Ashopkeeper sold a certain number ( a two-digit number)of toys all priced at a certain value (also a two-digit number when expressed in rupees). by mistake he reversed the digits of both , the number of items sold and the price of each item. in doing so he found that his stock left at the end of the day showed 72 items more than what it actually was. what could be the actual number of toys?

The answer is 91 toys sold, make the number ab where a is the 10th digit and b is the first digit. The value is 10a + b that can expressed as 10 (3) + 4 = 34

Let the price of each item: xy

10x + y

He accidentally reversed the digits to: 10b + a toys sold at 10y + x rupees per toy. To get use the formula, he sold 10a + b toys but thought he sold 10b + a toys. The number of toys that he thought he left over was 72 items more than the actual amount of toys left over. So he sold 72 more toys than he thought:

10a + b =10b + a +72

9a = 9b + 72

a = b + 8

The only numbers that could work are a = 9 and b = 1 since a and b each have to be 1 digit numbers. He reversed the digits and thought he sold 19 toys. So the actual number of toys sold was 10a + b = 10 (9) + 1 = 91 toys sold. By checking, he sold 91 – 19 = 72 toys more than the amount that he though the sold. As a result, the number of toys he thought he left over was 72 more than the actual amount left over as was stated in the question.

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]