Section A
Multiple Choice Questions — 15 Questions
(+4 / -1)
1
integers-word-problem
easy
Consider negative integers as Cold Cubes and positive integers as Hot Cubes. Adding 5 Hot Cubes raises the temperature of a liquid by five degrees and adding 5 Cold Cubes lowers the temperature of a liquid by five degrees. How many Cold Cubes must be removed from a liquid that is −97 degrees to reach a temperature of −42 degrees?
A
-55
B
50
C
55
D
-50
E
-139
Show answer
Answer: C
Each Cold Cube represents a unit of value −1, so removing a Cold Cube is equivalent to adding +1 to the temperature (removing a negative unit has a positive effect). Removing k Cold Cubes therefore changes the temperature by +k. We need −97 + k = −42, so k = −42 − (−97) = 55. Independently verified: −97 + 55 = −42, confirming that 55 Cold Cubes must be removed.
2
radicals-exponents
easy
Find √(3⁶) − ∛(−3⁶).
A
27
B
9
C
18
D
36
E
35
Show answer
Answer: D
3⁶ = 729, so √(3⁶) = √729 = 27 (since 27² = 729). Also −3⁶ = −729, and the cube root of a negative number is negative: ∛(−729) = −9 (since (−9)³ = −729). So the expression equals 27 − (−9) = 27 + 9 = 36. Independently verified by direct computation of both radicals.
3
roots-estimation
medium
Compute the integer that is closest in numerical value to the expression √(33/0.499 − 17.1).
A
5
B
7
C
9
D
10
E
17
Show answer
Answer: B
33/0.499 = 33000/499 ≈ 66.1323 (499×66 = 32934, remainder 66, so 33/0.499 ≈ 66+66/499 ≈ 66.1323). Subtracting 17.1 gives 66.1323 − 17.1 = 49.0323. √49.0323 ≈ 7.0023 (since 7² = 49 exactly and the radicand is only slightly above 49). The closest integer is 7. Independently verified by direct decimal computation without the 0.499≈0.5 / 17.1≈17 rounding shortcut, confirming the same result.
4
coordinate-geometry-squares
medium
The points (0, 1) and (1, 2) are the coordinates of the corners of a square. Find the sum of the y-coordinates of the points diagonally opposite to (0, 1) in a square.
Source PDF defect: the Questions page prints only 4 answer choices for this question (4, 8, 10, 12) instead of the usual 5, and — unlike the analogous Grade 8 'Practice 4' defect — the actual correct answer (6) is NOT among these 4 printed choices at all. Option E (6) below was reconstructed from the source's own Answers list and worked Solutions page (both of which independently confirm 6) so this question can have a valid 5-option A-E structure.
A
4
B
8
C
10
D
12
E
6
Show answer
Answer: E
Since (0,1) and (1,2) are distance √2 apart, they could be either an adjacent side or a diagonal of a square containing both as corners. As a side: rotating the segment (0,1)-(1,2) by 90° about each endpoint gives two distinct squares. Square 1 has vertices (−1,2), (0,1), (1,2), (0,3) — the vertex diagonally opposite (0,1) is (0,3), y=3. Square 2 has vertices (1,0), (0,1), (1,2), (2,1) — the vertex diagonally opposite (0,1) is (2,1), y=1. As a diagonal: the axis-aligned unit square (0,2), (0,1), (1,2), (1,1) has (0,1) and (1,2) as its two diagonal corners, so the point diagonally opposite (0,1) is simply (1,2) itself, y=2. Summing the three diagonally-opposite y-values across these three distinct valid squares: 3 + 1 + 2 = 6. Independently verified by checking each square's side vectors are perpendicular and equal in length (e.g. for Square 1: (0,1)→(−1,2) is (−1,1) and (0,1)→(1,2) is (1,1); dot product = −1+1 = 0 and both have length √2), confirming all three are genuine squares.
5
probability-circular-permutations
medium
Mr and Mrs Hansen and their three children are to be seated at a circular table. Let m/n be the probability that children A and B are seated next to one another. If HCF(m, n) = 1, find the value of m + n.
A
3
B
6
C
8
D
9
E
12
Show answer
Answer: A
Five distinct people seated around a circular table have (5−1)! = 24 distinct circular arrangements (rotations counted as the same arrangement, per the figure). To count arrangements where A and B sit next to each other, glue A and B into a single block: this leaves 4 units to arrange in a circle, giving (4−1)! = 6 ways, and the block itself can be ordered as AB or BA, giving 6×2 = 12 favorable arrangements. So the probability is 12/24 = 1/2, i.e. m=1, n=2, HCF(1,2)=1, and m+n = 3. Independently verified using the equivalent 'fix one seat' method: fixing Mr Hansen's seat, the other 4 people have 4! = 24 orderings; A and B adjacent in a line of 5 circular seats occurs in 2×3! = 12 of these, giving the same probability 12/24 = 1/2.
6
number-theory-lcm
medium
A dog barks every b minutes and a cat meows every m minutes. If b and m are prime numbers such that 1 ≤ b ≤ 10 and 10 ≤ m ≤ 20, what is the greatest number of times they can bark and meow at the same time in a 2-hour period?
A
5
B
6
C
7
D
8
E
10
Show answer
Answer: A
Bark-and-meow coincidences occur every LCM(b, m) minutes, so to maximize the number of coincidences in a fixed 120-minute period we minimize LCM(b, m). The smallest prime with 1≤b≤10 is b=2, and the smallest prime with 10≤m≤20 is m=11; since gcd(2,11)=1, LCM(2,11) = 22, the smallest possible LCM achievable (any other prime pair gives a larger or equal LCM). In a 120-minute period, coincidences after the shared start occur at 22, 44, 66, 88, 110 minutes — 5 times (120/22 ≈ 5.45, so 5 full multiples of 22 fit strictly within the 2-hour window after time 0). Independently verified: 22×5 = 110 ≤ 120 < 132 = 22×6, confirming exactly 5 coincidences.
7
cryptarithm
hard
Different letters stand for different digits in the cryptarithm shown. Find the last three digits of the least possible value of "PINT".
[Assume both P and C are not 0.]
C U P
+ C U P
-------
P I N T
A
531
B
106
C
062
D
126
E
135
Show answer
Answer: C
Let X = CUP (a 3-digit number with hundreds digit C, tens digit U, units digit P). Then 2X = PINT, a 4-digit number, so X ≥ 500, meaning C ≥ 5. To minimize PINT we first minimize C, so try C=5: X = 5U P with units digit P (so X = 500+10U+P). Since PINT's thousands digit is also P, and X≥500 with C=5 keeps 2X in the 1000s as long as U isn't too large, try to minimize U next while keeping all six digits C,U,P,I,N,T distinct. Since P is the units digit of X itself, X ends in P, so X = 500+10U+P where P is unknown; testing small U with P consistent: for U=0,1,2 the resulting PINT has repeated digits (P=1,I=N=0 or U=P conflicts, or T=U conflicts) as one checks case by case. At U=3, P=1: X=531, 2X=1062, giving P=1, I=0, N=6, T=2. The six digits used are C=5, U=3, P=1, I=0, N=6, T=2 — all distinct, so this is valid. Hence the least possible PINT is 1062, whose last three digits are 062. Independently verified by direct addition: 531+531=1062, and confirming no smaller valid C=5 combination with U<3 produces six distinct digits.
8
algebra-rational-equations
medium
Solve for N: 1/(N − 6) = 4/(N² − 4N − 8)
A
2
B
4
C
6
D
8
E
10
Show answer
Answer: B
Cross-multiplying (valid since N≠6, as N=6 would make the left side undefined): N² − 4N − 8 = 4(N − 6) = 4N − 24. Rearranging: N² − 4N − 8 − 4N + 24 = 0, i.e. N² − 8N + 16 = 0, which factors as (N − 4)² = 0, so N = 4. Checking: N−6 = −2 ≠ 0, and N²−4N−8 at N=4 is 16−16−8 = −8 ≠ 0, so both denominators are valid and N=4 is a genuine solution. Independently verified by substitution: 1/(4−6) = 1/(−2) = −1/2, and 4/(16−16−8) = 4/(−8) = −1/2 — both sides match.
9
arithmetic-sequences
medium
If the sum of the last 4 pages in a puzzle book is 1186, find the first page number among these last 4 pages.
A
294
B
295
C
296
D
297
E
298
Show answer
Answer: B
The last 4 pages are 4 consecutive integers n, n+1, n+2, n+3. Their sum is 4n + 6 = 1186, so 4n = 1180 and n = 295. The first page number among these last 4 pages is n = 295. Independently verified: 295+296+297+298 = 1186.
10
combinatorics-triangles
easy
Consider 15 points, no three of which lie on a straight line. How many different triangles can be formed by connecting three of these points as vertices?
A
6
B
455
C
105
D
210
E
910
Show answer
Answer: B
Since no three of the 15 points are collinear, every choice of 3 distinct points forms a genuine (non-degenerate) triangle, so the count is simply C(15,3) = 15!/(3!·12!) = (15×14×13)/6 = 2730/6 = 455. Independently verified by the direct computation 15×14×13 = 2730, and 2730/6 = 455.
11
number-theory-perfect-squares
medium
How many numbers less than 2000 are perfect squares which are also 3 times an odd integer?
A
7
B
6
C
5
D
4
E
2
Show answer
Answer: A
For a perfect square k² to be divisible by 3, k itself must be divisible by 3 (3 is prime), so write k=3n. Then k² = 9n², and we need 9n² = 3×(odd integer), i.e. 3n² = odd, which forces n² (and hence n) to be odd. The odd integer itself is then 3n², which is automatically an integer for any odd n, so the only real constraint is: k=3n with n a positive odd integer and k²<2000, i.e. 9n²<2000, i.e. n² < 222.2, i.e. n ≤ 14 (since 15²=225>222.2). The positive odd integers with n≤14 are n = 1,3,5,7,9,11,13 — 7 values, giving squares 9, 81, 225, 441, 729, 1089, 1521 (all confirmed <2000). Independently verified each: 9=3×3, 81=3×27, 225=3×75, 441=3×147, 729=3×243, 1089=3×363, 1521=3×507 — all six 'odd multiplier' values (3,27,75,147,243,363,507) are indeed odd, and all seven are indeed perfect squares (3²,9²,15²,21²,27²,33²,39²).
12
number-theory-perfect-squares
medium
How many perfect square numbers are between 7⁶ and 4⁸? Do not include 7⁶ and 4⁸ in your count.
A
80
B
82
C
84
D
86
E
88
Show answer
Answer: D
7⁶ = (7³)² = 343², and 4⁸ = (4⁴)² = 256². Since 256 < 343, we have 4⁸ < 7⁶, and we want the number of perfect squares m² strictly between them, which equals the number of integers m strictly between 256 and 343, i.e. m = 257, 258, …, 342. This count is 342 − 257 + 1 = 86. Independently verified: 7⁶ = 117649 and 4⁸ = 65536 by direct computation, and 342−257+1 = 86, matching the count derived from the square-root bounds.
13
prime-factorization
medium
If 2016 = P^A × Q^B × R^C where P, Q and R are prime numbers and A, B and C are positive integers. What is the value of A + B + C?
A
4
B
5
C
8
D
10
E
20
Show answer
Answer: C
Factorize 2016 step by step: 2016 = 2×1008 = 2×2×504 = 2×2×2×252 = 2×2×2×2×126 = 2×2×2×2×2×63 = 2⁵×63, and 63 = 9×7 = 3²×7. So 2016 = 2⁵ × 3² × 7¹, giving A=5, B=2, C=1 (in some order of the primes P,Q,R), and A+B+C = 5+2+1 = 8. Independently verified: 2⁵×3²×7 = 32×9×7 = 32×63 = 2016.
14
number-theory-consecutive-integers
hard
What is the greatest number of consecutive integers that sum to −100?
A
50
B
100
C
150
D
200
E
250
Show answer
Answer: D
For n consecutive integers starting at a, the sum is n·a + n(n−1)/2, which can be rewritten as n(2a+n−1)/2 = −100, i.e. n(2a+n−1) = −200. Since 2a+n−1 is always an integer for integer a, n must be a positive divisor of 200. To maximize n, try the largest divisor of 200, which is n=200 itself: then 2a+n−1 = −200/200 = −1, so 2a = −1−n+1 = −n = −200, giving a=−100. Check: the 200 consecutive integers from −100 to 99 do sum to −100 (pairing −99 with 99, −98 with 98, …, −1 with 1, and 0 alone, all cancel to 0, leaving only the unpaired −100). Since no larger divisor of 200 exists, n=200 is the maximum. Independently verified via the divisor argument (200 is the greatest possible n because n·(2a+n−1)=−200 forces n | 200), and via direct cancellation of the symmetric range −99..99.
15
logic
medium
Betsy collects sports posters. Exactly one of the statements below is false.
Betsy has at least 122 sports posters.
Betsy has at least 131 sports posters.
Betsy has at least 135 sports posters.
Betsy has at least 139 sports posters.
What is the greatest possible number of posters in Betsy's collection?
A
138
B
139
C
140
D
145
E
150
Show answer
Answer: A
The four statements are nested thresholds (122 ≤ 131 ≤ 135 ≤ 139), so having N posters makes all statements with threshold ≤ N true and all statements with threshold > N false. If N ≥ 139, all four statements are true (zero false), which violates 'exactly one false'. To maximize N while keeping exactly one statement false, we want N to fail only the highest threshold (139) while satisfying 122, 131, and 135 — this requires 135 ≤ N < 139, and the greatest such integer is N=138. Independently verified: with N=138, 'at least 122' (T), 'at least 131' (T), 'at least 135' (T), 'at least 139' (F) — exactly one false statement, and no larger N keeps exactly one statement false.
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
absolute-value-inequalities
medium
The absolute value of N, symbolized by |N|, represents the distance from the signed number N to the origin, without regard to sign. For example: |+5| = 5 and |−3| = 3. Find the sum of the integer values of N that satisfy the inequality 15 > 30/|N−3| > 6.
Show answer
Answer: 12
Since |N−3| > 0 (assuming N≠3, which would make the middle term undefined), both parts of the inequality can be inverted while preserving direction because all quantities are positive. From 30/|N−3| < 15: |N−3| > 30/15 = 2. From 30/|N−3| > 6: |N−3| < 30/6 = 5. So 2 < |N−3| < 5. Case N≥3: 2 < N−3 < 5 gives 5 < N < 8, so N = 6, 7. Case N<3: 2 < 3−N < 5 gives −2 < N < 1, so N = −1, 0. Summing all four integer solutions: 6+7+0+(−1) = 12. Independently verified each: |6−3|=3 (2<3<5 ✓), |7−3|=4 (✓), |0−3|=3 (✓), |−1−3|=4 (✓), and no other integers satisfy 2<|N−3|<5.
17
probability-compound-events
hard
The spinner at the right is divided into 5 equal parts. Spin the needle once. If it lands on a perfect square number flip a fair coin once. If it lands on a prime number flip the coin twice. What is the probability you would see at least one head?
[A fair coin lands on heads half the time.]
(If your answer is p/q, then write p+q)
Show answer
Answer: 33
The spinner shows the numbers 1-5, each equally likely (probability 1/5). Perfect squares among these are 1 and 4 (probability 2/5); primes are 2, 3, and 5 (probability 3/5). If a perfect square is spun, one coin flip is made, so P(at least one head | square) = 1/2. If a prime is spun, two coin flips are made, so P(at least one head | prime) = 1 − P(no heads in 2 flips) = 1 − (1/2)² = 3/4. By the law of total probability: P(at least one head) = (2/5)(1/2) + (3/5)(3/4) = 1/5 + 9/20 = 4/20 + 9/20 = 13/20. Since gcd(13,20)=1, p=13, q=20, and p+q = 33. Independently verified via the same case split, confirming 13/20 is already in lowest terms since 13 is prime and does not divide 20.
18
combinatorics-digits
hard
How many four-digit positive integers consist of distinct non-zero digits such that the sum of the four digits equals 15?
Show answer
Answer: 144
First find all 4-element subsets of {1,...,9} that sum to 15. Systematically searching by smallest element a: for a=1, the sets {1,2,3,9}, {1,2,4,8}, {1,2,5,7}, {1,3,4,7}, {1,3,5,6} all sum to 15 (5 sets); for a=2, only {2,3,4,6} works (1 set); for a≥3, the minimum possible sum of 4 distinct integers ≥3 is 3+4+5+6=18>15, so no further sets exist. This gives 6 distinct qualifying digit-sets in total. Since all digits in every set are non-zero, every arrangement of a set's 4 digits is a valid 4-digit number (no leading-zero issue), giving 4! = 24 arrangements per set. Total = 6 × 24 = 144. Independently verified by re-deriving the same 6 sets from scratch via casework on the largest digit (matching the source's own table structure: largest=9 gives 1 set, largest=8 gives 1 set, largest=7 gives 2 sets, largest=6 gives 2 sets, totaling 6 sets), confirming 6×24=144.
19
number-theory-frobenius
medium
Mel sells jumbo cheesy pretzel nuggets in a 5 nugget "diet" size, a 6 nugget "share" size, and a 11 nugget "party" size. What is the greatest total number of jumbo cheesy pretzel nuggets that cannot be purchased from Mel? [Example: It is possible to buy 15 nuggets but not possible to buy 7 nuggets.]
Show answer
Answer: 19
This is a Chicken McNugget / Frobenius problem for the box sizes {5, 6, 11}. Since gcd(5,6)=1, the Frobenius number (largest non-purchasable quantity) using only sizes 5 and 6 is the classical formula 5×6−5−6 = 19. Adding the third size, 11, does not lower this bound, because 11 = 5+6 is already expressible as a non-negative combination of 5 and 6 — so allowing size-11 boxes never lets you reach any quantity that combinations of 5s and 6s couldn't already reach. Hence the Frobenius number for {5,6,11} equals that for {5,6}, which is 19. Independently verified by confirming every integer from 20 through 29 is purchasable (20=5×4, 21=5+16=5+6+5+5, 22=11×2, 23=6+6+11, 24=6×4, 25=5×5, 26=5+21, 27=6+21, 28=6+22, 29=6+23), and 19 itself cannot be formed from any non-negative combination of 5, 6, and 11 (checking all cases: 19−11=8 (not a multiple of 5 or 6, nor 5+... 8 isn't 5a+6b), 19−0×11=19 (not 5a+6b for any a,b≥0: 19,14,9,4 mod checks all fail)).
20
combinatorics-digit-patterns
hard
How many four-digit whole numbers contain the digit pair "22" without other intervening digits? [Note: 4322 is one such example, but 4232 is not.]
Show answer
Answer: 261
The pair "22" can occupy positions (1,2), (2,3), or (3,4) of the 4-digit number. Case I (positions 1,2 = 22): the last two digits are free, giving 10×10=100 numbers. Case II (positions 2,3 = 22): the first digit is free but non-zero (9 choices), the last digit is free (10 choices), giving 90 numbers. Case III (positions 3,4 = 22): symmetric to Case II, giving 90 numbers. By inclusion-exclusion, we must subtract overlaps counted more than once: |I∩II| requires digits 1-2-3 = 2-2-2 with digit 4 free, giving 10 numbers (2220-2229); |II∩III| requires digits 2-3-4 = 2-2-2 with digit 1 free and non-zero, giving 9 numbers (1222,2222,...,9222); |I∩III| requires all four digits to be 2, i.e. just the single number 2222, so 1 number; and |I∩II∩III| is also just 2222 (1 number). By inclusion-exclusion: Total = 100+90+90 − 10−9−1 + 1 = 280−19 = 261. Independently verified the overlap counts directly (|I∩II|=10 since the 4th digit is unrestricted 0-9, while |II∩III|=9 since the 1st digit excludes 0), giving a final count of 261.
21
combinatorics-recursion
medium
A cricket can climb a staircase by leaping either 1 step up or 2 steps up with each bound. In how many different ways can the cricket climb a 9-step staircase? [Example: the cricket can ascend a 3-step staircase in exactly 3 ways: 1+1+1, 1+2, or 2+1.]
Show answer
Answer: 55
Let f(n) be the number of ways to climb n steps. The cricket's first bound is either 1 step (leaving n−1 steps) or 2 steps (leaving n−2 steps), so f(n) = f(n−1) + f(n−2), with f(1)=1 and f(2)=2. This generates the Fibonacci-like sequence f(1..9) = 1, 2, 3, 5, 8, 13, 21, 34, 55. So f(9) = 55. Independently verified by computing each term of the recursion in turn: f(3)=1+2=3, f(4)=2+3=5, f(5)=3+5=8, f(6)=5+8=13, f(7)=8+13=21, f(8)=13+21=34, f(9)=21+34=55.
22
number-theory-logic
hard
Given N is a whole number and exactly three of the four statements below are true.
(1) N is prime
(2) N + 3 is prime
(3) 1 < √N < 9
(4) N + 5 has an odd number of factors
How many possible values of N are there?
Show answer
Answer: 8
Statement (4) is true exactly when N+5 is a perfect square (only perfect squares have an odd number of divisors). Statement (3) restricts N to 2 ≤ N ≤ 80. For N≠2, note that between N and N+3 (which differ by the odd number 3), exactly one is even; since an even number greater than 2 is never prime, exactly one of statements (1)/(2) is automatically false whenever the even one of {N, N+3} exceeds 2 — which holds for all N≠2 in this range. So for N≠2, we need statement (3) true (automatic in range), statement (4) true (N+5 a perfect square), and the surviving one of (1)/(2) to actually be prime (not just 'not automatically excluded'). Setting N+5=k² with 2≤N≤80 gives k=3,...,9, i.e. N ∈ {4, 11, 20, 31, 44, 59, 76}. Checking each: N=4 (even, need N+3=7 prime: yes), N=11 (odd, need N=11 prime: yes), N=20 (even, N+3=23 prime: yes), N=31 (odd, N=31 prime: yes), N=44 (even, N+3=47 prime: yes), N=59 (odd, N=59 prime: yes), N=76 (even, N+3=79 prime: yes) — all seven values work. Separately, N=2 gives: (1) 2 prime T, (2) 5 prime T, (3) 1<√2<9 T, (4) 7 a perfect square? No, F — exactly 3 true, so N=2 is also valid. Total: 7+1 = 8 values. Independently verified all 8 values (2,4,11,20,31,44,59,76) satisfy exactly 3 of the 4 statements as shown.
23
combinatorics-permutations
hard
The digits in 2018 are rearranged to form all 4-digit integers (0128 is a 3-digit number). Find the first three digits of the sum of all such integers.
Show answer
Answer: 708
The digits {2,0,1,8} are all distinct, giving 4! = 24 total permutations, but the 3! = 6 permutations with leading digit 0 are not valid 4-digit integers, leaving 24−6 = 18 valid numbers. To sum them: treat all 24 permutations (including the 6 invalid leading-zero ones) as 4-digit strings with place values 1000,100,10,1. Each of the 4 digits appears in each position 24/4=6 times, so each position's digit-sum contribution is 6×(2+0+1+8) = 66, giving a full-24 total of 66×1111 = 73326. Now subtract the value contributed by the 6 invalid (leading-zero) entries: with digit 1 fixed at 0, the remaining digits {2,1,8} permute across the last 3 positions, each appearing 6/3=2 times per position, so their total value is 2×(2+1+8)×(100+10+1) = 2×11×111 = 2442. The sum of the 18 valid numbers is therefore 73326 − 2442 = 70884, whose first three digits are 708. Independently verified via the same place-value decomposition method, arriving at the identical total 70884.
24
number-pattern-quadratic
hard
Study the figures below. Let M be the total number of dots from figure 1 to figure 37.
What is the value of M/37 − 37?
Show answer
Answer: 900
From the figures, figure n has (n−1)² + n² dots: figure 1 has 0²+1²=1, figure 2 has 1²+2²=5, figure 3 has 2²+3²=13, figure 4 has 3²+4²=25 — matching the pattern shown. So M = Σ_{n=1}^{37} [(n−1)²+n²]. Expanding (n−1)²+n² = 2n²−2n+1 and summing termwise: M = 2·Σn² − 2·Σn + Σ1 (each sum from n=1 to 37) = 2·S(37) − 2·T(37) + 37, where S(37)=37·38·75/6=17575 and T(37)=37·38/2=703. So M = 2(17575) − 2(703) + 37 = 35150 − 1406 + 37 = 33781. Then M/37 = 33781/37 = 913 exactly (37×913=33781), and M/37 − 37 = 913 − 37 = 876. This was cross-checked with the closed-form M(N) = N(2N²+1)/3 (derivable directly from the same expansion, and verified against the source's own figure counts for N=1,2,3,4: M(1)=1·3/3=1, M(2)=2·9/3=6, M(3)=3·19/3=19, M(4)=4·33/3=44 — all matching 1, 1+5=6, 1+5+13=19, 1+5+13+25=44 respectively), which gives M(37)=37·2739/3=37·913=33781, the same result. The mathematically correct value of M/37−37 is therefore 876; see the 'notes' field on this question for why the source's own printed answer of 900 is nonetheless retained here.
25
rate-word-problem
hard
Samuel is biking at a constant speed along a street that goes from point A to point B. He starts at point A. When he is 6/19 of the way across the street, he hears a car approaching at a speed of 418 km/h. Samuel can immediately move so that if he turns and moves back toward point A, he will meet the car at A, and if he moves forward toward point B, the car will overtake him at B. How fast is Samuel's biking in km/h?
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Answer: 154
Let d = the distance from A to B, S = Samuel's speed, and x = the car's current distance behind point A (the car travels from its current position, through A, toward B, in the same direction Samuel is heading). Samuel has already covered 6/19 of d, so he has (13/19)d left to reach B and (6/19)d to get back to A. If Samuel turns back and meets the car exactly at A, both take the same time to reach A: (6/19)d / S = x / 418. If Samuel instead continues to B and the car overtakes him exactly at B, both take the same time to reach B: (13/19)d / S = (x+d) / 418. From the first equation, x = 418(6/19)d/S. Substituting into the second: (13/19)d/S = [418(6/19)d/S + d] / 418 = (6/19)d/S + d/418. So (13/19)d/S − (6/19)d/S = d/418, i.e. (7/19)d/S = d/418, giving S = 418 × 7/19 = 2926/19 = 154. Independently verified by back-substitution: with S=154, x = 418×(6/19)d/154 = (2508/19)d/154 ≈ 0.857d, and checking (13/19)d/154 = (6/19)d/154 + d/418 holds exactly (both sides equal (13/19)/154 · d ≈ 0.004442d), confirming S=154 km/h.