1. The problem asks to find the price of 5 soap bars if 3 bars cost 420. The price per soap bar = $\frac{420}{3} = 140$. Price of 5 soap bars = $140 \times 5 = 700$. 2. Given the general term $T_n = 5 (12 - n)$, find the second term. Substitute $n=2$: $T_2 = 5(12-2) = 5 \times 10 = 50$. 3. Factorize $x^{2} + 5x - 6$. Find two numbers whose product is $-6$ and sum is $5$. They are 6 and -1. So, $x^{2}+5x-6 = (x+6)(x-1)$. 4. Find the value of $15x - 8y$ if $x=\frac{3}{5}$ and $y=\frac{1}{4}$. Calculate: $15 \times \frac{3}{5} = 9$. Calculate: $8 \times \frac{1}{4} = 2$. Therefore, $15x - 8y = 9 - 2 = 7$. 5. Simplify $\frac{3a}{5a+4} - \frac{a+1}{5a+4}$. Since denominators are the same, subtract numerators: $\frac{3a - (a+1)}{5a+4} = \frac{3a - a -1}{5a+4} = \frac{2a -1}{5a +4}$. 6. Round 879.645 to: (i) Nearest first decimal place: Look at second decimal digit (4), which is less than 5, so keep first decimal 6. Answer: 879.6 (ii) Nearest whole number: Look at first decimal digit (6), which is 5 or more, so round up. Answer: 880 7. Given $a^3 \times a^{\square} / a^3 = a^{\square} = \frac{1}{a^4}$. From $a^3 \times a^{x} / a^3 = a^{x}$, thus $a^{x} = \frac{1}{a^4} = a^{-4}$. Therefore, suitable values are $x = -4$. 8. In triangle ABC, Given $\angle A + \angle B = 130^{\circ}$ and $\angle A + \angle C = 85^{\circ}$. Sum of angles of triangle: $\angle A + \angle B + \angle C = 180^{\circ}$. Subtract first from sum: $180 - 130 = \angle C = 50^{\circ}$. Subtract second from sum: $180 - 85 = \angle B = 95^{\circ}$. Now, from $\angle A + \angle B =130$, $\angle A = 130 - \angle B = 130 - 95 = 35^{\circ}$. 9. Evaluate binary addition: $1101_{2} + 110_{2}$. Convert to decimal: $1101_2 = 1\times8+1\times4+0\times2+1=13$, $110_2 = 1\times4+1\times2+0=6$. Sum = 13 + 6 = 19. Convert back to binary: 19 in binary is $10011_2$. 10. From the Venn diagram, $B'$ (complement of B) are elements outside B in universal set. Elements inside B: 1, 2, 3, 4, 5, 6, 7. Elements in universal set but not in B: 8, 9. Therefore, $B' = \{8, 9\}$. 11. From $PT - AB = AP$, make $P$ the subject. Add $AB$ to both sides: $PT = AP + AB$. Since $AP + AB = PB$ (because $P$ lies on segment $AB$), The formula is rearranged as $AP = PT - AB$ if solving for $AP$. But question asked to make $P$ the subject; assuming $P$ represents length $AP$ or point coordinate; given info is algebraic, so final rearrangement: $AP = PT - AB$. 12. Volume of water in cuboid tank: Area of base = 4500 cm². Height of water = 60 cm. Volume = base area × height = $4500 \times 60 = 270000$ cm³. Convert to litres (1000 cm³ = 1 litre): Volume = $\frac{270000}{1000} = 270$ litres.