8. We arrange numbers in rows as follows : Row 1 I Row 2.3 Row 3 4,5,6 Row 4 7,8,9, 10 If we continue then the last number in Row 20 will be : A. 20 B. 100 C. 171 D. 210 E. 231 9. Sam and Mpho had R150 berveen them. Sam gives R25 to Mpho. Now Mpho has twice as much moncy as Sam. How much money did Sam have originelly? A. R20 B. R80 C. R75 D. R150 E not possible to work out. 10. The sum of 5 -consecutive odd numbers is 245 . The number in the middle \( = \) ? A. 30 B. 49 C. 55 D. 57 E. 59 11. Thrce friends \( A, B, \& \) C went on a picnic, At tea time A took out 5 cup cakes, B took out 3 cup cakes. L said sorry she forgot to bring any so all the eight cupcaker wore cut into equal pieces so the three friends shared the pieces cqually. In appreciation C offered R4 From this R4, B crust get R : A. 0 B. 0,50 C. 1,50 D. 2 E. 3,50 12. In a certain sequence \( \mathrm{T}_{1}=3, \mathrm{~T}_{2}=7, \mathrm{~T}_{3}=11, \ldots, \mathrm{~T}_{51}=203 \) Now 203 is not a multiple of 11 . Which number just before 203, in this sequence. is divisible by 11 ? A. 143 B. 176 C. 187 D. 231 E. none of these 13. To deal with women affairs the council needs to form a sub-committee of 2 female and I male member to be chosen from 4 mele and 4 female members. How many different subcommitiess can be formed? A. 16 B. 18 C. 20 D. 22 E. 24 14. Did you know the notation n! (read as n factorial) means the product of all numbers from 1 to n. Like \( 4!=1 \times 2 \times 3 \times 4 \). Let \( P=501=1 \times 2 \times 3 \times 4 \times \ldots \times 50 \). How many zeroes, in front, will P have? A. 14 B. 13 C. 11 D. 10 E. 9 15. In a quadrilateral \( \mathrm{ABCD}, \mathrm{AB}=\mathrm{AC} \) and \( \mathrm{AD}=\mathrm{CD} \). Angle \( \mathrm{ABC}=75^{\circ} \) and angle \( \mathrm{ADC}=50^{\circ} \). angle \( \mathrm{BAD}= \) ? A. \( 30^{\circ} \) B. \( 85^{\circ} \) C. \( 95^{\circ} \) D. \( 125^{\circ} \) E. \( 140^{\circ} \)

Let's solve each problem step by step. ### Problem 8 We need to find the last number in Row 20. The pattern of the rows is as follows: - Row 1: 1 number (1) - Row 2: 2 numbers (2, 3) - Row 3: 3 numbers (4, 5, 6) - Row 4: 4 numbers (7, 8, 9, 10) The number of elements in each row corresponds to the row number. Therefore, the total number of elements up to Row \( n \) is given by the formula for the sum of the first \( n \) natural numbers: \[ S_n = \frac{n(n + 1)}{2} \] For Row 20, we calculate: \[ S_{20} = \frac{20(20 + 1)}{2} = \frac{20 \times 21}{2} = 210 \] Thus, the last number in Row 20 is **210**. ### Problem 9 Let \( S \) be the amount of money Sam originally had, and \( M \) be the amount Mpho originally had. We know: 1. \( S + M = 150 \) 2. After Sam gives R25 to Mpho, Mpho has twice as much as Sam: \[ M + 25 = 2(S - 25) \] Now we can solve these equations step by step. From the first equation, we can express \( M \): \[ M = 150 - S \] Substituting \( M \) into the second equation: \[ (150 - S) + 25 = 2(S - 25) \] Simplifying this: \[ 175 - S = 2S - 50 \] Combining like terms: \[ 175 + 50 = 2S + S \] \[ 225 = 3S \] \[ S = 75 \] Thus, Sam originally had **R75**. ### Problem 10 Let the five consecutive odd numbers be represented as: \[ x - 4, x - 2, x, x + 2, x + 4 \] The sum of these numbers is: \[ (x - 4) + (x - 2) + x + (x + 2) + (x + 4) = 5x \] We know this sum equals 245: \[ 5x = 245 \] \[ x = 49 \] Thus, the middle number is **49**. ### Problem 11 C has no cupcakes, so they share the 8 cupcakes equally. Each friend gets: \[ \frac{8}{3} \text{ cupcakes} \] B originally had 3 cupcakes, so the amount B contributed is: \[ \frac{8}{3} - 3 = \frac{8 - 9}{3} = -\frac{1}{3} \text{ (not possible)} \] Thus, B does not get any money from C. Therefore, B gets **R0**. ### Problem 12 The sequence is an arithmetic sequence where: \[ T_n = 3 + (n - 1) \cdot 4 = 4n - 1 \] To find the largest \( n \) such that \( T_n < 203 \): \[ 4n - 1 < 203 \implies 4n < 204 \implies n < 51 \] The last term before 203 is \( T_{50} \): \[ T_{50} = 4 \cdot 50 - 1 = 199 \] Now we check if 199 is divisible by 11: \[ 199 \div 11 \approx 18.09 \text{ (not divisible)} \] The next term is \( T_{49} \): \[ T_{49} = 4 \cdot 49 - 1 = 195 \] Checking divisibility: \[ 195 \div 11 \approx 17.73 \text{ (not divisible)} \] Continuing this way, we find \( T_{48} \): \[ T_{48} = 4 \cdot 48 - 1 = 191 \] Continuing until we find \( T_{44} \): \[ T_{44} = 4 \cdot 44 - 1 = 175 \] Finally, we find \( T_{43} \): \[ T_{43} = 4 \cdot 43 - 1 = 171 \] Checking divisibility: \[ 171 \div 11 = 15.545 \text{ (not divisible)} \] Continuing this way, we find \( T_{40} \): \[ T_{40} = 4 \cdot 40 - 1 = 159 \] Continuing until we find \( T_{39} \): \[ T_{39} = 4 \cdot 39 - 1 = 155 \] Continuing until we find \( T_{36} \): \[ T_{36} = 4 \cdot 36 - 1 = 143 \] Checking divisibility: \[ 143 \div 11 = 13 \text{ (divisible)} \] Thus, the number just before 203 that is divisible by 11 is **143**. ### Problem 13 To form a sub-committee of 2 females and 1 male from 4 males and 4 females, we calculate: \[ \text{Ways to choose 2 females} = \binom{4}{2} = 6 \] \[ \text{Ways to choose 1 male} = \binom{4}{1} = 4 \] Total combinations: \[ 6 \times 4 = 24 \] Thus, the number of different sub-committees that can be formed is **24**. ### Problem 14 To find the number of trailing zeros in \( P = 50! \), we use the formula: \[ \text{Number of trailing zeros} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor \] Calculating for \( n = 50 \): \[ \left\lfloor \frac{50}{5} \right\rfloor + \left\lfloor \frac{50}{25} \right\rfloor = 10 + 2 = 12 \] Thus, \( P \) has **12** trailing zeros. ### Problem 15 In quadrilateral \( ABCD \): - \( AB = AC \) (isosceles triangle) - \( AD = CD \) (isosceles triangle) - \( \angle ABC = 75^\circ \) - \( \angle ADC = 50^\