From a sample of 230 software engineers, determine the ratio of those proficient in Python to those proficient in Java using the given information:
Category
Number
Proficient in Python and Java
60
Proficient in Python only
70
Proficient in Java only
80
Proficient in neither language
20
A.11 : 12
B.13 : 14
C.13 : 11
D.13 : 7
Solution
Ratio from a data table
Total proficient in Python = only Python + both = 70 + 60 = 130.
Total proficient in Java = only Java + both = 80 + 60 = 140.
Ratio = 130 : 140 = 13 : 14 — option (b).
55
Sonu started a business by investing ₹40,000. After 4 months, Prem joined with ₹80,000. At the end of 8 months from the start, Alok joined with ₹1,20,000. If the total profit is ₹1,56,000 at the end of the year, find the share of Alok.
P and Q start a business. P invests ₹40,000 for 9 months, Q invests ₹60,000 for 6 months. What is Q's share of a ₹45,000 profit?
A.₹26,500
B.₹28,000
C.₹36,000
D.₹22,500
Solution
Partnership — two partners
Capital × time: P = 40000 × 9 = 3,60,000 and Q = 60000 × 6 = 3,60,000.
Profit ratio P : Q = 1 : 1.
Q's share = × 45000 = ₹22,500 — option (d).
57
What is the average of all integers between 100 and 252 that are exactly divisible by 11?
A.176
B.156
C.186
D.136
Solution
Average of multiples of 11
Multiples of 11 in the range run from 110 to 242.
They form an arithmetic progression, so the average = (first + last) ÷ 2.
(110 + 242) ÷ 2 = 176 — option (a).
58
The average of 15 numbers is 80. The average of the first 6 numbers is 72. The average of the next 6 numbers is 33.33% more than the average of the first 6 numbers. The 13th number is 8 more than the 15th number, and the 14th number is 10 less than the 15th number. What is the average of the 13th and 14th numbers?
A.63.89
B.85
C.63.67
D.63.65
Solution
Average — grouped numbers
Total = 15 × 80 = 1200; first 6 = 432; next 6 average = 72 × = 96, so their sum = 576. Sum of the last 3 = 1200 − 432 − 576 = 192.
Let the 15th number be x: then (x + 8) + (x − 10) + x = 192 ⇒ 3x = 194 ⇒ x = 64.67.
13th = 72.67 and 14th = 54.67; their average = = 63.67 — option (c).
59
A landlord bought a flat for ₹12,00,000. He wants to earn a 9% annual return on his investment after paying ₹2,500 per month for maintenance. What should be the monthly rent he charges?
A.₹11,800
B.₹10,500
C.₹11,500
D.₹11,700
Solution
Rent for a target return
Required annual return = 9% of 12,00,000 = ₹1,08,000.
Annual maintenance = 2500 × 12 = ₹30,000, so total rent needed = 1,08,000 + 30,000 = ₹1,38,000 per year.
Monthly rent = = ₹11,500 — option (c).
60
An amount is said to double in 10 years with compound interest. How many years will it take for the amount to grow to 8 times its original value?
A.15
B.30
C.31
D.18
Solution
CI — doubling to 8 times
8 = 2³, so the amount must double three times over.
Each doubling takes 10 years under compound interest.
Total time = 3 × 10 = 30 years — option (b).
61
A sum becomes ₹6,600 in 3 years and ₹7,920 in 4 years at compound interest. What is the original principal?
A.₹3,020.44
B.₹3,829.44
C.₹3,819.44
D.₹3,883.44
Solution
CI — principal from two amounts
The 4th-year growth factor = = 1.2, so the rate = 20% per annum.
Amount after 3 years: P × = 6600.
P = = ₹3,819.44 — option (c).
62
Kusum purchased 22 liters of milk at ₹42 per liter and another 15 liters of milk at ₹50 per liter. She combined both quantities and then sold the entire mixture at ₹48 per liter. What was her total profit or loss?
A toy manufacturer produced 1200 toy cars at a total cost of ₹91,000. He donated 200 cars to a charity event. For the rest, he announced a 10% discount on the market price of ₹120 per car. He also offered 2 toy cars free for every 8 toy cars purchased. If all 1200 toy cars were distributed, what is his overall gain or loss percentage?
A.5.05% loss
B.5.05% profit
C.6.05% loss
D.6.05% profit
Solution
Discount + free units — loss%
After donating 200 cars, 1000 remain; with 2 free per 8 purchased, they distribute in lots of 10 — so only × 1000 = 800 cars are actually paid for.
Selling price per paid car = 120 × 0.90 = ₹108, so revenue = 800 × 108 = ₹86,400.
Loss = 91000 − 86400 = ₹4,600 ⇒ × 100 ≈ 5.05% loss — option (a).
64
A retailer marks a car 80% above its cost price. He offers a first discount of 25% on the marked price. During a festive offer, an additional discount of 10% is applied on the already discounted price. If the final selling price is ₹19,440, what is the approximate cost price of the car?
A 80-liter mixture contains Milk and water in the ratio 5 : 3. How much water (in liters) must be added to this mixture to change the ratio of Milk to water to 2 : 3?
A.40 litres
B.45 litres
C.35 litres
D.30 litres
Solution
Mixture — adding water
In 80 litres (ratio 5 : 3): milk = 50 L, water = 30 L.
Let x litres of water be added: = .
150 = 60 + 2x ⇒ x = 45 litres — option (b).
66
Uttam is able to complete a task in 15 days, while Deepak takes 20 days to finish the same task. If they collaborate and work together for 5 days, what fraction of the work will still remain?
A.
B.
C.
D.
Solution
Work — remaining fraction
Combined daily work = + = .
Work done in 5 days = 5 × = .
Remaining work = 1 − = — option (c).
67
Two Lentil priced at ₹90 per kg and ₹150 per kg are blended and sold at ₹250 per kg, achieving a profit margin of 25%. What is the ratio of the two Lentil in the mixture?
A.5 : 11
B.5 : 12
C.5 : 2
D.15 : 1
Solution
Alligation — blending
Mean cost price of the blend = = ₹200 per kg.
The mean lies above both prices, so the alligation gaps are taken as magnitudes: |200 − 150| = 50 and |200 − 90| = 110.
Ratio = 50 : 110 = 5 : 11 — option (a).
68
Three pipes, P, Q, and R are capable of filling a tank in 12, 15, and 24 hours, respectively. When all three pipes are opened together, they operate for 1.5 hours before pipe R is closed. How much additional time will it take to completely fill the tank after that?
A.4 hours
B.4 hours
C.4 hours
D.4 hours
Solution
Pipes — partial closing
Take the tank = 120 units: P = 10, Q = 8, R = 5 units/hour.
In 1.5 hours all three fill (10 + 8 + 5) × 1.5 = 34.5 units; remaining = 120 − 34.5 = 85.5 units.
P and Q together fill 18 units/hour ⇒ time = = 4.75 = 4 hours — option (a).
69
A bullet train covers a fixed distance in 30 minutes at an average speed of 360 km/h. Due to track maintenance, it needs to be diverted, increasing the distance by 25%. If the train needs to arrive at its destination on time (i.e., in 30 minutes), what should its new average speed be in km/h?
A.560 km/h
B.450 km/h
C.620 km/h
D.520 km/h
Solution
Speed — increased distance
Original distance = 360 × = 180 km.
New distance = 180 × 1.25 = 225 km, to be covered in the same half hour.
New speed = 225 ÷ = 450 km/h — option (b).
70
Two cars, X and Y, start from points 320 km apart and travel at constant speeds. If they move towards each other, they meet in 8 hours. If they move in the same direction, they meet in 16 hours. What is the speed of car X? (Assume X is the faster car).
A.30 km/h
B.35 km/h
C.45 km/h
D.90 km/h
Solution
Relative speed — two cars
Towards each other: x + y = = 40 km/h.
Same direction: x − y = = 20 km/h.
Adding: 2x = 60 ⇒ x = 30 km/h — option (a).
71
A circular pizza having a radius of 21 cm. If 80% of it is eaten, what is the area of pizza remaining?
A.240.42 cm²
B.356.2 cm²
C.277.2 cm²
D.266.2 cm²
Solution
Circle — remaining area
Total area = × = 1386 cm².
If 80% is eaten, 20% remains.
Remaining area = 0.20 × 1386 = 277.2 cm² — option (c).
72
A ring-shaped disc has outer radius 10 cm and inner radius 8 cm. What is the approximate ratio of the ring's area to the whole outer circle?
A.9 : 25
B.6 : 25
C.9 : 20
D.3 : 15
Solution
Annulus — area ratio
Ring area = ( − ) = (100 − 64) = 36 cm².
Outer circle area = 100 cm².
Ratio = 36 : 100 = 9 : 25 — option (a).
73
A truck wheel having a radius of 42 cm. What percentage of its circumference is covered in a quarter turn? (Use π = )
A.35%
B.25%
C.40%
D.45%
Solution
Circumference — quarter turn
A quarter turn means the wheel rotates through one-fourth of a full revolution.
Distance covered = of the circumference, irrespective of the radius.
That is exactly 25% — option (b).
74
The line y = mx + 7 passes through (2, 9). Find m.
A.4
B.3
C.1
D.2
Solution
Line through a point
Substitute the point (2, 9) into the equation: 9 = 2m + 7.
2m = 2.
m = 1 — option (c).
Convert to decimals: = 0.545, = 0.571, = 0.600, = 0.625.
Ascending order of the values: 0.545 < 0.571 < 0.600 < 0.625.
So < < < — option (b).
Capital × time: Sonu = 40000 × 12 = 4,80,000; Prem = 80000 × 8 = 6,40,000; Alok = 120000 × 4 = 4,80,000.
Profit ratio = 480 : 640 : 480 = 3 : 4 : 3 (total 10 parts).
Alok's share = × 156000 = ₹46,800 — option (d).
Capital × time: P = 40000 × 9 = 3,60,000 and Q = 60000 × 6 = 3,60,000.
Profit ratio P : Q = 1 : 1.
Q's share = × 45000 = ₹22,500 — option (d).
Multiples of 11 in the range run from 110 to 242.
They form an arithmetic progression, so the average = (first + last) ÷ 2.
(110 + 242) ÷ 2 = 176 — option (a).
Required annual return = 9% of 12,00,000 = ₹1,08,000.
Annual maintenance = 2500 × 12 = ₹30,000, so total rent needed = 1,08,000 + 30,000 = ₹1,38,000 per year.
Monthly rent = = ₹11,500 — option (c).
8 = 2³, so the amount must double three times over.
Each doubling takes 10 years under compound interest.
Total time = 3 × 10 = 30 years — option (b).
The 4th-year growth factor = = 1.2, so the rate = 20% per annum.
Amount after 3 years: P × = 6600.
P = = ₹3,819.44 — option (c).
Total cost = 22 × 42 + 15 × 50 = 924 + 750 = ₹1,674.
Selling price = 37 litres × ₹48 = ₹1,776.
Profit = 1776 − 1674 = ₹102 — option (b).
Final SP = CP × 1.80 × 0.75 × 0.90.
That is SP = 1.215 × CP = 19440.
CP = = ₹16,000 — option (a).
In 80 litres (ratio 5 : 3): milk = 50 L, water = 30 L.
Let x litres of water be added: = .
150 = 60 + 2x ⇒ x = 45 litres — option (b).
Combined daily work = + = .
Work done in 5 days = 5 × = .
Remaining work = 1 − = — option (c).
Mean cost price of the blend = = ₹200 per kg.
The mean lies above both prices, so the alligation gaps are taken as magnitudes: |200 − 150| = 50 and |200 − 90| = 110.
Ratio = 50 : 110 = 5 : 11 — option (a).
Take the tank = 120 units: P = 10, Q = 8, R = 5 units/hour.
In 1.5 hours all three fill (10 + 8 + 5) × 1.5 = 34.5 units; remaining = 120 − 34.5 = 85.5 units.
P and Q together fill 18 units/hour ⇒ time = = 4.75 = 4 hours — option (a).
Original distance = 360 × = 180 km.
New distance = 180 × 1.25 = 225 km, to be covered in the same half hour.
New speed = 225 ÷ = 450 km/h — option (b).
Towards each other: x + y = = 40 km/h.
Same direction: x − y = = 20 km/h.
Adding: 2x = 60 ⇒ x = 30 km/h — option (a).
Total area = × = 1386 cm².
If 80% is eaten, 20% remains.
Remaining area = 0.20 × 1386 = 277.2 cm² — option (c).
Ring area = ( − ) = (100 − 64) = 36 cm².
Outer circle area = 100 cm².
Ratio = 36 : 100 = 9 : 25 — option (a).
A quarter turn means the wheel rotates through one-fourth of a full revolution.
Distance covered = of the circumference, irrespective of the radius.
That is exactly 25% — option (b).
Substitute the point (2, 9) into the equation: 9 = 2m + 7.
2m = 2.
m = 1 — option (c).