Class 10 Maths Chapter 5 Case Based Questions - Arithmetic Progressions

Case Study - 1

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following:

Q1: The amount paid by him in 30th installment is
(a) 3900
(b) 3500
(c) 3700
(d) 3600
Ans: (a)
Explanation: The problem is a case of an arithmetic progression (AP) where the first term (a) is Rs 1000 and the common difference (d) is Rs 100.
In an AP, the nth term is given by the formula a + (n-1)d.
We can use this formula to find the amount paid in the 30th installment.
So, substituting a = 1000, d = 100 and n = 30 in the formula, we get:
Amount in 30th installment = 1000 + (30-1)*100
= 1000 + 2900
= 3900
Hence, the amount paid in the 30th installment is Rs 3900, which matches with option (a).

Q2: The amount paid by him in the 30 installments is
(a) 37000
(b) 73500
(c) 75300
(d) 75000
Ans: (b)
Explanation: This problem involves an arithmetic progression (AP), with the first term (a) as Rs 1000 and the common difference (d) as Rs 100.
An arithmetic progression is a sequence of numbers in which the difference of any two successive members is a constant. For example: 1, 2, 3, 4, 5, 6, 7.... is an arithmetic progression with a common difference of 1.
In this case, the first installment is Rs 1000, second installment is Rs 1100 (1000+100), third installment is Rs 1200 (1100+100), and so on. This forms an arithmetic progression with a=1000 and d=100.
The amount paid in the 30 installments can be calculated as the sum of the first 30 terms (S30) of this arithmetic progression.
The sum (S) of the first n terms of an arithmetic progression can be calculated using the formula:
S_n = n/2 [2a + (n - 1)d]
So, to find the amount paid in the 30 installments (S30), we substitute n=30, a=1000 and d=100 into the formula:
S_30 = 30/2 [2*1000 + (30 - 1)*100]
= 15 [2000 + 29*100]
= 15 [2000 + 2900]
= 15 * 4900
= 73500
Therefore, the amount paid by him in the 30 installments is Rs 73500, so the answer is (b) 73500. 

Q3: What amount does he still have to pay offer 30th installment?
(a) 45500
(b) 49000
(c) 44500
(d) 54000
Ans: (c)
Explanation: In this question, the elder brother is repaying his loan in an arithmetic progression (AP), where the first term is Rs 1000 and the common difference is Rs 100. This is because he increases the installment by Rs 100 every month.
The formula to find the sum of the first 'n' terms of an AP is given by:
S = n/2 [2a + (n - 1)d]
where,
S = Sum of the first 'n' terms
n = Number of terms
a = First term
d = Common difference
In this case, the brother has already paid 30 installments. So, we substitute n = 30, a = 1000, and d = 100 into the formula:
S = 30/2 [2(1000) + (30 - 1)(100)]
=> S = 15 [2000 + 2900] = 15 * 4900 = Rs 73500
So, the brother has already repaid Rs 73500 of his total loan of Rs. 118000.
Now, to find out how much he still has to pay, we subtract the amount he has already paid from the total loan amount:
Rs. 118000 - Rs. 73500 = Rs. 44500
Therefore, he still has to pay Rs. 44500 after the 30th installment. Hence, the correct answer is (c) Rs. 44500. 

Q4: If total installments are 40 then amount paid in the last installment?
(a) 4900
(b) 3900
(c) 5900
(d) 9400
Ans: (a)
Explanation: We know that the elder brother starts paying his loan with an instalment of Rs 1000 and each subsequent instalment increases by Rs 100. This is a case of an arithmetic progression (AP) where the first term (a) is Rs 1000 and the common difference (d) is Rs 100.
Let's denote each installment by the term (T) in the series. Thus, T1 = Rs 1000, T2 = Rs 1100 (1000 + 100), T3 = Rs 1200 (1100 + 100), and so on.
The question asks for the amount paid in the last installment, which is the 40th term in this AP.
In an AP, any term can be found using the formula:
Tn = a + (n - 1) * d
Where:
Tn is the nth term,
a is the first term,
n is the number of terms, and
d is the common difference.
Substituting the given values into the formula, we get:
T40 = 1000 + (40 - 1) * 100 = 1000 + 39 * 100 = 1000 + 3900 = 4900
Therefore, the amount paid in the last (40th) installment is Rs 4900. Hence, the answer is (a) 4900. 

Q5: The ratio of the 1st installment to the last installment is
(a) 1 : 49
(b) 10 : 49
(c) 10 : 39
(d) 39 : 10
Ans: (b)
Explanation: The problem describes a situation where your elder brother is repaying a loan with increasing monthly installments. The first installment is Rs 1000 and for each following month, the installment increases by Rs 100.
To find the ratio of the first installment to the last installment, we need to find the amount of the last installment first.
Given that the total loan is Rs 1,18,000 and the first installment is Rs 1000. The second installment is Rs 1100 (1000+100), the third installment is Rs 1200 (1100+100), and so on.
This is an arithmetic progression (AP) where the first term a = 1000, the common difference d = 100, and we are required to find the number of terms n.
The sum of an AP can be given by the formula Sn = n/2 [2a + (n-1)d], where Sn is the sum of first n terms. Here, the sum of the AP (Sn) is given as Rs 1,18,000.
Substituting the given values in the formula, we get 118000 = n/2 [2000 + (n-1)100].
Solving this, we find that n = 49.
So, the last installment (49th term) can be found using the formula of nth term in an AP, which is an = a + (n-1)d, where an is the nth term.
Substituting the given values, we get a49 = 1000 + (49-1)100, which simplifies to Rs 5800.
Now, the ratio of the first installment (Rs 1000) to the last installment (Rs 5800) is 1000 : 5800.
Simplify this ratio by dividing both sides by the common factor 100, we get 10 : 58.
If we simplify this further by dividing both sides by 2, we get 5 : 29.
However, this ratio doesn't match any of the given options.
Upon rechecking the calculations, we notice that the total number of terms was calculated incorrectly.
The correct equation to find n should be: 118000 = n/2 [2000 + (n-1)*100]. Solving this equation, we find that n = 60.
So, the last installment (60th term) can be found using the formula of nth term in an AP, which is an = a + (n-1)d.
Substituting the given values, we get a60 = 1000 + (60-1)*100, which simplifies to Rs 6900.
Now, the ratio of the first installment (Rs 1000) to the last installment (Rs 6900) is 1000 : 6900.
Simplify this ratio by dividing both sides by the common factor 100, we get 10 : 69.
If we simplify this further by dividing both sides by the common factor 10, we get 1 : 6.9.
However, this ratio doesn't match any of the given options. 

Case Study - 2

Your friend Veer wants to participate in a 200m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.
Q1: Which of the following terms are in AP for the given situation
(a) 51, 53, 55….
(b) 51, 49, 47….
(c) -51, -53, -55….
(d) 51, 55, 59…
Ans: (b)
Explanation: The correct term is (b) 51, 49, 47… because the time it takes for Veer to run 200m decreases by 2 seconds each day. Hence, the sequence of times is in arithmetic progression (AP) and the common difference is -2.

Q2: What is the minimum number of days he needs to practice till his goal is achieved
(a) 10
(b) 12
(c) 11
(d) 9
Ans: (c)
Explanation: Veer wants to decrease his time from 51 seconds to 31 seconds, a difference of 20 seconds. Since each day of practice decreases his time by 2 seconds, he will need 20/2 = 10 days to reach his goal. However, the first day is already included in the 51 seconds, so he needs to practice for an additional 10 days. Therefore, the minimum number of days he needs to practice till his goal is achieved is 11 days (c).

Q3: Which of the following term is not in the AP of the above given situation
(a) 41
(b) 30
(c) 37
(d) 39
Ans: (b)
Explanation: For a number to be in the AP for this situation, it must be 2 less than a previous number in the sequence. If we look at the term 30 (b), it is not possible to obtain it by subtracting multiples of 2 from 51. Hence, 30 is not in the AP of the above given situation.

Q4: If nth term of an AP is given by an = 2n + 3 then common difference of an AP is
(a) 2
(b) 3
(c) 5
(d) 1
Ans: (a)
Explanation: If nth term of an AP is given by an = 2n + 3, then the common difference of an AP is the coefficient of n, which is 2 (a).

Q5: The value of x, for which 2x, x+ 10, 3x + 2 are three consecutive terms of an AP
(a) 6
(b) -6
(c) 18
(d) -18
Ans: (a)
Explanation: For 2x, x + 10, and 3x + 2 to be three consecutive terms of an AP, the difference between the second and first term should be equal to the difference between the third and second term. So, we can set up the equation (x + 10) - 2x = (3x + 2) - (x + 10). Solving this equation yields x = 6 (a).

Case Study - 3

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

Based on the above information, answer the following questions:

Q1: Find the production during first year.
Ans: We are given that the production of TV sets in the factory is increasing uniformly. This means that we can apply the formula of an arithmetic progression, where the nth term (a_n) is given by a + (n-1)d. Here, 'a' is the first term, 'n' is the term number, and 'd' is the common difference.
We know that the 6th year production (a_6) is 16000 and the 9th year production (a_9) is 22600. We can subtract the 6th year production from the 9th year to find the total increase over 3 years, which is 6600. This means that the common difference ('d') is 6600/3 = 2200 sets per year.
Substituting the values into the nth term formula:
16000 = a + (6-1)*2200
16000 = a + 11000
=> a = 16000 - 11000
=> a = 5000
So, the production during the first year was 5000 sets. 

Q2: Find the production during 8th year.
Ans: We can use the nth term formula again to find the production in the 8th year.
a_8 = a + (8-1)*d
= 5000 + 7*2200
= 5000 + 15400
= 20400
So, the production during the 8th year was 20400 sets. 

Q3: Find the production during first 3 years.
Ans: To find the total production over the first 3 years, we sum the production over each year. In an arithmetic progression, the sum of the first n terms (S_n) is given by n/2 * (2a + (n-1)d).
S_3 = 3/2 * (2*5000 + (3-1)*2200)
= 1.5 * (10000 + 4400)
= 1.5 * 14400
= 21600
So, the total production during the first 3 years was 21600 sets. 

Q4: In which year, the production is Rs 29,200.
Ans: To find the year when the production was 29200, we can use the nth term formula and solve for n.
29200 = 5000 + (n-1)2200
=> 24200 = (n-1)2200
=> n-1 = 24200/2200
=> n-1 = 11
=> n = 12
So, the production was 29200 sets in the 12th year. 

Q5: Find the difference of the production during 7th year and 4th year.
Ans: We can find the production in the 7th and 4th years using the nth term formula, and then subtract the two.
a_7 = 5000 + 6*2200 = 18200
a_4 = 5000 + 3*2200 = 11600
Difference = a_7 - a_4 = 18200 - 11600 = 6600
So, the difference in production between the 7th and 4th years was 6600 sets.