If x is positive and 6x=x24, what is the value of x?
(A) 144
(B) 12
(C) 26
(D) 4
(E) 23
Correct Answer: B
Solution 1:
Tip: Know the Properties of Proportions.
We solve for x by cross multiplying:
6xx2xx====x24144±1212givencross-multiplysquare root both sidesselect the positive answer
Solution 2:
Tip: Plug and check.
We plug each of the answer into the proportion and select the one that does not yield a contradiction.
(A) If x=144:
6x614424=x24=14424=61
Wrong choice.
(B) If x=12:
6x6122=x24=1224=2
This is the correct answer.
(C) If x=26:
6x62624=x24=2624=144
Wrong choice.
(D) If x=4:
6x6432=x24=424=6
Wrong choice.
(E) If x=23:
6x62312=x24=2324=144
Wrong choice.
Incorrect Choices:
(A)
If you solve 6x=24, you will get this wrong answer.
(C)
If you solve x=x24, you will get this wrong answer.
(D)
If you solve 6x=24, you will get this wrong answer.
(E)
If you solve for x, you will get this wrong answer.
How many pens can Alice buy with m dollars, if k pens cost n cents?
(A) 100kmn
(B) nkm
(C) 100mnk
(D) n100km
(E) 100kmn
Correct Answer: D
Solution 1:
Tip: Pay attention to units. m dollars =100m cents.
We set up the proportion:
nkn100km==100mxxcentspensmultiply both sides bym
Alice can purchase n100km pens with m dollars.
Solution 2:
Tip: Replace variables with numbers.
Let the price of k=1 pen be n=100 cents, and let Alice have m=2 dollars. Since 2 dollars =200 cents, with 200 cents she can buy 2 pens.
We check which of the choices yields 2 pens.
(A) 100kmn=100⋅1⋅2100=21=2. Wrong choice.
(B) nkm=1001⋅2=501=2. Wrong choice.
(C) 100mnk=100⋅2100⋅1=21=2. Wrong choice.
(D) n100km=100100⋅1⋅2=2. Correct answer.
(E) 100kmn=100⋅1⋅2⋅100=20,000. Wrong choice.
Incorrect Choices:
(A), (B), (C), and (E)
Solution 2 eliminates these choices by replacing the variables with numbers.
A
B
C
D
E
Which of the following is equivalent to 1.25:1.5?
(A) 2:3
(B) 6:5
(C) 10:12
(D) 25:5
(E) 125:15
Which of the following statements is always true?
I.II.III.a%ofb(a+b)%(ab)%===b%ofaa%+b%a%⋅b%
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II and III
Correct Answer: C
Solution 1:
We analyze each of the options.
I. a%ofb=100a⋅b=100ab=100b⋅a=b%ofa. This is true.
II. (a+b)%=100a+b=100a+100b=a%+b%. This is true.
Only options I and II are true statements, and therefore answer (C) is correct.
Solution 2:
Tip: Replace variables with numbers.
Let a=10 and b=20. For each option, we plug these in and check if the two sides of the equation yield the same answer.
I. Left Hand Side = a%ofb=10%of20=10010⋅20=2
and
Right Hand Side = b%ofa=20%of10=10020⋅10=2.
LHS = RHS and therefore this option is true.
II. LHS = (a+b)%=(10+20)%=10010+20=10030
and
RHS = a%+b%=10%+20%=10010+10020=10030.
LHS = RHS and therefore this options is true.
III. LHS = (ab)%=(10⋅20)%=30%=10030
and
RHS = a%⋅b%=10010⋅10020=1002.
But, 10030=1002, and therefore option III is false. The correct answer is choice (C).
Incorrect Choices:
(A), (B), (D), and (E)
Solution 1 shows why these choices are wrong. Solution 2 eliminates them by plugging and checking.
Review
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