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Standard IV

Mathematics

Encoding and Decoding Patterns

If Driver = 1...

Question

If Driver = 12, Pedestrian = 20, Accident = 16, then Car = ? (a) 3 (b) 6 (c) 8 (d) 10

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Solution

Clearly, we can find that each word is coded by the numeral which is twice the number of letters in the word
Driver = 6 x 2 = 12
Pedestrian = 10 x 2 = 20
Accident = 8 x 2 = 16
Car = 3 x 2 = 6
Therefore, car = 6
So, the correct option is (b)

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(ii)

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; find
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(ii)

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If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20},

C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find

(i) A – B

(ii) A – C

(iii) A – D

(iv) B – A

(v) C – A

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If Driver = 12, Pedestrian = 20, Accident = 16, then Car = ? (a) 3 (b) 6 (c) 8 (d) 10

Clearly, we can find that each word is coded by the numeral which is twice the number of letters in the wordDriver = 6 x 2 = 12Pedestrian = 10 x 2 = 20Accident = 8 x 2 = 16Car = 3 x 2 = 6Therefore, car = 6So, the correct option is (b)

Clearly, we can find that each word is coded by the numeral which is twice the number of letters in the word
Driver = 6 x 2 = 12
Pedestrian = 10 x 2 = 20
Accident = 8 x 2 = 16
Car = 3 x 2 = 6
Therefore, car = 6
So, the correct option is (b)