# Solution of SEAMO 2018 Paper B

**SEAMO 2018 Paper B**

SEAMO 2018 Paper B No. 1

Find the next number in the sequence5611, 111223, 232447, ….

A. 232495

B. 464793

C. 474895

D. 474896

E. None of the above

**Solution:** ** Show/Hide **

5, 6, 5+6 $\to $ 561111, 12, 11+12 $\to $ 111223

23, 24, (23+24) $\to $ 232447

47, 48, (47+48) $\to $ 474895

**Answer: C**

SEAMO 2018 Paper B No. 2

Find perimeter of the figure below.A. 38

B. 42

C. 44

D. 46

E. None of the above

**Solution:** ** Show/Hide **

$2\times (5+5+5+2+2+4)=46$**Answer: D**

SEAMO 2018 Paper B No. 3

6 footballs and 3 basketballs cost $\$$279. 2 footballs and 3 basketballs cost $\$$139. Find the prices of a football and a basketball, respectively.A. $\$$35 and $\$$21

B. $\$$35 and $\$$23

C. $\$$40 and $\$$21

D. $\$$40 and $\$$23

E. None of the above

**Solution:** ** Show/Hide **

For example, $x$ is the price of a football and $y$ is the price of a basketball. Then we get a system of equations in two variables.$\frac{\begin{align}6x+3y &= 279\,....\,(1) \\ 2x+3y &= 139\,....\,(2) \end{align}}{\begin{align}4x &= 140 \\ x &= 35 \end{align}}-$

substitute $x=35$ into equation (2),

$\begin{align}2x+3y &= 139 \\ 2(35)+3y &= 139 \\ 70+3y &= 139 \\ 3y &= 69 \\ y &= 23 \end{align}$

So, the prices of a football and a basketball are $\$$35 and $\$$23

**Answer: B**

SEAMO 2018 Paper B No. 4

How many cubes are there in the figure below?A. 20

B. 22

C. 23

D. 24

E. 25

**Solution:** ** Show/Hide **

From the front row to the back row we get the number of cubes in the picture is (6 + 8 + 8) = 22**Answer: B**

SEAMO 2018 Paper B No. 5

In the following addtions, each letter represents a different digit.Find $\overline{ABCD}$.

A. 8370

B. 9464

C. 8371

D. 9465

E. None of the above

**Solution:** ** Show/Hide **

In the first calculation:b + e $\to $ 0 then b + e = 10

1000a + 100b + 10c + d + 100e + 10f + g = 9063

1000a + 100(b+e) + 10(c+f) + d+g = 9000 + 60 + 3

1000a + 100(10) + 10(c+f) + d+g = 9000 + 60 + 3

1000a + 1000 + 10(c+f) + d+g = 9000 + 60 + 3

1000a + 10(c+f) + d+g = 8000 + 60 + 3

1000a = 8000 $\to $ a =8

In the second calculation:

a = 8

a + e cannot be 5 then

a + e = 15

8 + e = 15 $\to $ e = 7

100a + 10b + c + 1000d + 100e + 10f + g = 2529

1000d + 100(a+e) + 10(b+f) + c + g = 2000 + 500 + 20 + 9

1000d + 100(15) + 10(b+f) + c + g = 2000 + 500 + 20 + 9

1000d + 1500 + 10(b+f) + c + g = 2000 + 500 + 20 + 9

1000d + 10(b+f) + c + g = 1000 + 20 + 9

1000d = 1000 $\to $ d = 1

d + g = 3

1 + g = 3 $\to $ g = 2

c + g = 9

c + 2 = 9 $\to $ c = 7

b + e = 10

b + 7 = 10 $\to $ b = 3

ABCD = 8371

**Answer: C**

SEAMO 2018 Paper B No. 6

Pine trees are planted at equal intervals of 16 m along a straight road, including both ends of the road. If the road is 960 m long, how many pine trees are there altogether?A. 56

B. 58

C. 59

D. 60

E. 61

**Solution:** ** Show/Hide **

We can calculate the number of gaps between the trees by dividing the total length of the road by the distance between two trees:Number of gaps = $\frac{960}{16}$ = 60

Since there are 60 gaps, the number of trees will be one more than the number of gaps (because there is a tree at each end of the road):

Number of trees=60+1=61

Therefore, there are 61 pine trees altogether.

**Answer: E**

SEAMO 2018 Paper B No. 7

The numbers 1, 2, 3, … , 9 is to be filled in each circle, such that the sum of numbers along each line is S. Find S.A. 25

B. 26

C. 22

D. 34

E. None of the above

**Solution:** ** Show/Hide **

S + S – 5 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 92S – 5 = 45

2S = 50

S = 25

**Answer: A**

SEAMO 2018 Paper B No. 8

There are 14 children at a party. If each child shakes hands with each other exactly once, how many handshakes will take place?A. 75

B. 85

C. 87

D. 89

E. 91

**Solution:** ** Show/Hide **

To find the total number of handshakes when each child shakes hands with every other child exactly once, we can use the formula for combinations. Specifically, we want to find the number of ways to choose 2 children from 14 to shake hands. The formula for combinations is given by:$C(n,r)=\frac{n!}{r!(n-r)!}$

Where:

$n$ is the total number of children (14 in this case).

$r$ is the number of children shaking hands (2).

So we calculate:

$\begin{align}C(14,2) &= \frac{14!}{2!(14-2)!} \\ &= \frac{14!}{2!\times 12!} \\ &= \frac{14\times 13\times 12!}{2\times 1\times 2!} \\ &= \frac{182}{2} \\ &= 91 \end{align}$

Thus, the total of handshakes that will take place is 91.

**Answer: E**

SEAMO 2018 Paper B No. 9

When 925 is divided by $m$, the quotient is 35 and the remainder is 15. Find $m$.A. 23

B. 25

C. 26

D. 27

E. None of the above

**Solution:** ** Show/Hide **

Dividend = divisor $\times $ quotient + Remainder$\begin{align}925 &= 35\times m+15 \\ 925-15 &= 35\times m+15-15 \\ 910 &= 35\times m \\ \frac{910}{35} &= m \\ 26 &= m \end{align}$

Thus, the value of $m$ is 26.

**Answer: C**

SEAMO 2018 Paper B No. 10

There are 15 red, 20 white, 25 black and 40 yellow balls in bag. Without looking into the bag, find the least number of balls to be removed, without replacement, to obtain at least 4 balls of different colours.A. 86

B. 76

C. 75

D. 66

E. 67

**Solution:** ** Show/Hide **

In the worst-case scenario, you could first draw all the balls of three colors before finally getting a ball of the fourth color. To figure out the least number of balls to remove to guarantee 4 different colors, consider the following:You could draw all 40 yellow balls.

Then, you could draw all 25 black balls.

Next, you could draw all 20 white balls.

At this point, you have drawn:

40 (yellow) + 25 (black) + 20 (white) = 85 balls.

The next ball you draw must be red because you've already drawn all the yellow, black, and white balls. Therefore, the next ball will ensure you have at least one of the fourth color.

Thus, the total number of balls you need to remove to guarantee at least one ball of each color is 85 + 1 = 86 balls.

Therefore, the least number of balls to be removed to obtain at least 4 balls of different colors is 86.

**Answer: A**

SEAMO 2018 Paper B No. 11

Find the missing number in the number puzzle below.A. 5

B. 7

C. 4

D. 8

E. None of the above

**Solution:** ** Show/Hide **

$4\times (5+3)=32$$5\times (7+8)=75$

$\begin{align}?\times (6+5) &= 55 \\ ?\times 11 &= 55 \\ ? &= \frac{55}{11} \\ ? &= 5 \end{align}$

**Answer: A**

SEAMO 2018 Paper B No. 12

Peter has a total of 100 chickens and rabbits on his farm. If the animals have 260 legs altogether, how many chickens are there on the farm?A. 40

B. 50

C. 60

D. 70

E. 80

**Solution:** ** Show/Hide **

let $x$ be the number of chickens and $y$ be the number of rabbits.A chicken has 2 legs and a rabbit has 4 legs.

So we get a system of linear equations in two variables:

$x+y=100\,....\,(1)$

$\begin{align}2x+4y &= 260 \\ x+2y &= 130\,....\,(2) \end{align}$

Equation (2) minus equation (1) we get:

$\frac{\begin{align}x+2y &= 130 \\ x+y &= 100 \end{align}}{y=30}-$

Substitute y = 30 into equation (1)

$\begin{align}x+y &= 100 \\ x+30 &= 100 \\ x+30-30 &= 100-30 \\ x &= 70 \end{align}$

So, the number of chickens on the farm is 70.

**Answer: D**

SEAMO 2018 Paper B No. 13

Teacher Ali has a bag of sweets. If he gives each student 4 sweets, he will be left with 4 sweets. If he gives each student 5 sweets, he will be short of 7 sweets. How many sweets does Teacher Ali have?A. 45

B. 46

C. 47

D. 48

E. None of the above

**Solution:** ** Show/Hide **

For example, $x$ is the number of teacher Ali's sweets and $y$ is the number of students.We get 2 linear equations in two variables.

$x=4n+4$

$x=5n-7$

$\begin{align}5n-7 &= 4n+4 \\ n &= 11 \end{align}$

Substitute $n=11$ into the equation:

$\begin{align}x &= 4n+4 \\ &= 4\times 11+4 \\ &= 44+4 \\ x &= 48 \end{align}$

So, Teacher Ali has 48 sweets.

**Answer: D**

SEAMO 2018 Paper B No. 14

There is a total of 15 questions in a mathematics test. For every correct answer, 8 marks will be awarded. For every wrong answer, 4 marks are deducted. If John scored a total of 72 marks, how questions did he answer wrongly?A. 2

B. 3

C. 4

D. 5

E. 6

**Solution:** ** Show/Hide **

Let $x$ be the number of questions answered correctly and $y$ be the number of questions answered wrongly, then$x+y=15\to x=15-y$

$\begin{align}\text{Score} &= 8x-4y \\ 72 &= 8(15-y)-4y \\ 72 &= 120-8y-4y \\ 72 &= 120-12y \\ 72+12y &= 120-12y+12y \\ 12y+72 &= 120 \\ 12y+72-72 &= 120-72 \\ 12y &= 48 \\ y &= \frac{48}{12} \\ y &= 4 \end{align}$

So, the number of questions answered wrongly is 4.

**Answer: C**

SEAMO 2018 Paper B No. 15

Cindy’s father is 4 times older than her. Three years ago, both their ages added up to 49. How old are Cindy and her father, respectively, this year?A. 10 and 45

B. 11 and 44

C. 12 and 46

D. 9 and 36

E. None of the above

**Solution:** ** Show/Hide **

Let be Cindy's age this year and be Cindy's father's age now.Cindy’s father is 4 times older than her.

$y=4x\,....\,(1)$

Three years ago, both their ages added up to 49.

$\begin{align}(x-3)+(y-3) &= 49 \\ x+y &= 55\,....\,(2) \end{align}$

Substitute $y=4x$into equation (2):

$\begin{align}x+y &= 55 \\ x+4x &= 55 \\ 5x &= 55 \\ x &= \frac{55}{5} \\ x &= 11 \end{align}$

Substitute $x=11$ into equation (1):

$y=4x=4\times 11=44$

So, the ages of Cindy and her father are 11 and 44 years.

**Answer: B**

SEAMO 2018 Paper B No. 16

Mark had some stamps. He gave 10 less than half of his stamps of to Shaun. Then, he gave 6 more than half the remainder to Cindy. Finally, he gave 30 stamps to his cousin and was left with 40 stamps. How many stamps did he have at first?A. 284

B. 304

C. 322

D. 354

E. None of the above

**Solution:** ** Show/Hide **

Let the number of stamps Markus had initially be $x$.Markus gave 10 less than half of his stamps to Shaun.

The number of stamps he gave to Shaun: $\frac{1}{2}x-10$.

Stamps left after giving to Shaun:

$x-\left( \frac{1}{2}x-10 \right)=\frac{1}{2}x+10$

Markus gave 6 more than half of the remaining stamps to Cindy.

The number of stamps he gave to Cindy:

$\frac{1}{2}\left( \frac{1}{2}x+10 \right)+6$ = $\frac{1}{4}x+5+10$ = $\frac{1}{4}x+11$

Stamps left after giving to Cindy:

$\left( \frac{1}{2}x+10 \right)-\left( \frac{1}{4}x+11 \right)$ = $\frac{1}{4}x-1$

Then he gave 30 stamps to his cousin, and 40 stamps remained:

$\begin{align}\left( \frac{1}{4}x-1 \right)-30 &= 40 \\ \frac{1}{4}x-31 &= 40 \\ \frac{1}{4}x &= 71 \\ x &= 4\times 71 \\ x &= 284 \end{align}$

So, the number of stamps Markus initially had is 284 stamps.

**Answer: A**

SEAMO 2018 Paper B No. 17

The diagram below shows a rectangle of length 36 cm. It contains two shaded regions and 7 identical small rectangles. Find the area of one small rectangle, in $\text{cm}^2$.A. 98

B. 120

C. 144

D. 156

E. None of the above

**Solution:** ** Show/Hide **

$\frac{1}{4}y=x\to y=4x$

The length of the rectangle is 36 cm.

$\begin{align}y+x+x &= 36 \\ y+2x &= 36 \\ 4x+2x &= 36 \\ 6x &= 36 \\ x &= 6 \end{align}$

Substitute $x=6$ into:

$y=4x=4\times 6=24$

The area of one small rectangle:

$x\times y=6\times 24=144\,\text{cm}^2$

**Answer: C**

SEAMO 2018 Paper B No. 18

Train A left Town X for Town Y, at a speed of 60 km/h. 1h later, Train B left Town X for Town Y, at a speed of 75 km/h. How many hours did Train B take to catch up with Train A?A. 2 h 30 min

B. 3 h

C. 3 h 30 min

D. 3 h 45 min

E. 4 h

**Solution:** ** Show/Hide **

Train A:$t_A$ = t hours and $v_A$ = 60 km/h

$d_A=v_A\times t_A=60t$

Train B:

$t_B=t-1$ hours and $v_B$ = 75 kmk/h

$d_B=v_B\times t_B=75(t-1)$

Train B take to catch up with train A so the distance between them is the same.

$\begin{align}d_B &= d_A \\ 75(t-1) &= 60t \\ 75t-75 &= 60t \\ 15t &= 75 \\ t &= \frac{75}{15} \\ t &= 5 \end{align}$

$t_B$ = $t-1$ = 5 – 1 = 4 hours.

So, Train B took 4 hours to catch up with Train A.

**Answer: E**

SEAMO 2018 Paper B No. 19

In the diagram below, the topmost layer has 60 circles. How many circles are there altogether?A. 1750

B. 1780

C. 1830

D. 1880

E. None of the above

**Solution:** ** Show/Hide **

1 + 2 + 3 + … + 60= $\frac{60(60+1)}{2}=30\times 61=1830$

**Answer: C**

SEAMO 2018 Paper B No. 20

The diagram shows a square ABCD of side 10 cm and a shaded rectangle of area 6 $\text{cm}^2$. What is the area of EFGH?A. 49 $\text{cm}^2$

B. 51 $\text{cm}^2$

C. 62 $\text{cm}^2$

D. 53 $\text{cm}^2$

E. 58 $\text{cm}^2$

**Solution:** ** Show/Hide **

2(I + II + III + IV) + 6 = 10 10

2(I + II + III + IV) = 94

I + II + III + IV = 47

[EFGH]=I+II+III+IV+6 = 47+6 = 53 $\text{cm}^2$.

**Answer: D**

SEAMO 2018 Paper B No. 21

A new operation is defined as $a\odot b=5\times a-3\times b$. Given that $14\odot b=13$, find the value of $b$.**Solution:** ** Show/Hide **

$a\odot b=5\times a-3\times b$$\begin{align}14\odot b &= 13 \\ 5\times 14-3\times b &= 13 \\ 70-3b &= 13 \\ 70-3b-70 &= 13-70 \\ -3b &= -57 \\ \frac{-3b}{-3} &= \frac{-57}{-3} \\ b &= 19 \end{align}$

**Answer: 19**

SEAMO 2018 Paper B No. 22

Alan wrote on the whiteboard1, 2, 3, 4, 5, …, 173

How many times did he write the digit ‘7’ altogether?

**Solution:** ** Show/Hide **

7, 17, 27, 37, 47, 57, 67, 77, 87, 97, there are 10 times.70, 71, 72, 73, 74, 75, 76, 78, 79, there are 9 times.

107, 117, 127, 137, 147, 157, 167, there are 7 times.

170, 171, 172, 173, there are 4 times.

Total: 10 + 9 + 7 + 4 = 30

**Answer: 30**

SEAMO 2018 Paper B No. 23

Many years ago, there were 5 Saturdays in the month of February. On which day of the week was the last day of that year?**Solution:** ** Show/Hide **

For February to have 5 Saturdays, the month must start on a Saturday, and February must have 29 days, meaning it's a leap year.Let’s break it down:

- If February 1st is a Saturday in a leap year, February $29^{th} (the last day of the month) will be on a Saturday as well.
- This means March 1st will be a Sunday, and the rest of the days of the year will follow this sequence.

306 days ≡ 306 ÷ 7 = 43 weeks + 5 extra days.

Thus, if March $1^{st}$ is a Sunday, adding 5 days means December $31^{st}$ will fall on a Thursday.

So, the last day of that year was a Thursday.

**Answer: Thursday**

SEAMO 2018 Paper B No. 24

Jane walked from her house to Peter’s at a speed of 80 m/min. At the same time, Peter walked from his house to Jane’s at a speed of 60 m/min. They met 15 m from the mid-point of the two houses. How far is Jane’s house from Peter’s?**Solution:** ** Show/Hide **

$v_{Jane}$ = 80 m/min

$d_{Jane}=\frac{1}{2}d+15$

$t_{Jane}=\frac{v_{Jane}}{d_{Jane}}\to t_{Jane}=\frac{80}{\frac{1}{2}d+15}$

$v_{Peter}$ = 60 m/min

$d_{Peter}=\frac{1}{2}d-15$

$t_{Peter}=\frac{v_{Peter}}{d_{Peter}}\to t_{Peter}=\frac{60}{\frac{1}{2}d-15}$

$\begin{align}t_{Jane} &= t_{Peter} \\ \frac{80}{\frac{1}{2}d+15} &= \frac{60}{\frac{1}{2}d-15} \\ \frac{8}{\frac{1}{2}d+15} &= \frac{6}{\frac{1}{2}d-15} \\ 8\left( \frac{1}{2}d-15 \right) &= 6\left( \frac{1}{2}d+15 \right) \\ 4d-120 &= 3d+90 \\ 4d-3d &= 90+120 \\ d &= 210 \end{align}$

So, the distance from Jane's house to Peter's house is 210 m

**Answer: 210 m**

SEAMO 2018 Paper B No. 25

In the diagram below, ABCD is a square of side 4 cm. DEFG is a rectangle. EAF is a straight line. Given that CG = 3 cm and DG = 5 cm, find the length of DE in cm.**Solution:** ** Show/Hide **

$AF=x\to AE=5-x$

Consider triangle ABG:

$AG^2=AB^2+BG^2=4^2+1^2=17$

Consider triangle AFG:

$FG^2=AG^2-AF^2$

$FG^2=17-x^2$

Consider triangle AED:

$\begin{align}DE^2 &= AD^2-AE^2 \\ &= 4^2-(5-x)^2 \\ &= 16-(25-10x+x^2) \\ DE^2 &= -9+10x-x^2 \end{align}$

$\begin{align}DE &= FG \\ -9+10x-x^2 &= 17-x^2 \\ -9+10x &= 17 \\ 10x &= 26 \\ x &= \frac{26}{10} \\ x &= \frac{13}{5} \end{align}$

$\begin{align}DE^2 &= -9+10x-x^2 \\ &= -9+10\times \frac{13}{5}-\left( \frac{13}{5} \right)^2 \\ &= -9+26-\frac{169}{25} \\ &= 17-\frac{169}{25} \\ DE^2 &= \frac{256}{25} \\ DE &= \sqrt{\frac{256}{25}} \\ &= \frac{16}{5} \\ DE &= 3.2 \end{align}$

So the length of DE is 3.2 cm.

**Answer: 3.2 cm**

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