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The cost of 2 oranges and 3 apples is ₹ 28. If cost of
apple is doubled then cost of 3 oranges and 5 apples is ₹
75. The original cost of 7 oranges & 4 apples (in (₹)) is:
Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be?
apple is doubled then cost of 3 oranges and 5 apples is ₹
75. The original cost of 7 oranges & 4 apples (in (₹)) is:
Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be?
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The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is double...
Cost Calculation for Fruits
To determine the original cost of 7 oranges and 4 apples, we can set up a system of equations based on the information provided.
1. Setting up the equations
- Let the cost of one orange be \( x \) and the cost of one apple be \( y \).
From the first statement:
- \( 2x + 3y = 28 \) (Equation 1)
From the second statement, where the cost of apples is doubled (let the new cost be \( 2y \)):
- \( 3x + 5(2y) = 75 \) simplifies to \( 3x + 10y = 75 \) (Equation 2)
2. Solving the equations
- **From Equation 1:**
\( 2x + 3y = 28 \)
Rearranging gives:
\( 2x = 28 - 3y \)
\( x = 14 - 1.5y \)
- **Substituting \( x \) in Equation 2:**
\( 3(14 - 1.5y) + 10y = 75 \)
\( 42 - 4.5y + 10y = 75 \)
\( 5.5y = 33 \)
\( y = 6 \)
- **Finding \( x \):**
Substituting \( y \) back into Equation 1:
\( 2x + 3(6) = 28 \)
\( 2x + 18 = 28 \)
\( 2x = 10 \)
\( x = 5 \)
3. Calculating the final cost
Now, we can find the cost of 7 oranges and 4 apples:
- Cost = \( 7x + 4y = 7(5) + 4(6) = 35 + 24 = 59 \)
Thus, the original cost of 7 oranges and 4 apples is **₹ 59**.
Solution of the Equations
To solve the equations \( 2y - 5z + 4 = 0 \) and \( 2y + z - 8 = 0 \):
1. Rearranging the equations
- From the first equation:
\( 2y - 5z = -4 \)
\( 2y = 5z - 4 \) (Equation 3)
- From the second equation:
\( z = 8 - 2y \) (Equation 4)
2. Substituting to find solutions
- Substituting Equation 4 into Equation 3:
\( 2y = 5(8 - 2y) - 4 \)
\( 2y = 40 - 10y - 4 \)
\( 12y = 36 \)
\( y = 3 \)
- Finding \( z \):
Substituting \( y \) back into Equation 4:
To determine the original cost of 7 oranges and 4 apples, we can set up a system of equations based on the information provided.
1. Setting up the equations
- Let the cost of one orange be \( x \) and the cost of one apple be \( y \).
From the first statement:
- \( 2x + 3y = 28 \) (Equation 1)
From the second statement, where the cost of apples is doubled (let the new cost be \( 2y \)):
- \( 3x + 5(2y) = 75 \) simplifies to \( 3x + 10y = 75 \) (Equation 2)
2. Solving the equations
- **From Equation 1:**
\( 2x + 3y = 28 \)
Rearranging gives:
\( 2x = 28 - 3y \)
\( x = 14 - 1.5y \)
- **Substituting \( x \) in Equation 2:**
\( 3(14 - 1.5y) + 10y = 75 \)
\( 42 - 4.5y + 10y = 75 \)
\( 5.5y = 33 \)
\( y = 6 \)
- **Finding \( x \):**
Substituting \( y \) back into Equation 1:
\( 2x + 3(6) = 28 \)
\( 2x + 18 = 28 \)
\( 2x = 10 \)
\( x = 5 \)
3. Calculating the final cost
Now, we can find the cost of 7 oranges and 4 apples:
- Cost = \( 7x + 4y = 7(5) + 4(6) = 35 + 24 = 59 \)
Thus, the original cost of 7 oranges and 4 apples is **₹ 59**.
Solution of the Equations
To solve the equations \( 2y - 5z + 4 = 0 \) and \( 2y + z - 8 = 0 \):
1. Rearranging the equations
- From the first equation:
\( 2y - 5z = -4 \)
\( 2y = 5z - 4 \) (Equation 3)
- From the second equation:
\( z = 8 - 2y \) (Equation 4)
2. Substituting to find solutions
- Substituting Equation 4 into Equation 3:
\( 2y = 5(8 - 2y) - 4 \)
\( 2y = 40 - 10y - 4 \)
\( 12y = 36 \)
\( y = 3 \)
- Finding \( z \):
Substituting \( y \) back into Equation 4:
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The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be?
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The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be?.
The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be?.
Solutions for The cost of 2 oranges and 3 apples is ₹ 28. If cost of apple is doubled then cost of 3 oranges and 5 apples is ₹ 75. The original cost of 7 oranges & 4 apples (in (₹)) is: Q117. Solution of 2푦 − 5푧 + 4 = 0 & 2푦 + 푧 − 8 = 0 will be? in English & in Hindi are available as part of our courses for UPSC.
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