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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Given :tail to be as long as its head plus a quarter of the length of the bodylength of the body be ¾ of the total length.

Head 4 inches long

To Find :total length of this whale128

99

136

142

Solution:Head = 4 inch

Body = B inch

Tail = Head + (1/4) Body

= 4 + (1/4)B

= 4 + B/4 inches

Total Length = Head + Body + Tail

= 4 + B + 4 + B/4

= 8 + 5B/4

Body = (3/4) Total Length

=> B = (3/4)(8 + 5B/4)

=> 4B = 3( 32 + 5B)/4

=> 16B = 96 + 15B

=> B = 96

Total Length = 8 + 5B/4 = 8 + 5 * 96/4

= 8 + 5 * 24

= 8 + 120

= 128 inches

Head = 4 inches

Body = 96 inches

Tail = 28 inches

Total Length = 128 inches

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## Answer :-

Let us assume

tailto bet,headto beh,length of bodybexand thetotallengthbeL.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 :-

[tex]\implies\sf t = 4 + \dfrac{1}{4}x \: \: \: \: \: \: -i[/tex]

[tex]\implies\sf x = \dfrac{3}{4} L \: \: \: \: \: \: -ii[/tex]

[tex]\implies\sf L = x + 4 + t \: \: \: \: \: \: -iii[/tex]

Substituting the value offrom equation i in equation iii :-

t[tex]\implies\sf L = x + 4 + t[/tex]

[tex]\implies\sf L = x + 4 + 4 + \dfrac{1}{4}x[/tex]

[tex]\implies\sf L = 8 + x + \dfrac{1}{4}x\: \: \: \: \: \: -iv[/tex]

Substituting the value offrom equation ii in equation iv :-

x[tex]\implies\sf L = 8 + \dfrac{3}{4} L + \dfrac{1}{4} \times \dfrac{3}{4} L[/tex]

[tex]\implies\sf L = 8 + \dfrac{3}{4} L + \dfrac{3}{16} L[/tex]

[tex]\implies\sf L = 8 + \dfrac{12}{16} L + \dfrac{3}{16} L[/tex]

[tex]\implies\sf L = 8 + \dfrac{15}{16} L[/tex]

[tex]\implies\sf L – \dfrac{15}{16} L = 8[/tex]

[tex]\implies\sf \dfrac{1}{16} L = 8[/tex]

[tex]\implies\sf L = 8 \times 16[/tex]

[tex]\implies\sf L = 128[/tex]

## Verification :-

From equation ii :-

[tex]\implies\sf x = \dfrac{3}{4}L[/tex]

[tex]\implies\sf x = \dfrac{3}{4} \times 128[/tex]

[tex]\implies\sf x = 96[/tex]

Substituting the value in equation i :-

[tex]\implies\sf t = h + \dfrac{1}{4} x[/tex]

[tex]\implies\sf t = 4 + \dfrac{1}{4} \times 96[/tex]

[tex]\implies\sf t = 4 + 24[/tex]

[tex]\implies\sf t = 28[/tex]

Substituting the value of,,andin equation iii :-

txhL[tex]\implies\sf L = t + x + h[/tex]

[tex]\implies\sf LHS \longrightarrow L = 128[/tex]

[tex]\implies\sf RHS \longrightarrow t + x + h [/tex]

[tex]\implies\sf RHS = 28 + 96 + 4[/tex]

[tex]\implies\sf RHS = 128[/tex]

LHS = RHS

Hence verified.