10 questions. Try each one, then open the answer to check yourself and read the explanation.
Obtain the product of 11002 and 1012
Answer: A (1111002)
11002 = 12 and 1012 = 5, product 60. 60 = 32 + 16 + 8 + 4 = 1111002. (22205 = 310 and 11447 = 424, so they are wrong.)
Evaluate log8128 + log39
Answer: C (13/3)
128 = 87/3, so log8128 = 7/3. log39 = 2. Total 7/3 + 2 = 13/3.
Find the value of y if ½ log3y = 2
Answer: D (81)
log3y = 4, so y = 34 = 81.
The universal set U consists of all integers. Subsets of U are defined as: A = {y : y ≤ 3}, B = {y : −5 < y < 12}, C = {y : −2 ≤ y < 5}. A∩ (B∪C)' is
Answer: A ({y < −4})
B∪C = {−5 < y < 12}, so (B∪C)' = {y ≤ −5 or y ≥ 12}. Intersecting with A (y ≤ 3) leaves y ≤ −5, which for integers is {y < −4}.
Make k is the subject of the formula m = 2nk/p + k/(2p)
Answer: B (k = 2mp/(4n + 1))
Multiply by 2p: 2mp = 4nk + k = k(4n + 1). So k = 2mp/(4n + 1).
Which of the following is a perfect square?
Answer: D (x2 + 2x + 1)
x² + 2x + 1 = (x + 1)²; none of the others factorise as a square.
What is the highest possible value of 8/(1 + x2) if 0 ≤ x ≤ 3?
Answer: A (8)
The fraction is largest when the denominator is smallest, at x = 0: 8/1 = 8.
The fifth term in the progression 9, 27, 81, … is
Answer: C (729)
a = 9 and r = 3: T5 = 9 × 34 = 729.
The interior angles of an hexagon are 120°, 100°, 80°, 150°, x and 130°. The value of x is
Answer: D (140°)
A hexagon's angles add to 720°. 120 + 100 + 80 + 150 + 130 = 580, so x = 140°.
If the bearing of a town B from A is 145°, the bearing of A from B is
Answer: B (325°)
A back bearing differs by 180°: 145° + 180° = 325°.