GMAT - Problem Solving Aufgabe 5
Mit der folgenden Übungsaufgabe im Bereich Problem Solving können Sie Ihr Wissen für den GMAT trainieren und eine höherer Punktzahl erreichen:
1.) Oil revenue in the amount of x million dollars, x being an even number, is divided among 11 Russian oligarchs, giving each a fixed sum plus a remainder of 10 million for the government. Which of the following, when added to x, produces the sum which is divisible by 22?
(A) 3
(B) 5
(C) 7
(D) 12
(E) 17
2.) Three consecutive numbers are drawn from integers, which are strictly greater than 9 and strictly less than 20. Suppose w is the product of the numbers drawn, which of the following must be true?
I. w is an integer multiple of 3.
II. w is an integer multiple of 4. w is an integer multiple of 6.
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
3.) The ratio of South Korean aircraft to North Korean aircraft was 1 to 3. After the North announced it would acquire nuclear weapons, the U.S. transfenced 60 aircraft to South Korea. Then, the ratio became 3 to 5. What was the total number of the aircraft on the Korean peninsula after the U.S. aircraft transfer?
(A) 120
(B) 135
(C) 225
(D) 300
(E) 360
Lösung
Haben Sie die Aufgabe gelöst? Dann schauen Sie hier, ob Sie den optimalen Lösungsweg gefunden haben.
1.) Let y be the fixed sum that each oligarch receives. Then, x enjoys the following simple representation
x = 11y + 10
Note that y must be even. If it were odd, then x would be odd as well since a product of two odd number is an odd numbel and adding 10, an even number, would still keep x odd. Now, proceed by elimination noticing that lly is an even number and if an odd number is added to lly + 10, the result is odd - not divisible by 22, an even number. This leads us directly tc 12. Therefore, the best answer is (D).
2.) The key here is to recall that any integer can be decomposed into a product of primes. A necessary and sufficient conditior for a product of numbers to be divisible by (or to be a multiple of) any integer is that the prime number decomposition of this product contains that integer. For example, 24 = 23 x 3. So, it is divisible by (or is a multiple of) 2, 4, 6, and 8. Let us star with I. Do all consecutive triples between 10 and 19 have a 3 in their prime number decomposition? To be a multiple of z 3, a triple (by a triple, henceforth, we mean a consecutive triple) must contain one of the following: 12, 15, 18. It is easy tc see that every triple contains one of above numbers. So, I is true. How about II? To be a multiple of 4, a triple must contair either two even numbers or one of the following: 12, or 16. 13 x 14 x 15 has neither of the above. So, (B), (D), and (E) not true. To choose between (A) and (C), we must check if III is true. A triple is a multiple of 6, if it contains an even numbe and a multiple of 3 such as 15 or one of the following: 12, 18. It is a trivial, but not an easy under time pressure check, tha every triple satisfies the above condition. So, III is true, and (C) is the answer.
It is also convenient to consider the table below.
10 11 12
11 12 13
12 13 14
13 14 15
14 15 16
15 16 17
16 17 18
17 18 19
If each row above is a representation of the product w, then it is easy to see that each row contains a multiple of 3. It is al easy to find that the product 13 x 14 x 15 is not divisible by 4. As for option III, the first 3 rows contain 12. The next 3 rows contain the multiple of both 2 and 3, and the last two rows contain 18. Therefore, the best answer is (C).
3.) The answer may be obtained in a number of ways. The first one is by elimination. If (D) were the answer, then before transfer South Korea would have had
1/4 (300 — 60) = 60
