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Answers
Solution :
In the figure we can have a look at 3 different species that is • Tortoise (Shown in the first row) • Crab (Shown in the second row) and • Octopus (Shown in the third row)
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Let's assume the value of Tortoise be x, Crab be y and Octopus be z
Starting From the First Row :
Coming to the Second Row :
Coming to the Third Row :
We've taken value of Octopus as z but in this row there are 2 Octopuses. So, We'll take it as 2z
Now, We've to find the value of Tortoise + Crab × Octopus
Hence,Required Value is 60
Answer :--
Lets Assume That :-
→ Tortoise = x
→ Crab = y
→ Octopus = z.
Than, we Have Given That :-
→ x + x + x = 30
→ 3x = 30
Dividing both sides by 3 ,
→ x = 10 --------- Equation (1).
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Now, we have :-
→ x + y + y = 20
→ x + 2y = 20
Putting value from Equation (1) here,
→ 10 + 2y = 20
→ 2y = 20 - 10
→ 2y = 10
Dividing both sides by 2,
→ y = 5 ----------- Equation (2)
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In Last , we have Given :-
→ 2*Z + y = 25
Putting value from Equation (2),
→ 2z = 25 - 5
→ 2z = 20
Dividing both sides by 2,
→ z = 10 -------------- Equation (3).
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Now, we Have To Find :-
→ x + y * z
Putting values From Equation (1), (2) & (3),
→ 10 + 5 * 10
→ 10 + 50