Free SAT Maths Questions with Solutions and Explanations  Sample 2
Another set of SAT maths questions, corresponding to those in sample 2, are presented along with detailed solutions and explanations. Answers at the bottom of the page.

In the number 12ab, a and b are digits. Find a and b such that the number 12ab is divisible by 2, 5 and 7.
Solution
For the number 12ab to be divisible by 2, b must be 0, 2, 4, 6 or 8. For the same number to be divisible by 5, b must be 0 or 5. For 12ab to be divisible by 2 and 5, b must be equal to 0. Hence
b = 0
We now write the number 12ab as follows
12ab = 1200 + 10 a + b
The division of 1200 by 7 gives a quotient of 171 and a remainder of 3 and b = 0. Hence
12ab = 7*(171) + 3 + 10 a + 0
For 12ab to be divisible by 7, 10a + 3 must be divisible by 7. Let us try all possible values of a.
a = 0 , 10 a + 3 = 0 + 3 = 3 , not divisible by 3
a = 1 , 10 a + 3 = 10 + 3 = 13 , not divisible by 3
a = 2 , 10 a + 3 = 20 + 3 = 23 , not divisible by 3
a = 3 , 10 a + 3 = 30 + 3 = 33 , not divisible by 3
a = 4 , 10 a + 3 = 40 + 3 = 43 , not divisible by 3
a = 5 , 10 a + 3 = 50 + 3 = 53 , not divisible by 3
a = 6 , 10 a + 3 = 60 + 3 = 63 . 63 is divisible by 7
Hence for the number 12ab to be divisible by 2, 5 and 7, a and b must the values
a = 6 and b = 0

What is the average (arithmetic mean) of all numbers multiples of 6 from 6 to 510 inclusive?
Solution
The average A of all multiples of 6 from 6 to 510 is given by
A = S / n
where S = 6 + 12 + ... + 510 and n is the number of multiples from 6 to 510.
S = (1/2) n (6 + 510)
A is given by
A = (1/2)(6 + 510) = 258

Which of the following numbers can be expressed as the product of 3 different integers greater than 1?
I) 24
II) 27
III) 48
A) I and III only
B) I and II only
C) I, II and III
D) I only
E) III only
Solution
We first write the three numbers using power of prime numbers as follows
24 = 2^{3} * 3
27 = 3^{3}
48 = 2^{4} * 3
We now write, if possible, each of the given numbers as the product of 3 different integers
24 = 2 * 4 * 3
27 = 3 * 9 (not possible with 3 different integers greater than 1)
48 = 2 * 8 * 3
The answer is A): I and III only can be written as the product of 3 different integers greater than 1

If 60% of a is equal to 120% of b, then
2a  4b = ?
Solution
"60% of a is equal to 120% of b" is mathematically translated as
60% a = 120% b
Simplify to
60 a = 120 b
divide both sides by 60
a = 2 b
2a = 4b
2a  4b = 0

(4^{1000})(8^{2000}) / (16^{2000}) =
Solution
Note the following
4 = 2^{2}
8 = 2^{3}
16 = 2^{4}
Substitute 4, 8 and 16 by the above powers
(4^{1000})(8^{2000}) / (16^{2000})
= ( (2^{2})^{1000})( (2^{3})^{2000}) / ( (2^{4})^{2000})
Use rules of exponents to simplify
= ( 2^{2000})( 2^{6000}) / ( 2^{8000})
= 2^{8000} / 2^{8000} = 1

If n is a positive integer divisible by 2, 5 and 7, which of the following is true?
I) n^{2} + 100 is divisible by 100
II) n + 28 is divisible by 14
III) n^{2} + 1 is divisible by 2
A) I and II only
B) I, II and III
C) None
D) I and III only
E) II and III only
Solution
Since n is divisible by 2 and 5, it is a multiple of 10 and n^{2} is a multiple of 100. Hence the statement "n^{2} + 100 is divisible by 100" is true.
Since n is divisible by 2 and 7, it is a multiple of 14. Hence the statement "n + 28 is divisible by 14" is true.
Since n is divisible by 2, n^{2} + 1 is odd and is not divisible by 2 and therefore the statement "n^{2} + 1 is divisible by 2" is false.
I and II only are true. Answer A).

A spherical block of metal weighs 12 pounds. What is the weight, in pounds, of another block of the same metal if its radius is 3 times the radius of the 12pound block?
Solution
The weight W of an object of mass M is given by.
W = M g , where g is constant.
The mass M of an object of volume V is given by.
M = d V , where d is the density
The weight w of a spherical block of density d and radius r is given by.
w = Volume d g = (4/3)Pi r^{3} d g
The weight W of a spherical block of density d and radius 3 r is given by.
W = Volume d g = (4/3)Pi (3r)^{3} d g = 27 w = 27 * 12 = 325 pounds

If the ages of students in a class vary between 18 and 22 inclusive, which of the following describes all possible ages x in this class?
A) x  18≤4
B) x + 20≤2
C) x  21≤1
D) x  22≤0
E) x  20≤2
Solution
The age x varies between 18 and 22 inclusive is mathematically translated as
18 ≤ x ≤ 22
Subtract 20 from all parts of the inequality
18  20 ≤ x  20 ≤ 22  20
Simplify
2 ≤ x  20 ≤ 2
Using absolute value inequalities, the above inequality may be written as follows
x  20 ≤ 2

If r% of y is A, what is y?
Solution
r% of y is A is mathematically translated as
r% × y = A
Write r% as a fraction and solve for y
y (r / 100) = A
y r = 100 A
y = 100 A / r

Five years ago Ben was 4 times as old as Julie. If Julie is 10 years old now, how old is Ben?
Solution
Let x be Ben's age now. Five years ago Ben was x  5 and Julie was 10  5 = 5 years old. Five years ago "Ben was 4 times as old as Julie"; hence
x  5 = 4 * 5
x = 25 , Ben's age now

If (x + 2) / 7 is an integer greater than 2, then the remainder, when x is divided by 7 is
Solution
Let (x + 2) / 7 = k where k is an integer than 2. Hence
x + 2 = 7 k
x = 7k  2
Let us write the above as follows
x = 7(k  1) + 7  2
x = 7(k  1) + 5
We now divide x by 7
x / 7 = 7(k  1) / 7 + 5 / 7
Since k is greater than 2, x = 7k  2 is greater than 12. We now divide x by 7
x / 7 = k  1 + 5/7
Since x is an integer greater than 12, the remainder of the division of x by 7 is 5.

For how many positive integer values of n will the value of the expression 2n^{2} + 1 be an integer greater than 2 and smaller than 400?
Solution
Solve the inequality
2 < 2n^{2} + 1 < 400
1 / 2 < n^{2} < 399 / 2
Since n is positive
√ (1 / 2) < n^{2} < √ (399 / 2)
Since n is an integer
1 ≤ n ≤ 14
The number of integers between 1 and 14 inclusive is 14.

If x + y = √22 and x  y = √10, then xy =
Solution
Square both sides of the given equations as follows
(x + y)^{2} = 22 and (x  y)^{2} = 10
Expand
x^{2} + y^{2} + 2xy = 22 and x^{2} + y^{2}  2xy = 10
Subtract the right hand terms and the left hand terms of the equations to obtain
(x^{2} + y^{2} + 2xy)  (x^{2} + y^{2}  2xy) = 22  10
Simplify
4 xy = 12
xy = 3

In the figure below, the 3 circles have equal radii and are tangent to each other and to the sides of the rectangle. The width of the rectangle is 20 feet long. What is the area of the shaded (red) region outside and enclosed by all three circles?
.
Solution
The above diagram is reproduced below where the centers of the circles are connected to make an isosceles triangle as shown below. The radius r of each circle is given by
4r = 20 and r = 5.
The area of the enclosed region (red) is equal to the area of the equilateral triangle of side 10 minus 3 times the area of one sector (in blue).
area of triangle = (1/2) sin(t) *10 *10 = 50 sin(60 degrees) = 25 √ 3
area of one sector = (1/2)*r^{2}*t = (1/2) 25 (Pi/3)
The area of the enclosed region (red) is equal
25 √ 3  3[ (1/2) 25 (Pi/3) ] = 25(2√3  Pi) / 2

If 2x + 1 = x + 3, what is the value of
(2/3)x + 1/3?
Solution
Solve the given equation for x
2x + 1 =  x + 3
3x = 2
x = 2/3
We now substitute x in (2/3)x + 1/3 by 2/3
(2/3)x + 1/3 = (2/3)*(2/3) + 1/3 = 4/9 + 3/9 = 7/9

If a < b < c < d and the average (arithmetic mean) of a, b, c and d is m, which of the following is true?
I) a + b < c + d
II) m < d and a < m
III) b < m < c
A) II only
B) I only
C) III only
D) I and II only
E) II and III only
Solution
Using the fact that a < b < c < d, we can deduce the following inequalities
a < c and b < d
Add the right hand side and left hand side of the inequalities to obatin a new inequality
a + b < c + d , statement in I is true
Using the same inequality given above, we can write that
4 a < a + b + c + d and 4d > a + b + c + d
Divide both sides of the two inequalities above, we obtain
a < (a + b + c + d) / 4 and d > (a + b + c + d) / 4
Note that the mean of a, b, c and d is m = (a + b + c + d) / 4. Hence
a < m and d > m , statement in II is true.
Chose a = 3, b = 10, c = 11 and d = 12 and calculate m
m = (3 + 10 + 11 + 12) = 9
m = 9 is not greater than b = 10. Hence statement III is not true.
Only statement I and II are always true. Answer is D)

If x, y and z are positive numbers such that 5x = y/4, y/4 = z/5 and y + z = x/k, what is the value of k?
Solution
Equation 5x = y/4 may be written as
y = 20 x
Equation y/4 = z/5 may be written as
z = (5/4) y
Substitute y by 20 x to get
Z = (5/4) 20x = 25 x
Substitute y by 20 x and z by 25 x in equation y + z = x / k , divide all terms by x and solve for k
20 x + 25 x = x / k
20 + 25 = 1 / k
k = 1 / 45

Two square tables have sides of 10 and 15 inches respectively. The area of the larger table is what percent more than the area of the smaller table?
Solution
Find the two areas
Large = 15^{2} = 215
small = 10^{2} = 100
Percentage
(215  100) / 100 = 125%

Find a + b  c + d if
(x  1/5)(x + 1)(x + 3/7) = a x^{3} + bx^{2} + c x + d
for all real values of x.
Solution
Since the two sides of the given equation are equal for all values of x, then the two sides are also equal for x =  1. Hence
(1  1/5)(1 + 1)(1 + 3/7) = a(1)^{3} + b(1)^{2} + c(1) + d
Simplify to obtain
 a + b  c + d = 0

In the figure below line l is parallel to line k and line g is parallel to line h. What is the value of y?
.
A) 10
B) 20
C) 5
D) 15
E) 35

Line l has equation y  2x = 2. What is the equation of line p which is the reflection of line l on the line y = x?
A) y = 0.5x  1
B) y = 2x  1
C) y = x + 2
D) y = 2x  2
E) y = x

Mike drove 30 miles, at a constant speed, for t hours and then drove y miles, at another constant speed, for 1 hour and 15 minutes. What was his average speed, in miles per hour, for the whole journey?
A) (30/t + y/1.25) / 2
B) (30 + y) / (t + 1.25)
C) (30 + y) / t
D) (30 + y) / 1.25
E) (30 + y) / 2(t + 1.25)

If M and N are negative integers and 3M + 4N = 10, which of the following could be the value of N?
A) 2
B) 4
C) 6
D) 7
E) 9

If x^{6} = 20, what is x^{8}?
A) 400^{3}√(20)
B) 10√(20)
C) 20√(20)
D) 20^{3}√(20)
E) ^{3}√(20)

n is an integer chosen at random from the set {2,5,6} and p another integer chosen at random from the set {6,9,10}. What is the probability that the two numbers n and p are even?
A) 2/9
B) 5/9
C) 4/9
D) 1/9
E) 7/9

Three collinear points A, B and C are such that point B is between points A and C. The distance from A to B is 8 more than 4 times the distance from B to C and the distance from A to C is 5 times the distance from B to C. What is the distance from A to C?
A) 4
B) 20
C) 16
D) 24
E) 12
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 B
 A
 C
 D
 A
 A
 E
 E
 B
 B
 C
 B
 E
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