Detailed solutions to
new sat math questions.

Which positive real number is equal to the quarter of its cube root?
Solution
Let x be the number to find and use the information above to write the equation
x = (1 / 4) ^{3}√ x
Multiply all terms of the equation by 4 and rewrite the equation as
4 x = ^{3}√ x
Cube both sides of the equation and simplify
(4 x) ^{3} = (^{3}√ x) ^{3}
4 ^{3} x ^{3} = x
Rewrite equation with zero on the right, factor and solve
4 ^{3} x ^{3}  x = 0
x(4 ^{3} x ^{2}  1) = 0
Solutions: x = 0
and 4 ^{3} x ^{2}  1 = 0 , x ^{2} = 1 / 64 gives : x = ~+mn~ 1 / 8
We are looking for a positive number, hence positive solution only
Answer: x = 1 / 8

If the points with coordinates (a , b) and (c , d) lie on the the line with equation 2 y + 3 x = 4 and a  c = 3, then what is the value
of d  b?
Solution
Use the coordinates of the two points on the line to find the slope m of the line.
m = (d  b) / (c  a)
The slope m of the line can also be calculated using the equation of the line rewritten in the form y = m x + k.
2 y + 3 x = 4 can be written as 2y =  3x + 4
Divide all terms by 2 to obtain y = (3/2) x + 2
The slope m is equal to 3 / 2
Equate the expression of the slope found above to 3 / 2
3 / 2 = (d  b) / (c  a)
Use the above to find d  b
d  b = 3 / 2 (c  a) = (3 / 2)((a  c)) = 9 / 2
Answer: 9 / 2

Given the system of equations $$\dfrac{1}{3}x^2  \dfrac{1}{3}y^2 = 7 $$
$$0.01x + 0.01y = 0.05$$
What is x  y?
Solution
Multiply all terms of the first equation by 3 to obtain
x ^{2}  y ^{2} = 21
Multiply all terms of the second equation by 100 to obtain
x + y = 5
Factor the left side of equation x^{2}  y^{2} = 21 and rewrite as
(x + y)(x  y) = 21
Susbtitute x + y by 5 in the above and solve for x  y
5(x  y) = 21
x  y = 21/5
Answer: 21/5

Function f is given by f(x) = x ^{2} + a x + b, where a and b are real numbers. What are the values of a and b if the division of f(x) by x  1 gives a remainder equal to 2 and the division of f(x) by x + 2 gives a remainder equal to  5?
Solution
Remainder theorem: The remainder of the division of a polynomial f(x) by a linear polynomial x  a is equal to f(a). We use it twice in the division of f(x) by x  1 and x + 2.
1) f(1) = (1) ^{2} + a(1)+ b = 2 gives the equation: a + b = 3
2) f(2) = (2) ^{2} + a(2)+ b = 5 gives the equation: 2a + b = 9
Solve the above system of equations in a and b. Subtract the second equation from the first to obtain.
3 a = 6 , a = 2
Use the equation a + b = 3 to find b.
2 + b =  3 , gives b = 5
Answer: a = 2 and b = 5

What are the values of the real numbers a, b and c if the equation  4 x(x + 5)  3(4x + 2) = a x ^{2} + b x + c is true for all values of x?
Solution
Expand the left side of the given equation and group like terms.
 4x ^{2}  20x  12 x  6 = a x ^{2} + b x + c
 4x ^{2}  32x  6 = a x ^{2} + b x + c
The two polynomials on each side of the equation are equal if the corresponding coefficients (of the same power of x) are equal. Hence
Answer: a =  4 , b =  32 and c =  6

What is the simplified form of the expression x  10 + x  12 for values of x such that 10 < x < 12?
Solution
For 10 < x < 12, x  10 is positive and x  12 is negative. Using the definition of the absolute value, we have
x  10 = x  10 and x  12 = (x  12)
Hence
x  10 + x  12 = x  10  (x  12) = 2
Answer: x  10 + x  12 = 2 for values of x in the interval 10 < x < 12?

A function is defined by the formula y = (2x  1) / (x + 3). What is the value of x for y =  1 / 4?
Solution
Set y =  1 / 4 in the equations and solve for x
 1 / 4 = (2x  1) / (x + 3)
Cross multiply the above equations
 1(x + 3) = 4(2x  1)
Multiply and solve for x.
 x  3 = 8x  4
 9x =  1
Answer: x = 1 / 9

Which of the graphs below may be that of equation 3 x + 3 y = 3?
.
Solution
First write the equation 3 x + 3 y = 3 in slope intercept form by solving it for y.
3 y = 3 x + 3
Divide all terms by 3.
y = x + 1
From the equation we deduce that the graph of y = x + 1 has a positive slope equal to 1 and a y intercept equal to 1 which corresponds to line L.
Answer: line L

Which of the graphs below may be that of equation 2 y  2 (x  2)^{2}  2 = 0?
.
Solution
Solve the given equation for y.
2 y = 2 (x  2)^{2} + 2
Divide all terms by 2.
y = (x  2)^{2} + 1
The graph of the above equation is that of a parabola shifted 2 units to the right and 1 unit up which corresponds to graph M.
Answer: Parabola M

What is the solution set for the equation x  3 = √x + 17?
Solution
Square both sides of the given equation.
(x  3) ^{2} = (√x + 17) ^{2}
Simplify using the fact that: x ^{2} = x ^{2} and (√x) ^{2} = x
(x  3) ^{2} = x + 17
Expand the left side of the above equation and rewrite it with right side equal to zero.
x ^{2}  7 x  8 = 0
Factor and solve the above quadratic equation.
(x + 1)(x  8) = 0
Answer: Solution set: {1 , 8}

Find the solution set for the equations x ^{2} = 7  x   10
Solution
Rewrite it with right side equal to zero.
x ^{2}  7  x  + 10 = 0
Let u = x and u ^{2} = x ^{2} = x^{2}. Hence using u, the above equation may be written as
u ^{2}  7 u + 10 = 0
Factor the above equation.
(u  5)(u  2)=0
Solve the above equation for u.
u = 5 and u = 2
Let u = x and solve for x.
u = 5 = x gives solutions  5 and 5.
u = 2 = x gives solutions 2 and 2.
Answer: solution set: {5, 2 ,2 ,5}

Write the inequality 3/2 ≤ x ≤ 5/2 using one inequality symbol only.
Solution
The number that is at equal distance from 3/2 and 5/2 on a number line is given by.
(1/2)(3/2 + 5/2) = 2
The distances between 2 and 3/2 and 2 and 5/2 are equal and given by
3/2  2 = 0.5 and 5/2  2 = 0.5
Using the a absolute value, the given ineqality can be written as follows
Answer: x  2 ≤ 0.5

What is the solution set for the equations (x  2)(x^{2}  7 x + 13)  x + 2 = 0?
Solution
Expanding here will make the equation more difficult to solve. BUT note that x  2 may be factored ouy as follows.
(x  2)(x^{2}  7 x + 13)  (x  2) = 0
(x  2)(x^{2}  7 x + 13  1) = 0
(x  2)(x^{2}  7 x + 12) = 0
Solve by equating each of the factors to zero.
x  2 = 0 gives x = 2
x^{2}  7 x + 12 = 0
Factor the second equation.
(x  3)(x  4) = 0 gives the solutions 3 and 4
Answer: Solution set: {2, 3, 4}

If f is a function, which of the functions defined below must have a graph symmetric with respect the y axis?
a) g(x) = (f(x))^{2}
b) h(x) = f(x)
c) i(x) = f(x^{2})
d) j(x) = f(x)
Solution
For a graph of a function to be symmetric with respect to the y axis, the function must be even. An even function must satisfy the condition: f(x) = f(x). Of all the functions in parts a) b) c) and d), only the function in c) satisfies the condition for even function for any function f.
i(x) = f((x) ^{2}) = f(x ^{2}) = i(x)
f(x) = x  1 is an example which proves that g(x) = (x 1) ^{2}, h(x) = x  1 and j(x) =  x  1 are not even and therefore their graphs are not symmetric with respect to the y axis.
Answer: c)

Find all values of k for which the the equation  2 x  4  2 = k + 1 have two solutions?
Solution
Rearrange the equation so that the term with absolute value is on one side.
 2 x  4  2 = k + 1
x  4 = (k + 3)/(2)
For the above equation to have two solutions, the term (k + 3)/(2) must be positive. Hence solve the inequality
(k + 3)/(2) > 0
Multiply both side by 2 and change symbol of inequality
k + 3 < 0
k <  3
Any value of k less than  3 in the given equation yields an equation with two solutions .
Answer: k <  3

Find the value of x if (x + 2)^{2} + 2(x + 2) + y =  2 and y  2 = x.
Solution
Rewrite the second equation as follows.
y = x + 2
Let u = x + 2 and rewrite both equations in terms of u and y as follows:
u ^{2} + 2u + y =  2 and y = u
Substitute y by u in the equation u ^{2} + 2u + y =  2 and solve for u.
u ^{2} + 2u + u =  2
u ^{2} + 3 u + 2 = 0
(u + 1)(u + 2) = 0
u =  1
u =  2
u = x + 2 =  1 gives x =  3
u = x + 2 =  2 gives x =  4
Answer: two values of x: 3 and 4

Find the ratio r = \( \dfrac{f(x + h)  f(x)}{h} \) in terms m if f(x) = m x + b , where m and b are constant real numbers.
Solution
Find f(x + h).
f(x + h) = m (x + h) + b
Substitute in r and simplify.
r = \( \dfrac{m (x + h) + b  (mx + b)}{h} = \dfrac{m x + m h + b  mx  b}{h} = m \)
Answer: r = m

What is the solution set for the equations \( \dfrac{2}{w + 2} = \dfrac{4}{w+3}  \dfrac{1}{3} \)?
Solution
Find the lowest common multiple (lcm) to all the denominators in the equation
lcm = 3(w + 2)(w + 3)
Multiply all terms of the equation by the lcm.
\( \color{red}{3(w + 2)(w + 3)}\dfrac{2}{w + 2} = \color{red}{3(w + 2)(w + 3)}\dfrac{4}{w+3}  \color{red}{3(w + 2)(w + 3)}\dfrac{1}{3} \)
Simplify.
6(w + 3) = 12(w + 2)  (w + 2)(w + 3)
Expand, group and write equation with right side equal to zero.
w ^{2}  w = 0
Factor w out and solve.
w(w  1) = 0
w = 0
w  1 = 0 gives w = 1
Answer: Solution set: {0 , 1}

The equations of the two parabolas shown below are of the form: y = x ^{2} + A x + B and y =  x ^{2} + M x + N. The two parabolas are tangent (touches at one point) and A  M = 2. What is the value of B  N?
.
Solution
The point at which the two parabolas are tangent is found by solving the system of equations of the two parabolas.
y = x ^{2} + A x + B and y =  x ^{2} + M x + N
One way to solve the above system is to equate the right sides and solve for x.
x ^{2} + A x + B =  x ^{2} + M x + N
Rewrite the above equation with right side equal to zero.
2 x ^{2} + x(A  M) + (B  N) = 0
If D is the discriminant of the above quadratic equation, then the two parabolas intersect at two points if D > 0, are tangent and touches at one point if D = 0 and do not intersect if D < 0. We are interested in the case where the parabolas are tangent. Hence
D = (A  M)^{2}  (4)(2)(B  N) = 0
(A  M)^{2} = 8(B  N)
B  N = (A  M)^{2} / 8 = 1 / 2
Answer: B  N = 1 / 2

If the lines with equations A x + B y = C and M x + N y = P are perpendicular and M / N = 5, what is the value of A / B ?
Solution
Rewrite each of the given equations in slope intercept and find its slope.
A x + B y = C may be rewritten as y =  ( A / B ) x + C / B , slope =  A / B
M x + N y = P may be rewritten as y = (M / N) x + P / N , slope =  M / N
If two lines are perpendicular, the product of their slopes is equal to 1. Hence
(  A / B)( M / N ) =  1
Substitute M / N by 5 and find A / B
Answer: A / B =  1 / 5

Solve for x in terms of K, L, M, N and P the equation $$\dfrac{K x  L}{M x  N } = P $$.
Solution
Multiply the two sides of the equation by M x  N
$$\dfrac{K x  L}{M x  N } (M x  N) = P (M x  N)$$
Simplify
 K x + L = P M x  P N
Group all terms containing x on one side.
L + P N = P M x + K x
Factor x out and solve.
L + P N = x (P M + K)
Answer: x = (L + P N) / (P M + K)

The square root of a real number plus twice the same number is equal to 10. What is the number?
Solution
Translate the given information into an equation.
√ x + 2 x = 10
Let u = √ x and u^{2} = x and rewrite the equation using u
u + 2 u^{2} = 10 or 2 u^{2} + u  10 = 0
Solve the above equation for u.
u = 2 and u = 5 / 2
Find x.
√ x = u = 2 , gives x = 4.
√ x = u = 5 / 2 , has no solution since the square root of a real number cannot be negative.
Answer: number is 4

The graph of f(x) = x^{2} + a and the line y = x  2 are shown. The two graphs intersect at a point that is on the x  axis. Find a.
.
Solution
Set y = 0 to find the x intercept of the graph of y = x  2
0 = x  2 , x = 2
We now use the fact that xintercept found above lies on the parabola with equation f(x) = x^{2} + a
0 = (2)^{2} + a
Solve the above for a to find
Answer: a = 4

For what values of x is the function $$f(x) = \dfrac{x+2}{\sqrt{x24}} $$ not a real number?
Solution
The given function is not real if the denominator is zero or the quantity under the square root is negative. Hence the condition to find the values of x for which f is not real is
x  2  4 ≤ 0
Solve the inequality
x  2 ≤ 4
4 ≤ x  2 ≤ 4
2 ≤ x ≤ 6
f is not a real number for any value of x in the interval
Answer: [2 , 6]

What is the value of 0.25 x + 0.15 y if 5 x + 3 y = 2?
Solution
Note that
0.25 / 5 = 0.15 / 3 = 0.05
Hence multiply all terms of the given equation 5 x + 3 y = 2 by 0.05.
0.05(5 x + 3 y) = 0.05(2)
Simplify
Answer: 0.25 x + 0.15 y = 0.1

If the complex number (8  16i) / (2  2i) is written in the form a + ib , where i = √(1), then what is the value of b?
Solution
Multiply and divide (8  16i)/(2  2i) by the conjugate of the denominator: 2 + 2i
(8  16i)/(2  2i) = (8  16i)(2 + 2i)/ ( (2  2i)(2 + 2i) )
Simplify
(16 + 16i  32i  32i^{2}) / (4  4i^{2})
Use i^{2} =  1 and further simplify
= (48  16i) / 8 = 6  2i
Answer: b =  2

Given the system of equations 0.2 (x + y)^{2} = 4 and 0.5 (x  y)^{2} = 3, find the value of the product xy.
Solution
One way to answer this question is to solve the system, but that takes time. A faster method is to rewrite the equations as follows
(x + y)^{2} = 4 / 0.2 = 20
and
(x  y)^{2} = 3 / 0.5 = 6
Subtract the second from the first equations obtained above and simplify
(x + y)^{2}  (x  y)^{2} = 20  6
Expand
x^{2} + y^{2} + 2xy  (x^{2} + y^{2}  2xy) = 14
Simplify
4 xy = 14
Answer: x y = 7/2

There are 200 liters of water in a tank which started leaking at the rate of 0.25 liters per minute for about one hour. Then the rate at which water is leaking from the tank increases to 0.4 liter per minute. What is the quantity q of water left in the tank t hours after the tank started leaking with t > 1?
Solution
The quantity of water leaked during the first hour at the rate of 0.25 l/mn is given by
60mn × 0.25 l/mn = 15 liters
The quantity q of water leaked for t > 1 is given by
15 + 60 (t  1) 0.4
15 liters leaked during the first hour and 60(t  1) is the number of minutes after the first hour mutliplied by the rate 0.4 liters/minute. The quantity q left in the tank after t hours is given by
Answer:
q = 200  15  60 (t  1) 0.4

Zoe has to write an essay of 30 pages. On average, she writes 11 pages every 2 hours and 35 minutes. Hour many hours would it take Zoe to finish the essay?Round answer to the nearest hour.
Solution
Ley us first find Zoe's writing rate in pages/minute.
2 hours and 35 minutes = 120 + 35 = 155 minutes
r = 11/155 pages per minute
The time T to write 30 pages is given by
T = 30 / r = 30 × 155 / 11 = 422.72 minutes ≈ 7 hours.
Answer: ≈ 7 hours

A new car was bought at $50,000. The price of the car decreased at a rate of $4000 a year for the first two years and it has been decreasing continously at a constant rate of $6,000 since. Write a formula for the price P in $ as a function of time t in years with t = 0 corresponding 2 years after the car was bought.
Solution
2 years after the car was bought, the price was:
50,000  2×4000 = $42,000
For t > 0, after the first two years, the car started decreasing at a rate of $6000 per year. Hence the formula for the price P is given by
Answer: P(t) = 42,000  6000 t
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