CBSE Class 12 · 2014 · MatricesLong Answer

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Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. School A awards Rs.

x

, Rs.

y

, Rs.

z

each for the three values to 3, 2 and 1 students respectively with a total of Rs. 1,600. School B awards 4, 1 and 3 students (same per-value amounts) with a total of Rs. 2,300. If the total award for one prize on each value is Rs. 900, using matrices find the award money for each value. Suggest one more value for award.

Answer: Sincerity Rs. 200, Truthfulness Rs. 300, Helpfulness Rs. 400

Solution

The conditions give

3 x + 2 y + z = 1600, 4 x + y + 3 z = 2300, x + y + z = 900.

In matrix form

D X = E

with

D = 341211131

E = 16002300900

∣ D ∣ = - 5 \neq = 0

. Then

X = D^{- 1} E

x y z = 200300400 .

So Sincerity Rs.

200

, Truthfulness Rs.

300

, Helpfulness Rs.

400

. One more value: Punctuality.

x = 200, y = 300, z = 400

CBSE Class 12 Board · 2014 · Subjective

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CBSE Class 12 · 2014 · MatricesLong Answer

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