304250295-SASMO-2014-Round-1-Primary-Solutions.pdf
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SASMO 2014 Round 1 Primary 2 Solutions Section A [1 mark for each question] 1.
What is 2014 + 2 0 1 4 equal to? (a) 2014 (b) 2016 (c) 2021 (d) 2022 (e) None of the above Solution 2014 + 2 0 1 4 = 2014 + 0 = 2014 (a)
2.
Ten lampposts are equally spaced along a straight line. The distance between two consecutive lampposts is 40 m. What is the distance between the first and the last lampposts? (a) 360 m (b) 380 m (c) 400 m (d) 420 m (e) None of the above Solution Distance between the first and the last lampposts = 40 m 9 gaps = 360 m (a)
3.
Find the next term of the following sequence: 1, 1, 2, 3, 5, … (a) 6 (b) 7 (c) 8 (d) 9 (e) 10 Solution From the third term onwards, the next term is obtained by adding the previous two terms. the next term is 3 + 5 = 8 (c)
4.
Jane wrote the word STUDENTS thrice. How many times did she write the letter S? (a) 2 (b) 4 (c) 6 (d) 8 (e) 10
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Solution ‘Thrice’ means 3 times. Jane wrote the letter S a total of 2 3 = 6 times (c) 5.
What number between 37 and 47 is exactly divisible by both 2 and 3? (a) 38 (b) 39 (c) 42 (d) 44 (e) 45 Solution Method 1 Numbers between 37 and 47 that are exactly divisible by 2 are: 38, 40, 42, 44 and 46. Of these 5 numbers, only 42 is exactly divisible by 3. the number between 37 and 47 that is exactly divisible by both 2 and 3 is 42 (c). Method 2 A number that is exactly divisible by both 2 and 3 must also be exactly divisible by 6. he only number between 37 and 47 that is exactly divisible by 6 is 42. the number between 37 and 47 that is exactly divisible by both 2 and 3 is 42 (c).
6.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Ali has enough money to buy only 7 sweets. What is the biggest number of sweets that he can get from the shop? (a) 7 (b) 8 (c) 9 (d) 10 (e) 11 Solution 7 sweets 7 wrappers 2 sweets and 1 wrapper 3 wrappers 1 sweet biggest no. of sweets = 7 + 2 + 1 = 10 (d)
7.
There are 14 children playing “The eagle catches the chicks.” One of them is the ‘eagle’ while another child is the ‘mother hen’ whose job is to protect the ‘chicks’. The rest of the children are the ‘chicks’. After a while, the ‘eagle’ has caught 5 ‘chicks’. How many ‘chicks’ are still running around? (a) 6 (b) 7 (c) 8 (d) 9 (e) 10 2
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Solution No. of ‘chicks’ still running around = 14 1 (eagle) 1 (mother hen) 5 = 7 (b) 8.
Find the number A such that the following statement is true: 7 A = 3 8 + 4 8. (a) 3 (b) 4 (c) 5 (d) 7 (e) 8 [Ans] Solution Method 1 7 A = 3 8 + 4 8 = 24 + 32 = 56 A = 56 7 = 8 (e) Method 2 7 A = 3 8 + 4 8 = (3 + 4) 8 = 7 8 A = 8 (e)
9.
Two $1 coins and ten 50¢ coins are randomly distributed among 4 children such that each child receives the same number of coins. What is the difference between the biggest amount and the smallest amount a child can receive? (a) 50¢ (b) $1 (c) $1.50 (d) $2 (e) None of the above Solution There are a total of 2 + 10 = 12 coins. So each child receives 12 4 = 3 coins. Biggest amount a child can receive = $1 + $1 + 50¢ = $2.50 Smallest amount a child can receive = 50¢ + 50¢ + 50¢ = $1.50 difference between biggest amount and smallest amount = $2.50 $1.50 = $1 (b)
10.
Tim is 8 years old and Sally is 4 years old. How old will Sally be when Tim is 14 years old? (a) 7 (b) 8 (c) 9 (d) 10 (e) None of the above
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Solution Method 1 Tim will be 14 years old in 14 8 = 6 years’ time. Sally will be 6 + 4 = 10 years old (d). Method 2 Difference in age between Tim and Sally = 8 4 = 4 years when Tim is 14 years old, Sally will be 14 4 = 10 years old (d).
Section B [3 marks for each question] 11.
In the following alphametic, all the different letters stand for different digits. Find P and I. I I +
I P
I
Solution I 3 = _I By guess and check, the only possible solution for I is 5. 5 + 5 + 5 = 15, i.e. P = 1 and I = 5. 12.
A box contains 4 balls of different colours (red, green, yellow and blue) lying in a row. The green ball is not the second ball. The red ball is neither the first nor the last ball. The yellow ball is neither next to the red ball nor next to the blue ball. What is the order of the balls in the box from first to last? Solution The red ball is neither the first nor the last ball. Suppose the red ball is the third ball: _____, _____, __R__ , _____ The yellow ball is neither next to the red ball nor next to the blue ball. This means that the yellow ball is the first ball, and the blue ball is the last ball: __Y__, _____, __R__ , __B__ So the green ball is the second ball, which is a contradiction. 4
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Method 3 The pattern for the no. of white squares is 3 3 1, 3 4 2, 3 5 3, … (total no. of squares minus no. of black squares in each figure) no. of white squares that will surround one row of 50 black squares = 3 52 50 = 106 (e) Method 4 The pattern for the no. of white squares is 8, 10, 12, …, which is equal to 2 4, 2 5, 2 6, … no. of white squares that will surround one row of 50 black squares = 2 53 = 106 (e) 15.
The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of the next segment must coincide with the endpoint of the previous segment) to pass through all the 9 points.
Solution If you try to draw the line segments within the region bounded by the dots, you will realise that you need at least 5 consecutive line segments. you must draw some of the line segments outside the region as shown:
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SASMO 2014 Round 1 Primary 3 Solutions Section A [1 mark for each question] 1.
Jane is 9 years old and John is 5 years old. How old will John be when Jane is 15 years old? Solution Method 1 Jane will be 15 years old in 15 9 = 6 years’ time. John will be 6 + 5 = 11 years old. Method 2 Difference in age between Jane and John = 9 5 = 4 years when Jane is 15 years old, John will be 15 4 = 11 years old.
2.
A textbook is opened at random. To what pages is it opened if the product of the facing pages is 110? Solution Since 10 10 = 100, try 10 11 = 110. the pages are 10 and 11.
3.
Find the number B such that the following statement is true: 8 B = 3 9 + 5 9. Solution Method 1 8 B = 3 9 + 5 9 = 27 + 45 = 72 A = 72 8 = 9 Method 2 8 B = 3 9 + 5 9 = (3 + 5) 9 = 8 9 A=9
4.
It is given that a b = a b + a b. For example, 2 3 = 2 3 + 2 3 = 5. Find the value of 4 3 3 4. Solution Method 1 4 3 = 4 3 + 4 3 = 13 1
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3 4 = 3 4 + 3 4 = 11 4 3 3 4 = 13 11 = 2 Method 2 43=43+43=43+1 34=34+34=341 4 3 3 4 = 1 (1) = 2 5.
Jane has a rope of length 23 cm. She wants to cut the rope so that she can form the biggest possible square, where the length of each side, in cm, is a whole number. What is the length of the rope that she must cut to form the square? Solution Since 5 cm 4 = 20 cm, and 6 cm 4 = 24 cm, then the biggest possible square that she can form has a length of 5 cm. length of rope that she must cut to form the square = 5 cm 4 = 20 cm
6.
Find the missing term in the following sequence: 1, 2, 6, 24, _____, 720. Solution The pattern is as follows: 1, 2
2,
6, 3
24, _____, 720 4
5 6
the missing term is 24 5 = 120. 7.
On National Day, 39 soldiers lined up in a straight row on opposite sides of Stadium Street to welcome Prime Minister Lee. A soldier stands on each end of Stadium Street. The distance between two adjacent soldiers on either side was 20 m. The soldiers on one side were arranged such that each soldier filled the gap between two other soldiers on the opposite side. How long was Stadium Street? Solution There are 20 soldiers on one side and 19 soldiers on the other side. length of Stadium Street = 20 m (20 – 1) gaps = 380 m
8.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Sharon has enough money to buy only 11 sweets. What is the biggest number of sweets that she can get from the shop? Solution
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11 sweets 11 wrappers 3 sweets and 2 wrappers 5 wrappers 1 sweet and 2 wrappers 3 wrappers 1 sweet biggest no. of sweets = 11 + 3 + 1 + 1 = 16 [Common mistakes: 14, 15] 9.
At a workshop, there are 10 participants. Each of them shakes hand once with one another. How many handshakes are there? The first participant will shake hand with 9 other participants; the second participant will shake hand with 8 other participants; the third participant will shake hand with 7 other participants; etc. Thus total no. of handshakes = 9 + 8 + 7 + … + 3 + 2 + 1 1 + 9 = 10 2 + 8 = 10 3 + 7 = 10 4 + 6 = 10 5
4 pairs
total no. of handshakes = 10 4 + 5 = 45 10.
Ali uses identical square tiles to make the following figures. If he continues using the same pattern, how many tiles will there be in the 15th figure?
Solution Method 1 The two corner tiles are the same for all figures. the 15th figure will have 15 3 + 2 = 47 tiles. Method 2 The tiles in the top row have this pattern: 3, 4, 5, 6, … the 15th figure will have 17 + 2 15 = 47 tiles. Method 3 The tiles in each of the vertical column have this pattern: 2, 3, 4, 5, … the 15th figure will have 16 2 + 15 = 47 tiles. Method 4 The no. of tiles in each figure is equal to the “area of the rectangle” minus the “area of the hole in the middle”. 3
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The pattern for the “area of the rectangle” is 2 3, 3 4, 4 5, 5 6, … The pattern for the “area of the hole in the middle” is 1 1, 2 2, 3 3, 4 4, … the 15th figure will have 16 17 15 15 = 272 225 = 47 tiles. 11.
What is the least number of cuts required to cut 16 identical sausages so that they can be shared equally among 24 people? Solution Fraction of sausage each person will get =
=
This means that there must be at least 16 cuts since no one will get one whole sausage. Cut each of the 16 sausages at the -mark. Then 16 people will get one
sausage
each, and the remaining 8 people will get two sausages each. least no. of cuts = 16 12.
A vending machine accepts 10¢ coins, 20¢ coins, 50¢ coins and $1 coins only. Ivy wants to buy a can of drinks that costs $1.60. She has eight 10¢ coins, three 20¢ coins, two 50¢ coins and one $1 coin. If she wants to get rid of as many coins as possible, what is the combination of coins that she should put inside the vending machine? Solution To get rid of as many coins as possible, we try to use as many coins with the smallest value (i.e. the 10¢ coins) as possible. If Ivy uses all the eight 10¢ coins, then what is left is $1.60 80¢ = 80¢. Unfortunately, Ivy has only three 20¢ coins, which is not enough. So she has to use a 50¢ coin. But 50¢ + 20¢ = 70¢, which is 10¢ short. In other words, she can’t use all the eight 10¢ coins. she has to use seven 10¢ coins, one 50¢ coin and two 20¢ coins.
13.
The total cost of a pen and a pencil is $2.90. The pen costs 60¢ more than the pencil. What much does the pen cost? Solution Method 1 (Systematic Guess and Check) Make a systematic list, starting with a random guess: pencil costs $1 and pen costs $1.60. Cost of Pencil $1 $1.10 $1.15
Cost of Pen $1.60 $1.70 $1.75
Total Cost $2.60 $2.80 $2.90
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the pen costs $1.75 Method 2 (Model Method) Pen
60¢ $2.90
Pencil 2 units = $2.90 60¢ = $2.30 1 unit = $1.15 the pen costs $1.15 + 60¢ = $1.75 Method 3 (Algebraic Method) Let the cost of the pencil be $x. Then the cost of the pen is $(x + 0.6). So x + (x + 0.6) = 2.9 2x + 0.6 = 2.9 2x = 2.3 x = 1.15 the pen costs $(1.15 + 0.60) = $1.75 14.
If the three-digit number 3N3 is divided by 9, the remainder is 1. Find N. Solution Since 3N3 gives a remainder of 1 when divided by 9, then 3N3 1 = 3N2 is divisible by 9. Using the divisibility test for 9, 3 + N + 2 = N + 5 is also divisible by 9. N = 4.
15.
Charles has 16 marbles. He divides them into 4 piles so that each pile has a different number of marbles. Find the smallest possible number of marbles in the biggest pile. Solution For each pile to have a different number of marbles, and the biggest pile to have the smallest possible number of marbles, put 1 marble in the 1st pile, 2 marbles in the 2nd pile, 3 marbles in the 3rd pile and 4 marbles in the 4th pile. So the biggest pile is the 4th pile, but there are only 1 + 2 + 3 + 4 = 10 marbles. The 11th marble will have to go to the 4th pile so that each pile will have a different number of marbles. The 12th marble cannot go to the 4th pile because we want to find the smallest possible number of marbles in the biggest pile, so the 12th marble will have to go to the 3rd pile. Similarly, the 13th and 14th marbles will go to the 2nd and 1st piles respectively. The 15th marble will go to the 4th pile again, and the 16th marble to the 3rd pile. the largest pile (which is the 4th pile) will contain 4 + 1 + 1 = 6 marbles.
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Section B [2 marks for each question] 16.
In the following alphametic, all the different letters stand for different digits. Find the three-digit sum SEE. A +
S A
S
E
E
Solution Since the addition of two digits will give a maximum of 18, or a maximum of 19 if there is a carryover (or renaming) of 1, this means that the maximum carryover is 1. So A = 9 and there is a carryover of 1 for A to give 10, i.e. S = 1 and E = 0. 91 + 9 = 100, i.e. the three-digit sum SEE is 100. 17.
Find the total number of triangles in the diagram.
1
Solution Label the vertices as shown in the diagram. These are the vertices that form a triangle: 123, 124, 125, 126, 134, 135, 145, 156, 157, 158, 167, 168, 236, 256, 356, 457, 458, 568, 578, 678 there is a total of 20 triangles in the diagram.
2
3
4
5
7 8 6
18.
A teacher has a bag of sweets to treat her class. If she gave 5 sweets to each student, then she would have 40 sweets left. If she gave 7 sweets to each student, then she would have 6 sweets left. How many students and how many sweets are there? Solution If the teacher gave 5 sweets to each student, then she would have 40 sweets left. From the 40 sweets left, if she gave 2 more sweets to each student so that each student has 7 sweets, then she would have 6 left. This means that she gave a total of 40 6 = 34 sweets from the 40 sweets left to the students. Since each student receives only 2 more sweets (i.e. the 6th and 7th sweets), then there are 34 2 = 17 students. 6
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19.
there are 17 7 + 6 = 125 sweets. What are the last 2 digits of the sum 1 + 11 + 111 + … + 111…111? 50 digits Solution Last digit: 50 0 (and carry 5) Second last digit: 49 + 5 (carry) = 54 4 the last 2 digits are 40.
20.
Alvin tells the truth on Monday, Tuesday, Wednesday and Thursday. He lies on all other days. Doris tells the truth on Monday, Friday, Saturday and Sunday. She lies on all other days. One day they both said, “Yesterday I lied.” When was that ‘one day’? Solution
Mon Tue Wed Thur Alvin Doris indicates the person telling the truth
Fri
Sat
Sun
If Alvin tells the truth on that ‘one day’ that he lied ‘yesterday’, then that ‘one day’ must be Monday. If Alvin tells the lie on that ‘one day’ that he lied ‘yesterday’, then he must be telling the truth ‘yesterday’ and so that ‘one day’ must be Friday. based on Alvin, it has to be either Monday or Friday. If Carol tells the truth on that ‘one day’ that she lied ‘yesterday’, then that ‘one day’
must be Friday. If Carol tells the lie on that ‘one day’ that she lied ‘yesterday’, then she must be telling the truth ‘yesterday’ and so that ‘one day’ must be Tuesday. based on Carol, it has to be either Tuesday or Friday. Hence, that ‘one day’, when they both said that they lied ‘yesterday’, has to be Friday.
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SASMO 2014 Round 1 Primary 4 Solutions Section A [1 mark for each question] 1.
Find the missing term in the following sequence: 1, 4, 9, 16, _____, 36. Solution Observe that 1 = 12, 4 = 22, 9 = 32, 16 = 42 and 36 = 62. the missing term is 52 = 25.
2.
Find the smallest whole number between 1 and 100 that is divisible by 5 and by 7. Solution Smallest whole number between 1 and 100 that is divisible by 5 and by 7 = 5 7 = 35
3.
A frog fell into a drain that was 20 cm deep. After one hour, it mastered enough energy to make a jump of 6 cm but it then slid down 4 cm. If it continued in this manner after every one hour, how many hours will it take to get out of the drain? Solution After 7 hours, the frog would have jumped 7 2 = 14 cm. In the next hour, it would have jumped the last 6 cm and out of the drain. it takes the frog 7 + 1 = 8 hours to get out of the drain.
4.
The diagram shows a circle with centre O and radius 5 cm. OABC is a rectangle. What is the length of AC? B
A O
C
Solution AC = OB (diagonal of rectangle) = 5 cm (radius of circle) 5.
A textbook is opened at random. To what pages is it opened if the product of the facing pages is 420? Solution Since 20 20 = 400, try 20 21 = 420. the pages are 20 and 21. 1
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6.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Kelvin has enough money to buy only 19 sweets. What is the biggest number of sweets that he can get from the shop? Solution 19 sweets 19 wrappers 6 sweets and 1 wrappers 7 wrappers 2 sweets and 1 wrapper 3 wrappers 1 sweet biggest no. of sweets = 19 + 6 + 2 + 1 = 28
7.
A clock takes 6 seconds to make 3 chimes. Assuming that the rate of chiming is constant, and the duration of each chime is negligible, how long does the clock take to make 4 chimes? Solution Time taken by the clock between 2 chimes = 6 s / 2 gaps = 3 s time taken by the clock to make 4 chimes = 3 s 3 gaps = 9 s
8.
The diagram shows a square being divided into four rectangles. If the sum of the perimeter of the four rectangles is 32 cm, find the area of the square.
Solution Perimeter of the four rectangles = 8 length of square = 32 cm So length of square = 4 cm Area of square = 16 cm2 9.
Alvin, Betty and Cheryl scored a total of 2014 points during a competition. Betty scored 271 points less than Alvin. Betty scored 3 times as many points as Cheryl. How many points did Betty score? Solution Method 1 (Model Method) Cheryl Betty Alvin
2014 271
7 units = 2014 271 = 1743 2
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Method 2 There are 1 white square on the left side and 1 white square on the right side of the row of black squares. The pattern for the no. of white squares is 3 2 + 2, 4 2 + 2, 5 2 figure number with 2014 white squares = (2014 2) 2 2 = 1004 Method 3 The pattern for the no. of white squares is 3 3 1, 3 4 2, 3 5 3, … (total no. of squares minus no. of black squares in each figure) Thus no. of white squares that will surround one row of n black squares = 3(n + 2) n = 2n + 6 figure number with 2014 white squares = (2014 6) 2 = 1004 Method 4 The pattern for the no. of white squares is 8, 10, 12, …, which is equal to 2 4, 2 5, 2 6, … figure number with 2014 white squares = 2014 2 3 = 1004 13.
Find the total number of squares in a 4 4 square grid.
Solution No. of 1 1 squares = 16 = 42 No. of 2 2 squares = 9 = 32 No. of 3 3 squares = 4 = 22 No. of 4 4 squares = 1 = 12 total no. of squares in a 4 4 square grid = 12 + 22 + 32 + 42 = 30 14.
How many digits are there before the fifteen 7 of the following number? 37337333733337333337… Solution No. of 7s before the fifteen 7 of the number = 14 No. of 3s before the fifteen 7 of the number = 1 + 2 + 3 + … + 15
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Method 1 1 + 15 = 16 2 + 14 = 16 3 + 13 = 16
7 pairs
7 + 9 = 16 8 total no. of digits before the fifteen 7 of the number = 14 + (7 16) + 8 = 134 Method 2 Since 1 + 2 + 3 + … + n = = 134
the number = 14 + 15.
, then total no. of digits before the hundredth 7 of
A box contains 30 coloured pens: 15 blue, 10 red and 5 black. Alice takes some pens from the box without looking at the colours of the pens. What is the least number of pens she must take so that she has at least 8 pens of the same colour? Solution Worst case scenario for all the pens taken out by Alice to be of different colours: 7 blue, 7 red and 5 black, i.e. a total of 19 pens. Then the next pen taken out has to be either blue or red, i.e. Alice will have at least 8 pens of the same colour, either 8 blue or 8 red pens. least no. of pens Alice must take so that she has at least 8 pens of the same colour = 19 + 1 = 20
Section B [2 marks for each question] 16.
In the following alphametic, all the different letters stand for different digits. Find the four-digit sum PEEL.
+ P
A
M
L
A
P
E
E
L
Solution Since the addition of two digits will give a maximum of 18, or a maximum of 19 if there is a carryover (or renaming) of 1, this means that the maximum carryover is 1. So L = 9 and there is a carryover of 1 for L to give 10, i.e. P = 1 and E = 0. the four-digit sum PEEL is 1009.
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17.
A teacher has a bag of sweets to treat her class. If she gave 4 sweets to each student, then she would have 82 sweets left. If she gave 9 sweets to each student, then she would have 2 sweets left. How many students and how many sweets are there? Solution If the teacher gave 4 sweets to each student, then she would have 82 sweets left. From the 82 sweets left, if she gave 5 more sweets to each student so that each student has 9 sweets, then she would have 2 left. This means that she gave a total of 82 2 = 80 sweets from the 82 sweets left to the students. Since each student receives only one 5 more sweets (i.e. the 5th, 6th, 7th, 8th and 9th sweets), then there are 80 5 = 16 students. there are 16 9 + 2 = 146 sweets.
18.
When Amy, Betty and Cheryl eat out, each orders either beef or chicken. a. If Amy orders beef, Betty orders chicken. b. Either Amy or Cheryl orders beef, but not both. c. Betty and Cheryl do not both order chicken. Who could have ordered beef yesterday and chicken today? Solution If Amy orders beef, Condition a says that Betty orders chicken, and Condition b says that Cheryl also orders chicken, contradicting Condition c.
Beef Chicken
Amy (a)
Betty
Cheryl
(a)
(b)
Thus Amy always orders chicken. Then Condition b says that Cheryl always orders beef. Amy Beef Chicken
Betty
Cheryl (b)
So the only person who can change her order is Betty. Betty ordered beef yesterday and chicken today. 19.
What are the last 3 digits of the sum 1 + 11 + 111 + … + 111…111? 70 digits Solution Last digit: 70 0 (and carry 7) Second last digit: 69 + 7 (carry) = 76 6 (and carry 7) Third last digit: 68 + 7 (carry) = 75 5 the last 2 digits are 560. 6
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20.
The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of the next segment must coincide with the endpoint of the previous segment) to pass through all the 9 points.
Solution If you try to draw the line segments within the region bounded by the dots, you will realise that you need at least 5 consecutive line segments. you must draw some of the line segments outside the region as shown:
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SASMO 2014 Round 1 Primary 5 Solutions 1.
If n is a whole number, for what values of n is
also a whole number?
Solution is a whole number if n is a factor of 24. 24 = 1 24 = 2 12 =38 =46 2.
is a whole number if n = 1, 2, 3, 4, 6, 8, 12, 24.
A textbook is opened at random. To what pages is it opened if the product of the facing pages is 600? Solution Since 25 25 = 625, try 24 25 = 600. the pages are 24 and 25.
3.
The diagram shows a quadrant OAB of a circle with centre O. OPQR is a rectangle. Given that PR = 7 cm, find the length of OA. B Q
R
O
P
A
Solution OA = OQ (radii of quadrant) = PR (diagonal of rectangle) = 7 cm 4.
Find an even number between 300 and 400 that is divisible by 5 and by 7. Solution Since 2, 5 and 7 are relatively prime, if a number is divisible by 2, by 5 and by 7, then the number is divisible by 2 5 7 = 70, i.e. the number is a multiple of 70. an even number between 300 and 400 that is divisible by 5 and by 7 is 350.
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5.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Navin has enough money to buy only 29 sweets. What is the biggest number of sweets that he can get from the shop? Solution 29 sweets 29 wrappers 9 sweets and 2 wrappers 11 wrappers 3 sweets and 2 wrappers 5 wrappers 1 sweet and 2 wrappers 3 wrappers 1 sweet biggest no. of sweets = 29 + 9 + 3 + 1 + 1 = 43
6.
Find the next term of the following sequence: 2, 3, 4, 10, 38, … Solution From the third term onwards, the next term is obtained by multiplying the previous two terms and then subtracting 2. the next term is 10 38 2 = 378.
7.
The percentage passes in an exam for two classes are 80% and 60%. The numbers of students in the two classes are 20 and 30 respectively. Find the overall percentage pass for the two classes. Solution No. of students in the first class who pass the exam = 80% 20 = 16 No. of students in the second class who pass the exam = 60% 30 = 18 Total no. of students in both classes who pass the exam = 16 + 18 = 34 overall percentage pass for the two classes = 34 / 50 100% = 68% Note: Common mistake is 70%. But class size is different: the weaker class will pull down the percentage because it has more students than the better class. So you can’t average percentages unless the base (in this case, the number of students in each class) is the same.
8.
A clock takes 9 seconds to make 4 chimes. Assuming that the rate of chiming is constant, and the duration of each chime is negligible, how long does the clock take to make 3 chimes? Solution Time taken by the clock between 2 chimes = 9 s / 3 gaps = 3 s time taken by the clock to make 3 chimes = 3 s 2 gaps = 6 s
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9.
1 1 1 1 Evaluate 1 1 1 1 . 2 3 4 2014
Solution 1 1 2 3 2013 1 1 1 1 1 1 1 = 2 3 4 2014 2 3 4 2014 1 = 2014
10.
A frog fell into a drain that was 50 cm deep. After one hour, it mastered enough energy to make a jump of 6 cm but it then slid down 4 cm. If it continued in this manner after every one hour, how many hours will it take to get out of the drain? Solution After 22 hours, the frog would have jumped 22 2 = 44 cm. In the next hour, it would have jumped the last 6 cm and out of the drain. it takes the frog 22 + 1 = 23 hours to get out of the drain.
11.
A farmer’s chickens produced 4028 eggs one day. Was he able to pack all the eggs in full cartons of one dozen eggs each? Solution A dozen is equal to 12. If a number is divisible by 12 (= 3 4), then it is also divisible by 3 and 4, since 3 and 4 are relatively prime. Using the divisibility test for 4, 4028 is divisible by 4 since the last two digits 28 is divisible by 4. Using the divisibility test for 3, 4028 is not divisible by 3 since 4 + 0 + 2 + 8 = 14 is not divisible by 3, i.e. 4028 is not divisible by 12. the farmer was not able to pack all the eggs in full cartons of one dozen eggs.
12.
A farmer wants to find out the number of sheep and ducks that he has. He counted a total of 40 heads and 124 legs. How many sheep and how many ducks does he have? Solution Method 1 (Supposition) Suppose all the animals are ducks. Then there would be 40 2 = 80 legs. So the remaining 124 80 = 44 legs must belong to the “extra 2 legs” of the sheep. there are 44 2 = 22 sheep and 40 22 = 18 ducks.
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Method 2 (Systematic Guess and Check) Make a systematic list, starting with a random guess: 20 sheep and 20 ducks. No. of Sheep
No. of Ducks
20 21 22
20 19 18
No. of Sheep Legs 80 84 88
No. of Duck Legs 40 38 36
Total No. of Legs 120 122 124
there are 22 sheep and 18 ducks. Method 3 (Algebra) Let the no. of sheep be x. Then there are 40 x ducks. Total no. of legs = 4x + 2(40 x) = 124 4x + 80 2x = 124 2x = 44 x = 22 there are 22 sheep and 40 22 = 18 ducks. 13.
Jaime puts some blue and red cubes in a box. The ratio of the number of blue cubes to the number of red cubes is 2 : 1. She adds 12 more red cubes in the box and the ratio becomes 4 : 5. How many blue cubes are there in the box? Solution Method 1 (Model Method) Before Blue Red After Blue Red 12 From the diagram, 3 units = 12 1 unit = 4 4 units = 16 there are 16 blue cubes.
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Method 2 (Algebraic Method) Let the no. of red cubes at the start be x. Then the no. of blue cubes will be 2x. After adding 12 more red cubes, there are x + 12 red cubes. Ratio of blue cubes to red cubes is now 4 : 5, i.e.
.
Then 10x = 4x + 48 6x = 48 x=8 2x = 16 there are 16 blue cubes. 14.
The diagram shows a rectangle with its two diagonals. What percentage of the rectangle is shaded?
Solution Let the length and breadth of a rectangle be l and b respectively. Then area of rectangle = lb and area of shaded triangle = base height = b = . percentage of the rectangle that is shaded = 100% = 25% 15.
Given that a b = 2014, and a and b are whole numbers such that a < b, how many possible pairs (a, b) are there? Solution 2014 = 2 19 53, where 2, 19 and 53 are prime numbers. 2014 = 1 2014 = 2 1007 = 19 106 = 38 53 there are 4 possible pairs (a, b).
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16.
Amy had 3 times as much money as Betty. After they had spent $60 each, Amy had 4 times as much money as Betty. How much money did Amy have at first? Solution Method 1 (Model Method) After Amy Betty Before $60
Amy Betty
$60
$60
$60
From the Before diagram, where Amy had 3 times as much money as Betty, 1 unit = $60 + $60 = $120 4 units = $480 Amy had $480 + $60 = $540 at first. Method 2 (Algebraic Method) Let the amount of money Betty had at first be $x. Then the amount of money Alice had at first was $3x. After they had spent $60 each, Betty had $(x 60) and Alice had $(3x 60). Then $(3x 60) = 4 $(x 60) 3x 60 = 4x 240 x = 180 3x = $540 Amy had $540 at first.
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17.
Billy uses identical square tiles to make the following figures. If he continues using the same pattern, in which figure will there be 6044 tiles?
Figure 1
Figure 2
Figure 3
Figure 4
Solution Method 1 The two corner tiles are the same for all figures. figure number with 6044 tiles = (6044 2) 3 = 2014 Method 2 The tiles in the top row have this pattern: 3, 4, 5, 6, … Thus the nth figure will have (n + 2) + 2 n = 3n + 2 tiles. figure number with 6044 tiles = (6044 2) 3 = 2014 Method 3 The tiles in each of the vertical column have this pattern: 2, 3, 4, 5, … Thus the nth figure will have (n + 1) 2 + n = 3n + 2 tiles. figure number with 6044 tiles = (6044 2) 3 = 2014 Method 4 The no. of tiles in each figure is equal to the “area of the rectangle” minus the “area of the hole in the middle”. The pattern for the “area of the rectangle” is 2 3, 3 4, 4 5, 5 6, … The pattern for the “area of the hole in the middle” is 1 1, 2 2, 3 3, 4 4, … Thus the nth figure will have (n + 1) (n + 2) n n = 3n + 2 tiles. figure number with 6044 tiles = (6044 2) 3 = 2014 18.
What is the least number of cuts required to cut 12 identical sausages so that they can be shared equally among 20 people? Solution Fraction of sausage each person will get =
=
This means that there must be at least 12 cuts since no one will get one whole sausage. Cut each of the 12 sausages at the -mark. Then 12 people will get one
sausage
each. 7
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There are 12 times sausages left. But each person must get of a sausage. This means that 4 of the
sausages must be cut into half each so that the remaining 8
people will get one sausage and one sausage each. least no. of cuts = 12 + 4 = 16 19.
In the following alphametic, all the different letters stand for different digits. Find the two-digit sum PI.
+
I
S
I
S
I
S
I
S
P
I
Solution If I 3, then the sum will be a 3-digit number. So I = 1 or 2. In the ones column, S + S + S + S = 4 S = _I. Since 4 S is even, then I must be even, so I = 2. But 4 S = _2 implies that S = 3 or 8. If S = 8, then 28 4 is a 3-digit number, so S = 3. Thus 23 + 23 + 23 + 23 = 92, i.e. P = 9. the two-digit sum PI is 91. 20.
A teacher has a bag of sweets to treat her class. If she gave 6 sweets to each student, then she would have 5 sweets left. If she gave 7 sweets to each student, then she would have 30 sweets short. How many students and how many sweets are there? Solution If the teacher gave 6 sweets to each student, then she would have 5 sweets left. But 5 sweets left are not enough to give one more sweet to each student since she would have 30 sweets short. So if she had 30 more sweets, then she could give 30 + 5 = 35 sweets to the students so that each student has one more sweet, i.e. 6 sweets each. Since each student receives only 1 more sweet (i.e. the 6th sweet), then there are 35 students. there are 35 6 + 5 = 215 sweets.
21.
A box contains 80 coloured pens: 36 black, 24 blue, 12 red and 8 green. Alice takes some pens from the box without looking at the colours of the pens. What is the least number of pens she must take so that she has at least 20 pens of the same colour?
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Solution Worst case scenario for all the pens taken out by Alice to be of different colours: 19 black, 19 blue, 12 red and 8 green, i.e. a total of 58 pens. Then the next pen taken out has to be either black or blue, i.e. Alice will have at least 20 pens of the same colour, either 20 black or 20 blue pens. least no. of pens Alice must take so that she has at least 20 pens of the same colour = 58 + 1 = 59 22.
Find the value of
.
Solution Observe the following pattern:
23.
.
The diagram shows a square being divided into four rectangles. If the sum of the perimeter of the four rectangles is 40 cm, find the area of the square.
Solution Perimeter of the four rectangles = 8 length of square = 40 cm So length of square = 5 cm Area of square = 25 cm2 24.
Given that 5! means 5 4 3 2 1, find the last digit of 2014!. Solution
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Since 2014! contains the factor 10, and any number multiplied by 10 will have 0 as the last digit, then the last digit of 2014! is 0. 25.
Two women, Ann and Carol, and two men, Bob and David, are athletes. One is a swimmer, a second is a skater, a third is a gymnast, and a fourth is a tennis player. On a day they were seated around a square table: a. The swimmer sat on Ann’s left. b. The gymnast sat across from Bob. c. Carol and David sat next to each other. d. A woman sat on the skater’s left. Who is the tennis player? Solution There are only two possibilities to satisfy the first two conditions: Bob Bob swimmer
swimmer
gymnast
Ann gymnast
Ann
Possibility 1
Possibility 2
Only Possibility 2 satisfies Condition c. Condition d says that Ann cannot be the skater since Bob is on her left. Ann must be the tennis player.
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SASMO 2014 Round 1 Primary 6 Solutions 1.
A clock takes 3 seconds to make 4 chimes. Assuming that the rate of chiming is constant, and the duration of each chime is negligible, how long does the clock take to make 8 chimes? Solution Time taken by the clock between 2 chimes = 3 s / 3 gaps = 1 s time taken by the clock to make 8 chimes = 1 s 7 gaps = 7 s
2.
If n is a whole number, for what values of n is
also a whole number?
Solution is a whole number if n is a factor of 36. 36 = 1 36 = 2 18 = 3 12 =49 =66 3.
is a whole number if n = 1, 2, 3, 4, 6, 9, 12, 18, 36.
Find the smallest whole number between 1 and 100 that is divisible by 12 and by 30. Solution Smallest whole number between 1 and 100 that is divisible by 12 and by 30 = LCM(12, 30) = 60
4.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more sweet. Catherine has enough money to buy only 26 sweets. What is the biggest number of sweets that she can get from the shop? Solution Method 1 26 sweets 26 wrappers 8 sweets and 2 wrappers 10 wrappers 3 sweets and 1 wrapper 4 wrappers 1 sweet and 2 wrappers biggest no. of sweets = 26 + 8 + 3 + 1 = 38 [Common mistakes: 34, 37] Method 2 26 sweets 8 sweets and 2 wrappers 2 sweets and 4 wrappers 2 sweets biggest no. of sweets = 26 + 8 + 2 + 2 = 38 [Common mistakes: 34, 36]
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Method 2 (Model Method) 4 kg Duck
1.2
Chicken The idea of average is to “even out”. This means that the 4-kg mark will divide 1.2 kg into 2 equal parts. mass of duck = 4 kg + 0.6 kg = 4.6 kg Method 3 (Algebra) Let the mass of the chicken be x kg. Then the mass of the duck is (x + 1.2) kg. Average mass of the duck and the chicken =
=4 2x + 1.2 = 8 2x = 6.8 x = 3.4
mass of duck = 3.4 + 1.2 = 4.6 kg 8.
A whole number is between 50 and 100. When it is divided by 5, the remainder is 3. When it is divided by 7, the remainder is 5. Find the number. Solution Since the number leaves a remainder of 3 when divided by 5, then it must end with 3 or 8, i.e. the possible values of the number are 53, 58, 63, 68, …, 98. Since the number leaves a remainder of 3 when divided by 5, we can either check each of the above possible values one by one, or we can start with 7 7 + 5 = 54, and then add 7 until we get a number ending in 3 or 8, i.e. 54, 61, 68, … the number is 68.
9.
A man travelled at 120 km/h for the first half of a 12-km journey. Then he travelled at 60 km/h for the rest of his journey. What is his average speed for the whole journey? Solution Time taken for first half of the 12-km journey = 6 km / 120 km/h = 0.05 h Time taken for second half of the 12-km journey = 6 km / 60 km/h = 0.1 h average speed for whole journey = 12 km / 0.15 h = 80 km/h Note: Common mistake is 90 km/h. Since speed = distance / time, if you want to average speeds, then the base (in this case, the time taken) must be the same. But instead the distance travelled for both parts of the journey is the same, so you cannot average speeds.
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10.
Amy buys an item for a 20% discount during a sale, but she still needs to pay a 5% GST (Goods and Services Tax). She is given two options. Option A: 20% discount first, then add 5% GST Option B: Add 5% GST first, then 20% discount Which option is cheaper for Amy? Or does it not matter? Solution Option A: 0.8 1.05 Sales Price Option B: 1.05 0.8 Sales Price both options are the same, so it does not matter. Note: Common mistake is Option A. If you pay 5% GST first (Option B), although this GST is bigger than the GST in Option A, this GST is also discounted 20%, so it ends up the same as if you pay the GST after the discount (Option A). Notice that the base is the same (in this case, the selling price).
11.
The digits 3, 4, 5, 6 and 7 are all used to write a five-digit number ABCDE where the three-digit number ABC is divisible by 4, BCD is divisible by 5 and CDE is divisible by 3. Find all the possible values of the five-digit number ABCDE. Solution Since BCD divisible by 5, then D = 5. Since ABC divisible by 4, then BC is also divisible by 4, i.e. BC = 36 or 64. Since CDE is divisible by 3, then C + D + E = C + 5 + E is also divisible by 3. If BC = 36, then C + 5 + E = 6 + 5 + E = 11 + E is also divisible by 3, i.e. E = 4 or 7, so A = 7 or 4 respectively. If BC = 64, then C + 5 + E = 4 + 5 + E = 9 + E is also divisible by 3, i.e. E = 3, so A = 7. the possible values of ABCDE are 43657, 73654 and 76453.
12.
At a workshop, there are 20 participants. Each of them shakes hand once with one another. How many handshakes are there? Solution The first participant will shake hand with 19 other participants; the second participant will shake hand with 18 other participants; the third participant will shake hand with 17 other participants; etc. Thus total no. of handshakes = 19 + 18 + 17 + … + 3 + 2 + 1 Method 1 1 + 19 = 20 2 + 18 = 20 3 + 17 = 20
9 pairs
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9 + 11 = 20 10 total no. of handshakes = 20 9 + 10 = 190 Method 2 Since 1 + 2 + 3 + … + n = 13.
, then total no. of handshakes =
= 190
In the following cryptarithm, different letters represent different digits and an asterisk * may represent any digit. The product MATH is a four-digit number less than 5000. What number does MATH represent? E
H
E
H
*
*
6
*
8
M A
T
+
*
H
Solution In the ones column, H H = _6 implies H = 6. In the tens column, E H = E 6 = _8 implies E = 3 or 8. If E = 8, then 86 86 > 5000. So E = 3. MATH = 36 36 = 1296. 14.
There are 100 buns to be shared among 100 monks (consisting of senior and junior monks). The senior monks get 3 buns each and every 3 junior monks share 1 bun. How many senior monks are there? Solution Method 1 (By Observation) Observe that if you group one senior monk and 3 junior monks together, they will get a total of 4 buns. And 25 such groups will give a total of 100 monks and 100 buns. there are 25 senior monks. Method 2 (Algebra) Let the no. of senior monks be x. Then the no. of junior monks is 100 x. Total no. of buns = 3x +
= 100
9x + (100 x) = 300 5
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8x = 200 x = 25 there are 25 senior monks. 15.
What is the least number of cuts required to cut 5 identical sausages so that they can be shared equally among 9 people? Solution Fraction of sausage each person will get = This means that there must be at least 5 cuts since no one will get one whole sausage. Cut each of the 5 sausages at the -mark. Then 5 people will get one sausage each. There are 5 times sausages left. But each person must get of a sausage. This means that one of the
sausages must be cut into 4 equal parts each (i.e. 3 cuts
each) so that the remaining 4 people will get one sausage and one sausage each. least no. of cuts = 5 + 3 = 8 16.
How many digits are there before the 50th ‘8’ of the following number? 85855855585555855555… Solution No. of ‘8’s before the 50th ‘8’ of the number = 49 No. of ‘5’s before the 50th ‘8’ of the number = 1 + 2 + 3 + … + 49 Method 1 1 + 49 = 50 2 + 48 = 50 3 + 47 = 50
24 pairs
24 + 26 = 50 25 total no. of digits before the 50th ‘8’ = 49 + 50 24 + 25 = 1274 Method 2 Since 1 + 2 + 3 + … + n = number = 49 +
, then total no. of digits before the 50th ‘8’ of the
= 1274
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17.
Find the total number of triangles in the diagram.
Solution
1
Label the vertices as shown in the diagram. These are the vertices that form a triangle: 123, 124, 125, 126, 134, 135, 137, 145, 156, 157, 158, 167, 168, 236, 256, 347, 356, 357, 367, 457, 458, 568, 578, 678 there is a total of 24 triangles in the diagram. 18.
3
2
4
5
7 8 6
A teacher has a bag of sweets to treat her class. If she gave 5 sweets to each student, then she would have 13 sweets left. If she gave 8 sweets to each student, then she would have 20 sweets short. How many students and how many sweets are there? Solution If the teacher gave 5 sweets to each student, then she would have 13 sweets left. But 13 sweets left are not enough to give 3 more sweets to each student since she would have 20 sweets short. So if she had 20 more sweets, then she could give 20 + 13 = 33 sweets to the students so that each student has 3 more sweets, i.e. 8 sweets each. Since each student receives only 3 more sweets (i.e. the 6th, 7th and 8th sweets), then there are 33 3 = 11 students. there are 11 5 + 13 = 68 sweets.
19.
Find the value of
.
Solution Method 1 Observe the following pattern:
. 7
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Method 2 Since
20.
for all n > 1, then
Susie has 23 coins. She divides them into 5 piles so that each pile has a different number of coins. Find the smallest possible number of coins in the biggest pile. Solution For each pile to have a different number of coins, and the biggest pile to have the smallest possible number of coins, put 1 coin in the 1st pile, 2 coins in the 2nd pile, 3 coins in the 3rd pile, 4 coins in the 4th pile and 5 coins in the 5th pile. So the biggest pile is the 5th pile, but there are only 1 + 2 + 3 + 4 + 5 = 15 coins. The 16th coin will have to go to the 5th pile so that each pile will have a different number of coins. The 17th coin cannot go to the 5th pile because we want to find the smallest possible number of coins in the biggest pile, so the 17th coin will have to go to the 4th pile. Similarly, the 18th, 19th and 20th coins will go to the 3rd, 2nd and 1st piles respectively. The 21st coin will go to the 5th pile again, the 22nd coin to the 4th pile, and the 23rd coin to the 3rd pile. the largest pile (which is the 5th pile) will contain 5 + 1 + 1 = 7 coins.
21.
Find the next term of the following sequence: 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, … Solution Observe that the first 3 terms are perfect squares: 12, 22 and 32. But subsequent terms are not perfect squares. However, if you compare the given sequence with perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, …), you will observe that the n-th term can be found by squaring n and then adding the digits continuously until a single-digit number is obtained. the next term, which is the 11th term, is obtained by: 112 = 121 1 + 2 + 1 = 4.
22.
Given that 5! means 5 4 3 2 1, find the last two digits of 2014!. Solution Since 2014! contains the factor 100, and any number multiplied by 100 will have 00 as the last two digits, then the last two digits of 2014! are 00.
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23.
There were 210 students in the hall. 2 of the boys and 1 of the girls wore T-shirts. If 5
3
there were 78 students in the hall who wore T-shirts, how many boys were in the hall? Solution Method 1 (Model Method) The first model shows 2 of the boys and 1 of the girls wearing T-shirts (indicated by 5
3
solid boxes, not the dotted ones). We will call an unshaded box from the boys as Bbox, and a shaded box from the girls as G-box.
Boys
B1
B1
B2
B2 210
Girls
G1
G2
The 2 solid B-boxes and the one solid G-box (i.e. B1 + B1 + G1) represent a total of 78 students. If you take 2 dotted B-boxes and 1 dotted G-box (i.e. B2 + B2 + G2), the 3 dotted boxes also represent a total of 78 students. But you can’t do anything with the remaining 2 dotted boxes. So we double all the boxes as shown in the second model, where the total no. of students is 210 2 = 420.
Boys
B1
Girls
G1
B1
B2
B2
B3
B3
B4
B4
B5
B5 420
G2
G3
G4
G5
From the second model, we are able to form 5 groups of 2 B-boxes and 1 G-box (i.e. from B1 and G1 to B5 and G5), leaving behind only one G-box, where each group of 2 B-boxes and 1 G-box represent a total of 78 students. This means that the last G-box represents 420 5 78 = 420 390 = 30 students. there are 30 3 = 90 girls and 210 90 = 120 boys in the hall. Method 2 (Algebra) Let the no. of boys be x. Then the no. of girls is 210 x. Total no. of students who wore T-shirts = 2 x + 1 (210 x) = 78 5
3
6x + 5(210 x) = 78 15 6x + 1050 5x = 1170 x = 120
there are 120 boys. 9
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24.
A farmer wants to plant 10 trees in 5 rows such that there are exactly 4 trees in each row. Draw a diagram to show how the trees should be planted. Solution If you start with 10 trees and then try fitting 5 lines onto them, you will realise that the lines must overlap because there are not enough trees, and it’s not easy to fit 5 lines onto 10 trees. try to draw 5 overlapping lines first, and a common figure with 5 overlapping lines is the following star:
Then put in the trees and yes, it works. 25.
Frank knows 5 women: Amy, Betty, Cheryl, Doris and Elaine. a. 3 women are under 30 and the other 2 women are over 30. b. 3 women are nurses and the other 2 women are teachers. c. Amy and Cheryl are in the same age bracket. d. Doris and Elaine are in different age brackets. e. Betty and Elaine have the same occupation. f. Cheryl and Doris have different occupations. g. Of the 5 women, Frank will marry the teacher over 30. Who will Frank marry? Solution Conditions a, c and d Amy and Cheryl are under 30, and Betty is over 30. Conditions b, e and f Betty and Elaine are nurses, and Amy is a teacher. 3 women under 30 Amy Cheryl ?
2 women over 30 Betty ?
3 Nurses Betty Elaine ?
2 Teachers Amy ?
So Amy is a teacher under 30. Conditions b and g the other teacher must be over 30. Condition f Cheryl or Doris is the other teacher over 30. But Cheryl is under 30, so Doris is the other teacher over 30. Frank will marry Doris.
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