SAT Maths Questions with Solutions and Explanation  Sample 1
The detailed solutions and explanations of the 25 SAT Maths questions in sample 1, are presented.

What is the area, in square feet, of the triangle whose sides have lengths equal to 10, 6 and 8 feet?
Solution
If given the three sides of a triangle and asked to find the area of the triangle, we normally use Heron's formula. However the triangle given here is a special triangle. Note that the three sides of the given triangle are related as follows
√(6^{2} + 8^{2}) = 10
which means that the triangle is a right triangle and its hypotenuse is 10 and legs 6 and 8. The area A is given by
A = (1/2)(6)(8) = 24 feet^{2}

3^{102} + 9*3^{100} + 3^{103}/3 = ?
Solution
Rewrite 9*3^{100} as
9*3^{100} = 3^{2}*3^{100} = 3^{102}
Rewrite 3^{103}/3 as
3^{103}/3 = 3^{1} * 3^{103} = 3^{102}
Substitute the above into the given expression
3^{102} + 9*3^{100} + 3^{103}/3 = 3^{102} + 3^{102} + 3^{102} = 3*3^{102} = 3^{103}

Of the 80 students in class, 25 are studying German, 15 French and 13 Spanish. 3 are studying German and French; 4 are studying French and Spanish; 2 are studying German and Spanish; and none is studying all 3 languages at the same time. How many students are not studying any of the three languages?
Solution
We start by drawing a Venn including all the given information. 25 Students study German including 3 studying also French and 2 Spanish which means 20 study German only. 15 Studying French with 3 studying also German and 4 studying also Spanish which means 8 study French only. 13 Study Spanish including 4 studying French and 2 German meaning that 7 study Spanish only. The number of students that are not studying any language is given by
80  (20 + 3 + 8 + 4 + 2 + 7) = 36
.

In the figure below, AB is a diameter of the large circle. The centers C1 and C2 of the smaller circles are on AB. The two small circles are congruent and tangent to each other and to the larger circle. The circumference of circle C1 is 8 Pi. What is the area of the large circle?
.
Solution
Given the circumference of C1, its diameter is given by
8 Pi / Pi = 8
The radius R of the large circle equal to the diameter of C1 and is equal to 8. The area A of the large circle is given by
A = Pi * 8^{2} = 64 Pi

Round (202)^{2} to the nearest hundred.
Solution
Evaluate (202)^{2}
(202)^{2} = 40804
Rounded to the nearest hundreds
40800

If w workers, working at equal rates, can produce x toys in n days, how many days it takes c workers, working at same equal rates, to produce y toys?
Solution
The rate r as number of toys per worker per day is given by
r = x / (n * W)
If N is the number of days needed to produce y toys by c workers at the same rate, then
r = y / (N * c) = x / (n * W)
Solve the above for N
N = (y * n * W) / (c * x)

A number of the form 213ab, where a and b are digits, has a reminder less than 10 when divided by 100. The sum of all the digits in the above number is equal to 13. Find the digit b.
Solution
213ab may be written as
213ab = 21300 + 10a + b
When 213ab is divide by 100, the reminder is 10a + b. The only way for the remainder to be less that 10 is that a = 0. Hence the sum of the digits is
2 + 1 + 3 + 0 + b = 13
Solve for b
b = 7

Find a negative value of x that satisfies the equation
[(x+1)^{2}  (2x + 1)]^{1/2} + 2x  6 = 0
Solution
We expand and simplify the left hand side of the equation and then solve it
[x^{2} + 2x + 1  (2x + 1)]^{1/2} + 2x  6 = 0
[x^{2}]^{1/2} + 2x  6 = 0
x + 2x  6 = 0
3x = 6
x = 2
Solve to obtain two solutions and select the negative one which is 2

The equation 1/a + 1/a = 0 has
A) an infinite number of solutions
B) no solutions
C) 1 solution only
D) 2 solutions only
E) 3 solutions only
Solution
Let us consider the cases when a is negative, positive or zero.
case 1) a = 0 is not a solution because the division by zero is not allowed.
case 2) let a be positive. hence a = a and the equation is written as
1/a + 1/a = 0 or 2/a = 0 , this equation has no solution.
case 3) let a be negative. Hence a =  a and the equation is written as
1/a + 1/a = 0 , 0 = 0. All negative real numbers are solutions to the given equation and therefore it has An infinite number of solutions.

The inequality x^{2}  2x + 1 ≤ 0 has
A) no solutions
B) a set of solutions
C) 1 solution only
D) 2 solutions only
E) 3 solutions only
Solution
Rewrite the left hand side of the equation as a square
(x  1)^{2} ≤ 0
The above inequality has one solution only x = 1 which makes the left hand side zero.

In the figure below, AC is parallel to DE. AE, FG and CD intersect at the point B. FG is perpendicular to AC and DE. The length of DE is 5 inches, the length of BG is 8 inches and the length of AC is 6 inches. What is the area, in square inches, of triangle ABC?
.
Solution
Since AC and DE are parallel, triangles ABC and DBE are similar. Hence the lengths of DE, AC, BG and FB are related by
DE / AC = BG / FB
Solve for FB
FB = BG*AC / DE = 48 / 5
FB is the height and AC is the base of triangle ABC, hence the area of triangle ABC is given by
(1/2)(48/5)6 = 28.8 square inches.

Points A, B and C are defined by their coordinates in a standard rectangular system of axes. What positive value of b makes triangle ABC a right triangle with AC its hypotenuse?
.
Solution
Use Pythagora's theorem to write that
AB^{2} + BC^{2} = AC^{2}
Use coordinates of points to rewrite above as
(4  1)^{2} + (b  1)^{2} + (6  4)^{2} + (1  b)^{2} = (6  1)^{2} + (1  1)^{2}
Simplify above as
13 + 2 (b  1)^{2} = 25
Solve for b
b = 1  √6 and b = 1 + √6
Select positive solution b = 1 + √6

Line L passes through the points (2,0) and (0,a). Line LL passes through the points (4,0) and (6,2). What value of a makes the two lines parallel?
Solution
Find slopes first
Slope of L = (a  0) / (0  (2)) = a / 2
Slope of LL = (2  0) / (6  4) = 1
For L and LL to be parallel, their slopes must be equal
a / 2 = 1 , a = 2

a, b, c and d are numbers of different values such that
a + b = d
a*b*c = 0
Which of the 4 numbers MUST be equal to 0?
Solution
Since a, b and d are different and a + b = d, then neither a nor b is equal to zero. Now since a*b*c = 0, c must be equal to zero for the product to be zero. Hence c must be equal to 0.

10^{4}(5^{4}  2^{4}) / 21 =
Solution
Factor (5^{4}  2^{4}) as follows.
(5^{4}  2^{4}) = (5^{2}  2^{2})(5^{2} + 2^{2}) = (25  4)(25 + 4) = 21 * 29
Substitute in the main expression and simplify.
10^{4}(5^{4}  2^{4}) / 21 = 10^{4} * 21*29 / 21
= 290,000

For what value of k will the two equations 2x + 4 = 4(x  2) and x + k = 2x  1 have the same solution?
Solution
Solve the first equation 2x + 4 = 4(x  2) to obtain .
x = 6
Substitute x by 6 (same solution) in the second equation and solve for k.
 6 + k = 2(6)  1
solve for k.
k = 17

An object travels at fifteen feet per minute. How many feet does it travel in 24 minutes and 40 seconds?
Solution
Convert speed from feet/minute into feet/seconds.
15 feet / mn = 15 feet / 60 sec
In 24 minutes 40 seconds, there are
24*60 + 40 = 1480 seconds
We now multiply speed by time to find distance
1480*15/60 = 370 feet

4 / (√20  √12) =
Solution
Multiply numerator and denominator by (√20 + √12) and simplify.
4 / (√20  √12) = 4 (√20 + √12) / [ (√20  √12)(√20 + √12) ]
= 4 (√20 + √12) / [20  12]
= 4 (2 √5 + 2 √3) / 8
= √5 + √3

In the figure below, DE is parallel to CB and the ratio (length of AE / length of EB) is 4. If the area of triangle AED is 20 square inches, what is the area, in square inches, of triangle ABC?
.
Solution
Since DE is parallel to CB, the triangles AED and ABC are similar. Hence the ratio of any side or altitude of one triangle to the corresponding side or altitude of the second triangle is constant. Let us first find the ratio AB / AE.
AE / EB = 4 (given)
may be written
AE = 4 EB
AB / AE = (AE + EB) / (4 EB) = (4 EB + EB) / (4 EB) = 5 / 4
The ratio of the area A1 of triangles ABC and area A2 of triangle AED is given by
A1 / A2 = [ (1/2)(h1*CB) ] /[ (1/2)h2*DE] (h1 and h2 are the altitude of the triangles)
= (h1/h2)*(CB/DE) = (5/4)(5/4) = 1.5625
Area of triangle ABC = A1 = 1.5625 * area of triangle AED
= 31.25 square inches

If f(n) = n + √n, where n is a positive integer, which of the following would be a value of f(n)?
I) 5
II) 10
III) 12
Solution
We are given f(n), we need to solve the following equation equations for n and select only those that give n positive integer.
n + √n = 5 , n + √n = 10 , n + √n = 12
Let x = √n and rewrite the above equations in terms of x as follows
x^{2} + x  5 = 0 , x^{2} + x  10 = 0 , x^{2} + x  12 = 0
Since adding n and √n must give a positive integer such as 5, 10 or 12 and n is a positive integer, √n must be a positive integer and therefore x must be a positive integer. The only equation that has a solution that is a positive integer is the third one: x = 3. If x = 3 then n = 9. Check
n + √n = 9 + √9 = 12
12 is the only possible value for f(n)

If a and b are both even numbers, which of the following COULD be and odd integer?
A) (a + b)^{2}
B) a^{2} + b^{2}
C) (a + 1)^{2} + (b + 1)^{2}
D) (a + 1)*(b + 1)  1
E) (a + 1) / (b + 1)
Solution
Since a and b are even integers, they may be written in the forms: a = 2 n and b = 2 m where m and n are intgers. Let us substitute and expand the given expression
A) (a + b)^{2} = (2 n + 2 m)^{2} = 4(n + m)^{2} even
B) a^{2} + b^{2} = (2n)^{2} + (2m)^{2}
= 4(n^{2} + m^{2}) even
C) (a + 1)^{2} + (b + 1)^{2} = (2n + 1)^{2} + (2m + 1)^{2}
= 4 n^{2} + 4 n + 1 + 4 m^{2} + 4 m + 1
= 2(2 n^{2} + 2n + 2 m^{2} + 2 m + 1) even
D) (a + 1)*(b + 1)  1 = (2 n + 1)*(2 m + 1)  1
= 4 n m + 2n + 2 m + 1  1 = 2(2 n m + n + m) even
E) (a + 1) / (b + 1) : let a = 8 and b = 2
(a + 1) / (b + 1) = (8 + 1) / (2 + 1) = 9/3 = 3 odd
(a + 1) / (b + 1) is the only expression that could be and odd intger.

If x is a negative number, which of the following must be true?
I) x^{5} < x
II) x < √(x)
III) x  1/x < 0
Solution
I) Since x is negative x^{5} is also negative. x is positive and therefore x^{5} < x is true
II) Since x is negative √(x) is positive. Hence x < √(x) is true.
III) Since x is negative and 1 / x is positive, x  1/x < 0 is also true.

If n is a positive integer such that n! / (n  2)! = 342, find n.
Solution
Let expand n! as follows
n! = n*(n  1)*(n  2)!
n! / (n  2)! = n*(n  1)*(n  2)! = n(n  1) = 342
Solve the equation n(n  1) = 342 and select the positive integer solution which is n = 19

What is the sum of the reciprocals of the solutions to the equation
x^{2}  (3/5)x = 11/3
Solution
Let us write the given quadratic equation in standard form
x^{2}  (3/5)x + 11/3 = 0
If x1 and x2 are the solutions of the given equation, then the sum S of their reciprocal is given by
S = 1/x1 + 1/x2 = (x1 + x2) / (x1*x2)
In order to determine S we need x1 + x2 and x1*x2 which for a quadratic equation are given by
x1 + x2 = b/a = 3/5 , x1*x2 = c/a = 11/3
Hence
S = (3/5) / (11/3) = 9 / 55

A number is given as 987562153ab where a and b are digits. Which values of a and b, such that
a + b = 11 and a < b,
would result in 987562153ab being divisible by 4?
Solution
For an integer to be divisible by 4, the number formed by its last two digits must be divisible by 4. The number formed by the last two digits of 987562153ab is given by
10a + b
For 10a + b to be divisible by 4 it must of the form
10a + b = 4 k , where k is a positive integer
a and b are such that a + b = 11 or b = 11  a. Substitute b by 11  a in 10a + b = 4 k
10a + 11  a = 4 k
Solve the above for a
a = (4k  11) / 9 and b = 11  a
Substituting k by 3, 4, 5... and calculate a and b. Select the values of a and b such that a is less than b.
For k = 14, a = 5 and b = 11  a = 6 and so the values of a and b are: a = 5 and b = 6
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