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Value-Based Questions: Arithmetic Progressions Notes | Study Class 10 Mathematics by VP Classes - Class 10

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Q1. Savita has two options to buy a house:
(a) She can pay a lumpsum amount of ₹ 22,00,000
Or
(b) She can pay 4,00,000 cash and balance in 18 annual instalments of ₹ 1,00,000 plus 10% interest on the unpaid amount.
She prefers option (i) and donates 50% of the difference of the costs in the above two options to the Prime Minister Relief Fund.
(i) What amount was donated to Prime Minister Relief Fund?
(ii) Which mathematical concept is used in the above problem?
(iii) By choosing to pay a lumpsum amount and donating 50% of the difference to the Prime Minister Relief Fund, which value is depicted by Savita?
Sol. (a) Total cost of the house = ₹ 22,00,000
(b) Cash payment = ₹ 4,00,000
Balance = ₹ 22,00,000 – ₹ 4,00,000 = ₹ 18,00,000
1st instalment = ₹ [1,00,000 + 10% of balance]
Value-Based Questions: Arithmetic Progressions Notes | Study Class 10 Mathematics by VP Classes - Class 10
= ₹ [1,00,000 + 1,80,000]   = ₹ 2,80,000
Balance after 1st instalment = ₹ [18,00,000 – 1,00,000]  = ₹ 17,00,000
2nd instalment = ₹ [1,00,000 + 10% of 17,00,000]
= ₹ [1,00,000 + 1,70,000]  = ₹ [2,70,000]
Balance after 2nd instalment = ₹ 17,00,000 – ₹ 1,00,000  = ₹ 16,00,000
∵ 3rd instalment = ₹ [1,00,000 + 10% of 16,00,000]
= ₹ [1,00,000 + 1,60,000]  = ₹ 2,60,000
... and so on.
∵ Total amount in instalments = ₹ 2,80,000 + ₹ 2,70,000 + ₹ 2,60,000 + ..... to 18 terms
Value-Based Questions: Arithmetic Progressions Notes | Study Class 10 Mathematics by VP Classes - Class 10 where a = 2,80,000,  d = – 10,000, n = 18
Value-Based Questions: Arithmetic Progressions Notes | Study Class 10 Mathematics by VP Classes - Class 10
= ₹ 9 [560,000+17(-10,000)
= ₹ 9 [560,000 - 170,000]
= ₹ 9 = [390,000] = ₹ 35,10,000
∴ Total cost of house = ₹ 35,10,000 + 4,00,000  = ₹ 39,10,000
Difference in costs of the house in two options
= ₹ 39,10,000 – ₹ 22,00,000  = ₹ 17,10,000
∴ (i) Amount donated towards Prime Minister Relief Fund = 50% of ₹ 17,10,000
Value-Based Questions: Arithmetic Progressions Notes | Study Class 10 Mathematics by VP Classes - Class 10
(ii) Arithmetic Progressions
(iii) National Loyalty

Q2. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. Which value is shown in this question?
Sol. ∴ There are 12 classes in all.
Each class has 2 sections.
∴ Number of plants planted by class I = 1 x 2 = 2
Number of plants planted by class II = 2 x 2 = 4
Number of plants planted by class III = 3 x 2 = 6
Number of plants planted by class IV = 4 x 2 = 8
......................................................................................................
......................................................................................................
Number of plants planted by class XII = 12 x 2 = 24
The numbers 2, 4, 6, 8, ........................ 24 forms an A.P.
Here, a = 2, d = 4 – 2 = 2
∵ Number of classes = 12
∴ Number of terms (n) = 12
Now, the sum of n terms of the above A.P., is given by Sn = n/2 [2a+(n-1)d]
∴ S12 = 12/2 [2(2) -(12)-1) 2]
= 6 [4 + (11 x 2)]
= 6 x 26 = 156
Thus, the total number of trees planted = 156
Value shown: To enrich polution free environment.

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