A man first sold \(\frac23\) rd of his total quantity of rice and 100

1.	A man first sold \(\frac23\) rd of his total quantity of rice and 100 kg. Again he sold \(\frac12\) of the remaining quantity and 100 kg. If the total remaining quantity of the stock is 150 kg. Then, what was the original stock of rice ? (a) 2100 kg (b) 1800 kg (c) 2400 kg (d) 2000 kg
Answer» (b) 1800 g Let the original stock of rice be x kg. Parts of the stock sold first time = \(\bigg(\frac{2x}{3}+100\bigg)\) kg ∴ Remaining stock = \(\bigg[x-\bigg(\frac{2x}{3}+100\bigg)\bigg]\) kg = \(\big(\frac{x}{3}-100\big)\)kg Part of the stock sold second time = \(\bigg[\frac12\bigg(\frac{x}3-100\bigg)+100\bigg]\) = \(\bigg(\frac{x}{6}-50+100\bigg)\) kg = \(\big(\frac{x}{6}+50\big)\) kg ∴ Remaining stock = \(\big(\frac{x}{3}-100\big)\) - \(\big(\frac{x}{6}+50\big)\) = \(\big(\frac{x}{3}-\frac{x}6-100-50\big)\) kg = \(\big(\frac{x}{6}-50\big)\) kg Given, \(\frac{x}{6} - 150 = 150 ⇒ \frac{x}{6}=300 ⇒x=\) 1800 kg.

A man first sold \(\frac23\) rd of his total quantity of rice and 100 kg. Again he sold \(\frac12\) of the remaining quantity and 100 kg. If the total remaining quantity of the stock is 150 kg. Then, what was the original stock of rice ? (a) 2100 kg (b) 1800 kg (c) 2400 kg (d) 2000 kg

Answer»

(b) 1800 g

Let the original stock of rice be x kg.

Parts of the stock sold first time = \(\bigg(\frac{2x}{3}+100\bigg)\) kg

∴ Remaining stock = \(\bigg[x-\bigg(\frac{2x}{3}+100\bigg)\bigg]\) kg

= \(\big(\frac{x}{3}-100\big)\)kg

Part of the stock sold second time

= \(\bigg[\frac12\bigg(\frac{x}3-100\bigg)+100\bigg]\)

= \(\bigg(\frac{x}{6}-50+100\bigg)\) kg = \(\big(\frac{x}{6}+50\big)\) kg

∴ Remaining stock = \(\big(\frac{x}{3}-100\big)\) - \(\big(\frac{x}{6}+50\big)\)

= \(\big(\frac{x}{3}-\frac{x}6-100-50\big)\) kg

= \(\big(\frac{x}{6}-50\big)\) kg

Given, \(\frac{x}{6} - 150 = 150 ⇒ \frac{x}{6}=300 ⇒x=\) 1800 kg.