A dishonest dealer professes to sell his goods at their original price- But he uses a false load while measuring and gains 6 18-47- What is the actual weight of the load which is marked 1 kg -A- 940 gmsB- 947 gmsC- 960 gmsD- 965 gmsE- 953 gms

The correct option is A 940 gms Let the error = x gms. Then, x/1000 x× 100 = 618/47⇔100x/1000 x = 300/47 ↔ 47x= 3 (1000 x) ↔ x = 60 ∴ Weight used = (1000 ...

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Question

A dishonest dealer professes to sell his goods at their original price. But he uses a false load while measuring and gains $6 \frac{18}{47} %$ .
What is the actual weight of the load which is marked 1 kg?

940 gms

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947 gms

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953 gms

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960 gms

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965 gms

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Solution

The correct option is A
940 gms

Let the error = $x$ gms. Then, $\frac{x}{1000 - x} \times 100 = 6 \frac{18}{47} \Leftrightarrow \frac{100 x}{1000 - x} = \frac{300}{47}$

$\leftrightarrow 47 x = 3 (1000 - x)$

$\leftrightarrow x = 60$

$∴$ Weight used = (1000-60) = 940 gms

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A dishonest dealer professes to sell his goods at their original price. But he uses a false load while measuring and gains 61847%. What is the actual weight of the load which is marked 1 kg?

The correct option is A 940 gms Let the error = x gms. Then, x1000−x×100=61847⇔100x1000−x=30047 ↔ 47x=3(1000−x) ↔ x=60 ∴ Weight used = (1000-60) = 940 gms

A dishonest dealer professes to sell his goods at their original price. But he uses a false load while measuring and gains $6 \frac{18}{47} %$ .
What is the actual weight of the load which is marked 1 kg?

The correct option is A
940 gms

Let the error = $x$ gms. Then, $\frac{x}{1000 - x} \times 100 = 6 \frac{18}{47} \Leftrightarrow \frac{100 x}{1000 - x} = \frac{300}{47}$

$\leftrightarrow 47 x = 3 (1000 - x)$

$\leftrightarrow x = 60$

$∴$ Weight used = (1000-60) = 940 gms