Metrobank-MTAP-DepEd Math Challenge 2013 Individual Finals • Grade 7 • Category A • Answer Key Part I 1. 5
4. 42
7. 3
10. 23
13. 75
2. 5
5. 4
8. 6 and 17
11. 1968
14. 24 24π π m2
3. 10 years years old old
6. 32 pints
9. 8
12. 9
15. 5 square square units
Part II 1. radius of the sphere sphere =
18 2
= 9cm (1 9cm (1 pt); pt);
volum olumee of the the sphe sphere re = 34 πr 3 = 34 π(93 ) = 972π 972π cm2 (2 pts)
2. Let x be one side of the square cross section. section. By Pythagorean Pythagorean Theorem, we have x2 + x2 = 202 . (1 pt) Area of the square cross section = x = x 2 = 200cm2 (2 pts) 3. (x + 6)2 + (−2 − 4)2 = 102 (1 pt); pt);
(x + 6)2 = 64 64 (1 (1 pt); pt);
x = 2 or x =
14 (1 (1 −14
pt)
5+4 = 3 (1 pt) 2+1 2−5 −3 slope of the line joining (2, (2, 5) and (k, (k, 2): = (1 pt) k−2 k−2 −3 Equating the two slopes: = 3 =⇒ k = 1 (1 pt) k−2
4. slope of the line joining joining (−1, −4) and (2, (2, 5):
5. Let a be the length of edge A. A . Then edge B edge B has length 2a 2a, while edge C edge C has has length 3a 3a + 1. (1 pt) Volume of rectangular prism = a = a(2 (2a a)(3a )(3a + 1) = 180 =⇒ a = 3 the only real value (1 value (1 pt) Dimensions of the prism: 3 by 6 by 10 cm (1 cm (1 pt) Part III 1. Let W ( W (−5, 9), X (10, (10, 9), Y (10 Y (10,, 0), and Z (−5, 0). Then W Then W XY X Y Z is is a rectangle, (1 rectangle, (1 pt) and area(ABCDE area(ABCDE ) = area(W area(W X Y Z ) − area(AW area(AW B ) − area(C area(C XD XD)) − area(DY area(DY E ) − area(AZE area(AZE )) (1 pt) pts) = (9)(15) − 21 (7)(4) − 21 (4)(8) − 21 (10)(1) − 21 (5)(5) (2 pts) = 87 87..5 square square units. units. (1 pt) 2. Let V V be the volume of the water, V 1 the volume of the water in the lower triangular part of the pool, and V 2 the volume of the water in the upper rectangular part of the pool. Then V = V 1 + V 2 . V 1 = (width of the pool) × (area of the triangular cross-section) = 30 × 12 (50)(2) (1pt) 1pt) = 1500
(1pt) 1pt)
V 2 = (width of the pool) × (length of the pool) × (height of the water) = 30 × 50 × 0.8
(1pt) 1pt)
= 1200 (1pt) 1pt) Thus, V Thus, V = 1500 + 1200 1200 = 2700 m3.
(1 pt)
3. For a ≥ 1, we must have 1 ≤ b ≤ 9.
(1 pt)
Solving for a, we get a = (72 − 7b)/4, which implies that 72 − 7b must be divisible by 4 because a is an integer. (1 pt) Among the possible values of b, b , only b only b = 4 and b and b = 8 satisfy this last condition.
(2 pts)
And these values of b give (11, (11, 4) and (4, (4, 8) as the only ordered pairs of positive integers satisfying the equation. (1 pt)
