**SHSAT Math Sample Test Questions and Answers :**

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

**Question 11 :**

If N = √(36 + 49), then N is

**Solution :**

N = √(36 + 49)

N = √85

The square root of 81 is 9.

But root 85 is greater than 81, the number between 9 and 10.

**Question 12 :**

Susan is 5 years older than Phen is now. In N years, Susan will be twice as old as Phen is now. If Susan is now 22 years old, what is the value of N?

**Solution :**

Let "x" be Phen's age

x + 5 be Susan's age

x + 5 + N = 2x ---(1)

Now Susan's age = 22 years

x + 5 = 22

x = 22 - 5 = 17

By applying the value of x in (1)

17 + 5 + N = 2(17)

22 + N = 34

N = 34 - 22

N = 12

**Question 13 :**

The number of integer values of n for which 1 ≤ √n ≤ 3 is

**Solution :**

** 1 ≤ √n ≤ 3**

**√1 = 1**

**√2 = 1.41....**

**√3 = 1.73**

**√4 = 2**

**√5 = 2....**

**√9 = 3**

**Hence 9 is number of integer values of n.**

**Question 14 :**

In right triangle ABC, angle ACB is 90°. The number of degrees in angle BEC is

**Solution :**

In triangle ACB :

<ACB + <CBA + <BAC = 180

90 + <CBA + 20 = 180

<CBA = 180 - 110 = 70

Now consider the triangle CEB,

<CEB + <CBE + <BCE = 180

<CEB + 70 + 70 = 180

<CEB = 180 - 140

<CEB = 40

**Question 15 :**

If it is now 12:00 noon, what time was it 40 hours ago?

**Solution :**

40 + 12 = 52

By dividing 52 by 12, we get the quotient 4 and remainder as 4.

Now, we have to move the clock 4 hours back word from 12 : 00 noon. So we get 8 : 00 AM.

**Question 16 :**

The mean of all the odd integers between 6 and 24 is

**Solution :**

First let us list out the odd integers between 6 and 24.

7, 9, 11, 13, 15, 17, 19, 21, 23

So, there are 9 odd numbers lies between 6 and 24.

Mean = (7 + 9 + 11 + 13 + 15 + 17 + 21 + 23)/9

= 135/9

= 15

**Question 17 :**

Let x be an element of the set {.2, 1.2, 2.2, 3.2, 4.2}. For how many values of x is 10x/3 an integer?

**Solution :**

Let f(x) = 10x/3

x = 0.2 = 10(0.2)/3 = 2/3 Not integer |
x = 1.2 = 10(1.2)/3 = 12/3 = 4 = integer |
x = 2.2 = 10(2.2)/3 = 22/3 = Not integer |

x = 3.2 = 10(3.2)/3 = 32/3 = Not integer |
x = 4.2 = 10(4.2)/3 = 42/3 = integer |

Hence for 2 values of x, we get integer.

**Question 18 :**

George has just enough money to buy 3 chocolate bars and 2 ice cream cones. For the same amount money, he could buy exactly 9 chocolate bars. For the same amount of money, how many ice cream cones could George buy?

**Solution :**

Let "x" and "y" be the cost of one chocolate bar and one ice cream cone.

3x + 2y = f(x) ---(1)

Cost of 9 chocolate bars = 9x = f(x) ----(2)

(1) = (2)

3x + 2y = 9x

2y = 9x - 3x

2y = 6x

y = 3x ==> x = y/3

By applying y = 3x in (1), we get

3(y/3) + 2y = f(x)

y + 2y = f(x)

f(x) = 3y

Hence, we can buy 3 ice cream cone for the same amount.

**Question 19 :**

ABCD and PQRS are squares, as shown. The area of PQRS is

**Solution :**

To find the side length pf RS, we have to use Pythagorean theorem.

In triangle DSR,

SR^{2} = SD^{2} + DR^{2}

SR^{2} = 1^{2} + 2^{2 }= 5

SR = √5

Area of the square PQRS = a^{2}

= (√5)^{2 } = 5

**Question 20 :**

If x = 10 and y = 8, what is the value of y (3x – 2y)?

**Solution :**

= y(3x – 2y)

x = 10, y = 8

= 8(3(10) – 2(8))

= 8 (30 - 16)

= 8 (14)

= 112

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