FTTS MATH CHALLENGE 2013 Grades 9 and 10 – Solutions and Answers

1. Solve

a. b.

c. d.

Solution:

The terms are alternating wherein the odd terms have a common ratio of

, and the common

ratio of the even terms is

, then:

For the odd terms infinite sum: For the even terms infinite sum:

Where is the first odd term and is the first even term; is the common ratio of the odd

terms and is the common ratio of the even terms. Hence, the final answer is

Answer: b

2. What is the sum of the interior angles of a polygon with 20 sides?

a. b. c. d. e.

Solution:

The sum of the measures of the interior angles of a polygon with sides is . If the

polygon has 20 sides, then

Answer: a

3. If

, then find the value of

.

a. 1 b. 2 c. 3 d. 4

Solution:

Given

, then

Also,

then

Then,

Answer: c

4. Solve for the value of x.

a. b. c. d.

Solution: To solve for

To solve for

is a straight line then

A B

D

C G F

E x

Then,

Answer: d

5. The symbol means ”older than”. For example if Grace Liz, then

Grace is older than Liz. Based on the illustration below, who is the oldest?

a. Ali b. Patrick c. Muktar d. Qas

Solution:

We can rearrange the figure again to

Hence, Qas is the oldest.

Answer: d

6. Fatima wants to make a bracelet, composed of beads, for her friend Nor. If she has blue, yellow,

red, orange, and green beads how many ways can they be arranged in the bracelet?

a. 3 b. 6 c. 9 d.12

Solution:

The formula for the circular permutation of distinct objects is

Then, the answer is

ways. We divide it by 2 because there are two different points

of view, from the front and from the back.

Answer: d

Patrick Ali Qas

Mae

Muktar

7. If one root of is 2 then find the sum of the other two roots.

a. 2 b. 1 c. 0 d. -1

Solution:

If is the given root, then

The equation now is

Since

Answer: d

8. Calculate

a. √ √

b.

√ √

c.

d.

Solution:

Using the Sum and Difference Formula, then

√

√

√

√ √

Answer: a

9. In a certain city, all telephone numbers are composed of 7 digits and always start with either 9

or 8. If a telephone has its 2 and 5 button defective, how many phone numbers can it dial?

a. 40320 b. 524288 c. 10080 d. 2097152

Solution:

⏟⏞

⏟

⏞

Answer: b

10. If and

a.

b.

c.

d.

Solution:

and

Then,

.

.

.

Thus,

Answer: c

11. In the figure, |O1O2| = 5 cm and r1 + r2 = 17 cm. Find the value of r1 and r2.

a. b.

c. d.

Solution:

If two circles are tangent internally, if | |=| |=5cm and if r1 + r2 = 17 cm, then the only

possible values of and is 6cm and 11cm respectively since and

Answer: b

12. Calculate if .

a. 5 b. 25 c. 32 d. 16

Solution:

Answer: b

13.

What is the value of if all the squares are congruent and of side 1 unit?

a. 4 units b. 3 units c. 2 units d. 1 unit

Solution:

Since these two angles are just equal

,then

.

Answer: c

14. How many two-digit positive integers can satisfy the equation (x – 2)2 + (y -1)2 = 0

a. 4 b. 3 c. 2 d. 1

Solution:

For any values of , is always positive except if and

for any values of , is always positive except if .

Hence, and are the only solutions that satisfy then the only two-

digit number is 21 so the answer is 1.

Answer: d

x

x

15. Solve for if

a. 0 b. 1 c. 2 d.3

Solution:

Then,

Answer: c

16. A ball is dropped from a height of 5 meters. It falls and bounces by 3/5 of its previous height. On which bounce will it first not rise to more than of 1/2 meter?

a. 4th bounce b. 5th bounce c. 6th bounce d. 7th bounce

Solution:

Since this is a geometric sequence then, we

need to find at which bounce the value is

less than

with the common ration of

.

The sequences are,

.

We will stop at

since

. Then

on the fifth (5th) bounce it will first not rise to a

height of

.

Answer: b

m m

m 3m

5m

m

5th

bounce

3rd

bounce

2nd

bounce

1st

bounce

4th

bounce

17. (5k – 4)x – ky + 3 = 0 and y = (k+1)x – 3 are equations of lines. If the lines are parallel, what is the value of k?

a. √ b. √ c. d.

Solution:

Let and be the slopes of the equations of lines, then since the lines are parallel and

given the equations of the line is in the form of Hence,

and

Answer: d

18. Cesar went to a mall to buy school supplies he needed for the entire school year. He bought

three types of school supplies: notebooks, pencils, and erasers.

All he bought except 18 are notebooks.

All he bought except 21 are pencils.

All he bought except 19 are erasers.

How many notebooks did he buy?

a. 8 b. 9 c. 10 d. 11

Solution:

Let be the number of notebooks, for the number of pencils, and for the number of pencils. Then,

Adding we have

Since , then

Answer: d

19. Find the remainder when is divided by 15. a. 3 b. 6 c. 9 d. 12

Solution:

We will use modular arithmetic.

We can notice that the remainders have a sequence of 3, 9, 12 and 6.

{

3 - 2013th remainder.

Answer: a

20. Solve √ √ √

.

a. 0 b. -1 c. -2 d. undefined

Solution:

Using the formula for infinite radical √ √ √ √

√

then, √ √ √

√

√

.

Answer: c

21. In the figure, B and D are centers of two circles. If ABCD is a square and the area of the shaded

region is 36 cm2, what is the length |DE|?

a. √ b. √ c. √ d.

Solution:

Since the area of the shaded region is , then we can

solve for The shaded region is

of a circle then,

| | | | | | | | | | .

We can solve now for | | using Pythagorean Theorem,

| | | | | | | | √ .

Let | | | | | |, then

| | | | √

Hence, | | √

Answer: a

22. Solve for if

.

a. -3 b. 3 and -3 c. -2 and -1 d. -1

Solution:

is a geometric sequence with a common ratio of

Using the formula for the

infinite sum of a geometric sequence

, where is the first term of the sequence and is the

common ratio, we have,

Solving for the possible values of , . But since

then we will just take the

value of .

x

B A

P E

D F C

12cm

12cm

12cm

Answer: a

23. Find the area of the triangle.

a. √ b. √ c. √ d. √

Solution:

Let be the sides of the triangle. Using Heron’s Formula

√ where

. Solving for

.

√ √

Answer: a

24.

a. 19 b. 24 c. 25 d. 28

Solution: Using the pattern of the given circle,

Then,

Answer: b

2b

2a+1

a

b

c

3c

11

10

C

B A

9

5

11 3

4 2

C 7 12

4 15

3 B

4 6

2

2 1

A

25. Kath has a broken calculator. If the + (addition) key is pressed, it adds 46; when the –

(subtraction) key is pressed, it adds –46; if the x (multiplication) key is pressed, it adds 69, and

when the ÷ (division) key is pressed, it adds – 69. If the other keys do not function, what number

closest to 2013 can he get from the calculator which when turned on it displays 0?

a. 2001 b. 2012 c. 2018 d. 2024

Solution:

Let A be the number closest to 2013 that we can get from the calculator. Since the + and – keys add 46

and subtract 46 respectively, and the x and ÷ keys add 69 and subtract 69 respectively, then we have the

equation

46x + 69y = A, where x and y are real numbers.

Both 46 and 69 are divisible by 23, then we have

23(2x + 3y) = A.

This means that A must be a number divisible by 23. The numbers close to 2013 which are divisible by

23 are 2001 and 2024; 2024 is the closest, therefore A = 2024.

Answer: d

26. If (

)

, find

a.

b. c.

d.

Solution:

Let

. We have

(

)

(

) (

)

(

) (

) (

)

(

) (

) [(

) ] then,

. Hence,

⌈ ⌉

Answer: b

27. Sorted numbers are positive numbers, with two or more digits, such that the digits from left to

right are strictly in increasing order. 34, 123 and 479 are examples of sorted numbers but 122,

63, and 294 are not. If the sorted numbers are listed in increasing order, what is the 64th sorted

number?

a. 149 b. 169 c. 189 d. 239

Solution:

We list the sorted numbers.

For two – digit numbers:

First digit Second digit No. of sorted numbers

1 2 to 9 8

2 3 to 9 7

3 4 to 9 6

4 5 to 9 5

5 6 to 9 4

6 7 to 9 3

7 8 to 9 2

8 9 1

9 none 0

Total : 36

For three – digit numbers :

First digit Second digit Third digit No. of Sorted Numbers

1 2 3 to 9 7

1 3 4 to 9 6

1 4 5 to 9 5

1 5 6 to 9 4

1 6 7 to 9 3

1 7 8 to 9 2

1 8 9 1

1 9 none 0

Total: 28

Since the total of the two – digit and three – digit sorted numbers is 36 + 28 = 64, then the 64th sorted

number is the last sorted number on the list, 189.

Answer: c

28. In the figure below, the circles are arranged in such a way that 1/6 of the circumference of one

circle is inside the other circles. If the four circles are of radius 1, what is the total area of the

shaded regions?

a.

√

b.

√ c. d.

√

Solution:

Let us consider one circle.

√

Area of the segment (shaded area) = Area of the sector – Area of the triangle.

=

=

√

=

√

Thus,

Total area of the shaded regions = 8 shaded areas

= 8

√

=

√

Answer: b

29. On a dartboard, the probability of hitting a certain region is proportional to its corresponding

area. In the figure below, the outer circle has a radius of 4 and the inner circle has a radius of 2.

The two diameters divide each circle into 4 equal parts. If two darts hit the board, what is the

probability that the score is even?

a.

b.

c.

d.

Solution:

First we solve for the areas of the dart board.

a. Area of the outer circle =

=

b. Area of the inner circle =

=

c. Area of annulus =

=

d. The inner circle is divided into four regions.

Area of each region =

.

e. The annulus is also divided into four regions.

Area of each region =

=

Next, we solve for the probability of scoring 1 or 2 on the dart board.

a. Probability of scoring 1.

In the dartboard, 1 is found on 1 region of the inner circle and 2 regions of the annulus. Thus,

the probability of scoring 1 =

=

b. Probability of scoring 2.

On the dartboard, 2 is found 3 regions of the inner circle and 2 regions of the annulus. Thus,

the probability of scoring 2 =

.

To have an even score using two darts we must score either 1 and 1 or 2 and 2; 1 + 1 = 2 or 2 + 2 = 4.

Hence,

a. Probability of scoring 1 and 1 =

b. Probability of scoring 2 and 2 =

Therefore,

Probability of scoring even = Probability of scoring 1 and 1 or 2 and 2

=

=

=

.

Answer: d

30. Maria, Rai and Sharmaene have their birthday today. The sum of their ages is 30 and the

product of their ages is 270 more than the product of their ages on their birthday last year.

What is the sum of the squares of their ages?

a. 269 b. 299 c. 302 d. 332

Solution:

Let m, r ,and s be the ages of Maria, Rai and Sharmaene respectively.

a. m + r + s = 30

b. mrs = (m – 1) (r – 1) (s – 1) + 270

mrs = mrs – mr – ms – rs + m + r + s – 1 + 270

0 = – (mr + ms + rs ) + 30 + 269

mr + ms + rs = 299

c.

Answer : c