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 body
length of the body be ¾ of the total length.
Head 4 inches long
To Find : total length of this whale
128
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
total length of this whale 128 inches
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Answer :-
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 :-
[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 of t from equation i in equation iii :-
[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 of x from equation ii in equation iv :-
[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 t , x , h and L in equation iii :-
[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.