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lesantik [10]
3 years ago
12

Jim, Carla and Tommy are members of the same family. Carla is 5 years older than Jim. Tommy is 6 years older than Carla. The sum

of their three ages is 31 years. How old is each one them?
Mathematics
2 answers:
Masja [62]3 years ago
8 0
So,

We'll solve it with only one placeholder, since that's simpler.

Jim's age = J
Carla's age = C
Tommy's age = T

J = base
Carla's age = J + 5
Tommy's age = Carla's age + 6 or J + 5 + 6 or J + 11

Sum of ages = 31
(J) + (J + 5) + (J + 11) = 31

3J + 16 = 31

Subtract 16 from both sides
3J = 15

Divide both sides by 3
J = 5

Carla's age = J + 5
Carla's age = 10

Tommy's age = Carla's age + 6
Tommy's age = 10 + 6 or 16

5 + 10 + 16 = 31
15 + 16 = 31
31 = 31 This checks.

Jim is 5.
Carla is 10.
Tommy is 16.
yKpoI14uk [10]3 years ago
4 0
Lets
Jim  x years old
Carla (x+5) years old
Tommy (x+5)+6 years old
sum is x+(x+5)+(x+5)+6=31 years old
3x+16=31
3x=15
x=5 - age of Jim
x+5=5+5=10 - age of Carla
x+5+6=16  - age of Tommy
Answer: 5, 10 and 16 y.o


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Answer:

$8,000 in Treasury bills

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Step-by-step explanation:

Let the amount to be invested in treasury bills be $x , the amount to be invested in treasury bonds be $y

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So mathematically, this will be $(x - 3000)

So therefore, the amount invested in each will be;

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For treasury bills;

5% = 5/100 * x = 5x/100

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For Treasury bonds 7%

7% = 7/100 * y = 7y/100

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5x/100 + 7y/100 + (x-3000)/10 = 1390

Multiply through by 100

5x + 7y + 10(x-3000) = 1390 * 100

5x + 7y + 10(x-3000) = 139,000

5x + 7y + 10x -30,000 = 139,000

15x + 7y = 139,000 + 30,000

15x + 7y = 169,000 •••••••••(ii)

So we have two equations to solve simultaneously;

From i, y = 23,000 - 2x

Put this into ii

15x + 7(23,000 -2x) = 169,000

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x + 161,000 = 169,000

x = 169,000 - 161,000

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y = 23,000 - 2(8,000)

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So the last investment amount is x-3000 = 8,000 -3,000 = $5,000

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3 years ago
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