OMTEX CLASSES: Practice Set 3.4

Welcome to the complete Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we provide a detailed, step-by-step conceptual guide to solving Practice Set 3.4 from Chapter 3: Arithmetic Progression. These word problems are crucial for board exams, so we have structured every solution with clear Given data, Formulas, and Calculations to help you maximize your scores!

Practice Set 3.4 Questions and Answers

Question 1

On \( 1^{\text{st}} \) Jan 2016, Sanika decides to save ₹10, ₹11 on the second day, ₹12 on the third day. If she decides to save like this, then on \( 31^{\text{st}} \) Dec 2016 what would be her total saving?

Given: Sanika's daily savings form an Arithmetic Progression (A.P.): \( 10, 11, 12, \dots \)
First term \( a = 10 \)
Common difference \( d = 11 - 10 = 1 \)
Total number of days \( n = 366 \) (since 2016 is a leap year).
Formula:
Key Formula (Sum of an A.P.):
\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: \[ \begin{aligned} S_{366} &= \frac{366}{2}[2(10) + (366 - 1)1] \\ &= 183(20 + 365 \times 1) \\ &= 183(20 + 365) \\ &= 183 \times 385 \\ S_{366} &= 70455 \end{aligned} \]
Final Answer: Sanika's total saving on \( 31^{\text{st}} \) December 2016 would be ₹70,455.

Question 2

A man borrows ₹8000 and agrees to repay with a total interest of ₹1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹40. Find the amount of the first and last instalment.

Given: The instalments are in an A.P.
Total amount repaid in 12 instalments (\( S_{12} \)) = Amount borrowed + Total interest = \( 8000 + 1360 = 9360 \).
Number of instalments \( n = 12 \).
Each instalment is less than the preceding one by ₹40, therefore \( d = -40 \).
Formula:
Formulas Used:
Sum formula: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( n^{\text{th}} \) term formula: \( t_{n} = a + (n-1)d \)
Substitution & Calculation: Finding the first instalment (\( a \)): \[ \begin{aligned} S_{12} &= \frac{12}{2}[2a + (12 - 1)(-40)] \\ 9360 &= 6[2a + (11)(-40)] \\ 9360 &= 6(2a - 440) \\ \frac{9360}{6} &= 2a - 440 \\ 1560 &= 2a - 440 \\ 1560 + 440 &= 2a \\ 2000 &= 2a \\ a &= \frac{2000}{2} \\ a &= 1000 \end{aligned} \] Finding the last instalment (\( t_{12} \)): \[ \begin{aligned} t_{12} &= 1000 + (12 - 1)(-40) \\ &= 1000 + 11(-40) \\ &= 1000 - 440 \\ t_{12} &= 560 \end{aligned} \]
Final Answer: The amount of the first instalment is ₹1000 and that of the last instalment is ₹560.

Question 3

Sachin invested in a national saving certificate scheme. In the first year he invested ₹5000, in the second year ₹7000, in the third year ₹9000 and so on. Find the total amount that he invested in 12 years.

Given: Amounts invested by Sachin each year: \( 5000, 7000, 9000, \dots \)
This sequence is an A.P.
First term \( a = 5000 \)
Common difference \( d = 7000 - 5000 = 2000 \)
Number of years \( n = 12 \).
Formula: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: \[ \begin{aligned} S_{12} &= \frac{12}{2}[2(5000) + (12 - 1)2000] \\ &= 6(10000 + 11 \times 2000) \\ &= 6(10000 + 22000) \\ &= 6(32000) \\ S_{12} &= 192000 \end{aligned} \]
Final Answer: The total amount invested by Sachin in 12 years is ₹1,92,000.

Question 4

There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the \( 15^{\text{th}} \) row and also find how many total seats are there in the auditorium?

Given: The number of seats arranged row-wise are: \( 20, 22, 24, \dots \)
This is an A.P. where:
First term \( a = 20 \)
Common difference \( d = 22 - 20 = 2 \)
Total rows \( n = 27 \).
Formula:
\( n^{\text{th}} \) term: \( t_{n} = a + (n-1)d \)
Sum of n terms: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: Finding seats in the \( 15^{\text{th}} \) row (\( t_{15} \)): \[ \begin{aligned} t_{15} &= 20 + (15 - 1)2 \\ &= 20 + 14 \times 2 \\ &= 20 + 28 \\ t_{15} &= 48 \end{aligned} \] Finding total seats (\( S_{27} \)): \[ \begin{aligned} S_{27} &= \frac{27}{2}[2(20) + (27 - 1)2] \\ &= \frac{27}{2}(40 + 26 \times 2) \\ &= \frac{27}{2}(40 + 52) \\ &= \frac{27}{2} \times 92 \\ &= 27 \times 46 \\ S_{27} &= 1242 \end{aligned} \]
Final Answer: The number of seats in the \( 15^{\text{th}} \) row is 48 and the total seats in the auditorium are 1242.

Question 5

Kargil's temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was \( 5^{\circ}\text{C} \) more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was \( -30^{\circ}\text{C} \) then find the temperature on the other five days.

Given: Let the temperatures from Monday to Saturday in A.P. be:
\( a,\ a+d,\ a+2d,\ a+3d,\ a+4d,\ a+5d \)
Condition 1: Sum of Monday & Saturday = Sum of Tuesday & Saturday + \( 5^{\circ} \)
Condition 2: Temperature on Wednesday = \( -30^{\circ}\text{C} \)
Substitution & Calculation: From the first condition: \[ \begin{aligned} (a) + (a+5d) &= (a+d) + (a+5d) + 5 \\ 2a + 5d &= 2a + 6d + 5 \\ 5d - 6d &= 5 \\ -d &= 5 \\ d &= -5^{\circ} \end{aligned} \] From the second condition (Wednesday is the \( 3^{\text{rd}} \) term): \[ \begin{aligned} a + 2d &= -30^{\circ} \\ a + 2(-5^{\circ}) &= -30^{\circ} \\ a - 10^{\circ} &= -30^{\circ} \\ a &= -30^{\circ} + 10^{\circ} \\ a &= -20^{\circ} \end{aligned} \] Finding the temperatures for all days:
- Monday (\( a \)): \( -20^{\circ}\text{C} \)
- Tuesday (\( a+d \)): \( -20^{\circ} - 5^{\circ} = -25^{\circ}\text{C} \)
- Wednesday (\( a+2d \)): \( -30^{\circ}\text{C} \) (Given)
- Thursday (\( a+3d \)): \( -20^{\circ} + 3(-5^{\circ}) = -20^{\circ} - 15^{\circ} = -35^{\circ}\text{C} \)
- Friday (\( a+4d \)): \( -20^{\circ} + 4(-5^{\circ}) = -20^{\circ} - 20^{\circ} = -40^{\circ}\text{C} \)
- Saturday (\( a+5d \)): \( -20^{\circ} + 5(-5^{\circ}) = -20^{\circ} - 25^{\circ} = -45^{\circ}\text{C} \)
Final Answer: The temperatures on the other five days are \( -20^{\circ}\text{C} \), \( -25^{\circ}\text{C} \), \( -35^{\circ}\text{C} \), \( -40^{\circ}\text{C} \), and \( -45^{\circ}\text{C} \).

Question 6

On the world environment day, a tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Given: The number of trees planted row-wise are: \( 1, 2, 3, \dots \)
This sequence is an A.P.
First term \( a = 1 \)
Common difference \( d = 2 - 1 = 1 \)
Total number of rows \( n = 25 \).
Formula: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: \[ \begin{aligned} S_{25} &= \frac{25}{2}[2(1) + (25 - 1)1] \\ &= \frac{25}{2}(2 + 24) \\ &= \frac{25}{2} \times 26 \\ &= 25 \times 13 \\ S_{25} &= 325 \end{aligned} \]
Final Answer: The total number of trees planted in the 25 rows is 325.

Maharashtra Board Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.4 Solutions