SHSAT MATH TEST ONLINE

Question 6 :

A square of side 4 is topped by a semicircle, as shown.  Point P is the center of the arc at the top.  Then the shaded area is equal to

(A)  π + 2  (B)  (π/2) + 2  (C)  2π + 2 

(D)  2π – 4  (E)  2π + 4

Solution :

Shaded region  =  Area of quadrant + Area of square semi square - Area of triangle Radius of circle  =  2, side of square  = 4 =  (1/4) πr2 + (1/2)a2  - (1/2) base (height) =  (1/4) 4π + (1/2)42  - (1/2) 2 (4 + 2) =  π + 8 - 6 =  π + 2 Hence the answer is π + 2.

Question 7 :

(4⁶ x 3⁴ x 2²) / (4² x 3⁴ x 2⁶) =

         (A)  1  (B)  2  (C)  8  (D)  16  (E)  24

Solution :

  =   (4⁶ x 3⁴ x 2²) / (4² x 3⁴ x 2⁶)

  =  4^6-2 x 3^4-4 x 2^2-6

  =  4⁴ x 3⁰ x 2^-4

  =  2⁸ x 3⁰ x 2^-4

  =  2^8-4

  =  2⁴  =  16

Hence the answer is 16.

Question 8 :

Which one of the following numbers is less than 1/3?

(A)  1111/3333  (B)  299/895  (C)  .1 x 4 

(D)  .4 x 1  (E)  5! / 6!

Solution :

5! / 6!  =  5! / 6 ⋅ 5!  =  1/6

  =  0.1666

Hence the answer is 5! / 6!.

Question 9 :

The ratio of a to b is 3 to 4.  The ratio of b to c is 5 to 6.  The ratio of a to c is

(A)  9:10  (B)  5:8  (C)  1:2  (D)  2:5  (E)  none of these

Solution :

a : b  =  3 : 4

Multiply the ratio by 5

a : b  =  15 : 20

b : c  =  5 : 6

Multiply the ratio by 4

b : c  =  20 : 24

a : b : c  ==>  15 : 20 : 24

The ratio of a to c is 15 : 24

Hence the required ratio is 5  : 8.

Question 10 :

(All lines shown are either vertical or horizontal)  If x and y are integers, and the shaded area is 52, then the perimeter of the figure is

         (A)  20  (B)  24  (C)  52  (D)  56  (E)  196

Solution :

If we complete the square whose sides are length x, we see that the shaded are is x² - y^2. 

(x + y) (x - y)  =  52

x + y  =  2 ----(1)      x - y  =  26 ----(2)

(1) + (2)

2x  =  28

x  =  14 and y = 12

Perimeter of the shape = 2(x + y)

  =  2(14 + 12)

  =  2(26)

  =  52  

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