Question

There are 3 students in a group. If the weight of any student is added to the average weight of the other two the sums received are 48 kg, 52 kg, and 59 kg. The average weight (in kg) of the three students is:

• A. 27
• B. 26.5
• C. 27.5
• D. 28

Answer 1

Answer:

B. 26.5

Solution: Given: The weight of any student is added to the average weight of the other two the sums received are 48 kg, 52 kg, and 59 kg. Formula used: Average = Sum of all the values/Total number value Calculation: Let the weight (in kg) of the three students is x, y and z According to the question: [x + (y + z)/2] = 48 ⇒ (2x + y + z) = 96 ----(1) Again, [y + (x + z)/2] = 52 ⇒ (2y + x + z) = 104 ----(2) Now, [z + (x + y)/2] = 59 ⇒ (2z + x + y) = 118 ----(3) Multiply by 2 in equation (1); ⇒ (2x + y + z) × 2 = 96 × 2 ⇒ (4x + 2y + 2z) = 192 ----(4) Now, Subtracting from equation(4) to equation(2): [(4x + 2y + 2z) – (2y + x + z)] = (192 – 104) ⇒ [(4x + 2y + 2z – 2y – x – z)] = 88 ⇒ (3x + z) = 88 ----(a) Multiply by 2 in equation (3): ⇒ (2z + x + y) × 2 = 118 × 2 ⇒ 4z + 2x + 2y = 236 ----(5) Now, Subtracting from equation(5) to equation(2): [(4z + 2x + 2y) – (2y + x + z)] = 236 – 104 ⇒ x + 3z = 132 ----(b) Multiply by 3 in equation (a): ⇒ (3x + z) × 3 = 88 × 3 ⇒ (9x + 3z) = 264 ----(c) Now, Subtracting from equation(c) to equation(b) ⇒ [(9x + 3z) – (x + 3z)] = 264 – 132 = 132 ⇒ [9x + 3z – x – 3z] = 132 ⇒ 8x = 132 ⇒ x = 132/8 = 16.5 From equation (b): ⇒ (3x + z) = 88 ⇒ (3 × 16.5 + z) = 88 ⇒ 49.5 + z = 88 ⇒ z = 88 – 49.5 = 38.5 Putting the value of x and z in equation(1): (2 × 16.5 + y + 38.5) = 96 ⇒ 33 + y + 38.5 = 96 ⇒ y = (96 – 71.5) = 24.5 Average = (x + y + z)/3 ⇒ (16.5 + 24.5 + 38.5)/3 ⇒ 79.5/3 = 26.5 ∴ The average weight (in kg) of the three students is 26.5. Alternate Method: Given: The weight of any student is added to the average weight of the other two the sums received are 48 kg, 52 kg, and 59 kg. Formula used: Average = Sum of all the values/Total number value Calculation: Let the weight (in kg) of the three students is x, y and z According to the question: [x + (y + z)/2] = 48 ⇒ (2x + y + z) = 96 ----(1) Again, [y + (x + z)/2] = 52 ⇒ (2y + x + z) = 104 ----(2) Now, [z + (x + y)/2] = 59 ⇒ (2z + x + y) = 118 ----(3) Adding equation(1),equation(2) and equation(3); ⇒ (2x + y + z) + (2y + x + z) + (2z + x + y) = (96 + 104 + 118) ⇒ (4x + 4y + 4z) = 318 ⇒ 4(x + y + z) = 318 ⇒ (x + y + z) = 318/4 = 159/2 Now, Average = (x + y + z)/3 ⇒ [159/(2 × 3)] ⇒ 159/6 = 26.5 ∴ The average weight (in kg) of the three students is 26.5.