# SHSAT MATH PRACTICE

SHSAT Math Practice :

Here we are going to see some practice questions questions for SHSAT exams.

## SHSAT Math Practice - Practice Questions with Solution

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 + 3+ 35 is

(A)  3 (B)  315  (C)  95  (D)  9125  (E)  45

Solution :

35 + 3+ 3 =  3 (35

=  3(5 + 1)

=  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

=  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. 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

After having gone through the stuff given above, we hope that the students would have understood, how to solve problems in SHSAT exams.

Apart from the stuff given in this sectionif you need any other stuff in math, please use our google custom search here.

You can also visit our following web pages on different stuff in math.

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