Ganita Manjari | Grade 9 | Part II
Exercise Set 10.1 — Complete Solutions with Straight Fractions
Question 1
The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has 30 students and Section B has 25 students?
Solution:
Given Data:
• Section A: Number of students (n₁) = 30, Average score (x̄₁) = 72
• Section B: Number of students (n₂) = 25, Average score (x̄₂) = 76
Formula: Combined Average (x̄) =
n₁ · x̄₁ + n₂ · x̄₂n₁ + n₂
Step-by-Step Calculation:
1. Total score of Section A = 30 × 72 = 2160
2. Total score of Section B = 25 × 76 = 1900
3. Total students = 30 + 25 = 55
4. Combined Average =
2160 + 190055
=
406055
5. Simplifying fraction =
40655
= 73.82
Final Answer:
40655
(or 73.82)
Question 2
A farmer mixes three equal quantities of fertilisers. The first one contains
110
nitrogen, the second contains
950
nitrogen, and the third contains
360
nitrogen. What is the fraction of nitrogen in the mixture?
Solution:
Let the quantity of each fertiliser be 1 unit (Total quantity = 1 + 1 + 1 = 3 units).
Formula: Fraction of Nitrogen =
Total Amount of NitrogenTotal Mixture Quantity
Step-by-Step Calculation:
1. Nitrogen from each fertiliser: f₁ =
110
,
f₂ = 950
,
f₃ = 360
= 120
2. Total Nitrogen =
110
+
950
+
120
3. Taking LCM of denominators (10, 50, 20) = 100:
Total Nitrogen =
10 + 18 + 5100
=
33100
4. Fraction in combined mixture =
33 / 1003
=
33300
=
11100
Final Answer:
11100
Question 3
(Śrīdharācārya, Pāṭīgaṇita, c. 750 CE) In ancient India, Varna was the measure of gold purity. A purity of 16 varna meant pure gold; in general, a purity of k varna meant that the gold-alloy was
Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varna, 5 units at 10 varna, and 17 units at 11 varna. Find the purity in varna of the combined gold.
k16
gold and the rest impurities.Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varna, 5 units at 10 varna, and 17 units at 11 varna. Find the purity in varna of the combined gold.
Solution:
Given Data:
• Piece 1: Weight (w₁) = 9, Purity (v₁) = 12
• Piece 2: Weight (w₂) = 5, Purity (v₂) = 10
• Piece 3: Weight (w₃) = 17, Purity (v₃) = 11
Formula: Weighted Average Purity =
w₁v₁ + w₂v₂ + w₃v₃w₁ + w₂ + w₃
Step-by-Step Calculation:
1. Total Varna Units = (9 × 12) + (5 × 10) + (17 × 11)
Total Varna Units = 108 + 50 + 187 = 345
2. Total Weight = 9 + 5 + 17 = 31 units
3. Combined Purity =
34531
varna
Final Answer:
34531
varna (or ≈ 11.13 varna)
Question 4
The average rainfall per day in the months of May, June, and July in a certain location are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression that gives their combined average.
Solution:
Given Data:
• May: Days (d₁) = 31, Average (r₁) = 3.5 mm
• June: Days (d₂) = 30, Average (r₂) = 10 mm
• July: Days (d₃) = 31, Average (r₃) = 8.7 mm
Formula: Combined Average Rainfall =
d₁r₁ + d₂r₂ + d₃r₃d₁ + d₂ + d₃
Step-by-Step Calculation:
1. Total days = 31 + 30 + 31 = 92 days
2. Expression =
(31 × 3.5) + (30 × 10) + (31 × 8.7)92
3. Calculation =
108.5 + 300 + 269.792
=
678.292
4. Value = 7.37 mm/day
Expression:
Value: ≈ 7.37 mm/day
(31 × 3.5) + (30 × 10) + (31 × 8.7)92
Value: ≈ 7.37 mm/day
Question 5
Calculate the concentration of spice mix in these two scenarios:
(i) A 100 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 300 mL one with 15% spice mix are combined.
(ii) A 300 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 100 mL one with 15% spice mix are mixed.
(i) A 100 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 300 mL one with 15% spice mix are combined.
(ii) A 300 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 100 mL one with 15% spice mix are mixed.
Solution:
Formula: Concentration (%) =
Total Spice Mix VolumeTotal Kashayam Volume
× 100%
Scenario (i):
• Spice mix in 100 mL = 5 mL
• Spice mix in 200 mL = 20 mL
• Spice mix in 300 mL = 45 mL
• Total Volume = 100 + 200 + 300 = 600 mL
• Total Spice Mix = 5 + 20 + 45 = 70 mL
• Concentration (%) =
70600
× 100 =
706
% = 11.67%
Scenario (ii):
• Spice mix in 300 mL = 15 mL
• Spice mix in 200 mL = 20 mL
• Spice mix in 100 mL = 15 mL
• Total Volume = 300 + 200 + 100 = 600 mL
• Total Spice Mix = 15 + 20 + 15 = 50 mL
• Concentration (%) =
50600
× 100 =
506
% = 8.33%
Answers: (i)
353
% (11.67%) | (ii)
253
% (8.33%)
