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Num of Equations

Let $a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$ . What is $a + b + c + d$ ?

$ext{(A)} -5 qquad ext{(B)} -10/3 qquad ext{(C)} -7/3 qquad ext{(D)} 5/3 qquad ext{(E)} 5$

The sum of three numbers is $20$ . The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

$mathrm{(A) } 28qquad mathrm{(B) } 40qquad mathrm{(C) } 100qquad mathrm{(D) } 400qquad mathrm{(E) } 800$

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The sum of three numbers is $20$ . The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

$mathrm{(A) } 28qquad mathrm{(B) } 40qquad mathrm{(C) } 100qquad mathrm{(D) } 400qquad mathrm{(E) } 800$

A school store sells 7 pencils and 8 notebooks for $mathdollar 4.15$ . It also sells 5 pencils and 3 notebooks for $mathdollar 1.77$ . How much do 16 pencils and 10 notebooks cost?

$ext{(A)}mathdollar 1.76 qquad ext{(B)}mathdollar 5.84 qquad ext{(C)}mathdollar 6.00 qquad ext{(D)}mathdollar 6...$

Suppose that $frac{2}{3}$ of $10$ bananas are worth as much as $8$ oranges. How many oranges are worth as much as $frac{1}{2}$ of $5$ bananas?

$mathrm{(A)} 2qquadmathrm{(B)} frac{5}{2}qquadmathrm{(C)} 3qquadmathrm{(D)} frac{7}{2}qquadmathrm{(E)} 4$

The product of two positive numbers is 9. The reciprocal of one of these numbers is 4 times the reciprocal of the other number. What is the sum of the two numbers?

$extbf{(A)} frac{10}{3}qquad extbf{(B)} frac{20}{3}qquad extbf{(C)} 7qquad extbf{(D)} frac{15}{2}qquad extbf{...$

The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?

$extbf{(A)} 4qquad extbf{(B)} 5qquad extbf{(C)} 6qquad extbf{(D)} 7qquad extbf{(E)} 8$

10.

1.

2.

3.

Let $x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$ . Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have $4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20$ . Rearranging, we have $3(a+b+c+d)=-10$ , so $a+b+c+d=frac{-10}{3}$ . Thus, our answer is $oxed{ ext{(B)} -10/3}$ .

Solution 1

Let the numbers be $x$ , $y$ , and $z$ in that order. The given tells us that

$egin{eqnarray*}y&=&7zx&=&4(y+z)=4(7z+z)=4(8z)=32zx+y+z&=&32z+7z+z=40z=20z&=&frac{20}...$

Therefore, the product of all three numbers is $xyz=16cdotfrac{7}{2}cdotfrac{1}{2}=28 Rightarrow mathrm{(A)}$ .

Solution 2

Alternatively, we can set up the system in matrix form:

$egin{eqnarray*}1x+1y+1z&=&201x-4y-4z&=&0�x+1y-7z&=&0end{eqnarray*}$

Or, in matrix form
$egin{bmatrix}1 & 1 & 1 1 & -4 & -4 � & 1 & -7end{bmatrix}egin{bmatrix}x y z end{bmatr...$

To solve this matrix equation, we can rearrange it thus:

$egin{bmatrix}x y z end{bmatrix}= egin{bmatrix}1 & 1 & 1 1 & -4 & -4 � & 1 & -7end{bma...$

Solving this matrix equation by using inverse matrices and matrix multiplication yields

$egin{bmatrix}x y z end{bmatrix} =egin{bmatrix}frac{1}{2} frac{7}{2} 16 end{bmatrix}$

Which means that $x = frac{1}{2}$ , $y = frac{7}{2}$ , and $z = 16$ . Therefore, $xyz = frac{1}{2}cdotfrac{7}{2}cdot16 = 28$

We let $p =$ cost of pencils in cents, $n =$ number of notebooks in cents. Then

$egin{align*}7p + 8n = 415 &Longrightarrow 35p + 40n = 20755p + 3n = 177 &Longrightarrow 35p + 21n = 1239end{...$

Subtracting these equations yields $19n = 836 Longrightarrow n = 44$ . Backwards solving gives $p = 9$ . Thus the answer is $16p + 10n = 584 mathrm{(B)}$ .

6.

If $frac{2}{3}cdot10 ext{bananas}=8 ext{oranges}$ , then $frac{1}{2}cdot5 ext{bananas}=left(frac{1}{2}cdot 5 ext{bananas}<br /> ight)cdotleft(frac{8 ext{oranges}}{frac{2}...$ .