If 20% of (A + B) = 60% of (A − B), then find the ratio A : B.
A.1 : 2
B.2 : 1
C.1 : 3
D.3 : 1
Solution
Percentage equality
20(A + B) = 60(A − B) ⇒ (A + B) = 3(A − B).
A + B = 3A − 3B.
4B = 2A ⇒ = 2.
A : B = 2 : 1 — option (b).
54
A bag contains ₹5, ₹2, and ₹1 coins in the ratio 2 : 3 : 5. If the total amount of money in the bag is ₹252, find the number of coins of each kind.
A.16 coins of ₹5, 24 of ₹2, and 40 of ₹1
B.20 coins of ₹5, 30 of ₹2, and 50 of ₹1
C.24 coins of ₹5, 36 of ₹2, and 60 of ₹1
D.18 coins of ₹5, 27 of ₹2, and 45 of ₹1
Solution
Coins in ratio
Let the counts be 2x, 3x and 5x.
Value: 5(2x) + 2(3x) + 1(5x) = 10x + 6x + 5x = 21x.
21x = 252 ⇒ x = 12.
Coins = 24, 36 and 60 — option (c).
55
A shopkeeper marked a mobile phone at ₹800. He sold it at a 15% discount. If his profit on the sale was 20%, what was the mobile phone's purchase price?
A.₹550.36
B.₹555.47
C.₹566.67
D.₹570.34
Solution
Discount and profit
SP = 800 × = ₹680.
This SP gives 20% profit, so SP = 1.20 × CP.
CP = .
≈ ₹566.67 — option (c).
56
A product is sold at a 20% discount on its marked price. This transaction results in a profit of 8% on its cost price. If the profit earned is ₹224, what would have been the selling price if the product was sold at a 33.33% discount on its marked price?
A.₹2587.50
B.₹3000.20
C.₹2520.00
D.₹2510.00
Solution
Discount change
Profit 8% = ₹224, so CP = × 100 = ₹2,800 and SP = 2,800 + 224 = ₹3,024.
That SP is 80% of the marked price, so MP = 3,024 × = ₹3,780.
A 33.33% discount leaves of the MP: 3,780 × .
= ₹2,520 — option (c).
57
If the cubic equation a³ − 8a² + 10a − 4 = 0 has the roots x, y, z then find the value of x + y + z.
A.5
B.6
C.7
D.8
Solution
Sum of roots
For a cubic, the sum of the roots = −.
Here the coefficients are 1 and −8.
Sum = −.
= 8 — option (d).
58
Amar borrowed ₹6,000 from Shankar on March 15, 2025, at 12% per annum simple interest. Amar decided to repay the loan on September 15, 2025. What is the total amount Amar must repay Shankar?
A.₹6200.90
B.₹6362.96
C.₹6162.66
D.₹6862.56
Solution
SI over exact days
Count the days: 16 + 30 + 31 + 30 + 31 + 31 + 15 = 184 days, i.e. year.
SI = ≈ ₹362.96.
Amount = 6,000 + 362.96.
= ₹6,362.96 — option (b).
59
Gaurav invested a total of ₹4,400 in three different simple interest schemes at 4%, 5%, and 10% per annum for one year. At the end of one year, he received the same interest from all three schemes. How much money (in ₹) did he invest in the scheme with a 5% interest rate?
A.₹1000
B.₹1600
C.₹2000
D.₹800
Solution
Equal interest schemes
For equal interest, principal ∝ .
Rates 4 : 5 : 10 give principals : : = 5 : 4 : 2 (11 parts).
11 parts = ₹4,400 ⇒ 1 part = ₹400.
5% scheme = 4 parts = ₹1,600 — option (b).
60
A man standing on the top of a 150 m high tower observes the angles of depression of two cars on the same side of the tower to be 45° and 60°. What is the distance between the two cars?
A.150( − 1) m
B.150 m
C.150(1 − ) m
D.150(1 + ) m
Solution
Angles of depression
For the 45° car: tan45° = ⇒ OC = 150 m.
For the 60° car: tan60° = ⇒ OB = m.
Distance between them = 150 − .
= 150(1 − ) m — option (c).
61
A vertical tree PQ has its base on the ground at Q. From a point A 20 m away from Q, the angle of elevation of P is x°. From another point B 60 m away from Q (and in the same line as A), the angle of elevation of P is y°. If the height of the tree is 20 m, what is the value of x° + y°?
A.60°
B.75°
C.90°
D.120°
Solution
Two elevation angles
From A: tan x° = = ⇒ x° = 60°.
From B: tan y° = = ⇒ y° = 30°.
x° + y° = 60° + 30°.
= 90° — option (c).
62
Two identical solid hemispheres, each with a radius of 27 cm, are melted and recast into a single sphere. What is the radius of the new sphere formed?
A.21 cm
B.27 cm
C.24 cm
D.29 cm
Solution
Two hemispheres → sphere
Two hemispheres of radius r have volume 2 × πr³ = πr³.
That is exactly the volume of a sphere of the SAME radius r.
So the new sphere has radius 27 cm.
Hence option (b).
63
A right circular cone is cut parallel to its base, forming a frustum. The height of the original cone is 20 cm, and the cut is made 5 cm from the top. What is the ratio of the volume of the frustum to the volume of the entire cone?
A.63 : 64
B.62 : 63
C.16 : 19
D.14 : 15
Solution
Frustum : cone
The small cone cut off is similar to the whole cone with height ratio = .
So its volume is ()³ = of the whole.
Frustum = 1 − = of the cone.
Ratio = 63 : 64 — option (a).
64
Each side of a regular octagon is of the side of a regular hexagon. If the perimeter of the hexagon is 24 cm, what will be the perimeter of the octagon?
A.18 cm
B.20 cm
C.24 cm
D.16 cm
Solution
Octagon perimeter
Hexagon side = = 4 cm.
Octagon side = × 4 = 3 cm.
Perimeter = 8 × 3.
= 24 cm — option (c).
65
A decorative column is made in the shape of a regular hexagonal prism. The side length of its base is 8 cm. The column is built from multiple sections, and their heights form an arithmetic progression: the first section is 8 cm tall, and each subsequent section is 3 cm taller than the previous one. If the column has 4 sections, what is its total volume?
A.4200 cm³
B.4800 cm³
C.5500 cm³
D.4600 cm³
Solution
Hexagonal prism volume
Heights: 8, 11, 14, 17 — total height = 50 cm.
Area of a regular hexagon = a² = × 64 = 96 cm².
Volume = base area × total height = 96 × 50.
= 4800 cm³ — option (b).
66
Find the midpoint of the segment joining (1, 2) and (5, 8).
Diagonals of a decagon
Number of diagonals = .
For a decagon n = 10: .
= .
= 35 — option (a).
68
Triangle ABC has its centroid at point G(6, 7), and vertex A is located at (3, 5). If point D is the midpoint of side BC, what are the coordinates of D?
A.(8.5, 9.5)
B.(9.1, 10.5)
C.(7.5, 8.0)
D.(6.5, 9.5)
Solution
Centroid and midpoint
The centroid divides the median AD in the ratio 2 : 1, so G = (, ).
x: 6 = ⇒ 18 = 3 + 2x₂ ⇒ x₂ = 7.5.
y: 7 = ⇒ 21 = 5 + 2y₂ ⇒ y₂ = 8.
D = (7.5, 8.0) — option (c).
69
Given + = 9, then find the value of x.
A.18
B.20
C.16
D.12
Solution
Surd equation
Both roots must be whole numbers adding to 9 — try the options.
Take x = 20: + = 5 + 4.
= 9 ✓ (LHS = RHS).
x = 20 — option (b).
70
The distance between the centres of two circles is 20 cm. The length of the internal common tangent joining both circles is 12 cm. If the radius of the larger circle is 10 cm, what will be the radius of the smaller circle?
A.6 cm
B.8 cm
C.5 cm
D.7 cm
Solution
Internal common tangent
Internal common tangent: L = .
12 = ⇒ 144 = 400 − (10 + R₂)².
(10 + R₂)² = 256 ⇒ 10 + R₂ = 16.
R₂ = 6 cm — option (a).
71
In a cyclic quadrilateral PQRS, PR is the diameter of the circle. If ∠PRS = 55°, then what is the measure of ∠QPS?
A.40°
B.35°
C.45°
D.90°
Solution
Cyclic quadrilateral
PR is a diameter, so the angle in the semicircle ∠PSR = 90°.
In △PRS: ∠RPS = 180° − 90° − 55°.
= 35°.
Hence option (b).
72
If sin⁴A − cos⁴A = , find the value of cos²A.
A.
B.
C.
D.
Solution
sin⁴A − cos⁴A
Factorise: (sin²A − cos²A)(sin²A + cos²A) = .
Since sin²A + cos²A = 1, we get sin²A − cos²A = .
Replace sin²A by 1 − cos²A: 1 − 2cos²A = .
cos²A = — option (b).
73
If cotA = x + , find sec²A.
A.1 + (1 + )²
B.2 + (x + )²
C.1 + (x + )⁻²
D.2 + (x + )⁻²
Solution
sec²A from cotA
sec²A = 1 + tan²A.
tanA is the reciprocal of cotA, so tan²A = .
= (x + )⁻².
sec²A = 1 + (x + )⁻² — option (c).
74
What is the value of (0.11³ + 0.08³) ÷ (0.22³ + 0.16³)?
A.0.125
B.0.5
C.0.25
D.1
Solution
Cubes — common factor
Let a = 0.11 and b = 0.08, so 0.22 = 2a and 0.16 = 2b.
Denominator = (2a)³ + (2b)³ = 8(a³ + b³).
So the ratio = = .
= 0.125 — option (a).
Cubes — ratio
Let x = 0.006 and y = 0.002; then the terms are (100x)³ + (10x)³ + x³ and (100y)³ + (10y)³ + y³.
= — the bracket cancels.
= ()³ = 3³.
= 27 — option (c).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Expand: ()² + 3² + 2 × × 3.
= 5 + 9 + 6 = 14 + 6.
Comparing with ? + 6.
? = 14 — option (d).
Let the counts be 2x, 3x and 5x.
Value: 5(2x) + 2(3x) + 1(5x) = 10x + 6x + 5x = 21x.
21x = 252 ⇒ x = 12.
Coins = 24, 36 and 60 — option (c).
SP = 800 × = ₹680.
This SP gives 20% profit, so SP = 1.20 × CP.
CP = .
≈ ₹566.67 — option (c).
Profit 8% = ₹224, so CP = × 100 = ₹2,800 and SP = 2,800 + 224 = ₹3,024.
That SP is 80% of the marked price, so MP = 3,024 × = ₹3,780.
A 33.33% discount leaves of the MP: 3,780 × .
= ₹2,520 — option (c).
For equal interest, principal ∝ .
Rates 4 : 5 : 10 give principals : : = 5 : 4 : 2 (11 parts).
11 parts = ₹4,400 ⇒ 1 part = ₹400.
5% scheme = 4 parts = ₹1,600 — option (b).
For the 45° car: tan45° = ⇒ OC = 150 m.
For the 60° car: tan60° = ⇒ OB = m.
Distance between them = 150 − .
= 150(1 − ) m — option (c).
From A: tan x° = = ⇒ x° = 60°.
From B: tan y° = = ⇒ y° = 30°.
x° + y° = 60° + 30°.
= 90° — option (c).
Two hemispheres of radius r have volume 2 × πr³ = πr³.
That is exactly the volume of a sphere of the SAME radius r.
So the new sphere has radius 27 cm.
Hence option (b).
The small cone cut off is similar to the whole cone with height ratio = .
So its volume is ()³ = of the whole.
Frustum = 1 − = of the cone.
Ratio = 63 : 64 — option (a).
Heights: 8, 11, 14, 17 — total height = 50 cm.
Area of a regular hexagon = a² = × 64 = 96 cm².
Volume = base area × total height = 96 × 50.
= 4800 cm³ — option (b).
Midpoint = (, ).
x-coordinate = = 3.
y-coordinate = = 5.
= (3, 5) — option (c).
Number of diagonals = .
For a decagon n = 10: .
= .
= 35 — option (a).
The centroid divides the median AD in the ratio 2 : 1, so G = (, ).
x: 6 = ⇒ 18 = 3 + 2x₂ ⇒ x₂ = 7.5.
y: 7 = ⇒ 21 = 5 + 2y₂ ⇒ y₂ = 8.
D = (7.5, 8.0) — option (c).
Internal common tangent: L = .
12 = ⇒ 144 = 400 − (10 + R₂)².
(10 + R₂)² = 256 ⇒ 10 + R₂ = 16.
R₂ = 6 cm — option (a).
PR is a diameter, so the angle in the semicircle ∠PSR = 90°.
In △PRS: ∠RPS = 180° − 90° − 55°.
= 35°.
Hence option (b).
Factorise: (sin²A − cos²A)(sin²A + cos²A) = .
Since sin²A + cos²A = 1, we get sin²A − cos²A = .
Replace sin²A by 1 − cos²A: 1 − 2cos²A = .
cos²A = — option (b).
Let a = 0.11 and b = 0.08, so 0.22 = 2a and 0.16 = 2b.
Denominator = (2a)³ + (2b)³ = 8(a³ + b³).
So the ratio = = .
= 0.125 — option (a).
Let x = 0.006 and y = 0.002; then the terms are (100x)³ + (10x)³ + x³ and (100y)³ + (10y)³ + y³.
= — the bracket cancels.
= ()³ = 3³.
= 27 — option (c).