Question

Q20-23 with steps pls

Answer 1

20: Let $x$ be the amount of money. If this amount is shared between 8 people, each person will get x/8 dollars. But there actually are 6 people, so everyone is getting x/6 dollars. We know that this difference results in 30 dollars more, so we have

$\dfrac{x}{6}=\dfrac{x}{8}+30 \iff \dfrac{x}{6}-\dfrac{x}{8}=30$

Rearrange the left hand side as

$\dfrac{4x-3x}{24}=30$

And multiply both sides by 24 to get

$x=720$

21: Let $M$ and $J$ be the number of marbles owned by Mike and Judy, respectively. At the beginning, we have

$M=3J+11$

If they both get 9 more marbles, Mike will have $M+9$ marbles, and Judy will have $J+9$ marbles. So, we have

$M+J+18=93 \iff M+J=75 \iff M=75-J$

Plug this value in the first equation and we have

$75-J=3J+11 \iff 4J=64 \iff J=16$

And we deduce

$M=16\cdot 3 + 11=59$

So, the difference is  

$M-J=59-16=43$

22: Let $x,y,z$ be the number of $10, $20 and $50 coupon, respectively. We're given:

$\begin{cases}x=2y+3\\z=\frac{1}{2}y\\x+y+z=38\end{cases} We can write the third equation by substituting the expressions for x and z: [tex]x+y+z=(2y+3)+y+\left(\dfrac{1}{2}y\right)=38$

Rearrange as follows:

$(2y+3)+y+\left(\dfrac{1}{2}y\right)=38 \iff \dfrac{7}{2}y=35 \iff 7y=70 \iff y=10$

And now we can deduce the number of the other coupons:

$x=2\cdot 10+3=23,\quad z=\dfrac{1}{2}\cdot 10=5$

So, the total value is

$23\cdot 10+10\cdot 20+5\cdot 50=230+200+100=530$

23:  a) In order to find the first three terms of the sequence, you just need to plug n=1,2,3:

$a_1=4\cdot 1+3=7,\quad a_2=4\cdot 2+3=11,\quad a_3=4\cdot 3+3=15$

b) We have

$a_r = 4r+3=71 \iff 4r=68 \iff r=\dfrac{68}{4}=17$

c) 105 is a term of the sequence if and only if there exists an integer k such that

$a_k=4k+3=105 \iff 4k = 102 \iff k=\dfrac{102}{4}=25.5$

So, 105 is not a term of the sequence.