ACSI 2022 Y3 Core Mathematics Paper 1 Solution

1. [Maximum mark: 3] Express 2 24 11 2 3 2 2 x x x x +−+ − − as a single fraction in its simplest form. ( )( ) ( )( ) ( )( ) ( ) ( )( )( ) ( )( )( ) ( )( ) 2 24 2 2 22 11 2 3 2 2 11 23 2 1 1 11 3 1 2 1 1 2 2 3 2 1 3 1 1 2 1 3 1 1 2 1 3 x x x x x xx xx x x x x xx x x x x x x x xx +−+ − − +=− +− +− =− + − + − + − += + + − −= + + − ++ 2. [Maximum mark: 4] Given that 3 2 5 132,3 4 1 2 1 0 a a b b     −=           find the value of a and of .b 2 5 __(1) 3 4 13 __(2) (2) (1) 2 : 3, 1 ab ab ab −= −= −  = =− • Students should not try to combine the fractions without even factorising the denominators. • 2 1x + should be cancelled out immediately to keep the working simple. • Some students forgot about the 2 in the denominator when combining the two fractions. • As the question asked for a fraction in its simplest form, 1x− must be cancelled in the final step. • Some students ended up with 3 2 13.ab−= • Careless mistakes were made when solving. These could be rectified by substituting the answers into the original equations for checking.

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 2 3. [Maximum mark: 5] A right cone has base diameter 6 cm and height of 4 cm. (a) Find, in terms of , the total surface area of the cone. [3] (b) A solid hemisphere has the same surface area as the cone. Find the radius of the hemisphere in the form 2a cm, where a is a constant. [2] (a) Slant height 223 4 5= + = cm Total surface area of cone 2(3 ) (3)(5) 24  = + = (b) 23 ( ) 24r = 8 2 2r== cm 2a= 4. [Maximum mark: 5] (a) Solve 5 4127 81 .3 x  = [2] (b) Simplify 4 2log log 64 log 4.aa− [3] (a) 5 4 35 127 813 3 (3 ) 3 35 2 x x x x −  = = −= =− (b) 4 2log log 64 log 4 lg lg 64 4lg 4 lg 3lg 4 4lg 4 1 aa a a − =  − =− =− • The diameter was given in the question, not the radius. Students need to divide by 2 first. • As this is a solid hemisphere, the base area must be included in the calculations. • Many students used the wrong formulas for this question, sometimes even mixing up with volume. • Many misconceptions in the use of indices laws were observed such as: 3 1 23 (3 ) 3 .xx− = • Another common mistake was: 127 9.3 x  = • If students want to use shortcuts like 1log log a b b a= or 2 2log log ,a abb= they must be aware of potential pitfalls. For example, 4 13log 4 . 3log a a • Change of base law must be clearly shown in working. It cannot be taken for granted that 4log 4 log 1.a a=

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 3 5. [Maximum mark: 6] (a) Solve the simultaneous equations 16xy = and 3.yx= [3] (b) On the same axes, sketch the graphs of 16xy = and 3,yx= labelling the axes-intercept(s) and point(s) of intersection cl