THE METHOD OF AGE GRADATION 107 



good capacities in the higher levels. To be correct, 

 we must credit more difficult tests (those lying in 

 higher levels) with a larger fractional value than 

 the tests normal to the age in question when we fig- 

 ure in these higher tests for addition to a lower age- 

 level. We may propose a method of calculation for 

 this purpose, that is not too complicated and that, 

 like the mental quotient, takes account of the rela- 

 tion of the several years to each other. A test from 

 a higher level used to supplement a failure in a lower 

 level shall be counted not merely as one test, but as 

 a quotient of the two years in question. 



In our example just given, then, Child A would be figured out 

 thus : basal point, mental age of 6 years ; the tests from the four 

 following years would be counted in this way : 



Level 7 is formed by 2 tests from the 7th year (each of these 

 counted therefore as "1 test") and by 3 tests from the 8th year, 

 each of which are to be counted as 8/7 test). The 8th year would 

 be formed by 3 tests from the 9th year (each counting 9/8 test) 

 and 2 tests from the 10th year (each counting 10/8 test). We get, 

 therefore, as total additional credits : 



2 x 1 = 2 te sts 



8 



3 X = 3 - 4 tests 



7 



9 



3 X = 3 - 4 tests 

 8 



10 



2 x = 2 - 5 tests 

 8 



11.3 tests 



Since every 5 tests are worth one mental year, the above value 

 indicates a supplement of 11.3 -^- 5 = 2.3 mental years : so Child A 

 gets a mental age of 8.3 years. 



With Child B it works out thus : the 2 tests from age 9 serve to 



