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Given it takes x hrs for A to produce bolts and 2x/5 hrs for A&B to produce same no. of bolts
Let 'b' be number of hrs for B to produce same no. of bolts
Hence

1/x + 1/b = 5/2x

Solving be get b = 2x/3 Option C
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Bunuel
Machine A takes x hours to produce a batch of bolts. It takes \(\frac{2x}{5}\) hours for machine A and B working simultaneously to produce the same batch of bolts. In terms of \(x\), how many hours would it take machine B to produce the same batch of bolts working at its individual rate?

A) \(\frac{ x}{5}\)

B) \(\frac{x}{4}\)

C) \(\frac{2x}{3}\)

D) \(\frac{3x}{5}\)

E) \(\frac{x}{2}\)

­
Let machine A take x hours to produce a batch of bolts.

Let machines A and B together produce a batch of bolts in 2x/5 hours.

Machine B working alone produces ?

1/x + 1/B = 1/(2x/5)

1/B = 5/2x - 1/x

1/B = 3/2x

The time taken by B is 2x/3 hours.
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Machine A takes x hours to produce a batch of bolts. It takes 2x/5 hours for machine A and B working simultaneously to produce the same batch of bolts. In terms of x, how many hours would it take machine B to produce the same batch of bolts working at its individual rate?

Let, machine B takes b hours to produce same batch of bolts individually.
(1/x)+1/b)= 1/(2x/5)
(1/b)= (5/2x)-(1/x)= (5-2/2x) = 3/2x
b= 2x/3

Answer: C
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Direct formula: Combined time = T1*T2/(T1+T2)

Let time taken by b be y

\(\frac{2x}{5}\) = \(\frac{xy}{(x+y)}\)
\(\frac{2}{5}\) = \(\frac{y}{(x+y)}\)
2x+2y = 5y
2x = 3y
y = \(\frac{2x}{3}\)

Bunuel
Machine A takes x hours to produce a batch of bolts. It takes \(\frac{2x}{5}\) hours for machine A and B working simultaneously to produce the same batch of bolts. In terms of \(x\), how many hours would it take machine B to produce the same batch of bolts working at its individual rate?

A) \(\frac{ x}{5}\)

B) \(\frac{x}{4}\)

C) \(\frac{2x}{3}\)

D) \(\frac{3x}{5}\)

E) \(\frac{x}{2}\)

­
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