Question: When was the life expectancy in a country, whose population was 55,300 in 1989, 66.96341 years ?
Choices:
1885-12-01
2011-10-24
1989-12-01
1842-01-01
2000
1975-05-19
1967-04-10
2000-05-01
1774
1991-12-01
Answer: 2000
SPARQL: SELECT DISTINCT ?qpv WHERE { ?e <pred:instance_of> ?c . ?c <pred:name> "country" . ?e <population> ?pv_1 . ?pv_1 <pred:unit> "1" . ?pv_1 <pred:value> "55300"^^xsd:double . [ <pred:fact_h> ?e ; <pred:fact_r> <population> ; <pred:fact_t> ?pv_1 ] <point_in_time> ?qpv_1 . ?qpv_1 <pred:year> 1989 . ?e <life_expectancy> ?pv . ?pv <pred:unit> "year" . ?pv <pred:value> "66.96341"^^xsd:double . [ <pred:fact_h> ?e ; <pred:fact_r> <life_expectancy> ; <pred:fact_t> ?pv ] <point_in_time> ?qpv .  }
Program:
function: FindAll
function: FilterNum
dependencies: 0
inputs: population
55300
=
function: QFilterYear
dependencies: 1
inputs: point in time
1989
=
function: FilterConcept
dependencies: 2
inputs: country
function: QueryAttrQualifier
dependencies: 3
inputs: life expectancy
66.96341 year
point in time