1
In an examination of ‘n’ questions, a student replied 15 out of the first 20 questions correctly. Of the remaining questions, he answered one-third correctly. All the questions have the same credit. If the student gets 50% marks, the value of ‘n’ is
A.
B.
C.
D.
Answer & Solution
Answer:
50 15 +
1/3 (n – 20) = n × 50%
Þ 45 + n – 20 = n/2
Þ 90 + 2n – 40 = 3n
Þ n = 50.
2
In an examination, there are three papers and a candidate has to get 35% of the total to pass. In one paper, he gets 62 out of 150 and in the second 35 out of 150. How much must he get, out of 180, in the third paper to just qualify for a pass?
A.
B.
C.
D.
Answer & Solution
Answer:
71Let the marks required be x.
Then, 62 + 35 + x = (150 + 150 + 180) × 35%
Þ 97 + x = 480 × 35%
Þ x = 168 – 97 = 71.
3
A clothing supplier stores 800 coats in a warehouse, of which 15 percent are full-length coats. If 500 of the short-length coats are removed from the warehouse, then what percent of the remaining coats are full length?
A.
B.
C.
D.
Answer & Solution
Answer:
40%Number of short length coats before removal
= (100 – 15)% of
800 = 800 × 85% = 680
Number of short length coats after removal =
680 – 500 = 180
Total number of coats after removal = 800 –
500 = 300
Number of full length coats after removal =
300 – 180 = 120
\ % of remaining coats
of full length = (120/300) × 100 = 40%
4
At the college entrance examination, each candidate is admitted or rejected according to whether he has passed or failed the tests. Of the candidates who are really capable, 80% pass the tests and of the incapable, 25% pass the test. Given that 40% of the candidates are really capable, the proportion of capable college students is about
A.
B.
C.
D.
Answer & Solution
Answer:
68%Capable candidates: 40% of total.
80% of them pass →
0.8×40=32 out of 100 pass.
Incapable candidates: 60% of total.
25% of them pass →
0.25×60=15 out of 100 pass.
Total students who pass = 32+15=47
\ Proportion of capable students among those who pass:
(32/47) × 100 = 0.681×100 ≈ 68.10%.
5
In a History examination, the average for the entire class was 80 marks. If 10% of the students scored 95 marks and 20% scored 90 marks, what was the average marks of the remaining students of the class?
A.
B.
C.
D.
Answer & Solution
Answer:
75Let, the number of students in the class be
100 & required average be x.
Then, (10 × 95) + (20 × 90) + (70 × x) = 100 ×
80
Þ 70x = 8000 – (950 +
1800)
Þ 70x = 5250
Þ x = 75.
6
A scored 30% marks and failed by 15 marks. B scored 40% marks and obtained 35 marks more than those required to pass. The pass percentage is
A.
B.
C.
D.
Answer & Solution
Answer:
33%Let
the total marks in the examination be x and the passing marks be P.
A
scored 30% of total marks and failed by 15 marks:
A’s Marks=0.3x
Since
A failed by 15 marks, the passing marks are:
P=0.3x+15P
B
scored 40% of total marks and obtained 35 marks more
than the passing marks.
So, B’s Marks=0.4x
Since
B scored 35 marks more than passing marks, we have:
0.4x=P+35
Substituting
P=0.3x+15 into 0.4x=P+35:
0.4x=(0.3x+15)+35
Þ 0.4x=0.3x+50
Þ 0.4x−0.3x=50
Þ 0.1x=50
Þ x=500
Now,
Passing Marks and Pass Percentage
P=0.3(500)+15=150+15=165
\ The pass
percentage is:
(P/x)×100=(165/500)×100=33%.
7
In a recent survey, 40% houses contained two or more people. Of those houses containing only one person, 25% were having only a male. What is the percentage of all houses, which contain exactly one female and no males?
A.
B.
C.
D.
Answer & Solution
Answer:
45%Since
40% of houses have two or more people, the remaining 60% must be houses with
only one person.
Of these one-person houses (60%), 25% contain only a
male:
Houses with only a male=25%×60%=15%
Since
the one-person houses must contain either only a male or only
a female, the percentage of houses with only a female is:
Houses with only a female=60%−15%=45%.
8
In an area, of the total people 40% were women and 45% coffee drinkers. One-third of the males are coffee drinkers. Suppose the total number of persons in the area is 100, then the number of female non-coffee drinkers is
A.
B.
C.
D.
Answer & Solution
Answer:
15Number of women = 100 × 40% = 40
Number of men = 100 – 40 = 60
Number of coffee-drinkers = 100 × 45% = 45
Number of male coffee-drinkers = 60 × 1/3 = 20
Number of female coffee-drinkers = 45 – 20 =
25
\ Number of female
non-coffee drinkers = 40 – 25 = 15
9
37% of the candidates in an examination were girls, 75% of the boys and 62% of the girls passed and 342 girls failed. The number of boys failed was
A.
B.
C.
D.
Answer & Solution
Answer:
383Let
the total number of candidates be N.
Then, number of girls = 0.37N
Given
that 62% of the girls passed, so the percentage of girls who failed is:
Failed girls = 100% − 62% = 38%
Since
38% of the girls failed and their count is given as 342:
0.38 × (0.37N) = 342
Þ 0.1406N = 342
Þ 0.1406N = 342
Þ N = 2432
Since
63% of the candidates were boys:
Number of boys = 0.63N = 0.63
× 2432 = 1532
75%
of the boys passed, so the percentage of boys who failed is:
Failed boys
= 100% − 75% = 25%
\ Number of boys who failed = 0.25
× 1532 = 383.
10
Of the 50 researchers in a workgroup, 40% will be assigned to Team A and the remaining 60% to Team B. However, 70% of the researchers prefer Team A and 30% prefer Team B. What is the least possible number of researchers who will not be assigned to the team they prefer?
A.
B.
C.
D.
Answer & Solution
Answer:
15Number of researchers who prefer Team A = 50 ×
70% = 35
Number of researchers who prefer Team B = 50 –
35 = 15
Number of researchers assigned to Team A = 50
× 40% = 20
Number of researchers assigned to Team B = 50
– 20 = 30
To find the least possible number of
researchers who will not be assigned to the team they prefer, it can be assumed
that the maximum number of researchers get the Team they prefer.
So, number of researchers who are assigned to
the team they prefer
= 20 Team A + 15 Team B = 35
Hence, required number = 50 – 35 = 15.
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