Question

Solve a and b of question 7

Answer 1

Answer:

(a) 90 is the included number.

(b) (c - d)[b(c - d) - a + 3]

OR

(c - d)[bc - bd - a + 3]

Step-by-step explanation:

7. (a)

Let the included number be x.

Now,

We know that,

Mean = Sum of observations ÷ No. of observations

Let the Sum be S

So,

50 = S/9

S = 50 × 9

S = 450

Hence,

The sum of all the 9 numbers is 450.

Now,

We are adding another number x,

So,

Sum = 450 + x

No. of observations = 9 + 1 = 10

Hence,

Mean = (450 + x)/10

But, we are given,

Mean = 54

So,

54 = (450 + x)/10

450 + x = 54 × 10

450 + x = 540

x = 540 - 450

x = 90

Hence,

The included number was 90.

7. (b)

We have,

b(c - d)² + a(d - c) + 3c - 3d

b(c - d)² + a(d - c) + 3(c - d)

If we multiply (-1) to (d - c), we get,

(-1)(d - c) = -d + c = (c - d)

I did this so that I can factor out (c - d)

So,

b(c - d)² + a(-1)(d - c) + 3(c - d)

But if I do this to d - c then that would give us a wrong factorization because I multiplied (-1), but, if I multiply (-1) again I will become 1.

That is,

(-1) × (-1) = 1

So,

b(c - d)² + (1)a(d - c) + 3(c - d)

b(c - d)² + (-1 × -1)a(d - c) + 3(c - d)

Using Distributive Property,

b(c - d)² + (-1)a(-1)(d - c) + 3(c - d)

b(c - d)² - a(c - d) + 3(c - d)

Now, its correct because I only multiplied 1 to the equation.

Actually, we dont need all this process, I included it so that others who unable to understand, can understand faster.

b(c - d)² - a(c - d) + 3(c - d)

Expanding the Square,

b(c - d)(c - d) - a(c - d) + 3(c - d)

Now I can factor out (c - d)

(c - d)[b(c - d) - a + 3]

Now, if you want you can keep it this way or simplify more.

(c - d)[b(c - d) - a + 3]

= (c - d)[bc - bd - a + 3]

Hope it helped and believing you understood it........All the best

Answer 2

QUESTION A :

The mean of 9 numbers is 50. If one number is included, their mean becomes 54. Find the included number.

$\red {\tt Mean \ = \ \dfrac {Sum \ of \ all \ observations}{number \ of \ observations}}$

Put the given values in the above mean formula,

We get,

$\implies \tt 50 \ = \ \dfrac {Sum \ of \ all \ observations}{9}$

$\implies \tt 50 \times 9 \ = \ 450$

$\therefore$ Sum of all observations = 450.

If the number is added = 55.

Now,

The total number of observations = 9 + 1

The total number of observations = 10.

Let the number be y.

$\implies \tt 55 \ = \ Sum \ of \ all \ earlier \ observations \ + \ \dfrac {y}{10}$

$\implies \tt 55 \ = \ 450 \ + \ \dfrac {y}{10}$

$\implies \tt 550 \ = \ 450 \ + \ y$

$\implies \tt y \ = \ 550 \ - \ 450$

$\large \qquad {\boxed {\tt y \ = \ 100}}$

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QUESTION B :

Factorise the expression : $\tt b(c - d)^{2} \ + \ a(d - c) \ + \ 3c \ - \ 3d$

$\implies \tt b(c - d) (c - d) - a(c - d) \ + \ 3c \ - \ 3d$

$\implies \tt (c - d) [b (c - d) - a + 3]$

$\implies \tt (c - d) (bc - bd - a + 3)$

$\large \qquad {\boxed {\tt (c - d) (bc - bd - a + 3)}}$

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