**SHSAT Math Practice Test :**

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

SHSAT Math Practice Test

**Question 11 :**

In a diagram, each small box is a square whose side is 3. What is the area of the shaded figure ?

**Solution : **

Base of the triangle = 3

height = 4

Area of triangle = (1/2) ⋅ base ⋅ height

= (1/2) ⋅ 3 ⋅ 4

= 6 square units

**Question 12 :**

When expressed in scientific notation, the number 1, 230, 000, 000 is 1.23 x 10^{B}. The value of B is

**Solution :**

Since we move the decimal point 9 digits to the left side, the scientific notation of the given number is

1, 230, 000, 000 = 1.23 x 10^{9}

Hence the value of B is 9.

**Question 13 :**

In a diagram, lines m and n are parallel. What is the value of x ?

**Solution :**

Here l and m are parallel lines.

In a triangle, sum of angles = 180

right angle + 70 + third angle = 180

90 + 70 + third angle = 180

third angle = 180 - 160 = 20

Since l and m are parallel lines, the missing angle is 20 degree.

**Question 14 :**

If p = √(2 + √10), Q = √(10 + √2) and R = √(5 + √5)

**Solution :**

**Approximate value of P :**

The approximate value of √10 is 3.....

By adding 3.... and 2, we get 5......... By taking √5......, we get 2.......

**Approximate value of Q :**

The approximate value of √2 is 1.....

By adding 1.... and 10, we get 11......... By taking √11......, we get 3.......

**Approximate value of R :**

The approximate value of √5 is 2......

By adding 2.... and 5, we get 7......... By taking √7......, we get 2.......

P = √5......, Q = √11...... and R = √7......

Hence the answer is P < R < Q.

(A) P < Q < R (B) P < R < Q (C) Q < R < P

(D) R < Q < P (E) P = Q

**Question 15 :**

The sides of a triangle all have integer lengths. Two side have lengths 7 and 10. If the largest and smallest possible perimeters for the triangle are L and S, the value of L + S.

**Solution :**

The length of 2 sides of a triangle is 7 and 10. According to triangle inequality property, the sum of length of two sides will be less than to the third side.

So, length of third side is less than 17.

The longest third side could be 16. L = 7 + 10 + 16 = 33

The shortest the third side could be 4., because a side of 3 plus a side of 7 is not greater than a side of 10.

S = 7 + 10 + 4

S = 21

L + S = 33 + 21 = 54.

**Question 16 :**

(1/2) + (1/3) - (1/4) =

**Solution :**

L.C.M (2, 3 and 4) = 12

In order to convert the denominator of first fraction as 12, we have to multiply both numerator and denominator by 6.

(1/2)⋅(6/6) = 6/12 ----(1)

The second fraction by 4

(1/3)⋅(4/4) = 4/12 ----(2)

The third fraction by 3

(1/4)⋅(3/3) = 3/12 ----(3)

(1) + (2) - (3)

= (6/12) + (4/12) - (3/12)

= (6 + 4 - 3)/12

= 7/12

**Question 17 :**

In the diagram, AC = BD = 17 and BC = 3. The length of segment AD is

**Solution :**

To find the length of AD, we have to add the lengths of AB, BC and CD.

AC = 17 AB = AC - BC = 17 - 3 AB = 14 |
BD = 17 CD = BD - BC = 17 - 3 CD = 14 |

AD = AB + BC + CD

= 14 + 3 + 14

= 31

Hence the answer is 31.

**Question 18 :**

What is the value of |x - y| + |y - x| if x = -3 and y = -7 ?

**Solution :**

|x - y| = |-3 - (-7)| = |-3 + 7| = 4 |
|y - x| = |-7 - (-3)| = |-7 + 3| = 4 |

|x - y| + |y - x| = 4 + 4 = 8

Hence the answer is 8.

**Question 19 :**

The product of the first ten prime numbers must be divisible by

**Solution :**

First 10 prime numbers are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Since it is multiple of 22 (2⋅11), it is divisible by 22.

**Question 20 :**

4^{3} +4^{3} + 4^{3} + 4^{3 }=

**Solution :**

4^{3} +4^{3} + 4^{3} + 4^{3 }= 4(4^{3})

= 4^{(1 + 3)}

= 4^{4}

(A) 4^{4 }(B) 4^{9 }(C) 4^{12 }(D) 16^{3 }(E) 16^{12}

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