7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Quiz 1\",\"description\":\"SAT Math Advanced Quizzes Quiz 1\",\"url\":\"https://theschoolofmathematics.com/quiz/sat-math-advanced-quizzes-quiz-1\",\"educationalLevel\":\"Advanced\",\"numberOfQuestions\":10,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Advanced Quizzes\"}}"}}],["$","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 1\",\"item\":\"https://theschoolofmathematics.com/quiz/sat-math-advanced-quizzes-quiz-1\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","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 1"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

SAT Math Advanced Quizzes Quiz 1

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":["$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":2304,"prismaId":"cmq0jyslk01cfjgodxo7mm822","slug":"sat-math-advanced-quizzes-quiz-1","title":{"rendered":"Quiz 1"},"content":{"rendered":"

SAT Math Advanced Quizzes - Quiz 1.

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

SAT Math Advanced Quizzes Quiz 1

"},"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":"cmq0jyslp01chjgodkxmsfpoq","texte_question":"

The graph of the quadratic function $$f(x)=ax^2+bx+c$$ has vertex at $$(4,-18)$$ and passes through the point $$(7,9)$$. What is the value of $$a+b+c$$?

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

Use vertex form.

$$f(x)=a(x-4)^2-18$$

The graph passes through $$(7,9)$$, so substitute $$x=7$$ and $$f(x)=9$$.

$$9=a(7-4)^2-18$$

$$9=9a-18$$

$$27=9a$$

$$a=3$$

So the function is:

$$f(x)=3(x-4)^2-18$$

Expand:

$$f(x)=3(x^2-8x+16)-18$$

$$f(x)=3x^2-24x+48-18$$

$$f(x)=3x^2-24x+30$$

Therefore:

$$a=3$$, $$b=-24$$, $$c=30$$

Find $$a+b+c$$.

$$3+(-24)+30=9$$

Final answer: $$9$$

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

$$-15$$

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

$$-9$$

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

$$3$$

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

$$9$$

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyslu01cnjgodh2i12pq0","texte_question":"

The function $$f$$ is defined by $$f(x)=a(x-h)^2+k$$ where $$a$$, $$h$$, and $$k$$ are constants. The graph of $$y=f(x)$$ has vertex at $$(-5,2)$$. Additionally, it is known that $$f(-2)$$ > $$f(-8)$$. Which of the following must be true?

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

The vertex is $$(-5,2)$$, so: $$h=-5$$

The function is:

$$f(x)=a(x+5)^2+2$$

Now compare $$f(-2)$$ and $$f(-8)$$.

$$f(-2)=a(-2+5)^2+2$$

$$f(-2)=a(3)^2+2$$

$$f(-2)=9a+2$$

Also:

$$f(-8)=a(-8+5)^2+2$$

$$f(-8)=a(-3)^2+2$$

$$f(-8)=9a+2$$

So: $$f(-2)=f(-8)$$

Final answer: $$f(-2)=f(-8)$$

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

$$f(1)$$ > $$f(3)$$

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

$$f(4)$$ > $$f(6)$$

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

$$f(-1)$$ > $$f(-3)$$

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

$$f(-2)=f(-8)$$

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyslz01ctjgod8kb83k4p","texte_question":"

For the quadratic equation $$kx^2-(14w)x+49k=0$$ suppose it has exactly one real solution and $$kw=400$$. What is the value of $$|k+w|$$?

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

A quadratic equation has exactly one real solution when its discriminant equals $$0$$.

For:

$$kx^2-(14w)x+49k=0$$

we have:

$$a=k$$, $$b=-14w$$, $$c=49k$$

Use the discriminant.

$$b^2-4ac=0$$

Substitute:

$$(-14w)^2-4(k)(49k)=0$$

$$196w^2-196k^2=0$$

Divide by $$196$$.

$$w^2-k^2=0$$

Factor:

$$(w-k)(w+k)=0$$

So: $$w=k$$

or: $$w=-k$$

Since $$kw=400$$ is positive, $$k$$ and $$w$$ must have the same sign.

Therefore: $$w=k$$

Then:

$$k^2=400$$

$$k=20$$ and $$w=20$$,

Or $$k=-20$$ and $$w=-20$$.

In either case: $$|k+w|=40$$

Final answer: $$40$$

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

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jysm401cwjgodhjskz3d2","texte_question":"

The system of equations below has two distinct solutions $$(x_1,y_1)$$ and $$(x_2,y_2)$$.

$$y=x^2-6x+13$$

$$y=kx+9$$

If $$x_1+x_2=14$$, what is the value of $$k$$?

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

Set the two equations equal.

$$x^2-6x+13=kx+9$$

Move all terms to one side.

$$x^2-(6+k)x+4=0$$

For a quadratic equation:

$$ax^2+bx+c=0$$

The sum of the solutions is: -$$\\frac{b}{a}$$

Here: $$a=1$$, $$b=-(6+k)$$

So:

$$x_1+x_2=6+k$$

Given:

$$x_1+x_2=14$$

Therefore:

$$6+k=14$$

$$k=8$$

Final answer: $$8$$

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

$$6$$

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

$$8$$

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

$$10$$

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

$$12$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jysm901d2jgodbd7b2v1k","texte_question":"

In the system:

$$\\frac{a}{6}x+\\frac{7}{12}=\\frac{b}{6}y-\\frac{5}{12}$$

$$-30x+8y=9-4y$$

where $$a$$ and $$b$$ are constants, the system has no solution. The ratio $$a:b$$ is equivalent to $$18:k$$. What is the value of $$k$$?

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

Rewrite the first equation.

$$\\frac{a}{6}x+\\frac{7}{12}=\\frac{b}{6}y-\\frac{5}{12}$$

Move the variable terms to the left and the constants to the right.

$$\\frac{a}{6}x-\\frac{b}{6}y=-1$$

Multiply by $$6$$.

$$ax-by=-6$$

Rewrite the second equation.

$$-30x+8y=9-4y$$

Add $$4y$$ to both sides.

$$-30x+12y=9$$

For the system to have no solution, the lines must be parallel but different. So the coefficients of $$x$$ and $$y$$ must be proportional.

Compare: $$ax-by=-6$$ and $$-30x+12y=9$$

Thus: $$a:b=30:12$$

Simplify:

$$a:b=5:2$$

The ratio is equivalent to: $$18:k$$

Set up the proportion. $$\\frac{5}{2}=\\frac{18}{k}$$

Cross multiply.

$$5k=36$$

$$k=\\frac{36}{5}$$

Final answer: $$\\frac{36}{5}$$

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

$$\\frac{36}{5}$$

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

$$\\frac{7}{5}$$

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

$$\\frac{12}{7}$$

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

$$7$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jysmd01d8jgodoxpuoeao","texte_question":"

Consider the system of equations:

$$(x-4)^2+(y+2)^2=49$$

$$y=3x-k$$

where $$k$$ is a constant. If the system has exactly one real solution $$(x,y)$$, what is the sum of all possible values of $$k$$?

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

The circle is:

$$(x-4)^2+(y+2)^2=49$$

So, the center is: $$(4,-2)$$ and the radius is: $$7$$

The line is: $$y=3x-k$$

Rewrite it in standard form.

$$3x-y-k=0$$

For the system to have exactly one real solution, the line must be tangent to the circle. So, the distance from the center to the line must equal the radius. Use the point-to-line distance formula.

$$\\frac{|3(4)-(-2)-k|}{\\sqrt{3^2+(-1)^2}}=7$$

Simplify.

$$\\frac{|12+2-k|}{\\sqrt{10}}=7$$

$$\\frac{|14-k|}{\\sqrt{10}}=7$$

$$|14-k|=7\\sqrt{10}$$

So:

$$14-k=7\\sqrt{10}$$

$$14-k=-7\\sqrt{10}$$

Thus:

$$k=14-7\\sqrt{10}$$ or $$k=14+7\\sqrt{10}$$

Find the sum.

$$(14-7\\sqrt{10})+(14+7\\sqrt{10})=28$$

Final answer: $$28$$

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

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jysmh01dbjgodqhgluc6d","texte_question":"

A project requires between 540 and 900 labor-hours, inclusive. A team consists of 8 senior workers and 5 junior workers. Each senior worker contributes between 32 and 48 hours per week. Each junior worker contributes between 20 and 30 hours per week. If the project is completed in exactly $$w$$ weeks, which inequality represents the possible values of $$w$$?

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

Find the minimum weekly labor-hours.

Senior workers:

$$8(32)=256$$

Junior workers:

$$5(20)=100$$

Minimum weekly labor-hours:

$$256+100=356$$

Find the maximum weekly labor-hours.

Senior workers:

$$8(48)=384$$

Junior workers:

$$5(30)=150$$

Maximum weekly labor-hours:

$$384+150=534$$

In $$w$$ weeks, the team can complete between: $$356w$$ and $$534w$$ labor-hours.

The project requires between $$540$$ and $$900$$ labor-hours.

For the work to be possible:

$$534w\\ge540$$ and $$356w\\le900$$

Solve the first inequality.

$$w\\ge\\frac{540}{534}$$

$$w\\ge\\frac{90}{89}$$

Solve the second inequality.

$$w\\le\\frac{900}{356}$$

$$w\\le\\frac{225}{89}$$

Final answer: $$\\frac{90}{89}\\le w\\le\\frac{225}{89}$$

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

$$\\frac{90}{89}\\le w\\le\\frac{225}{89}$$

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

$$\\frac{45}{59}\\le w\\le\\frac{45}{29}$$

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

$$\\frac{15}{29}\\le w\\le\\frac{90}{59}$$

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

$$\\frac{90}{59}\\le w\\le\\frac{15}{29}$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jysmm01dhjgod2ycspie5","texte_question":"

From an external point $$P$$, two tangent segments, $$PA$$ and $$PB$$, are drawn to a circle with center $$O$$. The radius of the circle is $$18$$. The length of chord $$AB$$ is $$24$$. What is the distance between point $$O$$ and point $$P$$?

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

Let $$M$$ be the midpoint of chord $$AB$$.

Since $$AB=24$$, half the chord is: $$AM=12$$

The radius is: $$OA=18$$

In right triangle $$OMA$$:

$$OM^2+AM^2=OA^2$$

Substitute:

$$OM^2+12^2=18^2$$

$$OM^2+144=324$$

$$OM^2=180$$

$$OM=6\\sqrt{5}$$

For a chord of contact from an external point, the distance from the center to the chord satisfies:

$$OM=\\frac{r^2}{OP}$$

Substitute:

$$6\\sqrt{5}=\\frac{18^2}{OP}$$

$$6\\sqrt{5}=\\frac{324}{OP}$$

Solve for $$OP$$.

$$OP=\\frac{324}{6\\sqrt{5}}$$

$$OP=\\frac{54}{\\sqrt{5}}$$

Rationalize.

$$OP=\\frac{54\\sqrt{5}}{5}$$

Final answer: $$\\frac{54\\sqrt{5}}{5}$$

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

$$18$$

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

$$24$$

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

$$\\frac{54\\sqrt{5}}{5}$$

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

$$36$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jysmr01dnjgoddtmoy2ug","texte_question":"

A circle in the $$xy$$-plane has the equation $$x^2+y^2-4x+8y-20=0$$ What is the length of a tangent segment drawn from the point $$(11,-4)$$ to the circle?

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

Rewrite the circle equation in standard form.

$$x^2+y^2-4x+8y-20=0$$

Group the $$x$$-terms and $$y$$-terms.

$$x^2-4x+y^2+8y=20$$

Complete the square.

$$x^2-4x=(x-2)^2-4$$

$$y^2+8y=(y+4)^2-16$$

Substitute:

$$(x-2)^2-4+(y+4)^2-16=20$$

$$(x-2)^2+(y+4)^2=40$$

So, the center is: $$(2,-4)$$ and the radius is: $$\\sqrt{40}$$

The external point is: $$(11,-4)$$

Find the distance from the point to the center.

$$d=11-2$$

$$d=9$$

The tangent length $$t$$ satisfies:

$$t^2=d^2-r^2$$

Substitute:

$$t^2=9^2-40$$

$$t^2=81-40$$

$$t^2=41$$

$$t=\\sqrt{41}$$

Final answer: $$\\sqrt{41}$$

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

$$5$$

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

$$\\sqrt{41}$$

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

$$4\\sqrt{2}$$

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

$$3\\sqrt{5}$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jysmw01dtjgodef6lqvem","texte_question":"

In the $$xy$$-plane, a circle has its center at the origin. An external point $$P$$ is located at $$(16,0)$$. Two lines tangent to the circle are drawn from point $$P$$, touching the circle at points $$A$$ and $$B$$. The measure of the angle formed by these two tangent lines, $$\\angle APB$$, is $$120^\\circ$$. What is the radius of the circle?

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

The center is the origin, so: $$OP=16$$

The two tangent segments from $$P$$ are symmetric, so $$OP$$ bisects $$\\angle APB$$.

Since:

$$\\angle APB=120^\\circ$$

we have:

$$\\angle APO=60^\\circ$$

A radius drawn to a tangent point is perpendicular to the tangent line. So triangle $$OPA$$ is a right triangle with: $$\\angle OAP=90^\\circ$$

In triangle $$OPA$$, the radius $$OA$$ is opposite the $$60^\\circ$$ angle, and $$OP$$ is the hypotenuse.

Use sine.

$$\\sin60^\\circ=\\frac{OA}{OP}$$

$$\\frac{\\sqrt{3}}{2}=\\frac{r}{16}$$

Solve for $$r$$.

$$r=16\\cdot\\frac{\\sqrt{3}}{2}$$

$$r=8\\sqrt{3}$$

Final answer: $$8\\sqrt{3}$$

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

$$4$$

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

$$8$$

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

$$8\\sqrt{3}$$

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

$$16$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Advanced Quizzes"],"date":"2026-06-05T06:38:06.920Z","modified":"2026-06-05T06:38:06.920Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"sat-math-advanced-quizzes-quiz-1"}]}]]}]}]]

Quiz 1

Comments

Leave a comment