A dog eats $7$ cans of food in $3$ days. At this rate, how many cans of food does the dog eat in $3 + d$ days?

\frac{7}{3} + d

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\frac{7}{3} + \frac{d}{3}

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\frac{7}{3} + \frac{7}{3 d}

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7 + \frac{d}{3}

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7 + \frac{7 d}{3}

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Solution

The correct option is E $7 + \frac{7 d}{3}$
We can see that:
Let the number of cans be $x$
Now as given:
No. of cans No. of days
$7$ $3$
$x$ $3 + d$
By direct proportion,
$7 \times (3 + d) = 3 \times x$
$∴ x = \frac{21 + 7 d}{3}$
$∴ x = 7 + \frac{7 d}{3}$ cans
Answer is option E

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A dog eats 7 cans of food in 3 days. At this rate, how many cans of food does the dog eat in 3+d days?

The correct option is E 7+7d3We can see that:Let the number of cans be x Now as given:No. of cans No. of days 7 3 x 3+d By direct proportion,7×(3+d)=3×x∴x=21+7d3∴x=7+7d3 cansAnswer is option E

A dog eats $7$ cans of food in $3$ days. At this rate, how many cans of food does the dog eat in $3 + d$ days?

The correct option is E $7 + \frac{7 d}{3}$
We can see that:
Let the number of cans be $x$
Now as given:
No. of cans No. of days
$7$ $3$
$x$ $3 + d$
By direct proportion,
$7 \times (3 + d) = 3 \times x$
$∴ x = \frac{21 + 7 d}{3}$
$∴ x = 7 + \frac{7 d}{3}$ cans
Answer is option E