SHSAT MATH TEST WITH SOLUTION

Question 1:

The sum of 7 unequal positive integers is 61.  What is the largest possible value that any of those integers can have?

(A)  55  (B)  40  (C)  39  (D)  13  (E)  9

Solution :

If the the largest number be 55, then the sum of other 6 numbers be 11. So, it is not possible.

If the the largest number be 40, then the sum of other 6 numbers be 21. 

The possible 6 numbers are 1, 2, 3, 4, 5, 6 which are unequal and gives the sum 21.

Question 2 :

In the set {1, 5, x, 10, 15}, the integer x is the median.  The mean of the five numbers is one less than x.  The value of x is

(A)  6  (B)  7  (C)  8  (D)  9  (E)  none of these.

Solution :

Mean of 5 numbers  =  (1 + 5 + x + 10 + 15)/5

x - 1  =  (31 + x)/5

5(x - 1)  =  31 + x

  5x - 5  =  31 + x

5x - x  =  31 + 5

4x  =  36

x  =  9

Hence the required value of x is 9.

Question 3 :

If 7/50 < 1/x < 8/51 where x is an integer, then x =

(A)  8  (B)  7  (C)  6  (D)  5  (E)  4

Solution :

7/50 < 1/x < 8/51 

By combining these two inequalities, we get

6  3/8 < x < 7  1/7

Since x must be an integer, the value of x is 7.

Question 4 :

Half of 2⁸ is

(A)  2⁷ (B)  2⁶ (C)  2⁴ (D)  1⁸  (E)  8

Solution :

  =  Half of 2⁸

  =  (1/2) ⋅ 2⁸

  =  2⁷

Hence the answer is 2⁷.

Question 5 :

If x is greater than 1/2 but less than 1, which of the following has the largest value?

(A)  2x – 1  (B)  x²  (C)  1/x  (D)  1 – x  (E)  x³ 

Solution :

1/2 < x < 1

Let us test option A :

Multiply 2 through out the inequality

1 < 2x < 2

Subtract 1,

0 < 2x - 1 < 1

The value of 2x - 1 lies between 0 and 1.

Option B :

1/2 < x < 1

Taking squares,

1/4 < x² < 1

Option C :

1/2 < x < 1

Taking reciprocals,

2 < 1/x < 1

The value of 1/x lies between 1 and 2.

Option D :

1/2 < x < 1

Subtract 1,

1 - 1/2 < 1 - x  < 1 - 1

1/2 < 1 - x < 0

The value of 1 - x lies between 0 and 1/2.

Option E :

1/2 < x < 1

1/8 < x³ < 1

Hence the largest value is between 1 and 2. So, option C is correct.

Question 6 :

For the function f, the graph of y = f(x) in the xy-plane has a slope of 3 and passes through the point (0, −8). Which equation defines f ?

A)  f(x) = 3x      B) f(x) = 3x - 8    C) f(x) = 3x + 5

D) f(x) = 3x + 11

Solution :

Slope = 3

Passes through the point (0, -8)

(y - y₁) = m(x - x₁)

(y - (-8)) = 3(x - 0)

y + 8 = 3x 

y = 3x - 8

So, option B is correct.

Question 7 :

The triangle shown has a perimeter of 22 units. If x = 9
units and y = 7 units, what is the value of z, in units?

A) 6      B) 7    C) 9    D) 16

Solution :

Perimeter = 22 units

x + y + z = 22

Applying the value of x and y, we get

9 + 7 + z = 22

16 + z = 22

z = 22 - 16

z = 6

So, the possible value of z is 6 units. Option A is correct.

Question 8 :

The function h is defined by h (x) = 3x − 7. What is the value of h(−2) ?

A) −13    B) −10    C) 10    D) 13

Solution :

h (x) = 3x − 7

Applying x = -2, we get

h(-2) = 3(-2) - 7

= -6 - 7

= -13

So, option A is correct.

Question 9 :

The y-intercept of the graph of 12x + 2y = 18 in the xy-plane is (0, y). What is the value of y ?

Solution :

12x + 2y = 18

Applying the point (0, y), we get

12(0) + 2y = 18

2y = 18

y = 18/2

y = 9

So, the value of y is 9.

Question 10 :

A model predicts that a certain animal weighed 241 pounds when it was born and that the animal gained 3 pounds per day in its first year of life. This model is defined by an equation in the form

f (x) = a + bx

where f(x) is the predicted weight, in pounds, of the animal x days after it was born, and a and b are constants. What is the value of a ?

Solution :

The animal weighed 241 pounds at birth. This means when x=0 (the day the animal was born), the weight f(x) should be equal to 241 pounds.

According to the function f(x) = a + bx, when x=0:

f(0) = a + b(0) = a.

Since the weight on the day the animal was born is 241 pounds, we have:

a = 241

Therefore, the value of a, which represents the initial weight of the animal, is 241.

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