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Three pieces of different lengths are cut from cardboard. The length of third piece is 9/5 of the first. The average length of the first and the second is 11.25 cm. The length of the first piece is 5 times the even prime number. If L is the length of the cardboard, which of the following is/are correct? A. L + 4.5 is a multiple of 9 B. L – 1.5 = 40 C. L + 8.5 is a perfect square D. L/5 = 7

Three pieces of different lengths are cut from cardboard. The length of third piece is 9/5 of the first. The average length of the first and the second is 11.25 cm. The length of the first piece is 5 times the even prime number. If L is the length of the cardboard, which of the following is/are correct? A. L + 4.5 is a multiple of 9 B. L – 1.5 = 40 C. L + 8.5 is a perfect square D. L/5 = 7 Correct Answer A and C

Given:

Length of 3rd piece = 9/5 of length of 1st

Average of length of 1st and 2nd = 11.25 cm

Length of 1st piece = 5 × Even prime number

Formula Used:

Average = Sum of values/Number of values

Calculation:

Only even prime number is 2.

Length of 1st piece = 5 × 2

⇒ 10

Length of 3rd piece = 9/5 × 10

⇒ 18 cm

Sum of Length of 1st and 2nd = Average of length of 1st and 2nd × 2

⇒ Length of 1st + Length of 2nd = 11.25 × 2

⇒ 10 + Length of 2nd = 22.5

⇒ Length of 2nd = 12.5 cm

Length of the cardboard = 10 + 12.5 + 18

⇒ 40.5 cm

⇒ L = 40.5 cm

A. L + 4.5 is a multiple of 9

40.5 + 4.5 = 45

⇒ 5 × 9

⇒ A is correct

B. L – 1.5 = 40

40.5 – 1.5 = 39

⇒ B is not correct

C. L + 8.5 is a perfect square

40.5 + 8.5 = 49

⇒ 72

⇒ C is correct

D. L/5 = 7

40.5/5 = 8.1

⇒ D is not correct

∴ A and C are correct

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