10th CBSE math solution for exercise 5.1 part 1

This page 10th CBSE math solution for exercise 5.1 part 1 is going to provide you solution for every problems that you find in the exercise no 5.1 part 1

10th CBSE math solution for exercise 5.1 part 1

(1) In which of the following situations,does the list of numbers involved make an arithmetic progression,and why?

(i) The taxi fare each km when the fare is Rs.15 for the first km and Rs.8 for each additional km.

Solution:

From the above given information we come to know that,

Taxi fare for 1st km = 15

Taxi fare for 2nd km = 15 + 8 = 23

Taxi fare for 3rd km = 23 + 8 = 31

    15,23,31,.................

Every term of this sequence is 8 more than the previous term. This sequence clearly form an A.P


(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Solution:

Let "x" be the amount of air present in a cylinder

In each stroke vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

In other words, after every stroke,only 1-(1/4) = (3/4) th part of air will remain.

Therefore the volumes will be x,(3x/4),(3x/4)²,,,,,,,,,

Therefore the common difference are not same in the above sequence.So this sequence is not A.P


(iii) The cost of digging a well after every meter of digging,when it costs Rs.150 for first meter and rises by Rs.50 for each subsequent meters.

Solution:

Cost of digging for the first meter = Rs.150

Cos t of digging for the second meter = 150 + 50

                                                       = Rs. 200

Cost of digging for the third meter = 200 + 50 

                                                     = Rs. 250

150,200,250,..............

Every term is 50 more than the previous term,since the common difference is same. It is A.P          


(iv) The amount of money in the account every year,when Rs.10000 is deposited at compound interest at 8% per annum.

Solution:

We know that if Rs P is deposited at r % compound interest per annum for n years, our money will be p[1+(r/100)]^n after n years.

Therefore, after every year, our money will be

10000(1.08),10000(1.08)²,10000(1.08)³,..............

Clearly, adjacent terms of this series do not have the same difference

between them. Therefore, this is not an A.P.




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