45 questions. Try each one, then open the answer to check yourself and read the explanation.
A rectangular picture 6cm by 8cm is enclosed by a frame ½cm wide, calculate the area of the frame.
Answer: A (15sq.cm)
With the frame the outer rectangle is 7cm by 9cm = 63sq.cm. The picture is 6 × 8 = 48sq.cm, so the frame is 63 − 48 = 15sq.cm.
Multiply (x + 3y + 5) by (2x2 + 5y + 2)
Answer: B (2x3+6yx2+5xy+15y2+31y+10x2+2x+10)
Multiply each term of the first bracket by each term of the second: x gives 2x³ + 5xy + 2x, 3y gives 6x²y + 15y² + 6y, and 5 gives 10x² + 25y + 10. Collecting the y terms gives 31y.
In the figure below, DE//BC, DB//FE, DE = 2cm, FC = 3cm, AE = 4cm. Determine the length of EC
Answer: B (6cm)
DBFE is a parallelogram, so BF = DE = 2cm and BC = 5cm. Triangles ADE and ABC are similar, so AE/AC = DE/BC: 4/(4 + EC) = 2/5, giving 4 + EC = 10 and EC = 6cm.
In the figure below, AB is parallel to CD then x + y + z =
Answer: E (360°)
Draw a line through the vertex z parallel to AB and CD. The co-interior angles on each side add to 180°, so x + y + z = 180° + 180° = 360°.
Arrange 3/5, 9/16, 34/59 and 71/97 in ascending order of magnitude.
Answer: C (9/16, 34/59, 3/5, 71/97)
As decimals: 9/16 = 0.5625, 34/59 ≈ 0.576, 3/5 = 0.6 and 71/97 ≈ 0.732. Smallest to largest gives option C.
The sum of the progression is 1 + x + x2 + x3 + . . . =
Answer: A (1/(1 − x))
This is a geometric progression with first term 1 and ratio x. For |x| < 1 the sum to infinity is a/(1 − r) = 1/(1 − x).
The number of telephone calls N between two cities A and B varies directly as the population PA, PB in A and B respectively and inversely as the square of the distance D between A and B. Which of the following equations represents this relation?
Answer: B (N = KPAPB/D2)
Varying directly as both populations puts PAPB on top, and inversely as the square of the distance puts D² underneath, with one constant K.
Find a square root of 170 − 20√30
Answer: C (2√5 − 5√6)
(2√5 − 5√6)² = 20 + 150 − 2(2√5)(5√6) = 170 − 20√30, so 2√5 − 5√6 is a square root (the other one is 5√6 − 2√5).
If x2 + 4 = 0, then x =
Answer: E (none of these.)
x² = −4, and no real number has a negative square, so none of the given values works.
In the figure below, FD//AC. The area of AEF = 6sq.cm, AE = 3cm, BC = 3cm, CD = 5cm, <BCD is an obtuse angle. Find the length of BD.
Answer: D (2√13)
½ × 3 × h = 6 gives the distance between the parallel lines h = 4cm. From CD = 5cm, D is 3cm beyond C along the line, so BD is the hypotenuse of a right triangle with sides 4 and 3 + 3 = 6: BD = √52 = 2√13.
What is the number whose logarithm to base 10 is 2̄.3482?
Answer: E (0.02229.)
The mantissa .3482 gives the digits 2229 from antilog tables. A characteristic of bar 2 (−2) puts the decimal point to give 0.02229.
If y = 2x2 + 9x − 35, find the range of values for which y < 0.
Answer: A (−7 < x < 5/2)
2x² + 9x − 35 = (2x − 5)(x + 7), which is zero at x = −7 and x = 5/2. The parabola opens upwards, so it is negative between the roots.
Father reduced the quantity of food bought for the family by 10% when he found that the cost of living had increased by 15%. Thus the fractional increase in the family food bill is now
Answer: D (7/200)
New bill = 90% of the quantity at 115% of the price = 0.9 × 1.15 = 1.035 of the old bill. The increase is 0.035 = 7/200.
Given that a*b = ab + a + b and that a°b = a + b + 1, find an expression (not involving * or °) for (a*b)°(a*c) if a, b, c, are real numbers and the operations on the right are ordinary addition and multiplication of numbers.
Answer: E (ab + ac + 2a + b + c + 1)
a*b = ab + a + b and a*c = ac + a + c. Joining them with ° adds them and adds 1: ab + ac + 2a + b + c + 1.
If a circular paper disc is trimmed in such a way that its circumference is reduced in the ratio 2:5, in what ratio is the surface area reduced?
Answer: D (4:25)
Circumference is proportional to the radius and area to the radius squared, so the areas are in the ratio 2² : 5² = 4 : 25.
If the four interior angles of a quadrilateral are (P + 10)°, (P − 30)°, (2P − 45)°, and (P + 15)°, then P is
Answer: B (82°)
The angles of a quadrilateral add up to 360°: 5P − 50 = 360, so 5P = 410 and P = 82°.
Simplify (a − b)/(a + b) − (a + b)/(a − b)
Answer: C (−4ab/(a2 − b2))
Over the common denominator (a + b)(a − b) = a² − b², the top is (a − b)² − (a + b)² = −4ab.
In the figure above, O is the centre of the circle ABC <CED = 30°, <EDA = 40°. What is the size of <ABC?
Answer: D (125°)
The lines EC and DA cross at O, so in triangle EOD the angle at O is 180° − 30° − 40° = 110°, and the vertically opposite angle AOC is 110°. The angle at the circumference on the minor arc is 180° − ½(110°) = 125°.
A force of 5 units acts on a particle in the direction to the east and another force of 4 units acts on the particle in the direction north-east. The resultant of the two forces is
Answer: C (√(41 + 20√2) units)
The angle between the forces is 45°, so R² = 5² + 4² + 2(5)(4)cos 45° = 41 + 20√2.
The minimum point on the curve y = x2 − 6x + 5 is at
Answer: E ((3,−4).)
dy/dx = 2x − 6 = 0 gives x = 3, and y = 9 − 18 + 5 = −4.
A canal has rectangular cross section of width 10cm and breadth 1m. If water of uniform density 1gm cm-3 flows through it at a constant speed of 1000 mm per minute, the adjacent sea is...
Answer: A (100 000)
The cross-section is 10cm × 100cm = 1000cm² and the water moves 100cm a minute, so 100 000cm³ (100 000g) of water reaches the sea each minute.
A man runs a distance of 9km at a constant speed for the first 4km and then 2km/h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4km is
Answer: B (8km/h)
With speed v: 4/v + 5/(v + 2) = 1, which gives v² − 7v − 8 = 0, so v = 8km/h.
A pyramid is constructed on a cuboid. The figure has
Answer: E (Sixteen edges.)
The cuboid has 12 edges and the pyramid on top adds 4 slant edges, giving 16 edges (it has 9 faces and 9 vertices).
In a geometric progression the first term is 153 and the sixth term is 17/27. The sum of the first four terms is
Answer: B (680/3)
153r⁵ = 17/27 gives r⁵ = 1/243, so r = 1/3. S₄ = 153(1 − 1/81)/(1 − 1/3) = 153 × 80/81 × 3/2 = 680/3.
An arithmetic progression has first term 11 and fourth term 32. The sum of the first nine terms is
Answer: A (351)
11 + 3d = 32 gives d = 7. S₉ = 9/2 × (2 × 11 + 8 × 7) = 9/2 × 78 = 351.
A triangle has angles 30°, 15° and 135°. The side opposite to the angle 30°, has length 6cm. The side opposite to the angle 135° is equal to
Answer: C (6√2cm)
By the sine rule, x/sin 135° = 6/sin 30°, so x = 12 × sin 45° = 12 × (√2/2) = 6√2cm.
A regular hexagon is constructed inside a circle of diameter 12cm. The area of the hexagon is
Answer: C (54√3 cm2)
The hexagon is six equilateral triangles of side 6cm (the radius). Area = 6 × (√3/4) × 36 = 54√3cm².
In a soccer competition in one season, a club had scored the following goals: 2,0,3,3,2,1,4,0,0,5,1,0,2,2,1,3,1,4,1 and 1. The mean, median and mode are respectively
Answer: B (1.8, 1.5 and 1)
The 20 scores add up to 36, so the mean is 1.8. In order, the 10th and 11th scores are 1 and 2, so the median is 1.5, and 1 occurs most often (6 times).
If sec2 θ + tan2 θ = 3, then the angle θ is equal to
Answer: B (45°)
sec²θ = 1 + tan²θ, so 1 + 2tan²θ = 3, tan²θ = 1 and θ = 45°.
A hollow right prism of equilateral triangular base of side 4cm is filled with water up to a certain height. If a sphere of radius 1/2cm is immersed in the water then the rise of water is
Answer: C (π/(24√3) cm)
The sphere's volume is (4/3)π(½)³ = π/6. The base area is (√3/4) × 16 = 4√3, so the rise is (π/6) ÷ 4√3 = π/(24√3)cm.
The set of values of x and y which satisfies the equations x2 − y − 1 = 0 and y − 2x + 2 = 0 is
Answer: A (1,0)
From the second equation y = 2x − 2. Substituting: x² − 2x + 1 = 0, so x = 1 and y = 0.
A solid sphere of radius 3cm, a solid right cone of radius 3cm and height 12cm and a solid right circular cylinder of radius 3cm and height 4cm. Which of the following statements is true?
Answer: C (The total surface area of the cone is greater than that of the sphere)
All three volumes equal 36π, so A and B are false. Surface areas: sphere 36π, cylinder 2π(3)(4) + 18π = 42π, cone π(3)(√153) + 9π ≈ 46π. Only C is true.
The quantity (x + y) is a factor of
Answer: D (2x3+2x2y−xy+3x−y2+3y)
(x + y) is a factor when putting x = −y gives 0. For D: −2y³ + 2y³ + y² − 3y − y² + 3y = 0; the others do not give 0.
Two triangles have the same area if
Answer: B (Three sides in one triangle are equal to three sides in the other)
Three equal sides (SSS) make the triangles congruent, so their areas must be equal. The other conditions allow triangles of different sizes or shapes.
If (25)x − 1 = 64 (5/2)6, then x has the value
Answer: B (4)
64 × (5/2)⁶ = 64 × 5⁶/64 = 5⁶. So 5^(2x − 2) = 5⁶, 2x − 2 = 6 and x = 4.
Assuming loge 4.4 = 1.4816 and loge7.7 = 2.0142 then the value of loge7/4 is
Answer: A (0.5326)
7/4 = 7.7/4.4, so log(7/4) = log 7.7 − log 4.4 = 2.0142 − 1.4816 = 0.5326.
The locus of all points having a distance of l units from each of the two fixed points a and b is:
Answer: E (a circle in a plane perpendicular to ab and centre at the mid-point of the line ab.)
Points at distance l from a lie on a sphere of radius l about a, and likewise for b. Two equal spheres meet in a circle centred at the mid-point of ab, in the plane perpendicular to ab.
If x4 − kx3 + 10x2 + lx − 3 is divisible by (x − 1) and if when it is divided by (x + 2) the remainder is 27, find the constants k and l.
Answer: A (k = −7, l = −15)
f(1) = 0 gives l − k = −8. f(−2) = 27 gives 8k − 2l = −26, i.e. 4k − l = −13. Solving: k = −7 and l = −15.
Evaluate without using tables sin(−1290°)
Answer: E (½.)
sin(−1290°) = −sin 1290°. 1290° − 3 × 360° = 210°, and sin 210° = −½, so the value is ½.
The angle of elevation of the top of a vertical tower from a point A on the ground is 60°. From a point B, 2 units of distance further away from the foot of the tower, the angle of elevation of the top of the tower is 45°. Find the distance of A from the foot of the tower.
Answer: D (1 + √3)
If A is x from the foot, the height is x tan 60° = x√3 and also (x + 2) tan 45° = x + 2. So x(√3 − 1) = 2 and x = 2/(√3 − 1) = √3 + 1.
The vectors a and b are given in terms of two perpendicular unit vectors i and j on a plane by a = 2i − 3j, b = −i + 2j. Find the magnitude of the vector a + 3b
Answer: E (√10.)
a + 3b = (2 − 3)i + (−3 + 6)j = −i + 3j, whose magnitude is √(1 + 9) = √10.
In the figure below ΔABC and ΔAED are in adjacent planes. AB = AC = 5cm, BC = 6cm and <ADE = 60°, then AE is equal to
Answer: B (2√3)
D is the mid-point of BC, so AD = √(5² − 3²) = 4cm. Triangle AED is right-angled at E, so AE = AD sin 60° = 4 × √3/2 = 2√3.
Add the same number to the numerator and denominator of 3/18. If the resulting fraction is 1/2, then the number added is
Answer: D (12)
(3 + x)/(18 + x) = 1/2 gives 6 + 2x = 18 + x, so x = 12.
In a circle of radius 10cm, a cord of length 10cm is xcm from its centre where x is
Answer: B (5√3)
The perpendicular from the centre bisects the chord, so x² = 10² − 5² = 75 and x = 5√3cm.
The smallest number such that when it is divided by 8 has a remainder of 6 and when it is divided by 9, has a remainder of 7 is
Answer: C (502)
Such a number is 2 less than a multiple of 72 (70, 142, …, 502). Checking the options, only 502 = 8 × 62 + 6 = 9 × 55 + 7 works.