The NSAA is an important part of the Natural Science and Veterinary Medicine admission β so understanding the Admission Test is essential.
Section 1 of the NSAA is split into four parts, with candidates required to answer Part A and one from B, C and D.
We will go over everything you need to know for Part A, which is all about Mathematics.
What Is Section 1A of the NSAA?
Section 1A of the Natural Science Admissions Assessment is all about Mathematics.
Being confident with Maths is extremely important for the NSAA. Many students find that improving their numerical and algebraic skills usually results in big improvements in their scores.
Even good students who are studying Maths at A2 can struggle with certain NSAA topics because they are typically glossed over at school.
Being confident with Maths is extremely important for the NSAA. Many students find that improving their numerical and algebraic skills usually results in big improvements in their scores.
What Maths Knowledge Do You Need For NSAA Section 1A?
It is important to know what knowledge you will be assumed to know to ensure you are using your revision time effectively.
You will not be given a formula sheet for the NSAA so you must know:
| 2D Shapes | 3D Shapes | |||
|---|---|---|---|---|
| Area | Surface Area | Volume | ||
| Circle | ππ2 | Cuboid | Sum of all 6 faces | Length x width x height |
| Parallelogram | π΅ππ π Γ ππππ‘ππππ hπππht | Cylinder | 2 ππ2 + 2πππ | ππ2 π₯ π |
| Trapezium | 0.5 Γ h Γ (π + π) | Cone | πππ2 + πππ | ππ2 π₯ (h/3) |
| Triangle | 0.5 Γ πππ π Γ hπππhπ‘ | Sphere | 4 ππ2 | (4/3) ππ3 |
In addition to this is the topics that you will be required to know:
Quadratic Formula
The solutions for a quadratic equation in the form ππ₯Β² + ππ₯ + π = 0 are given by: π₯ = βπΒ±βπ2β4ππ/2π
Remember that you can also use the discriminant to quickly see if a quadratic equation has any solutions:
πΌπ π2 β 4ππ < 0: ππ π πππ’π‘ππππ
πΌπ π2 β 4ππ = 0: πππ π πππ’π‘πππ
πΌπ π2 β 4ππ > 2: ππ€π π πππ’π‘ππππ
Completing the Square
If a quadratic equation cannot be factorised easily and is in the format ππ₯Β² + ππ₯ + π = 0 then you can rearrange it into the form π(π₯+ π /2π )Β²+[πβπΒ² /4π ]=0
This looks more complicated than it is β remember that in the NSAA you are extremely unlikely to get quadratic equations where a > 1 and the equation does not have any easy factor.
Difference between 2 Squares
If you are asked to simplify expressions and find that there are no common factors but it involved square numbers β you might be able to factorise by using the βdifference between two squaresβ.
For example, x2-25 can also be expressed as (π₯+5)(π₯-5).
Once you have mastered the above it would be worth also revising the following areas which cover a much broader knowledge of the A-Level Maths curriculum.
- Algebra
- Graphing Functions
- Law of Logarithms
- Differentiation
- Integration
- Geometry
- Series
- Trigonometry
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10 NSAA Section 1A Questions and Worked Solutions
Question 1
What is the circumference of a circle with an area of 10Ο?
- 2Οβ10
- Οβ10
- 10Ο
- 20Ο
- β10
- More information needed.
Answer: 1
π΄ = ππΒ², therefore 10π = ππ
Thus, = β10
Therefore, the circumference is 2Οβ10
Question 2
If a.b = (ab) + (a + b) the calculate the value of (3.4).5
- 19
- 54
- 100
- 119
- 132
Answer: 4
3.4= 12 +(3+4)=19
19.5= 95 +(19+5)=119
Question 3
Which of the following graphs do not intersect?
y=x
y=x2
y=1-x2
y=2
- 1 and2
- 2 and 3
- 3 and 4
- 1 and 3
- 1 and 4
- 2 and 4
Answer: 3
Whilst you definitely need to solve this graphically, it is necessary to complete the square for the first equation to allow you to draw it more easily:
(π₯ + 2)Β² = π₯Β²+ 4π₯ + 4
Thus,π¦=(π₯+2)Β² +10= π₯Β² +4π₯+14
This is now an easy curve to draw (y = xΒ² Β that has moved 2 units left and 10 units up). The turning point of this quadratic is to the left and well above anything in xΒ³, so the only solution is the first intersection of the two curves in the upper right quadrant around (3.4, 39).
Question 4
Calculate the product of 897,653 and 0.009764.
- 87646.8
- 8764.68
- 876.468
- 87.6468
- 8.76468
- 0.876468
Answer: 2
Notice that youβre not required to get the actual values β just the numberβs magnitude. Thus, 897.653 can
be approximated to 900,000 and 0.009764 to 0.01. Therefore, 900,000 x 0.01 = 9,000, which represents the same magnitude of 8764.68.
Question 5
Solve the equation π₯2 β 10π₯ β 100 = 0
- β5 Β±5β5
- β5Β±β5
- 5 Β±5β5
- 5Β±β5
- 5 Β±5β125
- β5Β±β125
Answer: 3
The easiest way to do this is to βcomplete the squareβ:
(π₯β5)Β² =π₯Β²β10π₯+25
Thus,(π₯β 5)Β²β 125 = π₯Β²β10π₯β100=0
Therefore, (π₯ β 5)Β² = 125
π₯β5 = Β±β125= Β±β25β5=Β±5β5
π₯ = 5 Β± 5β5
Question 6
The aspect ratio of my television screen is 4:3 and the diagonal is 50 inches. What is the area of my television screen?
- 1,200 inchesΒ²
- B. 1,000 inchesΒ²
- C. 120 inchesΒ²
- D. 100 inchesΒ²
- E. More information needed.
Answer: 1
Let the width of the television be 4x and the height of the television be 3x. Then by Pythagoras: (4π₯)Β² + (3π₯)Β² = 50Β²
Simplify: 25π₯Β² =Β 2500
Thus: π₯ = 10. Therefore: the screen is 30 inches by 40 inches, i.e. the area is 1,200 inchesΒ² .
Question 7
The two inequalities π₯ + π¦ β€ 3 πππ π₯3 β π¦2 < 3 define a region on a plane. Which of the following points is inside the region?
- (2,1)
- (2.5, 1)
- (2.5, 1)(1,2)
- (3, 5)
- (1,2.5)
- None of the above.
Answer: 3
This is fairly straightforward; the first inequality is the easier one to work with: B and D and E violate it, so we just need to check A and C in the second inequality.
C:1Β³ -2Β² <3, but A:2Β³ -1Β² >3
Question 8
How many times do π¦ = π₯3 πππ π¦ = π₯ intersect?
- 0
- 1
- 2
- 3
- 4
Answer: 4
Itβs better to do this algebraically as the equations are easy to work with and you would need to sketch very accurately to get the answer. Intersections occur where the curves have the same coordinates. Thus: π₯Β³ = π₯
π₯Β³β π₯ = 0
Thus: π₯(π₯Β² β 1) = 0
Spot the βdifference between two squaresβ: π₯(π₯ + 1)(π₯ β 1) = 0
Thus there are 3 intersections: at π₯= 0,1πππβ1
Question 9
Fully factorise: 3a3 β30a2 + 75a
- 3a(aβ3)Β³
- a(3a β 5)Β²
- 3a(aΒ²Β β10a+25)
- 3a(a β 5)Β²
- 3a(a+5)Β²
Answer: 4
This is one of the easier maths questions. Take 3a as a factor to give: 3a(aΒ²Β β10a+25) =3a(aβ5)(aβ5) =3a(aβ5)Β²
Question 10
How many solutions are there for: 2(2(π₯Β² β 3π₯)) = β9 ?
- 0
- 1
- 2
- 3
- Infinite solutions.
Answer: 2
Expand the brackets to give: 4π₯Β² β 12π₯ + 9 = 0.
Factorise: (2π₯ β 3)(2π₯ β 3) = 0.
Thus, only one solution exists, x = 1.5.
Note that you could also use the fact that the discriminant, πΒ² β 4ππ = 0 to get the answer.
Conclusion
Hopefully, this has given you a better understanding of what is in store with Section 1A of the NSAA.
What is important to remember is that all the knowledge you already know, it is about being able to apply it in new ways.Β
The questions provided should hopefully give you the opportunity to the test, as long as you do not peek at the answer too early as then you are only hindering yourself.Β
Best of luck for the NSAA!
Maximise your NSAA score through effective NSAA preparation.
The NSAA is a vital component of your Natural Science and Veterinary Science application so scoring highly can mean the difference between an offer or rejection. At UniAdmissions, we are experts at boosting your NSAA score and maximising your chances of gaining a place.
Discover ourΒ NSAA ProgrammeΒ by clicking the button below toΒ enrol and triple your chances of success.
