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1) An old man had some coins to give to his friends. If he gave each friend 6 coins,he would have 4 coins too many. But if he gave each friend 7 coins,he would need another 5 coins. How many friends did he have?

2) 45 teachers and pupils were at an exhibition. There were 35 more teachers than pupils. If there were 6 fewer male teachers than female teachers, how many male teachers were there?

2007-02-17 23:00:16 · 13 answers · asked by keeshbubbles 1 in Science & Mathematics Mathematics

I need help for the maths but please not in algebra form but model form.Thank you in advance.

2007-02-19 23:56:04 · update #1

13 answers

The First :
you have 2 equations on the basis of division theorem :
1- Coins=6*Friends+4
2- Coins = 7*Friends-5

now you have 2 equation and 2 parametrs
bysolving them you have : Friends = 9 and coins = 58

The second :
MaleTeachers=mt
FemaleTeaxhers= ft
students=s

now we have 3 equations with 3 parametrs :
1- mt + ft + s = 45
2- mt + ft = s +35
3- mt = ft - 6

by solving these you have ft=23 and mt=23-6 = 17 and students are 5

2007-02-17 23:14:15 · answer #1 · answered by Anonymous · 0 0

1) the sum of four coins too many and 5 coins short is 9 so he had 9 friends 9 x 6 = 54 + 4 too many = 58
9 x 7 = 63 -5 = 58

2) ok 40 teachers and 5 students (35 more teachers, total of 45)
40 teachers minus 6 = 34
34 divided by 2 = 17
17 male teachers
17 + 6 = 23 female teachers
total 40 teachers

2007-02-24 23:49:51 · answer #2 · answered by sydneygal 6 · 0 0

1) method: Guess and check

You guess the number of friends, then multiply it by 6, then add 4 to the answer because he will have 4 coins too many.

You multiply the number of friends by 7, then minus 5 from the answer, as he will need 5 more coins.

So he should have 9 friends.

2)
45-35=10
10 divide by 2 = 5
5+35=40
40-6=34

There were 34 male teachers

2007-02-22 00:59:54 · answer #3 · answered by Anonymous · 0 0

Question 1:
Let the number of friends be f and the number of coins be c.
Then c - 6f = 4 (1)
c - 7f = -5 (2)
Subtract (2) from (1):
f = 9
He had 9 friends.

Question 2:
Let m be the number of male teachers,
f be the number of female teachers,
p be the number of pupils.
Then m + f + p = 45 (1)
m + f - p = 35 (2)
f - m = 6 (3)
Add (1) and (2) to eliminate p:
2m + 2f = 80
m + f = 40 (4)
Subtract (3) from (4) to eliminate f:
2m = 34
m = 17
There were 17 male teachers.

2007-02-17 23:31:01 · answer #4 · answered by Anonymous · 0 0

1)he has 9 friends and 50 coins
let's suppose
x= number of friends
y= number of coins

If he gave each friend 6 coins,he would have 4 coins too many. shall give the equation
6x = y-4.................(1)
if he gave each friend 7 coins,he would need another 5 coins.
7x= y+5...................(2)

now solve both equtaions, you will get the respective answers


2) 17 male teachers

oh! 35 more teacher's than pupil so remaing 10 contaions 5 students and 5 teachers.........
now the total teachers are 40
then he says 6 fewer male teachers than female.
so 40 -6 = 34 is the number of teacher's where male and female teachers are equal so dividing by 2 shall give me the number of teachers
34/2 = 17 Ans
]
take care

2007-02-17 23:15:36 · answer #5 · answered by ghost07 2 · 0 0

t = teachers, p = pupils
t + p = 45 and t - p = 35
t = 45 - p
(45 - p) - p = 35
45 - 2p = 35
-2p = -10, p = 5 pupils and 40 teachers
f = females and m = males
f + m = 40
f - m = 6, f = m + 6
(m + 6) + m = 40
2m + 6 = 40
2m = 34, m = 17 males and 23 females

2007-02-25 16:22:02 · answer #6 · answered by DJ Fox 1 · 0 0

Here is my answer on Q2 in model form.

Let T = Teacher & P = Pupil

T |--1unit--|------------------|
P |--1unit--|<--35 more-->=45

2 units + 35 = 45
2 units = 45 - 35 =10
2 units -> 10
1 unit -> 10 / 2 = 5
Teacher, T = 1 unit + 35 =5+35 = 40.
There were 40 Teachers.

Let F = Female Teacher, M = Male Teacher
F|------1unit--------|-----------|
M|-------1unit-------|<6less> =40
2 units + 6 = 40
2 units = 40-6 = 34
2 units -> 34
1 unit -> 34 / 2 = 17
Male Teacher, M = 1 unit = 17.
There were 17 male Teachers

2007-02-25 20:50:31 · answer #7 · answered by Mimi 1 · 0 0

Showing work:

1)
y=6X+4 y=7X-5
6X+4 = 7X-5
X=9
Y =58

2)
45 = s+t
t=35+s
45=2s+35
s=5
5 students 40 teachers
m+f=40
m=f-6
2f-6=40
2f=46
f=23
m=17

17 male teachers.

2007-02-25 15:50:52 · answer #8 · answered by Anonymous · 0 0

1) 6x-4=0
7x+5=0
6x-4=7x+5
x=-9
I do not know how he has -9 friends but he has 9 friends,though.

2) (35+x)+x=45
35+2x=45
x=5

Therefore, there were 5 students and 40 teachers.
(40-6)/2=17 male teachers.

I hoep this helps. :)

2007-02-17 23:17:46 · answer #9 · answered by Juni Mccoy 3 · 0 0

You can already work this out from the first bit If it travels at 40km/h and arrives one hour later, the distance is 40km but it doesn't agree with the second bit. If it travels at 50km/h and arrives 48 minutes earlier, ie. 12minutes The distance should be 10km

2016-05-24 01:29:38 · answer #10 · answered by Anonymous · 0 0

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