Source: CBSE · Competency Focused Practice Questions, Mathematics (Volume 3), Grade 10, pp. 17–31. Questions and answers are reproduced verbatim from the official CBSE document.
Try the question yourself before revealing the answer — that's how marks stick.
Q11 markMCQp. 17
In a game, a player must gather 20 flags positioned 5 meters apart in a straight line. The starting point is 10 meters away from the first flag. The player starts from the starting point, collects the 20 flags and comes back to the starting point to complete one round.What will be the total distance covered by a player upon completing one round?
Reveal official answer
Correct option: 2 — 210 m
Q21 markMCQp. 17
Shown below are some squares whose sides form an arithmetic progression (AP).(Diagram: four squares with side lengths 6 cm, 9 cm, 12 cm and 15 cm respectively.)(Note: The figures are not to scale.)Which of these are also in AP?i) The areas of these squares.ii) The perimeters of these squares.iii) The length of the diagonals of these squares.
Reveal official answer
Correct option: 3 — only ii) and iii)
Q31 markMCQp. 17
Given below is an arithmetic progression. X and Y are unknown.4¼, 6¾, X, 11¾, Y, 16¾Which of these are X and Y?
Reveal official answer
Correct option: 3 — X = 9¼, Y = 14¼
Q41 markMCQp. 18
Which of the following are in Arithmetic progression?i) 2, 12, 22, 32, 42, ...ii) 1, 2, 4, 7, 11, 16, ...iii) 7, 6.5, 6, 5.5, 5, ...
Reveal official answer
Correct option: 3 — only i) and iii)
Q51 markMCQp. 18
Given below is a pattern.−¾, −⅝, −½, −⅜, −¼, ...If the pattern is extended, what would be the 41st term?
Reveal official answer
Correct option: 3 — 17/4
Q61 markMCQp. 18
Vanshika decided to plant a certain number of seeds every month as a part of a gardening project. In the first month, she planted 5 flower seeds, and in the final month, she planted 50 flower seeds. Every month, she planted 3 more seeds than the previous month.How many flower seeds did Vanshika plant in total?
Reveal official answer
Correct option: 4 — 440
Q71 markMCQp. 18
A construction company is working on construction of new floors in an old building which already had 6 floors. During the first week, they completed 5 floors. Each subsequent week, they completed 3 more floors.If this progression continues for 12 weeks, how many floors will the building have in total?
Reveal official answer
Correct option: 2 — 44
Q81 markMCQp. 18
Which term of the arithmetic progression (AP) 21, 18, 15, ... is 0?
Reveal official answer
Correct option: 3 — 8th term
Q91 markWrittenp. 18
Write the first four terms of an Arithmetic Progression, whose first term is 3.75, and the common difference is (−1.5).
Reveal official answer
• Writes the first four terms as: 3.75, 2.25, 0.75, −0.75 [1]
Q101 markWrittenp. 18
If the first term of an arithmetic progression (AP) is 5 and the common difference is (−3), then the nth term of the progression is given by Tₙ = 5n − 3.Is the above statement true or false? Justify your answer.
Reveal official answer
• Writes false and justifies the answer. For example, writes that the nth term of an AP is: Tₙ = 5 + (−3)(n − 1) = 8 − 3n [1]
Q112 marksWrittenp. 19
In a library, the arrangement of bookshelves follows a pattern where the number of books on each successive shelf increases by 10 books. The first shelf has 30 books, and the last shelf has 160 books.i) How many shelves are there in the library?ii) How many total books are there in the library?Show your work.
Reveal official answer
• i) Assumes the total number of shelves in the bookshelf as n and writes the equation as: 160 = 30 + 10(n − 1) [0.5]
• Solves the above equation to find the value of n as 14. [0.5]
• ii) Finds the total number of books in the shelf as: 14/2 × (30 + 160) = 1330 [1]
Q122 marksWrittenp. 19
The common difference of an arithmetic progression is 5/2. The 9th term is 17.i) Find the first term.ii) Find the 101th term.Show your work.
Reveal official answer
• i) Assumes the first term of the arithmetic progression to be a and forms the equation: 17 = a + (9 − 1) × 5/2. Solves the above equation to find the value of a as (−3). [1]
• ii) Finds the 101th term as: (−3) + (101 − 1) × 5/2 = 247 [1]
Q132 marksWrittenp. 19
Sameer is saving up to buy a bike, which costs Rs 46,000. He plans to save money each month. In the first month, he saves Rs 1,000 and every subsequent month, he saves Rs 250 more than the previous month.After how many months will he be able to buy the bike? Show your work.
Reveal official answer
• Assumes the number of months to be n and writes the equation: 46000 = n/2)} [(2 × 1000) + (n − 1) × 250] [sic — printed thus in the official PDF] [1]
• Solves the above equation to get n as 16 or −23. Writes that the number of months cannot be negative and hence after 16 months, he will be able to buy the bike. [1]
Q142 marksWrittenp. 19
The nth term of an arithmetic progression (AP) is Tₙ = (2n + 1)² − 3.Determine the sum of the first 10 terms of the AP. Show your work.
Reveal official answer
• Finds the 1st term of the AP as: (2 × 1 + 1)² − 3 = 6 [0.5]
• Finds the 10th term of the AP as: (2 × 10 + 1)² − 3 = 438 [0.5]
• Finds the sum of first 10 terms of the AP as: 10/2 (6 + 438) = 2220 [1]
Q152 marksWrittenp. 19
John is renovating his house. He began by painting one wall, which took him 2 hours on the first day. Each subsequent day, he spends an additional 30 min on the renovation project.On which day will he spend 12 hours of his day on the renovation? Show your work.
Reveal official answer
• Finds the first term of the progression as 2 × 60 = 120 min and writes the common difference as 30 min. [0.5]
• Finds the time spent on the nth day as 12 × 60 = 720 min. [0.5]
• Writes the equation for the nth day as: 720 = 120 + (n − 1) × 30. Solves the above equation to find that John will spend 12 hours of his day on the 21st day. [1]
Q162 marksWrittenp. 19
How many terms of the arithmetic progression 5, 7½, ... add up to 50? Show your work.
Reveal official answer
• Writes the equation for the sum of n terms of an arithmetic progression as: 50 = n/2 [2 × 5 + (n − 1) × 2½] [0.5]
• Solves the above equation to get the values of n as 5 or (−8). Writes that the number of terms cannot be negative and hence n = 5. [1.5]
Q172 marksWrittenp. 19
Given below are 2 arithmetic progressions (AP):AP₁: 5, 9, 13, 17, ...AP₂: 30, 40, 50, 60, ...The xth term of AP₁ is the same as the yth term of AP₂.Find the relationship between x and y. Show your work.
Reveal official answer
• Writes the equation for the xth term of AP₁ as: 5 + (x − 1) × 4 [0.5]
• Writes the equation for the xth term of AP₂ as: 30 + (y − 1) × 10 [sic — printed thus in the official PDF] [0.5]
• Equates the above two equations and writes: 5 + (x − 1) × 4 = 30 + (y − 1) × 10 => 4x − 10y = 19 [1]
Q183 marksWrittenp. 20
A theatre charges Rs 350 for the first ticket and Rs 20 less for every subsequent ticket. The offer is valid for 12 tickets only.i) Find the discounted price for the first four tickets.ii) How much would someone pay for 8 tickets?iii) What would be the discounted price of the 12th ticket?Show your work.
Reveal official answer
• i) Finds the price for first ticket as Rs 350 and the subsequent 3 tickets as Rs 330, Rs 310, and Rs 290. [1]
• ii) Writes the equation for the price of 8 tickets as: 8/2 × [(2 × 350) + (7 × (−20))] [0.5]
• Solves the above equation to get total price of 8 tickets as Rs 2240. [0.5]
• iii) Finds the discounted price of 12th ticket as: 350 + 11 × (−20) = Rs 130 [1]
Q193 marksWrittenp. 20
How many three-digit numbers are smaller than 200 and divisible by 8? Find sum of these numbers. Show your work.
Reveal official answer
• Writes the sequence of 3-digit numbers less than 200 divisible by 8 as 104, 112, 120, ..., 192 and mentions that it forms an arithmetic progression (AP). [0.5]
• Assumes that the AP has n terms and writes the equation for the last term as: 192 = 104 + (n − 1)8 [0.5]
• Solves the above equation to find the total number of terms in the AP as 12. [1]
• Finds the sum of all terms of the AP as: 12/2 (104 + 192) = 1776 [1]
Q203 marksWrittenp. 20
In an arithmetic progression, the sum of the first n terms is given by Sₙ = 2n² − 5n.Determine the first term and the common difference of this arithmetic progression. Show your work.
Reveal official answer
• Finds the first term (T₁) of the arithmetic progression as: S₁ = 2(1)² − 5(1) = (−3) [1]
• Finds the second term (T₂) of the arithmetic progression as: T₁ + T₂ = S₂ = 2(2)² − 5(2) = (−2) ⇒ T₂ = (−2) − (−3) = 1 [1.5]
• Finds the common difference as: T₂ − T₁ = 1 − (−3) = 4 [0.5]
Q213 marksWrittenp. 20
In a new school, student enrolments occured over a period of 30 days, with 5 students joining each day than the previous day. The first day started with an enrolment of 12 students.i) After how many days did the school have a total of 110 students?ii) How many students were enrolled in the 30 days?Show your work.
Reveal official answer
• i) Writes that the first term of the arithmetic progression (AP) is 12, common difference is 5. Assumes the required number of days as n and writes the equation for 110 students as: 110 = n/2 × (24 + (n − 1) × 5) [0.5]
• Solves the above equation to find the values of n as 5 or (−8.8). Writes that after 5 days, the school had a total of 110 students. [1]
• ii) Finds the total number of students enrolled in 30 days as: 30/2 × (24 + (30 − 1) × 5) = 2535 [1.5]
Q223 marksWrittenp. 20
In a construction project of making chairs, the team adds 3 chairs every day. On the first day, they added 4 chairs.i) After how many days will the office have a total of 40 chairs?ii) Calculate the total number of chairs after 30 days.iii) If they added 5 chairs instead of 3 chairs each day, find the minimum number of days after which there will be more than 150 chairs.Show your work.
Reveal official answer
• i) Finds the first term (a) as 4 and common difference (d) as 3. Using the formula to determine the number of days (n), 40 = 4 + (n − 1) × 3 => n = 13. Concludes that after 13 days, there would be total of 40 chairs in office. [1]
• ii) Finds the total number of chairs after 30 days as: 4 + (30 − 1) × 3 = 91 [1]
• iii) Finds the new common difference to be 5. Assumes the minimum number of days as n and writes the equation for the number of days after which there will be more than 150 chairs as: 4 + (n − 1) × 5 > 150 => n > 30.2 ≅ 31. After 31 days there will be more than 150 chairs. [1]
Q233 marksWrittenp. 21
A librarian wanted to add more books to a library that had a current collection of 150 books. He added 5 books every week.i) How many books were there in the library after 11 weeks?ii) Determine the total number of new books added in the 11 weeks.iii) If the library has a maximum capacity of 300 books, after how many weeks would the library reach its limit?Show your work.
Reveal official answer
• i) Writes that the number of books added forms an arithmetic progression with first term 150 and common difference 5. Finds the number of books in the library after 11 weeks as: 150 + (11 − 1) × 5 = 200 [1]
• ii) Finds the total number of new books added in the 11 weeks as 200 − 150 = 50. [0.5]
• iii) Assumes that after n weeks, there were 300 books. Writes the equation as: 300 = 150 + (n − 1) × 5 [1]
• Solves the above equation for n and finds the required number of weeks as 31. [0.5]
Q243 marksWrittenp. 21
The difference between the 5th and 10th terms of an arithmetic progression (AP) is 15.If the first term is 4, find the common difference and the 15th term of the AP. Show your work.
Reveal official answer
• Writes the 5th and 10th term of the arithmetic progression as (a + 4d) and (a + 9d), where a is the first term and d is the common difference of the AP. [0.5]
• Writes the difference of both the terms as 5d or (−5d) and equates it with 15 to get the common difference as (3) or (−3). [0.5]
• Finds the 15th term of the AP as 46 or (−38). The working may look as follows: case i) when a = 4, n = 15 and d = 3: T₁₅ = 4 + (15 − 1) × 3 = 46. case ii) when a = 4, n = 15 and d = −3: T₁₅ = 4 − (15 − 1) × 3 = −38 [2]
Q253 marksWrittenp. 21
The difference between the 2nd and 4th term of an arithmetic progression (AP) is 6.Find the common difference of the AP. Show your work.
Reveal official answer
• Represents the 2nd and 4th term of the AP as (a + d) and (a + 3d) with the first term as a and common difference as d. [1]
• Finds the difference of 2nd and 4th term as (a + 3d) − (a + d) = 2d or (a + d) − (a + 3d) = (−2d). [1]
• Concludes that the common difference can either be 3 or (−3). [1]
Q265 marksWrittenp. 21
The cannon fires every 2 minutes, with the first shot occurring 10 minutes after the start of the fight. Additionally, the weight of each cannonball increases by 0.5 kg with each successive shot, starting with the first ball weighing 0.5 kg.i) How many balls are fired after the first 30 minutes of fight?ii) What is the ball's weight when the 12th ball is fired?iii) After how much time will the ball of 10 kg be fired?Show your work.
Reveal official answer
• i) Finds the first term (a) = 10 and common difference (d) = 2. Assumes n as the number of balls fired. 30 = 10 + (n − 1) × 2. Finds the value of n as 11 and hence 11 balls have been fired after the first 30 minutes of fight. [1]
• ii) Finds the first term (a) = 0.5 and common difference (d) = 0.5. Required weight = 0.5 + (12 − 1) × 0.5. Thus concludes weight of the 12th ball fired is 6 kg. [1.5]
• iii) Assumes that after nth ball, the 10 kg ball is fired and writes the equation as: 10 = 0.5 + (n − 1) × 0.5. Solves the equation to find n as 20 and hence after the 20th ball, the ball would weigh 10 kg. [1]
• Uses the above n to evaluate the time as: 10 + (20 − 1) × 2. Concludes that after 48 mins of the fight starting, 10 kg ball will be fired. [1.5]
Q275 marksWrittenp. 21
A car covers 55 km in the first hour and increases its speed by 10 km/hr every hour.i) Find the total distance covered in 6 hours.ii) How long will the car take to cover 1000 km?iii) Find the speed of the car in the 9th hour.Show your work.
Reveal official answer
• i) States that speed of car forms an arithmetic progression with common difference, d = 10 and first term, a = 55. [1]
• Finds the total distance covered after 6 hours as 480 km. The working may look as follows: 6/2 {2 × 55 + (6 − 1) × 10} = 480 km [1]
• ii) Uses the equation of sum of first n terms of an arithmetic progression and finds that the car will cover the distance of 1000 km in 11 hours [sic — printed thus in the official PDF]. The working may look as follows: 1000 = n/2 {2 × 55 + (n − 1) × 10} => n² + 10n − 200 = 0 => n = 10 or (−20). Concludes that n = 10 since negative value of time is not possible. [2]
• iii) Uses the equation of nth terms of an AP to find the 9th term and states that the speed will be 135 km/h. The working may look as follows: T₉ = 55 + (9 − 1) × 10 = 135 km/h [1]
Q281 markWrittenp. 22
Answer the questions based on the given information.Isha is planning to grow her orchard. She wants to plant rows of fruit trees in a way that each row has more trees than the one before, following a specific pattern. Given below are the details of her plan:i) The first row will have 5 trees.ii) Each new row will have 3 more trees than the one before.iii) There will be a total of 10 rows of trees.Calculate the number of trees in the 10th row of the orchard. Show your work.
Reveal official answer
• Writes that the first row contains 5 trees, and each subsequent row has 3 more trees than the previous row. Concludes that the given pattern is in AP, and identifies a as 5 and d as 3. [0.5]
• Finds the number of trees in the 10th row as: 5 + (10 − 1) × 3 = 32 [0.5]
Q292 marksWrittenp. 22
Answer the questions based on the given information.Isha is planning to grow her orchard. She wants to plant rows of fruit trees in a way that each row has more trees than the one before, following a specific pattern. Given below are the details of her plan:i) The first row will have 5 trees.ii) Each new row will have 3 more trees than the one before.iii) There will be a total of 10 rows of trees.What will be the total number of trees in the orchard after all 10 rows are planted? Show your work.
Reveal official answer
• Uses the sum of an arithmetic series formula and writes: 10/2 × (2 × 5 + (10 − 1) × 3). Solves the above equation to get the total number of trees in the orchard after all 10 rows are planted as 185. [2]
Q303 marksWrittenp. 22
Answer the questions based on the given information.Isha is planning to grow her orchard. She wants to plant rows of fruit trees in a way that each row has more trees than the one before, following a specific pattern. Given below are the details of her plan:i) The first row will have 5 trees.ii) Each new row will have 3 more trees than the one before.iii) There will be a total of 10 rows of trees.Isha changed her plan by not planting in rows 5 and 6 to create a pathway for walking, without altering the pattern for the rows. All rows will have the same number of trees as before.Calculate the number of trees now. Show your work.
Reveal official answer
• Forms two APs such as: AP₁: 1st, 2nd, 3rd, 4th row. AP₂: 7th, 8th, 9th, 10th row. [0.5]
• Finds the total number of trees in AP₁ as: 4/2 × (2 × 5 + (4 − 1) × 3) = 38 [0.5]
• Calculates the number of trees in the 7th row as: 5 + (7 − 1) × 3 = 23 [0.5]
• Finds total number of trees in AP₂ as: 4/2 × (2 × 23 + (4 − 1) × 3) = 110 [1]
• Finds the total number of trees as 38 + 110 = 148 trees. (Award full marks if students calculate total number of trees and subtract number of trees in Row 5 and 6.) [0.5]