12 multiple-choice questions, progressively harder.
Let a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1 and a⋄b=a2−ba \diamond b = a^2 - ba⋄b=a2−b. What is (3⋆1)⋄2(3 \star 1) \diamond 2(3⋆1)⋄2?
Solution
Correct answer: A
Evaluate the inner star first.
3⋆1=2(3)+1−1=63 \star 1 = 2(3) + 1 - 1 = 63⋆1=2(3)+1−1=6
Then apply the diamond, which squares its first input.
6⋄2=62−2=36−2=346 \diamond 2 = 6^2 - 2 = 36 - 2 = 346⋄2=62−2=36−2=34
Using the same scoring rule a⊙b=5a−2ba \odot b = 5a - 2ba⊙b=5a−2b, turn X has 333 hits and 111 miss, and turn Y has 111 hit and 333 misses. Which scores higher?
Correct answer: B
Score each turn.
X:5(3)−2(1)=13,Y:5(1)−2(3)=−1\text{X}: 5(3) - 2(1) = 13, \qquad \text{Y}: 5(1) - 2(3) = -1X:5(3)−2(1)=13,Y:5(1)−2(3)=−1
Since 13>−113 > -113>−1, turn X scores higher. The rule rewards hits and penalizes misses, and X has more hits.
Let a⊗b=a+b−4a \otimes b = a + b - 4a⊗b=a+b−4. What is its identity element (the value eee with a⊗e=aa \otimes e = aa⊗e=a)?
Correct answer: C
Set a⊗e=aa \otimes e = aa⊗e=a and solve for eee.
a+e−4=a ⟹ e−4=0 ⟹ e=4a + e - 4 = a \;\Longrightarrow\; e - 4 = 0 \;\Longrightarrow\; e = 4a+e−4=a⟹e−4=0⟹e=4
Check: a⊗4=a+4−4=aa \otimes 4 = a + 4 - 4 = aa⊗4=a+4−4=a.
Let a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1 and a⊕b=a+b+aba \oplus b = a + b + aba⊕b=a+b+ab. What is ((−1)⊕2)⋆3((-1) \oplus 2) \star 3((−1)⊕2)⋆3?
Evaluate the inner circle-plus first.
(−1)⊕2=−1+2+(−1)(2)=−1(-1) \oplus 2 = -1 + 2 + (-1)(2) = -1(−1)⊕2=−1+2+(−1)(2)=−1
Then apply the star with 333.
(−1)⋆3=2(−1)+3−1=0(-1) \star 3 = 2(-1) + 3 - 1 = 0(−1)⋆3=2(−1)+3−1=0
A tariff charges aaa dollars of base plus bbb minutes as a△b=a+b2a \triangle b = a + \dfrac{b}{2}a△b=a+2b dollars. What does a plan with 101010 base dollars and 303030 minutes cost?
Substitute a=10a = 10a=10 and b=30b = 30b=30 into the rule.
10△30=10+302=10+15=2510 \triangle 30 = 10 + \frac{30}{2} = 10 + 15 = 2510△30=10+230=10+15=25
The plan costs 252525 dollars.
For which of these operations does the order of the inputs never matter?
Correct answer: D
Order never matters exactly when the operation is commutative. The rule a+b+aba + b + aba+b+ab is symmetric in aaa and bbb.
b⊕a=b+a+ba=a+b+ab=a⊕bb \oplus a = b + a + ba = a + b + ab = a \oplus bb⊕a=b+a+ba=a+b+ab=a⊕b
Each of the other three treats its first input differently (doubling, squaring, or tripling it), so swapping the inputs changes the result.
Let a⊕b=a+b+aba \oplus b = a + b + aba⊕b=a+b+ab. What is (1⊕1)⊕1(1 \oplus 1) \oplus 1(1⊕1)⊕1?
1⊕1=1+1+1=31 \oplus 1 = 1 + 1 + 1 = 31⊕1=1+1+1=3
Then apply the operation again.
3⊕1=3+1+3=73 \oplus 1 = 3 + 1 + 3 = 73⊕1=3+1+3=7
An operation is defined by a⊕b=a+b+aba \oplus b = a + b + aba⊕b=a+b+ab. Which value changes nothing, so that a⊕e=aa \oplus e = aa⊕e=a for every aaa?
Solve the identity condition.
a+e+ae=a ⟹ e(1+a)=0 ⟹ e=0a + e + ae = a \;\Longrightarrow\; e(1 + a) = 0 \;\Longrightarrow\; e = 0a+e+ae=a⟹e(1+a)=0⟹e=0
The identity element is 000, since a⊕0=a+0+0=aa \oplus 0 = a + 0 + 0 = aa⊕0=a+0+0=a.
Let a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1. Simplify (x⋆1)⋆0(x \star 1) \star 0(x⋆1)⋆0.
Simplify the inner star first.
x⋆1=2x+1−1=2xx \star 1 = 2x + 1 - 1 = 2xx⋆1=2x+1−1=2x
Then apply the star with 000 as the second input.
(2x)⋆0=2(2x)+0−1=4x−1(2x) \star 0 = 2(2x) + 0 - 1 = 4x - 1(2x)⋆0=2(2x)+0−1=4x−1
To evaluate any defined operation a∗ba \ast ba∗b, what should you do?
A defined operation is just a rule written with familiar operations.
a∗b=(the given formula, with the inputs substituted in)a \ast b = (\text{the given formula, with the inputs substituted in})a∗b=(the given formula, with the inputs substituted in)
So you replace aaa and bbb by the inputs and simplify with the ordinary order of operations. The symbol is the only new thing.
Let a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1, a⋄b=a2−ba \diamond b = a^2 - ba⋄b=a2−b, and a⊕b=a+b+aba \oplus b = a + b + aba⊕b=a+b+ab. What is (2⋄3)⊕(1⋆1)(2 \diamond 3) \oplus (1 \star 1)(2⋄3)⊕(1⋆1)?
Evaluate each parenthesis first.
2⋄3=4−3=1,1⋆1=2+1−1=22 \diamond 3 = 4 - 3 = 1, \qquad 1 \star 1 = 2 + 1 - 1 = 22⋄3=4−3=1,1⋆1=2+1−1=2
Then combine with circle-plus: 1⊕2=1+2+2=51 \oplus 2 = 1 + 2 + 2 = 51⊕2=1+2+2=5.
The main lesson of defining new operations is that a made-up symbol like ⋆\star⋆ or ⋄\diamond⋄ is:
A custom operation only looks mysterious because its symbol is unfamiliar.
a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1
The right side uses operations you already know, so you evaluate it by substituting the inputs and simplifying. The notation is new; the arithmetic is not.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.