ENGLISH AS MEDIUM OF INSTRUCTION ASSIGNMENT

Exercise and Solution about Smart Solution

By:

Shofia Hidayah (110210101036)

Dina Rizki A. (110210101081)

DEPARTMENT OF MATHEMATICS EDUCATION

UNIVERSITY OF JEMBER

1

NUMBER THEORY

• Exercise

1. Calculate the value of 20062−20052 +20042−20032 + ...+42−32 +22−12!

2. Calculate the value of 13 − 23 + 33 − 43 + 53 − 63 + ... + 20053!

3. Determine the value of (1− 12)(1− 1

3)(1− 1

4)...(1− 1

2004)!

4. Determine the value of 11.2

+ 12.3

+ 13.4

+ ... + 12004.2005

!

5. Determine the value of 1√1+√

2+ 1√

2+√

3+ 1√

3+√

4... + 1√

9999+√

10000!

• Solution

1. 20062 − 20052 + 20042 − 20032 + ... + 42 − 32 + 22 − 12

=(20062 − 20052) + (20042 − 20032) + ... + (42 − 32) + (22 − 12)

=(2006+2005)(2006−2005)+(2004+2003)(2004−2003)+ ...+(4+3)(4−

3) + (2 + 1)(2− 1)

=(2006 + 2005)1 + (2004 + 2003)1 + ... + (4 + 3)1 + (2 + 1)1

=(2006 + 2005 + 2004 + 2003 + ... + 4 + 3 + 2 + 1)

=20062

.(2006 + 1)

=2013021

2. 13 − 23 + 33 − 43 + 53 − 63 + ... + 20053

=(13 + 23 + 33 + ... + 20053)− 2(23 + 33 + ... + 20043)

=(13 + 23 + 33 + ... + 20053)− 2.23(13 + 23 + 33 + ... + 10023)

=(12.2005.2006)2 − 16(1

2.1002.1003)2

=10032(20052 − (4.501)2)

2

=10032(20052 − 20042)

=10032(2005 + 2004)(2005− 2004)

=10032.4009

=2019060063

3. (1− 12)(1− 1

3)(1− 1

4)...(1− 1

2004)

=12.23.34...2002

2003.20032004

= 12004

4. 11.2

+ 12.3

+ 13.4

+ ... + 12004.2005

1k(k+1)

= 1k− 1

k+1

11.2

+ 12.3

+ 13.4

+ ... + 12004.2005

=(11− 1

2) + (1

2− 1

3) + (1

3− 1

4) + ... + ( 1

2004− 1

2005)

=1− 12005

=20042005

5. 1√1+√

2+ 1√

2+√

3+ 1√

3+√

4... + 1√

9999+√

10000!

=(−√

1 +√

2) + (−√

2 +√

3) + (−√

3 +√

4) + ... + (−√

9999 +√

10000)

=(−√

1 +√

10000)

=(−1) + 100

=99

3

ALGEBRA

• Exercise

1. For each of Real Number a,b prove that a2 + b2 ≥ 2ab !

2. For each of Real Number a,b if a ≥ 0 and b ≥ 0 prove that a+b2≥√

ab !

3. If a,b,c and d in Positive Number prove that√

(a + c)(b + d) ≥√

ab+√

cd!

4. If a and b in Real Positive Number prove that a3+b3

2≥ (a+b

2)3 !

5. If a = 12

1+ 22

3+ 32

5+ ... + 10022

2003and b = 12

3+ 22

5+ 32

7+ ... + 10022

2005, determine

the integers that closest from a-b !

• Solution

1. (a− b)2 ≥ 0

a2 − 2ab + b2 ≥ 0

a2 + b2 ≥ 2ab

2. (√

a−√

b)2 ≥ 0

a− 2√

ab + b ≥ 0

a + b ≥ 2√

ab

a+b2≥ ab

3. (√

ad−√

bc)2 ≥ 0

ad− 2√

adbc + bc ≥ 0

ad + bc ≥ 2√

adbc

ad + bc + ab + cd ≥ 2√

adbc + ab + cd

(a + c)(b + d) ≥ (√

ab +√

cd)2√(a + c)(b + d) ≥

√ab +

√cd

4

4. (a− b)2 ≥ 0

(a + b)(a− b)2 ≥ 0

a3 + b3 − a2b− ab2 ≥ 0

a3 + b3 ≥ a2b + ab2

3a3 + 3b3 ≥ 3a2b + 3ab2

(3a3 + 3b3) + (a3 + b3) ≥ (3a2b + 3ab2) + (a3 + b3)

4a3 + 4b3 ≥ a3 + b3 + 3a2b + 3ab2

4a3 + 4b3 ≥ (a + b)3

4(a3 + b3) ≥ (a + b)3

a3 + b3 ≥ (a+b)3

4

a3 + b3 ≥ (a+b)3

22

a3+b3

2≥ (a+b)3

22.2

a3+b3

2≥ ( (a+b)

2)3

5. a− b = (12

1+ 22

3+ 32

5+ ... + 10022

2003)− (12

3+ 22

5+ 32

7+ ... + 10022

2005)

a− b = 12

1+ (22

3− 12

3) + (32

5− 22

5) + ... + (10022

2003− 10012

2003)− 10022

2005

a− b = 1− 10022

2005+ (1 + 1 + 1 + ... + 1)

a− b = 1002− 10022

2005

a− b = 1002(2005−1002)2005

a− b = 1002.10032005

a− b = 501

5

GEOMETRY

• Exercise

1. A beam has a size comparison long : width : high = 6 : 3 : 2. If the length

of the space diagonal is 28cm, determine the volume of the beam !

2. A cone is made of a pie-shaped piece of cardboard circle with central angle

288 degrees and the radius of 10 cm. Calculate the volume of a cone is

formed! (use π = 3.14) !

3. A tube with a diameter of 20 cm of water half full. If a ball diameter of

6 cm is inserted into the tube, what is the water level rising? Look at the

picture !

6

4. A pendulum consists of a tube and a half sphere with radius 6 cm as shown

below. If the total structural height of 15 cm and π = 227. Calculate the

volume of the pendulum !

5. Picture below is a tube with a top and bottom half of the ball. If the tube

diameter and 8.4 cm high and 20 cm tube pi = frac 22 7, determine the

surface area of the shaded tube!

7

• Solution

1. Let l = 6x, w = 3x, h = 2x so,

282 = (6x)2 + (3x)2 + (2x2)

784 = 49.x2 ⇐⇒ x = 4

So the volume is V = l . w . h = 24.12.8 = 2304

2. For a cone made of pie, then the area will be equal to the broad segment

covers the cone, and the radius of pie will be a line of cones painter.

Area of cardboard pie is:

A = sudut360o × πr2

A = 288o

360o × π102

A = 80π cm2

Broad blanket cone = Area of pie carton=80π cm2

Painter line = cone radius = 10 cm

Broad blanket cone = πrs

80π = πr10

80 = 10r

r = 8 cm

Next find the cone height using Pythagorean formula

t2 = s2 − r2 = 102 − 82 = 100− 64 = 36

t =√

36 = 6 cm

So the volume of the cone is:

V = 13× πr2 × t

8

V = 13× π82 × 6

V = 13× π64× 6

V = 128π cm

3. First we find volume of a sphere.

Vball = 43πr3

Vball = 43× π × 33

Vball = 4× π × 32

Vball = 36π cm3

The volume of water that rise is equal to the volume of a sphere. Search

the rising water level by using the volume of water that rose on the tube.

Vwater = π × r2 × t

36π = π × 102 × t

36 = 100.t

t = 36100

t = 0.36 cm

So the water level rise was 0.36 cm.

4. high of the cone = high cone entirely - fingers bulb

t = 15 cm− 6 cm = 9 cm

Conical of pendulum Volume = Volume + Volume hemispherical

V = 13πr2t + 1

2× 4

3πr3

V = 13× π × 62 × 9 + 2

3× π × 63

V = π × 36× 3 + 2× π × 72

V = 108π + 144π

9

V = 252π

V = 252× 227

V = 36× 22

V = 792 cm3

5. Tube broad shaded = Area of cylinder - 2 Broad half-ball (without lid)

L = π d t− 2× 12× 4πr2

L = π × 8, 4× 20 − 4× π × (4, 2)2

L = 168π − 4× π × 17.64

L = 168π − 70, 56π

L = 97, 44π

L = 97, 44× 227

L = 13, 92× 22

L = 306, 24 cm2

10

PROBABILITY

• Exercise

1. Number of lines that can be made from 8 points are available, with no 3

point line is ...

2. A box contains 5 red marbles and 3 white marbles. Two marbles are drawn

one by one in which marbles are taken first returned again in the box. The

first chance of picking first and second marbles are red is ....

3. Two dice thrown one-eyed six times simultaneously. Probability the emer-

gence of the dice number 5 or number on the dice 10 is ....

4. By how many ways can 4 people sit in chairs around the circular table?

5. In a meeting consisting of 6 people in a circular position. If the chairman

and representatives should always sit next to each other, how many sit

formations that can be formed?

• Solution

1. 8 C3 = 8!(8−3)!.3!

8 C3 = 8.7.6.5!5!.3!

8 C3 = 8.7.63.2.1

8 C3 = 56

2. P (A ∩B) = P (A).P (B) = 58.58

= 2564

11

3. Probability the emergence of the dice number 5 is 436

Probability the emergence of the dice number 10 is 336

So probability the emergence of the dice number 5 or number on the dice

10 is P (A) + P (B) = 436

+ 336

= 736

4. 4 people sit in chairs around the circular table with (4− 1)! = 3! = 3.2.1 =

6 ways

5. Because of the chairman and deputy should always sit next to each other,

then we think of as a person, so (5− 1)! = 4! = 4.3.2.1 = 24

For the position of chairman and vice= 2! = 2.

Thus, the formation of which can be formed = 24 x 2 = 48.

12