Grade 7 MTAP 2015 Questions with Solutions Part 5

This is the fifth part of the Metrobank-MTAP Math Challenge Questions for Grade 7. In this post, we discuss the solutions to number 41-50. You can also read the solutions to 1-10, 11-20, 21 – 30, and 31-40.

41.) Each square in the figure has side length 1. The curve each square is a circular arc with center at a corner of the square. What is the area of the shaded region?

Solution: will be discussed in a separate post.

Answer: 8 square units

42.) What is the least common multiple of 168 and 420? 

Solution

Left as an exercise.

Answer: 840

43.) A salesman receives a basic salary of Php10, 000 and a commission of 5% on all sales. Find his total sales in a month when he earned Php25, 000?

Solution

10, 000 + .05x= 25, 000

0.05x = 15, 000

x = 300 000

Answer: Php 300, 000.00

44.) What is the absolute value of at ?

Solution

Answer:

45.) Find all real numbers for which (7 – x)/2 > (4x + 3)/4

Solution

Multiplying both sides by 4, we have

2(7 – x) > 4x + 3

14 – 2x > 4x + 3

11 > 6x

11/6 > x

x < 11/6

46.) For what values of x is (3x – 1)/4 < (2x + 5)/3?

Solution

Multiplying both sides by 12, we obtain

3(3x – 1) < 4(2x + 5)

9x – 3 < 8x + 20

x < 23

47.) Write an equation of the line through (-2, 3) and (1,4).

Solution

The slope of the line is m = (4 – 3)/(1-(-2)) = 1/3.

We use the point slope form. That is,

y – 4 = (1/3)(x – 1)

y – 4 = 1/3x – 1/3

3y = x + 11

y = (x + 11)/3

Answer: y = (x + 11)/3

48.) Solve for x and y: 7x – 3y = 23 and x + 2y = 13.

Solution

Multiplying x + 2y = 13 by -7, we have

-7x – 14y = -91 (*)

7x – 3y = 23 (**)

Adding (*) and (**), we obtain

-17y = -68

y = 4.

Substituting y to x + 2y = 13,

x + 2(4) = 13

x = 5

49.) Pete is 12 years old and His grandfather 63 years old. And how many years we’ll be age be one fourth of his grandfather?

Solution    

Let x be the number of years to be added to both ages.

(1/4)(63 + x) = 12 + x

Multiplying both sides by 4,

63 + x = 48 + 4x

3x = 15

x = 5

Answer: 5 years

50.) ABC has its coordinates at A(a, -1), B(8, -1), and C(5,4). If the base is twice the height, find a.

Solution

If we let AB be the base, then the height is the difference between y-coordinate of C and -1. That is, |4 -(-1)| = 5.

Now, |a – 8| = 2(5) = 10

a – 8 = 10
a = 18
-(a – 8) = 10
8 – a = 10
–a = 2
a = -2

Answers: a = -2 or a = 18