If the simple interest on a certain sum of money for 2 years at 5% per annum is ₹500, what would be the simple interest on the same sum for 5 years at 6% per annum?
A.₹950
B.₹1,500
C.₹1,150
D.₹1,200
Solution
Simple interest
SI = , so 500 = .
P = ₹5,000.
New SI = .
= ₹1,500 — option (b).
53
Anshu travelled a certain distance by bike at the rate of 50 km/h and walked back at the rate of 8 km/h. If the whole journey took 5 hours 48 minutes, what was the distance?
A.45 km
B.42 km
C.40 km
D.38 km
Solution
To and fro journey
Total time = 5 + = hours.
For equal distances, distance = × total time.
= × = × .
= 40 km — option (c).
54
The angles of elevation of the top of a tower from two points on the ground, 400 m apart and on opposite sides of the tower, are 30° and 45°. What is the height of the tower?
A.100( + 1) m
B.200( − 1) m
C.100( − 1) m
D.200 m
Solution
Two angles of elevation
Let the height be h. From the 45° point the distance is h; from the 30° point it is h.
The two points are on opposite sides, so h + h = 400.
h( + 1) = 400 ⇒ h = .
Rationalising: h = 200( − 1) m — option (b).
55
A cuboid has a volume of 1400 cm³ and its length is twice its height. If the breadth is 7 cm, find its dimensions.
A.18 cm × 7 cm × 9 cm
B.12 cm × 7 cm × 6 cm
C.20 cm × 7 cm × 10 cm
D.10 cm × 7 cm × 5 cm
Solution
Cuboid dimensions
Volume = l × b × h, so l × h = = 200.
Given l = 2h, we get 2h² = 200.
h² = 100 ⇒ h = 10 cm and l = 20 cm.
20 cm × 7 cm × 10 cm — option (c).
56
A starts a business with ₹4,000. B joins after 5 months with ₹8,000. If the profit at the end of the year is ₹5,800, what is B's share (approximately)?
A.₹3,173.07
B.₹2,123.07
C.₹3,123.07
D.₹4,173.07
Solution
Partnership
A invests for the full 12 months: 4,000 × 12 = 48,000.
B joins after 5 months, so he invests for 7: 8,000 × 7 = 56,000.
Ratio = 48 : 56 = 6 : 7, a total of 13 parts.
B’s share = × 5,800 ≈ ₹3,123.07 — option (c).
57
The sum of the interior angles of a regular polygon is five times the sum of its exterior angles. Determine the number of diagonals in the polygon.
A.63
B.72
C.54
D.97
Solution
Diagonals of a polygon
The exterior angles of any polygon add up to 360°, so the interior sum is 5 × 360 = 1800°.
(n − 2) × 180 = 1800 ⇒ n − 2 = 10 ⇒ n = 12.
Number of diagonals = .
= = 54 — option (c).
58
A trader marks his goods 20% above the cost price and gives a discount of 10%. If he gains ₹72 on an item, what is the cost price of the item?
A.₹500
B.₹900
C.₹400
D.₹600
Solution
Markup and discount
Take CP = 100 units, so MP = 120 units.
After a 10% discount, SP = 120 × 0.9 = 108 units.
Gain = 8 units, and 8 units = ₹72, so 1 unit = ₹9.
CP = 100 × 9 = ₹900 — option (b).
59
A square plot has an area of 576 m². If a footpath 1 m wide is constructed around it, what is the total area including the footpath?
A.729 m²
B.676 m²
C.784 m²
D.756 m²
Solution
Path around a plot
Side of the plot = = 24 m.
The path runs all around, so each side grows by 1 m on both ends.
New side = 24 + 2 = 26 m.
Total area = 26² = 676 m² — option (b).
60
A bag contains red, blue and green marbles in the ratio 4 : 5 : 6. If 6 red marbles are removed and 3 green marbles are added, the new ratio becomes 2 : 5 : 7. How many red marbles were originally in the bag?
A.12
B.14
C.16
D.20
Solution
Ratio after change
Let the marbles be 4x, 5x and 6x. The blue count does not change, so compare red with blue.
= ⇒ 20x − 30 = 10x ⇒ x = 3.
Check the green: = = ✓.
Red originally = 4 × 3 = 12 — option (a).
61
A train covers 3 equal distances at speeds in the ratio 2 : 3 : 5. What is the ratio of the average speed over the entire journey to the lowest speed?
A.15 : 14
B.60 : 31
C.45 : 31
D.60 : 13
Solution
Average speed ratio
For three equal distances, average speed = .
+ + = = .
Average = 3 × = .
Ratio to the lowest speed 2 = : 2 = 45 : 31 — option (c).
62
A reduction of 20% in the price of sugar enables a person to buy 5 kg more for ₹100. What is the original price per kg?
A.₹20
B.₹10
C.₹5
D.₹15
Solution
Price reduction
Let the original price be ₹P per kg; the new price is 0.8P.
− = 5.
= 5 ⇒ = 5.
P = ₹5 per kg — option (c).
63
The average temperature of a city from Monday to Friday is 30°C. If the temperature on Saturday is 55°C, what will be the average temperature from Monday to Saturday?
A.41.17°C
B.34°C
C.34.17°C
D.41°C
Solution
New average
Total for five days = 5 × 30 = 150.
Adding Saturday: 150 + 55 = 205.
Average over six days = .
≈ 34.17°C — option (c).
64
An online shopkeeper offers a ₹180 discount on an item originally priced at ₹3,600. What percentage discount is being offered?
A car travels the first 200 km at 50 km/h and the next 200 km at 25 km/h. What is the average speed for the total journey?
A.33.33 km/h
B.37.5 km/h
C.36 km/h
D.40.33 km/h
Solution
Average speed
Time for the first part = = 4 hours.
Time for the second = = 8 hours.
Average speed = = .
≈ 33.33 km/h — option (a).
67
77 women can do a piece of work in 89 days. Find the number of days required by 7 women to complete th of the work.
A.75
B.77
C.89
D.79
Solution
Work and days
Total work = 77 × 89 woman-days.
One-eleventh of it = = 7 × 89 woman-days.
With 7 women, days = .
= 89 days — option (c).
68
A rectangle has a length that is four times its width. If the perimeter is 200 cm, what is the area of the rectangle?
A.1400 cm²
B.1500 cm²
C.1540 cm²
D.1600 cm²
Solution
Rectangle area
Let the width be w, so the length is 4w.
Perimeter = 2(4w + w) = 10w = 200 ⇒ w = 20 cm.
Length = 80 cm.
Area = 80 × 20 = 1600 cm² — option (d).
69
If sinA = , then what is cotA in terms of x?
A.
B.x
C.
D.
Solution
cot A in terms of x
Take the perpendicular as 2x and the hypotenuse as 1 + x².
Base = = = 1 − x².
cotA = .
= — option (d).
70
A tank contains 150 litres of liquid, 30% of which is alcohol. If 20 litres of this mixture is replaced by another liquid containing 40% alcohol, what is the percentage of alcohol in the resulting solution?
A.31.33%
B.35%
C.36.33%
D.31%
Solution
Replacement mixture
Alcohol at the start = 30% of 150 = 45 litres.
Removing 20 litres of the mixture takes away 30% of 20 = 6 litres, leaving 39 litres.
The 20 litres added brings in 40% of 20 = 8 litres, so the alcohol is 39 + 8 = 47 litres in 150.
× 100 ≈ 31.33% — option (a).
If tan x = cot y and cot x = tan y, then x + y equals:
A.45°
B.60°
C.90°
D.180°
Solution
Complementary angles
For complementary angles, tanθ = cot(90° − θ).
So tan x = cot y means x = 90° − y.
Therefore x + y = 90°.
Hence option (c).
73
A triangle has a perimeter of 40 cm and an inradius of 2 cm. What is its area?
A.10 cm²
B.20 cm²
C.30 cm²
D.40 cm²
Solution
Area from inradius
Semiperimeter s = = 20 cm.
For any triangle, area = inradius × semiperimeter.
= 2 × 20.
= 40 cm² — option (d).
74
In a circle, there is a chord measuring 18 cm that is located 13 cm away from the centre. What is the approximate radius of the circle?
A.10 cm
B.18 cm
C.16 cm
D.25 cm
Solution
Radius from a chord
The perpendicular from the centre bisects the chord, so half-chord = 9 cm.
Radius = .
= = .
≈ 15.8, that is about 16 cm — option (c).
75
A tangent is drawn to a circle with a radius of 8 cm. At the point of tangency, what is the angle between the tangent and the radius?
A.45°
B.60°
C.90°
D.180°
Solution
Tangent and radius
The radius drawn to the point of contact is always perpendicular to the tangent.
This holds for a circle of any size, so the value 8 cm makes no difference.
The angle is therefore a right angle.
= 90° — option (c).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Work inside the root first: 8 × 447 = 3576.
3576 ÷ 24 = 149.
149 + 75 − 28 = 196.
= 14 — option (b).
SI = , so 500 = .
P = ₹5,000.
New SI = .
= ₹1,500 — option (b).
Total time = 5 + = hours.
For equal distances, distance = × total time.
= × = × .
= 40 km — option (c).
Let the height be h. From the 45° point the distance is h; from the 30° point it is h.
The two points are on opposite sides, so h + h = 400.
h( + 1) = 400 ⇒ h = .
Rationalising: h = 200( − 1) m — option (b).
Volume = l × b × h, so l × h = = 200.
Given l = 2h, we get 2h² = 200.
h² = 100 ⇒ h = 10 cm and l = 20 cm.
20 cm × 7 cm × 10 cm — option (c).
A invests for the full 12 months: 4,000 × 12 = 48,000.
B joins after 5 months, so he invests for 7: 8,000 × 7 = 56,000.
Ratio = 48 : 56 = 6 : 7, a total of 13 parts.
B’s share = × 5,800 ≈ ₹3,123.07 — option (c).
The exterior angles of any polygon add up to 360°, so the interior sum is 5 × 360 = 1800°.
(n − 2) × 180 = 1800 ⇒ n − 2 = 10 ⇒ n = 12.
Number of diagonals = .
= = 54 — option (c).
Take CP = 100 units, so MP = 120 units.
After a 10% discount, SP = 120 × 0.9 = 108 units.
Gain = 8 units, and 8 units = ₹72, so 1 unit = ₹9.
CP = 100 × 9 = ₹900 — option (b).
Side of the plot = = 24 m.
The path runs all around, so each side grows by 1 m on both ends.
New side = 24 + 2 = 26 m.
Total area = 26² = 676 m² — option (b).
Let the marbles be 4x, 5x and 6x. The blue count does not change, so compare red with blue.
= ⇒ 20x − 30 = 10x ⇒ x = 3.
Check the green: = = ✓.
Red originally = 4 × 3 = 12 — option (a).
For three equal distances, average speed = .
+ + = = .
Average = 3 × = .
Ratio to the lowest speed 2 = : 2 = 45 : 31 — option (c).
Let the original price be ₹P per kg; the new price is 0.8P.
− = 5.
= 5 ⇒ = 5.
P = ₹5 per kg — option (c).
Total for five days = 5 × 30 = 150.
Adding Saturday: 150 + 55 = 205.
Average over six days = .
≈ 34.17°C — option (c).
Time for the first part = = 4 hours.
Time for the second = = 8 hours.
Average speed = = .
≈ 33.33 km/h — option (a).
Total work = 77 × 89 woman-days.
One-eleventh of it = = 7 × 89 woman-days.
With 7 women, days = .
= 89 days — option (c).
Let the width be w, so the length is 4w.
Perimeter = 2(4w + w) = 10w = 200 ⇒ w = 20 cm.
Length = 80 cm.
Area = 80 × 20 = 1600 cm² — option (d).
Take the perpendicular as 2x and the hypotenuse as 1 + x².
Base = = = 1 − x².
cotA = .
= — option (d).
Alcohol at the start = 30% of 150 = 45 litres.
Removing 20 litres of the mixture takes away 30% of 20 = 6 litres, leaving 39 litres.
The 20 litres added brings in 40% of 20 = 8 litres, so the alcohol is 39 + 8 = 47 litres in 150.
× 100 ≈ 31.33% — option (a).
Radians = degrees × .
= 160 × .
= .
= — option (a).
Semiperimeter s = = 20 cm.
For any triangle, area = inradius × semiperimeter.
= 2 × 20.
= 40 cm² — option (d).
The perpendicular from the centre bisects the chord, so half-chord = 9 cm.
Radius = .
= = .
≈ 15.8, that is about 16 cm — option (c).
The radius drawn to the point of contact is always perpendicular to the tangent.
This holds for a circle of any size, so the value 8 cm makes no difference.
The angle is therefore a right angle.
= 90° — option (c).