To find questions from 1 to 5, please visit the page "SHSAT Math Practice for 9th Graders"
To find questions from 6 to 10, please visit the page "SHSAT Math Test with Answers"
To find questions from 11 to 15, please visit the page "SHSAT Math Test with Solution"
Question 16 :
Maria, Anoki, and Boris are teenagers, and the sum of their ages now is 49. The sum of their ages 8 years from now will be F, and the sum of their ages 10 years ago was P. The value of F – P is
(A) 54 (B) 18 (C) 17 (D) 6 (E) 2
Solution :
Sum of ages of Maria, Anoki and Boris = 49
Sum of their ages 8 years from now = 49 + 8 + 8 + 8
F = 73
Sum of their ages 10 years ago = 49 - 10 - 10 - 10
= 49 - 30
P = 19
F - P = 73 - 19
= 54
Hence the value of F - P is 54.
Question 17 :
If (1/4) + (1/x) = 1, then x =
(A) 1/8 (B) 2/3 (C) 4/3 (D) 2 (E) 3
Solution :
(1/4) + (1/x) = 1
(x + 4) / 4x = 1
x + 4 = 4x
Subtract x on both sides
-3x + 4 = 0
Subtract 4 on both sides,
-3x = -4
x = 4/3
Hence the value of x is 4/3.
Question 18 :
When the integer N is divided by 7, the quotient is Q and the remainder is 5. When N + 24 is divided by 7, the remainder is ‘
(A) 4 (B) 3 (C) 2 (D) 1 (E) 0
Solution :
Let us choose a number which gives the remainder 5 by dividing it by 7.
Let N = 12, by dividing 12 by 7, we get 5 as remainder.
By adding 12 by 24, we get 6. Now let us divide 36 by 7, we get the remainder 1.
Hence 1 is the answer.
Question 19 :
Point B is on line segment AD, and point E is on line segment BC. The value of x is
(A) 20 (B) 40 (C) 50 (D) 60 (E) 70
Solution :
In triangle ABC,
<A + <ABC + <C = 180
50 + <ABC + 70 = 180
<ABC = 180 - 120 = 60
In triangle EBD,
<EBD + <BDE + <DEB = 180
120 + 20 + x = 180
x = 180 - 140
x = 40
Hence the required angle is 40.
Question 20 :
Three-fourths of a number is equal to L. What is three-halves of that original number, in terms of L?
(A) 4/3L (B) 3/2L (C) 2L (D) 3L (E) 4L
Solution :
Let the number be "N"
(3/4) of N = L --(1)
= (3/2) of N
Form (1), N = 4L/3
= (3/2) (4L/3)
= 2 L
Hence the answer is 2 L.
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