Question 11 :
In a diagram, each small box is a square whose side is 3. What is the area of the shaded figure ?
(A) 108 (B) 54 (C) 36 (D) 18 (E) 6
Solution :
Base of the triangle = 3 units => 3 x 3 ==> 9
height = 4 units ==> 4 x 3 ==> 12
Area of triangle = (1/2) ⋅ base ⋅ height
= (1/2) ⋅ 9 ⋅ 12
= 54 square units
Question 12 :
When expressed in scientific notation, the number
1,230,000,000 is 1.23 x 10B.
The value of B is
(A) 2 (B) 7 (C) 8 (D) 9 (E) 10
Solution :
Scientific notation of the given number
1,230,000,000 = 1.23 x 109
Since we move the decimal 9 digits to the left side, we use positive sign.
Hence the value of B is 9.
Question 13 :
In a diagram, lines m and n are parallel. What is the value of x ?
(A) 20 (B) 40 (C) 45 (D) 60 (E) 70
Solution :
Here l and m are parallel lines.
In a triangle, sum of angles = 180
<ABC + <BAC + <BCA = 180
<ABC + 90 + 70 = 180
<ABC + 160 = 180
<ABC = 180 - 160
<ABC = 20 = x (alternate interior angles)
Hence the value of x is 20.
Question 14 :
If p = √(2 + √10), Q = √(10 + √2) and R = √(5 + √5)
(A) P < Q < R (B) P < R < Q (C) Q < R < P
(D) R < Q < P (E) P = Q
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.
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.
(A) is 20 (B) is 51 (C) is 52 (D) is 54
(E) cannot be determined from the information given
Solution :
Any side of a triangle has to be greater than the difference of the other 2, and less than the sum of the other 2
c > 10 - 7
c > 3
c < 10 + 7
c < 17
The smallest integer value is 4, and the largest is 16.
S = 7 + 10 + 4 = 21
L = 7 + 10 + 16 = 33
L + S = 54
Question 16 :
Evaluate (1/2) + (1/3) - (1/4) =
(A) 1/20 (B) 7/12 (C) 3/4 (D) 1 (E) 13/12
Solution :
L.C.M (2, 3 and 4) = 12
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
(A) 20 (B) 24 (C) 28 (D) 31 (E) 34
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 ?
(A) -14 (B) -6 (C) 0 (D) 8 (E) 20
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 sum 210 + 110 + (1/2)10 + (1/3)10 + (1/4)10 is closest to
(A) 21 (B) 32 (C) 513 (D) 1025 (E) 2000
Solution :
210 = 1024
210 = 1
Other than the above two, the values of other terms will be too small.
Hence the sum is closest to 1025.
Question 20 :
43 + 43 + 43 + 43 =
(A) 44 (B) 49 (C) 412 (D) 163 (E) 1612
Solution :
43 + 43 + 43 + 43 = 4(43)
= 4(1 + 3)
= 44
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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