1. A cylinder 45 cm high has a circumference of 15 cm. A string makes exactly 4 complete turns round the cylinder while its two ends touch the cylinder's top and bottom. How long is the string in cm?
Solution
Consider 1 revolution. The height of the new cylinder would be 1
4 of the height of the original cylinder, i.e. 45
4 cm. If we cut the cylinder along the line of the string we’d get the following shape:
Let x be the length of the string for 1 revolution. By Pythagoras,
2
2 2 45
15 4
x = + ⎜ ⎟⎛⎝ ⎞⎠ . Hence 75
x= 4 . For 4 revolutions, the length of the string is 75
4 75
× 4 = cm.
ANS:75 cm 2. What is the sum of all integers between 500 and 1500 which are divisible neither by 2 nor by
5?
Solution
The sum is
501+503+507+509+511+513+517+519+…+1491+1493+1497+1499
=(501+503+507+509)+(511+513+517+519)+…+(1491+1493+1497+1499)
=2020+2060+2100+…+5980
=(2020 5980) 100 2
+ ×
=400000
ANS:400000 3. The symbol n! is used to represent the product n× − × − × × × ×(n 1) (n 2) L 3 2 1. For example,
4! 4 3 2 1= × × × . Find n satisfying n! 2= 17× × × × × × ×39 54 7 11 13 17 193 . Solution
Since the largest prime factor of n! is 19, n=19, 20, 21 or 22. Since n! has 7 as a factor, then 3 n is at least 21. Because 11 is not a factor of n! and 11 is a factor of n!, n is less than 22. So 2 n=21.
ANS:21 4. In the figure below, the rectangle at the corner measures 3 cm by 6 cm. What is the radius of
the circle in cm?
Solution
Using Pythagoras’ theorem for the right-angled triangle in the figure alongside,
r2 = (r – 3)2 + (r – 6)2 r2 – 18r + 45=0 (r – 15)(r – 3)=0 r = 15 or r = 3
Clearly r≠3, so r = 15 cm.
ANS:15 cm
45 cm 45
4 15
6 3
r r – 6 r – 3
5. Divide 2008 marbles into a number of bags so that I can ask for any number of marbles from 1 to 2008, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?
Solution
We can divide 2008 marbles into 11 bags as follows:
20=1, 21=2, 22=4, 23=8, 24=16, 25=32, 26=64, 27=128, 28=256, 29=512, 210=1024.
Thus when we need n marbles, we just need to convert n from a base-10 integer numeral to its base-2 (binary) equivalent.
ANS:11 6. Let a be a real number such that 3
3a 1 0
− + =a . What is the value of 3 13 3 a −a + ? Solution
Since 3
3a 1 0
− + =a , 3
3a 1
− = −a , i.e. 1 1 a 3
− = −a . Because
3 3
3
1 3 1
3
a a a
a a a
⎛ − ⎞ = − + −
⎜ ⎟
⎝ ⎠ ,
3 3
3
1 1 3 1 28
3 1
27 27
a a a
a a a
⎛ ⎞
− =⎜ − ⎟ + − = − − = −
⎝ ⎠ . Hence 3 13 53
3 27 a −a + = .
ANS: 53 27 7. In an office, there are 14 desks of four types: one-drawer, two-drawer, three-drawer and
four-drawer respectively. There are 33 drawers altogether in those desks. How many
one-drawer desks are there, if it is known that there are as many of them as the two-drawer and three-drawer desks altogether?
Solution
Let w, x, y and z be the numbers of one-drawer, two-drawer, three-drawer and four-drawer desks, respectively. Then we have
14 (1)
2 3 4 33 (2)
(3) w x y z
w x y z
w x y + + + =
⎧⎪ + + + =
⎨⎪ = +
⎩
From (1) and (3), we obtain 2w+z=14. Hence z=14－2w. Therefore w+2x+3y+4z =w+2(x+y)+y+4z
=w+2w+y+4(14－2w)
=56+y－5w
Hence 56+y－5w=33 in view of (2). Hence y=5w－23.
Since y>0, we have w≥5. Since z>0 and z=14－2w, we obtain w≤6. Therefore w=5 or w=6.
If w=6, then z=14－2w=2, y=5w－23=7 and x=w－y=-1 which is impossible.
If w=5, then z=14－2w=4, y=5w－23=2 and x=w－y=3. Thus the number of one-drawer desks is 5.
ANS: 5
8. In a circumference a right triangle ∆ABC with hypotenuse AB is inscribed. On the longer leg BC is chosen a point D so that AC = BD. Find the angle EDC, if E is the midpoint of the arc ACB.
Solution
Since E is the middle of the arc ACB (AB is a diameter of the circle ), the triangle ∆AEB is isosceles and right, which means that AE=BEand ∠EAB=45°. ∠CAE= ∠DBE, since they are inscribed and responding to one and the same arc. Therefore ∆ECA≅ ∆EDB, which means that CE=DE. Considering the fact that
45o ECB EAB
∠ = ∠ = we obtain that the triangle ∆CED is isosceles with two angles of 45°. Hence the angle at the vertex E is right i.e. ∠DEC= °90 .
Since CE=ED, we have ∠EDC =45°
E
D C
A B
