A wooden box has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2.5 m apart, what will be the length of wood required for the rungs?

2.8

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3.8

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Solution

The correct option is C
3.8

Gap between two consecutives rungs = 25cm
No of rungs = $\frac{250}{25} + 1 = 11$
Hence AP will form due to uniform decrement in length
a =45, $3_{11} = 25$
Length of the wood required = $s_{11} = \frac{11}{2} (45 + 25)$
So, $s_{11} = 3.8 m$

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A wooden box has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2.5 m apart, what will be the length of wood required for the rungs?

The correct option is C 3.8Gap between two consecutives rungs = 25cm No of rungs =25025+1=11 Hence AP will form due to uniform decrement in length a =45, 311=25 Length of the wood required =s11=112(45+25) So, s11=3.8m

The correct option is C
3.8

Gap between two consecutives rungs = 25cm
No of rungs = $\frac{250}{25} + 1 = 11$
Hence AP will form due to uniform decrement in length
a =45, $3_{11} = 25$
Length of the wood required = $s_{11} = \frac{11}{2} (45 + 25)$
So, $s_{11} = 3.8 m$