1. What is the place value of the underlined digit in 789,505?
Ans.: hundred thousands Sol.: The given number is equal to 7(100,000) + 8(10,000) + 9(1,000) + 5(100) + 0(10) + 5(1). Hence the place value of the the underlined digit is hundred thousands. Ans. Sol. Hide Ans./Sol.

2. Write five hundred ninety nine thousand two hundred four using Hindu-Arabic numerals.
Ans.: 599,204 Sol.: The given number is equal to 599(1,000) + 2(100) + 0(10) + 4(1) = 599,204. Ans. Sol. Hide Ans./Sol.

3. How many common divisors (including 1) do 60 and 200 have?
Ans.: 6 Sol.: The number of common divisors (or factors) of any two counting numbers is as many as the divisors of their greatest common factor (GCF). The GCF of 60 and 200 is 20 and this GCF has 6 divisors.
Observe that the prime factorization of 200 = $ 2^{3}5^{2} $ and the prime factorization of 60 = $ 2^{2}3^{1}5^{1} $.
It follows that their GCF is $ 2^{2}5^{1} = 20 $, which has (2 + 1)(1 + 1) = 6 divisors. Ans. Sol. Hide Ans./Sol.

4. Find two integers whose sum is 19 and whose product is 88.
Ans.: 8 and 11 Sol.: Since the sum of the two numbers is positive and the product is also positive, each number must be positive.
So we can try number pairs such as 1 and 18, 2 and 17, 3 and 16, 4 and 15, 5 and 14, among others. One can show that the choice of 10 and 9 is too big since 90 $>$ 88. However, the choice of 8 and 11 is fine since 11+8=19 and (11)(8)=88. Ans. Sol. Hide Ans./Sol.

5. How many 8ths are in 5/2?
Ans.: 20 Sol.: This answer is obtained by dividing 5/2 by 1/8 giving (5/2)/(1/8) = (5/2)(8) = (5)(8/2) = (5)(4) = 20.
One can speedily verify that 20(1/8) = 5/2. Ans. Sol. Hide Ans./Sol.

6. Covert the mixed fraction $ 7 \frac{3}{4} $ into an equivalent improper fraction in lowest terms.
Ans.: 31/4 Sol.: $ 7 + \frac{3}{4} = 7(\frac{4}{4}) + \frac{3}{4} = \frac{(7)(4)}{4} + \frac{3}{4} = \frac{(7)(4)+3}{4} = \frac{31}{4} $
or
$ a + \frac{b}{c} = \frac{a(c) + b}{c} $.
Here, $ a=7$, $ b=3$ and $ c=4$. So that $ 7 \frac{3}{4} = \frac{(7)(4)+3}{4} = \frac{28+3}{4} = \frac{31}{4} $. Ans. Sol. Hide Ans./Sol.

7. If $\frac{15}{10} = \frac{N}{100}$, what is the value of $ N$ ?
Ans.: 150 Sol.: Notice that the left-hand side ratio $\frac{15}{10} = \frac{5(3)}{5(2)} = \frac{3}{2} $ in lowest terms, and that the denominator of the right-hand side ratio, 100, is obtained from the denominator of the left-hand side ratio by multiplying the denomiator of the left-hand side (lowest terms form) by 50. Hence, also multiplying the left-hand side numerator (lowest terms form), 3, by 50 should give us the numerator of the right-hand side: (3)(50) = 150. Ans. Sol. Hide Ans./Sol.

8. Mr. Chef bought exactly $ 2 \frac{1}{4}$ dozen eggs and used $ 1 \frac{1}{6}$ to cook a certain egg dish. How many of the eggs that he bought for cooking this egg dish remained (that is, were not used for cooking the dish)?
Ans.: 13 eggs Sol.: Since there are 12 eggs in a dozen of eggs, the answer is $ 12(2 \frac{1}{4} - 1 \frac{1}{6}) = 12(\frac{8+1}{4} - \frac{6+1}{6}) = 12(\frac{9}{4} - \frac{7}{6}) = 13 $ eggs. Ans. Sol. Hide Ans./Sol.

9. A certain Grade 4 class contains 40 students. If 8 of the students are boys, what is the ratio of boys to girls?
Ans.: 1:4 or 1/4 Sol.: If 8 of the students are boys, then 40 - 8 = 32 of the students are girls. So the ratio of the number of boys to girls for this class is $\frac{8}{40-8} = \frac{8}{32} = \frac{2^3}{2^3(2^2)} = \frac{1}{2^2} = \frac{1}{4} $. Ans. Sol. Hide Ans./Sol.

10. A certain Grade 4 class contains 40 students. If 8 of the students are boys, what percentage of the class are girls?
Ans.: 80 percent Sol.: If 8 of the students are boys, then 40 - 8 = 32 of the students are girls. So the percentage of girls in the class is $\frac{32}{40}\times 100\% = 80\% $. Ans. Sol. Hide Ans./Sol.

11. Which has the smaller angle measure: right angle or obtuse angle?
Ans.: right Sol.: Recall that an acute angle has a measure that is less than $ 90^\circ$ while an obtuse angle has a measure that is more than $ 90^\circ$; a right angle measures exactly $ 90^\circ$. Ans. Sol. Hide Ans./Sol.

12. The length of a rectangle is twice its width. If the perimeter of the rectangle is 102 meters, find the measurement of the width.
Ans.: width = 17 meters or 17 m Sol.: Let $ w$ and $ l$ be the width and length, respectively, of the rectangle. Since the length is twice the width, the the ratio of the length to the width is 2:1, or that $\frac{l}{w}=\frac{2}{1}$.
Since the perimeter is twice the sum of the length and the width, then the sum $ l + w = 102/2 = 51.$
So the width must be $ \frac{1}{2+1}(l+w) = \frac{1}{3}(l+w) = \frac{51}{3} = 17$ m. Ans. Sol. Hide Ans./Sol.

13. The length of a rectangle is twice its width. If the length of the rectangle is 34 meters, find the area of the rectangle.
Ans.: 578 square meters (or 578 m$ {}^2$) Sol.: Area is length times width or $ A=l\times w$. Since the length of the given rectangle is twice its width, then its width $ w = \frac{l}{2} = \frac{34}{2} = 17$ m. So its areas is $ A = l\times w = (34)(17) = 578$ square meters (m$ {}^2$). Ans. Sol. Hide Ans./Sol.

14. Trees were planted 5 meters apart along a straight line segment so that the distance between the first tree and the last tree planted is 130 meters. How many trees were planted along this straight line segment?
Ans.: 27 Sol.: For a line segment 1(5)=5 m long, two trees; for a line segment 2(5)=10 m long, three trees; for a line segment 3(5)=15 m long, 4 trees; ...; for a line 26(5)=130 m long, 27 trees. So, in general, divide the length of the line segment that was planted with trees by the constant distance between two adjacent trees, then add 1, i.e., 130/5 + 1 = 26 + 1 = 27. Ans. Sol. Hide Ans./Sol.

15. The base of a rectagular prism has a width of 6 meters and a length of 6 meters. If the height of the prism is 2 meters, find its total surface area.
Ans.: 120 square meters Sol.: The total surface area of a rectangular prism is equal to the sum of the areas of its six faces. These six faces are all rectangles. Two opposite faces each has an area of length times width $ (l\times w)$ square meters. Another pair of opposite faces each has an area of length times height $ (l\times h)$ square meters while the third pair of opposite faces each has an area of width times height $ (w\times h)$ square meters. Hence, the total surface area $ S$ of any rectangular prism is $ S = 2( l\times w) + 2(l\times h) + 2(w\times h)$. For this particular problem, length $ l=6$ meters, width $ w=6$ meters, and height $ h=2$ meters. So the total surface area is $ S = 2(6)(6) + 2(6)(2) + 2(6)(2) = 72 + 24 + 24 = 120$ square meters. Ans. Sol. Hide Ans./Sol.