Eton King's Scholarship 2016 A

Solutions

1ai) $27x-12-11x=20$
    $27x-11x=20+12$
        $16x=32$
          $x=2$

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1aii) Multiply the whole equation by 20 to get
   $\frac{20x}{2}+\frac{60}{4}=\frac{20x}{5}$
    $10x+15=4x$
   $10x+15-4x=0$
   $10x-4x=-15$
      $6x=-15$
        $x=-\frac52$

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1bi) $\frac{72}{65}\times1\frac{1}{12}=\frac{72}{65}\times\frac{13}{12}$

By cancelling the top left and botttom right by $12$ and cancelling the top right and bottom left by $13$ you get:

$=\frac65$
$=1\frac15$

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1bii) $2016\frac23-2014\frac34$
$=2016+\frac23-2014-\frac34$
$=2016-2014+\frac23-\frac34$
$=2+\frac{8}{12}-\frac{9}{12}$
$=2-\frac{1}{12}$
$=1\frac{11}{12}$

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1ci) Multiply the $9.9$ by $10$ to get $99$, making the calculation easier.
Remember to divide by $10$ at the end!

$13\times99$
$=13\times100-13\times1$
$=1300-13$
$=1287$

Divide by $10$ to get the answer.
$1287\div10=128.7$

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1cii) To find the total score of the first class:
$12\times12.4=12\times12+12\times0.4$
$=144+12\times\frac25$
$=144+\frac{24}{5}$
$=144+4\frac45$
$=148\frac45$

To find the total score of the second class:
$13\times9.9=128.7$ (from 1ci)
$=128\frac{7}{10}$

To find the total score for both classes, add these numbers:
$148\frac45+128\frac{7}{10}$
$=276+\frac{8}{10}+\frac{7}{10}$
$=266+1\frac{5}{10}$
$=277\frac12$

To find the mean of the $25$ children, divide the result by $25$
$277\frac12\div25=11.1$ (using the bus stop method).
So the answer is $11.1$

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1d) Angle BFD= angle EDF=$25$° (alternate angles on parallel lines AB and DF and CE)

triangle FDB and triangle DFE are isosceles triangles as they have two equal sides

angle DBF=angle DEF=$25$° (base angles are equal on isosceles triangles)

angle BFD=angle EFD=$180$°$-25$°$-25$° (angles in triangles add to $180$°)
$=130$°

X$=360$°$-130$°$-130$° (angles around a point add to $360$°
$=100$°

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1e) $4y=x-13$ ①
$2y=5x-16$ ②

②$\times2$: $4y=10x-32$ ③
①  $4y=x-14$
③$-$①  $0=9x-18$
   $9x=18$
   $x=2$

Substitute $x=2$ into 1:

$4y=2-14$ $4y=-12$   $y=-3$

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1fi) Let the length of rectangle A be $x$.
Let the width of rectangle B be $y$.

Area A = $xy$

If Rectangle B is increase length by $12\%$ this is the same as multiplying the length by $1.12$ so:

Rectangle B length = $x\times1.12$

If Rectangle B is decrease width by $10\%$ this is the same as multiplying the length by $0.9$ so:

Rectangle B width = $y\times0.9$

Area B = $x\times1.12\times y\times0.9$
Area B = $xy\times\frac{28}{25}\times\frac{9}{10}$
Area B = $xy\times\frac{126}{125}$

Difference in Areas is Area B $-$ Area A
Difference = $\frac{126}{125}xy-xy$
Difference = $\frac{1}{125}xy$

Need to convert $\frac{1}{125}$ into a percentage. To do this multiply by $100\%$

$\frac{1}{125}\times100\%$
=$\frac{100}{125}\%$
=$\frac{4}{5}\%$
=$0.8\%$

So Area B is $0.8\%$ bigger than the area of A

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1fii) Let the length of rectangle A be $x$.
Let the width of rectangle B be $y$.

Area A = $xy$

If Rectangle C is increase length by $25\%$ this is the same as multiplying the length by $1.25$ so:

Rectangle C length = $x\times1.25$

If Rectangle C is decrease width, this is the same as multiplying the length by an unknown $z$ so:

Rectangle C length = $y\times z$

Area C = $x\times1.25\times y\times z$
Area C = $x\times\frac54\times y\times z$
Area C = $\frac54xyz$

As they are equal areas Area A = Area C so:

$\frac54xyz=xy$
$\frac54z=1$
$z=1\div\frac54$
$z=1\times\frac45$
$z=\frac45$
Multiplying by $\frac45$ is the same as multiplying by $0.8$ which is the same as decreasing by $20\%$. Therefore the width was decreased by $20\%$

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1gi) $x=\frac{\frac32}{2\times3^2}+\frac{6}{\frac94}$
 $=\frac32\times\frac{1}{18}+6\times\frac49$
 $=\frac{1}{12}+\frac{24}{9}$
 $=\frac{1}{12}+\frac83$
 $=\frac{2}{24}+\frac{64}{24}$
 $=\frac{66}{24}$
 $=\frac{11}{4}$

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1gii) if $b=a^2$ then $b^2=a^4$

$6=\frac{a}{2a^4}+\frac{2a^2}{a^2}$
$6=\frac{1}{2a^3}+2$
$4=\frac{1}{2a^3}$
$\frac14=2a^3$
$\frac18=a^3$
$\frac12=a$

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1hi) Say you need to subtract $x$:

$p-2q+3r-x=p+2q-3r$
$p-p-2q-2q+3r+3r-x=0$
$-4q+6r=x$

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1hii) Say you need to subtract $x$:

$p-2q+3r-x=3p-2q+3r$
$p-3p-2q+2q+3r-3r-x=0$
$-2p=x$

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1i)

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1ji) $-7x-4<4-3x$
    $-4<4-3x+7x$
  $-4-4<-3x+7x$
    $-8<4x$
    $-2      $x>-2$

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1jii) Multiply the whole equation by $15$ to eliminate fractions

    $\frac{30}{3}(4-3x)-\frac{15(8-6x)}{5}>15$
$10(4-3x)-3(8-6x)>15$
     $40-30x-24+18x>15$
         $16-12x>15$
        $16-15>12x$
           $1>12x$
            $\frac{1}{12}>x$
              $x<\frac{1}{12}$

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1k) Let the number be $x$

$\frac{\frac{x}{5}+4}{0.6}=$ whole number
$\frac{\frac{x}{5}+4}{\frac35}=$ whole number
$({{\frac{x}{5}+4}})\times{\frac53}=$ whole number
$({{\frac{x}{5}+4}})\times{\frac53}=$ whole number
${\frac{5x}{15}+\frac{20}{3}}=$ whole number
${\frac{x}{3}+\frac{20}{3}}=$ whole number
$\frac{x+20}{3}=$ whole number

So $x+20$ has to be a multiple of $3$ so that when divided by $3$ it equals a whole number.

First multiple of $3$ above $20$ is $21$ so $x$ has to be $1$.
The next multiple of $3$ is $24$ so $x$ has to be $4$.

Continue this pattern $x=1,4,7,...2014$

There are $671$ numbers in this pattern $({\frac{2014-1}{3}})$

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2ai)

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2aii)

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2b)

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3a) $(4\div5)\div20$
$=\frac45\times\frac{1}{20}$
$=\frac15\times\frac15$
$=\frac{1}{25}$

$4\div(5\div20)$
$=4\div\frac{5}{20}$
$=4\div\frac{1}{4}$
$=4\times\frac{4}{1}$
$=4\times4$
$=16$

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3b) In $(a\div b)\div c$:

$\frac ab\div c$
$\frac ab\times \frac1c$
$\frac {a}{bc}$

In $a\div (b\div c)$:

$a\div \frac bc$
$\frac a1\times \frac cb$
$\frac {ac}{b}$

So when $(a\div b)\div c=a\div (b\div c)$

$\frac {a}{bc}=\frac {ac}{b}$

So for them to be equal it must mean dividing and multiplying by $c$ has the same result. This is only true when $c=1$ or $c=-1$

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3ci) Through logic:

$a\div((b\div c)\div d)$
$a\div(b\div( c\div d))$
$(a\div(b\div c))\div d$

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3cii) $a\div(b\div( c\div d))$
$=a\div({b \div \frac cd})$
$=a\div({b \times \frac dc})$
$=a\div{\frac {bd}{c}}$
$=a\times{\frac {c}{bd}}$
$=\frac a1\times{\frac {c}{bd}}$
$=\frac {ac}{bd}$

$(a\div(b\div c))\div d$
$=({a\div \frac bc})\div d$
$=({a\times \frac cb})\div d$
$=\frac{ac}{b}\div d$
$=\frac{ac}{b}\times \frac1d$
$=\frac{ac}{bd}$

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4ai) $1+99=100$
$2+98=100$
$3+97=100$
 ...
$49+51=100$

So there is $49$ times $100$ plus an extra $50$.

$49\times100+50=4950$

$4950\div99=\frac{4950}{99}=\frac{550}{11}=50$

So the mean is $50$

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4aii) $2,3,4,5,6,......100$ is the same sequence as
$1,2,3,4,5,....99$ but subtract $1$ and plus $100$

From 4ai, the total of the numbers was $4950$
So the total of the new sequence is $4950-1+100=5049$

Mean = $5049\div99=\frac{5049}{99}=\frac{561}{11}=51$

So the mean is $51$

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4b) As this is an explain question, the answer and solution are the same.

From part 4ai, the sum of the numbers $1,2,3,4,5,...99$ is $4950$

This must mean the sum of the numbers $1,2,3,4,5,....,99,100$ is $100$ greater so their sum is $4950+100=5050$

If we delete a number $n$ then the new sum is $5050-n$

Out of the original 100 numbers, after taking one away there is now $99$ numbers. So the mean can be calculated as:

$\frac{5050-n}{99}$

For the mean to be whole numbers, $5050-n$ must be a multiple of $99$. Only when $n=1$ or $n=100$ is this true, as only $5049$ or $4950$ are divisible by $99$.

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4c) Following on from 4b, them mean of these numbers, subtract $m$ is:

$\frac{5151-m}{100}$

As this needs to be a whole number, $5151-m$ must be divisible by $100$. This is only possible when $m=51$, therefore we get $\frac{5100}{100}=51$.

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5ai) Two cases:

$x$ is odd:

If $x$ is odd then $y$ must be even to make $(x+y)$ odd. Therefore $(x-y)$is an odd number subtract an even number which is an odd number.

$x$ is even:

If $x$ is even then $y$ must be odd to make $(x+y)$ odd. Therefore $(x-y)$is an even number subtract an odd number which is an odd number.

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5aii) Two cases:

$x$ is odd:

If $x$ is odd then $y$ must be odd to make $(x+y)$ even. Therefore $(x-y)$is an odd number subtract an odd number which is an even number.

$x$ is even:

If $x$ is even then $y$ must be even to make $(x+y)$ even. Therefore $(x-y)$is an even number subtract an even number which is an even number.

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5b) $(x-y)(x+y)$
=$x^2-xy+xy-y^2$
=$x^2-y^2$

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5c)

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6a) As this is a show that question, the answer and solution are the same.

01111 use Z on underlined

0111010 use Y on underlined

01010 use T on underlined

110 use Y on underlined

0

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6bi)

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6bii)

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