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MA_775_DIABLO [31]
3 years ago
9

A dishonest shopkeeper has two false balances. One balance weighs 10% more while

Mathematics
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

Shopkeepers gain just by weighing is 22.22%

Step-by-step explanation:

Explanation:

Assume that the shopkeeper bought x units of goods i.e. actual weight of goods.

As the balance used to buy goods weighs 10% more, he actually bought x×110100=1.1x weight units i.e. although the shopkeeper paid for x units, he actually bought 1.1x units by weight.

Now he has 1.1x units of goods, but for selling he uses another balance, which actually measures 10% less i.e. 1−0.10=0.9 units for each unit weight of goods sold.

So while selling 1.1x becomes 1.1x0.90=1.2222... units.

Hence, shopkeepers gain just by weighing is 22.22%

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\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

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Thus, there are 24+18+12+6=\boxed{60} numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

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