A clerk in a bank engaged in counting one rupee currency note. For the first 10 minutes quickly at a speed of Rs.150 per minute then he slows down and begins to count at a speed of Rs. 2 less
etery minute than previous minute. Find what time will he take to count a sum of Rs. 4500.

The time it would take the clerk to count a sum of Rs. 4,500 is 34 minutes

Step-by-step explanation:

The given parameters are;

The rate at which the clerk in the bank counted one rupee currency notes for the first 10 minutes =  Rs. 150/minute

After the first 10 minutes, the rate at which the clerk counted = Rs. 2 less than the previous minute

The time it will take the clerk to count Rs. 4,500 is given as follows;

The number of one rupee currency notes the clerk counts in the first 10 minutes = 10 × Rs. 150 = Rs. 1,500

The amount in one rupee currency notes the clerk counted at a decreasing rate of Rs. 2 less every minute = Rs. 4,500 - Rs. 1,500 = Rs. 3,000

From the 11th minute the number of rupees the clerk counts = Rs. 150 - Rs. 2 = Rs. 148

We have an arithmetic progression with the first term, a = 148, and the common difference, d = -2

The sum of n terms of an AP,

Which gives;

3000 = 149·t - t²

Which gives;

t² - 149·t + 3,000 = 0

(t - 24) × (t - 125) = 0, therefore, t = 24 minutes or t = 125 minutes

The time it would take the clerk to count a sum of Rs. 4,500 = 10 + 24 = 34 minutes.

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.

well what did you do it matters what you did : )

step-by-step explanation: