Constructing, solving simultaneous lin. eqns (marbles, ages)

[yumm]

hello i need help on these questions, its from a year 11 textbook. i need to know the working out for it..
thanks!

1- Two children had 110 marbles between them. After one child has lost half her marbles and the other had lost 20 they had an equal number. how many marbles did each child start with and how many did they finis with?

answer- started with 60 and 50, ended with 30

2-in four years time a mother will be three times as old as her son. four years ago she was five times as old as her son. find their present age

answer- 44 and 12
attempt.

let mother be x, let son be y

in four years time will be 3 times as old as son- mother is 3(x+4) ?
four years ago she was five times as old as her son. - 5(x-4) ?
i am unsure about the 2 equation i have acquired.

3- a party was organized for thirty people at which they could have either a hamburger or a pizza. if there was five times as many hamburgers as pizzas calculate the number of each.
x be hamburger, y be pizza
5x..

answer- 5 pizzas and 25 hamburgers

4-a shopkeeper sold his entire stock of shirts and ties in a sale for $10 000. The shirts were priced at 3 for $100 and the ties $20 each. if he had sold only half the shirts and two-thirds of the ties he would have received $6000. how many of each did he sell in the sale?

answer- 120 shirts and 300 ties

attempt.
shirts -x ties - y
x+y= 10 000

3x= $100
y=$20
if he had sold only half the shirts and two-thirds of the ties he would have received $6000 ?
shirts x/2 ties 2y/3 ?? so x/2 + 2y/3= 6000 ??

the answers are from he back of the book but i want to now how it go to it.

THANKS !

 

[tkhunny]

1- Two children had 110 marbles between them. After one child has lost half her marbles and the other had lost 20 they had an equal number. how many marbles did each child start with and how many did they finis with?

You must show what you have done. I do not believe you have NO idea.

Step #1 - Name Stuff.

"how many marbles did each child start with and how many did they finis with"

x = # of Marbles for Child 1 at the Start

y = # of Marbles for Child 2 at the Start

Step #2 - Translate the Problem Statement - One piece at a time.

"had 110 marbles between them"

x + y = 110

"one child has lost half her marbles"

x - (x/2) = x/2

"the other had lost 20"

y - 20

"they had an equal number"

(x/2) = y - 20

There it is.

x + y = 110
(x/2) = y - 20

 

[Denis]

2-in four years time a mother will be three times as old as her son. four years ago she was five times as old as her son. find their present age.
attempt.

let mother be x, let son be y

in four years time will be 3 times as old as son- mother is 3(x+4) ?
four years ago she was five times as old as her son. - 5(x-4) ?
i am unsure about the 2 equation i have acquired.

C'mon yumm, you can do better than that :shock:

in 4 years: mother = x+4 and son = y+4 ; so x+4 = 3(y+4) : they BOTH will be 4 years older, kapish?
4 years ago: mother = x-4 and son = y-4 ; so x-4 = 5(y-4)

You now have 2 equations:
x+4 = 3(y+4) ; x + 4 = 3y + 12 ; x - 3y = 8 [1]
x-4 = 5(y-4) ; x - 4 = 5y - 20 ; x - 5y = -16 [2]

Solve 'em ! Hint: start by subtracting one equation from the other...Go Man Go!

 

[masters]

hello i need help on these questions, its from a year 11 textbook. i need to know the working out for it..
thanks!

3- a party was organized for thirty people at which they could have either a hamburger or a pizza. if there was five times as many hamburgers as pizzas, x=5y

calculate the number of each.
x be hamburger, y be pizza
5x..

answer- 5 pizzas and 25 hamburgers

THANKS !

Let x = # hamburgers
Let y = # pizzas
Let 30 = # people
Since there were 5 times as many hamburgers (x) as pizzas (y) then x = 5y.
Since there were 20 people total, then x + y = 30

Substituting x=5y into x+y=30, we have

\(\displaystyle 5y + y = 30 \Longrightarrow 6y = 30 \Longrightarrow y = 5\)

If y=5, then,

\(\displaystyle x=5(y) \Longrightarrow x=5(5) \Longrightarrow x = 25\)

 

[yumm]

wow thanks guys you really helped me alot !!
in the future ill need more help on my maths work, and i know where to go now
thanks

 

[masters]

hello i need help on these questions, its from a year 11 textbook. i need to know the working out for it..
thanks!

4-a shopkeeper sold his entire stock of shirts and ties in a sale for $10 000. The shirts were priced at 3 for $100 and the ties $20 each. if he had sold only half the shirts and two-thirds of the ties he would have received $6000. how many of each did he sell in the sale?

answer- 120 shirts and 300 ties

attempt.
shirts -x ties - y
x+y= 10 000

3x= $100
y=$20
if he had sold only half the shirts and two-thirds of the ties he would have received $6000 ?
shirts x/2 ties 2y/3 ?? so x/2 + 2y/3= 6000 ??

the answers are from he back of the book but i want to now how it go to it.

THANKS !

Here's the logic:

Each shirt (x) cost $100/3 and each tie (y) cost $20. Total cost is $10,000. Hence, first equation:

\(\displaystyle \boxed{\frac{100}{3}x + 20y=10000}\Longrightarrow\boxed{100x+60y=30000}\)

Cost of 1/2 of the shirts is \(\displaystyle \frac{100}{3}\left(\frac{x}{2}\right) \ \ or \ \ \frac{50x}{3}\)
Cost of 2/3 of the ties \(\displaystyle 20\left(\frac{2y}{3}\right) \ \ or \ \ \frac{40y}{3}\)

Total cost under these conditions is $6,000. Hence, the second equation:

\(\displaystyle \boxed{\frac{50x}{3}+\frac{40y}{3}=6000}\Longrightarrow\boxed{50x+40y=18000}\)

Now solve the system.