Work Life & Job Hunting · SPI Prep Lab
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In a class of 40, 25 like mathematics, 18 like English and 10 like both. How many like at least one of them?
25+18 counts the 10 who like both twice, so subtract one of those counts: 25+18−10=33. Answering 43 just adds and forgets to subtract the 10; it exceeds the 40 in the class, which shows at once that it is wrong. 15 is the number who like only mathematics, and 7 is the number who like neither.
Of 50 staff at a company, 30 use a bus to get to work, 22 use a train and 14 use both. How many use neither?
Those using at least one are 30+22−14=38, so 50−38=12. Answering 38 gives the number using at least one. 16 is those using only the bus, 30−14, and 8 is those using only the train, 22−14; both answer a different category from the one asked about.
Of 60 candidates, 35 passed qualification A, 28 passed qualification B and 12 passed both. How many passed A only?
The 35 for A include the 12 who passed both, so 35−12=23. Answering 35 gives all who passed A. 16 is those who passed B only, 28−12. 51 is those who passed at least one, 35+28−12.
60 people sat a test. 35 answered question A correctly, 28 answered question B correctly, and 5 got neither right. How many answered both correctly?
Those getting at least one right are 60−5=55, so those getting both right are 35+28−55=8. Answering 3 uses the whole of 60, 35+28−60, forgetting to take out the 5 who got neither right first. 27 is those right on A only and 20 those right on B only.
80 people were asked about their use of two products. 18 use both, 27 use A only, and 20 use neither. How many use B?
Those using B only are 80−18−27−20=15. Those using B are that plus the 18 who use both: 15+18=33. Answering 15 gives the B-only figure and forgets to add those using both. 45 is the number using A, 27+18. 60 is the number using at least one, 80−20.
In a survey of 200 people, 60% knew product A, 45% knew product B and 25% knew both. How many knew neither?
The proportion knowing at least one is 60+45−25=80%, which is 160 people, so 200−160=40. Answering 160 gives the number knowing at least one. 70 is the number knowing A only, 200×0.35. 50 is the number knowing both, 200×0.25.
In a survey of 100 people, 70 answered yes to question 1 and 60 answered yes to question 2. What is the smallest possible number who answered yes to both?
70+60=130 exceeds 100, so at least 130−100=30 must overlap. And 30 is achievable: 30 saying yes to both, 40 to question 1 only, 30 to question 2 only and 0 to neither satisfies the conditions, so 30 is the minimum. Answering 0 imagines no overlap, but 130 answers cannot fit into 100 people. 10 is the difference 70−60, and 60 is the largest possible overlap.
Of 120 visitors, 90 took a leaflet and 75 answered a survey. What is the largest possible number who did both?
The overlap is largest when all 75 who answered the survey also took a leaflet, giving 75. That is achievable: 75 doing both, 15 taking a leaflet only and 30 doing neither. Answering 90 gives the larger of the two figures, but only 75 answered the survey, so it is impossible. 45 is 90+75−120, which is the minimum. 30 is 120−90.
Of 300 members, 40% use service X, 35% use service Y and 12% use both. How many members use at least one of them?
40+35−12=63%, and 300×0.63=189. Answering 225 uses 40+35=75% and forgets to subtract the overlap. 111 is the number using neither, 300×0.37. 36 is the number using both.
At an event, 48 people attended the morning session, the afternoon session or both; 30 attended the afternoon session and 11 attended both. How many attended the morning session only?
Taking the 30 who attended the afternoon away from the 48 who attended at least one leaves those who attended the morning only: 48−30=18. Answering 7 subtracts the 11 as well, 48−30−11, taking out those who attended both twice over. 19 is those who attended the afternoon only, 30−11. 29 is the total who attended the morning, 18+11.
In a survey, 30 people fitted A, 25 fitted B, 20 fitted C, 12 fitted both A and B, 8 fitted both B and C, 10 fitted both A and C, and 5 fitted all three. How many fitted at least one?
30+25+20−12−8−10+5=50. Answering 75 just adds without subtracting the overlaps. Answering 45 forgets to add back the 5 who fit all three. Answering 80 adds the 5 without subtracting the pairwise overlaps, 75+5.
70 people were asked about their use of three media. 30 read newspapers, 25 read magazines, 40 used the internet, 15 did both newspapers and magazines, 12 both magazines and internet, 18 both newspapers and internet, and 8 all three. How many used none of them?
Those using at least one are 30+25+40−15−12−18+8=58, so 70−58=12. Answering 58 gives the number using at least one. 29 is the number using exactly one and 21 the number using exactly two; both answer a different category from the one asked about.
30 people fit A, 25 fit B and 20 fit C; the overlap of A and B is 12, of B and C is 8, and of A and C is 10, while 5 fit all three. How many fit exactly one of them?
A only is 30−12−10+5=13, B only is 25−12−8+5=10, and C only is 20−8−10+5=7, making 30 in all. Answering 50 gives those fitting at least one. 15 is the number fitting exactly two. 20 is the number fitting two or more, 15+5.
In a survey of media use, 30 read newspapers, 25 read magazines, 40 used the internet, 15 did both newspapers and magazines, 12 both magazines and internet, 18 both newspapers and internet, and 8 all three. How many used exactly two?
From each pairwise overlap take away those who used all three: (15−8)+(12−8)+(18−8)=7+4+10=21. Answering 45 just adds the overlaps, 15+12+18. Answering 37 subtracts the 8 only once. Answering 29 gives those using two or more, 21+8.
Of 50 people, 40 fit at least one. A is 28, B is 24, C is 20, A and B is 14, B and C is 10, and A and C is 12. How many fit all three?
Putting this into the at-least-one formula, 40=28+24+20−14−10−12+x. Since 72−36=36, x=40−36=4. Answering 10 gives those fitting none, 50−40. Answering 0 simply assumes there is no overlap. Answering 24 gives those fitting exactly two.
Of 100 staff, 80 took training A, 70 took B and 60 took C. What is the smallest possible number who took all three?
Those who missed a course are 20 for A, 30 for B and 40 for C, which comes to only 90 in total. At most 90 people missed at least one, so at least 100−90=10 took all three. As a formula, 80+70+60−2×100=10. And it is achievable: the A-and-B overlap 50, the B-and-C overlap 30, the A-and-C overlap 40 and 10 in all three can be fitted into 100 people. 0 is impossible, 60 is the maximum, and 30 comes from 100−70.
40 pupils were asked their strong subjects: 18 said Japanese, 20 mathematics, 15 English, 8 both Japanese and mathematics, 6 both mathematics and English, 5 both Japanese and English, and 3 all three subjects. How many are strong in at least one subject?
18+20+15−8−6−5+3=37. Answering 53 just adds without subtracting the overlaps, and it exceeds the 40 pupils. Answering 34 forgets to add back the 3. Answering 24 gives those strong in exactly one subject.
90 people were asked about their use of three places. 45 used the cafe, 38 the bookshop, 30 the gym, 20 both cafe and bookshop, 12 both bookshop and gym, 15 both cafe and gym, and 7 all three. How many used none of them?
Those using at least one are 45+38+30−20−12−15+7=73, so 90−73=17. Answering 73 gives the number using at least one. 40 is the number using exactly one and 26 the number using exactly two.
In a survey of strong subjects, 18 said Japanese, 20 mathematics, 15 English, 8 both Japanese and mathematics, 6 both mathematics and English, 5 both Japanese and English, and 3 all three. How many are strong in exactly two subjects?
(8−3)+(6−3)+(5−3)=5+3+2=10. Answering 19 just adds the overlaps, 8+6+5, forgetting to take out those strong in all three. Answering 13 gives those strong in two or more, 10+3. Answering 24 gives those strong in exactly one subject.
In a survey of place use, 45 used the cafe, 38 the bookshop, 30 the gym, 20 both cafe and bookshop, 12 both bookshop and gym, 15 both cafe and gym, and 7 all three. How many used exactly one?
Cafe only is 45−20−15+7=17, bookshop only is 38−20−12+7=13, and gym only is 30−12−15+7=10, making 40 in all. Answering 73 gives those using at least one. 26 is the number using exactly two. 47 is 73−26, which forgets to take out the 7 who used all three.
A survey of opinion gave: men 48 in favour and 32 against, women 36 in favour and 44 against. How many were in favour overall?
Add the 'in favour' column down: 48+36=84. Answering 80 gives the men's total, 48+32, mixing up the row and the column. 76 is the total against, 32+44. 160 is the total number of respondents.
In a survey of 300 people, 120 were in their 20s, and 72 of those had used the service. 150 people in all had used the service. What percentage of those in their 20s had used it?
The denominator is the 120 people in their 20s: 72/120=0.6, that is 60%. Answering 24% uses all 300 respondents as the denominator, 72/300, which is the wrong denominator. 48% is the share of the 150 users who are in their 20s, 72/150. 40% is the share of those in their 20s who had not used it, 48/120.
Of 200 respondents, 120 were men, 110 were in favour, and 70 were men in favour. How many women were against?
The women number 200−120=80, and the women in favour are 110−70=40, so the women against are 80−40=40. As a formula, 200−120−110+70=40. Answering 50 gives the men against, 120−70. 90 is the total against, 200−110. 80 is the total number of women.
Of 400 staff, 60% are in the sales department. 30% of the sales department and 20% of those outside it hold a certain qualification. How many staff hold it?
The sales department is 400×0.6=240, of whom 240×0.3=72 hold it. Outside sales there are 160, of whom 160×0.2=32 hold it. 72+32=104. Answering 100 averages 30% and 20% into 25% and takes 400×0.25; the two groups differ in size, so their percentages cannot be averaged. 120 is 30% of the whole. 72 counts the sales department only.
A department of 40 people has 24 men and 16 women. Those holding a qualification are 15 men and 9 women, and of those, the managers are 6 men and 2 women. How many staff hold the qualification and are not managers?
For men it is 15−6=9 and for women 9−2=7, making 16. Answering 24 gives all who hold the qualification, 15+9, forgetting to take out the managers. 8 is the total of qualified managers, 6+2. 32 subtracts those 8 managers from the 40, which wrongly counts staff without the qualification as well.
Of 500 employees at a company, 60% are men. When 40% of the men and 25% of the women are managers, what percentage of all employees are managers?
There are 300 men, of whom 120 are managers, and 200 women, of whom 50 are. The managers number 170, and 170/500=0.34, that is 34%. Answering 32.5% takes the plain average of 40 and 25; the two groups differ in size, so they cannot be averaged. 40% simply repeats the men's rate. 24% divides only the 120 male managers by the whole, 120/500.
A shop had 120 visitors on Monday, an unknown number on Tuesday, 85 on Wednesday, 110 on Thursday and 145 on Friday, with 578 over the five days. How many visitors were there on Tuesday?
578−(120+85+110+145)=578−460=118. Answering 263 forgets Friday's 145, 578−315. 460 is the total for the four days other than Tuesday. 348 forgets Wednesday and Friday, 578−120−110.
Of 100 people, 60 subscribe to service A and 45 to service B. When 30 subscribe to A only, how many subscribe to neither?
Those subscribing to both are 60−30=30, so those subscribing to at least one are 60+45−30=75, and 100−75=25. Answering 30 gives those subscribing to both. 15 is those subscribing to B only, 45−30. 55 is those not subscribing to B, 100−45.
A shop's sales were 1,200,000 yen for product A and 800,000 yen for product B in April, and 1,320,000 yen for product A and 880,000 yen for product B in May. How did product A's share of total sales change from April to May?
April is 120/200=0.6, that is 60%, and May is 132/220=0.6, also 60% (working in units of ten thousand yen), so the share does not change. Saying it rose by 10 points confuses the 10% rise in product A's sales, from 1,200,000 yen to 1,320,000 yen, with a change in its share of the whole. Product B rose by the same 10%, from 800,000 yen to 880,000 yen, so the composition does not move. Saying it rose by 6 points keeps May's denominator at April's 2,000,000 yen and compares 132/200=66%. Saying it fell by 10 points sees the same 10% rise but gets the direction of change wrong.
In a class of 50, 32 travel to school by bicycle and 28 belong to a club. Which is the possible range for the number who do both?
The minimum is 32+28−50=10, achievable as 10 doing both, 22 cycling only, 18 in a club only and 0 doing neither. The maximum is when all 28 club members cycle, giving 28, achievable as 28 doing both, 4 cycling only, 0 in a club only and 18 doing neither. 'From 0' is impossible because 32+28=60 will not fit into 50 people. 'To 32' is impossible because there are only 28 club members. 'From 4' mistakes the difference 32−28 for the minimum: with 4 doing both there would be 28 cycling only and 24 in a club only, making 56, which will not fit into 50.
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