# SHSAT PRACTICE TEST QUESTIONS

Question 1 :

Find the next term in the series :

0, 1, 1, 2, 4, 7, 13, 24,.......

(A)  28  (B)  37  (C)  44  (D)  48  (E)  81

Solution :

By adding the preceding three terms, we get the next term.

4th term  =  0 + 1 + 1  ==>  2

5th term  =  1 + 1 + 2  =  4

6th term  =  1 + 2 + 4  =  7

In this way, we will get the

9th term  =  7 + 13 + 24  ==>  44

Hence the required term is 44.

Question 2 :

Which is true about A and B ?

A  =  √65 - 9

B  =  √50 - 8

(A)  A = B   (B)  A > B  (C)  A > 1  (D)  B > A  (E)  A + B > 0

Solution :

√65  =  8.07

√65 - 9  = 8.06 - 9  =  -0.9377

√50  =  7.07

√50 - 8  =  7.07 - 8     =   -0.9289

Hence B > A is the true statement.

Question 3 :

Which is the solution of the following equation ?

(x2 + x - 6)/(x - 2)  =  0

(A)  -2  (B)  -3  (C)  0  (D)  2  (E)  3

Solution :

By factoring the numerator, we get

(x - 2)(x + 3)/(x - 2)  =  0

x + 3  =  0

x  =  -3

Hence the solution is -3.

Question 4 :

John is stacking boxes directly upon each other. He stacks as 1 box per minute initially and every minute after that, his staking rate increases by 1 box per minute. If each box is 5 inches tall, in how many minutes will the stack be 30 inches tall.

(A)  2 min  (B)  3 min  (C)  5 min  (D)  6 min  (E)  30 min

Solution :

If we place the second box over the first box, the height will become 10 inches.

In the same way, by placing the third over the first two boxes the height of the stack will become 15 inches.

Total height of the stack  =  30 inches

Number of minutes taken  =  30/5

=  6 minutes

So we need 6 minutes to build a stack of height 30 inches.

Question 5 :

A Dilob has a mass of 20.2 milligrams. What is the Dilob's mass in grams ?

(A)  0.0202 g  (B)  0.202 g  (C)  2.02 g

(D)  20.2 g  (E)  202 g

Solution :

1000 milligrams  =  1 gram

1 milligram  =  1/1000 gram

20.2 milligram  =  20.2/1000

=  0.0202 g

Hence Dilob's mass in grams is 0.0202 g.

Question 6 :

Jesse goes to store. He buys a magazine for \$8. The he sells it for \$10 and buys it back again for \$11. He finally sells it for \$12. What was his profit ?

(A)  -\$1    (B)  \$0  (C)  \$2  (D)  \$3  (E)  \$5

Solution :

Cost price of magazine  =  8 + 11

=  \$19

Selling price  =  10 + 12

=  \$22

Profit  =  Selling price - Cost price

=  22 - 19

=  \$3

Hence his profit is \$3.

Question 7 :

After a 10% increase, a population was 55. What was the population before the increase ?

(A)  54.45  (B)  55.45  (C)  50  (D)  56  (E)  10

Solution :

Let x be the original population

After 10% increase, percentage of new population will be 110% of x.

110% of x  =  55

1.10 x  =  55

x  =  55/1.10

x  =  50

So, before increase the population was 50.

Question 8 :

A substance's length doubles every hour. At 2 PM it was 3 meters. What was the length at 12 PM that same day?

(A)  0.375 meters  (B)  0.5 meters  (C)  0.75 meters

(D)  1 meter  (E)  1.5 meters

Solution :

Given :

At 2 pm  =  3 meters

At 1 pm  =  Half of the previous length

1.5 meter  =  3/2

=  (1/2) x 3

At 12 pm  =  Half of the previous length

=  (1/2) x (1/2) x 3

=  (1/2)x 3

=  3/4

3/4  =  0.75 meters

Question 9 :

Which is true of these three functions ? (A)  f = g = h  (B)  f = g  (C)  h = g

(D)  f  =  g + h  (E)  f ≠  g

Solution :

f  =  x x^4

When we have power raised to another power, we can multiply powers.

g  =  x4x

h  =  x 4^x

Hence we can say ≠ g.

Question 10 :

Linda scored  66, 82, 81 and 92 on her English exams. What score must Linda obtain on the next math test, to have an average of exactly 84 ?

(A)  84  (B)  87  (C)  95  (D)  99  (E)  100

Solution :

Linda's marks are 66, 82, 81 and 92.

Average of 5 marks  =  84

Let "x" be the unknown score in English

(66 + 82 + 81 + 92 + x)/5  =  84

321 + x  =  84 (5)

321 + x  =  420

x  =  420 - 321

x  =  99

Hence Linda has to score 99 marks in the next test to get the the average 84.

Question 11 :

If today is Saturday, what day of the week will it be in 365 days from now?

(A)  Monday  (B)  Tuesday  (C)  Thursday

(D)  Friday   (E)  Sunday

Solution :

The problem starts with Saturday, every 7 days from Saturday, it will be Saturday.

By dividing 365 by 7, we get 1 as remainder. So the answer will be 1 day after Saturday.

Question 12 :

What is the remainder of 7, 700, 000, 000, 202 divided by 9 ?

(A)  0  (B)  1  (C)  2  (D)  3  (E)  8

Solution :

If the sum of digits is divisible by 9, then the given number is divisible by 9.

7 + 7 + 2 + 2  =  18

Since the sum of the digits is divisible 9, the given number is divisible by 9.

Hence the remainder is 0.

Question 13 : In the figure above AB  =  BC  =  DE  =  EF  and the side of the square is 12. Point B and E are the midpoints of the square. Find the area of the shaded region.

(A)  12π (B)  24π  (C)  36π  (D)  72π  (E)  144π

Solution :

AC  =  12, AE  =  6

There are four semicircles.

Area of 4 semicircles with radius 6  =  πr2/2

= 4(πr2/2)  =  2 πr2

=  2π(6)2

=  2π(36)

=  72π

Question 14 :

8x + 3  =  643x, what is the value of x ?

(A)  1/2  (B)  3/5  (C)  5/3  (D)  2  (E)  3

Solution :

x + 3  =  64 3x

In order to compare the two terms, we have to convert with them having same base.

x + 3  =  (823x

x + 3  =  8 6x

6x  =  x + 3

6x - x  =  3

5x  =  3

x  =  3/5

Question 15 :

If A ⊏ ⊐ B  =  AB (A/B) - 23, then what is the value of 3 ⊏ ⊐ (6 ⊏ ⊐ 2)

(A)  1   (B)  3   (C)  5   (D)  2   (E)  7

Solution :

First let us calculate the value of (6 ⊏ ⊐ 2), here instead of "A" we have "6" and instead of "B" we have 2.

(6 ⊏ ⊐ 2)  =  6(2) (6/2) - 23

=  12(3) - 8

=  36 - 8

(6 ⊏ ⊐ 2)  =  28

3 ⊏ ⊐ 28  =  3(28) (3/28) - 23

=  84(3/28) - 8

=  3(3) - 8

=  9 - 8

=  1

Question 16 :

If the average of 11 numbers is 15, then what is the sum of these 11 numbers ?

(A)  324   (B)  350   (C)  155   (D)  165   (E)  174

Solution :

Average of 11 numbers  =  15

Sum of 11 numbers/11  =  15

Sum of 11 numbers  =  15(11)  =  165.

Hence the required answer is 165.

Question 17 :

What is the average of

(9b - 7) + (7 - 3b) - (-3 -3b) + (6 + 3b)

(A)   3b + 9/4   (B)   9b/4 + 3   (C)  3b + 2

(D)  2b + 3  (E)  3b + (4/9)

Solution :

=   (9b - 7) + (7 - 3b) - (-3 -3b) + (6 + 3b)

=  9b - 7 + 7 - 3b + 3 + 3b + 6 + 3b

=  9b + 3b + 9

=  12 b + 9

Average of four terms  =  (12b + 9)/4

=  3b + 9/4

Question 18 :

(y-1⋅y3⋅y4⋅y5⋅y6) / (y6⋅y5⋅y4)

(A)   y2  (B)   y1  (C)  1  (D)   y5   (E)   y-3

Solution :

=  (y-1⋅y3⋅y4⋅y5⋅y6) / (y6⋅y5⋅y4)

(y-1⋅y18) / (y15)

=  y-1 ⋅y3

=  y (-1 + 3)

=  y2

Question 19 :

If the pattern continues, what will be the 50th symbol ? Solution :

Each group consists of 6 different symbols. A group ends with the symbol upside, then the new starts with triangle.

Let us divide 50 by 6, we get the remainder 2. The symbol triangle must be filled in the 49th place and 50th place should be heart.

So, option B is correct.

Question 20 :

A trapezoid has base length in the ratio of 2 : 6. If the area of a trapezoid is 260 and altitude is 5, then what is the length of longer side ?

(A)  13  (B)  20  (C)  26  (D)  65  (E)  78

Solution :

Area of trapezoid  =  (1/2) h (a + b)

a  =  2x, b = 6x

(1/2) 5 (2x + 6x)  =  260

(5/2)(8x)  =  260

x  =  260/20

x  =  13

2x  =  2(13)  =  26 and 6x  =  6(13)  =  78

Hence the longer side is 78. More Practice Test Papers

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SHSAT math practice test - Paper 11

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