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ivanzaharov [21]

2 years ago

15

At a concession stand, seven hot dog(s) and four hamburger(s) cost $21:25, four hot dog(s) and seven hamburger(s) cost $22.75 Fi

nd the cost of one hot dog and the cost of one hamburger

2 answers:

Alina [70]2 years ago

7 0

Step-by-step explanation:

If 4 hamburger = $21.25

1 hamburger = x

Cross multiply

4 x x = $21.25 x 1

4x = $21.25

Devide both side with the coefficient of x

4x/4 = $21.25/4

X = $5.3125

Therefore 1 hamburger cost $5.31

If 4 hotdogs = $22.75

1 hotdog = y

Cross multiply

4 x y = $22.75 x 1

4y = $22.75

Devide both sides with the coefficient of y

4y/4 = $22.75/4

Y = $5.6875

Therefore one hotdog cost $5.69

scoundrel [369]2 years ago

6 0

Answer:

See below ↓↓↓

Step-by-step explanation:

<u>Given</u> :

7 hot dogs + 4 hamburgers = $21.50
4 hot dogs + 7 hamburgers = $22.75

<u>Let's set some variables.</u>

hot dogs = x
hamburgers = y

<u>So, our equations are</u> :

7x + 4y = 21.50
4x + 7y = 22.75

<u>Mulitply equation #1 by 7 and equation #2 by 4.</u>

49x + 28y = 150.50 (Equation 3)
16x + 28y = 91.00 (Equation 4)

<u>Subtract equation 4 from equation 3.</u>

49x - 16x + 28y - 28y = 150.5 - 91
25x = 59.5
x = <u>$2.38 [cost of one hot dog]</u>

<u></u>

<u>Substitute value of x in equation 4.</u>

16(2.38) + 28y = 91
38.08 + 28y = 91
28y = 52.92
y = <u>$1.89 [cost of one hamburger]</u>

<u />

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1. an alloy contains zinc and copper in the ratio of 7:9 find weight of copper of it had 31.5 kgs of zinc.

m_a_m_a [10]

Answer:

Step-by-step explanation:

Question (1). An alloy contains zinc and copper in the ratio of 7 : 9.

If the weight of an alloy = x kgs

Then weight of copper = $\frac{9}{7+9}\times (x)$

= $\frac{9}{16}\times (x)$

And the weight of zinc = $\frac{7}{7+9}\times (x)$

= $\frac{7}{16}\times (x)$

If the weight of zinc = 31.5 kg

31.5 = $\frac{7}{16}\times (x)$

x = $\frac{16\times 31.5}{7}$

x = 72 kgs

Therefore, weight of copper = $\frac{9}{16}\times (72)$

= 40.5 kgs

2). i). 2 : 3 = $\frac{2}{3}$

4 : 5 = $\frac{4}{5}$

Now we will equalize the denominators of each fraction to compare the ratios.

$\frac{2}{3}\times \frac{5}{5}$ = $\frac{10}{15}$

$\frac{4}{5}\times \frac{3}{3}=\frac{12}{15}$

Since, $\frac{12}{15}>\frac{10}{15}$

Therefore, 4 : 5 > 2 : 3

ii). 11 : 19 = $\frac{11}{19}$

19 : 21 = $\frac{19}{21}$

By equalizing denominators of the given fractions,

$\frac{11}{19}\times \frac{21}{21}=\frac{231}{399}$

And $\frac{19}{21}\times \frac{19}{19}=\frac{361}{399}$

Since, $\frac{361}{399}>\frac{231}{399}$

Therefore, 19 : 21 > 11 : 19

iii). $\frac{1}{2}:\frac{1}{3}=\frac{1}{2}\times \frac{3}{1}$

$=\frac{3}{2}$

$\frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times \frac{4}{1}$

= $\frac{4}{3}$

Now we equalize the denominators of the fractions,

$\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$

And $\frac{4}{3}\times \frac{2}{2}=\frac{8}{6}$

Since $\frac{9}{6}>\frac{8}{6}$

Therefore, $\frac{1}{2}:\frac{1}{3}>\frac{1}{3}:\frac{1}{4}$ will be the answer.

IV). $1\frac{1}{5}:1\frac{1}{3}=\frac{6}{5}:\frac{4}{3}$

$=\frac{6}{5}\times \frac{3}{4}$

$=\frac{18}{20}$

$=\frac{9}{10}$

Similarly, $\frac{2}{5}:\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}$

$=\frac{4}{15}$

By equalizing the denominators,

$\frac{9}{10}\times \frac{30}{30}=\frac{270}{300}$

Similarly, $\frac{4}{15}\times \frac{20}{20}=\frac{80}{300}$

Since $\frac{270}{300}>\frac{80}{300}$

Therefore, $1\frac{1}{5}:1\frac{1}{3}>\frac{2}{5}:\frac{3}{2}$

V). If a : b = 6 : 5

$\frac{a}{b}=\frac{6}{5}$

$=\frac{6}{5}\times \frac{2}{2}$

$=\frac{12}{10}$

And b : c = 10 : 9

$\frac{b}{c}=\frac{10}{9}$

Since a : b = 12 : 10

And b : c = 10 : 9

Since b = 10 is common in both the ratios,

Therefore, combined form of the ratios will be,

a : b : c = 12 : 10 : 9

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