Answered Sample GENERAL MATHEMATICS SKILLS TEST
1.
𝟎. 𝟎𝟑𝟓 =?
(𝑨) 𝟕/𝟐𝟎
(𝑩) 𝟑𝟓/𝟏𝟎𝟎
(𝑪) 𝟕/𝟏𝟎𝟎𝟎
(𝑫) 𝟕/𝟐𝟎𝟎
Solution:
0.035 = 35/1000. Divide top and bottom by 5 = 7/200
2.
(Answer: Option D)
If 𝟒𝒙 < 𝟎 then?
(𝑩) − 𝒙 < 𝟎
(A) 𝟒𝒙 > 𝒙
(𝑪) 𝒙 < −𝒙
(D) None of these
Solution:
Divide both sides by 4 so x < 0
Take any valid number for x that should be less than zero, say x = -1 then apply it to all given options
(A) 𝟒𝒙 > 𝒙
4(−1) > −1
(B) −𝒙 < 𝟎
−1(−1) < 0
(C) 𝒙 < −𝒙
−1 < −(−1)
−4 > −1
1 < 0
−1 < 1
Which is incorrect
Which is incorrect
Which is correct
(Answer: Option C)
3.
If 𝟑 < 𝒙, then?
(𝑨) − 𝟑 < −𝒙
(𝑩) − 𝒙 < 𝟎
(𝑪) − 𝟑 = −𝒙
(D) None of these
Solution:
3 < 𝑥 𝑠𝑜 𝑥 > 3 which can be read as x is larger than 3, Try x = 4 and apply it to the given options
-3 < −4
Which is incorrect
−4 < 0
− 3 = −4
Which is Correct
Which is incorrect
(Answer Option B)
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4.
Tom takes 20 coins from a box, and he gets 12 red coins and 8 black coins. If the whole box
contains 10000 coins, about how many of them are black?
(A)8000
(B)4000
(C)400
(D)6000
Solution:
This is a question about ratios. The ratio of red to black is 12: 8 giving 12 + 8 = 20 part
There are 10,000 coins so each part is 10,000/ 20 = 500 coins
The number of black coins is therefore 500 x 8 = 4,000
(Answer :Option B)
5.
Roger is looking at a map where the scale is 2km =4.5cm. If he measures the distance from his
house to the center of town on the map as 12.6cm, how many kilometers is the distance from the
center?
(A) 2.8
(B) 5.6
(C) 6
(D) 7.6
Solution:
Another question about ratios.
4.5 cm represents 2 km.
1 cm. represents 2/4.5 km.
So, 12.6 cm represents 𝟏𝟐. 𝟔 ∗ (𝟐/𝟒. 𝟓) 𝑲𝒎. = 𝟓. 𝟔 𝑲𝒎.
6.
(Answer: Option B)
For which of the following inequalities is the point (-7,3) a solution?
(𝑰)𝒙 – 𝟑 ≤ 𝟒𝒚
(A) 𝑰 𝒐𝒏𝒍𝒚
(𝑰𝑰) − 𝟑𝒙 > 𝟏𝟓 + 𝟐𝒚
(𝑩)𝑰𝑰 𝒐𝒏𝒍𝒚
(𝑰𝑰𝑰) – 𝒙 – 𝟒 ≥ 𝟑𝒚
(𝑪)𝑰 𝒂𝒏𝒅 𝑰𝑰 𝒐𝒏𝒍𝒚
(𝑫) 𝑰𝑰 𝒂𝒏𝒅 𝑰𝑰𝑰 𝒐𝒏𝒍𝒚
Solution:
Since (𝑥 = −7, 𝑦 = 3), for each of the three options we substitute (-7) for x and (3) for y in the provided
equations.
𝒙 – 𝟑 ≤ 𝟒𝒚
−7 – 3 ≤ 4 ∗ 3
−𝟑𝒙 > 𝟏𝟓 + 𝟐𝒚
– 𝒙 – 𝟒 ≥ 𝟑𝒚
−3(−7) > 15 + 2(3)
−(−7) – 4 ≥ 3(3)
-10 ≤ 12
21 > 15 + 6
7– 4 ≥ 9
-10 ≤ 12
21 > 21
3 ≥ 9
Which is incorrect (21=21)
Which is incorrect
Which is correct
(Answer :Option A)
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7.
(𝒙 + 𝟐𝒚)𝟐 =?
(𝑨) 𝟐𝒙𝟐 + 𝒚𝟐
(𝑩)𝟒𝒙𝟐 + 𝒚𝟐
(𝒄)𝟒𝒙𝟐 + 𝟒 𝒙𝒚 + 𝒚𝟐
(𝑫)𝒙𝟐 + 𝟒𝒙𝒚 + 𝟒𝒚𝟐
Solution:
(𝑥 + 2𝑦)2 = (𝑥 + 2𝑦)(𝑥 + 2𝑦) = 𝑥(𝑥 + 2𝑦) + 2𝑦(𝑥 + 2𝑦)
= 𝑥 2 + 2𝑥𝑦 + 2𝑦𝑥 + 4𝑦 2
= 𝑥 2 + 4𝑥𝑦 + 4𝑦 2
8.
(Answer: Option D)
If 𝒚 = 𝒙𝟐 − 𝟑𝒙 + 𝟖, what is the value of y when 𝒙 = 𝟐
(A)6
(B)8
(C)10
(D)12
Solution:
Here we substitute the value of x with 2 to find the value of y.
𝑦 = 𝑥 2 – 3𝑥 + 8 = (2)2 – 3 ∗ (2) + 8 = 4 – 6 + 8 = 6
(Answer: Option A)
9. What is the equation of the line that passes through the points (5,1) and (3,5)?
(𝑩)𝟐𝒙 + 𝒚 = 𝟏𝟏
(A) 𝟐𝒙 + 𝒚 = 𝟐𝟐
(𝑪)𝒙 + 𝟐𝒚 = 𝟐𝟐
(𝑫)𝒙 + 𝟐𝒚 = 𝟏𝟏
Solution:
To find the equation of a line we need to find the slope (Gradient)
𝑆𝑙𝑜𝑝𝑒 𝑚 =
(𝑦2 – 𝑦1)
(𝑥2 – 𝑥1)
=
(5 −1)
(3 −5)
=
4
−2
= −2
The equation of the line is 𝑦 – 𝑦1 = 𝑚 (𝑥 – 𝑥1) (We can use either of the two points. If they give different
answers, then our slope is wrong.)
𝑆𝑜, 𝑦 – 1 = −2(𝑥 – 5)
𝑦 − 1 = −2𝑥 + 10 𝑜𝑟 2𝑥 + 𝑦 = 11
(Answer: Option B)
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10.
If John can run 9 km in y minutes, how many km can he run in 10 minutes?
(𝑨) 𝟗𝒚⁄𝟏𝟎
(𝑪) 𝟏⁄𝟗𝟎𝒚
(𝑩)𝟗𝟎𝒚
(𝑫) 𝟗𝟎⁄𝒚
Solution:
In y minutes John can run 9 km.
In 1 minute, John can run 9/y km.
So, in 10 minutes John can run ( 𝟗 ∗ 𝟏𝟎/𝒚) km= 𝟗𝟎/𝒚 km.
11.
The simplified form of
(A)
𝑥
1
1+𝑦
𝒚
𝟏
𝟏+𝒙
=?
(B)
Solution:
To simplify this equation, we can multiply by
𝑦
1
1+𝑥
𝑥
𝑥𝑦
𝑥
𝑥
𝑥+𝑥
∗ =
=
(Answer Option D)
𝑥
𝑥
𝑦
𝑥+𝑦
(C)
𝑥𝑦
𝑥+1
(D)
𝑥+𝑦
𝑥𝑦
which is similar of multiplying by 1
𝑥𝑦
(Answer: Option C)
𝑥+𝟏
12.. At which points does the graph of 𝒚 = 𝒙𝟐 – 𝟔𝒙 + 𝟖 cross the x-axis?
(𝑨)(𝟔, 𝟎) & (𝟐, 𝟎)
(𝑩)(−𝟒, 𝟎) & (−𝟐, 𝟎)
(𝑪)(𝟒, 𝟎) & (𝟐, 𝟎)
(𝑫) (−𝟔, 𝟎) & (−𝟐, 𝟎)
Solution:
A graph will always cross the x-axis when y = 0 so we need to solve the equation for y=0:
𝑥 2 – 6𝑥 + 8 = 0
factorize the left side
(𝑥 – 4) (𝑥 − 2) = 0
If either of these two brackets 0 then the whole left side becomes 0. This happens when 𝑥 = 2 𝑜𝑟 𝑥 = 4 Our
solution is therefore (2,0) 𝑎𝑛𝑑 (4,0) as y = 0
(Answer: Option C)
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𝟓
13. At which point does the graph of 𝒚 = (𝒙 – 𝟑) + 𝟔𝟎 cross the y axis?
(𝑨)(𝟎, 𝟔𝟎)
(𝑩)(𝟎, 𝟑𝟎𝟑)
(𝑪)(𝟎, −𝟏𝟖𝟑)
(𝑫) (𝟎, 𝟓𝟕)
Solution:
This is the same as the previous question, only this time 𝑥 = 0
𝑦 = (0 – 3) 5 + 60 = (−3)5 + 60 = −243 + 60 = − 183
(Answer: Option C)
14. Find the solutions of the following quadratic equation 𝒙𝟐 – 𝒙 – 𝟏𝟎 = 𝟐
(𝑨)𝟒 𝒂𝒏𝒅 − 𝟑
(𝑩) − 𝟒 𝒂𝒏𝒅 𝟑
(𝑪)𝟏 𝒂𝒏𝒅 − 𝟏𝟐
(𝑫) − 𝟏 𝒂𝒏𝒅 𝟏𝟐
Solution:
Rearranging the equation so we have zero on one side:
𝑥 2 – 𝑥 − 10 − 2 = 0
𝑥 2 – 𝑥 − 12 = 0
Now we need to factorize
(𝑥 − 4) (𝑥 + 3) = 0
If either bracket is 0 then the whole left side becomes 0. This happens when 𝑥 = 4 𝑜𝑟 𝑥 = −3
(Answer: Option A)
15. Find the solutions of the following equation −𝟑𝒙𝟐 + 𝟏𝟐𝒙 = 𝟎
(𝑨)𝟒
(𝑩)𝟎
(𝑪)𝟎 𝒂𝒏𝒅 𝟒
(𝑫) 𝒏𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
Solution:
We can simplify this equation by dividing by -3 giving us 𝑥 2 – 4𝑥 = 0. Now we need to factorize so x (x – 4) =0
and as before this can happen when x = 0 and x = 4
(Answer: Option C)
16.Find the solutions of the following quadratic equation 𝒙𝟐 = 𝟑𝟔
(𝑨)𝟔
(𝑩) − 𝟔
(𝑪)𝟑𝟔
(𝑫) 𝟔 𝒂𝒏𝒅 − 𝟔
Solution:
This is known as the difference of two squares
x2 – 36 = 0
Now we factorize as before (𝑥 + 6) (𝑥 – 6) = 0 𝑠𝑜 𝑥 = +6 𝑜𝑟 𝑥 = −6
(Answer: Option D)
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𝒙𝟑 = −𝟖
17. Find the solutions of the following equation
(𝑨)𝟖
(𝑩) − 𝟖
(𝑪)𝟐
(𝑫) − 𝟐
Solution:
Here we are looking for a number multiplied by itself and multiplied by itself again which will give us -8. In this
case it is -2 because −2 ∗ −2 ∗ −2 = −8 , or simply take the cube root of both sides.
(Answer: Option D)
18. Find the solution of the following equation. (𝒙 – 𝟒) (𝒙 + 𝟑) (𝒙 + 𝟏) = 𝟎
(𝑨)𝟏, 𝟑, −𝟒
(𝑩) − 𝟏, −𝟑, 𝟒
(𝑪) − 𝟏, 𝟑, −𝟒
(𝑫)𝟏, −𝟑, 𝟒
Solution:
This is the same as question 14 after it has been factorized. if any of the three brackets is 0 then the whole left
side becomes 0. This happens when x = 4 or -3 or -1
(Answer: Option B)
19. Find the solution of the following equation
(𝑨)𝟑, 𝟏𝟎
𝟑(𝒙 + 𝟏𝟎)𝟐 = 𝟎
(𝑩)𝟑, −𝟏𝟎
(𝑪)𝟏𝟎
(𝑫) − 𝟏𝟎
Solution:
Divide both sides by 3 which gives us
Take the square root of both sides
20. Let 𝒂 =
𝟓
𝟔
and 𝒃
=
−𝟑
𝟓
(𝑨) 𝟐
(𝑥 + 10)2 = 0
𝑥 + 10 = 0
𝑥 = −10
Then compute
(𝑩) 𝟏/𝟐
(Answer: Option D)
ab + 1
(𝑪) − 𝟏/𝟐
(𝑫) 𝟐
Solution:
By substituting the values for a and b in the given equation:
𝑎𝑏 + 1 =
𝟓 −𝟑
−𝟏𝟓
∗
+𝟏=
+𝟏=
𝟔 𝟓
𝟑𝟎
−𝟏
𝟏
+𝟏=𝟐 #
𝟐
(Answer: Option B)
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21. Let 𝑎 =
1
4
and 𝒃 =
(A)12
−𝟏
𝟑
Then compute
1
𝑎+𝑏
(B) 1/12
(C)-12
(D)-1/12
Solution:
𝟏
𝟏
=
= ; multiply each fraction by the denominator of the other
𝟏
𝟏
𝒂+𝒃
−
𝟒
𝟑
𝟏
𝟏
𝟏𝟐
=
=
= −𝟏𝟐
𝟑
𝟒
−𝟏 −𝟏
𝟏𝟐 − 𝟏𝟐 𝟏𝟐
(Answer Option C)
22. The sum of angles in a triangle is 𝟏𝟖𝟎𝒐 degrees. If the first angle is 𝟖𝟎𝒐 and the second angle is
three times the third angle, find the second angle.
(𝑨)𝟏𝟎𝟎𝒐
(𝑩)𝟕𝟓𝒐
(𝑪)𝟓𝟎𝒐
(𝑫)𝟐𝟓𝒐
Solution:
If the first angle is 𝟖𝟎𝒐 then the other two angles add up to 100𝑜
These two angles are in the ratio of 1:3 (4 parts)
1 part =
100𝑜
4
= 25𝑜 and so the larger one is 100𝑜 − 25𝑜 = 75𝑜
(Answer: Option B)
23. A new Company is formed by Paul and Jane. The total investment by them is AED 60,000. It is
known that Paul’s investment is twice as much as Jane’s. What is Jane’s investment in AED?
(A) 20,000
(B) 30,000
(C)40000
Solution:
Here is another ratio question. Investments are in the ratio 1:2 giving us 3 parts
Jane’s investment is 60,000/3 = AED 20,000
(D) 15,000
(Answer: Option A)
24. The edge of square A is three times that of square B. Then the area of square A is how many times the
area of square B?
(A) three times
(B) nine times
(C) one-third
(D) one-ninth
Solution:
This is a slightly different version of ratios. If side B is 1 unit long, then side A is 3 units long.
The area of A is 9 square units and area of B is one square unit, so square A is nine times the size of square B
(Answer: Option B)
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25. In the system of equations 𝟑𝒙 + 𝟓𝒚 = 𝟓 and
equations is:
(A)0
−𝟔𝒙 – 𝟓𝒚 = −𝟏𝟕, the value of 𝒙 which satisfies both
(B)1
(C)2
(D)4
Solution:
We solve this problem using simultaneous equations:
−6𝑥 – 5𝑦 = −17
3𝑥 + 5𝑦 = 5
Adding the two equations together
−3𝑥 = −17 + 5
−3𝑥 = −12
𝑥 = 4
(Answer: Option D)
26. Determine the domain of the function 𝟏⁄
√(𝒙 + 𝟏)
(𝑨)(∞, −𝟏)𝑼(𝟏 , ∞)
(𝑩)[−𝟏, ∞)
(𝑪)(∞, −𝟏]𝑼[𝟏, ∞)
(𝑫) (−𝟏, ∞)
Solution:
The value of x is specified by two conditions:
1- The denominator should not equal to 0
2- The square root of (x+1) needs to be positive
Therefore, x must be greater than -1 Which is (−𝟏, , ∞) ≡ 𝒙 > −𝟏
(Answer: Option D)
𝟏
27. 𝑰𝒇 𝒇(𝒙) = 𝒙 + 𝟒 𝒂𝒏𝒅 𝒈(𝒙) = 𝒙𝟐 − 𝟒 , 𝒕𝒉𝒆𝒏 𝒘𝒉𝒂𝒕 𝒊𝒔 𝒇(𝒈( −𝟐))?
(𝑨)𝟏
(𝑩)𝟒
(𝑪) 𝟏/𝟒
(𝑫) 𝟏/𝟖
Solution:
To find 𝑓(𝑔(𝑥)) we substitute the value of 𝑔(𝑥) into 𝑓(𝑥)
𝑓(𝑔(𝑥)) = 2
1
1
𝒙 +4–4
= 2
𝒙
Then to find 𝑓(𝑔(−2)) we now substitute -2 for x
1
1
𝑓(𝑔(−2)) = 22 = 4
(Answer: Option C)
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28. 𝑰𝒇 𝒇(𝒙) = 𝒙𝟐 + 𝒌𝒙 – 𝟓, 𝒂𝒏𝒅 𝒇 (𝟑) = 𝟏𝟎, 𝒕𝒉𝒆𝒏 𝒌 = ?
(𝑨) − 𝟐
(𝑩) − 𝟒
(𝑪)𝟐
(𝑫)𝟒
Solution:
𝐼𝑓 𝑓(𝑥) = 𝑥 2 + 𝑘𝑥 − 5
then 𝑓(3) = 10 = 32 + 3𝑘 – 5
10 = 9 + 3𝑘 – 5
Therefore 3𝑘 + 4 = 10
3𝑘 = 6
𝑘 = 2
(Answer: Option C)
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