Question 1 :
If 3x + 2y = 19 and 2x + 3y = 91, what is the value of x + y ?
(A) 8 (B) 9 (C) 21 (D) 22 (E) 72
Solution :
In order to solve these equations, we may use elimination method.
3x + 2y = 19 -------(1)
2x + 3y = 91 -------(2)
(1) ⋅ 3 ==> 9x + 6y = 57
(1) ⋅ 2 ==> 4x + 6y = 182
(-) (-) (-)
------------------
5x = -125
x = -125/5 = -25
By applying x = -25 in (1), we get
3(-25) + 2y = 19
-75 + 2y = 19
2y = 19 + 75
2y = 94
y = 94/2 = 47
x + y = -25 + 47 = 22
Hence the value of x + y is 22.
Question 2 :
Kenny buys candy bars at 9 for $1 and sells them at 3 for $1. How many candy bars must he sell in order to make a profit of exactly $10.
(A) 27 (B) 30 (C) 45 (D) 60 (E) 90
Solution :
Let x be the number of candies that he has to sell to make the profit of $10.
Cost price of 9 candy bars = $1
Cost price of 1 candy = 1/9
Cost price of x candy bars = x/9
Selling price of 3 candy bars = $1
Selling price of 1 candy = 1/3
Selling price of x candy bars = x/3
Profit = Selling price - Cost price
10 = (x/3) - (x/9)
10 = (3x - x)/9
90 = 2x
x = 45
Hence he has to sell 45 candies to make the profit of $10.
Question 3 :
Juan travels at the rate of 30 miles per hour for 4 hours. He then returns over the same route in 3 hours. What was his average rate for the return trip, in mile per hour ?
(A) 22 1/2 (B) 34 2/7 (C) 35 (D) 36 (E) 40
Solution :
Speed = 30 miles per hour
Time taken for travelling = 4 hours
Time = Distance / speed
4 = Distance/30
Distance = 30(4) = 120 miles
Now, we have to find the speed taken by him to cover the the same distance that is 120 miles in 3 hours.
3 = 120/Speed
Average speed = 120/3 = 40 miles per hour
Question 4 :
The value of 35 + 35 + 35 is
(A) 36 (B) 315 (C) 95 (D) 9125 (E) 45
Solution :
35 + 35 + 35 = 3 (35)
= 3(5 + 1)
= 36
Hence the answer is 36.
Question 5 :
Lindsay has P dollars and mark has $9 less than Lindsay. If mark receives an additional $11, how many dollars will mark now have, in terms of P?
(A) P - 20 (B) 20 - P (C) P + 2 (D) 2 - P (E) P + 11
Solution :
Number of dollars that Lindsay has = P
Number of dollars mark has = P - 9
Number of dollars received by mark additionally = $11
= P - 9 + 11
= P + 2
Hence the number of dollars received by mark is P + 2
Question 6 :
R = 2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 11 and S = 3 ⋅ 7 ⋅ 13 ⋅ 17, what is the greatest common factor of R and S ?
(A) 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 7 ⋅ 11 ⋅ 11 ⋅ 13 ⋅ 17
(B) 2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17
(C) 3 ⋅ 7 (D) 3 ⋅ 3 ⋅ 7 ⋅ 7 (D) 3 ⋅ 3 ⋅ 3 ⋅ 7 ⋅ 7
Solution :
Greatest common factor = 3 ⋅ 7
Question 7 :
The counting numbers are placed in order in the chart, as shown. Assuming the pattern continues, in which column will the 200 appear ?
(A) Q (B) S (C) T (D) U (E) V
Solution :
If we write 1 from the first column, we get multiples of 7 in vth column.
First let us find how many 7's are in 200.
![]() |
![]() |
7 (28) = 196
From the picture given above, we know that 200 will be in the position of Sth column. Since we started 1 from the column R, we have to move 200 two columns towards the right.
So, 200 will be in Uth column.
Question 8 :
The equation 2(3x + 6) = 3(2x + 4) is satisfied by
(A) No value of x (B) Only negative values of x
(C) Only x = 0
(D) only positive values of x (E) all values of x
Solution :
2(3x + 6) = 3(2x + 4)
6x + 12 = 6x + 12
All values of x is the answer.
Question 9 :
Wai ling averaged 84 on her first three exams and 82 on her next 2 exams. What grade must she obtain on her sixth test in order to average 85 for all six exams.
(A) 96 (B) 94 (C) 90 (D) 89 (E) 86
Solution :
By writing the marks scored in six exams
Let "x" be the required mark in sixth subject.
84, 84, 84, 82, 82, x
Average mark = 85
[3(84) + 2(82) + x]/6 = 85
252 + 164 + x = 85(6)
416 + x = 510
x = 510 - 416
x = 94
Hence Wai has to score 94 mark in 6th exam.
Question 10 :
How many prime numbers between 8 and 60 leave a remainder of 2 when divided by 6 ?
(A) 0 (B) 1 (C) 4 (D) 6 (E) 7
Solution :
Write the multiples of 6 lies between 8 and 60.
12, 18, 24, 30, 36, 42, 48, 54, 60
If a number is a multiple of 6, then it should be a even number. 2 is the only even prime number.
Hence we will not have a prime number which is divisible by 6 and leaves the remainder 2.
So, the answer is 0.
More Practice Test Papers
SHSAT math practice test - Paper 1
SHSAT math practice test - Paper 2
SHSAT math practice test - Paper 3
SHSAT math practice test - Paper 4
SHSAT math practice test - Paper 5
SHSAT math practice test - Paper 6
SHSAT math practice test - Paper 7
SHSAT math practice test - Paper 8
SHSAT math practice test - Paper 9
SHSAT math practice test - Paper 10
SHSAT math practice test - Paper 11
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
If you have any feedback about our math content, please mail us :
v4formath@gmail.com
We always appreciate your feedback.
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Trigonometry word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits