Question

A dishonest shopkeeper uses 990 g weight instead of 1 kg. Neha went to the shopkeeper and asked him to give $\dfrac{5}{2}$ kg sugar. Fraction of sugar he cheated of is $\dfrac{1}{100}$.
If true then enter 1 and if false then enter 0.

Answer 1

Hint: We start solving the problem by recalling the conversion of kilograms to grams. We then find the weight of sugar cheated by the shopkeeper for every 1kg of sugar. We then find the total weight of sugar that needs to be present in $\dfrac{5}{2}$ kg of sugar. We then calculate the total amount of weight of sugar cheated by the shopkeeper. We take the fraction using the fact that the fraction of a in b is determined as $\dfrac{a}{b}$ to get the desired result.

Complete step-by-step answer:
According to the problem, we have a dishonest shopkeeper who is using 990 g weight instead of 1 kg. Neha asked shopkeeper to give her $\dfrac{5}{2}$ kg of sugar. We need to find whether the fraction of sugar cheated by shopkeeper is $\dfrac{1}{100}$.
We know that 1 kg is equal to 1000 g. We have shopkeeper using 990 g weight instead of 1 kg.
So, the weight cheated by the shopkeeper per kg is $\left( 1000-990 \right)$ g.
The weight cheated by the shopkeeper per kg is 10 g ---(1).
Neha bought $\dfrac{5}{2}$ kg of sugar from the shopkeeper. Total weight of sugar bought by Neha is $\left( \dfrac{5}{2}\times 1000 \right)$.
$\Rightarrow \left( 5\times 500 \right)$g
$\Rightarrow 2500$g ---(2).
So, Neha bought 2500g of sugar.
But the shopkeeper is using 990 g in place of 1 kg. So, the weight of sugar that is cheated by the shopkeeper is $\left( \dfrac{5}{2}\times 10 \right)$g.
$\Rightarrow \left( 5\times 5 \right)$g.
$\Rightarrow 25$g ---(3).
So, the shopkeeper cheated 25g.
We need to know what is the fraction of 25g in a total of 2500g of sugar.
We know that the fraction of a in b is determined as $\dfrac{a}{b}$. We use this to find the required fraction.
So, the required fraction is $\dfrac{25}{2500}$.
The required fraction is $\dfrac{1}{100}$.
We have found the fraction of weight cheated by the shopkeeper as $\dfrac{1}{100}$.
So, we enter 1.

Note: We can also find the weight of sugar cheated by calculating the actual weight of sugar bought by Neha and subtracting it from 2500 grams. We need to make sure that the conversion of weights is done perfectly. We can also find the percentage of weight cheated by shopkeeper by multiplying the obtained fraction with 100. Similarly, we can expect problems to find the actual weight of sugar taken home by Neha, percentage of weight cheated.