Linear Equation
- The cost of 2 sarees and 4 shirts is ₹ 16000 while 1 saree and 6 shirts cost the same. The cost of 12 shirts is
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Let cost of one saree and shirt be x and y, respectively.
2x + 4y = 16000 ....(i)
x + 6y = 16000 ....(ii)
Solve above equations and find y
And finally cost of 12 shirts = 12yCorrect Option: B
Let cost of one saree and shirt be x and y, respectively.
2x + 4y = 16000 ....(i)
x + 6y = 16000 ....(ii)
On multiplying Eq. (ii) by 2 and subtracting from Eq. (i). we get
2x + 4y = 16000
2x + 12y = 32000
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-8y = -16000
∴ y = 2000
Putting the value of y in Eq. (ii), we get
x +6 x 2000 = 16000
∴ x = 4000
∴ Cost of 12 shirts = 12y
= 12 x 2000 = ₹ 24000
- The system of equations 3x + y - 4 = 0 and 6x + 2y - 8 = 0 has
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Given equations of system
3x + y = 4 ...(i)
x + 2y = 8 ...(ii)
Here, a1 = 3 , b2 = 2 and c2 = B
∵ a1/a2 = b1/b2 = c1/c2 = 1/2
So, the system of equations has infinite solutions, because it represents a parallel line.Correct Option: D
Given equations of system
3x + y = 4 ...(i)
x + 2y = 8 ...(ii)
Here, a1 = 3 , b2 = 2 and c2 = B
∵ a1/a2 = b1/b2 = c1/c2 = 1/2
So, the system of equations has infinite solutions, because it represents a parallel line.
- If 6x - 10y = 10 and x / (x + y) = 5/7, then (x - y) = ?
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Given, 6x - 10y = 10 ..........(i)
and x/(x + y) = 5/7
⇒ 7x = 5x + 5y
⇒ 2x - 5y = 0 ...(ii)
Multiplying Eq. (ii) by 2 and subtracting from Ed.(i),Correct Option: D
Given, 6x - 10y = 10 ..........(i)
and x/(x + y) = 5/7
⇒ 7x = 5x + 5y
⇒ 2x - 5y = 0 ...(ii)
On multiplying Eq. (ii) by 2 and subtracting from Ed.(i), we get
6x - 10y = 10
4x - 10y = 0
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2x = 10
∴ x = 5
Putting the value of x in Eq. (i), we get
30 - 10y = 10
⇒ 10y = 20
⇒ y = 2
∴ (x - y) = 5 - 2 = 3
- The system of equations 2x + 4y = 6 and 4x + 8y = 6 has
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Given equations 2x + 4y = 6 and 4x + 8y = 6
then,
a1/a2 = 2/4 = 1/2;
b1/b2 = 4/8 = 1/2;
c1/c2 = 6/6 = 1
∴ a1/b2 = b1/b2 ≠ c1/c2Correct Option: B
Given equations 2x + 4y = 6 and 4x + 8y = 6
then,
a1/a2 = 2/4 = 1/2;
b1/b2 = 4/8 = 1/2;
c1/c2 = 6/6 = 1
∴ a1/b2 = b1/b2 ≠ c1/c2
So there is no solution for these equations.
- In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1 : 2, Based on the information; the total number of coins in the collection now becomes.
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Let the number of gold coins initially be x and the number of non-gold coins be y.
According to the question,
3x = y
When 10 more gold coins, total number of gold coins becomes x + 10 and the number of non-gold coins remain the same at y.
Now, we have 2(x + 10) = y
Solving these two equations, we get
x = 20 and y = 60.Correct Option: A
Let the number of gold coins initially be x and the number of non-gold coins be y.
According to the question,
3x = y
When 10 more gold coins, total number of gold coins becomes x + 10 and the number of non-gold coins remain the same at y.
Now, we have 2(x + 10) = y
Solving these two equations, we get
x = 20 and y = 60.
Total number of coins in the collection at the end is equal to
x + 10 + y = 20 + 10 + 60 = 90.