A shopkeeper mixes two varieties of pulses - one costing ₹50 per kg and the other costing ₹90 per kg, in the ratio 3 : 1. If he sells the mixed variety at ₹96 per kg, find his gain or loss percent.
A.60% Gain
B.40% Gain
C.45% Loss
D.60% Loss
Solution
Mixture — gain%
CP of the mixture = = = ₹60 per kg.
SP = ₹96 per kg, so gain = ₹36 per kg.
Gain% = × 100.
= 60% gain — option (a).
54
A and B invest ₹60,000 and ₹80,000 respectively. A remains in business for 8 months, B for 9 months. If the total profit is ₹40,000, what is A's share?
The average weight of 40 students is 70 kg. If one student is removed, the average drops to 69.5 kg. What is the weight of the removed student?
A.84.5 kg
B.89.5 kg
C.80.5 kg
D.86.5 kg
Solution
Average — student removed
Total of 40 = 40 × 70 = 2800 kg.
Total of the remaining 39 = 39 × 69.5 = 2710.5 kg.
Removed student = 2800 − 2710.5.
= 89.5 kg — option (b).
56
In a group of 30 boys, the average height was initially determined to be 150 cm. However, it was later discovered that the height of one boy was incorrectly recorded as 154 cm when it should have been 145 cm. Calculate the accurate average height.
A.149.7 cm
B.150.4 cm
C.159.7 cm
D.179.5 cm
Solution
Average — correction
The recorded total is too high by 154 − 145 = 9 cm.
Spread over 30 boys, the average falls by = 0.3 cm.
Correct average = 150 − 0.3.
= 149.7 cm — option (a).
57
If ₹8,000 amounts to ₹9,680 over 2 years with compound interest, what is the annual interest rate?
A.10%
B.9%
C.8%
D.11%
Solution
CI — find rate
9680 = 8000(1 + )².
= = (1 + )².
Taking the square root: = 1 + .
R = 10% — option (a).
58
A fruit vendor sells 20 kg of oranges for ₹900, thereby gaining the cost price of 5 kg of oranges. What is his profit percentage?
A.28.65%
B.25%
C.26%
D.31.56%
Solution
Gain = CP of 5 kg
Profit = SP − CP, and the profit equals the CP of 5 kg.
So CP of 20 kg + CP of 5 kg = SP of 20 kg ⇒ 25 CP-units = 20 SP-units.
Thus CP : SP for equal quantity = 20 : 25.
Profit% = × 100 = 25% — option (b).
59
A refrigerator has a marked price of ₹15,000. It is sold with two successive discounts. If the second discount is 30% and the final selling price is ₹9,000, what is the percentage of the first discount?
A.12.5%
B.12%
C.14.28%
D.15%
Solution
Successive discounts
Apply the 30% discount to the marked price first: 15,000 × = ₹10,500.
The remaining fall from ₹10,500 to ₹9,000 is the other discount.
Discount = × 100 = × 100.
= 14.28% — option (c).
60
A barrel contains a mixture of alcohol and water in the ratio 5 : 3 respectively. If 24 litres of water are added to it, the ratio of alcohol to water becomes 5 : 9. Find the initial total quantity of the mixture in the barrel.
A.32 litres
B.40 litres
C.52 litres
D.60 litres
Solution
Mixture — add water
Alcohol is unchanged, so keep its share fixed at 5 units: before 5 : 3, after 5 : 9.
The water rose by 9 − 3 = 6 units, and that equals 24 litres ⇒ 1 unit = 4 litres.
Initial mixture = 5 + 3 = 8 units.
8 × 4 = 32 litres — option (a).
61
A furniture store marks its items at 60% above the cost price. They offer a discount of 20% on the marked price. If a customer receives an additional loyalty discount of ₹221, and the store still makes a 15% profit on the cost price, what is the cost price of the furniture item?
A.₹1,700
B.₹2,500
C.₹1,500
D.₹2,200
Solution
Markup, discount, loyalty cut
Let CP = 100 units ⇒ MP = 160 units.
After 20% discount: 160 × = 128 units.
Final SP gives 15% profit = 115 units, so 128 − 115 = 13 units = ₹221 ⇒ 1 unit = ₹17.
CP = 100 units = ₹1,700 — option (a).
62
A spherical ball is submerged in water in a cylindrical container. The radius of the cylindrical container is 6 cm, and the height is 35 cm. What is the volume of water displaced by the ball if its radius is 6 cm?
A.π cm³
B.π cm³
C.π cm³
D.π cm³
Solution
Water displaced by sphere
A fully submerged body displaces water equal to its own volume.
Volume of the sphere = πr³ = π(6)³.
= π = π cm³ (that is 288π).
Hence option (d).
63
A hemispherical dome covers a circular area of diameter 12 m. What is the approximate curved surface area of this dome?
A.226 m²
B.220 m²
C.408 m²
D.762 m²
Solution
Hemisphere CSA
Diameter 12 m ⇒ radius r = 6 m.
CSA of a hemisphere = 2πr².
= 2 × × 36 ≈ 226.29.
≈ 226 m² — option (a).
64
A hollow hemisphere has uniform thickness. Its inner radius is r, and outer radius is R. If R = 4r, find the ratio of outer to inner curved surface areas.
A.16 : 1
B.8 : 5
C.6 : 5
D.2 : 5
Solution
Hollow hemisphere — CSA ratio
CSA of a hemisphere = 2π × (radius)².
Outer = 2π(4r)² = 32πr²; inner = 2πr².
Ratio = 32πr² : 2πr².
= 16 : 1 — option (a).
65
A square pyramid has its lateral surface area equal to twice the area of its base. If the side of the base is 8 cm, find its slant height.
A.8 cm
B.9 cm
C.5 cm
D.2 cm
Solution
Pyramid — slant height
Base area = 8² = 64 cm², so LSA = 2 × 64 = 128 cm².
LSA of a square pyramid = 2 × side × slant height = 2al.
2 × 8 × l = 128.
l = 8 cm — option (a).
66
A circular garden has a diameter of 42 m. If a gardener wants to fence 25% of the garden's circumference, what length of fencing is needed? (Use π = )
A.28 m
B.33 m
C.32 m
D.31 m
Solution
Quarter of circumference
Radius = 21 m, so circumference = 2 × × 21 = 132 m.
25% means one quarter.
132 × .
= 33 m — option (b).
67
In an equilateral triangle with side length 'b', what is the ratio of the inradius to the circumradius?
A.2 : 1
B.3 : 1
C.4 : 1
D.1 : 2
Solution
Inradius : circumradius
For an equilateral triangle, inradius r = and circumradius R = .
Ratio = ÷ .
= × = .
= 1 : 2 — option (d).
68
A triangle ABC is inscribed in a circle. A tangent is drawn to the circle at point A, intersecting the extension of side BC at D. If AD = 12 cm and CD = 8 cm, what is the length of side BC?
A.10 cm
B.8 cm
C.12 cm
D.18 cm
Solution
Tangent-secant
Tangent-secant relation: AD² = CD × BD.
12² = 8 × BD ⇒ BD = = 18 cm.
BC = BD − CD = 18 − 8.
= 10 cm — option (a).
69
In a trapezoid PQRS with PQ parallel to SR, the diagonals PR and QS intersect at T. What is the ratio of the area of △PQT to the area of △SRT?
A.The ratio of PQ to SR squared.
B.The ratio of PQ to SR.
C.The ratio of the perimeter of △PQT to the perimeter of △SRT.
D.The ratio of the area of △PQT to the area of △SRT.
Solution
Trapezium — area ratio
PQ ∥ SR makes △PQT and △SRT similar (alternate angles at T are equal).
In similar triangles, areas vary as the SQUARE of corresponding sides.
So ar(△PQT) : ar(△SRT) = ()².
Hence option (a).
70
A circle with radius 2x touches another circle with radius 4x externally. What is the length of a direct common tangent?
A.2x
B.3x
C.4x
D.3x
Solution
Direct common tangent
Touching externally means d = 2x + 4x = 6x.
Direct common tangent L = .
= = .
= 4x — option (c).
71
Two circles have radii 5 cm and 3 cm, and their centres are 8 cm apart. How many common tangents can be drawn between them, and what type(s) of tangents are they?
A.4 tangents: 2 direct and 2 transverse
B.3 tangents: 2 direct and 1 transverse
C.2 tangents: only direct tangents
D.0 tangents: one circle lies completely inside the other
Solution
Tangents — touching circles
Here r₁ + r₂ = 5 + 3 = 8 = d, so the circles touch each other EXTERNALLY.
Externally touching circles have 2 direct common tangents plus 1 at the point of contact.
Total = 3 tangents.
Hence option (b).
72
Two chords PQ and RS intersect at a point T inside a circle. If PT = 8 cm, TQ = 10 cm, and RT = 6 cm, what is the length of ST?
CP of the mixture = = = ₹60 per kg.
SP = ₹96 per kg, so gain = ₹36 per kg.
Gain% = × 100.
= 60% gain — option (a).
Capital × time: A = 60,000 × 8 = 4,80,000; B = 80,000 × 9 = 7,20,000.
Ratio = 48 : 72 = 2 : 3 (5 parts).
A’s share = × 40,000.
= ₹16,000 — option (a).
Total of 40 = 40 × 70 = 2800 kg.
Total of the remaining 39 = 39 × 69.5 = 2710.5 kg.
Removed student = 2800 − 2710.5.
= 89.5 kg — option (b).
The recorded total is too high by 154 − 145 = 9 cm.
Spread over 30 boys, the average falls by = 0.3 cm.
Correct average = 150 − 0.3.
= 149.7 cm — option (a).
9680 = 8000(1 + )².
= = (1 + )².
Taking the square root: = 1 + .
R = 10% — option (a).
Apply the 30% discount to the marked price first: 15,000 × = ₹10,500.
The remaining fall from ₹10,500 to ₹9,000 is the other discount.
Discount = × 100 = × 100.
= 14.28% — option (c).
Alcohol is unchanged, so keep its share fixed at 5 units: before 5 : 3, after 5 : 9.
The water rose by 9 − 3 = 6 units, and that equals 24 litres ⇒ 1 unit = 4 litres.
Initial mixture = 5 + 3 = 8 units.
8 × 4 = 32 litres — option (a).
Let CP = 100 units ⇒ MP = 160 units.
After 20% discount: 160 × = 128 units.
Final SP gives 15% profit = 115 units, so 128 − 115 = 13 units = ₹221 ⇒ 1 unit = ₹17.
CP = 100 units = ₹1,700 — option (a).
A fully submerged body displaces water equal to its own volume.
Volume of the sphere = πr³ = π(6)³.
= π = π cm³ (that is 288π).
Hence option (d).
Diameter 12 m ⇒ radius r = 6 m.
CSA of a hemisphere = 2πr².
= 2 × × 36 ≈ 226.29.
≈ 226 m² — option (a).
CSA of a hemisphere = 2π × (radius)².
Outer = 2π(4r)² = 32πr²; inner = 2πr².
Ratio = 32πr² : 2πr².
= 16 : 1 — option (a).
Base area = 8² = 64 cm², so LSA = 2 × 64 = 128 cm².
LSA of a square pyramid = 2 × side × slant height = 2al.
2 × 8 × l = 128.
l = 8 cm — option (a).
Radius = 21 m, so circumference = 2 × × 21 = 132 m.
25% means one quarter.
132 × .
= 33 m — option (b).
For an equilateral triangle, inradius r = and circumradius R = .
Ratio = ÷ .
= × = .
= 1 : 2 — option (d).
Tangent-secant relation: AD² = CD × BD.
12² = 8 × BD ⇒ BD = = 18 cm.
BC = BD − CD = 18 − 8.
= 10 cm — option (a).
PQ ∥ SR makes △PQT and △SRT similar (alternate angles at T are equal).
In similar triangles, areas vary as the SQUARE of corresponding sides.
So ar(△PQT) : ar(△SRT) = ()².
Hence option (a).
Touching externally means d = 2x + 4x = 6x.
Direct common tangent L = .
= = .
= 4x — option (c).
Intersecting chords: PT × TQ = RT × TS.
8 × 10 = 6 × TS.
TS = .
= cm — option (a).
The perpendicular from the centre bisects the chord, so half-chord = 8 cm.
r² = 8² + 6² = 64 + 36.
r² = 100.
r = 10 cm — option (b).
Note that 0.03 = 0.1 × 0.3 and 0.002 = 0.1 × 0.02.
So the numerator = (0.1)³ × (0.3³ + 0.02³).
The bracket cancels, leaving (0.1)³.
= 0.001 — option (d).
Numerator = 0.8³ + 0.04³ = 4³(0.2³ + 0.01³).
Denominator = 1.6³ + 0.08³ = 8³(0.2³ + 0.01³).
The common bracket cancels: = .
= — option (c).