Basic Math Solution Set

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SOLUTION SET (MATH I) 1.

( 800 ÷ 40 ) + [9 2 ÷ (17 − 8) ] − ( 2 2 + 3 2 ) = 20 + ( 81 ÷ 9) − ( 4 + 9) = 20 + 9 − 13 = 16

Answer: B 2.

( 58 + 42) − 67 = 100 − 67 = 33 Answer: A

3.

[ 72,846 ÷ ( 2 × 3) ] − ( 75)( 75) = ( 72,846 ÷ 6) − 5,625 = 12,141 − 5,625 = 6,516 Answer: B

4. A. ( 2 + 2 + 2) 2 = 6 2 = 36

B. ( 2 2 ) + 2 2 + 2 = 4 2 + 4 + 2 = 16 + 6 = 22 Answer: D 2

C. ( 2 × 2 × 2) 2 = 8 2 = 64

[

D. ( 2 + 2) 2

] = (4 ) 2

2 2

= 16 2 = 256

5. 5719x to be divisible by 2, 3, and 6 must follow the divisibility rule for 2, 3, and 6. Divisibility rule for 2: last digit must be an even numbers (0, 2, 4, 6, 8) Divisibility rule for 3: sum of all digits must be divisible by 3. Divisibility rule for 6: must follow the divisibility rule for 2 and 3. Applying the divisibility rule for 6, we now get the sum of all digits to check if this number is divisible by 3. 5+7+1+9+x = 22+x, since we need an even number for the last digit so that it can be divisible by 2, x = 2 so that 22 + 2 = 24, 24 is divisible by 3. Answer: B 1  −1 6. A. 6( 8 − 8) = 5 ⋅ 5 − 1 ⇒ 6( 0) =  ⋅ 5  − 1 ⇒ 0 = 1 − 1 ⇒ 0 = 0 statement is true. 5  B. 3( − 3) + 3( 3) = −9 + 9 = 0 statement is true. 1 10 −1 ≠ 0 statement is false. C. 3 + 3 = 3 + = 3 3 Answer: C

(

)

5 1 7 11 5 1 7 11 , , , ⇒ = 0.416, = 0.111, = 0.875, = 0.611 not in ascending order. 12 9 8 18 12 9 8 18 3 1 2 5 3 1 2 5 , , , ⇒ = 0.187, = 0.143, = 0.400, = 0.833 not in ascending order. B. 16 7 5 6 16 7 5 6 6 7 3 4 6 7 3 4 , , , ⇒ = 0.316, = 0.583, = 0.750, = 0.800 in ascending order. C. 19 12 4 5 19 12 4 5 13 1 3 5 13 1 3 5 , , , ⇒ = 0.867, = 0.250, = 0.375, = 0.833 not in ascending order. D. 15 4 8 6 15 4 8 6 Answer: C

7. A.

8. let x = original length of the rope 6 12  1  3   2  3  x1 −   = 6 ⇒ x   = 6 ⇒ x = 6 ⇒ x = 6 ⋅ ⇒ x = 12 12 6  3  4   3  4  Answer: D 9. let x = total no. of people in Rocky’s party

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4 10  1  1   1  4  x1 − 1 −  = 16 ⇒ x   = 16 ⇒ x = 16 ⇒ x = 16 ⋅ ⇒ x = 4 ⋅ 10 ⇒ x = 40 10 4  2  5   2  5  Answer: C 10. let x = original number of packages before the 1st delivery 4 1 4 1 8−5 3 10 90 x −9 = x ⇒ x − x = 9⇒ x = 9 ⇒ x = 9 ⇒ x = 9⋅ ⇒ x = ⇒ x = 30 5 2 5 2 10 10 3 3 Answer: C 11. ( 0.143 + 0.27 + 0.852 + 0.009 ) − ( 0.235 + 0.51 + 0.006 ) ⇒1.274 − 0.751 = 0.523 Answer: C 12. 87.5 ×0.01 ÷1,000 ⇒0.875 ÷1,000 = 0.000875 ⇒8.75 ×10 −4 Answer: B 13. ( 2.45 × 0.06) + ( 0.057 ÷ 0.3) ⇒ 0.147 + 0.19 = 0.337 Answer: D 14. k = 0.02, 1 = 1 k

0.02

50 ⇒0.02 1 ⇒2 100

Answer: A 15. 2:3, ratio of men to women. Let k = constant which we will use as multiplier to the ratio 350 2k + 3k = 350 ⇒ 5k = 350 ⇒ k = ⇒ k = 70 the total number of women in the concert is (70)(3) = 210 5 Answer: B 16. 6:5:4 – ratio of apples, banana, & peaches respectively. Let k = constant which we will use as multiplier to the ratio 225 6k + 5k + 4k = 225 ⇒ 15k = 225 ⇒ k = ⇒ k = 15 15 weight of apple = (6)(15) = 90, weight of banana = (5)(15) = 75, weight of peaches = (4)(15) = 60 weight of apple – weight of peaches = 90 – 60 = 30 grams Answer: A 17. 24 hours – 9 hours (sleeping) = 15 hours (awake) 30% of 15 hours = 0.3 x 15 = 4.5 hours per day reading a novel 4.5 x 14 days/2 weeks = 63 hours reading novels in 2 weeks 63 ÷ 7 hours/novel = 9 novels she can read in 2 weeks Answer: B 18. let OP = original price discounted price = OP – 0.1OP = 0.9OP tax = 0.9OP x 0.1 = 0.09OP amount paid = discounted price + tax = 0.9OP + 0.09OP = 0.99OP Answer: C 19. let P = original cost of car selling price of car = P + 0.25P = P +

1 P 4

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1 1  1  P = P 24  8 Answer: B tax =

20. let x = total number of participants 63 ⇒ x = 210 (30%)x = 63 ⇒ 0.3 x = 63 ⇒ x = 0.3 Answer: B 21. first day of the fall: the reservoir is 60% of full capacity during heavy rains, the reservoir raises 40% of 60% full capacity of the reservoir after the rain = 60% + ( 0.4 x 0.6) = 60% + 24% = 84% full capacity of the reservoir Answer: C 22.

83 + 86 + 89 + 90 + x 348 + x = 85 ⇒ = 85 ⇒ 348 + x = 85 ⋅ 5 ⇒ 348 + x = 425 ⇒ x = 425 − 348 ⇒ x = 77 5 5 Answer: B

23. five consecutive integers: x, x +1, x + 2, x +3, x + 4 x + ( x + 1) + ( x + 2) + ( x + 3) + ( x + 4) = 35 ⇒ 5 x + 10 = 35 ⇒ 5 x = 35 − 10 ⇒ 5 x = 25 ⇒ x =

25 ⇒ x=5 5

so the five consecutive integers are: 5, 6, 7, 8, 9 let check if their sum is 35: 5 + 6 + 7 + 8 + 9 = 11 + 7 + 8 + 9 = 18 + 8 + 9 = 26 + 9 = 35 the prime numbers are 5 and 7. Answer: C 24. six consecutive integers: x, x +1, x + 2, x +3, x + 4, x + 5 sum of first three terms: x + ( x + 1) + ( x + 2) = 27 ⇒ 3 x + 3 = 27 ⇒ 3x = 24 ⇒ x =

24 ⇒ x=8 3

so the six consecutive integers are: 8, 9, 10, 11, 12, 13. The sum of last three integers: 11 + 12 + 13 = 36 Answer: D 25. n is any integer, let n = 1 for odd numbers & n = 2 for even numbers. If n = 1, A. n, n + 1, n + 3 (1), (1) + 1, (1) + 3 1, 2, 4 not a consecutive odd integers B. n, n + 2, n + 4 (1), (1) + 2, (1) + 4 1, 3, 5 a consecutive odd integers C. 2n + 1, 2n + 2, 2n + 3 2(1) + 1, 2(1) + 2, 2(1) + 3 3, 4, 5 not a consecutive odd integers D. 2n + 1, 2n + 3, 2n + 5 2(1) + 1, 2(1) + 3, 2(1) + 5 3, 5, 7 a consecutive odd integers If n = 2, A. n, n + 1, n + 3 (2), (2) + 1, (2) + 3 2, 3, 5 not a consecutive odd integers B. n, n + 2, n + 4 (2), (2) + 2, (2) + 4 2, 4, 6 not a consecutive odd integers but consecutive even integers C. 2n + 1, 2n + 2, 2n + 3 2(2) + 1, 2(2) + 2, 2(2) + 3 5, 6, 7 not a consecutive odd integers D. 2n + 1, 2n + 3, 2n + 5 2(2) + 1, 2(2) + 3, 2(2) + 5 5, 7, 9 a consecutive odd integers D satisfy both sample value for n. Answer: D 26. Given three consecutive integers: r, s, t and r > s > t. let r = 3, s = 2, t = 1

( r − s )( s − t )( r − t ) ⇒ ( 3 − 2)( 2 −1)( 3 −1) ⇒ (1)(1)( 2) ⇒ 2

Answer: D

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27. x and y are negative integers, x > y. let x = -1 and y = -2 2 2 2 A. − ( xy ) ⇒ −( ( − 1)( − 2) ) ⇒ −( 2) ⇒ −4 B. x 2 y ⇒ ( − 1) 2 ( − 2) ⇒ (1)( − 2) ⇒ −2 C. xy ⇒ ( − 1)( − 2) ⇒ 2 D. y − x ⇒ ( − 2 ) − ( − 1) ⇒ −2 + 1 ⇒ −1 Answer: C

28. x and y are postive integers, x > y. let x = 2 and y = 1 (1) 2 1 1 1 y2 y A. 2 > ⇒ 2 > ⇒ > not true x x ( 2) 2 4 2

x x2 2 ( 2) 4 B. > 2 ⇒ > 2 ⇒ 2 > ⇒ 2 > 4 not true y y 1 (1) 1 2

(1) ( 2) y2 x2 1 4 1 > 2 ⇒ 2 > 2 ⇒ > ⇒ > 4 not true 2 4 1 4 x y ( 2) (1) 2

C.

2

( 2) 2 4 2 x2 x D. 2 > ⇒ 2 > ⇒ > ⇒ 4 > 2 true y y (1) 1 1 1 Answer: D 2

29. a and b are odd numbers, let a = 1 and b = 3 A. ab + 2 ⇒ (1)( 3) + 2 ⇒ 3 + 2 ⇒ 5 not an even number B. 2a + b ⇒ 2(1) + ( 3) ⇒ 2 + 3 ⇒ 5 not an even number C. a + b ⇒ (1) + ( 3) ⇒ 4 an even number D. a + b + 1 ⇒ (1) + ( 3) + 1 ⇒ 5 not an even number Answer: C 30. 5y – {3y + (2y – 5) – [3 – (2 + 4y)]} 5y – {3y + (2y – 5) – [3 – 2 – 4y)]} 5y – {3y + 2y – 5 – 1 + 4y} 5y – (9y – 6) 5y – 9y + 6 6 – 4y Answer: B

5y – {3y + 2y – 5 – (1 – 4y)}