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Find the result of the following expression if, M denotes modulus operation, R denotes round-off, T denotes truncation: M(373,5)+R(3.4)+T(7.7)+R(5.8)
A20
B29
C19
D39
Answer: Option C
Explanation:Modulus gives a remainder: M(373,5) = 3, since 5*74 = 370 Rounding rounds to the nearest integer: so R(3.4) = 3, and R(5.8) = 6 and Truncation 'rounds' to an integer by removing any decimal fractions, so T(7.7) = 7 adding the answers above gives 3+3+6+7 = 19
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2 / 70
A student's grade in a course is determined by 6 quizzes and one examination. If the examination counts thrice as much as each of the quizzes, what fraction of final grade is determined by the examination?
A\(\frac{1}{6}\)
B\(\frac{1}{5}\)
C\(\frac{1}{3}\)
D\(\frac{1}{4}\)
Answer: Option C
Explanation:Solution: Let the marks allotted for one quiz be 10 marks so acc. to the question...
Marks allotted for the examination will be its three times i.e. 30 marks
So total marks of 6 quizzes and 1 examination will be 60+30=90
Fraction of examination marks out of total marks is: 30/90=1/3
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3 / 70
In a library, there are two racks with 40 books per rack. On a given day, 30 books were issued. What fraction remained in the racks?
A\(\frac{5}{8}\)
B\(\frac{3}{8}\)
C\(\frac{1}{8}\)
D\(\frac{2}{3}\)
Answer: Option A
Explanation:Solution: On two racks there are 80 books in total.
Number of books issued on the day = 30
Fraction of books remaining: 50/80 =5/8
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4 / 70
If a salesman's average is a new order every other week, he will break the office record of the year. However, after 28 weeks, he is six orders behindschedule. In what proportion of the remaining weeks does he have to obtain a new order to break the record?
A\(\frac{3}{5}\)
B\(\frac{3}{4}\)
C\(\frac{2}{3}\)
DNone of these
Answer: Option B
Explanation:Number of weeks in a year = 52
Total orders in a year = \(\frac{52}{2}\) = 26
So after 28 weeks no. of orders = \(\frac{28}{2}\) = 14
Originally completed orders in 28 weeks = 14 - 6 = 8 orders (Because 6 orders behind)
next 24 (52- 28) weeks, orders needed to collect = 26 - 8 = 18
So, required proportion for remaining weeks = \(\frac{18}{24} = \frac{3}{4}\)
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5 / 70
An internet recently hired 8 new network,in advertisement 20 network already employed.All new network cam from university A.In addition 75%of computer advertisement came from same university A.what fraction of original 20 network advertisement came from same university A?
A\(\frac{13}{20}\)
B\(\frac{11}{20}\)
C13
D11
Answer: Option A
Explanation:8 new n/w,20 is already employed
Total n/w = 20+8 = 28 n/w
so that 75% came from same university then \(28 \times \frac{75}{100} = 21\)
we all know 8 n/w is recently hired then 21-8 = 13
=> the fraction is = \(\frac{13}{20}\)
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6 / 70
In a library there are 5 computer cluster open for public use.Each computer cluster consist 33 laptop, in first week 75 laptop are used, in second week 155 laptop used ,in third week 30 laptop and in fourth week 160 laptop used. What fraction of library computer are in used in month?
A\(\frac{3}{5}\)
B\(\frac{2}{7}\)
C\(\frac{2}{3}\)
D\(\frac{1}{14}\)
E\(\frac{1}{4}\)
Answer: Option C
Explanation:Given computer cluster open for public = 5
So, Total laptops = 33 * 5 = 165
In the first week used 75 means 75 out of \(165 = \frac{75}{165} = \frac{15}{33}\)
2nd week \(\frac{155}{165} = \frac{31}{33}\)
3rd week \(\frac{30}{165} = \frac{6}{33}\)
4th week \(\frac{160}{165} = \frac{32}{33}\)
So. Total no of laptops used = \(\frac{\left(15+6+31+32\right)}{33} = 88\)
The fraction of library computer is in used = \(\frac{88}{33} \times \frac{1}{4} = \frac{2}{3}\)
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7 / 70
After loading a dock, each worker on the night crew loaded \(\frac{3}{4}\) as many boxes as each worker on the day of the crew. If the night crew has \(\frac{4}{5}\) as many workers as the day crew, what fraction of all the boxes loaded by two crews did the day crew load?
A\(\frac{1}{2}\)
B\(\frac{2}{5}\)
C\(\frac{4}{5}\)
D\(\frac{5}{8}\)
Answer: Option D
Explanation:If there are x workers in day and each loaded y boxes
then boxes loaded in day = xy
boxes loaded in night = \(\frac{3}{4} \times y \times \frac{4}{5} \times x = \frac{3}{5} \times xy\)
Total boxes loaded = \(xy + \frac{3}{5} \times xy = \frac{8}{5} xy\)
fraction of all the boxes loaded by two crews did the day crew load = \(\frac{xy}{(\frac{8}{5} \times xy)} = \frac{5}{8}\)
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8 / 70
\(\frac{2}{3}\)rd of the balls in a bag are blue, the rest are pink. if \(\frac{5}{9}\)th of the blue balls and \(\frac{7}{8}\)th of the pink balls are defective, find the total number of balls in the bag given that the number of non-defective balls is 146?
A216
B649
C438
D578
Answer: Option C
Explanation:Let total no of balls = x blue = \(\frac{2x}{3}\)
pink = \(\frac{x}{3}\)
Total no of defective balls = \(\frac{10x}{27} + \frac{7x}{24}\)
= \(\frac{143x}{216}\)
Non defective balls = \(x - \frac{143x}{216} = 146\)
=> x = 432
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How many \(\frac{1}{5}\) are there in \(\frac{243}{5}\)?
A429
B529
C629
D729
Answer: Option D
Explanation:243 is 3 raise to 5 ;
729 is 3 raise to 6.
both can be multiple of \(\frac{1}{5}\), as multiple is what you say how much part of this is there is that.
so the ans is 729.
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10 / 70
Arrange the fractions \(\frac{3}{5},\frac{4}{7},\frac{8}{9} and \frac{9}{11}\) in their descending order.
A\(\frac{2}{9}>\frac{6}{11}>\frac{4}{5}>\frac{3}{7}\).
B\(\frac{8}{9}>\frac{6}{11}>\frac{4}{5}>\frac{3}{7}\).
C\(\frac{8}{9}>\frac{9}{11}>\frac{4}{5}>\frac{3}{7}\).
D\(\frac{8}{9}>\frac{9}{11}>\frac{3}{5}>\frac{4}{7}\).
Answer: Option D
Explanation:clearly, \(\frac{3}{5}=0.6,\frac{4}{7}=0.571,\frac{8}{9}=0.88,\frac{9}{11}=0.818\).
now, \(0.88>0.818>0.6>0.571\).
so,\(\frac{8}{9}>\frac{9}{11}>\frac{3}{5}>\frac{4}{7}\).
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