SHSAT MATH PRACTICE

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

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

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