Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
Show answer
Answer: C — $140.
Show hint
Hint 1
The 7 friends together covered Judi's share: 7 × $2.50. That's one-eighth of the total.
Show solution
Approach: Judi's share × 8
Judi's share = 7 × $2.50 = $17.50.
Everyone paid the same, so total = 8 × $17.50 = $140.
The Incredible Hulk can double the distance it jumps with each succeeding jump. If its first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will it first be able to jump more than 1 kilometer (1,000 meters)?
Show answer
Answer: C — 11th jump.
Show hint
Hint 1
Jump n is 2n−1 meters. Find the smallest n with 2n−1 > 1000.
The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
Show answer
Answer: C — 139 birds.
Show hint
Hint 1
If every animal had 4 legs, you'd see 800 legs. The shortage (800 − 522) comes from the 2-leg birds, each missing 2 legs.
Show solution
Approach: leg-shortage shortcut
If all 200 had 4 legs: 200 · 4 = 800 legs.
Actual: 522 ⇒ shortage = 800 − 522 = 278 legs.
Each bird is short 2 legs ⇒ birds = 278 / 2 = 139.
The top of one tree is 16 feet higher than the top of another tree. The heights of the two trees are in the ratio 3 : 4. In feet, how tall is the taller tree?
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Answer: B — 64 feet.
Show hint
Hint 1
The ratio 3:4 means the difference (1 part) corresponds to 16 feet. So 1 part = 16 ft.
Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?
A sequence of numbers starts with 1, 2, and 3. The fourth number of the sequence is the sum of the previous three numbers in the sequence: 1 + 2 + 3 = 6. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?
Show answer
Answer: D — 68.
Show hint
Hint 1
Just iterate the rule: each new term = sum of the previous three.
Pick two consecutive positive integers whose sum is less than 100. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?
Show answer
Answer: C — 79.
Show hint
Hint 1
(x+1)2 − x2 = 2x + 1 = sum of the two integers. The sum < 100 and is odd.
Show solution
Approach: factor the difference
(x+1)2 − x2 = 2x + 1 = (x) + (x+1).
Difference equals the sum of the two integers — less than 100, and odd.
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?
Show answer
Answer: D — 32.
Show hint
Hint 1
Five-term arithmetic sequence with common difference 6. Middle term = average = 100/5 = 20.
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?
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Answer: C — 5 tricycles.
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Hint 1
Imagine all 7 children on bicycles first, then see how many extra wheels you still need.
Show solution
Approach: assume all bicycles, then add the extra wheels
If all 7 rode bicycles, that's 7 × 2 = 14 wheels.
There are 19 − 14 = 5 extra wheels, and each tricycle adds exactly one extra wheel, so there are 5 tricycles.
Another way — solve the system:
With b bicycles and t tricycles: b + t = 7 and 2b + 3t = 19.
Subtract twice the first from the second: t = 19 − 14 = 5.
Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?
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Answer: C — 45 coins.
Show hint
Hint 1
Set up two equations for the same gold-coin count: g = 9(n − 2) and g = 6n + 3.
Show solution
Approach: two equations, same g
Let n = chests, g = coins. Nine-per-chest leaves 2 empty: g = 9(n − 2). Six-per-chest with 3 left over: g = 6n + 3.
Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read 1/5 of the pages plus 12 more, and on the second day she read 1/4 of the remaining pages plus 15 pages. On the third day she read 1/3 of the remaining pages plus 18 pages. She then realized that there were only 62 pages left to read, which she read the next day. How many pages are in this book?
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Answer: C — 240 pages.
Show hints
Hint 1 of 2
Work backwards from the 62 pages left. Day 3 left her with 62 pages after reading (1/3)R + 18 from R; so 2R/3 − 18 = 62.
Still stuck? Show hint 2 →
Hint 2 of 2
Repeat the inversion for day 2 and day 1.
Show solution
Approach: work backwards day by day
After day 3 there are 62 pages left. If R3 = pages at start of day 3: (2/3)R3 − 18 = 62 ⇒ R3 = 120.
In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
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Answer: C — 7.
Show hints
Hint 1 of 2
Imagine she got all 10 right first, then see what each wrong answer costs.
Still stuck? Show hint 2 →
Hint 2 of 2
Turning one correct answer into a wrong one drops the score by 5 + 2 = 7.
Show solution
Approach: start from a perfect score and subtract
All 10 correct would score 5 Ć 10 = 50.
Each wrong answer (instead of right) costs 5 + 2 = 7 points, and 50 ā 29 = 21 = 3 Ć 7.
So 3 were wrong and 7 were correct.
Another way — set up an equation:
Let x = number correct, so 10 ā x are wrong: 5x ā 2(10 ā x) = 29.
A special key on a calculator replaces the displayed number x with 1 Ć· (1 ā x). (For example, from 2 it gives 1 Ć· (1 ā 2) = ā1.) If the calculator shows 5 and the key is pressed 100 times in a row, the calculator will display
Show answer
Answer: A — ā0.25.
Show hints
Hint 1 of 2
Press the key a few times and watch for a repeating cycle.
Still stuck? Show hint 2 →
Hint 2 of 2
Once you know the cycle length, reduce 100 by it.
Show solution
Approach: find the repeating cycle
Starting at 5: 5 ā ā0.25 ā 0.8 ā 5, a cycle of length 3.
Since 100 = 3Ā·33 + 1, after 100 presses the display matches one press: ā0.25.
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?
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Answer: E — 995 dollars.
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Hint 1 of 2
Let n be the number of employees and write the fund two ways.
Still stuck? Show hint 2 →
Hint 2 of 2
Fund = 50n ā 5 (just short of $50 each) and fund = 45n + 95 (after the $45 bonuses).
Show solution
Approach: set two expressions for the fund equal
The fund is 50n ā 5 (5 short of $50 each) and also 45n + 95 (gave $45 each, kept $95).
Setting them equal: 50n ā 5 = 45n + 95 gives n = 20, so the fund was 45Ā·20 + 95 = $995.
Pauline can shovel snow at the rate of 20 cubic yards for the first hour, 19 cubic yards for the second, 18 for the third, and so on, always shoveling one cubic yard less per hour than the previous hour. If her driveway is 4 yards wide, 10 yards long, and covered with snow 3 yards deep, then the number of hours it will take her to shovel it clean is closest to
Show answer
Answer: D — 7.
Show hints
Hint 1 of 2
First find the total volume of snow: 4 Ć 10 Ć 3.
Still stuck? Show hint 2 →
Hint 2 of 2
Add 20 + 19 + 18 + ā¦ until you reach that volume.
Show solution
Approach: accumulate the decreasing hourly amounts
The snow is 4 Ć 10 Ć 3 = 120 cubic yards.
Running totals: 20, 39, 57, 74, 90, 105, 119 ā after 7 hours she's at 119, just shy of 120, so the time is closest to 7 hours.
Distance from home rises while going out, stays flat while shopping, then falls coming back.
Still stuck? Show hint 2 →
Hint 2 of 2
Because the speed changes (gentle in city, steep on highway), each side of the graph bends rather than staying a single straight line.
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Approach: match each changing-speed leg to the graph's slope
Going out, distance rises gently (city) then steeply (highway), so the climb curves and gets steeper; it stays flat for the hour at the mall; coming home it falls steeply (highway) then gently (city).
Graph B shows these two different slopes on each side ā the straight-sided trapezoid (A) would mean a single constant speed each way.
The distance between the 5th and 26th exits on an interstate highway is 118 miles. If any two exits are at least 5 miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the 5th and 26th exits?
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Answer: C — 18 miles.
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Hint 1 of 2
There are 21 gaps between the 5th and 26th exits.
Still stuck? Show hint 2 →
Hint 2 of 2
To stretch one gap as far as possible, make all the others as small as allowed (5 miles).
Show solution
Approach: shrink the other gaps to free up room
There are 26 ā 5 = 21 gaps. Making 20 of them the minimum 5 miles uses 100 miles.
Many calculators have a reciprocal key 1/x that replaces the current number displayed with its reciprocal. For example, if the display is 00004 and the 1/x key is pressed, then the display becomes 000.25. If 00032 is currently displayed, what is the fewest positive number of times you must depress the 1/x key so the display again reads 00032?
Show answer
Answer: B — 2.
Show hints
Hint 1 of 2
Try the operation once, then again ā does anything change after that?
Still stuck? Show hint 2 →
Hint 2 of 2
The reciprocal of the reciprocal is the original number.
The sale ad read: "Buy three tires at the regular price and get the fourth tire for three dollars." Sam paid 240 dollars for a set of four tires at the sale. What was the regular price of one tire?
Show answer
Answer: D — 79 dollars.
Show hint
Hint 1
Subtract the $3 promo tire first, then split what's left across three tires.
Show solution
Approach: isolate the three full-price tires
Three full-price tires cost 240 ā 3 = 237 dollars.
A calculator has a squaring key x² which replaces the current number displayed with its square. For example, if the display is 000003 and the x² key is depressed, then the display becomes 000009. If the display reads 000002, how many times must you depress the x² key to produce a displayed number greater than 500?
Nine copies of a certain pamphlet cost less than $10.00 while ten copies of the same pamphlet (at the same price) cost more than $11.00. How much does one copy of this pamphlet cost?
Show answer
Answer: E — $1.11.
Show hints
Hint 1 of 2
9p < 10 and 10p > 11 give a narrow interval for p.
Still stuck? Show hint 2 →
Hint 2 of 2
10ā9 ā 1.111 and 11ā10 = 1.10.
Show solution
Approach: bracket p from both inequalities
9p < 10 ā p < 10ā9 ā 1.111. 10p > 11 ā p > 1.10. So 1.10 < p < 1.111.
Jami picked three equally spaced integers on the number line. The sum of the first and second is 40, and the sum of the second and third is 60. What is the sum of all three numbers?
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Answer: B — 75.
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Hint 1 of 2
For equally spaced numbers, the middle one is the average of the other two.
Still stuck? Show hint 2 →
Hint 2 of 2
Add the two given sums and see how many times the middle number appears.
Show solution
Approach: the middle number carries everything
Since the numbers are equally spaced, first + third = 2 Ć second. Adding the two given sums: 40 + 60 = first + 2Ā·second + third = 4Ā·second, so the middle is 25.
Three positive integers are equally spaced on a number line. The middle number is 15 and the largest number is 4 times the smallest number. What is the smallest of these three numbers?
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Answer: C — 6.
Show hints
Hint 1 of 2
Equally spaced means the middle equals the average of the outer two.
Still stuck? Show hint 2 →
Hint 2 of 2
Let smallest = x, largest = 4x. Their average = 5x/2 = 15.
Show solution
Approach: middle = average of outers
Let smallest = x, so largest = 4x. Equally spaced ⇒ middle = (x + 4x)/2 = 5x/2.
Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?
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Answer: C — 8072 cents.
Show hints
Hint 1 of 2
Trade pennies for nickels: each swap adds 4 cents. So the total is a linear function of the nickel count n.
Still stuck? Show hint 2 →
Hint 2 of 2
Total = p + 5n = (p+n) + 4n = 2020 + 4n. With 1 ≤ n ≤ 2019, difference is 4 × (2019 − 1).
Show solution
Approach: rewrite total as 2020 + 4n
Total cents = p + 5n = (p + n) + 4n = 2020 + 4n.
Constraints: p ≥ 1 and n ≥ 1 with p + n = 2020 ⇒ n ∈ [1, 2019].
Maximum vs minimum total: 4(2019) − 4(1) = 4 × 2018 = 8072.
On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 20 days?
Show answer
Answer: D — 400 widgets.
Show hint
Hint 1
The first 20 positive odd numbers: 1, 3, 5, …, 39. The sum of the first n odd numbers is n2.
Show solution
Approach: sum of first n odd numbers = n^2
Day k sales: 2k − 1. Days 1 to 20 sum: 1 + 3 + 5 + … + 39.
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 13 of all the ninth graders are paired with 25 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Show answer
Answer: B — 4/11.
Show hints
Hint 1 of 2
Pairs are one-to-one, so the number of paired ninth-graders equals the number of paired sixth-graders. Set that up as one equation and pick a convenient size.
Still stuck? Show hint 2 →
Hint 2 of 2
Let there be 15 ninth-graders. Then 5 of them are paired (one-third). So 5 sixth-graders are paired too, which is two-fifths of all sixth-graders ⇒ 12 sixth-graders total.
Show solution
Approach: pick concrete sizes so the fractions are whole numbers
Pairs are one-to-one, so paired ninth-graders = paired sixth-graders. Pick small numbers that make both fractions whole: ninth = 6, sixth = 5.
Then paired ninth = (1/3)(6) = 2 and paired sixth = (2/5)(5) = 2. ✓
Total students = 6 + 5 = 11. Buddied students = 2 + 2 = 4. Fraction = 4/11.
Another way — algebra:
Let n = ninth-graders, s = sixth-graders. (1/3)n = (2/5)s ⇒ 5n = 6s.
Buddied / total = ((1/3)n + (2/5)s) / (n + s) = (2 · (1/3)n) / (n + s) (since the two numerators are equal).
Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
Show answer
Answer: D — 9 miles.
Show hints
Hint 1 of 2
Set the two distance expressions equal: rush-hour distance = no-traffic distance. Convert 20 min and 12 min to hours (1/3 and 1/5).
Still stuck? Show hint 2 →
Hint 2 of 2
Solve s · (1/3) = (s + 18) · (1/5) for the rush-hour speed s, then plug back.
Show solution
Approach: equate the two distance expressions
Let rush-hour speed be s mph. Distance: s · (1/3) = (s + 18) · (1/5).
Multiply by 15: 5s = 3(s + 18) ⇒ 2s = 54 ⇒ s = 27.
Ralph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost $1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost $4 a pair. If he bought at least one pair of each type, how many pairs of $1 socks did Ralph buy?
Show answer
Answer: D — 7 pairs.
Show hints
Hint 1 of 2
Let a, b, c be the counts of $1, $3, $4 pairs. Write the two equations and subtract.
Still stuck? Show hint 2 →
Hint 2 of 2
After subtracting, you get 2b + 3c = 12 with b, c ≥ 1. Use parity (2b is even) to pin down c.
Show solution
Approach: subtract the two equations
a + b + c = 12 and a + 3b + 4c = 24.
Subtract: 2b + 3c = 12. So 3c is even, meaning c is even; and 0 < c < 4 ⇒ c = 2.
Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
Show answer
Answer: C — 87431.
Show hint
Hint 1
Higher place values dominate. To maximize the sum, put the two biggest digits in the ten-thousands place (one each), the next two in the thousands place, etc.
Show solution
Approach: split digits into matched pairs by place value
Pair digits by descending size for each place: {9, 8}, {7, 6}, {5, 4}, {3, 2}, {1, 0}.
Each number gets one from each pair. So the digits of each number (left to right) come from {9, 8}, {7, 6}, {5, 4}, {3, 2}, {1, 0}.
Only 87431 matches: 8∈{9,8}, 7∈{7,6}, 4∈{5,4}, 3∈{3,2}, 1∈{1,0}. ✓
The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is 30% of the perimeter. What is the length of the longest side?
Show answer
Answer: E — 11 inches.
Show hint
Hint 1
Let the smallest side be s. Then perimeter = 3s + 3. Set s = 0.3 · (3s + 3).
Show solution
Approach: translate the percent condition
Sides: s, s+1, s+2. Perimeter: 3s + 3.
s = 0.3(3s + 3) ⇒ s = 0.9s + 0.9 ⇒ 0.1s = 0.9 ⇒ s = 9.
The hundreds digit of a three-digit number is 2 more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
Show answer
Answer: E — 8.
Show hints
Hint 1 of 2
Let units = u, tens = t, hundreds = u + 2. Original − reversed simplifies to a constant.
Still stuck? Show hint 2 →
Hint 2 of 2
Reversing flips the hundreds and units digits. Difference = 99 · (hundreds − units).
Andy and Bethany have a rectangular array of numbers with 40 rows and 75 columns. Andy adds the numbers in each row. The average of his 40 sums is A. Bethany adds the numbers in each column. The average of her 75 sums is B. What is the value of AB?
Show answer
Answer: D — 15/8.
Show hint
Hint 1
Both Andy and Bethany sum the same array total. So 40A = sum of array = 75B.
On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought 400 jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?
Show answer
Answer: B — 28 students.
Show hint
Hint 1
Each boy gets b beans, so all boys get b2. Similarly girls get g2. b2 + g2 = 394.
A ball is dropped from a height of 3 meters. On its first bounce it rises to a height of 2 meters. It keeps falling and bouncing to 23 of the height it reached in the previous bounce. On which bounce will it rise to a height less than 0.5 meters?
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Answer: C — 5th bounce.
Show hint
Hint 1
After the nth bounce, height = 3 · (2/3)n. Test small n until it drops below 1/2.
Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than 100 pounds or more than 150 pounds. So the boxes are weighed in pairs in every possible way. The results are 122, 125 and 127 pounds. What is the combined weight in pounds of the three boxes?
Show answer
Answer: C — 187 pounds.
Show hint
Hint 1
Each box is in 2 of the 3 pair-sums. Adding all three pair-sums double-counts each weight.
Show solution
Approach: sum the pair-sums and halve
Sum of pair-sums: 122 + 125 + 127 = 374 = 2(a + b + c).
Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of 50 units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
Show answer
Answer: D — 132.
Show hint
Hint 1
l + w = 25, both positive integers. Area = l(25 − l); max near l = 12 or 13, min at l = 1.
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and 3/4 of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
Show answer
Answer: B — 17.
Show hints
Hint 1 of 2
Let p be the common count passing. Boys = (3/2)p, girls = (4/3)p; total = (17/6)p.
Still stuck? Show hint 2 →
Hint 2 of 2
Total must be a positive integer; smallest p making it integer is p = 6.
Show solution
Approach: introduce a common-count variable
Boys = (3/2)p, girls = (4/3)p. Total = (3/2 + 4/3)p = (17/6)p.
Smallest positive integer total requires p = 6 ⇒ total = 17.
Before district play, the Unicorns had won 45% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
Show answer
Answer: A — 48 games.
Show hint
Hint 1
Let pre-district games be x. Pre-district wins: 0.45x. Final wins: 0.45x + 6 = (x + 8)/2.
Show solution
Approach: set up an equation
0.45x + 6 = (x + 8)/2.
Multiply by 10: 4.5x + 60 = 5x + 40 ⇒ 0.5x = 20 ⇒ x = 40.
A sequence of squares is made of identical square tiles. Each square's edge is one tile longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
Show answer
Answer: C — 13.
Show hints
Hint 1 of 2
A square that is n tiles on a side uses n Ć n tiles.
Still stuck? Show hint 2 →
Hint 2 of 2
Compare 7 Ć 7 with 6 Ć 6.
Show solution
Approach: the nth square uses nĀ² tiles
A square that is n tiles on a side uses nĀ² tiles.
So the seventh needs 7Ā² = 49 and the sixth needs 6Ā² = 36: the difference is 49 ā 36 = 13.
Another way — difference of squares (no squaring needed):
What is the smallest result that can be obtained from the following process? Choose three different numbers from the set {3, 5, 7, 11, 13, 17}, add two of them, then multiply their sum by the third number.
Show answer
Answer: C — 36.
Show hints
Hint 1 of 2
To make a product small, multiply by the smallest number.
Still stuck? Show hint 2 →
Hint 2 of 2
Then add the two next-smallest numbers.
Show solution
Approach: multiply by the smallest, add the next two
Use 3 as the multiplier and add the next two smallest, 5 and 7: (5 + 7) Ć 3 = 36.
Any other choice (such as (3 + 7) Ć 5 = 50) is larger, so the smallest is 36.
Let x be the number 0.00…01, where there are 1996 zeros after the decimal point before the 1. Which of the following expressions represents the largest number?
Show answer
Answer: D — 3/x.
Show hints
Hint 1 of 2
x is an extremely tiny positive number.
Still stuck? Show hint 2 →
Hint 2 of 2
Dividing by something tiny makes the result enormous.
Show solution
Approach: compare sizes when x is tiny
Since x is a tiny positive number, 3 + x, 3 ā x, and 3Ā·x are all near 3 (or near 0), and x/3 is tiny.
But 3/x divides by something tiny, giving a gigantic number, so 3/x is largest.
For every 3Ā° rise in temperature, the volume of a certain gas expands by 4 cubic centimeters. If the volume of the gas is 24 cubic centimeters when the temperature is 32Ā°, what was the volume in cubic centimeters when the temperature was 20Ā°?
Show answer
Answer: A — 8.
Show hints
Hint 1 of 2
From 32Ā° down to 20Ā° is a 12Ā° drop ā how many 3Ā° steps is that?
Still stuck? Show hint 2 →
Hint 2 of 2
Each step removes 4 cmĀ³.
Show solution
Approach: count the 3Ā° steps and subtract
A 12Ā° drop is four 3Ā° steps, each shrinking the gas by 4 cmĀ³, for 16 cmĀ³ less.
Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?
Show answer
Answer: A — 4/7.
Show hints
Hint 1 of 2
Sarika eats on turns 1, 4, 7, … — every third turn. What fraction does each Sarika-bite eat?
Still stuck? Show hint 2 →
Hint 2 of 2
Sarika's bites are 1/2, then 1/16, then 1/128, … — a geometric series with ratio 1/8.
Show solution
Approach: sum a geometric series
Each turn eats half of what's left, so the cheese remaining after turn n is 1/2n. Sarika eats at turns 1, 4, 7, …, taking 12, 116, 1128, … of the original block.
Geometric series with first term a = 12 and ratio r = 18.
A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was 3 : 1. Then 3 green frogs moved to the sunny side and 5 yellow frogs moved to the shady side. Now the ratio is 4 : 1. What is the difference between the number of green frogs and yellow frogs now?
Show answer
Answer: E — 24 frogs.
Show hints
Hint 1 of 2
Express both populations in one variable (yellow), then write the new ratio as an equation.
Still stuck? Show hint 2 →
Hint 2 of 2
After moves: green = 3y + 2, yellow = y − 2. Set (3y+2)/(y−2) = 4.
Show solution
Approach: let y = initial yellow, then use both ratios
Let y = initial yellow count, so initial green = 3y.
After the moves: green = 3y + 5 − 3 = 3y + 2 (5 in, 3 out). Yellow = y + 3 − 5 = y − 2.
New ratio: 3y + 2y − 2 = 4 ⇒ 3y + 2 = 4y − 8 ⇒ y = 10. So initial green = 30, yellow = 10.
After Euclid High School's last basketball game, it was determined that 14 of the team's points were scored by Alexa and 27 were scored by Brittany. Chelsea scored 15 points. None of the other 7 team members scored more than 2 points. What was the total number of points scored by the other 7 team members?
Show answer
Answer: B — 11 points.
Show hints
Hint 1 of 2
Let T = total. Then 1/4 and 2/7 of T are integers ⇒ T is a multiple of lcm(4,7) = 28.
Loki, Moe, Nick, and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money, and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
Show answer
Answer: B — 1/4.
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Hint 1 of 2
The three gifts are equal ā pick a convenient size for that common gift, like $1.
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Hint 2 of 2
Then Moe started with $5, Loki $4, Nick $3; passing money around never changes the group total.
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Approach: set the equal gift to a convenient $1
Let each gift be $1. Since $1 is Moe's fifth, Loki's fourth, and Nick's third, they began with $5, $4, and $3.
Handing money over doesn't change the total: $5 + $4 + $3 = $12, and Ott now holds the three gifts, $3.
Terri builds a sequence of positive integers by these rules: if the integer is less than 10, multiply it by 9; if it is even and greater than 9, divide it by 2; if it is odd and greater than 9, subtract 5. Find the 98th term of the sequence that begins 98, 49, … .
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Answer: D — 27.
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Hint 1 of 2
Generate terms until they start repeating.
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Hint 2 of 2
Once you spot the repeating block, use its length to jump ahead to the 98th term.
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Approach: find the repeating cycle, then index into it
The sequence runs 98, 49, 44, 22, 11, 6, 54, 27, 22, 11, ā¦ ā from the 4th term on it cycles 22, 11, 6, 54, 27 with length 5.
From the 4th term, (98 ā 4) = 94 steps and 94 Ć· 5 leaves remainder 4, landing on the 5th entry of the cycle: 27.
Three generous friends redistribute their money as follows: Amy gives Jan and Toy enough to double each of their amounts; then Jan gives Amy and Toy enough to double theirs; finally Toy gives Amy and Jan enough to double theirs. Toy had $36 at the beginning and $36 at the end. What is the total amount the three friends have?
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Answer: D — $252.
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Hint 1 of 2
The total never changes ā you only need to find it at one moment.
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Hint 2 of 2
Toy's money doubles in the first two rounds, so track it up to just before his own turn.
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Approach: track Toy's doubling, then read off the total
Toy's $36 doubles in each of the first two rounds: 36 ā 72 ā 144 just before his turn.
He ends with $36, so he gave away 144 ā 36 = 108, which exactly doubled Amy and Jan ā meaning they held $108 then.
The total, unchanged throughout, is 144 + 108 = $252.
A list of 8 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three numbers are 16, 64, 1024.
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Answer: B — 1/4.
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Hint 1 of 2
Each term is the product of the two before it, so divide to step backward.
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Hint 2 of 2
From 16, 64, 1024 work back: the term before 16 is 64 Ć· 16, and so on.
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Approach: divide to walk back to the start
Since 1024 = 16 Ā· 64, the term before 16 is 64 Ć· 16 = 4, then 16 Ć· 4 = 4, then 4 Ć· 4 = 1, then 4 Ć· 1 = 4, and finally 1 Ć· 4 = 1/4.
A fifth number, n, is added to the set {3, 6, 9, 10} to make the mean of the set of five numbers equal to its median. The number of possible values of n is
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Answer: C — 3.
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Hint 1 of 2
The median depends on where n falls in the sorted list; split into three cases.
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Hint 2 of 2
In each case, set mean = median and solve for n.
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Approach: case on where n sits in the sorted list
Mean = (28 + n)ā5. Case n ā¤ 6: median = 6 ā n = 2. Case 6 ā¤ n ā¤ 9: median = n ā 28 + n = 5n ā n = 7. Case n ā„ 9: median = 9 ā n = 17.
Each value of n is consistent with its case, so the answers are 2, 7, 17 ā 3 values.
A multiple choice examination consists of 20 questions. The scoring is +5 for each correct answer, ā2 for each incorrect answer, and 0 for each unanswered question. John's score on the examination is 48. What is the maximum number of questions he could have answered correctly?
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Answer: D — 12.
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Hint 1 of 2
Let c = correct, w = wrong; 5c ā 2w = 48 with c + w ā¤ 20 and c, w ā„ 0 integers.
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Hint 2 of 2
5c must be even, so c must be even ā that limits c to even values.
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Approach: maximize c subject to 5c ā 2w = 48 and c + w ā¤ 20
From 5c ā 2w = 48, c must be even (so 5c is even); from c + w ā¤ 20, c is at most about 12.6.
The biggest even c that fits is c = 12 (w = 6, total 18).
Total of all 6 numbers 10 + 11 + āÆ + 15 = 75. Adding the three side-sums gives each vertex twice and each midpoint once, so 3S = (vertex sum) + 75. Maximize by putting the three biggest at the vertices: 13 + 14 + 15 = 42.
3S = 42 + 75 = 117 ā S = 39 (with midpoints 12, 10, 11 between the matching vertex pairs).
