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ANTONII [103]
3 years ago
6

Rachel has 60 scones. She sells them to Robbie, Cameron, Louis, Tom and

Mathematics
1 answer:
Nutka1998 [239]3 years ago
3 0
We can start by writing all that we know about the problem. Let us write down how many scones each boy has:
R_{o} : x
C_{a} : x+k
L_{o} : x+2k
T_{o} : x+3k
C_{h} : x+4k
Note that k is the increase, we know it's a constant. This is not a system of equations, please keep that in mind. We will use rest of the information given to build the system. We have two variables x (the number of cookies the first boy bought) and k( the increase), that means we need to have two equations to solve this problem. 
We know that total amount of scones is 60, so we can use that information to build our first equation. We sum all of the scones and we get 60. 
R_{o}+C_{a}+L_{o}+T_{o}+C_{h}=60
5x+10k=60
We know that Robbie and Cameron combined have 3/7 of <span>Louis, Tom and Charlie combined. We can use this information to build our second equation.
</span>R_o+C_a= \frac{3}{7} (L_o+T_o+C_h)
2x+k= \frac{3}{7} (x+2k+x+3k+x+4k)
2x+k= \frac{3}{7} (3x+9k)
14x+7k=9x+27k
5x=20k
x=4k
Now this piece of information alows us to solve the first equation that we made and this will solve our problem. We simply plug in x=4k into 5x+10k=60.
5(4k)+10k=60
30k=60
k=2
Now we can plug this information into the same equation 5x+10k=60 and get x.
5x+10(2)=60
5x=40
x=8
And that's it. Now we can write out how many cookies each boy has bought.
R_{o} : x=8
C_{a} : x+k=10
L_{o} : x+2k=12
T_{o} : x+3k=14
C_{h} : x+4k=16
If you add them all up you get 60.
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Step-by-step explanation:

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Step-by-step explanation:

<h2>[1]</h2>

  • SI = $250
  • Rate (R) = 12\sf \dfrac{1}{2} %
  • Time (t) = 4 years

\longrightarrow \tt { SI = \dfrac{PRT}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 12\cfrac{1}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times \cfrac{25}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 25 \times 2}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 50}{100} } \\

\longrightarrow \tt { 250 \times 100 = P \times 50} \\

\longrightarrow \tt { 25000 = P \times 50} \\

\longrightarrow \tt { \dfrac{25000}{50} = P } \\

\longrightarrow \underline{\boxed{ \green{ \tt { \$ \; 500 = P }}}} \\

Therefore principal is $500.

<h2>__________________</h2>

<h2>[2]</h2>

  • 2/7 of the balls are red.
  • 3/5 of the balls are blue.
  • Rest are yellow.
  • Number of yellow balls = 36

Let the total number of balls be x.

→ Red balls + Blue balls + Yellow balls = Total number of balls

\longrightarrow \tt{ \dfrac{2}{7}x + \dfrac{3}{5}x + 36 = x} \\

\longrightarrow \tt{ \dfrac{10x + 21x + 1260}{35} = x} \\

\longrightarrow \tt{ \dfrac{31x + 1260}{35} = x} \\

\longrightarrow \tt{ 31x + 1260= 35x} \\

\longrightarrow \tt{ 1260= 35x-31x} \\

\longrightarrow \tt{ 1260= 4x} \\

\longrightarrow \tt{ \dfrac{1260 }{4}= x} \\

\longrightarrow \underline{\boxed{  \tt { 315 = x }}} \\

Total number of balls is 315.

A/Q,

3/5 of the balls are blue.

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}x} \\

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}(315)} \\

\longrightarrow \tt{ Balls_{(Blue)} = 3(63)} \\

\longrightarrow \underline{\boxed{ \green {\tt { Balls_{(Blue)} = 189 }}}} \\

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