Although mishkil has answered 2 quizzes correctly and Neela has answered all 4, here are the 'official' answers to all 4, with explanations:
1.1: Let us name the 4 persons A, B, C & D who cross in 1, 2, 7 & 10 mins resply.
1. A & B cross in 2 mins, A returns in 1 min. Total time taken 3 mins with B across the river.
2. C & D cross in 10 mins, B returns in 2 mins. Total time taken 12 mins with C & D across the river.
3. A & B cross in 2 mins. All 4 are across the river. Total time: 3 + 12 + 2 = 17 mins.
1.2: Ladders: The alley is 8.6445 ft wide.
Ladders 1 and 2, denoted L1 and L2, respectively, will rest along two
walls (taken to be perpendicular to the ground), and they will
intersect at a point O = (a,s), a height s from the ground. Find the
largest s such that this is possible. Then find the width of the
alley, w = a+b, in terms of L1, L2, and s. This diagram is not to
scale.
(Best to use a fixed-width font for the figure below.)
B D
|\ L1 L2 /|
| \ / | BC = L1 = 11 ft
| \ / | AD = L2 = 13 ft
| \ O / | s = 4 ft
x| \ / |y A = (0,0)
| /|\ | AE = a
| m / | \ n | EC = b
| / |s \ | AO = m
| / | \ | CO = n
|/____|____\|
(A a E b C
Without loss of generality, let L2 >= L1.
Observe that triangles AOB and DOC are similar. Let r be the ratio of
similitude, so that x=ry. Consider right triangles CAB and ACD. By
the Pythagorean theorem, L1^2 - x^2 = L2^2 - y^2. Substituting x=ry,
this becomes y^2(1-r^2) = L2^2 - L1^2. Letting L= L2^2 -L1^2 (L>=0),
and factoring, this becomes
(*) y^2 (1+r)(1-r) = L
Now, because parallel lines cut L1 (a transversal) in proportion, r =
x/y = (L1-n)/n, and so L1/n = r+1. Now, x/s = L1/n = r+1, so ry = x =
s(r+1). Solving for r, one obtains the formula r = s/(y-s).
Substitute this into (*) to get
(**) y^2 (y) (y-2s) = L (y-s)^2
NOTE: Observe that, since L>=0, it must be true that y-2s>=0.
Now, (**) defines a fourth degree polynomial in y. It can be written
in the form (by simply expanding (**))
(***) y^4 - 2s_y^3 - L_y^2 + 2sL_y - Ls^2 = 0
L1 and L2 are given, and so L is a constant. How large can s be?
Given L, the value s=k is possible if and only if there exists a real
solution, y', to (***), such that 2k <= y' < L2. Now that s has been
chosen, L and s are constants, and (***) gives the desired value of
y. (Make sure to choose the value satisfying 2s <= y' < L2. If the
value of s is "admissible" (i.e., feasible), then there will exist
exactly one such solution.) Now, w = sqrt(L2^2 - y^2), so this
concludes the solution.
L1 = 11, L2 = 13, s = 4. L = 13^2-11^2 = 48, so (***) becomes
y^4 - 8_y^3 - 48_y^2 + 384_y - 768 = 0
Numerically find root y ~= 9.70940555, which yields w ~= 8.644504.
1.3: Coins: 4 coins of 2 & 5 cents; 3 coins of 1,3 and 10 cents
Since there are at least 3 coins of each type, they will cost
3* (1 + 2 + 3 + 5 + 10) = 63 cents. Since the total is a multiple
of 10 cents (dimes), it must be 70 cents. The only combination
of two coins costing 70 - 63 = 7 cents is 2 & 5 cents.
1.4: The Dating Game: Sam: 55, Sally: 81, Sue: 64
Let Sam, Sal and Sue's house no.s be A, B and C.
Squares < 100: 1,4,9,16,25,36,49,64,81
Cubes < 100: 1,8,27,64
A + B + C =2*n^2 < 297: 2,8,18,32,50,72,98,128,162,200,242,288
Since Sal thought she could correctly guess A based on Sam's two
answers, his answers to 'A = n^2 ?' and 'A > 50 ?' must be y & y.
(If A <> n^2, we have 90 possibilities, which get down to 33 &
47 if A <= 50 & A > 50 resp.ly.) If A = n^2 & A > 50, A must be
64 or 81. Since Sal went confidently to one of them, she must be
living in the other. Since A < B, B must be 81 and Sal must
have thought A = 64. We also know that A > 50.
Apply the same logic to Sue. Sam's answers to 'A = n^3 ?' and
'A > 25 ?' must be y & y, for then only do we have 27 & 64 as
two choices. Again, Sue went confidently to one of them, so she
must be living in the other. Again, A < C, so C must be 64 and
Sue thought A = 27.
Now B + C = 81 + 64 = 145. 50 < A < 64, so A + B + C must be 200
which means A = 55.
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