Math, asked by Zahrabee, 1 month ago

The allotment officer is trying to come up with a method to calculate fair division of funds across

various affected families so that the fund amount and amount received per family can be easily

adjusted based on daily revised numbers. The total fund allotted for a village is x x6 20 9 x 3 2 + + + .

The officer has divided the fund equally among families of the village and each family receives an

amount of x x2 2 2 + + . After distribution, some amount is left.

(i) How many families are there in the village?

(a) x + 4 (b) x - 3

(c) x - 4 (d) x + 3

(ii) If an amount of <1911 is left after distribution, what is value of x ?

(a) 190 (b) 290

(c) 191 (d) 291

(iii) How much amount does each family receive?

(a) 24490 (b) 34860

(c) 22540 (d) 36865

(iv) What is the amount of fund allocated?

(a) Rs 72 72 759 (b) Rs 75 72 681

(c) Rs 69 72 846 (d) Rs 82 74 888

(v) How many families are there in the village?

(a) 191 (b) 98

(c) 187 (d) 195​

Answers

Answered by amitnrw
13

Given  : The total fund allotted is formulated by the officer is x³ + 6x² + 20x + 9

divided the fund equally among families of the village

each family receives an amount of x² + 2x + 2  and After distribution, an amount of 10x+ 1  is left

To Find : How many families are there in the village  

an amount of ₹1911 is left after distribution, what is value of x

Solution :

                             x + 4

    x² + 2x + 2   _|  x³ + 6x² + 20x + 9 |_

                               x³ + 2x² + 2x

                              ___________

                                      4x² + 18x + 9

                                      4x² + 8x + 8

                              ________________

                                                10x + 1

x + 4  families are there in the village

10x + 1 = 1911

=> 10x = 1910

=> x = 191

amount each family receive         x² + 2x + 2

= 191²  + 2(191) + 2

= 36865

amount of fund allocated  = x³ + 6x² + 20x + 9

191³ + 6(191²) + 20(191)+ 9

= 71,90,586

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