set L is a multiple of 9.

1) no info on set L 2) no info on set K Both: we know the elements but we don't know what they are SD is based on the numbers Note: Consecutive multiples of 9 are more spread out from each other than consecutive multiples of 3. It might be tempting to assume that L is greater than K because of this. Be careful. We have no information about which multiples of 3 and 9 are in these data sets. They could even be identical, as all multiples of 9 are also multiples of 3! For example,\mathrm{ K=\{9,18,27,36\},\;\;\;\;\;L=\{9,18,27,36\} }K={9,18,27,36},L={9,18,27,36}

What is the value of x? (1) (x2 + 3)2 - 11(x2 + 3) + 28 = 0 (2) x2 + x - 2 = 0

1) set u = x^2+3 (easier way to handle calculs) u^2-11u+28 = 0 (u-4)(x-7) u = 4, 7 x^2+3 = 4 --> x = pos/neg 1 x^2+3 = 7 --> x = pos/neg 2 NS 2) (x+2)(x-1) x=-2,1 NS Both: -2, 1 are both solutions so NS E

Is x divisible by 12? (1) x is divisible by 27. (2) x is divisible by 6.

12 - 3*2*2 (need these factors to be divisible by 12) 1) 27 = 3*3*3 Not sufficient to say that x is divisible by 12 - could and could not be 2) 6 = 3*2 Not sufficient to say that x is divisible by 12 - could and could not be (doesn't have the additional 2) Both: 27 = 3*3*3 6 = 3*2 Still not sure If a number is divisible by both 6 and 27, that means we have accounted for prime factors of: 2 × 3³ This still doesn't mean something can be divisible by 12, though, because, as we said, 12 requires 2² rather than just a single 2. That is why both together are not sufficient.

If the average (arithmetic mean) of four different positive integers is greater than 3 and less than 4, what is the range of the four numbers? (1) One number is greater than 7 (2) The median of the four numbers is 2.5

12 < Sum < 16 so sum could be 13, 14, 15 1) one number = 7 doesn't give info about other numbers 2) median = 2.5 Potential sum: 1 and 4 but this is ruled out because that would mean a = 0 (problem says >0 integer) or 2 and 3 which means a = 1 (different integer but les) Total so far = 6 d= 7, 8, 9 and ranges are all different so not sufficient even when combined

2 − [1 − (1 − [2 − 3] − 2) + 3] =

2-3 = -1 1-(-1)-2=0 1-0+3 = 4 2-4 = -2

(8-x)/(x+1) = x x^2 + 2x - 3 =?

8-x = x^2+x 8 = x^2+2x x^2 + 2x - 3 = 8 - 3=5

If point is on the circle and has legs on diameter of circle, what can we assume? If not on circle but inside circle, can we still use same theorem?

90 degrees on that angle and Pythagorean theorem If not on circle, cannot assume 90 degree angle

A box contains bags of marbles. All of the bags hold the same number of marbles except one bag, which holds one marble more than each of the other bags hold. If the box contains a total of 2001 marbles, how many bags are in the box? (1) The number of bags is between 13 and 23 inclusive (2) There is an even number of bags, and there is an even number of marbles in the bag containing the extra marble.

B=bags M= marbles/bag BM+1 = 2001 BM = 2000 = 2*2*2*2*5*5*5 1) (16,125) or (20,100) not sufficient 2) bags must have odd number of marbles (even number in extra marble bag) (16,125) or (80,25) not sufficient Both: only one solution C

In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?

Combination problem - order does not matter Think about the problem this way: I have 3 identical green shirts and 3 identical red shirts to distribute to six children. If I simply designate the three who will get green shirts, that determines everything, because once I know which three get the green shirts, I automatically know the other three get the red shirts. So, the number of possible outcomes is just the number of ways we could choose a combination of 3 from the set of six. That's the combination number nCr: 6C3 = 20. Counting as if all six shirts were different leads to major overlap and redundancies, which is why the answer, 720, is A LOT bigger than the answer, 20. The factorial rule for permutations, n!, only works if all n items are unique and different from one another. If some of the items are the same --- which they are in this case --- we need to use the MISSISSIPPI rule (explained in the Counting with Identical Items lesson video) to remove the redundancies.

k is an integer from 1 to 9 inclusive. If N = 29736 + k , what is the value of k ? (1) N is divisible by 9 (2) N is divisible by 5

Divisibility rules 1) N = 29736 + k divisible by 9 2+9+7+3+6 = 27 (divisible by 9) If x and y are both divisible by d then (x+y) is also divisible by d k must be divisible by 9 too so it must be 9 2) divisible by 5 means two possible values so not sufficient 29736 + 4 = 29740 or 29736 + 9 = 29745

A container currently holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

Draw it out to see the problem better We start with a mixture containing 4 q alcohol and 4 q water. We're going to add some unknown amount of water to this mixture — let's call that x. Our beginning ratio of alcohol to water is 4:4. Then we add x quarts of water, so our final ratio will be 4:(4+x). We want that ratio to equal \dfrac{3}{5}​5​​3​​, so we can set these equal to each other and solve for x: Alcohol/water = 4 / (4+x) New proportion alcohol / water = 3/5 4 / (4+x) = 3/5 [Cross multiply] 20 = 3(4+x) 20 = 12+3x 8 = 3x 8/3 = x

The Cyclepaths Bicycle Factory has 300 bicycles in stock. The bicycles are either blue or red, and they come in two sizes: large and small. If 175 of the bicycles are blue, and 140 are large, how many are both blue and large? (1) There 90 small red bicycles. (2) There are twice as many small blue bicycles as there are large red bicycles.

Draw out the double matrix 1) enough info to get large blue 2) blue small = 2x large red = x red small = 125-x 2x +125-x = 160 can find x and get large blue sufficient D

At a certain high school, there are three sports: baseball, basketball, and football. Some athletes at this school play two of these three, but no athlete plays in all three. At this school, the ratio of (all baseball players) to (all basketball players) to (all football players) is 15:12:8. How many athletes at this school play baseball? (1) 40 athletes play both baseball and football, and 75 play football only and no other sport (2) 60 athletes play only baseball and no other sport

Draw venn diagram and label the sets! So, in that diagram a = folks who play baseball only b = folks who play football only c = folks who play basketball only d = folks who play baseball and football only e = folks who play baseball and basketball only f = folks who play football and basketball only g = folks who play all three sports At the outset, we know g = 0, but that's still six unknowns!! Now, notice: everyone who plays baseball = a + d + e everyone who plays basketball = c + e + f everyone who plays football = b + d + f So, the ratio given in the problem is: (a + d + e):(c + e + f):(b + d + f) = 15:12:8 Of course, these are all fractions, so we can't simplify. We can't simplify when there are the same terms in addition. In fact, there is no way to simplify this. We would like to find the total number of baseball players, a + d + e. Statement #1 tells us that d = 40 and b = 75. Thus total number of baseball players = a + 40 + e We don't have enough information to calculate this, and we don't have enough ratio information to solve. This statement, alone and by itself, is insufficient. Statement #2 tells us a = 60. Thus total number of baseball players = 60 + d + e We don't have enough information to calculate this. This statement, alone and by itself, is insufficient. Combined statements: Now we know a = 60, b = 75, and d = 40. total number of baseball players = 60 + 40 + e = 100 + e total number of football players = 75 + 40 + f = 115 + f We know the ratio of these two quantities, baseball to football, is 15:8. Unfortunately, that would give us only one equation for two unknowns, e & f. If we can't solve for these, then we can't solve for the total number of baseball players. Even with both statements combined, we cannot determine the answer. Combined, the statements are still insufficient. Both statements combined are insufficient. Answer = E

The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8 + 1+9 + 2+0 + 2+1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

Figure out how many numbers as ones and tens digits ex: 3s 10 numbers with 3 in ones digit: 3, 13,23, 33, 43,...93 10 numbers with 3 in tens digit: 30, 31, 32, 33...39 Total = 20 numbers*3 Sum = (0*20) + (1*20) +...(9*20) = 20 (1+2+3...+9) = 20(45) = 900

Two teachers, Ms. Ames and Mr. Betancourt, each had N cookies. Ms. Ames was able to give the same number of cookies to each one of her 24 students, with none left over. Mr. Betancourt was also able to give the same number of cookies to each one of his 18 students, with none left over. If N > 0, what is the value of N? (1) N < 100 (2) N > 50

GCF and LCM! Factor our 18 and 24: 18 = 2×9 = 2×3×3 = 6×3 24 = 3×8 = 2×2×2×3 = 6×4 From the prime factorizations, we see that GCF = 6, so the LCM is LCM = 6×3×4 = 72 and all other common multiples of 18 and 24 are the multiples of 72: {72, 144, 216, 288, 360, ...} Statement #1: if N < 100, the only possibility is N = 72. This statement, alone and by itself, is sufficient. Statement #2: if N > 50, then N could be 72, or 144, or 216, or etc. Many possibilities. This statement, alone and by itself, is not sufficient. Answer = (A)

If the hypotenuse of an isosceles right triangle has length of 8, then the area of the triangle is

It can be helpful to remember that the ratio of the sides are 1-1-\sqrt{2}√​2​​​. Even if you don't remember that, let AC = BC = x. Then, using the Pythagorean Theorem: x^2 + x2 = 82 2x^2 = 64 x^2 = 32 There's no reason to find the square root, because we are looking for the area. Area = (0.5)bh = (0.5)(BC)(AC) = (0.5)x2 = (0.5)(32) = 16

If x and y are positive integers, is x/y < (x+2) / (y+3)? 1) y>20 2) x<5

Know the rule of (a+p)/(b+q)! what x/y is Case 1: if it is <1, x/y < (x+p)/(y+q) Case 2: if it is >1, x/y > (x+p)/(y+q) Statement #1: We are adding 2 to the numerator and 3 to the denominator, so we know the resultant fraction will move closer to 2/3. If all we know is that the denominator of the starting fraction is greater than 20, then we have no idea what the size of the starting fraction is: it could be much greater than 2/3, or much smaller than 2/3, depending on the numerator, of which we have no idea. We can draw no conclusion right now. This statement, alone, by itself, is insufficient. Statement #2: Now, all we know is that the numerator of the starting fraction is less than 5 — it could be 4, 3, 2, or 1. We have no idea of the denominator. If y = 50, then we get a very small fraction. But if x = 4 and y = 1, the fraction equals 4, much larger than 2/3. In this statement, we have no information about the denominator, and since we know nothing about the denominator, we know nothing about the size of the starting fraction: it could be either greater or less than 2/3. Therefore, we can draw no conclusion. This statement, alone, by itself, is also insufficient. Combined Statements: We know y > 20 and x < 5. Well no matter what values we choose, we are going to have a denominator much bigger than the numerator. The largest possible fraction we could have under these constraints would be 4/21 (largest possible numerator with smallest possible denominator). The fraction 4/21 is much smaller than 1/2, so it's definitely smaller than 2/3. Any fraction with y > 20 and x < 5 will be less than 2/3. Therefore, adding 2 to the numerator and 3 to the denominator will move the resultant fraction closer to 2/3, which has the net effect of increasing its value. Therefore, the answer to the prompt question is "yes." Because we can give a definite answer to the prompt, we have sufficient information. Neither statement is sufficient individually, but together, they are sufficient. Answer = C.

8​^16​​+16^​13​​+4^​24​​=

Make into same base Factor out common numbers 2^48+2^52+2^48 = 2^52+2^1*2^48 = 9 * 2^49

Working together, Machine A and Machine B can produce a total of 200 widgets in 4 hours. How many hours would it take Machine A, working alone, to produce 200 widgets? (1) Working alone, Machine B takes 5 hours to produce 50 widgets. (2) Machine A can produce 4 widgets in the same amount of time it takes Machine B to produce 1 widget.

Problem: Total 50 widgets/hr by A+B working together How long does it take A to produce 200 widgets? 1) B produces 10 widgets/hr so that means 50 = A + 10 A = 40 widgets/hr --> 200 = 40*Time Time = 5 hrs (sufficient) 2) 4 Rate A = Rate B 4x + x = 50 widgets/hr x = 10 widgets/hr Rate A = 40 widgets/hr ---> can solve for time

Same algebra equation

No solution Can't solve

Peter invests $100,000 in an account that pays 12% annual interest: the interest is paid once, at the end of the year. Martha invests $100,000 in an account that pays 12% annual interest, compounding monthly at the end of each month. At the end of one full year, compared to Peter's account, approximately how much more does Martha's account have? Zero $68.25 $682.50 $6825.00 $68250.00

Not zero since compounded monthly > simple interest Peter gets the simple interest $12K Martha gets the $12K and 1% on every thousand (1/12 of 12%) The sum of the 10,20,30...looks closer to hundreds Month 1: 1000 Month 2: 1000+10 Month 3: 1000+20 Month 4: 1000+30

If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

Odd number set - median = mean k+2 = median and mean For any collection of evenly spaced numbers, the mean and median are always equal. Because this is set with seven members, an odd number of elements, the median is the middle number, the 4th number on the list: there are three numbers on the list below it and three above it. Thus, (k + 2) is the mean as well as the median, the middle or 4th number on the list. Since there are only seven numbers, we can simply write them out: {k - 1, k, k + 1, k + 2, k + 3, k + 4, k + 5} The least, three lower than the median, is (k - 1). The greatest, three higher than the median, is (k + 5). Their product is: product = (k - 1)(k + 5) = k^2 + 4k - 5

By abc we denote a three digit number with digits a, b, and c. Is abc divisible by 3? (1) The product of (a) times (b) is a number divisible by 3 (2) c = 3

Test numbers! Try to get a no answer abc divisible by 3 means the sum of a,b,c is divisible by 3 1) ab is divisible by 3 let ab = 12 131 - not divisible by 3 132 - divisible by 3 Not sufficient 2) c=3 No info on a or b so not sufficient Both: 131 - divisible by 3 333 - divisible by 3

If for some value of x, 5^x + 5^-x = B, then which of the following is equal to 25^x + 25^-x?

Square B 25^x +2(5^x)(5^-x) + (25^-x) 25^x + 25^-x = B^2 - 2

If problem looks like it will take too many calculations then what do you do?

Simplify find the trick to solve question without doing out all the calculations Try different sets of numbers!

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is 8 12 16 18 24

Simplify 1440 to factors = 2^5*3^2*5 k = product of consec numbers See how many numbers to get to the factors of 1440 k = 1*2*3*2*2*5*2*3*7*2*4 (n = 8)

Positive integers a, b, c, d and e are such that a < b < c < d < e. If the average (arithmetic mean) of the five numbers is 6 and d − b = 3, then what is the greatest possible range of the five numbers?

To get largest range, a has to be smallest Sum = 5*6=30 So a = 1, b=2, c=3, d=3+2 = 5, e= 30-sum Range = 19-1 = 18

% increase

Y/X-1

S is a set of n consecutive positive integers. Is the mean of the set an integer? (1) the range of S is an even integer (2) the median of S is an integer

mean = integer in set if number of items in set is odd Is n odd? 1) range = even Odd set: 1,2,3 | range = 3-1 = 2 (even) SUFFICIENT Even set: 1,2 | range = 2-1 = 1 (odd) 2) median = integer Odd number - integer (if even, number will not be integer)

In a set of twenty numbers, nineteen of the twenty numbers are between 40 and 50. Is the median greater than the mean? (1) the standard deviation is greater than 15 (2) the twentieth number is greater than 500

mean = median if symmetric (consec numbers) If outlier number, that number pulls mean to that number 1) SD > 15 implies outlier in 20th number (Range for other numbers are 10) but we don't know which direction (higher or lower) 2) 20th # >500 This pulls up mean and median stays the same so sufficient

What is the remainder when positive integer n is divided by 4? (1) When n is divided by 8, the remainder is 1. (2) When n is divided by 2, the remainder is 1.

n/some number = k(R) n/8 = k (1) n = 8k+1 = 4*2*k+1 Divide by 4 --> remainder 1 Try out numbers - n = 5; 5/4 = R1 n=7; 7/4 = R2 Not sufficient

x and y are positive integers such that x < y. If 6√​6​​​=x√​y​​​, then xy could equal

rewrite the radical 2*3*sqrt 6 = sqrt4*3*sqrt6 OR 2*3*sqrt 6 = 2*sqrt9*sqrt6 = 2*sqrt54 which means x = 2 and y=54--> xy=2*54=108

If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

xy=? 2x+2y=20 x+y=10 x^2+y^2 = 9^2 x^2+y^2 = 81 Square x+y x^2+2xy+y^2 = 100 Substitute in x^2+y^2 = 81 2xy = 100-81 =19 2xy = 19 xy=19/2


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