Download solutions - NLCS Maths Department

1.
2 × 4 + 2 = (2 × 4) + 2 = 10; the other options all give the answer 12.
2.
The five integers are “consecutive”.
The third term is the middle term and so is equal to the average 10, while the second and fourth
terms are equal to 10 − d and 10 + d, where d = 1 is the common difference.
NOTE: (a) There is no need to work out the terms of the sequence.
(b) The same idea will work for any AP: the sum of two terms the same
distance either side of the middle term is equal to twice the average.
3.
240 : 120 = 2 : 1 = 110 : 55; 300 : 180 = 5 : 3 = 350 : 210. But 320 : 200 = 8 : 5.
4.
The sum, 60, is 5 times the middle number. So the middle number is 12. The numbers are
8, 10, 12, 14, 16.
Note: consecutive even numbers form an arithmetic sequence and the middle term of such a
sequence with an odd number of terms is always the mean of all the terms.
5.
Of the thirty pupils, 8 are left-handed. In the extreme case, these will all be girls. So the smallest
number of girls who could be right-handed is 6.
6.
Multiplying the decimal part of each option by 8 gives 0.40, 1.20, 2, 2.8 and 3.6.
7.
The prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19.
8.
If January 1st (or, in a leap year, January 1st or January 2nd) falls on a Sunday, then there will
be 53 Sundays in that particular year.
9.
The two shaded triangles will
overlap when the ‘net’ is folded.
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10.
11.
12.
The units digit of the product is 3; therefore the remainder when that number is divided by 5 is
also 3.
1
2
= 0·012345 + 0·00000067…; therefore
= 0·024690 + 0·0000013…
81
81
= 0·0246913…= 0·024691 to six decimal places.
Annakin has 5 coins and Obi-Wan Kenobi has 7.
13.
The shaded area in Y is clearly half the area of the
hexagon and the diagram above demonstrates
why the shaded areas in X and Z are also both equal to
half the area of the hexagon.
14.
100 hours equal four days and four hours so the candle will burn out at 4pm on Thursday.
15.
Since 107 is not divisible by 11, then neither is 107 − 11 nor 107 + 11. Therefore, the correct
answer is either B or D.
107 − 1 = 9 999 999 = 11 × 909090 + 9. Therefore
107 + 1 = 11 × 909090 + 11. Hence D is the correct answer.
16.
There are six quadrilaterals shaped
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and six shaped
.
2
17.
9
a
b c n
d
e f
5
Let a, b, c, d, e, f be the numbers in the squares shown.
Then the sum of the numbers in the four lines is
1 + 2 + 3 + … + 9 + b + n + e since each of the
numbers in the corner squares appears in exactly
one row and one column.
So 45 + b + n + e = 4 × 13 = 52, that is b + n + e = 7.
Hence b, n, e are 1, 2, 4 in some order.
If b = 2 then a = 2; if b = 4 then a = 0. Both cases
are impossible, so b = 1 and a = 3.
This means that n = 2 or n = 4. However, if n = 2
then c = 10, so n = 4 and c = 8.
(The values of the other letters are e = 2, d = 7, f = 6.)
18.
To have lost the fewest games, there need to be as many draws as possible. However, since 21 is
odd, there needs to be at least one win. That is possible with 8 draws, leaving 11 games lost.
19.
Since the numbers offered are all odd, they need to be the sum of an even prime and an odd
prime. But 2 is the only even prime so we need a number which is not two more than a prime.
Since 13 = 2 + 11; 33 = 2 + 31; 43 = 2 + 41 and 73 = 2 + 71, we have 53 = 2 + 3 × 17 as the
answer.
20.
The area of a rhombus ABCD is equal to half the product of the diagonals AC.BD.
[A rhombus is a special parallelogram, so the diagonals AC and BD bisect each other − at M
say. The diagonal AC cuts the rhombus into two congruent isosceles triangles ABC and CDA.
The line joining the apex B of the isosceles triangle ABC to the mid-point M of the base AC is
perpendicular to AC;  ΔABC has base AC and height BD/2, so has area (AC/2).(BD/2).
 the rhombus ABCD has area (AC.BD)/2.]
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21.
The diameter of the smaller circle and the radius of the larger circle are equal. Letting the radius
1
of the smaller circle be r, the areas are πr2 and π (2r)2 = 4πr2 so the smaller circle is of the
4
larger.
Note: this is an example of the property that in an enlargement the area scale factor is the
square of the linear factor.
22.
17 hours and 21 minutes = 1041 minutes and the average speed is therefore slightly less than 40
decimal places per minute.
23.
x − y = 9; x + y = 99. Adding gives 2x = 108 and therefore x = 54 and hence y = 45.
The ratio x : y = 54 : 45 = 6 : 5
(As we are interested only in the ratio rather than the numbers themselves, the problem could be
reduced to finding the ratio of two numbers which differ by 1 and have sum 11.)
24.
The average time each man took per seat is approximately (27 × 60 × 60 ÷ 32000)
seconds = (27 × 9 ÷ 80) seconds ≈ 3 seconds.
25.
Let AB and AD be of length b and h respectively. Then the area of ABCD = bh and the area of
1  h h  5bh
5
ABQP = b   
. Thus the required ratio =
.
2  2 3  12
12
26.
The percentage increase =
27.
Since the four numbers are symmetrical about the mean of the group, they must have the same
mean, 21, giving their total as 84. Alternatively, it is possible to state that the original numbers
must be 15, 17, 19, 21, 23, 25, 27 (since the numbers are given as consecutive and odd). Hence
the required sum is 15 + 19 + 23 + 27 = 84.
28.
For each row that the apprentice plants, the gardener has planted one more in his previous row.
The apprentice plants 50 rows and in addition the gardener plants the last one.
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58.1
100  400% .
15.2
4
29.
2002 = 2 × 7 × 11 × 13; 22 + 72 + 112 + 132 = 343
30.
The only two-digit cubes are 27 and 64. If 1 across is 64, then 1 down must also be 64 (the only
square between 61 and 69 inclusive), but there must be a different digit in each square so this is
impossible. Hence 1 across is 27 and 1 down is 25. The possibilities for 2 down are now 71, 73
and 79 giving 51, 53 and 59 as possibilities for 3 across. Of these, only 53 = 22 + 72 can be
written as the sum of two squares, so 2 down is 73 and 3 across 53.
31.
The number of inches in a furlong is 220 × 36. The number of barleycorns in a furlong is 10 ×
10  44  54
44 × 54. So the number of barleycorns in one inch is
, which evaluates to give 3.
220  36
32.
Second prize was 18 × £4.25 − 18 × 25p = 18 × £4 = £72. So in 2000, second prize was £22
more than first prize!
33.
Point F must be reached directly from B, C or E. There is one route from A to B, one route from
A to C, and there are three routes from A to E. So the number of possible routes is 1 + 1 + 3 = 5.
34.
For ropes of the same material and same length, the weight is directly proportional to the
cross-sectional area and hence to the square of the diameter of the rope. So the weight, in kg, of
the second rope
112
121 121
 2.7  2  2.7 

4.
81 30
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35.
Let the fraction of her lolly that Milly has eaten be x. Then Molly has eaten
This means that the fractions they have left are (1 − x) and (1 –
So 1 –
36.
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x of her lolly.
2
1
x) respectively.
2
1
4
x = 3 (1 − x), that is 5x = 4, giving x  .
2
5
Let the weights of squares, triangles and circles be s, t, c respectively. Then c + 3t = 6s; and 4c +
2t = 8s, so 2c + t = 4s. Hence (c + 3t) + (2c + t) = 10s.
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37.
There are eight 2-digit numbers which satisfy the required condition (12, 23, … , 89), seven
3-digit numbers (123, 234, … , 789), six 4-digit numbers (1234, 2345, … , 6789) and one
5-digit number (12345).
38.
Four lengths of the indoor pool are equivalent to five lengths of the outdoor pool. So Rachel
swam four ninths of the 63 days, that is 28 days, in the indoor pool.
39.
The first large sheet of paper will hold pages 1, 2, 19 and 20; the second will hold pages 3, 4, 17
and 18; the third will hold pages 5, 6, 15 and 16.
40.
A square of area 4 cm2 has side 2 cm and therefore its perimeter is 8 cm. A square of area 1 cm2
has side 1 cm, so its perimeter is 4 cm.
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