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CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 2 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Numerical Applications
Quiz 2
If the arithmetic mean of $$6, 8, 5, 7, x$$ and $$4$$ is $$7,$$ then $$x$$ is
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0%
$$12$$
0%
$$6$$
0%
$$8$$
0%
$$4$$
Explanation
$$\textbf{Step - 1: Solving for x.}$$
$$\text{Observation given = }{6, 8, 5, 7, x, 4}$$
$$\text{Arithmetic mean = 7}$$
$$\text{Arithmetic mean = }\dfrac{\text{Sum of observations.}}{\text{Number of observations.}}$$
$$\Rightarrow \dfrac{6 + 8 + 5 + 7 + x + 4}{6} = 7$$
$$30 + x = 7 \times 6$$
$$x = 42 - 30$$
$$x = 12$$
$$\textbf{Hence, the value of x is 12. (Option A)}$$
A man working $$8$$ hours a day takes $$5$$ days to complete a project. How many hours a day must he work to complete it in $$4$$ days?
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0%
$$10$$ hours
0%
$$11$$ hours
0%
$$12$$ hours
0%
$$14$$ hours
Explanation
Number of Hours required to do the $$Work$$ in $$5$$ days
$$\Rightarrow 5\times 8=40$$
To do the Same $$Work$$ in $$4$$ days
$$\Rightarrow 4\times hours=40$$
$$\Rightarrow hours =10$$
Therefore Answer is $$(A)$$
If $$20$$ persons can do a piece of work in $$7$$ days, then the number of persons required to complete the work in $$28$$ days.
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0%
$$4$$
0%
$$5$$
0%
$$14$$
0%
$$10$$
Explanation
If A can do a piece of work in $$x$$ days then A's one days work is $$\dfrac{1}{x}$$
Hence $$20$$ people can do a piece of work in $$7$$ days
$$20$$ people's one days work is $$\dfrac{1}{7}$$
$$1$$ persons $$1$$ days work is $$\dfrac{1}{7\times20}=\dfrac{1}{140}$$
Therefore $$1$$ person can do a piece of work in $$140$$ days
In $$28$$ days$$\dfrac{140}{28}=5\ people$$ can work
X and Y together can do a piece of work in $$8$$ days, which X alone can do in $$12$$ days. In how many days can Y do the same work alone?
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0%
$$12$$ days
0%
$$24$$ days
0%
$$36$$ days
0%
$$16$$ days
Explanation
$$X$$'s one day's work $$ = \displaystyle\frac { 1 }{ 12 }$$
Let, $$Y$$ work for $$x$$ days
$$\therefore Y$$'s one day's work $$ = \displaystyle\frac { 1 }{ x }$$
One day work by $$X$$ and $$Y$$ together
$$\displaystyle {\frac{1}{12}\, +\, \frac{1}{x}\, =\, \frac{1}{8}}$$
$$\displaystyle {\frac{1}{x}\, =\, \frac{1}{8}\, -\, \frac{1}{12}\, =\, \frac{1}{24}}$$
$$\therefore$$ $$x = 24$$ days
$$\therefore Y $$ can complete work alone in 24 days.
If 1 January 2012 falls on Sunday which day will be 1 December 2012 ?
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0%
Sunday
0%
Saturday
0%
Friday
0%
Thursday
Which of the following angles is made between the two numbers of a clock?
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0%
$$\displaystyle 40^{\circ}$$
0%
$$\displaystyle 30^{\circ}$$
0%
$$\displaystyle20^{\circ}$$
0%
$$\displaystyle 10^{\circ}$$
Which of the following cannot be the first day of a century year ?
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0%
Monday
0%
Tuesday
0%
Wednesday
0%
Thursday
If in a non leap year it is Tuesday on 28th February then the day which will be on 28th march is ------
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0%
Monday
0%
Tuesday
0%
Wednesday
0%
Thursday
A tank can be filled by one tap in 209 minutes and by another in 25 minutes. Both the taps are kept open for 5
minutes and then the second is turned off. In how many minutes more is the tank completely filled?
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0%
$$6$$
0%
$$11$$
0%
$$12$$
0%
$$17\dfrac {1}{2}$$
Explanation
Let flow rate of first tank $$=xm^3/min$$. Flow rate of $$2^{nd}tap=ym^3/min$$.
Let volume of tank $$=m^3$$
$$20x=v$$
$$x=\dfrac {v}{20}$$
$$25y=v$$
$$Y=\dfrac {v}{25}$$
$$5t=\dfrac {v-5x-5y}{x}=11 min$$
22 times are the hands of a clocks coincide in a day.
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0%
True
0%
False
Explanation
Hands of a clock coincide 11 times in 12 hours $$\therefore$$ they will concide 22 times in a day
The day after tomorrow will be X-mas day. What will be the day on New-year-day if today is Monday?
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0%
Monday
0%
Tuesday
0%
Wednesday
0%
Sunday
Out of the following four choices which does not show the coinciding of the hour hand and minute hand :
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$$3 : 16 : 2$$
0%
$$6 : 32 : 43$$
0%
$$9 : 59 : 05$$
0%
$$5 : 27 : 16$$
Explanation
Clearly, at 9:59:05 the hour hand and the minute hand does not coincide.
Answer is Option C
If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month?
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0%
Sunday,
0%
Monday
0%
Wednesday
0%
Friday
Explanation
Seventh day of the month is three days earlier than Friday i.e. it is Tuesday.
The fourteenth day will also be a Tuesday.
Nineteenth day is five days ahead.
Therefore, nineteenth day will be a Sunday.
Answer is Option A
A clock is set right at 5 am The clock loses 16 minutes in 24 hours What will be the true time when the clock indicates 10 pm on 4th day?
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0%
9 pm
0%
10 pm
0%
11 pm
0%
12 pm
If 4th day of any month was Sunday, what will be the day on 27th day of the same month ?
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0%
Monday
0%
Tuesday
0%
Wednesday
0%
Saturday
A is thrice as efficient as B and B is twice as efficient as C If A, B and C work together how long will they take to complete a job which B completes in 10 days?
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0%
$$\displaystyle \frac{20}{9}$$ days
0%
$$\displaystyle \frac{11}{9}$$ days
0%
3 days
0%
None of these
Explanation
Let B finish the work in $$x$$ days
$$\therefore $$ A finishes the work in $$\frac{x}{3}$$ days
and C finishes the work in $$2x$$ days
Now,
$$x=10$$ days
Let the work get completed working together be $$y$$ days
$$\therefore \frac{1}{x}+\frac{3}{x}+\frac{1}{2x}=\frac{1}{y}$$
$$=>\frac{2+6+1}{2x}=\frac{1}{y}$$
$$=>\frac{9}{2x}=\frac{1}{y}$$
$$=>y=\frac{2x}{9}$$
$$=>y=\frac{2\times 10}{9}$$
$$=>y=\frac{20}{9}$$
At what time are the hands of the clocks together between $$6\;and\;7$$ ?
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0%
$$32\displaystyle\frac{8}{11}\,\text{minutes past 6}$$
0%
$$34\displaystyle\frac{8}{11}\,\text{minutes past 6}$$
0%
$$30\displaystyle\frac{8}{11}\,\text{minutes past 6}$$
0%
$$32\displaystyle\frac{5}{7}\,\text{minutes past 6}$$
Explanation
$$\boxed{\\{\text{Angle difference}\,(\theta)=\left|{\dfrac{11}{2}min\,-\,30\times H}\right|\,\,\,}\\}$$
$$\theta=0,\quad H=6,\quad min=?$$
$$\Rightarrow 0=30\times6\,-\,\dfrac{11}{2}min$$
$$\therefore min=32\frac{8}{11}$$
Find the arithmetic mean of $$24 $$ and $$36.$$
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0%
26
0%
28
0%
30
0%
32
Explanation
Arithmetic mean of two quantity $$a,b$$ $$= \dfrac{a+b}{2}$$
$$AM $$ of $$24, 36 = \dfrac{24+36 }{2} = \dfrac{60}{2} = 30$$
Smt. Indira Gandhi died on 31$$\displaystyle ^{_{st}}$$ October, 1984 The day of the week was ____________.
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0%
Monday
0%
Tuesday
0%
Wednesday
0%
Friday
Explanation
1600 years contain 0 odd day, 300 years contain 1 odd day
Also, 83 years contain 20 leap years and 63 ordinary years and therefore (40 + 0) odd days i.e., 5 odd days
∴
∴
1983 years contain (0+1+5) i.e., 6 odd days.
Number of days from Jan. 1984 to 31st Oct. 1984
= (31+29+31+30+31+30+31+31+30+31)
= 305 days = 4 odd days
∴
∴
Total number of odd days = 6 + 4 = 3 odd days
So, 31st Oct, 1984 was Wednesday.
$$3$$ men of $$7$$ women can do a piece of work in $$32$$ days. The number of days required by $$7$$ men and $$5$$ women to do a piece of work twice as large is
Report Question
0%
$$19$$
0%
$$21$$
0%
$$27$$
0%
$$36$$
Explanation
Work of 3 men $$=$$ work of 7 women
Work of 1 man $$=$$ work of $$\displaystyle\frac{7}{3}$$ women
Work of 7 men $$=$$ work of $$\displaystyle\frac{7\times7}{3}$$
$$\displaystyle=\frac{49}{3}$$ women
$$\therefore$$ Work of 7 men + 5 women
$$=$$ work of $$\displaystyle\frac{49}{3}+5$$
$$\displaystyle=\frac{64}{3}$$ women
$$\displaystyle\begin{matrix}Women&Days\\7&32\\\frac{64}{3}&x\end{matrix}$$
$$\displaystyle\therefore\frac{64}{3}:7::32:x$$
$$\displaystyle\therefore\frac{64}{3}\times x=7\times32$$
$$\displaystyle\therefore x=\frac{7\times32\times3}{64}=10.5\:days$$
$$\therefore$$ So, 7 men and 5 women can do piece of work twice as large in $$21$$ days.
12 men can complete a piece of work in 16 days. How many days will 4 men take to complete the task ?
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0%
60 days
0%
45 days
0%
54 days
0%
48 days
Explanation
$$\displaystyle Men\qquad Work$$
$$\displaystyle 12\downarrow \qquad 16\uparrow $$
$$\displaystyle 4\qquad \qquad x$$
$$\displaystyle \therefore \dfrac { 12 }{ 4 } =\dfrac { x }{ 16 } $$
$$\displaystyle \Rightarrow x=\dfrac { 12\times 16 }{ 4 } =48days$$
A monkey climbs five metres at the beginning of each minute and then slips back two metres in the next minute before he again starts climbing If he starts his ascent at 8.00 am at what time will he first touch a flag 17 metres from the ground?
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0%
8:10 am
0%
8:06 am
0%
8:09 am
0%
8:12 am
Explanation
Clearly the monkey ascends 3 metres in every two minutes and 12 metres in 8 minutes Now the remaining 5 metres he will ascend in one minute Thus it takes 9 minutes to touch the flag and so he touches the flag at 8.09 am.
What is date of birth of Sahil?
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$$15$$
0%
$$16$$
0%
$$17$$
0%
$$18$$
Explanation
30th November is Sunday. Second Sunday of November is 9th November. Sahil's birthday is 7 days after the anniversary of his parents. So his birthday falls on 16th November.
The arithmetic mean of $$12$$ and $$20$$ is :
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0%
$$12$$
0%
$$14$$
0%
$$16$$
0%
$$18$$
Explanation
The arithmetic mean of two numbers is their average.
So, the AM of $$12$$ and $$20$$ is $$ \dfrac { 12+ 20 }{ 2 }= 16$$
Mean of $$14, 17, 11, 13, 26, 21, 31$$, and $$19 $$
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0%
18
0%
19
0%
17
0%
None of these
Explanation
Sum of number $$\displaystyle =152$$
Required mean $$\displaystyle =152/8=19$$
By looking in a mirror, it appears that it is 6 : 30 in the clock What is the real time ?
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6 : 30
0%
5 : 30
0%
6 : 00
0%
4 : 30
Explanation
Clearly, fig (A) shows the time ( 6 : 30 ) in the clock as it appears in a mirror Then its mirror-image i.e. Fig ( B ) shows the actual time in the clock i.e. 5:30. You can solve it quickly if you remember that the sum of actual time and image time is always 12 hours.
Arrange the given words in the sequence in which they occur in the dictionary and then choose the correct sequence.
PagePaganPalisadePageant
Palate
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$$1,4,2,3,5$$
0%
$$2,4,1,3,5$$
0%
$$2,1,4,5,3$$
0%
$$1,4,2,5,3$$
Explanation
Words 1,2 and 4 have first 3 letter as "pag" and words "3 and 5 have first 3 letters as "pal". So words starting with "pal" will come later than words starting with "pag" as in Alphabet set the letter "g" comes after letter "a".
The fourth letter of word 5 is "a" and of word 3 is "i" hence word 3 will be the last in order.
So choice A and B are incorrect and the answer will be either C or D.
Simply arranging the 4th letter of words 1,2 and 4 will give the correct sequence choice as C.
The weights of 9 apples are 50,60,65,62,67,70,64,45,48 grams Their mean weight is
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0%
$$60.5\ gram$$
0%
$$60\ gram$$
0%
$$59\ gram$$
0%
$$62\ gram$$
Explanation
Mean =$$\dfrac{Total\ weight\ of\ 9\ apples}{Total\ no\ of\ apple}$$
$$\Rightarrow \dfrac{50+60+65+62+67+70+64+45+48}{9}$$
$$\Rightarrow \dfrac{531}{9}=59$$
What is the correct time between 3 and 4 a.m. when the two hands of a clock point in opposite directions?
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0%
$$\displaystyle 3-49\frac{1}{11}$$
0%
$$ \displaystyle 3-49\frac{2}{11}$$
0%
$$ \displaystyle 3-48\frac{2}{11}$$
0%
3-50'
Explanation
45
×
12
11
=
540
11
=
49
1
11
45×1211=54011=49111
∴
∴
Correct time =
3
−
49
1
11
If a regular square pyramid has a base of side $$8 cm$$ and height of $$30 cm$$, then its volume is
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0%
$$120 cm^3.$$
0%
$$240 cm^3.$$
0%
$$640 cm^3.$$
0%
$$900 cm^3.$$
Explanation
Given that:
Side $$a=8\ cm$$, Height $$h=30\ cm$$
As we know that
Volume of regular square pyramid
$$\Rightarrow a^2\dfrac{h}{3}$$
$$\Rightarrow 8^2\dfrac{30}{3}$$
$$\Rightarrow 64\times 10$$
$$\Rightarrow 640\ cm^3$$
This is the required solution.
The mean of the factors of 24 is;
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0%
$$\displaystyle \frac{12}{5}$$
0%
$$\displaystyle \frac{9}{5}$$
0%
$$\displaystyle \frac{15}{2}$$
0%
$$\displaystyle \frac{17}{5}$$
Explanation
The factor of 24 are 1,2,3,4,6,8,12,24
$$\therefore A.M=\dfrac{\sum x}{n}$$
$$\Rightarrow \dfrac{1+2+3+4+6+8+12+24}{8}$$
$$\Rightarrow \dfrac{60}{8}=\dfrac{15}{2}$$
The mean of first 8 natural numbers is
Report Question
0%
$$4.5$$
0%
$$5$$
0%
$$4$$
0%
$$5.5$$
Explanation
$$\textbf{Step 1: Finding the sum of observations and number of observations}$$
$$\text{The first 8 natural numbers are }1,\;2,\;3,\;4,\;5,\;6,\;7,\;8$$
$$\text{Sum of observations}=1+2+3+4+5+6+7+8=36$$
$$\text{Number of observations}=8$$
$$\textbf{Step 2: Finding the mean}$$
$$\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}$$
$$\text{Mean}=\dfrac{36}{8}=4.5$$
$$\textbf{Thus, the mean of the first 8 natural numbers is 4.5}$$
42 men take 25 days to dig a pond. If the pond would have to be dug in 14 days, then what is the number of men to be employed?
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0%
67
0%
75
0%
81
0%
84
Explanation
In 25 days, 42 men dig a pond.
$$\therefore$$ In 14 days, $$\displaystyle\frac{42\times20}{14}=75\:men$$ dig a pond.
At what time in minutes between 3 o'clock and 4 o'clock both the needles will coincide each other?
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0%
$$ \displaystyle 5\frac{1"}{11}$$
0%
$$ \displaystyle 12\frac{4"}{11}$$
0%
$$ \displaystyle 13\frac{4"}{11}$$
0%
$$ \displaystyle 16\frac{4"}{11}$$
Explanation
At 3 o'clock the minute hand is 15 minutes spaces a part from the hour hand
To be coincident it must gain 15 minutes spaces Hence 55 minutes are gained in 60 minutes Hence 15 minutes are gained in
60
55
×
15
=
16
4
11
6055×15=16411
min
Hence hands are coincidentat
16
4
11
16411
minutes past 3
12 men complete a work in 20 days If only 8 men are emplited, then the time required to complete the same work is
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0%
24days
0%
25days
0%
30days
0%
35days
Explanation
12 men completed the work in 20 days
$$\therefore$$ One man can complete the same work in $$12\times20$$ days
So 8 men will complete the same work in $$\dfrac{12\times20}{8}=30\;\; days$$
A ball is dropped from a height 64 m above the ground, and every time it hits the ground it rises to a height equal to half of the previous. What is the height attained after it hits the ground for the 16th time?
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0%
$$2^{-12}\:m$$
0%
$$2^{-11}\:m$$
0%
$$2^{-10}\:m$$
0%
$$2^{-9}\:m$$
Explanation
Since height attained after every hit is $$\displaystyle\frac{1}{2}$$ of the previous one so the height attained after hitting the ground for the 16th time
$$\displaystyle=64\times\left(\frac{1}{2}\right)^{16}=2^6\times2^{-16}=2^{-10}$$.
The mean of the following natural numbers $$1, 2,3 ...... 10$$ is
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0%
$$6.5$$
0%
$$4.5$$
0%
$$5.5$$
0%
$$5.4$$
Explanation
Mean $$ = \dfrac {1 + 2 + 3 + 4 + 5 +6 + 7 + 8 + 9 + 10}{10} = 5.5 $$
A clocks takes six seconds to strike four how long it will take to strike ten?
Report Question
0%
15 sec
0%
16 sec
0%
20 sec
0%
18 sec
Explanation
When the clock strikes four there are three
intervals between So three interval take 6
seconds and each interval takes 2 seconds
Now when it strikes 10 there are 9 intervals
and it takes 18 sec
If $$35$$ men can reap a field in $$8$$ days; in how many days can $$20$$ men reap the same field ?
Report Question
0%
$$14$$
0%
$$10$$
0%
$$7$$
0%
$$20$$
Explanation
Since Total work done is same.
$$35 \times 8=20 \times x$$
$$\Longrightarrow x=\dfrac{35 \times 8}{20}$$
$$\Longrightarrow x=14$$ days
A can do a piece of work in 40 days. He worked at it for 5 days, then B finished it in 21 days. The number of days that A and B take together to finish the work are
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0%
15 days
0%
14 days
0%
13 days
0%
10 days
Explanation
A csn do a piece of work in 40 days
Work of A in 1 day=1/40
Work of A in 5 days=5/40=1/8
1/8th part of work is done by A
Remaining work=$$\left( 1-\frac { 1 }{ 8 } \right) =\frac { 7 }{ 8 } $$
7/8th part is done by B in 21 days
Total work will be done by B in $$=\frac { 21 }{ \frac { 7 }{ 8 } } =\frac { 21 }{ 7 } \times 8=24days$$
Work done by B in one day=1/24
Work done by A and B together in one day will be$$\left( \frac { 1 }{ 40 } +\frac { 1 }{ 24 } \right) $$
$$\left( \frac { 3+5 }{ 120 } \right) =\frac { 8 }{ 120 } =\frac { 1 }{ 15 } $$
A and B will take 15 days to complete the work together.
Out of 100 students 50 fail in English and 30 in Maths. If 12 students fail in both English and Maths, then the number of students passing both the subjects is
Report Question
0%
26
0%
28
0%
30
0%
32
Explanation
Total number of students=100
Number of students fail in English=(50-12)=38
Number of students fail in Maths=(30-12)=18
Number of students fail in both=12
Therefore total failing students=(38+18+12)=68
Pass in both the subjects=(100-68)
=32 students
32 students have passed in both subjects
At the end of a business conference, the ten people present all shake hands with each other once. How many handshakes will there be altogether?
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0%
20
0%
45
0%
55
0%
90
Explanation
Total number of handshakes
=
10
×
9
2
=
45
\dAXisplaystyle=10×92=45$$
In 8 March, 1988, which is your date of birth, is a Monday, on what day of the week will your birthday fall in the year 1989?
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0%
Tuesday
0%
Sunday
0%
Monday
0%
Friday
Explanation
Since, 8 March, 1988 comes after 29 February the number of odd days between 8 March 1988 and 8 March 1989 is only one. Hence, in 1989 your birthday will be one day after Monday, i.e., on Tuesday.
A group of 1200 persons consisting of captains and soldiers is travelling in a train. For every 15 soldiers there is one captain. The number of captains in the group is:
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0%
85
0%
80
0%
75
0%
70
Explanation
Out of 16 men, there is a captain.
Number of captains in 1200 men = $$\displaystyle \\ 1200\div 16=75$$.
If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?
Report Question
0%
Sunday
0%
Saturday
0%
Tuesday
0%
Wednesday
Explanation
The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
$$\displaystyle \therefore $$
The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
$$\displaystyle \therefore $$ 6th March, 2004 is Sunday (1 day before to 6th March, 2005).
A garrison of '$$n$$' men had enough food to last for $$30$$ days. After $$10$$ days, $$50$$ more men joined them. If the food now lasted for $$16$$ days, what is the value of $$n$$?
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0%
$$200$$
0%
$$240$$
0%
$$280$$
0%
$$320$$
Explanation
After $$10$$ days, the food for n men is there for $$20$$ days.
This food can be eaten by $$(n + 50)$$ men in $$16$$ days.
$$\therefore 20n = 16(n + 50)$$
$$\therefore n = 200$$
A machine $$P$$ can print one lakh books in $$8$$ hours, machine $$Q$$ can print the same number of books in $$10$$ hours while machine $$R$$ can print them in $$12$$ hours. All the machines are started at $$9$$ A.M. while machine $$P$$ is closed at $$11$$ A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished?
Report Question
0%
$$11:30$$ A.M.
0%
$$12$$ noon
0%
$$12:30$$ P.M.
0%
$$1:00$$ P/M/
Explanation
$$(P+Q+R)$$'s $$1$$ hour's work $$=\left( \cfrac { 1 }{ 8 } +\cfrac { 1 }{ 10 } +\cfrac { 1 }{ 12 } \right) =\cfrac { 37 }{ 120 } $$
Work done by $$P.Q$$ and $$R$$ in $$2$$ hours $$=\left( \cfrac { 37 }{ 120 } \times 2 \right) =\cfrac { 37 }{ 60 } $$
$$(Q+R)$$'s $$1$$ hour's work$$=\left( \cfrac { 1 }{ 10 } +\cfrac { 1 }{ 12 } \right) =\cfrac { 11 }{ 60 } $$.
Now, $$\cfrac{11}{60}$$ work done is done by $$Q$$ and $$R$$ in $$1$$ hour.
So, $$\cfrac{23}{60}$$ work will be done by $$Q$$ and $$R$$ in $$\left( \cfrac { 60 }{ 11 } \times \cfrac { 23 }{ 60 } \right) =\cfrac { 23 }{ 11 }$$ hours $$ \approx 2$$ hours.
So, the work will be finished approximately $$2$$ hours after $$11$$ A.M i.e, around $$1$$ P.M.
Tara's three bowling scores in a tournament were $$167, 178$$, and $$186$$. What was her average score for the tournament?
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0%
$$176$$
0%
$$177$$
0%
$$178$$
0%
$$179$$
0%
$$180$$
Explanation
Three bowling scores of tournament are $$167, 178$$ and $$186$$.
Average score will be $$=\dfrac { 167+178+186 }{ 3 } =177$$
$$A$$ can finish a work in $$18$$ days and $$B$$ can do the same work in $$15$$ days. $$B$$ worked for $$10$$ days and left the job. In how many days, $$A$$ alone can finish the remaining work?
Report Question
0%
$$5$$
0%
$$5\cfrac{1}{2}$$
0%
$$6$$
0%
$$8$$
Explanation
$${\textbf{Step 1 : Write the values given in the question}}{\textbf{.}}$$
$$A{\text{ can finish a work in 18 days}}{\text{.}}$$
$$B{\text{ can do the work in 15 days}}{\text{.}}$$
$$B{\text{ worked for 10 days and left the job}}{\text{.}}$$
$${\textbf{Step 2 : Apply mathematics word problem rule}}{\textbf{.}}$$
$$B's{\text{ 10 day's work = }}\left( {\dfrac{1}{{15}} \times 10} \right) = \dfrac{2}{3}$$
$${\text{Remaining work = }}\left( {1 - \dfrac{2}{3}} \right) = \dfrac{1}{3}$$
$${\text{Now, }}\dfrac{1}{{18}}{\text{ work done by }}A{\text{ in 1 day}}{\text{.}}$$
$$\therefore \dfrac{1}{3}{\text{ work is done by }}A{\text{ in }}\left( {18 \times \dfrac{1}{3}} \right) = 6{\text{ days}}$$
$${\textbf{Hence, }}\mathbf{A}{\textbf{ can remaining work in 6 days}}{\textbf{.}}$$
It was Sunday on Jan 1,What was the day of the week Jan 1, 2010?
Report Question
0%
Sunday
0%
Saturday
0%
Friday
0%
Wednesday
Explanation
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴
∴
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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