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Q: The numerator of a fraction is 3 less than the denominator. If 2 is added to  

     both the numerator and the denominator, then the sum of the new and the

     original fraction is 29/20. Find the original  fraction.

                                                      SOLUTION.

Let the numerator of the fraction be x. Then, according to the question,

                                              Denominator = x + 3

Thus, the fraction is x  / (x+3). Now, adding 2 to both the Nr. And the Dr,

                         New fraction = (x+2) / (x+5).

Now, as per the question, we have,

                           [ x  / (x+3)] + [(x+2) / (x+5)] = 29 / 20. … Adding 2 to x and x+3

Simplifying, we have

                          ( 2 * x^2 + 10 * x + 6) / ( x^2 + 8 * x + 15) = 29/20

On further simplification after cross multiplying, we have

                             11 * x^2 - 32 * x – 315 = 0

Factorising the above quadratic into 2 linear factors, we have

                              11 * x * (x – 7) + 45 * (x – 7) = 0

•                  11x + 45 = 0 or x -7 = 0

Now, the first root is -45/11 and the second root is 7.

Taking x = 7 as the numerator, we have denominator as 10.

Therefore our fraction is 7/10.

Note: We ignore the first root as it is negative and does not yield the satisfactory answer.    

Q: Sum of the area of two squares is 400 square cm. If the difference of their perimeters is 16 cm, find the sides of the two squares.

                                                                  SOLUTION

Let the side of the first square be ‘a’ and that of the second square be ‘b’.  then, according to the problem,

                                         a^2 + b^2 = 400   …(1)

and                                  4a – 4b = 16          …(2)    Hint : Perimeter = 4 times the side of a square

Now, from (2), a = b + 4,

substituting this value of a in (1), we have

                             (4+b) ^2+b^2 = 400    

Expanding the above we get the following quadratic equation:

                                                b^2 + 4*b - 192 = 0             ….(3)

factoring the polynomial, we get,

                                               b^2 + 16b-12b-192 = 0        …. (4)

simplifying, we obtain,

                                              b(b+ 16)-12(b+ 16) = 0

• (b-12) (b+16) = 0

From which we get two solutions viz

                                             b = 12 or b = -16

Now, since the side of a square cannot be negative, thus the correct solution is b = 12, whence, a = b + 4.

Thus, a = 16.

Therefore, the sides of the square are 12 and 16 cm respectively.

                                            Arithmetic Progression Example.

Q:If  the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289,find the sum of its first n terms.

                                                                  Solution:

 We know that the Sum of the first n terms of an AP is given by:

                                   Sn = (n / 2) * ((2 * a + (n-1) * d)).

Therefore, as per the first requirement,

                                          49 = (7 / 2) * ((2 * a + (7-1) * d)).

• 98 = 14a + 42d
• 7 = a + 3 d                                    (1)

Similarly, as per the second requirement,

                                    289 = (17 / 2) * ((2 * a + (17-1) * d)).

• 578 = 34a + 272d
• 17 = a + 8d (2)

Now, from (1), we have , a = 7 – 3d. Substituting this value of a in (2), we have

                          17 = 7-3d + 8d

                         Whence, 10 = 5d,

                           Thus, d = 2.

And, then a is 7 – (3 * 2) = 7 – 6 = 1

Thus, we have, a = 1 and d = 2. Now, the sum of the first n terms is

                  Sn = (n / 2) * ((2 * a + (n-1) * d)).

• Sn = (n / 2) * ((2 * 1 + (n-1) * 2)). ---- Substituting the values of a and d obtained earlier
• Sn = (n / 2) * ((2 + 2n -2))
• Sn = (n / 2) * (2n)
• Sn = n^2

Thus,

                    The sum of its first n terms = n^2.

Q:  Two casks A and B were filled with 2 kinds of sherry, mixed in the cask A in the ratio of 2:7, and in the cask B in the ratio of 1:5. What quantity must be taken from each to form a mixture which shall consist of 2 gallons of one kind and 9 gallons of the other,

                                                                     SOLUTION.

Let us assume that we take ‘a’  gallons from the first cask and ‘b’ from the other cask. The, considering the first cask, we have :

                                             2a/9 parts(or, units) of First type of sherry,S1,and

                                             7a/9 parts(or, units) of First type of sherry,S2 .

And, considering the second cask, we have:

                                             b/6 parts (or, units) of First type of sherry, S1,and

                                             5b/6 parts (or, units) of First type of sherry, S2.

 In the mixture, therefore, we have 2a/9+b/6 units of S1 and 7a/9+5b/6 units of S2.

Now, 2a/9+b/6 = (4a+3b) / 18 and 7a/9+5b/6 = (14a+15b) / 18

Therefore, as per the question, the ratio of S1 to S2 in the mixture is,

                                              --    (  (4a+3b) / 18  )  /   ( (14a+15b) / 18 )     =  2 / 9

                                              --   (4a+3b) / (14a+15b)  =  2 / 9

From which,

                                              --        36a+27b = 28a+30b

                                             --           8a = 3b

From which, we have finally,

                                                              a/b = 3/8

Thus, we have to take 3 gallons from cask A and 8 gallons from cask B to form the mixture.