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2 years ago

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If there are tickets sold to children for $5 and $12 to adults, and there are only 400 tickets sold for $3925 then how many chil

dren tickets were sold?

1 answer:

Sever21 [200]2 years ago

4 0

125 children tickets were sold

<u>Solution</u>:

Given that

Price of 1 ticket sold to children = $5

Price of 1 ticket sold to adult = $12

Total number of ticket sold = 400

Amount generated by selling 400 tickets = $3925.

Need to calculate number of ticket sold to children’s

Let’s assume number of tickets sold to childrens = x

So number of tickets sold to adults = 400 – x

Amount generated from x ticket sold to children $=x \times \text { Price of } 1 \text { ticket sold to children }$

$x \times 5=5 x$

Amount generated from (400 – x) tickets sold to adults $=x \times \text { Price of } 1 \text { ticket sold to adult }$

$=(400-x) \times 12=4800-12 \mathrm{x}$

<em>Total amount generated from 400 tickets = Amount generated from children’s tickets + Amount generated from adults ticket </em>

= 5x + (4800 – 12x) = 4800 -7x

As given that amount generated = 3965

=> 3965 = 4800 – 7x

Solving above expression to get value of x, we get

7x = 4800 - 3965

=> 7x = 875

=> x = 125.

Hence number of children tickets sold = 125.

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The del operator is used for finding the divergence and the curl of a vector field.

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The given vector fields are:

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1) Verifying the identity: $\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

Consider L.H.S

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⇒ $\nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the dot product between these two vectors,

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(1)

Consider R.H.S

⇒ $a\nabla.F1+b\nabla.F2$

So,

$\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)$

⇒ $\nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}$

$\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)$

⇒ $\nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}$

Then,

$a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})$

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(2)

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Consider L.H.S

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Applying the cross product,

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Then,

$a\nabla\times F1+b\nabla\times F2$ =

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$\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

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Disclaimer: The given question on the portal is incomplete.

Question: Let $F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$ be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

$1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

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