Multiple choice Question Select One Answer Choice Only 24 If the probability

# Multiple choice question select one answer choice

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[Multiple-choice Question – Select One Answer Choice Only] 24. If the probability that Mike will miss at least one of the ten jobs assigned to him is 0.55, then what is the probability that he will do all ten jobs? (A) 0.1 (B) 0.45 (C) 0.55 (D) 0.85 (E) 1
Test 7—Solutions135Answers and Solutions Test 7: QuestionAnswer1. D 2. B 3. D 4. E 5. B 6. D 7. A 8. E 9. E 10. D 11. A 12. A 13. A 14. A 15. D 16. B, D, F 17. B, C, E 18. D 19. C, E 20. D 21. D 22. C 23. E, F 24. B 1. If a= –1, both columns equal –1. If a= –2, the columns are unequal. The answer is (D). 2. The function a* bis defined to be a/bb/afor any numbers aand b. Applying this definition to the columns gives Column A Column B 1m*1n=1m1n±1n1m=nm±mn=nn±mmmn=n2±m2mn1n*1m=1n1m±1m1n=mn±nm=mm±nnmn=m2±n2mnThe given inequality m> n> 0 indicates that both mand nare positive and therefore their product mnis positive. Multiplying both columns by mnto clear fractions yields n2m2m2n2
GRE Math Tests 136Adding n2and m2to both columns yields 2n22m2Finally, dividing both columns by 2 yields n2m2Since both mand nare positive and m > n, m2> n2. Hence, Column B is greater than Column A, and the answer is (B). 3. Suppose the two positive integers mand ndo not have a common factor, apart from 1. Then the LCM of mand nis mn. For example, when m± ²³ ´± µ K ¶· 0=3n± ²¸ ´± ¹ K ¸·º C74 \$»% 8B ²³ K ²¸¼ "= C78B 20B4ºthe LCM equals mn, and the Column A equals Column B. Now, suppose the two integers mand nhave at least one common factor (other than 1). Then the LCM uses the common factors only once, unlike mn. Hence, the LCM is less than mn. For example, suppose m = 10 ´± µ K ¸· 0=3n ± ²³ ´± µ K ¶·¼ !4A4ºmand nhave 2 as a common factor. The LCM of mand n8B µ K ¸ K ¶ ±¶½º 0=3mn[email protected];B µ K ¸ K2 K ¶ ± ²³½¼ +74 \$»% 383 =>C DB4 C74 D=34A;8=43 µ 8= C74 4E0;D0C8>=¼ !4=24º 74A4ºColumn A is less than Column B. Since this is a double case, the answer is (D). 4. The average of the five consecutive positive integers 1, 2, 3, 4, and 5 is (1 + 2 + 3 + 4 + 5)/5 = 15/5 = 3. After dropping 5 (the greatest number), the new average becomes (1 + 2 + 3 + 4)/4 = 10/4 = 2.5. The percentage drop in the average is Old average – New averageOld average²100=3±2.53²100=1006=16.66%The answer is (E). 5. We are given the two inequalities 2x+ 1 > 3x+ 2 5x+ 2 > 4xSubtracting 2x+ 2 from both sides of the top inequality and subtracting 4x+ 2 from both sides of the bottom inequality yields –1 > xx > –2 Combining these inequalities yields –1 > x > –2. Since any number between –1 and –2 is less than 1, Column B is greater than Column A. The answer is (B).
Test 7—Solutions²¹¶6. Column AIn ±ABC, ²B± ¶µJColumn B²A¶µJSumming the angles of ±ABCC> ²¾½J H84;3B²A+ ²B+ ²C= 180. Substituting the value of ²B´± ¶µJ·into this equation yields ²A¿ ¶µ ¿²C= 180. Solving the equation for ²Ayields ²A= 180 – ¶µ²C= 108 – ²C= Column A.