Sumit starts a business with ₹80,000. Three months later, Sachin joins with ₹1,00,000. Six months after starting, Sumit increases his investment by ₹40,000. Sachin leaves the business three months before the end of the year. What is their profit-sharing ratio?
A.7 : 4
B.2 : 1
C.1 : 4
D.1 : 2
Solution
Partnership — changing capital
Sumit: ₹80,000 for the first 6 months and ₹1,20,000 for the next 6 months.
Sumit = 80,000×6 + 1,20,000×6 = 4,80,000 + 7,20,000 = 12,00,000.
Sachin joined after 3 months and left 3 months early ⇒ 6 months: 1,00,000×6 = 6,00,000.
Ratio = 12 : 6 = 2 : 1 — option (b).
56
The average age of 21 students is 17 years. A teacher joins the group, and the average becomes 20 years. What is the teacher's age?
A.72
B.83
C.65
D.73
Solution
Average — teacher's age
Total age of 21 students = 21 × 17 = 357 years.
Total after the teacher joins = 22 × 20 = 440 years.
Teacher’s age = 440 − 357.
= 83 years — option (b).
57
The average salary of a team of 90 employees is ₹58,000. If the average salary of 40 of them is ₹68,000, what is the average salary of the remaining employees?
A.₹52,000
B.₹48,000
C.₹50,000
D.₹54,000
Solution
Average — remaining group
Total of 90 = 90 × 58,000 = 52,20,000.
Total of 40 = 40 × 68,000 = 27,20,000.
Remaining 50 share 52,20,000 − 27,20,000 = 25,00,000.
Average = = ₹50,000 — option (c).
58
A sum of money at compound interest doubles its original amount in 3 years. In how many years will it become 64 times its original amount?
A.16
B.20
C.14
D.18
Solution
CI — doubling time
Doubling takes 3 years, so after 3n years the money becomes 2ⁿ times.
64 = 2⁶, so n = 6 doublings are needed.
Time = 3 × 6.
= 18 years — option (d).
59
If the ratio between the loss and cost price of an article is 1 : 5, then the ratio between its selling price and cost price is ____.
A.9 : 10
B.4 : 5
C.2 : 5
D.7 : 10
Solution
Loss : CP → SP : CP
Let loss = 1 unit and CP = 5 units.
SP = CP − loss = 5 − 1 = 4 units.
SP : CP = 4 : 5.
Hence option (b).
60
An item costs ₹300. A customer gets a 20% discount on it. If the seller still makes a 25% profit, what is the cost price of the item?
A.₹192
B.₹209
C.₹186
D.₹159
Solution
Discount then profit
Marked price = ₹300; after 20% discount SP = 300 × 0.80 = ₹240.
This SP gives a 25% profit, so SP = 1.25 × CP.
CP = .
= ₹192 — option (a).
61
A web developer sells a premium website template to a distributor at a 25% discount. An additional 10% service fee is added to the discounted price. The distributor then sells the template for ₹9000 more, earning a 20% profit. What is the original marked price of the template fixed by the developer?
A.₹55,550.45
B.₹57,545.35
C.₹56,540.25
D.₹54,545.45
Solution
Discount, fee, profit → MP
Take MP = 100 units. After 25% discount: 75 units.
With the 10% service fee the distributor’s cost = 75 × = 82.5 units.
At 20% profit his SP = 82.5 × = 99 units, so the profit = 99 − 82.5 = 16.5 units = ₹9,000.
100 units = = ₹54,545.45 — option (d).
62
A 120-litre mixture contains milk and water in the ratio 5 : 3. If 'x' litres of milk and 'x' litres of water are added to the mixture, the new ratio of milk to water becomes 3 : 2. What is the value of 'x'?
A.12.33 litres
B.11.54 litres
C.15 litres
D.30 litres
Solution
Mixture — equal addition
Milk = 120 × = 75 L; water = 120 × = 45 L.
After adding x to each: = .
150 + 2x = 135 + 3x ⇒ x = 15.
Hence 15 litres — option (c).
63
A sphere is completely contained within a cylinder, with the height and diameter of the cylinder matching the diameter of the sphere. Given that the volume of the sphere is 288π cm³, what is the volume of the vacant space inside the cylinder?
A.122π cm³
B.169π cm³
C.144π cm³
D.196π cm³
Solution
Cylinder minus sphere
πr³ = 288π ⇒ r³ = 216 ⇒ r = 6 cm.
Cylinder: radius 6 and height = diameter = 12 ⇒ volume = π(6)²(12) = 432π.
Vacant space = 432π − 288π.
= 144π cm³ — option (c).
64
A solid hemisphere is melted to form a cone whose radius is half that of the hemisphere. Find the height of the resulting cone.
A.4R
B.6R
C.8R
D.10R
Solution
Hemisphere → cone
Volume is conserved: πR³ = π()²h.
R³ = × × h.
2R³ = ⇒ h = 8R.
Hence 8R — option (c).
65
Two solid hemispheres of radii 5 cm and 9 cm respectively are melted and recast into a single hemisphere. What is the approximate radius of the new hemisphere formed?
A.9.5 cm
B.10.2 cm
C.10.5 cm
D.11 cm
Solution
Two hemispheres → one
Volumes add, and each volume ∝ r³.
R³ = 5³ + 9³ = 125 + 729 = 854.
R = ∛854 ≈ 9.48.
≈ 9.5 cm — option (a).
66
A regular right pyramid has a square base with a side length of 16 cm. Its slant height is 10 cm. What is its volume?
A.512 cm³
B.640 cm³
C.768 cm³
D.800 cm³
Solution
Square pyramid volume
Height h = = = 6 cm.
Volume = × base area × height.
= × 16² × 6 = × 256 × 6.
= 512 cm³ — option (a).
67
The ratio of the radii of two circular tyres is 1 : . If the area of the smaller tyre is 120 cm², what is the area of the larger tyre?
A.240 cm²
B.360 cm²
C.320 cm²
D.500 cm²
Solution
Areas ∝ r²
Area varies as the square of the radius.
Ratio of areas = 1² : ()² = 1 : 3.
Larger area = 3 × 120.
= 360 cm² — option (b).
68
The angle bisectors of triangle PQR converge at the in-center 'O'. Given that the measure of ∠QOR is 125°, what is the measure of ∠QPR?
In a quadrilateral PQRS, PQ is parallel to RS, and PR is a diagonal. If PQ = RS, which of the following is true?
A.△PQR ≅ △RSP by SSS
B.△PQR ≅ △RSP by SAS
C.△PQR ≅ △RSP by ASA
D.The triangles are similar, but not necessarily congruent
Solution
Congruence — SAS
PQ = RS (given) and PR = PR (common side).
Since PQ ∥ RS, the alternate interior angles ∠QPR = ∠SRP.
Two sides and the INCLUDED angle match.
Congruent by SAS — option (b).
70
In triangle PQR, a line segment ST is drawn through the centroid O, parallel to side QR. S lies on PQ and T lies on PR. What is the ratio of the area of the smaller triangle PST to the area of the trapezoid STRQ?
A.3 : 1
B.1 : 2
C.4 : 5
D.4 : 9
Solution
Line through centroid
The centroid divides the median in the ratio 2 : 1 from the vertex, so ST cuts the sides in the ratio 2 : 3 from P.
△PST ~ △PQR with side ratio 2 : 3 ⇒ area ratio = .
Trapezium STRQ = 1 − = of the whole.
PST : STRQ = 4 : 5 — option (c).
71
Two circles of radii 9 cm and 5 cm have their centers 15 cm apart. What is the length of the internal common tangent?
A.12 cm
B. cm
C. cm
D. cm
Solution
Internal common tangent
Internal (transverse) common tangent: L = .
= = .
= .
Hence cm — option (c).
72
Two circles with radii r₁ = 14 cm and r₂ = 6 cm have their centers 40 cm apart. Find the length of the transverse common tangent between them.
A. cm
B.1200 cm
C. cm
D.13 cm
Solution
Transverse tangent
Transverse common tangent: L = .
= .
= .
= cm — option (a).
73
In a circle, two chords are 8 cm and 10 cm long. If the 10 cm chord is 6 cm away from the center, what can be said about the distance of the 8 cm chord from the center?
A.It is less than 3 cm.
B.It is equal to 3 cm.
C.It is greater than 3 cm.
D.It cannot be determined.
Solution
Chord distances
Same radius for both: ()² + d₁² = ()² + d₂².
5² + 6² = 4² + d₂² ⇒ 25 + 36 = 16 + d₂².
d₂² = 45 ⇒ d₂ = 3 ≈ 6.7 cm.
That is greater than 3 cm — option (c).
74
A circle is circumscribed about a triangle. If one of the angles of the triangle is 150°, and the radius of the circumscribed circle is R, what is the length of the side opposite the 150° angle?
A.R
B.2R
C.R
D.R
Solution
Sine rule — 150°
Sine rule: = 2R, so a = 2R sin A.
sin150° = sin(180° − 30°) = sin30° = .
a = 2R × .
= R — option (c).
75
Add ( + ) and express the result as a decimal.
A.0.09275
B.0.09375
C.0.09175
D.0.09475
Solution
Fractions → decimal
Take the LCD 32: = .
+ = .
3 ÷ 32 = 0.09375.
Hence option (b).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Let x be subtracted: = .
Try x = 2: = .
Both equal ✓.
Hence 2 — option (a).
Sumit: ₹80,000 for the first 6 months and ₹1,20,000 for the next 6 months.
Sumit = 80,000×6 + 1,20,000×6 = 4,80,000 + 7,20,000 = 12,00,000.
Sachin joined after 3 months and left 3 months early ⇒ 6 months: 1,00,000×6 = 6,00,000.
Ratio = 12 : 6 = 2 : 1 — option (b).
Total age of 21 students = 21 × 17 = 357 years.
Total after the teacher joins = 22 × 20 = 440 years.
Teacher’s age = 440 − 357.
= 83 years — option (b).
Total of 90 = 90 × 58,000 = 52,20,000.
Total of 40 = 40 × 68,000 = 27,20,000.
Remaining 50 share 52,20,000 − 27,20,000 = 25,00,000.
Average = = ₹50,000 — option (c).
Doubling takes 3 years, so after 3n years the money becomes 2ⁿ times.
64 = 2⁶, so n = 6 doublings are needed.
Time = 3 × 6.
= 18 years — option (d).
Let loss = 1 unit and CP = 5 units.
SP = CP − loss = 5 − 1 = 4 units.
SP : CP = 4 : 5.
Hence option (b).
Marked price = ₹300; after 20% discount SP = 300 × 0.80 = ₹240.
This SP gives a 25% profit, so SP = 1.25 × CP.
CP = .
= ₹192 — option (a).
Take MP = 100 units. After 25% discount: 75 units.
With the 10% service fee the distributor’s cost = 75 × = 82.5 units.
At 20% profit his SP = 82.5 × = 99 units, so the profit = 99 − 82.5 = 16.5 units = ₹9,000.
100 units = = ₹54,545.45 — option (d).
Milk = 120 × = 75 L; water = 120 × = 45 L.
After adding x to each: = .
150 + 2x = 135 + 3x ⇒ x = 15.
Hence 15 litres — option (c).
πr³ = 288π ⇒ r³ = 216 ⇒ r = 6 cm.
Cylinder: radius 6 and height = diameter = 12 ⇒ volume = π(6)²(12) = 432π.
Vacant space = 432π − 288π.
= 144π cm³ — option (c).
Volume is conserved: πR³ = π()²h.
R³ = × × h.
2R³ = ⇒ h = 8R.
Hence 8R — option (c).
Volumes add, and each volume ∝ r³.
R³ = 5³ + 9³ = 125 + 729 = 854.
R = ∛854 ≈ 9.48.
≈ 9.5 cm — option (a).
Height h = = = 6 cm.
Volume = × base area × height.
= × 16² × 6 = × 256 × 6.
= 512 cm³ — option (a).
Area varies as the square of the radius.
Ratio of areas = 1² : ()² = 1 : 3.
Larger area = 3 × 120.
= 360 cm² — option (b).
For the incentre: ∠QOR = 90° + ∠QPR.
125° = 90° + ∠QPR.
∠QPR = 35°.
∠QPR = 70° — option (b).
The centroid divides the median in the ratio 2 : 1 from the vertex, so ST cuts the sides in the ratio 2 : 3 from P.
△PST ~ △PQR with side ratio 2 : 3 ⇒ area ratio = .
Trapezium STRQ = 1 − = of the whole.
PST : STRQ = 4 : 5 — option (c).
Internal (transverse) common tangent: L = .
= = .
= .
Hence cm — option (c).
Same radius for both: ()² + d₁² = ()² + d₂².
5² + 6² = 4² + d₂² ⇒ 25 + 36 = 16 + d₂².
d₂² = 45 ⇒ d₂ = 3 ≈ 6.7 cm.
That is greater than 3 cm — option (c).
Sine rule: = 2R, so a = 2R sin A.
sin150° = sin(180° − 30°) = sin30° = .
a = 2R × .
= R — option (c).