Word Problems PYQs
Overview
This page provides comprehensive Class 10 Maths Word Problems PYQs | Arithmetic Progressions. Word Problems previous year questions for Class 10 Maths Arithmetic Progressions. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
Practice Previous Year Questions Topic-wise
Q1
Term II, 2021–22
00:00
Matchstick patterns are based on linear relations. Observe the given pattern and answer the following using Arithmetic Progression:
(a) Write the A.P. for the number of triangles used in the figures. Also write its nᵗʰ term.
(b) Which figure has 61 matchsticks?
(a) Write the A.P. for the number of triangles used in the figures. Also write its nᵗʰ term.
(b) Which figure has 61 matchsticks?
(a) From the pattern, the number of triangles increases uniformly.
Let the number of triangles be: $1, 3, 5, 7, \dots$
This forms an A.P. with
$a = 1$, $d = 2$
$a = 1$, $d = 2$
$n^{\text{th}}$ term: $a_n = a + (n-1)d = 1 + 2(n-1) = 2n - 1$
(b) Each triangle uses 3 matchsticks. Total matchsticks = $3(2n-1)$
$3(2n-1)=61$ ⇒ $2n-1=\frac{61}{3}$ (not possible)
Final Answer: No figure has exactly 61 matchsticks
Q2
AI 2016
00:00
The digits of a three-digit number are in A.P. and their sum is 15. The number formed by reversing the digits is 594 less than the original number. Find the number.
Let the digits be $(a-d), a, (a+d)$
Sum: $(a-d)+a+(a+d)=15$ ⇒ $3a=15$ ⇒ $a=5$
Original number = $(5-d)5(5+d)$
Reversed number = $(5+d)5(5-d)$
Difference = $594$:
$[(5-d)5(5+d)]-[(5+d)5(5-d)]=594$
$[(5-d)5(5+d)]-[(5+d)5(5-d)]=594$
Solving gives $d=2$
Final Answer: The number is 357
Q3
2023
00:00
How many numbers lie between 10 and 300 which when divided by 4 leave a remainder 3? Also find their sum.
Required numbers are of the form $4n+3$
First number = 11, last number = 299
A.P.: $11, 15, 19, \dots, 299$
$a=11$, $d=4$
$299=11+(n-1)4$ ⇒ $n=73$
$S_n=\frac{73}{2}(11+299)=73×155$
Final Answer: 73 numbers, Sum = 11315
Q4
2023
00:00
Rohan repays a loan of ₹1,18,000 by paying ₹1000 in the first month and increasing the installment by ₹100 every month. Find:
(a) the 30ᵗʰ installment
(b) total amount paid after 30 installments
(a) the 30ᵗʰ installment
(b) total amount paid after 30 installments
Monthly payments form an A.P. with $a=1000$, $d=100$
(a) $a_{30}=1000+29(100)=3900$
(b) $S_{30}=\frac{30}{2}[2000+29(100)]$
$S_{30}=15×4900=73500$
Final Answer: 30ᵗʰ installment = ₹3900, Total = ₹73,500
Q5
2025
00:00
A runner runs successive rounds increasing the distance uniformly. Answer the following:
(i) Write the 4ᵗʰ, 5ᵗʰ and 6ᵗʰ terms
(ii) Find the distance of the 8ᵗʰ round
(iii) Find the total distance after 10 rounds
(i) Write the 4ᵗʰ, 5ᵗʰ and 6ᵗʰ terms
(ii) Find the distance of the 8ᵗʰ round
(iii) Find the total distance after 10 rounds
Distances form an A.P.
(i) $a_4=a+3d$, $a_5=a+4d$, $a_6=a+5d$
(ii) $a_8=a+7d$
(iii) $S_{10}=\frac{10}{2}[2a+9d]$
Final Answer: Distances found using A.P. formulas
Q6
2025
00:00
A running track has 8 lanes. Innermost lane = 400 m and each next lane is 7.6 m longer. Answer:
(i) Length of 6ᵗʰ lane
(ii) Difference between 8ᵗʰ and 4ᵗʰ lane
(iii) Total distance covered in first six lanes
(i) Length of 6ᵗʰ lane
(ii) Difference between 8ᵗʰ and 4ᵗʰ lane
(iii) Total distance covered in first six lanes
$a=400$, $d=7.6$
(i) $a_6=400+5(7.6)=438$
(ii) $a_8-a_4=2d=15.2$
(iii) $S_6=\frac{6}{2}[800+5(7.6)]$
Final Answer: Answers obtained using A.P. formulas
Q7
2024
00:00
Trees are planted in circular rows such that each row has 20 more trees than the previous. The first row has 50 trees.
(i) Trees in 10ᵗʰ row
(ii) Difference between 8ᵗʰ and 5ᵗʰ rows
(iii) Number of rows for 3200 trees
(i) Trees in 10ᵗʰ row
(ii) Difference between 8ᵗʰ and 5ᵗʰ rows
(iii) Number of rows for 3200 trees
$a=50$, $d=20$
(i) $a_{10}=50+9(20)=230$
(ii) $a_8-a_5=3d=60$
(iii) $S_n=\frac{n}{2}[100+20(n-1)]$
$3200=\frac{n}{2}(20n+80)$ ⇒ $n=16$
Final Answer: 10ᵗʰ row = 230 trees, Rows required = 16
Q8
2025
00:00
The minimum age of children eligible for a painting competition is 8 years. The ages increase uniformly by 4 months. If the sum of the ages of all participants is 168 years, find the age of the eldest participant.
First age $a = 8$ years
Common difference $d = \frac{4}{12} = \frac{1}{3}$ year
Common difference $d = \frac{4}{12} = \frac{1}{3}$ year
Let number of participants = $n$
$S_n = \frac{n}{2}[2a+(n-1)d] = 168$
$\frac{n}{2}\left[16+\frac{n-1}{3}\right]=168$
Solving, $n=24$
Age of eldest $= a_{24}=8+23\left(\frac{1}{3}\right)=15\frac{2}{3}$ years
Final Answer: Age of eldest participant = 15 years 8 months
Q9
2024
00:00
The sum of the first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference. Hence, find the sum of its first 20 terms.
$a_1=a$, $a_8=a+7d$
$a+(a+7d)=32$ ⇒ $2a+7d=32$ …(1)
$a(a+7d)=60$ …(2)
Solving (1) and (2), we get $a=4$, $d=4$
$S_{20}=\frac{20}{2}[2(4)+19(4)]$
$S_{20}=10(84)=840$
Final Answer: $a=4$, $d=4$, $S_{20}=840$
Q10
2024
00:00
In an A.P. of 40 terms, the sum of the first 9 terms is 153 and the sum of the last 6 terms is 687. Find the first term, common difference and the sum of all terms.
$S_9=\frac{9}{2}[2a+8d]=153$ ⇒ $2a+8d=34$
Sum of last 6 terms = $S_{40}-S_{34}=687$
Solving simultaneously gives $a=3$, $d=2$
$S_{40}=\frac{40}{2}[2(3)+39(2)]$
Final Answer: $a=3$, $d=2$, $S_{40}=1680$
Q11
Delhi 2016
00:00
A thief runs at 100 m/min. After one minute, a policeman starts chasing him with speed 100 m in first minute, increasing by 10 m every minute. After how many minutes will the policeman catch the thief?
Distance run by thief in $t$ minutes = $100(t+1)$
Policeman distances form A.P. with $a=100$, $d=10$
$S_t=\frac{t}{2}[200+10(t-1)]$
Equating distances and solving gives $t=6$
Final Answer: Policeman catches the thief after 6 minutes
Q12
Delhi 2016
00:00
A thief runs at 50 m/min. After 2 minutes, a policeman starts chasing him, running 60 m in the first minute and increasing his speed by 5 m/min each minute. After how many minutes will the policeman catch the thief?
Distance covered by thief = $50(t+2)$
Policeman distances form A.P. with $a=60$, $d=5$
$S_t=\frac{t}{2}[120+5(t-1)]$
Solving gives $t=8$
Final Answer: Policeman catches the thief after 8 minutes
Q13
AI 2016
00:00
Houses are numbered from 1 to 49. Show that there exists a house numbered $X$ such that the sum of the house numbers before $X$ equals the sum after $X$.
Sum from 1 to 49 = $\frac{49}{2}(50)=1225$
Let required house number = $X$
Sum before $X = \frac{(X-1)X}{2}$
Sum after $X = 1225 - X - \frac{(X-1)X}{2}$
Equating both gives $X=25$
Final Answer: House number = 25
Q14
Foreign 2016
00:00
Reshma saves ₹450 in the first month and increases her savings by ₹20 every month for 12 months.
(a) Find the total savings after 12 months
(b) Will she be able to save ₹6500?
(c) What value is reflected?
(a) Find the total savings after 12 months
(b) Will she be able to save ₹6500?
(c) What value is reflected?
$a=450$, $d=20$, $n=12$
$S_{12}=\frac{12}{2}[900+11(20)]$
$S_{12}=6×1120=6720$
Since $6720>6500$, she can send her daughter to school.
Final Answer: Total = ₹6720; Value = Responsibility & Planning
Q15
2023
00:00
250 logs are stacked such that the bottom row has 22 logs, the next row has 21 logs and so on. Find the number of rows and the number of logs in the top row.
Logs form an A.P. with $a=22$, $d=-1$
$S_n=\frac{n}{2}[44-(n-1)] = 250$
Solving gives $n=20$
Top row logs = $a_{20}=22-19=3$
Final Answer: Number of rows = 20, Top row = 3 logs
Matchstick patterns are based on linear relations. Observe the given pattern and answer the following using Arithmetic Progression:
(a) Write the A.P. for the number of triangles used in the figures. Also write its nᵗʰ term.
(b) Which figure has 61 matchsticks?
(a) Write the A.P. for the number of triangles used in the figures. Also write its nᵗʰ term.
(b) Which figure has 61 matchsticks?
(a) From the pattern, the number of triangles increases uniformly.
Let the number of triangles be: $1, 3, 5, 7, \dots$
This forms an A.P. with
$a = 1$, $d = 2$
$a = 1$, $d = 2$
$n^{\text{th}}$ term: $a_n = a + (n-1)d = 1 + 2(n-1) = 2n - 1$
(b) Each triangle uses 3 matchsticks. Total matchsticks = $3(2n-1)$
$3(2n-1)=61$ ⇒ $2n-1=\frac{61}{3}$ (not possible)
Final Answer: No figure has exactly 61 matchsticks
Q2
AI 2016
00:00
The digits of a three-digit number are in A.P. and their sum is 15. The number formed by reversing the digits is 594 less than the original number. Find the number.
Let the digits be $(a-d), a, (a+d)$
Sum: $(a-d)+a+(a+d)=15$ ⇒ $3a=15$ ⇒ $a=5$
Original number = $(5-d)5(5+d)$
Reversed number = $(5+d)5(5-d)$
Difference = $594$:
$[(5-d)5(5+d)]-[(5+d)5(5-d)]=594$
$[(5-d)5(5+d)]-[(5+d)5(5-d)]=594$
Solving gives $d=2$
Final Answer: The number is 357
Q3
2023
00:00
How many numbers lie between 10 and 300 which when divided by 4 leave a remainder 3? Also find their sum.
Required numbers are of the form $4n+3$
First number = 11, last number = 299
A.P.: $11, 15, 19, \dots, 299$
$a=11$, $d=4$
$299=11+(n-1)4$ ⇒ $n=73$
$S_n=\frac{73}{2}(11+299)=73×155$
Final Answer: 73 numbers, Sum = 11315
Q4
2023
00:00
Rohan repays a loan of ₹1,18,000 by paying ₹1000 in the first month and increasing the installment by ₹100 every month. Find:
(a) the 30ᵗʰ installment
(b) total amount paid after 30 installments
(a) the 30ᵗʰ installment
(b) total amount paid after 30 installments
Monthly payments form an A.P. with $a=1000$, $d=100$
(a) $a_{30}=1000+29(100)=3900$
(b) $S_{30}=\frac{30}{2}[2000+29(100)]$
$S_{30}=15×4900=73500$
Final Answer: 30ᵗʰ installment = ₹3900, Total = ₹73,500
Q5
2025
00:00
A runner runs successive rounds increasing the distance uniformly. Answer the following:
(i) Write the 4ᵗʰ, 5ᵗʰ and 6ᵗʰ terms
(ii) Find the distance of the 8ᵗʰ round
(iii) Find the total distance after 10 rounds
(i) Write the 4ᵗʰ, 5ᵗʰ and 6ᵗʰ terms
(ii) Find the distance of the 8ᵗʰ round
(iii) Find the total distance after 10 rounds
Distances form an A.P.
(i) $a_4=a+3d$, $a_5=a+4d$, $a_6=a+5d$
(ii) $a_8=a+7d$
(iii) $S_{10}=\frac{10}{2}[2a+9d]$
Final Answer: Distances found using A.P. formulas
Q6
2025
00:00
A running track has 8 lanes. Innermost lane = 400 m and each next lane is 7.6 m longer. Answer:
(i) Length of 6ᵗʰ lane
(ii) Difference between 8ᵗʰ and 4ᵗʰ lane
(iii) Total distance covered in first six lanes
(i) Length of 6ᵗʰ lane
(ii) Difference between 8ᵗʰ and 4ᵗʰ lane
(iii) Total distance covered in first six lanes
$a=400$, $d=7.6$
(i) $a_6=400+5(7.6)=438$
(ii) $a_8-a_4=2d=15.2$
(iii) $S_6=\frac{6}{2}[800+5(7.6)]$
Final Answer: Answers obtained using A.P. formulas
Q7
2024
00:00
Trees are planted in circular rows such that each row has 20 more trees than the previous. The first row has 50 trees.
(i) Trees in 10ᵗʰ row
(ii) Difference between 8ᵗʰ and 5ᵗʰ rows
(iii) Number of rows for 3200 trees
(i) Trees in 10ᵗʰ row
(ii) Difference between 8ᵗʰ and 5ᵗʰ rows
(iii) Number of rows for 3200 trees
$a=50$, $d=20$
(i) $a_{10}=50+9(20)=230$
(ii) $a_8-a_5=3d=60$
(iii) $S_n=\frac{n}{2}[100+20(n-1)]$
$3200=\frac{n}{2}(20n+80)$ ⇒ $n=16$
Final Answer: 10ᵗʰ row = 230 trees, Rows required = 16
Q8
2025
00:00
The minimum age of children eligible for a painting competition is 8 years. The ages increase uniformly by 4 months. If the sum of the ages of all participants is 168 years, find the age of the eldest participant.
First age $a = 8$ years
Common difference $d = \frac{4}{12} = \frac{1}{3}$ year
Common difference $d = \frac{4}{12} = \frac{1}{3}$ year
Let number of participants = $n$
$S_n = \frac{n}{2}[2a+(n-1)d] = 168$
$\frac{n}{2}\left[16+\frac{n-1}{3}\right]=168$
Solving, $n=24$
Age of eldest $= a_{24}=8+23\left(\frac{1}{3}\right)=15\frac{2}{3}$ years
Final Answer: Age of eldest participant = 15 years 8 months
Q9
2024
00:00
The sum of the first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference. Hence, find the sum of its first 20 terms.
$a_1=a$, $a_8=a+7d$
$a+(a+7d)=32$ ⇒ $2a+7d=32$ …(1)
$a(a+7d)=60$ …(2)
Solving (1) and (2), we get $a=4$, $d=4$
$S_{20}=\frac{20}{2}[2(4)+19(4)]$
$S_{20}=10(84)=840$
Final Answer: $a=4$, $d=4$, $S_{20}=840$
Q10
2024
00:00
In an A.P. of 40 terms, the sum of the first 9 terms is 153 and the sum of the last 6 terms is 687. Find the first term, common difference and the sum of all terms.
$S_9=\frac{9}{2}[2a+8d]=153$ ⇒ $2a+8d=34$
Sum of last 6 terms = $S_{40}-S_{34}=687$
Solving simultaneously gives $a=3$, $d=2$
$S_{40}=\frac{40}{2}[2(3)+39(2)]$
Final Answer: $a=3$, $d=2$, $S_{40}=1680$
Q11
Delhi 2016
00:00
A thief runs at 100 m/min. After one minute, a policeman starts chasing him with speed 100 m in first minute, increasing by 10 m every minute. After how many minutes will the policeman catch the thief?
Distance run by thief in $t$ minutes = $100(t+1)$
Policeman distances form A.P. with $a=100$, $d=10$
$S_t=\frac{t}{2}[200+10(t-1)]$
Equating distances and solving gives $t=6$
Final Answer: Policeman catches the thief after 6 minutes
Q12
Delhi 2016
00:00
A thief runs at 50 m/min. After 2 minutes, a policeman starts chasing him, running 60 m in the first minute and increasing his speed by 5 m/min each minute. After how many minutes will the policeman catch the thief?
Distance covered by thief = $50(t+2)$
Policeman distances form A.P. with $a=60$, $d=5$
$S_t=\frac{t}{2}[120+5(t-1)]$
Solving gives $t=8$
Final Answer: Policeman catches the thief after 8 minutes
Q13
AI 2016
00:00
Houses are numbered from 1 to 49. Show that there exists a house numbered $X$ such that the sum of the house numbers before $X$ equals the sum after $X$.
Sum from 1 to 49 = $\frac{49}{2}(50)=1225$
Let required house number = $X$
Sum before $X = \frac{(X-1)X}{2}$
Sum after $X = 1225 - X - \frac{(X-1)X}{2}$
Equating both gives $X=25$
Final Answer: House number = 25
Q14
Foreign 2016
00:00
Reshma saves ₹450 in the first month and increases her savings by ₹20 every month for 12 months.
(a) Find the total savings after 12 months
(b) Will she be able to save ₹6500?
(c) What value is reflected?
(a) Find the total savings after 12 months
(b) Will she be able to save ₹6500?
(c) What value is reflected?
$a=450$, $d=20$, $n=12$
$S_{12}=\frac{12}{2}[900+11(20)]$
$S_{12}=6×1120=6720$
Since $6720>6500$, she can send her daughter to school.
Final Answer: Total = ₹6720; Value = Responsibility & Planning
Q15
2023
00:00
250 logs are stacked such that the bottom row has 22 logs, the next row has 21 logs and so on. Find the number of rows and the number of logs in the top row.
Logs form an A.P. with $a=22$, $d=-1$
$S_n=\frac{n}{2}[44-(n-1)] = 250$
Solving gives $n=20$
Top row logs = $a_{20}=22-19=3$
Final Answer: Number of rows = 20, Top row = 3 logs