# A rare whale fossil was uncovered under layers of sediment. When the excavator took measurements, she found the tail to be as long

By Maya

A rare whale fossil was uncovered under layers of sediment. When the excavator took measurements, she found the tail to be as long as its head plus a quarter of the length of the body. When the archaeologist came in, he measured the length of the body. He found it to be ¾ of the total length. They made arguments about how such a fish would swim in the water. To make proper predictions they measured the head size, and found it to be 4 inches long. What is the total length of this whale?

128

99

136

142

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### 2 thoughts on “A rare whale fossil was uncovered under layers of sediment. When the excavator took measurements, she found the tail to be as long”

Let us assume tail to be t, head to be h, length of body be x and the total length be L.

Tail = Head + ¼ of length of body

→ t = h + ¼ x

Length of body = ¾ of total length

→ x = ¾ L

Also,

Total length = Length of body + Head + tail

→ L = x + h + t

Now, the question has given us the length of head. Substituting the value :-

$$\implies\sf t = 4 + \dfrac{1}{4}x \: \: \: \: \: \: -i$$

$$\implies\sf x = \dfrac{3}{4} L \: \: \: \: \: \: -ii$$

$$\implies\sf L = x + 4 + t \: \: \: \: \: \: -iii$$

Substituting the value of t from equation i in equation iii :-

$$\implies\sf L = x + 4 + t$$

$$\implies\sf L = x + 4 + 4 + \dfrac{1}{4}x$$

$$\implies\sf L = 8 + x + \dfrac{1}{4}x\: \: \: \: \: \: -iv$$

Substituting the value of x from equation ii in equation iv :-

$$\implies\sf L = 8 + \dfrac{3}{4} L + \dfrac{1}{4} \times \dfrac{3}{4} L$$

$$\implies\sf L = 8 + \dfrac{3}{4} L + \dfrac{3}{16} L$$

$$\implies\sf L = 8 + \dfrac{12}{16} L + \dfrac{3}{16} L$$

$$\implies\sf L = 8 + \dfrac{15}{16} L$$

$$\implies\sf L – \dfrac{15}{16} L = 8$$

$$\implies\sf \dfrac{1}{16} L = 8$$

$$\implies\sf L = 8 \times 16$$

$$\implies\sf L = 128$$

### Verification :-

From equation ii :-

$$\implies\sf x = \dfrac{3}{4}L$$

$$\implies\sf x = \dfrac{3}{4} \times 128$$

$$\implies\sf x = 96$$

Substituting the value in equation i :-

$$\implies\sf t = h + \dfrac{1}{4} x$$

$$\implies\sf t = 4 + \dfrac{1}{4} \times 96$$

$$\implies\sf t = 4 + 24$$

$$\implies\sf t = 28$$

Substituting the value of t , x , h and L in equation iii :-

$$\implies\sf L = t + x + h$$

$$\implies\sf LHS \longrightarrow L = 128$$

$$\implies\sf RHS \longrightarrow t + x + h$$

$$\implies\sf RHS = 28 + 96 + 4$$

$$\implies\sf RHS = 128$$

LHS = RHS

Hence verified.