If 20% of A = 0.5 of B = of C, then what is A : B : C?
A.5 : 2 : 4
B.5 : 4 : 2
C.5 : 4 : 8
D.5 : 3 : 2
Solution
Ratio from percentages
Write each as a fraction: = = = k.
So A = 5k, B = 2k and C = 4k.
A : B : C = 5 : 2 : 4.
Hence option (a).
54
Two containers of equal capacity are filled with juice and water. The first contains juice and water in the ratio 2 : 3, and the second in the ratio 3 : 2. The mixtures are then combined. What is the ratio of juice to water in the resulting mixture?
A.1 : 1
B.3 : 4
C.4 : 3
D.5 : 5
Solution
Mixing two containers
Take 5 units in each container.
Juice = 2 + 3 = 5 units; water = 3 + 2 = 5 units.
Ratio = 5 : 5.
= 1 : 1 — option (a).
55
A pair of shoes is listed at ₹900. After two successive discounts, it is sold for ₹612. If the first discount offered was 20%, what was the second discount percentage?
A.16%
B.20%
C.15%
D.18%
Solution
Second discount
After the first discount: 900 × = ₹720.
Let the second discount be D%: 720 × = 612.
100 − D = = 85.
D = 15% — option (c).
56
A retailer marked a product 60% above its cost price. He provided two successive discounts of 25% and 10%. If he earned a profit of ₹240, what was the cost price of the product?
A.₹2000
B.₹1500
C.₹2700
D.₹3000
Solution
Markup and two discounts
Take CP = 100 units ⇒ MP = 160 units.
SP = 160 × × = 108 units.
Profit = 108 − 100 = 8 units = ₹240 ⇒ 1 unit = ₹30.
CP = 100 units = ₹3,000 — option (d).
57
If x + = 5, find the value of x³ + .
A.95
B.105
C.115
D.100
Solution
Cube identity
Use a³ + b³ = (a + b)³ − 3ab(a + b) with a = x and b = .
ab = 2, so x³ + = 5³ − 3(2)(5).
= 125 − 30.
= 95 — option (a).
58
An educational fund of ₹6,00,000 earns simple interest at 11% per annum. This annual interest is used to grant three scholarships. If the first and third scholarships are ₹18,000 and ₹22,000 respectively, what is the value of the second scholarship?
A.₹24,000
B.₹26,000
C.₹18,000
D.₹20,000
Solution
SI — scholarships
Annual interest = = ₹66,000.
The three scholarships share this amount.
Second = 66,000 − (18,000 + 22,000) = 66,000 − 40,000.
= ₹26,000 — option (b).
59
Suppose ₹P is invested, and after 2 years at 5% simple interest, it yields ₹Q interest. If ₹Q is then invested for 2 years at 5% simple interest, it yields ₹R interest. Which of the following is true?
A.Q = P + R
B.Q² = PR
C.P = Q + R
D.P² = Q + R
Solution
SI relation
Q = = .
R = = = .
Check: PR = P × = , and Q² = ()² = .
So Q² = PR — option (b).
60
An airplane flying at a height of 400 m notices the angles of depression to two boats located on opposite sides as 45° and 30°. Find the distance between the two boats.
A.400( − 1) m
B.400 m
C.400(1 + ) m
D.400( + 1) m
Solution
Angles of depression
For the 45° boat: tan45° = ⇒ BD = 400 m.
For the 30° boat: tan30° = ⇒ CD = 400 m.
The boats are on OPPOSITE sides, so the distances add.
400 + 400 = 400( + 1) m — option (d).
61
A statue 3.6 metres in height is positioned on top of a pedestal. From a point on the ground, the angle of elevation to the top of the statue is 60°, while the angle of elevation to the top of the pedestal from the same point is 45°. Determine the height of the pedestal.
A.1.5( − 1) m
B.1.6( − 1) m
C.1.5( + 1) m
D.1.8( + 1) m
Solution
Statue on a pedestal
Let the common base be 1 unit. Heights are then tan45° = 1 unit (pedestal) and tan60° = units (statue top).
The statue accounts for ( − 1) units = 3.6 m.
1 unit = × = .
Pedestal = 1.8( + 1) m — option (d).
62
The inner and outer diameters of a hollow hemispherical shell are 14 cm and 28 cm respectively. Find the approximate volume of the metal used.
A.4986.25 cm³
B.5030.66 cm³
C.4532.66 cm³
D.5530.25 cm³
Solution
Hollow hemisphere volume
Outer radius R = 14 cm and inner radius r = 7 cm.
Metal volume = π(R³ − r³) = × × (2744 − 343).
= × × 2401.
≈ 5030.66 cm³ — option (b).
63
A right circular cone is divided into two sections by a plane parallel to the base, with the volumes of the frustum and the smaller cone in the ratio 19 : 8. Determine the ratio of their heights.
A.1 : 1
B.1 : 2
C.5 : 6
D.2 : 1
Solution
Frustum : small cone
Volumes vary as the cube of heights: = .
8H³ = 27h³ ⇒ = .
Taking h = 2 and H = 3, the frustum’s height = 3 − 2 = 1.
Frustum : small cone = 1 : 2 — option (b).
64
If the side of a regular hexagon is tripled, what happens to its area?
A.Doubled
B.Tripled
C.Quadrupled
D.Increases 9 times
Solution
Hexagon area
Area of a regular hexagon = a² — it varies as the SQUARE of the side.
Tripling the side multiplies the area by 3².
= 9 times.
Hence option (d).
65
A cylindrical storage container has a base diameter of 6 cm. Its sections have heights in an arithmetic progression, starting from 2 cm for the bottom section and reaching 10 cm for the topmost, covering a total of 5 sections. What is the total volume of the container? (Use π ≈ 3.14)
A.853.6 cm³
B.847.8 cm³
C.842.8 cm³
D.613.4 cm³
Solution
Cylinder with AP heights
Radius = 3 cm. Heights: 2, 4, 6, 8, 10 — total 30 cm.
Volume = πr²h.
= 3.14 × 9 × 30.
= 847.8 cm³ — option (b).
66
Find the slope of the line 5x + 3y = 9.
A.−
B.
C.−
D.
Solution
Slope of a line
Write it as y = mx + c.
3y = −5x + 9 ⇒ y = −x + 3.
Comparing, the slope m = −.
Hence option (a).
67
What is the total sum of the interior angles of a convex nonagon (a polygon with 9 sides)?
A.1080°
B.1260°
C.1360°
D.1440°
Solution
Interior angles — nonagon
Sum of interior angles = (n − 2) × 180°.
For n = 9: (9 − 2) × 180°.
= 7 × 180°.
= 1260° — option (b).
68
In triangle ABC, medians AD, BE and CF intersect at the centroid G. If the area of triangle ABC is 96 cm², what is the area of triangle GBC?
A.16 cm²
B.24 cm²
C.48 cm²
D.32 cm²
Solution
Centroid divides area
The three medians divide a triangle into six equal parts; the centroid joins each vertex pair to form three equal triangles.
So ar(△GBC) = × ar(△ABC).
= × 96.
= 32 cm² — option (d).
69
Given = 2, then what is the value of x?
A.7
B.4
C.6
D.2
Solution
Surd equation
Try the options so that both roots come out whole.
Take x = 4: = 3 and = 1.
= = 2 ✓ = RHS.
x = 4 — option (b).
70
Two circles have radii in the ratio 2 : 1. The length of a direct common tangent is 4 cm, and the distance between their centres is 5 cm. What is the radius of the smaller circle?
A.1 cm
B.2 cm
C.3 cm
D.4 cm
Solution
Direct common tangent
Let the radii be 2r and r, with d = 5 cm.
Direct common tangent: L = ⇒ 4 = .
16 = 25 − r² ⇒ r² = 9.
r = 3 cm — option (c).
71
In a circle having a diameter of 20 cm, a chord is 6 cm away from the centre. Find the length of the chord.
A.16 cm
B.15 cm
C.13 cm
D.17 cm
Solution
Chord length
Radius = = 10 cm.
The perpendicular from the centre bisects the chord: 10² = 6² + (half-chord)².
Half-chord = = 8 cm.
Chord = 2 × 8 = 16 cm — option (a).
72
If tanB + cotB = 4 and tanB · cotB = 1, find tan²B + sec²B.
tanA from sinA
Take perpendicular = 2x and hypotenuse = 1 + x².
Base = = = 1 − x².
tanA = .
= — option (c).
74
Simplify: (0.16³ + 0.04³) ÷ (0.4³ + 0.1³)
A.0.64
B.0.064
C.0.6
D.0.006
Solution
Cubes — common factor
Note that 0.16 = 0.4 × 0.4 and 0.04 = 0.4 × 0.1.
So the numerator = (0.4)³ × (0.4³ + 0.1³).
The bracket cancels with the denominator, leaving (0.4)³.
= 0.064 — option (b).
75
Find the value of (0.02³ + 0.002³) ÷ (0.04³ + 0.004³)
A.
B.
C.
D.
Solution
Cubes — common factor
Let a = 0.02 and b = 0.002; then 0.04 = 2a and 0.004 = 2b.
Denominator = (2a)³ + (2b)³ = 8(a³ + b³).
Ratio = .
= — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Write each as a fraction: = = = k.
So A = 5k, B = 2k and C = 4k.
A : B : C = 5 : 2 : 4.
Hence option (a).
Take 5 units in each container.
Juice = 2 + 3 = 5 units; water = 3 + 2 = 5 units.
Ratio = 5 : 5.
= 1 : 1 — option (a).
After the first discount: 900 × = ₹720.
Let the second discount be D%: 720 × = 612.
100 − D = = 85.
D = 15% — option (c).
Take CP = 100 units ⇒ MP = 160 units.
SP = 160 × × = 108 units.
Profit = 108 − 100 = 8 units = ₹240 ⇒ 1 unit = ₹30.
CP = 100 units = ₹3,000 — option (d).
Use a³ + b³ = (a + b)³ − 3ab(a + b) with a = x and b = .
ab = 2, so x³ + = 5³ − 3(2)(5).
= 125 − 30.
= 95 — option (a).
Annual interest = = ₹66,000.
The three scholarships share this amount.
Second = 66,000 − (18,000 + 22,000) = 66,000 − 40,000.
= ₹26,000 — option (b).
For the 45° boat: tan45° = ⇒ BD = 400 m.
For the 30° boat: tan30° = ⇒ CD = 400 m.
The boats are on OPPOSITE sides, so the distances add.
400 + 400 = 400( + 1) m — option (d).
Let the common base be 1 unit. Heights are then tan45° = 1 unit (pedestal) and tan60° = units (statue top).
The statue accounts for ( − 1) units = 3.6 m.
1 unit = × = .
Pedestal = 1.8( + 1) m — option (d).
Outer radius R = 14 cm and inner radius r = 7 cm.
Metal volume = π(R³ − r³) = × × (2744 − 343).
= × × 2401.
≈ 5030.66 cm³ — option (b).
Volumes vary as the cube of heights: = .
8H³ = 27h³ ⇒ = .
Taking h = 2 and H = 3, the frustum’s height = 3 − 2 = 1.
Frustum : small cone = 1 : 2 — option (b).
Area of a regular hexagon = a² — it varies as the SQUARE of the side.
Tripling the side multiplies the area by 3².
= 9 times.
Hence option (d).
Radius = 3 cm. Heights: 2, 4, 6, 8, 10 — total 30 cm.
Volume = πr²h.
= 3.14 × 9 × 30.
= 847.8 cm³ — option (b).
The three medians divide a triangle into six equal parts; the centroid joins each vertex pair to form three equal triangles.
So ar(△GBC) = × ar(△ABC).
= × 96.
= 32 cm² — option (d).
Let the radii be 2r and r, with d = 5 cm.
Direct common tangent: L = ⇒ 4 = .
16 = 25 − r² ⇒ r² = 9.
r = 3 cm — option (c).
Radius = = 10 cm.
The perpendicular from the centre bisects the chord: 10² = 6² + (half-chord)².
Half-chord = = 8 cm.
Chord = 2 × 8 = 16 cm — option (a).
Take perpendicular = 2x and hypotenuse = 1 + x².
Base = = = 1 − x².
tanA = .
= — option (c).
Note that 0.16 = 0.4 × 0.4 and 0.04 = 0.4 × 0.1.
So the numerator = (0.4)³ × (0.4³ + 0.1³).
The bracket cancels with the denominator, leaving (0.4)³.
= 0.064 — option (b).
Let a = 0.02 and b = 0.002; then 0.04 = 2a and 0.004 = 2b.
Denominator = (2a)³ + (2b)³ = 8(a³ + b³).
Ratio = .
= — option (a).