Solutions – Middle Primary Division
1.
1. 20 20 + + 16 = 16 = 36,36,
hence (C). hence (C).
2.
2. (Also (Also UP1)UP1)
The numbers in order are 555, 556, 565, 566, 655, The numbers in order are 555, 556, 565, 566, 655,
hence (D). hence (D).
3.
3. The digit 3 is in thThe digit 3 is in the thousands posite thousands position, so it repreion, so it represents 30sents 3000,00,
hence (D). hence (D).
4.
4. I am I am 12 yea12 years old, so our combrs old, so our combineined ages are 6 + d ages are 6 + 12 = 1812 = 18,,
hence (C). hence (C).
5.
5. Here are all Here are all possible lpossible lines of syines of symmetry:mmetry:
Although the Z shape has a point of symmetry, it does not have a line of symmetry, Although the Z shape has a point of symmetry, it does not have a line of symmetry, hence (E). hence (E).
6.
6. (Also (Also UP2)UP2)
One pizza will have 4 quarters, so two pizzas will have 2
One pizza will have 4 quarters, so two pizzas will have 2××4 = 8 quarters,4 = 8 quarters,
hence (D). hence (D).
7.
7. AftAfter 30 miner 30 minutes iutes it is 5t is 5 pm, anpm, and after and after anotheother 15 minr 15 minutes iutes it is 5:15t is 5:15 pm,pm,
hence (E). hence (E).
8.
8. The nThe number oumber of wheelf wheels is s is 44××2 + 22 + 2 ××3 + 13 + 1 ××4 = 18,4 = 18,
hence (E). hence (E).
9.
9. (Also (Also UP6)UP6)
1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10
The opposite chair is both 5 places forward and 5 places back. The opposite chair is both 5 places forward and 5 places back. Five places back from chair 9 is chair 4,
Five places back from chair 9 is chair 4,
hence (D). hence (D).
10.
10. (Also UP5) (Also UP5)
Alterna
Alternative tive 11
In cents, 500
In cents, 500÷÷80 = 6r20 so that he buys 6 chocolates and has 20 cents left,80 = 6r20 so that he buys 6 chocolates and has 20 cents left,
hence (C). hence (C).
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Alternative 2 Alternative 2
Multiples of 80 are 80
Multiples of 80 are 80,,160160,,240240,,320320,,400400,,480480,,560. 560. FFrom thisrom this, he , he can afforcan afford 6 d 6 chchoco-oco-
lates but not 7, lates but not 7,
hence (C). hence (C).
11.
11. To make the total as large as possible, the large digits should have place value as To make the total as large as possible, the large digits should have place value as
large as possible. That is, 9 and 8 will be in the tens positions of the two numbers. large as possible. That is, 9 and 8 will be in the tens positions of the two numbers. Then 3 and 2 will be in the units positions.
Then 3 and 2 will be in the units positions.
Then the sum is either 93 + 82 = 175 or 92 + 83 = 175, both with total 175, Then the sum is either 93 + 82 = 175 or 92 + 83 = 175, both with total 175,
hence (A). hence (A).
12.
12. (Also J6) (Also J6)
Each time the paper is unfolded, the two parts will be reflections of each other through Each time the paper is unfolded, the two parts will be reflections of each other through the fold line.
the fold line.
hence (A). hence (A).
13.
13. (Also UP8) (Also UP8)
She either has a 50c coin or not. She either has a 50c coin or not.
If she has a 50c coin, then she has one other 10c coin: 50 + 10 = 60. If she has a 50c coin, then she has one other 10c coin: 50 + 10 = 60. If she has no 50c coins, then she either has 0, 1, 2 or 3 20c coins: If she has no 50c coins, then she either has 0, 1, 2 or 3 20c coins:
220 0 + + 220 0 + + 220 0 = = 660 0 220 0 + + 220 0 + + 110 0 + + 110 0 = = 6600 220 0 + + 10 10 + + 10 10 + + 10 10 + + 10 10 = = 60 60 110 0 + + 110 0 + + 110 0 + + 110 0 + + 110 0 + + 110 0 = = 6600 In all, there are 5 possibilities,
In all, there are 5 possibilities,
hence (D). hence (D).
14.
14. Since there are 3 colours, you can’t be sure that the first 3 beans include a pair. Since there are 3 colours, you can’t be sure that the first 3 beans include a pair.
However, with 4 beans, they can’t all be different colours, so there must be a pair of However, with 4 beans, they can’t all be different colours, so there must be a pair of the same colour.
the same colour.
So 4 beans (and no more) are needed to make sure you have a pair, So 4 beans (and no more) are needed to make sure you have a pair,
hence (B). hence (B).
15.
15. (Also UP11) (Also UP11)
Starting from the outer end of the spiral (the loop on the rope) the dark and light Starting from the outer end of the spiral (the loop on the rope) the dark and light sections are longest, and the light sections are of similar length to the dark sections. sections are longest, and the light sections are of similar length to the dark sections. As you move towards the other end of the rope, both dark and light sections get As you move towards the other end of the rope, both dark and light sections get shorter. Only rope (A) shows this,
shorter. Only rope (A) shows this,
hence (A). hence (A).
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16.
16. There are 3 + 6 + 2 + 1 + 4 = 16 students, so half the class is 8 students. There are 3 + 6 + 2 + 1 + 4 = 16 students, so half the class is 8 students.
The 6 with orange hats are in one-half of the class, and so the other two students in The 6 with orange hats are in one-half of the class, and so the other two students in that half of the class have black hats.
that half of the class have black hats.
So the only way to split the class into two equal groups is with orange and black in So the only way to split the class into two equal groups is with orange and black in one group and red, green and yellow in the other,
one group and red, green and yellow in the other,
hence (A). hence (A).
17.
17. The sum of The sum of the first five digithe first five digits is ts is 22. 22. TheThereforefore the re the sum of the sum of the laslast two digit two digits mustts must
be 12, as 34
be 12, as 34−−22 = 12. 22 = 12. TherThere are 7 e are 7 possipossibilbilitiities for the last twes for the last two digito digits: s: 39, 93, 48,39, 93, 48,
84, 57, 75, and 66, 84, 57, 75, and 66,
hence (B). hence (B).
18.
18. (Also UP12) (Also UP12)
Th
The e lalarge squrge square muare must havst have e siside 8de 8 cmcm. . TheThen n alall l of the of the corcornerners s of the of the pipieceeces s in thein the tan
tangragram lie on a m lie on a grigrid of d of 22 cmcm××22 cm squcm squaresares..
The shaded square has the same area as 2 of the grid squares, or 2
The shaded square has the same area as 2 of the grid squares, or 2××(2cm(2cm××22 cmcm) =) =
88 cmcm22,,
hence (D). hence (D).
19.
19. No matter how a single fold is made, there will be one of the original right-angle No matter how a single fold is made, there will be one of the original right-angle
cor
cornerners s thathat t is on is on the bounthe boundardary y of the shapeof the shape. . Of the 5 Of the 5 shshapesapes, , (E) does not hav(E) does not havee an
any right angly right angles, so it es, so it can’can’t bt be e madmade e witwith a h a sinsingle foldgle fold. . The other four figureThe other four figures cans can be made with a single fold:
be made with a single fold:
((AA)) ((BB)) ((CC)) ((DD))
hence (E). hence (E).
20.
20. Alternative 1 Alternative 1
There are 5 ways of making 12 from these numbers: There are 5 ways of making 12 from these numbers: 7 + 4 + 1 = 12
7 + 4 + 1 = 12,, 7 + 3 7 + 3 + 2 = 12+ 2 = 12,, 6 + 5 6 + 5 + 1 = 12+ 1 = 12,, 6 + 6 + 4 + 2 = 124 + 2 = 12,, 5 + 5 + 4 + 3 = 124 + 3 = 12
The number 4 is in 3 of these sums, and not in 1 + 5 + 6 and 2 + 3 + 7. The number 4 is in 3 of these sums, and not in 1 + 5 + 6 and 2 + 3 + 7. The number 4 could be either in a corner, a side or the centre:
The number 4 could be either in a corner, a side or the centre: (i)
(i) 44 (ii)(ii) 44 (iii)(iii)
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However, the highlighted lines in each can only be 1+5+6 and 2+3+7 which don’t However, the highlighted lines in each can only be 1+5+6 and 2+3+7 which don’t have a number in common, so (i) and (ii) don’t work. Placing 4 in the centre, there have a number in common, so (i) and (ii) don’t work. Placing 4 in the centre, there are several ways to arrange the other numbers:
are several ways to arrange the other numbers:
44 1 1 5 5 66 2 2 3 3 77 44 1 1 6 6 55 3 3 2 2 77 hence (D). hence (D). Alternative 2 Alternative 2 All sev
All seven en circcircles add les add to to 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 + 7 7 = = 28. 28. The top The top and bottom and bottom rorowsws together add to 2
together add to 2××12 = 24. So the middle circle must be 2812 = 24. So the middle circle must be 28−−24 = 4,24 = 4,
hence (D). hence (D).
21.
21. Since each team played three games, we can complete the record for the Eagles, Since each team played three games, we can complete the record for the Eagles,
Falcons and Condors. Falcons and Condors.
Pl
Plaayyed ed WWiin n DDraraw w LoLosss s PPoioinntsts E Eaaggllees s 3 3 3 3 0 0 0 0 99 H Haawwkks s 33 F Faallccoonns s 3 3 0 0 1 1 2 2 11 C Coonnddoorrs s 3 3 0 0 1 1 2 2 11 The Eagles won all their games, so the Hawks lost to the Eagles. The Eagles won all their games, so the Hawks lost to the Eagles.
The Falcons and the Condors had no wins, so the game between them must have The Falcons and the Condors had no wins, so the game between them must have been drawn
been drawn. . TherThereforefore e both lost to both lost to the Hawkthe Hawks.s.
The Hawks’ three games were 2 wins and 1 loss, for 6 points, The Hawks’ three games were 2 wins and 1 loss, for 6 points,
hence (C). hence (C).
22.
22. Label the points as shown and work down the diagram. Label the points as shown and work down the diagram. P P
Q Q A A B B C C X X Y Y Z Z
There is 1 route from
There is 1 route from P P toto AA, and also 1 route from, and also 1 route from P P toto X X .. Routes to
Routes to BB will come through either will come through either AA oror X X , which gives only 2, which gives only 2 routes. Likewise, there are 2 routes to
routes. Likewise, there are 2 routes to Y Y .. Routes to
Routes to C C will come through either will come through either BB oror Y Y . There are 2 routes. There are 2 routes to
to BB and 2 routes to and 2 routes to Y Y , so there are 4 routes to, so there are 4 routes to C C . Likewise, there. Likewise, there are 4 routes to
are 4 routes to Z Z .. Routes to
Routes to QQ are the 4 routes that come through are the 4 routes that come through C C plus the 4 plus the 4 routes that come through
routes that come through Z Z , making 8 routes,, making 8 routes,
hence (D). hence (D).
23.
23. (Also UP21) (Also UP21)
The discs are back in their original positions after 5 moves. The discs are back in their original positions after 5 moves.
rr b b gg y y oo 11 − −→→ gg y y oo rr b b 22 − −→→ oo rr b b gg y y 33 − −→→ b b gg y y oo rr 44 − −→→ y y oo rr b b gg 55 − −→→ rr b b gg y y oo
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They will be in their original positions again after 10, 15 and 20 moves. After 1 more They will be in their original positions again after 10, 15 and 20 moves. After 1 more move, blue will be on the bottom,
move, blue will be on the bottom,
hence (B). hence (B).
24.
24. (Also UP22) (Also UP22)
Alternative 1 Alternative 1
Replacing the leopard by another lion (of the same weight as the lion) would add Replacing the leopard by another lion (of the same weight as the lion) would add 90
90 kg, and replkg, and replaciacing the tiger by anothng the tiger by another lion wouer lion would add 50ld add 50 kg. kg. Then 3 lions weThen 3 lions weighigh 310
310 + 90 + 90 + 50 = 450+ 50 = 450 kg kg and 1 and 1 liolion n weweighs 450ighs 450÷÷3 = 1503 = 150 kgkg,,
hence (B). hence (B).
Alternative 2 Alternative 2
If the lion wei
If the lion weighs 100ghs 100 kg, then the leopkg, then the leopard weiard weighs 10ghs 10 kg and the tiger 50kg and the tiger 50 kg for a kg for a totatotall of 160
of 160 kgkg. . ThThis is 150is is 150 kg too ligkg too lightht. . AdAddiding 150ng 150÷÷3 = 503 = 50 kg to kg to eaceach weighh weight keeps thet keeps the
diff
differenerences in ces in weweighight t the same. the same. So the So the liolion n weweighighs s 150150 kg,kg,
hence (B). hence (B).
25.
25. Jane must have given away her $2 coin, otherwise Tom would have $4 or more. Tom Jane must have given away her $2 coin, otherwise Tom would have $4 or more. Tom
must end up with an even number of cents, so he must have given away his 5c coin. must end up with an even number of cents, so he must have given away his 5c coin. With just these two coins given to Angus, Jane has $1.85 and Tom has $3.80. So it With just these two coins given to Angus, Jane has $1.85 and Tom has $3.80. So it can’t be done with just 2 coins given to Angus.
can’t be done with just 2 coins given to Angus.
It can be done with 3 coins: if Tom gives away his 5c and his 10c he has $3
It can be done with 3 coins: if Tom gives away his 5c and his 10c he has $3..70, and70, and if Jane gives away her $2, she has $1.85,
if Jane gives away her $2, she has $1.85,
hence (B). hence (B).
26.
26. This table lists the numbers according to their first two digits. This table lists the numbers according to their first two digits.
F
Fiirrsst t SSeeccoonnd d ddiiggiitt d diiggiit t 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 C8 Coouunntt 1 1 11001 1 11112 2 11223 3 11334 4 11445 5 11556 6 11667 7 11778 8 11889 9 99 2 2 22002 2 22113 3 22224 4 22335 5 22446 6 22557 7 22668 8 22779 9 88 3 3 33003 3 33114 4 33225 5 33336 6 33447 7 33558 8 33669 9 77 4 4 44004 4 44115 5 44226 6 44337 7 44448 8 44559 9 66 5 5 55005 5 55116 6 55227 7 55338 8 55449 9 55 6 6 66006 6 66117 7 66228 8 66339 9 44 7 7 77007 7 77118 8 77229 9 33 88 880088 881199 22 9 9 990099 11 45 45 The total number of possibilities is 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45, The total number of possibilities is 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45,
hence (45). hence (45).
27.
27. Once the age of the younger daughter is known, the other four ages and the total can Once the age of the younger daughter is known, the other four ages and the total can
be calculated. Here are the first few possibilities: be calculated. Here are the first few possibilities:
Y
Yououngenger r dadaughughter ter OlOlder der daudaughghter ter YYounounger ger soson n OlOlder der soson n TTotaotall 0 0 22 44 77 1133 1 1 33 66 99 1199 2 2 44 88 1111 2255
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This pattern conti
This pattern continuesnues, with the , with the total going up in 6s, total going up in 6s, since for each +1 on since for each +1 on the youngerthe younger daughter, there is +1 on the older daughter and +2 on both sons.
daughter, there is +1 on the older daughter and +2 on both sons.
A total of 55 requires 30 more than in the last line above, which will happen 5 rows A total of 55 requires 30 more than in the last line above, which will happen 5 rows later. The final line shows:
later. The final line shows: Y
Yououngenger r dadaughughter ter OlOlder der daudaughghter ter YYounounger ger soson n OlOldeder r sonson Total Total 7
7 99 1188 2211 5555
hence (7). hence (7).
28.
28. If A and B are both chosen, then the third stamp is either D or E. (2 possibilities) If A and B are both chosen, then the third stamp is either D or E. (2 possibilities)
If A but not B is chosen, then D must be chosen, and either C or E. (2 possibilities) If A but not B is chosen, then D must be chosen, and either C or E. (2 possibilities) If B but not A is chosen, then E must be chosen, and either D or F. (2 possibilities) If B but not A is chosen, then E must be chosen, and either D or F. (2 possibilities) If neither A nor B is chosen, then the 3 stamps must be in a row of three: CDE or If neither A nor B is chosen, then the 3 stamps must be in a row of three: CDE or DEF. (2 possibilities)
DEF. (2 possibilities)
In all there are 8 possibilities, In all there are 8 possibilities,
hence (8). hence (8).
29.
29. (Also UP28) (Also UP28)
The first cube
The first cube useuses s 12 match12 matches, then each subsees, then each subsequenquent t cube uses 8 cube uses 8 matcmatches. hes. SinSincece 2016
2016−−12 = 2004 and 200412 = 2004 and 2004÷÷8 = 2508 = 250r4, ther4, there are 1re are 1 ++ 250 = 2250 = 251 cubes mad51 cubes made, wite, with 4h 4
matches left over, matches left over,
hence (251). hence (251).
30.
30. (Also J20) (Also J20)
Consider the prime factorisation of 2016 = 2 Consider the prime factorisation of 2016 = 255
×
×3322××7. The factors of 2016 under 107. The factors of 2016 under 10
are 1, 2, 3, 4, 6, 7, 8 and 9. are 1, 2, 3, 4, 6, 7, 8 and 9.
Only 7 has prime factor 7, so this must be one of the ages. Only 7 has prime factor 7, so this must be one of the ages. The 3
The 322
in the prime factorisation will either be from 3
in the prime factorisation will either be from 3××6 6 = = 22××3322 or from 9 = 3or from 9 = 322..
In the first case, the factorisation is 3
In the first case, the factorisation is 3××66××77××16, where 16 is too large.16, where 16 is too large.
In the second case, two of the ages are 7 and
In the second case, two of the ages are 7 and 9. 9. Then the remaininThen the remaining two ages mulg two ages multiplytiply to
to 2255
= 32, so they must be 4 and 8. = 32, so they must be 4 and 8.
Hence the ages are 4, 7, 8 and 9, which add to 28, Hence the ages are 4, 7, 8 and 9, which add to 28,
hence (28). hence (28).
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