Question

Seema has three balls A,B,and C.A is twice as heavy as B,B weighs 1/3 of C.The heaviest ball is
(i)A (ii)B (iii)C (iv)Can't determine​

Answer 1

Answer:

Step-by-step explanation:

The heaviest weight of ball is B

A is twice heavy=2

So we do multiply 1/3×2=2/6

Now compare time

1/3 2/6

3×6>1×2

=18>2

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Answer 2

Answer:

C is the heaviest of all the balls.

Step-by-step explanation:

Given: Seema has three balls.

A is twice as heavy as B.

B weighs 1/3 of C.

⇒ A = 2B                         …(1)

⇒ B = (1/3)C                    …(2)

From equation (1), we can deduce that:

(Weight of A) > (Weight of B)

And from equation (2):

(Weight of C) > (Weight of B)

Hence, B is not the heaviest ball as it weighs less than both the balls A and C.

Now, we will compare the weight of A and C.

⇒ A = 2B and B = (1/3)C

⇒ A = 2 × (1/3) × C

⇒ A = (2/3)C

⇒ C = (3/2)A

Hence,

(Weight of C) > (Weight of A)

Therefore, C is the heaviest ball.