7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Quiz 25\",\"description\":\"Advanced SAT Math Quiz 25\",\"url\":\"https://theschoolofmathematics.com/quiz/sat-math-advanced-quizzes-quiz-25\",\"educationalLevel\":\"Advanced\",\"numberOfQuestions\":13,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Advanced Quizzes\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Quiz 25\",\"item\":\"https://theschoolofmathematics.com/quiz/sat-math-advanced-quizzes-quiz-25\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","div",null,{"className":"container mx-auto px-4 py-8 md:py-12","children":[["$","div",null,{"className":"mb-12","children":["$undefined",["$","div",null,{"className":"mb-8","children":["$","h1",null,{"className":"text-4xl md:text-5xl lg:text-6xl font-bold text-gray-900 mb-4","children":"Quiz 25"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

Advanced SAT Math Quiz 25

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":["$undefined",["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M8.228 9c.549-1.165 2.03-2 3.772-2 2.21 0 4 1.343 4 3 0 1.4-1.278 2.575-3.006 2.907-.542.104-.994.54-.994 1.093m0 3h.01M21 12a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Questions"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":10}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M9 12l2 2 4-4m6 2a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Level"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Advanced"}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M7 7h.01M7 3h5c.512 0 1.024.195 1.414.586l7 7a2 2 0 010 2.828l-7 7a2 2 0 01-2.828 0l-7-7A1.994 1.994 0 013 12V7a4 4 0 014-4z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Category"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Advanced Quizzes"}]]}]]}]]}]]}]]}],["$","$L17",null,{"quiz":{"id":2470,"prismaId":"cmq0jyupc02epjgodcacbfdqk","slug":"sat-math-advanced-quizzes-quiz-25","title":{"rendered":"Quiz 25"},"content":{"rendered":"

Advanced SAT Math Quiz 25

"},"excerpt":{"rendered":"

Advanced SAT Math Quiz 25

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":"$undefined","niveau_difficulte":"Advanced","categorie":"Advanced Quizzes","nombre_questions":10,"score_minimum":70,"ordre_questions":"Fixe","questions":[{"id":"cmq0jyupj02erjgod7kgqxeeu","texte_question":"$18","type_question":"QCM","explication":"

Since the graph starts at $$(3,-2)$$, write the function as:

$$f(x)=a\\sqrt{x-3}-2$$

Use the point $$(12,4)$$.

$$4=a\\sqrt{12-3}-2$$

$$4=3a-2$$

$$6=3a$$

$$a=2$$

So:

$$f(x)=2\\sqrt{x-3}-2$$

Now evaluate $$f(28)$$.

$$f(28)=2\\sqrt{28-3}-2$$

$$f(28)=2\\sqrt{25}-2$$

$$f(28)=10-2$$

$$f(28)=8$$

Final answer: $$8$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$6$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$7$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$8$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$10$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyupy02exjgodlttia1jj","texte_question":"$19","type_question":"QCM","explication":"

The shaded region is to the left of the $$y$$-axis.

The solid decreasing line passes through $$(0,8)$$ and $$(\\frac{11}{2},0)$$.

Its slope is: $$\\frac{0-8}{\\frac{11}{2}-0}=-\\frac{16}{11}$$

So, its equation is: $$y=-\\frac{16}{11}x+8$$

The shaded region is below this solid line.

$$y\\le-\\frac{4}{3}x+8$$

The dashed increasing line passes through $$(0,2)$$ and $$(5,6)$$.

Its slope is: $$\\frac{6-2}{5-0}=\\frac{4}{5}$$

So, its equation is: $$y=\\frac{4}{5}x+2$$

Because the line is dashed, the boundary is not included.

The shaded region is above this dashed line.

$$y$$ > $$\\frac{4}{5}x+2$$

Final answer:

$$y\\le-\\frac{16}{11}x+8$$

$$y$$ > $$\\frac{4}{5}x+2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$y\\le-\\frac{3}{4}x+5$$

$$y$$ > $$\\frac{3}{2}x+2$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$y\\le-\\frac{16}{11}x+8$$

$$y$$ > $$\\frac{4}{5}x+2$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$y\\ge-\\frac{5}{4}x+5$$

$$y$$ < $$\\frac{3}{5}x+8$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$y\\ge-\\frac{11}{2}x+5$$

$$y$$ < $$\\frac{5}{4}x+8$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuq702f3jgodzyfs5dct","texte_question":"

A circle in the coordinate plane has its center on the line $$y=x-2$$. The circle is tangent to the $$x$$-axis at the point $$(5,0)$$. Which of the following is the equation of the circle?

","type_question":"QCM","explication":"

Since the circle is tangent to the $$x$$-axis at $$(5,0)$$, the center must be directly above or below $$(5,0)$$.

So, the center has $$x$$-coordinate $$5$$.

The center also lies on: $$y=x-2$$

Substitute $$x=5$$.

$$y=5-2$$

$$y=3$$

So, the center is: $$(5,3)$$

The radius is the distance from $$(5,3)$$ to $$(5,0)$$.

$$r=3$$

Therefore, the equation is:

$$(x-5)^2+(y-3)^2=9$$

Final answer: $$(x-5)^2+(y-3)^2=9$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(x-5)^2+(y-3)^2=9$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(x-5)^2+(y-3)^2=25$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$(x-5)^2+(y+3)^2=9$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$(x-3)^2+(y-5)^2=25$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuqb02f9jgodndql74x8","texte_question":"

A data set consists of $$21$$ distinct real numbers with mean $$\\mu$$ and standard deviation $$\\sigma$$. A new data set is formed by multiplying each number by $$-4$$ and then adding $$7$$. Which of the following must be true about the new data set?

","type_question":"QCM","explication":"

Multiplying every value by $$-4$$ multiplies the mean by $$-4$$.

Adding $$7$$ to every value adds $$7$$ to the mean.

So, the new mean is: $$-4\\mu+7$$

Standard deviation measures spread.

Multiplying every value by $$-4$$ multiplies the standard deviation by the absolute value of $$-4$$.

$$|-4|=4$$

Adding $$7$$ does not change the standard deviation.

So, the new standard deviation is: $$4\\sigma$$

Final answer: The new mean is $$-4\\mu+7$$ and the new standard deviation is $$4\\sigma$$.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"The new mean is $$-4\\mu+7$$ and the new standard deviation is $$4\\sigma$$.","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"The new mean is $$-4\\mu+7$$ and the new standard deviation is $$-4\\sigma$$.","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

The new mean is $$4\\mu+7$$ and the new standard deviation is $$4\\sigma$$.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

The new mean is $$-4\\mu$$ and the new standard deviation is $$4\\sigma-7$$.

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuqf02ffjgod48lcszop","texte_question":"

A regular octagon is inscribed in a circle of radius $$10$$. A smaller circle is inscribed in the octagon. What is the area of the region that lies inside the octagon but outside the smaller circle? Round to the nearest hundredth.

","type_question":"QCM","explication":"

The area of a regular octagon with circumradius $$R$$ is:

$$A=4R^2\\sin(45^\\circ)$$

Substitute $$R=10$$.

$$A=4(10)^2\\cdot\\frac{\\sqrt{2}}{2}$$

$$A=200\\sqrt{2}$$

The radius of the smaller inscribed circle is the apothem of the octagon.

$$r=10\\cos(22.5^\\circ)$$

So, the area of the smaller circle is:

$$\\pi r^2=\\pi(10\\cos(22.5^\\circ))^2$$

$$=100\\pi\\cos^2(22.5^\\circ)$$

Use:

$$\\cos^2(22.5^\\circ)=\\frac{2+\\sqrt{2}}{4}$$

So:

$$\\pi r^2=25\\pi(2+\\sqrt{2})$$

The required area is:

$$200\\sqrt{2}-25\\pi(2+\\sqrt{2})$$

$$\\approx14.70$$

Final answer: $$14.70$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$82.84$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$117.16$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$165.69$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$14.70$$

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuqj02fljgodyuzm7fc2","texte_question":"

The volume $$V$$ of a cone is given by $$V=\\frac{1}{3}\\pi r^2h$$ A company redesigns a cone-shaped container by increasing its radius by $$60\\%$$ and decreasing its height by $$25\\%$$. What is the ratio of the volume of the new container to the volume of the original container?

","type_question":"QCM","explication":"

The original volume is proportional to: $$r^2h$$

The new radius is increased by $$60\\%$$.

So, the new radius is: $$1.6r$$

The new height is decreased by $$25\\%$$.

So, the new height is: $$0.75h$$

The new volume is proportional to: $$(1.6r)^2(0.75h)$$

Simplify.

$$(1.6)^2(0.75)r^2h$$

$$=2.56(0.75)r^2h$$

$$=1.92r^2h$$

Convert $$1.92$$ to a fraction.

$$1.92=\\frac{192}{100}=\\frac{48}{25}$$

Final answer: $$48:25$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$48:25$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$64:25$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$96:25$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$192:25$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuqn02frjgodsm0fuls7","texte_question":"

In the coordinate plane, the vertices of $$\\triangle PQR$$ are $$P(1,4)$$, $$Q(7,1)$$, and $$R(x,13)$$. The area of $$\\triangle PQR$$ is $$42$$. If $$x$$ > $$7$$, what is the value of $$x$$?

","type_question":"TexteLibre","explication":"

Use the coordinate area formula.

$$A=\\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$$

Substitute

$$P(1,4)$$, $$Q(7,1)$$, and $$R(x,13)$$.

$$42=\\frac{1}{2}|1(1-13)+7(13-4)+x(4-1)|$$

Simplify inside the absolute value.

$$1(1-13)=-12$$

$$7(13-4)=63$$

$$x(4-1)=3x$$

So:

$$42=\\frac{1}{2}|-12+63+3x|$$

$$42=\\frac{1}{2}|51+3x|$$

Since $$x$$ > $$7$$,

$$51+3x$$ is positive.

$$42=\\frac{51+3x}{2}$$

$$84=51+3x$$

$$33=3x$$

$$x=11$$

Final answer: $$11$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuqq02fujgodl7dstxga","texte_question":"

In the coordinate plane, $$\\triangle ABC$$ has vertices $$A(0,0)$$, $$B(8,0)$$, and $$C(14,6)$$. Line $$m$$ passes through $$A$$ and is parallel to $$BC$$. Line $$n$$ passes through $$B$$ and is parallel to $$AC$$. The lines intersect at point $$D$$. What is the $$y$$-coordinate of point $$D$$?

","type_question":"TexteLibre","explication":"

Find the slope of $$BC$$.

$$m_{BC}=\\frac{6-0}{14-8}$$

$$m_{BC}=1$$

Since line $$m$$ passes through $$A(0,0)$$ and is parallel to $$BC$$: $$y=x$$

Find the slope of $$AC$$.

$$m_{AC}=\\frac{6-0}{14-0}$$

$$m_{AC}=\\frac{3}{7}$$

Line $$n$$ passes through $$B(8,0)$$ and has slope $$\\frac{3}{7}$$.

$$y=\\frac{3}{7}(x-8)$$

At the intersection point, set the equations equal.

$$x=\\frac{3}{7}(x-8)$$

Multiply by $$7$$.

$$7x=3x-24$$

$$4x=-24$$

$$x=-6$$

Since $$y=x$$:

$$y=-6$$

Final answer: $$-6$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

-6

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuqu02fxjgod62n68cco","texte_question":"

A data set consists of $$17$$ distinct numbers arranged in ascending order. The mean is $$24$$ and the median is $$19$$. A new data set is created by increasing each of the $$8$$ numbers greater than the median by $$k$$ and decreasing each of the $$8$$ numbers less than the median by $$k$$, where $$k$$ > $$0$$. Which of the following statements must be true?

","type_question":"QCM","explication":"

There are $$17$$ numbers, so the median is the $$9$$th number.

The $$8$$ numbers less than the median are each decreased by $$k$$.

The total decrease is: $$8k$$

The $$8$$ numbers greater than the median are each increased by $$k$$.

The total increase is: $$8k$$

So, the total sum does not change.

Since the number of values stays the same, the mean remains unchanged.

The median value itself is not changed.

Also, the values below the median move farther down and the values above the median move farther up, so the spread changes.

Therefore, both the mean and the median remain unchanged.

Final answer: Both the mean and the median remain unchanged.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Only the mean remains unchanged.","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"Only the median remains unchanged.","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

Both the mean and the median remain unchanged.

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

The standard deviation remains unchanged.

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyuqx02g3jgodies455so","texte_question":"

If $$a$$ and $$b$$ are positive numbers such that $$\\frac{a}{b}+\\frac{b}{a}=10$$, what is the value of $$\\frac{a^3}{b^3}+\\frac{b^3}{a^3}$$?

","type_question":"QCM","explication":"

Let: $$x=\\frac{a}{b}$$

Then:

$$\\frac{b}{a}=\\frac{1}{x}$$

So:

$$x+\\frac{1}{x}=10$$

We need:

$$x^3+\\frac{1}{x^3}$$

Use the identity:

$$x^3+\\frac{1}{x^3}=\\left(x+\\frac{1}{x}\\right)^3-3\\left(x+\\frac{1}{x}\\right)$$

Substitute $$10$$.

$$x^3+\\frac{1}{x^3}=10^3-3(10)$$

$$=1000-30$$

$$=970$$

Final answer: $$970$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$820$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$970$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$980$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$990$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyur202g9jgodpj1znnzf","texte_question":"

In $$\\triangle ABC$$, $$AD$$ is the angle bisector of $$\\angle BAC$$, with point $$D$$ on $$BC$$. Through $$D$$, a line parallel to $$AB$$ intersects $$AC$$ at point $$E$$. If $$\\angle ABC=75^\\circ$$ and $$\\angle ACB=45^\\circ$$, what is the measure of $$\\angle CED$$?

","type_question":"QCM","explication":"

Find $$\\angle BAC$$.

$$\\angle BAC=180^\\circ-75^\\circ-45^\\circ$$

$$\\angle BAC=60^\\circ$$

Since $$DE\\parallel AB$$, angle $$CED$$ is the angle between $$CE$$ and a line parallel to $$AB$$.

Because $$CE$$ lies on $$AC$$, this angle equals the angle between $$CA$$ and $$AB$$.

That is:

$$\\angle CAB=60^\\circ$$

Final answer: $$60^\\circ$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$45^\\circ$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$60^\\circ$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$75^\\circ$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$90^\\circ$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyur702gfjgodkmp1ffdx","texte_question":"

The polynomial $$P(x)=125x^3-nk^3$$ has a factor of $$5x-2k$$, where $$n$$ and $$k$$ are constants and $$k>0$$. What is the value of $$n$$?

","type_question":"QCM","explication":"

Since $$5x-2k$$ is a factor, the polynomial equals $$0$$ when:

$$5x-2k=0$$

Solve for $$x$$.

$$x=\\frac{2k}{5}$$

Substitute into the polynomial.

$$125\\left(\\frac{2k}{5}\\right)^3-nk^3=0$$

Simplify.

$$125\\cdot\\frac{8k^3}{125}-nk^3=0$$

$$8k^3-nk^3=0$$

Factor.

$$k^3(8-n)=0$$

Since $$k$$ > $$0$$, $$k^3\\ne0$$.

So:

$$8-n=0$$

$$n=8$$

Final answer: $$8$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$8$$

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyurc02gijgodevbtab4b","texte_question":"

Three machines, $$A$$, $$B$$, and $$C$$, produce electronic components. Machine $$A$$ produces components at a rate that is $$r$$ times the rate of Machine $$B$$. Machine $$C$$ produces components at a rate that is $$s$$ times the rate of Machine $$A$$. If Machine $$B$$ produces $$P$$ components in $$Q$$ hours, which expression represents the total number of components all three machines can produce working together for $$T$$ hours?

","type_question":"QCM","explication":"

Machine $$B$$ produces $$P$$ components in $$Q$$ hours.

So, the rate of Machine $$B$$ is: $$\\frac{P}{Q}$$

Machine $$A$$ produces at $$r$$ times the rate of Machine $$B$$.

So, the rate of Machine $$A$$ is: $$r\\frac{P}{Q}$$

Machine $$C$$ produces at $$s$$ times the rate of Machine $$A$$.

So, the rate of Machine $$C$$ is:

$$s\\left(r\\frac{P}{Q}\\right)=rs\\frac{P}{Q}$$

The total rate is:

$$\\frac{P}{Q}+r\\frac{P}{Q}+rs\\frac{P}{Q}$$

Factor.

$$\\frac{P}{Q}(1+r+rs)$$

For $$T$$ hours, the total number of components is: $$\\frac{PT}{Q}(1+r+rs)$$

Final answer: $$\\frac{PT}{Q}(1+r+rs)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{PT}{Q}(1+r+s)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{PT}{Q}(1+r+rs)$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$\\frac{PT}{Q}(r+s+rs)$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$\\frac{PT}{Q}(1+rs)$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Advanced Quizzes"],"date":"2026-06-05T06:38:09.648Z","modified":"2026-06-05T06:38:09.648Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L1a",null,{"targetType":"quiz","targetSlug":"sat-math-advanced-quizzes-quiz-25"}]}]]}]}]]

Quiz 25

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