8 apples and 10 oranges weigh 5 kg. 12 apples and 20 oranges weigh 9 kg. What is the weight (in kg, rounded off to the nearest integer) of 15 apples and 24 oranges?
11
To solve this problem, let's define variables for the weights of apples and oranges:
Let \( a \) be the weight of one apple, and \( o \) be the weight of one orange. We have the following system of equations based on the given conditions:
1. \( 8a + 10o = 5 \) (Equation 1)
2. \( 12a + 20o = 9 \) (Equation 2)
To solve for \( a \) and \( o \), we will use these equations. First, simplify Equation 2 by dividing all terms by 2:
\( 6a + 10o = 4.5 \) (Equation 3)
Now, subtract Equation 3 from Equation 1:
\((8a + 10o) - (6a + 10o) = 5 - 4.5\)
\(2a = 0.5\)
Hence, \( a = 0.25 \) kg per apple.
Substitute \( a = 0.25 \) back into Equation 1:
\(8(0.25) + 10o = 5\)
\(2 + 10o = 5\)
\(10o = 3\)
\(o = 0.3 \) kg per orange.
Now, let's find the combined weight of 15 apples and 24 oranges:
Total weight \( = 15a + 24o\)
\( = 15(0.25) + 24(0.3)\)
\( = 3.75 + 7.2\)
\( = 10.95\) kg
Rounding off 10.95 to the nearest integer gives
11 kg
Important Questions from Ratio and proportion
A’s marks in Mathematics are directly proportional to practice time. In 6 hours of practice, A gets 70 marks. What should be the practice time (approximately) to get 90 marks?
The average of the areas of 2 similar triangles is 706.5 m2 whose perimeters are in the ratio of 6 : 11. What is 20% of the difference (in m2) in areas of both triangles?
In a triangle ABC, D and E are two points on sides AB and AC, respectively, such that DE is parallel to BC and AD : DB = 3 : 5. If AC = 5.6 cm, then find the value (in cm) of AE.
In a triangle ABC, P and Q are two points on AB and AC, respectively, such that PQ is parallel to BC. If AC = 5QC, then the ratio PQ : BC is equal to:
Find the mean proportional between 25 and 81.
