kang chuen tat


{ City } penang
< Country > malaysia
* Profession * biochemical engineer
User No # 116977
Total Questions Posted # 219
Total Answers Posted # 220

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Answers / { kang chuen tat }

Question { 1151 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.29 : An engineering company that produces small biochemical tools applies Installment Sales Method in its accounting. Let A = Installment Sales, B = Cost of Installment Sales, C = Gross Profit, D = Gross Profit Ratio. In year 20X8, let A = $400, B = $250, C = $150. In year 20X9, let A = $450, B = $315, C = $135. If the value of D = 37.5 % in year 20X8 : (a) find the value of D for year 20X9; (b) calculate the realized gross profit = ED / 100, for year 20X8 when the cash collected from sales is E = $100.


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.29 : (a) By trial and error for year 20X8, D = 100 C / A = 100 ($150) / ($400) = 37.5 %. Applying the same formula for year 20X9 will produce D = 100 C / A = 100 ($135) / ($450) = 30 %. (b) Realized gross profit = ED / 100 = $100 (37.5 / 100) = $37.5. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1153 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.30 : A biochemical engineering sales contract adopts Cost Recovery Method in its company accounting. Gross profit is realised only when cash collections exceed the total cost of goods sold. Let Y = Original Cost Recovered, Z = Gross Profit Realised. Its property was sold at A = $500k with cost of B = $300k. Its engineering accountant receives cash of C = $240k in first year, D = $180k in second year and E in third year. (a) Find the values of Y and Z in : (i) first year; (ii) second year; (iii) third year. (b) What is the exact value of E if all cash has been fully received from sales?


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.30 : (a)(i) First year : Y = C = $240k, Z = 0 (since C < B). (ii) Second year : Y = B - C = $300k - $240k = $60k, Z = C + D - B = $240k + $180k - $300k = $120k [ since (C + D) > B ]. (iii) Third year : Y = 0, Z = E. (b) E = A - C - D = $500k - $240k - $180k = $80k. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1274 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.31 : In a biochemical engineering company, let the actual revenue generated = A, actual cost incurred = B. According to the standard budget, the revenue generated = C, cost incurred = D. (a) Find the profit according to : (i) actual situation; (ii) standard budget. (b) Calculate the Variance Due to Price and Efficiency in : (i) revenue generated; (ii) cost incurred; (iii) profit. (c) If E > F where > represents : more than, find the 4 pairs of values of E and F in favorable conditions.


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.31 : (a)(i) actual profit = A - B; (ii) budgeted profit = C - D; (b)(i) revenue variance = A - C or C - A; (ii) cost variance = B - D or D - B; (iii) profit variance = (A - B) - (C - D) = A + D - B - C or (C - D) - (A - B) = C + B - D - A. (c) Favorable conditions will generate more profits according to actual situation, then A > B, C > D, A > C, D > B. The paired values of E - F are : A - B, C - D, A - C, D - B. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1300 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.32 : In the calculation of the total cost of production of an engineering material, a linear equation Y = MX + C could be applied. Let T = total cost, F = fixed cost, V = variable cost, Q = quantity produced, W = variable cost per unit produced. Form a linear equation that relates : (a) T, F and V; (b) V, Q and W.


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.32 : (a) T = F + V; (b) V = QW + C where C = 0. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1218 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.33 : An engineering corporation that produces a type of healthy food has A number of shares of $B preferred stock and C number of shares of $H common stock outstanding. In a typical financial year, let the net income obtained = D. The amount retained by the corporation = E. (a) Calculate the amount of money distributed to the shareholders if : (i) D > E; (ii) D = E; (iii) D < E. (b) Let preferred dividend = F, common dividend = G, CH > G. Find the values of F and G, in term of A, B, D and / or E when : (i) (D - E) > AB; (ii) (D - E) < AB; (iii) (D - E) = AB.


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.33 : (a) Amount of money distributed : (i) D - E, if D > E. (ii) D - E = 0, if D = E. (iii) Undetermined, when D < E. (b)(i) F = AB, G = D - E - AB where [ (D - E) - AB ] > 0. (ii) F = D - E, G = 0 where [ (D - E) - AB ] < 0. (iii) F = AB, G = 0 where (D - E) = AB. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1229 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.34 : A biochemical engineering company has stock P = 20,000 shares of M % preferred stock of Q = $100 par and N number of shares of $20 par common stock. In year 1 : $10,000 is distributed as dividend. In year 2 : $25,000 is distributed as dividend. Let (Total Dividend) = (Dividend for Preferred Stock) + (Dividend for Common Stock). (a) In year 1, calculate the amount distributed as dividend for preferred stock, when no dividend was distributed for common stock. (b) In year 2, the amount distributed for common stock was $5000. Find the values of M and N, when the dividend per share for common stock is $0.05.


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.34 : (a) (Dividend for Preferred Stock) = (Total Dividend) - (Dividend for Common Stock) = $10,000 - 0 = $10,000. (b) N = (Dividend for Common Stock) / (Dividend Per Share for Common Stock) = ($5000) / ($0.05) = 100,000. (Dividend for Preferred Stock) = (Total Dividend) - (Dividend for Common Stock) = $25,000 - $5,000 = $20,000 = PQ (M / 100). Then M = (20,000)(100) / PQ = (2,000,000) / [ (20,000)(100) ] = 1. Finally M = 1, N = 100,000. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1264 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.35 : A voluntary accountant is performing Du Pont Analysis for a charitable engineering organization. References for 3-step Du Pont formula are : (ROE) = (Net Income $) / (Equity $), where ROE is Return On Equity. (ROS) = (Net Income $) / (Sales $), where ROS is Return On Sales. (Asset Turnover) = (Sales $) / (Assets $). (Leverage) = (Assets $) / (Equity $). From the balance sheet and income statement, the company has Assets of $3753.19 this year and $3216.86 last year. Sales for this year is $2292.34. Let ROE = 0.37 and ROS = 0.21. (a) Calculate the Asset Turnover of the organization. (b) Find the value of the organizational Leverage. (c) Calculate the Net Income $ of the organization.


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.35 : (a) (Assets $) = [ (Assets of this year) + (Assets of last year) ] / 2 = ($3753.19 + $3216.86) / 2 = $3485.025. Asset Turnover = (Sales $) / (Assets $) = ($2292.34) / ($3485.025) = 0.658. (b) From 3-step Du Pont formula, (Net Income $) = (ROE) (Equity $) = (ROS) (Sales $). Then (Equity $) = (ROS) (Sales $) / (ROE) = (0.21) ($2292.34) / (0.37) = $1301.0578. From answer (a), (Assets $) = $3485.025. (Leverage) = (Assets $) / (Equity $) = ($3485.025) / ($1301.0578) = 2.6786. (c) From 3-step Du Pont formula, (Net Income $) = (ROS) (Sales $) = (0.21) ($2292.34) = $481.3914. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1315 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.36 : Du Pont analysis is used to perform calculation on Return On Equity (ROE) for an engineering organization. Let Net Income = A; Earning Before Tax (EBT) = B; Earning Before Interest, Tax (EBIT) = C; Sales = D; Assets = E; Equity = F. In 5-step Du Pont formula, let Tax Burden = G = A / B; Interest Burden = H = B / C = 1.04; EBIT % = I = C / D = 0.27; Asset Turnover = J = D / E = 0.66; Leverage = K = E / F = 2.66; ROE = L = A / F. If (1 / G) = (4 / 3) : (a) find the value of L; (b) calculate the values of A, B, C, E and F when D = $1500; (c) verify the answer (b) is correct by using the answer (a).


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.36 : (a) (L = A / F) = (G = A / B) (H = B / C) (I = C / D) (J = D / E) (K = E / F) = (3 / 4) (1.04) (0.27) (0.66) (2.66) = 0.37. (b) When D = $1500 : C / D = 0.27, C = 0.27 D = 0.27 ($1500) = $405. B / C = 1.04, B = 1.04 C = 1.04 ($405) = $421.20. A / B = 3 / 4, A = 3 B / 4 = 3 ($421.20) / 4 = $315.90. D / E = 0.66, E = D / 0.66 = ($1500) / 0.66 = $2272.73. E / F = 2.66, F = E / 2.66 = ($2272.73) / 2.66 = $854.41. Finally : A = $315.90, B = $421.20, C = $405, E = $2272.73, F = $854.41. (c) From answer (a), L = 0.37. From answer (b), A = $315.90, F = $854.41. Verification : L = A / F = ($315.90 / $854.41) = 0.37. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1500 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.37 : A biochemical engineering professional applies 5-step Du Pont formula in the accounting. Let Net Income = A, Earning Before Tax (EBT) = B, Earning Before Interest, Tax (EBIT) = C, Net Sales = D, Total Assets = E, Shareholders Equity = F. Let Tax Burden = A / B = 0.75, Interest Burden = B / C = 1.05, EBIT Margin = C / D = 0.27, Asset Turnover = D / E = 0.66, Return on Equity = A / F = 0.37. (a) Calculate the value of Leverage = E / F. (b) If AB + BC + CD + DE + EF = $$ 141932, find the values of A, B, C, D, E and F.


Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.37 : (a) Leverage = E / F = (A / F) / [ (A / B) (B / C) (C / D) (D / E) ] = 0.37 / [ (0.75) (1.05) (0.27) (0.66) ] = 2.6366. (b) Let 6 simulteneous equations with 6 unknowns could be formed where A / B = 0.75, B / C = 1.05, C / D = 0.27, D / E = 0.66, AB + BC + CD + DE + EF = 141932 = G. Simple computer programming could be used to perform trial-and-error calculation. For example, DATA : What-if Analysis : Goal Seek in Excel 2013 is able to set the value of G to 141932 by changing the value of A, B, C, D, E or F. Finally A ≈ $48, B ≈ $64, C ≈ $61, D ≈ $226, E ≈ $342, F ≈ $129. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1098 }

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – QUESTION 1 : 12 tens and 34 hundreds is equal to - (A) 1234, (B) 3520, (C) 3412, (D) 34120, (E) 352.


Answer

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – ANSWER 1 : B, where 12 x 10 + 34 x 100 = 120 + 3400 = 3520.

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Question { 1100 }

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – QUESTION 2 : When copying a compact disc on a computer the dialogue box showed that the task was 60 % complete. If it had taken 12 minutes so far, how much longer will it take to complete the copying? (A) 6 mins, (B) 8 mins, (C) 12 mins, (D) 20 mins, (E) 40 mins.


Answer

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – ANSWER 2 : B, where 60 % is 12 mins, 10 % is 12 / 6 = 2 mins then 100 % is 2 x 10 = 20 mins. Difference : 100 – 60 = 40 % or 4 x 2 = 8 mins.

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Question { 1190 }

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – QUESTION 3 : Grenville thinks of 7 consecutive numbers. The sum of the smallest three is 33. What is the sum of the largest three? (A) 36, (B) 39, (C) 42, (D) 45, (E) 48.


Answer

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – ANSWER 3 : D. Sum of the smallest 3 consecutive numbers = X + (X + 1) + (X + 2) = 33, X = 10 = smallest number. So the 7 consecutive numbers are 10, 11, 12, 13, 14, 15, 16. Sum of largest 3 = 14 + 15 + 16 = 45.

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Question { 1082 }

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – QUESTION 4 : A 4 × 4 grid is made using 40 matches. How many matches are needed to construct a 10 × 10 grid?


Answer

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – ANSWER 4 : 220. For 10 x 10 grid : Every vertical line has 10 matches, then total vertical lines = 10 + 1 = 11. Every horizontal line has 10 matches, then total horizontal lines = 10 + 1 = 11. So total matches required = 2 x 10 x 11 = 220.

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Question { 1411 }

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – QUESTION 5 : If Karen starts counting from 1 and goes up in threes : 1, 4, 7, ..., and Anne starts counting at the same time from 101 and goes down in twos at the same rate : 101, 99, 97, ..., they will say the same number together. What is that number?


Answer

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – ANSWER 5 : 61, where Karen : Ann = 1 : 101, 4 : 99, 7 : 97, 10 : 95, 13 : 93, 16 : 91, 19 : 89, 22 : 87, 25 : 85, 28 : 83, 31 : 81, 34 : 79, 37 : 77, 40 : 75, 43 : 73, 46 : 71, 49: 69, 52 : 67, 55 : 65, 58 : 63, (61 : 61).

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Question { 1147 }

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – QUESTION 6 : In a shift-code, the letters of the alphabet are replaced by the numbers 1 to 26 starting at some letters. An example is : A = 25, B = 26, C = 1, D = 2, E = 3, F = 4, G = 5, H = 6, I = 7, J = 8, K = 9, L = 10, M = 11, N = 12, O = 13, P = 14, Q = 15, R = 16, S = 17, T = 18, U = 19,V = 20, W = 21, X = 22, Y = 23, Z = 24. Using a shift-code in which K = 15, what is the sum of the numbers which represent the letters in the word SKY?


Answer

VOLUNTEER TRAINING FOR DISADVANTAGED IN MATHS – ANSWER 6 : 41, where S = 23, K = 15, Y = 3, S + K + Y = 23 + 15 + 3 = 41.

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