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Miscellaneous Exercise · Q7

Q.A manufacturer produces three products x,y,zx, y, z which he sells in two markets. Annual sales are indicated below: Market I — xx: 10,000, yy: 2,000, zz: 18,000 Market II — xx: 6,000, yy: 20,000, zz: 8,000

(a) If unit sale prices of x,yx, y and zz are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Sikkim CbseNCERTSubjective· 5mImportance★★★★★
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Treating the sales table as a 2×32\times 3 matrix and the prices/costs as 3×13\times 1 columns: revenue is SP=[4600053000]S P = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}, and the gross profit (total revenue minus total cost) is ₹ 32,000\text{₹}\,32{,}000.

Multiplying a row of quantities by a column of prices sums each product's quantity times its price, so stacking both markets into one matrix gives all the totals in a single product.

1. Sales matrix (rows = Market I, Market II; columns = x,y,zx,y,z):

S=[100002000180006000200008000]S = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}

  1. Total revenue in each market. Price column P=[2.501.501.00]P = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}.

    R=SP=[10000(2.50)+2000(1.50)+18000(1.00)6000(2.50)+20000(1.50)+8000(1.00)]=[25000+3000+1800015000+30000+8000]=[4600053000]R = SP = \begin{bmatrix} 10000(2.50)+2000(1.50)+18000(1.00) \\ 6000(2.50)+20000(1.50)+8000(1.00) \end{bmatrix} = \begin{bmatrix} 25000+3000+18000 \\ 15000+30000+8000 \end{bmatrix} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}

    So revenue is ₹ 46,000\text{₹}\,46{,}000 in Market I and ₹ 53,000\text{₹}\,53{,}000 in Market II (total ₹ 99,000\text{₹}\,99{,}000).
  2. Gross profit. Cost column C=[2.001.000.50]C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} (50 paise =₹ 0.50=\text{₹}\,0.50). …

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