Q1-The-expansion-x-1-3x-12-i-find-the-term-independent-of-x-ii-find-the-term-x-4-Q2-There-are-some-nut-in-a-bag-when-2-ounces-of-peanut-are-added-to-the-mixture-the-percentage-of-

Question Number 54858 by Otchere Abdullai last updated on 13/Feb/19

(Q 1) T h e e x p a n s i o n {(x - \frac{1}{3 x})}^{12}

(i) f i n d t h e t e r m i n d e p e n d e n t o f x

(i i) f i n d t h e t e r m x^{4}

(Q 2) T h e r e a r e s o m e n u t i n a b a g . w h e n

2 o u n c e s o f p e a n u t a r e a d d e d t o t h e

m i x t u r e, t h e p e r c e n t a g e o f p e a n u t

b e c o m e s 20 % . R i c h a r d a d d e d 2

o u n c e s o f c a s h e w t o t h e m i x t u r e a n d

t h e p e r c e n t a g e o f c a s h e w n u t w a s

33 . 33 % . f i n d t h e p e r c e n t a g e o f

c a s h e w n u t t h a t w e r e t h e r e i n i t i a l l y .

p l e a s e s i r h e l p

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Feb/19

1) l e t (r + 1) t h t e r m i n d e p e n d e n t o f x [\bar{i e} x^{0}]

12 c_{r} {(x)}^{12 - r} {(- \frac{1}{3 x})}^{r}

12 c_{r} \times x^{12 - r} \times {(- 1)}^{r} \times \frac{1}{3^{r} x^{r}}

12 c_{r} \times {(- 1)}^{r} \times \frac{1}{3^{r}} \times x^{12 - r - r}

s o x^{12 - 2 r} = x^{0} \to 2 r = 12 s o r = 6

x i n d e p e n d e n t t e r m i s

12 c_{6} \times {(- 1)}^{6} \times \frac{1}{3^{6}} \times x^{12 - 6 - 6}

= \frac{12!}{6! 6!} \times 1 \times \frac{1}{3^{6}}

p l s c a l c u l a t e b y y o u r s e l f \dots

n e x t q u e s t i o n x^{4}

x^{12 - 2 r} = x^{4} \to 2 r = 8 r = 4

s o r e q u i r e d a n s w e r i s

12 c_{4} \times {(- 1)}^{4} \times \frac{1}{3^{4}} \times x^{12 - 4 - 4}

12 c_{4} \times 1 \times \frac{1}{3^{4}} \times x^{4} \to c a l c u l a t e t h e v a l u e b y y o u r s e l f

Commented by Otchere Abdullai last updated on 13/Feb/19

T h a n k y o u p r o f e s o r c h a u d h u n r y a m

m u c h g r e a t f u l

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Feb/19

2) p e a n u t : c a s h e w = p : c

i n i t i a l w t o f m i x t u r e = w o u n c e

p e a n u t = \frac{p}{p + c} \times w o u n c e

c a s h e w = \frac{c}{p + c} \times w o u n c e

n o w 2 o u n c e p e a n u t a d d e d \to (\frac{p w}{p + c} + 2) o u n c e p e a n u t

s o % a g e p e a n u t = \frac{\frac{p w}{p + c} + 2}{w + 2} \times 100 = 20

\frac{p w}{p + c} + 2 = \frac{w + 2}{5} \dots \dots (1)

\frac{\frac{c w}{p + c} + 2}{w + 2} \times 100 = 33.33

\frac{c w}{p + c} + 2 = \frac{w + 2}{3} \dots . (2)

\frac{5 p w}{p + c} + 10 = \frac{3 c w}{p + c} + 6

\frac{w}{p + c} (3 c - 5 p) = 4

i t h i n k d a t a i n s u f f i c i e n t \dots p l s r e c h e c k \dots