The investments of three partners A, B and C are initially in the ratio 5 : 6 : 7. After a year A adds ₹40,000, B adds ₹35,000 and C adds some amount. The new ratio of their investments becomes 7 : 8 : 9. Find the amount (in ₹) added by partner C.
A.₹20,000
B.₹25,000
C.₹30,000
D.₹36,000
Solution
Partnership — added capital
Let the old investments be 5k, 6k, 7k and the new ones 7m, 8m, 9m.
5k + 40,000 = 7m and 6k + 35,000 = 8m. Solving: 40k + 3,20,000 = 42k + 2,45,000 ⇒ k = 37,500.
Then 5(37,500) + 40,000 = 7m ⇒ m = 32,500; C’s new investment = 9m = ₹2,92,500.
C added 2,92,500 − 7(37,500) = ₹30,000 — option (c).
53
An employee contributes 10% of their salary to charity. From the remaining amount, 50% is allocated to investments. The rest is divided for food and education in the ratio 6 : 3. If the expense on education is ₹3,000, what is the employee’s total salary?
A.₹28,000
B.₹20,000
C.₹32,000
D.₹35,000
Solution
Salary split
Education is 3 parts of 9, so the leftover = 3,000 × = ₹9,000.
That leftover is 90% × 50% = 45% of the salary.
45% = 9,000 ⇒ salary = × 100.
= ₹20,000 — option (b).
54
A certain amount increases to 3 times its initial value after 10 years of compound interest. How many years will it take to grow to twenty-seven times its original amount?
A.32
B.30
C.19
D.18
Solution
CI — multiples
Under compound interest the multiple grows in powers: 3 times in 10 years.
27 = 3³, so we need the 3-times growth to happen three times over.
Time = 3 × 10.
= 30 years — option (b).
55
A software company developed 1600 licences for a new application at a total cost of ₹80,00,000. They provided 100 licences free to their top beta testers. For the remaining licences they offered a 30% discount on the market price of ₹12,000 per licence, and additionally gave 1 licence free for every 9 licences sold. What is the profit or loss percentage?
A.45.64% profit
B.41.75% profit
C.43.30% loss
D.43.64% loss
Solution
Licences — profit %
Licences left after the free 100 = 1600 − 100 = 1500.
1 free with every 9 means only of them are actually paid for, i.e. a further 10% off.
SP = 1500 × 12,000 × × = ₹1,13,40,000.
Profit% = × 100 = 41.75% — option (b).
56
The speeds of two motorcyclists are in the ratio 4 : 3. If the slower motorcyclist has a 27 km head start and his speed is 18 km/h, how far (in km) will the faster motorcyclist travel to overtake him?
A.80 km
B.90 km
C.108 km
D.72 km
Solution
Overtaking
The slower speed is the ratio-3 part, so the faster = 18 × = 24 km/h.
Relative speed = 24 − 18 = 6 km/h.
Time to close the 27 km gap = = 4.5 hours.
Distance covered by the faster rider = 24 × 4.5 = 108 km — option (c).
57
A shopkeeper took a loan of ₹7,000 from a bank on 1 April 2023 with simple interest at 8% per annum. He cleared the loan on 1 October 2023. How much did he pay back to the bank?
A.₹7,280.76
B.₹7,581.25
C.₹7,784.79
D.₹7,790.00
Solution
SI over exact days
From 1 April to 1 October = 30 + 31 + 30 + 31 + 31 + 30 = 183 days.
SI = .
≈ ₹280.76.
Amount = 7,000 + 280.76 = ₹7,280.76 — option (a).
58
Which of these is both an integer and a whole number?
A.−6
B.1
C.4.5
D.
Solution
Integer and whole number
Whole numbers are 0, 1, 2, 3, … — no negatives, no fractions.
Integers include negatives as well.
−6 is an integer but not a whole number; 4.5 and are neither.
1 satisfies both — option (b).
59
P, Q and R invested in a business in the ratio 4 : 3 : 5. After a year the total profit is ₹6,00,000. What is Q's share?
A solid cone is to be painted. If paint covers 3 m² per litre and the total surface area is 72 m², how much paint is required, including 10% extra for retouching?
An L-shaped floor is made of two rectangles (7 m × 4 m and 7 m × 6 m). Tiles cost ₹500/m², but a 10% discount is given if more than 60 m² is tiled. What is the total cost after discount?
A.₹31,500
B.₹39,160
C.₹30,240
D.₹31,540
Solution
L-shaped floor
Areas: 7 × 4 = 28 m² and 7 × 6 = 42 m²; total = 70 m².
70 > 60, so the discount applies.
Cost before discount = 70 × 500 = ₹35,000.
After 10% off: 35,000 × = ₹31,500 — option (a).
62
A large regular shape is made by combining 12 regular hexagons around a common centre. What is the ratio of the area of the combined shape to that of two hexagons?
A.4 : 1
B.6 : 1
C.3 : 1
D.5 : 1
Solution
Hexagons ratio
All the hexagons are identical, so area is simply proportional to their number.
Combined shape : two hexagons = 12 : 2.
Dividing both by 2.
= 6 : 1 — option (b).
63
A triangular traffic sign has sides measuring 9 m, 10 m and 17 m. If the cost to paint the sign is ₹7 per square metre, what will be the total expense to paint the entire sign?
A.₹422
B.₹380
C.₹252
D.₹462
Solution
Heron's formula
s = = 18 m.
Area = = .
= = 36 m².
Cost = 36 × 7 = ₹252 — option (c).
64
If tanθ = cotθ, what is the value of θ (in degrees)?
A.45°
B.60°
C.90°
D.55°
Solution
tanθ = cotθ
cotθ = , so tanθ = .
tan²θ = 1 ⇒ tanθ = 1.
In the first quadrant tan45° = 1.
θ = 45° — option (a).
65
The base area of a prism is increased by 35%, while the height remains the same. By what percentage does the volume increase?
A.10%
B.20%
C.35%
D.30%
Solution
Prism volume change
Volume of a prism = base area × height.
The height is unchanged, so volume varies directly with the base area.
A 35% rise in base area gives a 35% rise in volume.
Hence option (c).
66
A man and a boy are together filling a tank with water. The man pours 4 litres every 3 minutes, while the boy pours 3 litres every 4 minutes. How much time will it take to fill 200 litres in the tank?
Three numbers are in the ratio 6 : 9 : 5. If their average is 300, what is the difference between the largest and the smallest number?
A.180
B.900
C.800
D.700
Solution
Ratio and average
Let the numbers be 6x, 9x and 5x; their sum = 20x.
Average 300 ⇒ 20x = 300 × 3 = 900 ⇒ x = 45.
Largest = 9x = 405 and smallest = 5x = 225.
Difference = 4x = 4 × 45 = 180 — option (a).
68
A can contains 30 litres of oil at ₹50/litre. Another can contains 70 litres of oil at ₹50/litre. They are mixed, and 25 litres are sold at ₹60/litre. How much profit or loss does the seller make on this sale?
A.₹600 profit
B.₹250 loss
C.₹280 loss
D.₹250 profit
Solution
Mixture and sale
Both cans cost the same ₹50 per litre, so the mixture also costs ₹50 per litre.
Cost of 25 litres = 25 × 50 = ₹1,250.
Selling price = 25 × 60 = ₹1,500.
Profit = 1,500 − 1,250 = ₹250 — option (d).
69
If cosB = 1 − 2sin²B, then what is the value of cos2B?
A.sinB
B.1 − 2sin²B
C.secB
D.Cannot be determined
Solution
cos2B identity
The standard double-angle identity is cos2B = 1 − 2sin²B.
This holds for every value of B.
So cos2B equals the same expression.
= 1 − 2sin²B — option (b).
70
What is the equation of the line through (3, 4) and (5, 8)?
A.y = 3(x − 1)
B.y = 3x + 1
C.y = 4x − 2
D.y = 2(x − 1)
Solution
Line through two points
Slope = = = 2.
Point-slope form: y − 4 = 2(x − 3).
y = 2x − 6 + 4 = 2x − 2.
= 2(x − 1) — option (d).
71
If sin(2x + 30°) = cos(5x − 3°), find x.
A.27°
B.30°
C.12°
D.9°
Solution
Complementary angles
sinA = cosB holds when A and B are complementary: A + B = 90°.
(2x + 30) + (5x − 3) = 90.
7x + 27 = 90 ⇒ 7x = 63.
x = 9° — option (d).
72
The number of triangles formed by the diagonals drawn from one vertex of a hexagon is:
A.4
B.5
C.8
D.6
Solution
Triangles from one vertex
From one vertex of an n-sided polygon the diagonals cut it into (n − 2) triangles.
For a hexagon n = 6.
6 − 2 = 4.
Hence option (a).
73
A right triangle has vertices at A(0, 0), B(10, 0) and C(0, 8). What are the coordinates of its circumcentre?
A.(5, 4)
B.(4, 5)
C.(3, 4)
D.(6, 4)
Solution
Circumcentre
The right angle is at A, so BC is the hypotenuse.
For a right triangle the circumcentre is the MIDPOINT of the hypotenuse.
Midpoint of B(10, 0) and C(0, 8) = (, ).
= (5, 4) — option (a).
97² = 9409, so = 97.
43² = 1849, so = 43.
97 − 43.
= 54 — option (a).
Let the old investments be 5k, 6k, 7k and the new ones 7m, 8m, 9m.
5k + 40,000 = 7m and 6k + 35,000 = 8m. Solving: 40k + 3,20,000 = 42k + 2,45,000 ⇒ k = 37,500.
Then 5(37,500) + 40,000 = 7m ⇒ m = 32,500; C’s new investment = 9m = ₹2,92,500.
C added 2,92,500 − 7(37,500) = ₹30,000 — option (c).
Education is 3 parts of 9, so the leftover = 3,000 × = ₹9,000.
That leftover is 90% × 50% = 45% of the salary.
45% = 9,000 ⇒ salary = × 100.
= ₹20,000 — option (b).
Under compound interest the multiple grows in powers: 3 times in 10 years.
27 = 3³, so we need the 3-times growth to happen three times over.
Time = 3 × 10.
= 30 years — option (b).
Licences left after the free 100 = 1600 − 100 = 1500.
1 free with every 9 means only of them are actually paid for, i.e. a further 10% off.
SP = 1500 × 12,000 × × = ₹1,13,40,000.
Profit% = × 100 = 41.75% — option (b).
The slower speed is the ratio-3 part, so the faster = 18 × = 24 km/h.
Relative speed = 24 − 18 = 6 km/h.
Time to close the 27 km gap = = 4.5 hours.
Distance covered by the faster rider = 24 × 4.5 = 108 km — option (c).
From 1 April to 1 October = 30 + 31 + 30 + 31 + 31 + 30 = 183 days.
SI = .
≈ ₹280.76.
Amount = 7,000 + 280.76 = ₹7,280.76 — option (a).
Paint needed = = 24 litres.
Extra 10% = × 24 = 2.4 litres.
Total = 24 + 2.4.
= 26.4 litres — option (b).
Areas: 7 × 4 = 28 m² and 7 × 6 = 42 m²; total = 70 m².
70 > 60, so the discount applies.
Cost before discount = 70 × 500 = ₹35,000.
After 10% off: 35,000 × = ₹31,500 — option (a).
All the hexagons are identical, so area is simply proportional to their number.
Combined shape : two hexagons = 12 : 2.
Dividing both by 2.
= 6 : 1 — option (b).
s = = 18 m.
Area = = .
= = 36 m².
Cost = 36 × 7 = ₹252 — option (c).
Man's rate = L/min and boy's rate = L/min.
Combined = = L/min.
Time = 200 ÷ = 200 × .
= 96 minutes — option (d).
Let the numbers be 6x, 9x and 5x; their sum = 20x.
Average 300 ⇒ 20x = 300 × 3 = 900 ⇒ x = 45.
Largest = 9x = 405 and smallest = 5x = 225.
Difference = 4x = 4 × 45 = 180 — option (a).
sinA = cosB holds when A and B are complementary: A + B = 90°.
(2x + 30) + (5x − 3) = 90.
7x + 27 = 90 ⇒ 7x = 63.
x = 9° — option (d).
From one vertex of an n-sided polygon the diagonals cut it into (n − 2) triangles.
For a hexagon n = 6.
6 − 2 = 4.
Hence option (a).
The right angle is at A, so BC is the hypotenuse.
For a right triangle the circumcentre is the MIDPOINT of the hypotenuse.
Midpoint of B(10, 0) and C(0, 8) = (, ).
= (5, 4) — option (a).
= = 2.
= = 4.
2 + 4.
= 6 — option (a).
At the same moment the ratio of height to shadow is the same for all objects.
= .
h = × 35.
= 49 feet — option (b).