138 Question on Mathematical operations

Here are a few verbal reasoning questions related to mathematical operations:

1. Number Series Puzzle:

• Question: Look at this series:
  2, 6, 12, 20, 30, ?What number should come next in the series?
• Answer:
  The difference between consecutive numbers is increasing by 2 each time:
  6−2=46 – 2 = 46−2=4, 12−6=612 – 6 = 612−6=6, 20−12=820 – 12 = 820−12=8, 30−20=1030 – 20 = 1030−20=10.
  So, the next difference will be 10+2=1210 + 2 = 1210+2=12.
  Hence, 30+12=4230 + 12 = 4230+12=42.
  The next number is 42.

2. Mathematical Relationship Question:

• Question: If A + B = C and B + C = D, what is the value of C – A?
• Answer:
  From the first equation: A + B = C
  From the second equation: B + C = D
  Now subtract the first equation from the second:(B+C)−(A+B)=D−C(B + C) – (A + B) = D – C(B+C)−(A+B)=D−CSimplifying:C−A=D−CC – A = D – CC−A=D−CTherefore, C – A = D – C.

3. Percentage Operation:

• Question: A number is increased by 25% and then decreased by 20%. If the original number is 100, what is the final result?
• Answer:
  First, increase 100 by 25%:
  100+(100×0.25)=100+25=125100 + (100 \times 0.25) = 100 + 25 = 125100+(100×0.25)=100+25=125.
  Now, decrease 125 by 20%:
  125−(125×0.20)=125−25=100125 – (125 \times 0.20) = 125 – 25 = 100125−(125×0.20)=125−25=100.
  Therefore, the final result is 100.

4. Simple Equation:

• Question: Solve for X:
  3X+7=223X + 7 = 223X+7=22.
• Answer:
  Subtract 7 from both sides:
  3X=153X = 153X=15.
  Now, divide both sides by 3:
  X=5X = 5X=5.
  The value of X is 5.

5. Ratio and Proportion:

• Question: The ratio of the ages of Alex and Ben is 4:5. If Alex is 12 years old, how old is Ben?
• Answer:
  Let Ben’s age be xxx.
  The ratio of their ages is 12x=45\frac{12}{x} = \frac{4}{5}x12​=54​.
  Cross-multiply:
  12×5=4×x12 \times 5 = 4 \times x12×5=4×x,
  60=4×60 = 4×60=4x.
  Now, divide by 4:
  x=604=15x = \frac{60}{4} = 15x=460​=15.
  Ben is 15 years old.

6. Basic Arithmetic Operations:

• Question: If P is 8 times Q, and Q is 7 less than R, what is the value of P when R = 14?
• Answer:
  First, find Q:
  Q=R−7=14−7=7Q = R – 7 = 14 – 7 = 7Q=R−7=14−7=7.
  Now, find P:
  P=8×Q=8×7=56P = 8 \times Q = 8 \times 7 = 56P=8×Q=8×7=56.
  Therefore, the value of P is 56.

7. Word Problem on Basic Arithmetic:

• Question: Sarah bought 5 books for $12 each and 3 pencils for $1 each. How much did she spend in total?
• Answer:
  The cost of 5 books is:
  5×12=605 \times 12 = 605×12=60.
  The cost of 3 pencils is:
  3×1=33 \times 1 = 33×1=3.
  Total cost = 60+3=6360 + 3 = 6360+3=63.
  Therefore, Sarah spent $63 in total.

8. Average Calculation:

• Question: The average of 5 consecutive numbers is 15. What is the largest of these numbers?
• Answer:
  Let the 5 consecutive numbers be x,x+1,x+2,x+3,x+4x, x+1, x+2, x+3, x+4x,x+1,x+2,x+3,x+4.
  The average of these 5 numbers is given as 15:x+(x+1)+(x+2)+(x+3)+(x+4)5=15\frac{x + (x+1) + (x+2) + (x+3) + (x+4)}{5} = 155x+(x+1)+(x+2)+(x+3)+(x+4)​=15Simplifying the equation:5x+105=15\frac{5x + 10}{5} = 1555x+10​=15 5x+10=755x + 10 = 755x+10=75 5x=655x = 655x=65 x=13x = 13x=13So, the numbers are 13, 14, 15, 16, 17.
  The largest number is 17.

9. Time and Work Problem:

• Question: If A can complete a task in 8 hours and B can complete the same task in 12 hours, how long will it take for them to complete the task together?
• Answer:
  Work done by A in 1 hour = 18\frac{1}{8}81​.
  Work done by B in 1 hour = 112\frac{1}{12}121​.
  Total work done together in 1 hour = 18+112=324+224=524\frac{1}{8} + \frac{1}{12} = \frac{3}{24} + \frac{2}{24} = \frac{5}{24}81​+121​=243​+242​=245​.
  Time taken to complete the task together = 245=4.8\frac{24}{5} = 4.8524​=4.8 hours.
  Therefore, they will complete the task together in 4.8 hours or 4 hours and 48 minutes.

10. Profit and Loss Question:

• Question: A man buys a shirt for $120 and sells it for $150. What is the profit percentage?
• Answer: Profit = Selling Price – Cost Price = 150−120=30150 – 120 = 30150−120=30. Profit Percentage = 30120×100=25%\frac{30}{120} \times 100 = 25\%12030​×100=25%. Therefore, the profit percentage is 25%.

11. Work Efficiency:

• Question: If a machine can complete a task in 5 hours, and another machine can complete the same task in 8 hours, how long will it take for both machines to complete the task together?
• Answer: Work done by the first machine in 1 hour = 15\frac{1}{5}51​.
  Work done by the second machine in 1 hour = 18\frac{1}{8}81​.
  Total work done together in 1 hour = 15+18=840+540=1340\frac{1}{5} + \frac{1}{8} = \frac{8}{40} + \frac{5}{40} = \frac{13}{40}51​+81​=408​+405​=4013​.
  Time taken to complete the task together = 4013≈3.08\frac{40}{13} \approx 3.081340​≈3.08 hours.
  Therefore, it will take both machines about 3.08 hours or 3 hours and 5 minutes to complete the task.

12. Mixture Problem:

• Question: A container contains a mixture of two liquids in the ratio 3:4. If 18 liters of the first liquid are removed and replaced with 18 liters of the second liquid, the ratio becomes 2:5. How many liters of the first liquid were in the container initially?
• Answer: Let the initial quantity of liquid in the container be xxx.
  Initially, the amount of the first liquid = 37×x\frac{3}{7} \times x73​×x, and the amount of the second liquid = 47×x\frac{4}{7} \times x74​×x.
  After removing 18 liters of the first liquid and adding 18 liters of the second liquid, the new ratio of the first liquid to the second liquid is 2:5.
  So,
  37x−1847x+18=25\frac{\frac{3}{7}x – 18}{\frac{4}{7}x + 18} = \frac{2}{5}74​x+1873​x−18​=52​.
  Solving this equation gives the initial quantity x=63x = 63x=63.
  Therefore, there were 63 liters of liquid in the container initially.

13. Average Calculation:

• Question: The average of five consecutive even numbers is 38. What is the sum of the largest and smallest numbers in the sequence?
• Answer: Let the five consecutive even numbers be x,x+2,x+4,x+6,x+8x, x+2, x+4, x+6, x+8x,x+2,x+4,x+6,x+8.
  The average of these numbers is given as 38:x+(x+2)+(x+4)+(x+6)+(x+8)5=38\frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} = 385x+(x+2)+(x+4)+(x+6)+(x+8)​=38Simplifying:5x+205=38\frac{5x + 20}{5} = 3855x+20​=38 5x+20=1905x + 20 = 1905x+20=190 5x=1705x = 1705x=170 x=34x = 34x=34The numbers are 34,36,38,40,4234, 36, 38, 40, 4234,36,38,40,42.
  The smallest number is 34, and the largest number is 42.
  The sum of the largest and smallest numbers is 34+42=7634 + 42 = 7634+42=76.
  Therefore, the sum is 76.

14. Time and Distance:

• Question: A car travels 120 kilometers in 2 hours. How long will it take to travel 450 kilometers at the same speed?
• Answer: Speed = 1202=60\frac{120}{2} = 602120​=60 km/h.
  Time = 45060=7.5\frac{450}{60} = 7.560450​=7.5 hours.
  Therefore, it will take 7.5 hours to travel 450 kilometers.

15. Ratio Problem:

• Question: The ratio of the ages of Sarah and Rachel is 5:7. If Rachel is 28 years old, how old is Sarah?
• Answer: Let Sarah’s age be xxx.
  The ratio of their ages is x28=57\frac{x}{28} = \frac{5}{7}28x​=75​.
  Cross-multiply:7x=5×287x = 5 \times 287x=5×28 7x=1407x = 1407x=140 x=20x = 20x=20Therefore, Sarah is 20 years old.

16. Simple Interest Calculation:

• Question: If $1000 is invested at an annual interest rate of 5%, what is the interest earned after 3 years?
• Answer: Simple Interest I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​, where:
  • P=1000P = 1000P=1000,
  • R=5R = 5R=5,
  • T=3T = 3T=3.
  Substituting the values:I=1000×5×3100=150I = \frac{1000 \times 5 \times 3}{100} = 150I=1001000×5×3​=150Therefore, the interest earned after 3 years is $150.

17. Fractional Operation:

• Question: What is the value of 25×34\frac{2}{5} \times \frac{3}{4}52​×43​?
• Answer: Multiply the numerators and denominators:25×34=2×35×4=620=310\frac{2}{5} \times \frac{3}{4} = \frac{2 \times 3}{5 \times 4} = \frac{6}{20} = \frac{3}{10}52​×43​=5×42×3​=206​=103​Therefore, the value is 310\frac{3}{10}103​.

18. Profit Calculation:

• Question: A shopkeeper sells an item for $2500 at a profit of 20%. What is the cost price of the item?
• Answer: Let the cost price be CCC.
  The selling price is given by:Selling Price=C+20% of C=C+20100×C=1.2C\text{Selling Price} = C + 20\% \text{ of } C = C + \frac{20}{100} \times C = 1.2CSelling Price=C+20% of C=C+10020​×C=1.2CSo,1.2C=25001.2C = 25001.2C=2500 C=25001.2=2083.33C = \frac{2500}{1.2} = 2083.33C=1.22500​=2083.33Therefore, the cost price of the item is $2083.33.

19. Age Calculation:

• Question: The sum of the ages of two friends is 40 years. If one friend is 4 years older than the other, what are their ages?
• Answer: Let the age of the younger friend be xxx.
  Then the age of the older friend is x+4x + 4x+4.
  The sum of their ages is 40:x+(x+4)=40x + (x + 4) = 40x+(x+4)=40Simplifying:2x+4=402x + 4 = 402x+4=40 2x=362x = 362x=36 x=18x = 18x=18Therefore, the younger friend is 18 years old, and the older friend is 18+4=2218 + 4 = 2218+4=22 years old.

20. Profit and Loss Percentage:

• Question: A person sells a product for $900, gaining a profit of 25%. What was the cost price?
• Answer: Let the cost price be CCC.
  The selling price is given by:Selling Price=C+25% of C=C+25100×C=1.25C\text{Selling Price} = C + 25\% \text{ of } C = C + \frac{25}{100} \times C = 1.25CSelling Price=C+25% of C=C+10025​×C=1.25CSo,1.25C=9001.25C = 9001.25C=900 C=9001.25=720C = \frac{900}{1.25} = 720C=1.25900​=720Therefore, the cost price is $720.

21. Time and Speed Problem:

• Question: A train travels 120 kilometers in 2 hours. What will be the time taken for the same train to travel 180 kilometers at the same speed?
• Answer: The speed of the train is 1202=60\frac{120}{2} = 602120​=60 km/h.
  Time taken to travel 180 kilometers = 18060=3\frac{180}{60} = 360180​=3 hours.
  Therefore, the time taken will be 3 hours.

22. Average Calculation:

• Question: The average of five consecutive odd numbers is 21. What is the largest of these numbers?
• Answer: Let the five consecutive odd numbers be x,x+2,x+4,x+6,x+8x, x+2, x+4, x+6, x+8x,x+2,x+4,x+6,x+8.
  The average of these numbers is given as 21:x+(x+2)+(x+4)+(x+6)+(x+8)5=21\frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} = 215x+(x+2)+(x+4)+(x+6)+(x+8)​=21Simplifying the equation:5x+205=21\frac{5x + 20}{5} = 2155x+20​=21 5x+20=1055x + 20 = 1055x+20=105 5x=855x = 855x=85 x=17x = 17x=17So, the numbers are 17, 19, 21, 23, 25.
  The largest number is 25.

23. Work and Time Problem:

• Question: A worker can complete a task in 10 days. If another worker joins and together they complete the task in 6 days, how many days will it take for the second worker to finish the task alone?
• Answer: Let the rate of work of the first worker be 110\frac{1}{10}101​ (since the first worker completes the task in 10 days).
  Let the rate of work of the second worker be 1x\frac{1}{x}x1​, where xxx is the number of days the second worker takes to finish the task.
  Together, their combined rate of work is 16\frac{1}{6}61​.
  So,110+1x=16\frac{1}{10} + \frac{1}{x} = \frac{1}{6}101​+x1​=61​Solving for xxx:1x=16−110=530−330=230=115\frac{1}{x} = \frac{1}{6} – \frac{1}{10} = \frac{5}{30} – \frac{3}{30} = \frac{2}{30} = \frac{1}{15}x1​=61​−101​=305​−303​=302​=151​Therefore, the second worker will complete the task in 15 days.

24. Ratio Problem:

• Question: The ratio of the ages of Alice and Bob is 4:5. If Bob is 40 years old, how old is Alice?
• Answer: Let Alice’s age be xxx.
  The ratio of their ages is x40=45\frac{x}{40} = \frac{4}{5}40x​=54​.
  Cross-multiply:5x=4×405x = 4 \times 405x=4×40 5x=1605x = 1605x=160 x=32x = 32x=32Therefore, Alice is 32 years old.

25. Profit and Loss Problem:

• Question: A man sells a watch for $800 at a profit of 25%. What was the cost price of the watch?
• Answer: Let the cost price be CCC.
  The selling price is given by:Selling Price=C+25% of C=C+25100×C=1.25C\text{Selling Price} = C + 25\% \text{ of } C = C + \frac{25}{100} \times C = 1.25CSelling Price=C+25% of C=C+10025​×C=1.25CSo,1.25C=8001.25C = 8001.25C=800 C=8001.25=640C = \frac{800}{1.25} = 640C=1.25800​=640Therefore, the cost price of the watch is $640.

26. Simple Interest:

• Question: If $1500 is invested at a rate of 6% per annum for 2 years, what will be the simple interest?
• Answer: Simple Interest I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​, where:
  • P=1500P = 1500P=1500,
  • R=6R = 6R=6,
  • T=2T = 2T=2.
  Substituting the values:I=1500×6×2100=180I = \frac{1500 \times 6 \times 2}{100} = 180I=1001500×6×2​=180Therefore, the simple interest earned is $180.

27. Fractional Addition:

• Question: What is the value of 34+58\frac{3}{4} + \frac{5}{8}43​+85​?
• Answer: To add the fractions, first make the denominators the same.
  34=68\frac{3}{4} = \frac{6}{8}43​=86​, so:68+58=118\frac{6}{8} + \frac{5}{8} = \frac{11}{8}86​+85​=811​Therefore, the sum is 118\frac{11}{8}811​, which is 1 and 38\frac{3}{8}83​.

28. Age Problem:

• Question: The sum of the ages of two sisters is 36 years. If one sister is 4 years older than the other, how old are they?
• Answer: Let the age of the younger sister be xxx.
  Then, the age of the older sister is x+4x + 4x+4.
  The sum of their ages is 36:x+(x+4)=36x + (x + 4) = 36x+(x+4)=36Simplifying:2x+4=362x + 4 = 362x+4=36 2x=322x = 322x=32 x=16x = 16x=16Therefore, the younger sister is 16 years old, and the older sister is 16+4=2016 + 4 = 2016+4=20 years old.

29. Percentage Discount:

• Question: A jacket is originally priced at $150. If a 20% discount is applied, what is the sale price of the jacket?
• Answer: Discount = 20% of 150=20100×150=3020\% \text{ of } 150 = \frac{20}{100} \times 150 = 3020% of 150=10020​×150=30.
  Sale price = Original price – Discount = 150−30=120150 – 30 = 120150−30=120.
  Therefore, the sale price of the jacket is $120.

30. Speed and Distance Problem:

• Question: A person walks at a speed of 6 km/h for 2 hours, then at 4 km/h for 1 hour. What is his average speed for the entire trip?
• Answer: Distance covered in the first part of the trip = 6×2=126 \times 2 = 126×2=12 km.
  Distance covered in the second part of the trip = 4×1=44 \times 1 = 44×1=4 km.
  Total distance = 12+4=1612 + 4 = 1612+4=16 km.
  Total time = 2+1=32 + 1 = 32+1=3 hours.
  Average speed = Total distanceTotal time=163≈5.33\frac{\text{Total distance}}{\text{Total time}} = \frac{16}{3} \approx 5.33Total timeTotal distance​=316​≈5.33 km/h.
  Therefore, the average speed is 5.33 km/h.

31. Average Age:

• Question: The average of five consecutive even numbers is 30. What is the smallest number?
• Answer: Let the five consecutive even numbers be x,x+2,x+4,x+6,x+8x, x+2, x+4, x+6, x+8x,x+2,x+4,x+6,x+8.
  The average of these numbers is given as 30:x+(x+2)+(x+4)+(x+6)+(x+8)5=30\frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} = 305x+(x+2)+(x+4)+(x+6)+(x+8)​=30Simplifying:5x+205=30\frac{5x + 20}{5} = 3055x+20​=30 5x+20=1505x + 20 = 1505x+20=150 5x=1305x = 1305x=130 x=26x = 26x=26Therefore, the smallest number is 26.

32. Profit Loss and Discount:

• Question: A shopkeeper sells a book for $180 after offering a discount of 10%. What was the original price of the book?
• Answer: Let the original price be PPP.
  After a 10% discount, the selling price becomes P−10% of P=0.9PP – 10\% \text{ of } P = 0.9PP−10% of P=0.9P.
  So,0.9P=1800.9P = 1800.9P=180 P=1800.9=200P = \frac{180}{0.9} = 200P=0.9180​=200Therefore, the original price of the book was $200.

33. Ratio of Speeds:

• Question: Two cars start from the same point at the same time. The first car travels at 60 km/h, and the second car travels at 80 km/h. What is the ratio of the speeds of the two cars?
• Answer: The ratio of the speeds of the two cars is 6080=34\frac{60}{80} = \frac{3}{4}8060​=43​.
  Therefore, the ratio of the speeds is 3:4.

34. Simple Interest Calculation:

• Question: If $2000 is invested at an interest rate of 8% per annum for 5 years, what will be the total amount after the investment period?
• Answer: Simple Interest I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​, where:
  • P=2000P = 2000P=2000,
  • R=8R = 8R=8,
  • T=5T = 5T=5.
  Substituting the values:I=2000×8×5100=800I = \frac{2000 \times 8 \times 5}{100} = 800I=1002000×8×5​=800Total amount A=P+I=2000+800=2800A = P + I = 2000 + 800 = 2800A=P+I=2000+800=2800.
  Therefore, the total amount after 5 years is $2800.

35. Age Difference:

• Question: The sum of the ages of two friends is 36 years. If one friend is 6 years older than the other, how old is each friend?
• Answer: Let the age of the younger friend be xxx.
  Then, the age of the older friend is x+6x + 6x+6.
  The sum of their ages is 36:x+(x+6)=36x + (x + 6) = 36x+(x+6)=36Simplifying:2x+6=362x + 6 = 362x+6=36 2x=302x = 302x=30 x=15x = 15x=15Therefore, the younger friend is 15 years old, and the older friend is 15+6=2115 + 6 = 2115+6=21 years old.

36. Ratio Problem:

• Question: A fruit basket contains apples and oranges in the ratio 3:5. If there are 15 apples, how many oranges are there?
• Answer: Let the number of oranges be xxx.
  The ratio of apples to oranges is 15x=35\frac{15}{x} = \frac{3}{5}x15​=53​.
  Cross-multiply:3x=15×53x = 15 \times 53x=15×5 3x=753x = 753x=75 x=25x = 25x=25Therefore, there are 25 oranges.

37. Mixture Problem:

• Question: A container has a mixture of two liquids in the ratio 7:3. If 6 liters of the mixture are removed and replaced with 6 liters of the second liquid, the ratio becomes 1:1. How many liters of the first liquid were there initially?
• Answer: Let the total quantity of the mixture be xxx liters.
  Initially, the amount of the first liquid is 710×x\frac{7}{10} \times x107​×x, and the amount of the second liquid is 310×x\frac{3}{10} \times x103​×x.
  After removing 6 liters of the mixture, the amount of the first liquid is 710×x−6\frac{7}{10} \times x – 6107​×x−6, and the amount of the second liquid is 310×x+6\frac{3}{10} \times x + 6103​×x+6.
  The new ratio is 1:1, so:710x−6310x+6=1\frac{\frac{7}{10}x – 6}{\frac{3}{10}x + 6} = 1103​x+6107​x−6​=1Solving this equation:710x−6=310x+6\frac{7}{10}x – 6 = \frac{3}{10}x + 6107​x−6=103​x+6 710x−310x=12\frac{7}{10}x – \frac{3}{10}x = 12107​x−103​x=12 410x=12\frac{4}{10}x = 12104​x=12 x=30x = 30x=30Therefore, the initial quantity of the mixture was 30 liters, and the amount of the first liquid initially was 710×30=21\frac{7}{10} \times 30 = 21107​×30=21 liters.

38. Time and Work Problem:

• Question: If 6 workers can complete a task in 8 days, how many days will it take 12 workers to complete the same task?
• Answer: The total work done is inversely proportional to the number of workers.
  If 6 workers can complete the task in 8 days, then 12 workers will take half the time to complete the task.
  So,
  Time taken by 12 workers = 82=4\frac{8}{2} = 428​=4 days.
  Therefore, it will take 4 days for 12 workers to complete the task.

39. Speed and Distance Problem:

• Question: A cyclist travels 30 kilometers in 2 hours. How long will it take him to travel 75 kilometers at the same speed?
• Answer: Speed = 302=15\frac{30}{2} = 15230​=15 km/h.
  Time to travel 75 kilometers = 7515=5\frac{75}{15} = 51575​=5 hours.
  Therefore, it will take the cyclist 5 hours to travel 75 kilometers.

40. Percentage Calculation:

• Question: A student scored 75 marks out of 100 in an exam. What percentage did the student score?
• Answer: Percentage = 75100×100=75%\frac{75}{100} \times 100 = 75\%10075​×100=75%.
  Therefore, the student scored 75%.

41. Number Series:

• Question: What is the next number in the series: 2, 6, 12, 20, 30, __?
• Answer: The difference between consecutive numbers is increasing by 2:6−2=4,12−6=6,20−12=8,30−20=106 – 2 = 4, \quad 12 – 6 = 6, \quad 20 – 12 = 8, \quad 30 – 20 = 106−2=4,12−6=6,20−12=8,30−20=10So, the next difference will be 12.
  Therefore, the next number in the series is:30+12=4230 + 12 = 4230+12=42So, the next number is 42.

42. Simple Interest Calculation:

• Question: A sum of money amounts to $3000 in 2 years at a rate of 5% per annum. What is the principal amount?
• Answer: Using the formula for simple interest:A=P+P×R×T100A = P + \frac{P \times R \times T}{100}A=P+100P×R×T​Given, A=3000A = 3000A=3000, R=5%R = 5\%R=5%, T=2T = 2T=2.
  Substituting the values:3000=P+P×5×21003000 = P + \frac{P \times 5 \times 2}{100}3000=P+100P×5×2​ 3000=P+10P100=P+0.1P=1.1P3000 = P + \frac{10P}{100} = P + 0.1P = 1.1P3000=P+10010P​=P+0.1P=1.1P P=30001.1=2727.27P = \frac{3000}{1.1} = 2727.27P=1.13000​=2727.27Therefore, the principal amount is $2727.27.

43. Age Problem:

• Question: The present age of A is 4 years more than B. In 5 years, the sum of their ages will be 50 years. What are their current ages?
• Answer: Let the age of B be xxx, and the age of A will be x+4x + 4x+4.
  In 5 years, their ages will be x+5x + 5x+5 and (x+4)+5=x+9(x + 4) + 5 = x + 9(x+4)+5=x+9, respectively.
  The sum of their ages in 5 years is given as 50:(x+5)+(x+9)=50(x + 5) + (x + 9) = 50(x+5)+(x+9)=50Simplifying:2x+14=502x + 14 = 502x+14=50 2x=362x = 362x=36 x=18x = 18x=18Therefore, B’s current age is 18 years, and A’s age is 18+4=2218 + 4 = 2218+4=22 years.

44. Ratio and Proportion:

• Question: The ratio of the number of girls to boys in a class is 3:4. If there are 84 students in total, how many girls are there?
• Answer: Let the number of girls be 3x3x3x and the number of boys be 4x4x4x.
  The total number of students is 84:3x+4x=843x + 4x = 843x+4x=84 7x=847x = 847x=84 x=12x = 12x=12Therefore, the number of girls is 3x=3×12=363x = 3 \times 12 = 363x=3×12=36.

45. Percentage Problem:

• Question: In a class of 60 students, 30% are absent on a particular day. How many students are present?
• Answer: The number of students absent is:30% of 60=30100×60=1830\% \text{ of } 60 = \frac{30}{100} \times 60 = 1830% of 60=10030​×60=18Therefore, the number of students present is:60−18=4260 – 18 = 4260−18=42So, 42 students are present.

46. Time and Work Problem:

• Question: A machine can complete a job in 6 hours, while another machine takes 9 hours. If both machines work together, how long will it take them to complete the job?
• Answer: The rate of work of the first machine is 16\frac{1}{6}61​ jobs per hour, and the rate of work of the second machine is 19\frac{1}{9}91​ jobs per hour.
  When both machines work together, their combined rate of work is:16+19=318+218=518\frac{1}{6} + \frac{1}{9} = \frac{3}{18} + \frac{2}{18} = \frac{5}{18}61​+91​=183​+182​=185​Therefore, the time taken to complete the job is:Time=1518=185=3.6 hours\text{Time} = \frac{1}{\frac{5}{18}} = \frac{18}{5} = 3.6 \text{ hours}Time=185​1​=518​=3.6 hoursSo, the job will be completed in 3.6 hours.

47. Profit and Loss Problem:

• Question: A man buys a book for $60 and sells it for $72. What is the profit percentage?
• Answer: The profit is 72−60=1272 – 60 = 1272−60=12 dollars.
  The profit percentage is:ProfitCost Price×100=1260×100=20%\frac{\text{Profit}}{\text{Cost Price}} \times 100 = \frac{12}{60} \times 100 = 20\%Cost PriceProfit​×100=6012​×100=20%Therefore, the profit percentage is 20%.

48. Time and Speed Problem:

• Question: A train travels 120 km in 2 hours. How long will it take to cover 240 km at the same speed?
• Answer: The speed of the train is 1202=60\frac{120}{2} = 602120​=60 km/h.
  Time taken to cover 240 km is:24060=4 hours\frac{240}{60} = 4 \text{ hours}60240​=4 hoursTherefore, it will take the train 4 hours to cover 240 km.

49. Compound Interest:

• Question: If $1000 is invested at an interest rate of 10% per annum, compounded annually, what will be the amount after 2 years?
• Answer: The formula for compound interest is:A=P(1+R100)TA = P \left(1 + \frac{R}{100}\right)^TA=P(1+100R​)TWhere P=1000P = 1000P=1000, R=10%R = 10\%R=10%, and T=2T = 2T=2.
  Substituting the values:A=1000(1+10100)2=1000×1.12=1000×1.21=1210A = 1000 \left(1 + \frac{10}{100}\right)^2 = 1000 \times 1.1^2 = 1000 \times 1.21 = 1210A=1000(1+10010​)2=1000×1.12=1000×1.21=1210Therefore, the amount after 2 years is $1210.

50. LCM and HCF Problem:

• Question: Find the LCM and HCF of 12 and 18.
• Answer: The prime factorization of 12 is 22×32^2 \times 322×3, and for 18, it is 2×322 \times 3^22×32.
  To find the HCF (Highest Common Factor), take the lowest powers of the common factors:HCF=21×31=6\text{HCF} = 2^1 \times 3^1 = 6HCF=21×31=6To find the LCM (Least Common Multiple), take the highest powers of all the factors:LCM=22×32=36\text{LCM} = 2^2 \times 3^2 = 36LCM=22×32=36Therefore, the HCF is 6 and the LCM is 36.

51. Mixture Problem:

• Question: A container contains 60 liters of a solution with 25% alcohol. How much pure alcohol is in the solution?
• Answer: The amount of alcohol is:25100×60=15 liters\frac{25}{100} \times 60 = 15 \text{ liters}10025​×60=15 litersTherefore, the amount of pure alcohol in the solution is 15 liters.

52. Work and Time Problem:

• Question: A worker can complete a job in 12 hours, while another worker can complete the same job in 18 hours. If they work together, how long will it take them to complete the job?
• Answer: The rate of the first worker is 112\frac{1}{12}121​ of the job per hour, and the rate of the second worker is 118\frac{1}{18}181​ of the job per hour.
  When both workers work together, their combined rate of work is:112+118=336+236=536\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}121​+181​=363​+362​=365​Therefore, the time taken to complete the job is:Time=1536=365=7.2 hours\text{Time} = \frac{1}{\frac{5}{36}} = \frac{36}{5} = 7.2 \text{ hours}Time=365​1​=536​=7.2 hoursSo, it will take 7.2 hours to complete the job.

53. Simple Interest Problem:

• Question: A sum of money is invested at 6% per annum simple interest. If the total interest after 4 years is $240, what is the principal amount?
• Answer: Using the formula for simple interest:I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​Given, I=240I = 240I=240, R=6%R = 6\%R=6%, and T=4T = 4T=4. Substituting the values:240=P×6×4100240 = \frac{P \times 6 \times 4}{100}240=100P×6×4​ 240=24P100240 = \frac{24P}{100}240=10024P​ 240=0.24P240 = 0.24P240=0.24P P=2400.24=1000P = \frac{240}{0.24} = 1000P=0.24240​=1000Therefore, the principal amount is $1000.

54. Speed and Distance Problem:

• Question: A person walks 4 km in 40 minutes. How far will the person walk in 2 hours at the same speed?
• Answer: The speed of the person is 44060=6\frac{4}{\frac{40}{60}} = 66040​4​=6 km/h.
  In 2 hours, the person will walk:6×2=12 km6 \times 2 = 12 \text{ km}6×2=12 kmTherefore, the person will walk 12 km in 2 hours.

55. Discount Problem:

• Question: An item is sold for $500 after a 20% discount. What was the original price of the item?
• Answer: Let the original price be PPP.
  After a 20% discount, the selling price is 80%80\%80% of the original price, so:0.8P=5000.8P = 5000.8P=500 P=5000.8=625P = \frac{500}{0.8} = 625P=0.8500​=625Therefore, the original price of the item was $625.

56. Ratio Problem:

• Question: The ratio of the number of girls to boys in a class is 5:3. If there are 120 students in the class, how many girls are there?
• Answer: Let the number of girls be 5x5x5x and the number of boys be 3x3x3x.
  The total number of students is 120:5x+3x=1205x + 3x = 1205x+3x=120 8x=1208x = 1208x=120 x=15x = 15x=15Therefore, the number of girls is 5x=5×15=755x = 5 \times 15 = 755x=5×15=75.

57. Age Problem:

• Question: The sum of the ages of A and B is 60 years. In 5 years, A will be twice as old as B. How old are A and B currently?
• Answer: Let A’s age be xxx and B’s age be 60−x60 – x60−x.
  In 5 years, A’s age will be x+5x + 5x+5 and B’s age will be 60−x+5=65−x60 – x + 5 = 65 – x60−x+5=65−x.
  According to the problem, A’s age in 5 years will be twice B’s age:x+5=2(65−x)x + 5 = 2(65 – x)x+5=2(65−x)Simplifying:x+5=130−2xx + 5 = 130 – 2xx+5=130−2x 3x=1253x = 1253x=125 x=1253≈41.67x = \frac{125}{3} \approx 41.67x=3125​≈41.67Therefore, A is approximately 41.67 years old, and B is 60−41.67=18.3360 – 41.67 = 18.3360−41.67=18.33 years old.

58. Percentage Profit and Loss:

• Question: A shopkeeper buys a watch for $200 and sells it for $240. What is the percentage profit?
• Answer: The profit is 240−200=40240 – 200 = 40240−200=40 dollars.
  The profit percentage is:40200×100=20%\frac{40}{200} \times 100 = 20\%20040​×100=20%Therefore, the profit percentage is 20%.

59. Time and Work Problem:

• Question: If 8 men can complete a task in 12 days, how many men will be needed to complete the task in 6 days?
• Answer: The total work is 8×12=968 \times 12 = 968×12=96 man-days.
  To complete the task in 6 days, the number of men required will be:966=16 men\frac{96}{6} = 16 \text{ men}696​=16 menTherefore, 16 men will be needed.

60. Profit and Loss Problem:

• Question: A trader sells a product for $150, making a profit of 25%. What is the cost price of the product?
• Answer: Let the cost price be PPP.
  The selling price is P+25% of P=1.25PP + 25\% \text{ of } P = 1.25PP+25% of P=1.25P.
  Given, the selling price is $150, so:1.25P=1501.25P = 1501.25P=150 P=1501.25=120P = \frac{150}{1.25} = 120P=1.25150​=120Therefore, the cost price of the product is $120.

61. Speed and Time Problem:

• Question: A car travels 180 km in 3 hours. How long will it take to travel 450 km at the same speed?
• Answer: The speed of the car is:1803=60 km/h\frac{180}{3} = 60 \text{ km/h}3180​=60 km/hThe time taken to travel 450 km at the same speed is:45060=7.5 hours\frac{450}{60} = 7.5 \text{ hours}60450​=7.5 hoursTherefore, it will take 7.5 hours to travel 450 km.

62. Profit and Discount Problem:

• Question: A shopkeeper buys a watch for $500 and sells it for $650. What is the profit percentage?
• Answer: The profit is 650−500=150650 – 500 = 150650−500=150 dollars.
  The profit percentage is:150500×100=30%\frac{150}{500} \times 100 = 30\%500150​×100=30%Therefore, the profit percentage is 30%.

63. Age Problem:

• Question: A father is 30 years older than his son. In 10 years, the father will be twice as old as his son. What are their current ages?
• Answer: Let the son’s current age be xxx.
  The father’s current age is x+30x + 30x+30.
  In 10 years, the father’s age will be x+30+10=x+40x + 30 + 10 = x + 40x+30+10=x+40, and the son’s age will be x+10x + 10x+10.
  According to the problem:x+40=2(x+10)x + 40 = 2(x + 10)x+40=2(x+10)Simplifying:x+40=2x+20x + 40 = 2x + 20x+40=2x+20 40−20=2x−x40 – 20 = 2x – x40−20=2x−x x=20x = 20x=20Therefore, the son’s current age is 20 years, and the father’s current age is 20+30=5020 + 30 = 5020+30=50 years.

64. Ratio and Proportion:

• Question: The ratio of the number of boys to girls in a class is 4:5. If there are 180 students in total, how many boys are there?
• Answer: Let the number of boys be 4x4x4x and the number of girls be 5x5x5x.
  The total number of students is 180:4x+5x=1804x + 5x = 1804x+5x=180 9x=1809x = 1809x=180 x=20x = 20x=20Therefore, the number of boys is 4x=4×20=804x = 4 \times 20 = 804x=4×20=80.

65. Percentage Increase Problem:

• Question: The price of a shirt was $40 last month. This month, the price increased by 15%. What is the new price?
• Answer: The price increase is 15% of 40=15100×40=615\% \text{ of } 40 = \frac{15}{100} \times 40 = 615% of 40=10015​×40=6.
  Therefore, the new price is:40+6=4640 + 6 = 4640+6=46So, the new price of the shirt is $46.

Also read : Verbal reasoning question on Clock and Calendar

66. Time and Work Problem:

• Question: If 4 men can complete a job in 12 days, how many men will be required to complete the job in 6 days?
• Answer: The total work is 4×12=484 \times 12 = 484×12=48 man-days.
  To complete the job in 6 days, the number of men required will be:486=8 men\frac{48}{6} = 8 \text{ men}648​=8 menTherefore, 8 men will be required.

67. Time and Distance Problem:

• Question: A train travels at 80 km/h. How long will it take to cover 160 km?
• Answer: The time taken to cover 160 km is:16080=2 hours\frac{160}{80} = 2 \text{ hours}80160​=2 hoursTherefore, it will take 2 hours.

68. Number Series:

• Question: What is the next number in the series: 1, 4, 9, 16, 25, __?
• Answer: The numbers are perfect squares:12,22,32,42,521^2, 2^2, 3^2, 4^2, 5^212,22,32,42,52The next number in the series is:62=366^2 = 3662=36So, the next number is 36.

69. Simple Interest Problem:

• Question: A sum of money amounts to $2000 in 3 years at 5% simple interest. What is the principal amount?
• Answer: Using the simple interest formula:A=P+P×R×T100A = P + \frac{P \times R \times T}{100}A=P+100P×R×T​Given, A=2000A = 2000A=2000, R=5%R = 5\%R=5%, and T=3T = 3T=3. Substituting the values:2000=P+P×5×31002000 = P + \frac{P \times 5 \times 3}{100}2000=P+100P×5×3​ 2000=P+15P100=P+0.15P=1.15P2000 = P + \frac{15P}{100} = P + 0.15P = 1.15P2000=P+10015P​=P+0.15P=1.15P P=20001.15≈1739.13P = \frac{2000}{1.15} \approx 1739.13P=1.152000​≈1739.13Therefore, the principal amount is approximately $1739.13.

70. Fraction Problem:

• Question: What is 12+13\frac{1}{2} + \frac{1}{3}21​+31​?
• Answer: To add the fractions, we find the least common denominator:12+13=36+26=56\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}21​+31​=63​+62​=65​Therefore, the result is 56\frac{5}{6}65​.

71. Speed and Time Problem:

• Question: A cyclist covers 50 km in 2 hours. How far will the cyclist cover in 5 hours?
• Answer: The speed of the cyclist is:502=25 km/h\frac{50}{2} = 25 \text{ km/h}250​=25 km/hThe distance covered in 5 hours is:25×5=125 km25 \times 5 = 125 \text{ km}25×5=125 kmTherefore, the cyclist will cover 125 km in 5 hours.

72. Average Problem:

• Question: The average of 10 numbers is 20. If one of the numbers is excluded, the average becomes 18. What is the excluded number?
• Answer: The sum of all 10 numbers is:10×20=20010 \times 20 = 20010×20=200The sum of the remaining 9 numbers is:9×18=1629 \times 18 = 1629×18=162Therefore, the excluded number is:200−162=38200 – 162 = 38200−162=38So, the excluded number is 38.

73. Time and Distance Problem:

• Question: A car travels 100 km in 2 hours. How far will it travel in 5 hours?
• Answer: The speed of the car is:1002=50 km/h\frac{100}{2} = 50 \text{ km/h}2100​=50 km/hThe distance traveled in 5 hours is:50×5=250 km50 \times 5 = 250 \text{ km}50×5=250 kmTherefore, the car will travel 250 km in 5 hours.

74. Simple Interest Problem:

• Question: A sum of money amounts to $1500 in 2 years at 10% simple interest. What is the interest earned?
• Answer: Using the formula for simple interest:I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​Given, A=1500A = 1500A=1500, R=10%R = 10\%R=10%, and T=2T = 2T=2. The principal PPP is:P=15001+R×T100=15001+10×2100=15001+0.2=15001.2=1250P = \frac{1500}{1 + \frac{R \times T}{100}} = \frac{1500}{1 + \frac{10 \times 2}{100}} = \frac{1500}{1 + 0.2} = \frac{1500}{1.2} = 1250P=1+100R×T​1500​=1+10010×2​1500​=1+0.21500​=1.21500​=1250The interest earned is:1500−1250=2501500 – 1250 = 2501500−1250=250Therefore, the interest earned is $250.

75. Age Problem:

• Question: A mother is 24 years older than her daughter. In 6 years, the mother will be twice as old as the daughter. How old are they currently?
• Answer: Let the daughter’s age be xxx.
  The mother’s age is x+24x + 24x+24.
  In 6 years, the mother’s age will be x+24+6=x+30x + 24 + 6 = x + 30x+24+6=x+30, and the daughter’s age will be x+6x + 6x+6.
  According to the problem:x+30=2(x+6)x + 30 = 2(x + 6)x+30=2(x+6)Simplifying:x+30=2x+12x + 30 = 2x + 12x+30=2x+12 30−12=2x−x30 – 12 = 2x – x30−12=2x−x x=18x = 18x=18Therefore, the daughter is 18 years old, and the mother is 18+24=4218 + 24 = 4218+24=42 years old.

76. Ratio Problem:

• Question: The ratio of the number of apples to oranges in a basket is 3:5. If there are 48 fruits in total, how many apples are there?
• Answer: Let the number of apples be 3x3x3x and the number of oranges be 5x5x5x.
  The total number of fruits is 48:3x+5x=483x + 5x = 483x+5x=48 8x=488x = 488x=48 x=6x = 6x=6Therefore, the number of apples is 3x=3×6=183x = 3 \times 6 = 183x=3×6=18.

77. Profit and Loss Problem:

• Question: A shopkeeper bought an item for $300 and sold it for $450. What is the percentage profit?
• Answer: The profit is 450−300=150450 – 300 = 150450−300=150 dollars.
  The percentage profit is:150300×100=50%\frac{150}{300} \times 100 = 50\%300150​×100=50%Therefore, the profit percentage is 50%.

78. Work and Time Problem:

• Question: If 10 men can do a piece of work in 15 days, how many men will be needed to finish the work in 10 days?
• Answer: The total work is 10×15=15010 \times 15 = 15010×15=150 man-days.
  To complete the work in 10 days, the number of men required will be:15010=15 men\frac{150}{10} = 15 \text{ men}10150​=15 menTherefore, 15 men will be required.

79. Percentage Problem:

• Question: A student scored 80 marks out of 100 in an exam. What is his percentage?
• Answer: The percentage is:80100×100=80%\frac{80}{100} \times 100 = 80\%10080​×100=80%Therefore, the student’s percentage is 80%.

80. Speed and Distance Problem:

• Question: A train travels 150 km in 3 hours. How long will it take to travel 400 km at the same speed?
• Answer: The speed of the train is:1503=50 km/h\frac{150}{3} = 50 \text{ km/h}3150​=50 km/hThe time taken to travel 400 km at the same speed is:40050=8 hours\frac{400}{50} = 8 \text{ hours}50400​=8 hoursTherefore, it will take 8 hours to travel 400 km.

81. Mixture Problem:

• Question: A solution contains 30% salt. How much of the solution must be taken to obtain 90 grams of salt?
• Answer: Let the amount of solution to be taken be xxx.
  The amount of salt in the solution is 30%30\%30% of xxx, so:30100×x=90\frac{30}{100} \times x = 9010030​×x=90 0.3x=900.3x = 900.3x=90 x=900.3=300x = \frac{90}{0.3} = 300x=0.390​=300Therefore, 300 grams of the solution must be taken.

82. Age Problem:

• Question: A father is 40 years old and his son is 12 years old. How many years ago was the father three times as old as his son?
• Answer: Let the number of years ago be xxx.
  In xxx years, the father’s age was 40−x40 – x40−x, and the son’s age was 12−x12 – x12−x.
  According to the problem:40−x=3(12−x)40 – x = 3(12 – x)40−x=3(12−x)Simplifying:40−x=36−3×40 – x = 36 – 3×40−x=36−3x 40−36=−3x+x40 – 36 = -3x + x40−36=−3x+x 4=−2×4 = -2×4=−2x x=−2x = -2x=−2Therefore, 2 years ago, the father was three times as old as the son.

83. Simple Interest Problem:

• Question: A sum of money is invested at 8% simple interest for 4 years. If the total interest is $320, what is the principal amount?
• Answer: Using the formula for simple interest:I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​Given, I=320I = 320I=320, R=8%R = 8\%R=8%, and T=4T = 4T=4. Substituting the values:320=P×8×4100320 = \frac{P \times 8 \times 4}{100}320=100P×8×4​ 320=32P100320 = \frac{32P}{100}320=10032P​ 320=0.32P320 = 0.32P320=0.32P P=3200.32=1000P = \frac{320}{0.32} = 1000P=0.32320​=1000Therefore, the principal amount is $1000.

84. Time and Work Problem:

• Question: A piece of work can be completed by 8 men in 10 days. How many days will it take 5 men to complete the same work?
• Answer: The total work is 8×10=808 \times 10 = 808×10=80 man-days.
  To complete the work in xxx days, the number of men required will be:5×x=805 \times x = 805×x=80 x=805=16x = \frac{80}{5} = 16x=580​=16Therefore, it will take 16 days for 5 men to complete the work.

85. Ratio and Proportion Problem:

• Question: The ratio of the number of boys to girls in a class is 5:3. If the total number of students is 40, how many girls are there?
• Answer: Let the number of boys be 5x5x5x and the number of girls be 3x3x3x.
  The total number of students is 40:5x+3x=405x + 3x = 405x+3x=40 8x=408x = 408x=40 x=5x = 5x=5Therefore, the number of girls is 3x=3×5=153x = 3 \times 5 = 153x=3×5=15.

86. Percentage Problem:

• Question: The price of an article is reduced by 20%. If the original price is $250, what is the new price?
• Answer: The price reduction is 20% of 250=20100×250=5020\% \text{ of } 250 = \frac{20}{100} \times 250 = 5020% of 250=10020​×250=50.
  Therefore, the new price is:250−50=200250 – 50 = 200250−50=200So, the new price is $200.

87. Distance Problem:

• Question: A car travels 120 km in 3 hours. How long will it take to travel 200 km at the same speed?
• Answer: The speed of the car is:1203=40 km/h\frac{120}{3} = 40 \text{ km/h}3120​=40 km/hThe time taken to travel 200 km is:20040=5 hours\frac{200}{40} = 5 \text{ hours}40200​=5 hoursTherefore, it will take 5 hours to travel 200 km.

88. Simple Interest Problem:

• Question: A sum of money amounts to $1200 in 2 years at 10% simple interest. What is the principal amount?
• Answer: Using the simple interest formula:A=P+P×R×T100A = P + \frac{P \times R \times T}{100}A=P+100P×R×T​Given, A=1200A = 1200A=1200, R=10%R = 10\%R=10%, and T=2T = 2T=2. Substituting the values:1200=P+P×10×21001200 = P + \frac{P \times 10 \times 2}{100}1200=P+100P×10×2​ 1200=P+20P100=P+0.2P=1.2P1200 = P + \frac{20P}{100} = P + 0.2P = 1.2P1200=P+10020P​=P+0.2P=1.2P P=12001.2=1000P = \frac{1200}{1.2} = 1000P=1.21200​=1000Therefore, the principal amount is $1000.

89. Time and Work Problem:

• Question: A pipe can fill a tank in 12 hours, and another pipe can empty the tank in 15 hours. If both pipes are open, how long will it take to fill the tank?
• Answer: The first pipe fills 112\frac{1}{12}121​ of the tank per hour, and the second pipe empties 115\frac{1}{15}151​ of the tank per hour.
  The net rate of filling the tank is:112−115=560−460=160\frac{1}{12} – \frac{1}{15} = \frac{5}{60} – \frac{4}{60} = \frac{1}{60}121​−151​=605​−604​=601​Therefore, the time taken to fill the tank is 606060 hours.

90. Ratio Problem:

• Question: The ratio of the number of girls to boys in a class is 4:3. If the number of boys is 120, how many girls are there?
• Answer: Let the number of girls be 4x4x4x and the number of boys be 3x3x3x.
  Given, the number of boys is 120:3x=1203x = 1203x=120 x=40x = 40x=40Therefore, the number of girls is 4x=4×40=1604x = 4 \times 40 = 1604x=4×40=160.

91. Percentage Problem:

• Question: A price is reduced by 25% to $60. What was the original price?
• Answer: Let the original price be xxx.
  After a 25% reduction, the price becomes 75%75\%75% of the original price:0.75x=600.75x = 600.75x=60 x=600.75=80x = \frac{60}{0.75} = 80x=0.7560​=80Therefore, the original price was $80.

92. Simple Interest Problem:

• Question: A sum of money is invested at 5% simple interest for 3 years. The interest earned is $150. What is the principal amount?
• Answer: Using the formula for simple interest:I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​Given, I=150I = 150I=150, R=5%R = 5\%R=5%, and T=3T = 3T=3. Substituting the values:150=P×5×3100150 = \frac{P \times 5 \times 3}{100}150=100P×5×3​ 150=15P100150 = \frac{15P}{100}150=10015P​ 150=0.15P150 = 0.15P150=0.15P P=1500.15=1000P = \frac{150}{0.15} = 1000P=0.15150​=1000Therefore, the principal amount is $1000.

93. Age Problem:

• Question: The sum of the ages of a father and his son is 56 years. The father is four times as old as the son. How old are they?
• Answer: Let the son’s age be xxx.
  The father’s age is 4x4x4x.
  According to the problem:x+4x=56x + 4x = 56x+4x=56 5x=565x = 565x=56 x=11.2x = 11.2x=11.2Therefore, the son’s age is 11.2 years, and the father’s age is 4×11.2=44.84 \times 11.2 = 44.84×11.2=44.8 years.

94. Speed and Time Problem:

• Question: A person walks 60 meters in 12 seconds. How long will it take him to walk 120 meters?
• Answer: The speed of the person is:6012=5 meters per second\frac{60}{12} = 5 \text{ meters per second}1260​=5 meters per secondThe time taken to walk 120 meters is:1205=24 seconds\frac{120}{5} = 24 \text{ seconds}5120​=24 secondsTherefore, it will take 24 seconds to walk 120 meters.

95. Profit and Loss Problem:

• Question: A person buys a chair for $150 and sells it for $180. What is the profit percentage?
• Answer: The profit is 180−150=30180 – 150 = 30180−150=30 dollars.
  The profit percentage is:30150×100=20%\frac{30}{150} \times 100 = 20\%15030​×100=20%Therefore, the profit percentage is 20%.

96. Ratio Problem:

• Question: The ratio of the ages of two friends is 5:7. If the sum of their ages is 48 years, what are their ages?
• Answer: Let the ages of the two friends be 5x5x5x and 7x7x7x.
  The sum of their ages is 48:5x+7x=485x + 7x = 485x+7x=48 12x=4812x = 4812x=48 x=4x = 4x=4Therefore, the first friend’s age is 5x=5×4=205x = 5 \times 4 = 205x=5×4=20 years, and the second friend’s age is 7x=7×4=287x = 7 \times 4 = 287x=7×4=28 years.

97. Time and Work Problem:

• Question: If 12 workers can complete a task in 18 days, how many workers are needed to complete the task in 9 days?
• Answer: The total work is 12×18=21612 \times 18 = 21612×18=216 worker-days.
  To complete the task in 9 days, the number of workers required will be:2169=24\frac{216}{9} = 249216​=24Therefore, 24 workers will be required.

98. Distance and Speed Problem:

• Question: A person travels 120 km in 3 hours. How far will he travel in 5 hours at the same speed?
• Answer: The speed of the person is:1203=40 km/h\frac{120}{3} = 40 \text{ km/h}3120​=40 km/hThe distance traveled in 5 hours is:40×5=200 km40 \times 5 = 200 \text{ km}40×5=200 kmTherefore, the person will travel 200 km in 5 hours.

99. Simple Interest Problem:

• Question: A sum of money is invested at 6% simple interest for 4 years. The total interest earned is $480. What is the principal amount?
• Answer: Using the formula for simple interest:I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​Given, I=480I = 480I=480, R=6%R = 6\%R=6%, and T=4T = 4T=4. Substituting the values:480=P×6×4100480 = \frac{P \times 6 \times 4}{100}480=100P×6×4​ 480=24P100480 = \frac{24P}{100}480=10024P​ 480=0.24P480 = 0.24P480=0.24P P=4800.24=2000P = \frac{480}{0.24} = 2000P=0.24480​=2000Therefore, the principal amount is $2000.

100. Work and Time Problem:

• Question: If 5 men can complete a job in 20 days, how many days will it take for 8 men to complete the same job?
• Answer: The total work is 5×20=1005 \times 20 = 1005×20=100 man-days.
  To complete the job in xxx days, the number of men required will be:8×x=1008 \times x = 1008×x=100 x=1008=12.5x = \frac{100}{8} = 12.5x=8100​=12.5Therefore, it will take 12.5 days for 8 men to complete the job.

101. Time and Distance Problem:

• Question: A car travels 75 km in 1.5 hours. What is its average speed?
• Answer: The average speed is:751.5=50 km/h\frac{75}{1.5} = 50 \text{ km/h}1.575​=50 km/hTherefore, the average speed of the car is 50 km/h.

102. Age Problem:

• Question: The sum of the ages of two people is 50 years. In 5 years, the older person will be twice as old as the younger person. How old are they now?
• Answer: Let the younger person’s age be xxx.
  The older person’s age is 50−x50 – x50−x.
  In 5 years, the older person’s age will be 50−x+550 – x + 550−x+5, and the younger person’s age will be x+5x + 5x+5.
  According to the problem:50−x+5=2(x+5)50 – x + 5 = 2(x + 5)50−x+5=2(x+5)Simplifying:55−x=2x+1055 – x = 2x + 1055−x=2x+10 55−10=2x+x55 – 10 = 2x + x55−10=2x+x 45=3×45 = 3×45=3x x=15x = 15x=15Therefore, the younger person is 15 years old, and the older person is 50−15=3550 – 15 = 3550−15=35 years old.

103. Speed, Time, and Distance Problem:

• Question: A person runs at a speed of 12 km/h for 3 hours. How far does he run?
• Answer: The distance is given by:Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}Distance=Speed×TimeSubstituting the values:Distance=12×3=36 km\text{Distance} = 12 \times 3 = 36 \text{ km}Distance=12×3=36 kmTherefore, the person runs 36 km.

104. Mixture Problem:

• Question: A container contains 60 liters of a solution of alcohol and water in the ratio 2:3. How much alcohol is in the solution?
• Answer: The total parts in the solution are 2+3=52 + 3 = 52+3=5.
  The amount of alcohol in the solution is:25×60=24 liters\frac{2}{5} \times 60 = 24 \text{ liters}52​×60=24 litersTherefore, there are 24 liters of alcohol.

105. Profit and Loss Problem:

• Question: A merchant sold an article for $250, making a profit of 25%. What was the cost price?
• Answer: Let the cost price be xxx.
  The profit is 25%, so:x+0.25x=250x + 0.25x = 250x+0.25x=250 1.25x=2501.25x = 2501.25x=250 x=2501.25=200x = \frac{250}{1.25} = 200x=1.25250​=200Therefore, the cost price was $200.

106. Percentage Problem:

• Question: A student scored 72 marks out of 90 in a test. What is his percentage score?
• Answer: The percentage score is:7290×100=80%\frac{72}{90} \times 100 = 80\%9072​×100=80%Therefore, the student’s percentage score is 80%.

107. Work and Time Problem:

• Question: If 10 men can complete a job in 15 days, how long will it take 5 men to complete the same job?
• Answer: The total work is 10×15=15010 \times 15 = 15010×15=150 man-days.
  To complete the work in xxx days, the number of men required will be:5×x=1505 \times x = 1505×x=150 x=1505=30x = \frac{150}{5} = 30x=5150​=30Therefore, it will take 30 days for 5 men to complete the job.

108. Ratio and Proportion Problem:

• Question: The ratio of the number of boys to girls in a class is 3:5. If the total number of students is 40, how many girls are there?
• Answer: Let the number of boys be 3x3x3x and the number of girls be 5x5x5x.
  The total number of students is 40:3x+5x=403x + 5x = 403x+5x=40 8x=408x = 408x=40 x=5x = 5x=5Therefore, the number of girls is 5x=5×5=255x = 5 \times 5 = 255x=5×5=25.

109. Simple Interest Problem:

• Question: A sum of money is invested at 10% simple interest for 4 years. If the total interest is $200, what is the principal amount?
• Answer: Using the formula for simple interest:I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​Given, I=200I = 200I=200, R=10%R = 10\%R=10%, and T=4T = 4T=4. Substituting the values:200=P×10×4100200 = \frac{P \times 10 \times 4}{100}200=100P×10×4​ 200=40P100200 = \frac{40P}{100}200=10040P​ 200=0.4P200 = 0.4P200=0.4P P=2000.4=500P = \frac{200}{0.4} = 500P=0.4200​=500Therefore, the principal amount is $500.

110. Time and Work Problem:

• Question: If 8 men can complete a job in 16 days, how long will it take 4 men to complete the same job?
• Answer: The total work is 8×16=1288 \times 16 = 1288×16=128 man-days.
  To complete the work in xxx days, the number of men required will be:4×x=1284 \times x = 1284×x=128 x=1284=32x = \frac{128}{4} = 32x=4128​=32Therefore, it will take 32 days for 4 men to complete the job.

111. Speed and Distance Problem:

• Question: A car travels 120 km in 2 hours. What is its average speed?
• Answer: The average speed is:1202=60 km/h\frac{120}{2} = 60 \text{ km/h}2120​=60 km/hTherefore, the average speed is 60 km/h.

112. Work and Time Problem:

• Question: If 6 workers can complete a task in 12 days, how long will it take for 3 workers to complete the same task?
• Answer: The total work is 6×12=726 \times 12 = 726×12=72 worker-days.
  To complete the work in xxx days, the number of workers required will be:3×x=723 \times x = 723×x=72 x=723=24x = \frac{72}{3} = 24x=372​=24Therefore, it will take 24 days for 3 workers to complete the task.

113. Profit and Loss Problem:

• Question: A person bought a watch for $150 and sold it for $180. What is his profit percentage?
• Answer: The profit is 180−150=30180 – 150 = 30180−150=30 dollars.
  The profit percentage is:30150×100=20%\frac{30}{150} \times 100 = 20\%15030​×100=20%Therefore, the profit percentage is 20%.

114. Percentage Problem:

• Question: A store offers a 10% discount on a product that costs $250. What is the discount amount?
• Answer: The discount amount is:10100×250=25\frac{10}{100} \times 250 = 2510010​×250=25Therefore, the discount amount is $25.

115. Age Problem:

• Question: The sum of the ages of a mother and her daughter is 60 years. The mother is 5 times as old as her daughter. How old are they?
• Answer: Let the daughter’s age be xxx.
  The mother’s age is 5x5x5x.
  According to the problem:x+5x=60x + 5x = 60x+5x=60 6x=606x = 606x=60 x=10x = 10x=10Therefore, the daughter is 10 years old, and the mother is 5×10=505 \times 10 = 505×10=50 years old.

116. Simple Interest Problem:

• Question: A sum of money amounts to $1200 in 3 years at 8% simple interest. What is the principal amount?
• Answer: Using the simple interest formula:A=P+P×R×T100A = P + \frac{P \times R \times T}{100}A=P+100P×R×T​Given, A=1200A = 1200A=1200, R=8%R = 8\%R=8%, and T=3T = 3T=3. Substituting the values:1200=P+P×8×31001200 = P + \frac{P \times 8 \times 3}{100}1200=P+100P×8×3​ 1200=P+24P1001200 = P + \frac{24P}{100}1200=P+10024P​ 1200=P+0.24P=1.24P1200 = P + 0.24P = 1.24P1200=P+0.24P=1.24P P=12001.24=967.74P = \frac{1200}{1.24} = 967.74P=1.241200​=967.74Therefore, the principal amount is approximately $967.74.

117. Average Problem:

• Question: The average of five numbers is 20. If one number is removed, the average of the remaining four numbers becomes 18. What is the number that was removed?
• Answer: Let the sum of the five numbers be 5×20=1005 \times 20 = 1005×20=100.
  The sum of the remaining four numbers is 4×18=724 \times 18 = 724×18=72.
  Therefore, the number that was removed is:100−72=28100 – 72 = 28100−72=28Hence, the number removed is 28.

118. Profit and Loss Problem:

• Question: A man sells a product for $300, making a profit of 20%. What is the cost price of the product?
• Answer: Let the cost price be xxx.
  The selling price is 300300300, and the profit is 20%, so:x+0.20x=300x + 0.20x = 300x+0.20x=300 1.20x=3001.20x = 3001.20x=300 x=3001.20=250x = \frac{300}{1.20} = 250x=1.20300​=250Therefore, the cost price is $250.

119. Speed and Time Problem:

• Question: A car travels 180 kilometers in 3 hours. How much time will it take to travel 600 kilometers at the same speed?
• Answer: The speed of the car is:1803=60 km/h\frac{180}{3} = 60 \text{ km/h}3180​=60 km/hThe time taken to travel 600 kilometers is:60060=10 hours\frac{600}{60} = 10 \text{ hours}60600​=10 hoursTherefore, it will take 10 hours to travel 600 kilometers.

120. Ratio Problem:

• Question: The ratio of the number of men to women in a room is 3:4. If there are 63 people in total, how many women are there?
• Answer: Let the number of men be 3x3x3x and the number of women be 4x4x4x.
  The total number of people is 63:3x+4x=633x + 4x = 633x+4x=63 7x=637x = 637x=63 x=9x = 9x=9Therefore, the number of women is 4x=4×9=364x = 4 \times 9 = 364x=4×9=36.

121. Simple Interest Problem:

• Question: A sum of money invested at 8% simple interest for 5 years amounts to $850. What is the principal amount?
• Answer: Using the formula for simple interest:A=P+P×R×T100A = P + \frac{P \times R \times T}{100}A=P+100P×R×T​Given, A=850A = 850A=850, R=8%R = 8\%R=8%, and T=5T = 5T=5. Substituting the values:850=P+P×8×5100850 = P + \frac{P \times 8 \times 5}{100}850=P+100P×8×5​ 850=P+40P100850 = P + \frac{40P}{100}850=P+10040P​ 850=P+0.4P=1.4P850 = P + 0.4P = 1.4P850=P+0.4P=1.4P P=8501.4=607.14P = \frac{850}{1.4} = 607.14P=1.4850​=607.14Therefore, the principal amount is approximately $607.14.

122. Work and Time Problem:

• Question: If 12 men can complete a task in 10 days, how long will it take 6 men to complete the same task?
• Answer: The total work is 12×10=12012 \times 10 = 12012×10=120 man-days.
  To complete the work in xxx days, the number of men required will be:6×x=1206 \times x = 1206×x=120 x=1206=20x = \frac{120}{6} = 20x=6120​=20Therefore, it will take 20 days for 6 men to complete the task.

123. Percentage Problem:

• Question: A person gets a 15% discount on an article priced at $200. What is the amount of discount?
• Answer: The discount is:15100×200=30\frac{15}{100} \times 200 = 3010015​×200=30Therefore, the discount amount is $30.

124. Time and Distance Problem:

• Question: A train travels 120 kilometers in 2 hours. How long will it take to travel 360 kilometers at the same speed?
• Answer: The speed of the train is:1202=60 km/h\frac{120}{2} = 60 \text{ km/h}2120​=60 km/hThe time taken to travel 360 kilometers is:36060=6 hours\frac{360}{60} = 6 \text{ hours}60360​=6 hoursTherefore, it will take 6 hours to travel 360 kilometers.

125. Ratio Problem:

• Question: The ratio of the number of boys to girls in a school is 5:7. If there are 360 boys, how many girls are there?
• Answer: Let the number of girls be 7x7x7x, and the number of boys is given as 360.
  According to the ratio:57=3607x\frac{5}{7} = \frac{360}{7x}75​=7×360​ 7x=7×3605=5047x = \frac{7 \times 360}{5} = 5047x=57×360​=504Therefore, the number of girls is 504.

126. Work and Time Problem:

• Question: If 20 men can complete a job in 10 days, how many men will it take to complete the same job in 5 days?
• Answer: The total work is 20×10=20020 \times 10 = 20020×10=200 man-days.
  To complete the work in 5 days, the number of men required will be:200÷5=40200 \div 5 = 40200÷5=40Therefore, 40 men will be required to complete the job in 5 days.

127. Profit and Loss Problem:

• Question: A person bought an article for $500 and sold it for $600. What is his profit percentage?
• Answer: The profit is 600−500=100600 – 500 = 100600−500=100.
  The profit percentage is:100500×100=20%\frac{100}{500} \times 100 = 20\%500100​×100=20%Therefore, the profit percentage is 20%.

128. Simple Interest Problem:

• Question: A sum of money is invested at 12% simple interest for 2 years. The total interest is $240. What is the principal amount?
• Answer: Using the formula for simple interest:I=P×R×T100I = \frac{P \times R \times T}{100}I=100P×R×T​Given, I=240I = 240I=240, R=12%R = 12\%R=12%, and T=2T = 2T=2. Substituting the values:240=P×12×2100240 = \frac{P \times 12 \times 2}{100}240=100P×12×2​ 240=24P100240 = \frac{24P}{100}240=10024P​ 240=0.24P240 = 0.24P240=0.24P P=2400.24=1000P = \frac{240}{0.24} = 1000P=0.24240​=1000Therefore, the principal amount is $1000.

129. Distance and Speed Problem:

• Question: A person travels 100 kilometers at 20 km/h. How long will it take him to complete the journey?
• Answer: The time taken is:10020=5 hours\frac{100}{20} = 5 \text{ hours}20100​=5 hoursTherefore, it will take 5 hours to complete the journey.

130. Average Problem:

• Question: The average of five numbers is 16. If one of the numbers is 12, what is the average of the remaining four numbers?
• Answer: The sum of all five numbers is 5×16=805 \times 16 = 805×16=80.
  The sum of the remaining four numbers is 80−12=6880 – 12 = 6880−12=68.
  The average of the remaining four numbers is:684=17\frac{68}{4} = 17468​=17Therefore, the average of the remaining four numbers is 17.

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