MCQ Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 4

If base and height of a prism and pyramid are same, then the volume of a pyramid is:

0%
$\dfrac{1}{3}\times$ Volume of prism
0%
${3}\ \times$ Volume of prism
0%
$\dfrac{1}{2}\times$ Volume of prism
0%
$2\ \times$ Volume of prism

Explanation

If a pyramid and a prism have the same base and height, their volumes are always in the ratio of

$1:3\times$ Volume of prism.

General formula of volume of a prism is:

0%
Area of base $\times$ height
0%
Area of triangle $\times$ height
0%
Area of square $\times$ height
0%
Area of rectangle $\times$ height

Explanation

The volume of a general 3d figure is simply = Area of Base

$\times$ Its height.

This applies for prism also.

$8$ taps of the same size fill a tank in

$27$ minutes. If two taps go out of order, how long would the remaining taps take to fill the tank?

0%
$86$ minutes
0%
$56$ minutes
0%
$66$ minutes
0%
$36$ minutes

Explanation

Given that,

$8$ taps of the same size fill a tank in

$27$ minutes.

So,

$1$ tap can fill the tank in

$8 \times 27 = 216$ minutes

If two taps go out of order, then the remaining taps

$= 8 - 2 = 6$

Therefore, Time taken by

$6$ taps to fill the tank

$= \dfrac{216}{6} = 36$ minutes.

The mean of the following data

$12, 18, 15, 17, 11, 10$ is

$13.83$ . Is it true or false?

0%
True
0%
False

Explanation

Total given numbers

$=6$

$Mean = \cfrac{12+18+15+17+11+10}{6}$

$\implies \cfrac{83}{6}$

$\implies 13.833$

Answer: True

$3$ letters are posted in

$5$ letters boxes. If all the letters are not posted in the same box, then number of ways of posting is

0%
$120$
0%
$125$
0%
$130$
0%
$124$

Explanation

According to problem,

$3$ letters are posted

$5$ boxes

implies that,

we have a letter and can be posted in any of the

$5$ boxes.

Similarly next letter can be posted in 5 ways, and the all other follow same method

implies that

${\left(5\right)}^3$

$= 5 \times 5 \times 5$

Hence,

$125$ possible ways for posting in

$5$ boxes.

A _____ is an arrangement of all or part of set of object in a definite order.

0%
permutation
0%
function
0%
combination
0%
factorial

$85$ L of a mixture contains milk and water in the ratio

$27: 7$ How much more water is to be added to get a new mixture containing milk and water in the ratio

$3:1$ ?

0%
5 $\mathrm { L }$
0%
6.5 $\mathrm { L }$
0%
7.25 $\mathrm { L }$
0%
8 $L$

Explanation

$Milk:Water=27:7$

$Milk=27x$

$Water=7x$

$27x+7x=34x=85l$

$x=\dfrac{85}{34}$

$Milk=27x=\dfrac{27\times 85}{34}=67.5l$

$Water=7x=\dfrac{7\times 85}{34}=17.5l$

New ratio

$=3:1$

Let k l of water be added to the mixture.

$\dfrac{3}{1}=\dfrac{67.5}{17.5+k}$

$52.5+3k=67.5$

$3k=67.5-52.5$

$3k=15$

$k=5l$

Kaushlaya can do a work in

$20$ days, while Kaikeyi can do the same work in

$25$ days. They started the work jointly. A few days later Sumitra joined them and thus all of them completed the whole work in

$10$ days. All of them were paid a total

$Rs. 700$ . What is the share of Sumitra?

0%
$Rs. 130$
0%
$Rs. 185$
0%
$Rs. 70$
0%
Can't be determined

Explanation

Efficiency of Kaushalya

$= 5\%$

Efficiency of Kaikeyi

$= 4\%$

Thus, in

$10\ days$ working together they will complete only

$90\%$ of the work.

$[(5+4)\times 10] =90$

Hence, the remaining work will surely done by Sumitra, which is

$10\%$ .

Thus, Sumitra will get

$10\%$ of

$Rs.\ 700$ , which is

$Rs.\ 70$ .

Hence, this is the answer.

$\left( {n + 1} \right)! = 12 \times (n - 1)!\;then\;n =$

0%
3
0%
4
0%
2
0%
5

Explanation

We have,

$\left( {n + 1} \right)! = 12 \times (n - 1)!$

$(n + 1)\times n\times (n-1)! = 12 \times (n - 1)!$

$(n + 1)\times n = 12$

$n^2 +n- 12=0$

$n^2 +4n-3n- 12=0$

$(n+4)(n-3)=0$

$(n+4)\ne 0$

So,

$n-3=0$

$n=3$

Hence, this is the answer.

Find the mean of

$16,29,60,18,27$

0%
30
0%
40
0%
50
0%
10

Explanation

The given data is

$16,29,60,18,27$

The sum of observations is

$16+29+60+18+27=150$

The mean is given as

$\dfrac{150}{5}=30$

The mean of

$6,8,10,12,14,16,18$

0%
$12$
0%
$10$
0%
$9$
0%
$8$

Explanation

Given data

$6,8,10,12,14,16,18$

The sum of observations is

$6+8+10+12+14+16+18=84$

Mean

$= \dfrac{\text{Sum of given observations}}{\text{Total number of observations}}$

Total number of observations = 7

Mean =

$\dfrac {84}7=12$

The mean of

$33,53,32,35,47$ is

0%
$40$
0%
$56$
0%
$55$
0%
$6^{6}-6!$

Explanation

Given observations

$33,53,32,35,47$

no. of observations is

$5$

Sum of observations is

$33+53+32+35+47=200$

Mean is

$\dfrac{200}{5}=40$

Find the mean of

$22,16,19,12,26,32,87,58$

0%
34
0%
32
0%
43
0%
none of these

Explanation

given observations

$22,16,19,12,26,32,87,58$

No .of observations

$8$

Sum f observations

$22+16+19+12+26+32+87+58=272$

Mean is given as

$\dfrac{272}8=34$

If P (n, n) denotes the number of permutations of n different things taken all at a time then P (n, n ) is also identical to:

0%
$P ( n - 1 , n - 1 )$
0%
$P ( n , n - 1 )$
0%
$\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )$
0%
$( n - r ) \cdot P ( n , r )$

What was the day of the week on,

${16}^{th}$ July,

$1776$ ?

0%
Monday
0%
Tuesday
0%
Saturday
0%
Wednesday

Explanation

${16}^{th}\, July, \,1776 =\left(1775 \,years +\, Period \,from \,{1}^{st}\, January\, 1776\, to \,{16}^{th}\, July,\, 1776\right)$

Counting of odd days :

$1600$ years have

$0$ odd day.

$100$ years have

$5$ odd days.

$75$ years

$=\left(18 \,leap \,years + 57 \,ordinary\, years\right)$

$=\left[\left(18 \times 2\right)+\left(57 \times 1\right)\right]$ odd days

$= 93$ odd days

$=\left(13\, weeks + 2\, days\right) = 2$ odd days.

$1775$ years have

$\left(0 + 5 + 2\right)$ odd days

$= 7$ odd days

$= 0$ odd day.

$Jan+Feb+March+April+May+June+July$

$31 + 29 + 31 + 30 + 31 + 30 +16 = 198$ days

$=\left(28\, weeks + 2\, days\right) =2$ days

Total number of odd days

$=\left(0 + 2\right) = 2$ .

Required day was 'Tuesday'.

$\therefore\,$ Option

$(b)$ is correct

The mean of

$25,37,84$ is

0%
$25$
0%
$37$
0%
$84$
0%
None.

Explanation

Given observations

$25,37,84$

No .of observations

$3$

The mean of observation

$\dfrac{25+37+84}{3}=\dfrac {146}3=48.6$

A alone can do a piece in

$10$ days and

$B$ alone can so it in

$15$ days. In how many days will

$A$ and

$B$ together do the same work?

0%
$5$ days
0%
$6$ days
0%
$8$ days
0%
$9$ days

Explanation

We know that,
Number of days

$A$ required to do piece of work

$=10$ days

Number of days

$B$ required to do piece of work

$=15$ days

$\therefore$ We can calculate, work done

$A$ in

$1$ day

$=1/10$

Work done by

$B$ in

$1$ days

$=1/15$

Now, work done by both

$A$ and

$B$ together in

$1$ day

$=1/10+1/15=10/60=1/6$

$\therefore A$ and

$B$ can do work together in

$6$ days

The rates of working of

$A$ and

$B$ are in ratio

$3:4$ . The number of days taken by them to finish the work are in the ratio

0%
$3:4$
0%
$9:16$
0%
$4:3$
0%
$16:9$

Explanation

We know that the number days taken to work is the reciprocal of the rate of the work i.e they are inversely proportional.

So, we calculate the work done by

$A=1/3$

And the work done by

$B=\dfrac{1}{4}$

$\therefore$ ratio of the work done

$=\dfrac{1}{3}:\dfrac{1}{4}$ i.e.

$4:3$

In a

$800$ m race,

$A$ defeated

$B$ by

$15$ seconds. If

$A$ 's speed was

$8$ km/hour. What was

$B$ 's speed?

0%
$\displaystyle 7\dfrac {17}{25}$ km/hour
0%
$\displaystyle 2\frac {5}{18}$ km/hour
0%
$\displaystyle 2\frac {5}{21}$ km/hour
0%
$\displaystyle 7\frac {18}{25}$ km/hour

Explanation

Given, Distance (

$D$ )

$=$

$800$ m, Speed of

$A =$

$V_{A} = 8$

$km/hour$ and speed of

$B =$

$V_{B}$

Time taken by

$A$ to finish the race

$= \dfrac {0.8}{8}= 0.1$ hours

$= 6$ minutes

$=360$ seconds.

Time taken by

$B$

$= 360+15= 375$ seconds

Speed of

$B$

$=\dfrac{800}{375}$

$m/s$ or

$\dfrac{192}{25}$

$km/hour$ ....

$[1m/s = \dfrac{18}{5}$

$km/hour]$

$A$ can do a job in

$16$ days and

$B$ can do the same job in

$12$ days With the help of

$C$ , they can finish the job in

$6$ days only. Then,

$C$ alone can finish it in

0%
$34$ days
0%
$22$ days
0%
$36$ days
0%
$48$ days

Explanation

We know that,
Number of days

$A$ requires to do piece of work

$=16$ days

Number of days

$B$ requires to do piece of work

$=12$ days

Let us consider number of days

$C$ requires to do piece of work

$=x$ days

Number of days

$A, B$ and

$C$ together requires to do the piece of work

$=6$ days

$\therefore$ We can calculate, work done by

$A$ in

$1$ day

$=1/16$
Work done by

$B$ in

$1$ day

$=1/12$
Work done by

$C$ in

$1$ day

$=1/x$

Work done by

$A, B$ and

$C$ together in a

$1$ day

$=1/6$

Now, work done by

$A, B$ and

$C$ together in

$1$ day is also equal to the sum of work done by each of them in

$1$ day

$=1/16+1/12+1/x=7/48+1/x$

Therefore,

$7/48+1/x=1/6$

$(7x+48)/48x=1/6$

$42x+288=48x$

$6x=288$

$x=288/6=48$

$\therefore C$ can do the work in

$48$ days.

Mean can never be a fraction.

0%
True
0%
False

A and B can complete work together in 5 days. If A works at twice his speed and B at half of his speed this works can be finished in 4 days. How many days would it take for A alone to complete the job ?

0%
$10.0$
0%
$12.0$
0%
$15.0$
0%
$18.0$

Explanation

Let

$A$ 's one day work be

$\dfrac {1}{x}$ and

$B$ 's one day work

$\dfrac {1}{y}$ .

It is given that

$A$ and

$B$ can complete work together in

$5$ days, therefore, we have:

$A+B=\dfrac { 1 }{ x } +\dfrac { 1 }{ y } =\dfrac { 1 }{ 5 } \\ \Rightarrow \dfrac { y }{ xy } +\dfrac { x }{ xy } =\dfrac { 1 }{ 5 } \\ \Rightarrow \dfrac { y+x }{ xy } =\dfrac { 1 }{ 5 } \\ \Rightarrow 5(x+y)=xy\quad .....(1)$

$A$ worked at twice his speed means

$\dfrac {2}{x}$

$B$ worked at half of his speed means

$\dfrac {1}{2y}$

Together they complete work in

$4$ days, therefore, we have:

$\dfrac { 2 }{ x } +\dfrac { 1 }{ 2y } =\dfrac { 1 }{ 4 } \\ \Rightarrow \dfrac { 4y }{ 2xy } +\dfrac { x }{ 2xy } =\dfrac { 1 }{ 4 } \\ \Rightarrow \dfrac { 4y+x }{ 2xy } =\dfrac { 1 }{ 4 } \\ \Rightarrow 4(x+4y)=2xy\\ \Rightarrow 2(x+4y)=xy\\ \Rightarrow 2x+8y=xy\quad .....(2)$

Substitute equation 2 in 1:

$5x+5y=2x+8y\\ \Rightarrow 5x-2x=8y-5y\\ \Rightarrow 3x=3y\\ \Rightarrow x=y$

$x=y$ means that number of days worked by

$A$ alone is equal to

$B$ alone.

Now, substitute the value in equation 1:

$\dfrac { 1 }{ x } +\dfrac { 1 }{ x } =\dfrac { 1 }{ 5 } \\ \Rightarrow \dfrac { 1+1 }{ x } =\dfrac { 1 }{ 5 } \\ \Rightarrow \dfrac { 2 }{ x } =\dfrac { 1 }{ 5 } \\ \Rightarrow x=5\times 2\\ \Rightarrow x=10$

Hence, it would take

$10$ days for

$A$ alone to complete the job.

State true or false.

A.M. of any

$n$ positive numbers

$a_1, a_2, a_3, ...., a_n$ is

$A$ then

$A=\dfrac {a_1+a_2+....+a_n}{n}$

0%
True
0%
False

Explanation

Arithmetic mean is the average of all the numbers.

Hence for n positive numbers:

$a_1$ ,

$a_2$ ,

$a_3$ ,....

$a_n$ , the arithmetic mean is

$\dfrac{a_1+a_2+a_3+...+a_n}{n}$ .

Say true or false.

The A.M. between

$(a-b)^2$ and

$(a+b)^2$ is

$a^2+b^2$ .

0%
True
0%
False

Explanation

If the two terms are

$x$ and

$y$ , then their AM is

$\dfrac{x+y}{2}$
AM between

$(a-b)^2$ and

$(a+b)^2$

$=$

$\dfrac{(a-b)^2 + (a+b)^2}{2}$

$=$

$\dfrac{(a^2 +b^2 - 2ab + a^2+b^2 + 2ab)}{2}$

$=$

$a^2 + b^2$

Given:

$\dfrac {20!}{18!}=380$

0%
True
0%
False
0%
Either
0%
Neither

Explanation

$\cfrac { 20! }{ 18! } =?\\ 20!=20\times 19\times 18\times ........3\times 2\times 1\\ =20\times 19\times (18!)\\ \cfrac { 20! }{ 18! } =\cfrac { 20\times 19\times (18!) }{ 18! } =20\times 19=380$

Hence the value matches so the equation is true.

$20$ persons are invited for a party, then the number of ways in which they and the host be seated at a round table is

0%
20!
0%
21!
0%
22!
0%
2.20!

Explanation

Total no. of persons including host are

$(20+1)$
As they have to sit in round table, the required permutation is

$(n-1)!$

$=(20+1-1)!$

$=20!$

The mean of five numbers is

$30$ . If one number is excluded, their mean becomes

$28$ . The excluded number is:

0%
$28$
0%
$30$
0%
$35$
0%
$38$

Explanation

The mean of

$5$ numbers

$=30$
The sum of

$5$ numbers is

$=30\times5 = 150$

After excluding a number say,

$x$
Mean

$= 28$
Sum of four numbers

$= 28\times4 = 112$

$x = 150 -112 = 38$

$5$ workers can build a fence in

$65$ days, how long would it take

$13$ workers to build the same fence?

0%
$15$ days
0%
$20$ days
0%
$25$ days
0%
$35$ days

Explanation

More the workers, lesser will be the time it will take to build the fence.

Let

$x$ be the number of days required to build the fence.
In the equation form, we get

$5\times 65 = 13\times x$

$\therefore x = \cfrac{5\times 65}{13}$

$= 5\times 5 = 25$

Carpenter

$A$ can make a chair in

$6$ hours, carpenter

$B$ in

$7$ hours and carpenter

$C$ in

$8$ hours. If each carpenter works for

$8$ hours per day, how many chairs will be made in

$21$ day?

0%
$61$
0%
$67$
0%
$73$
0%
$79$

Explanation

$6$ hours, the number of chairs made by carpenter

$A$ is one.

Therefore, in

$(21\times 8)$ hours, number of chairs made by

$A$ is:

$\dfrac { 21\times 8 }{ 6 } =\dfrac { 168 }{ 6 } =28$

Similarly, number of chairs made by

$B$ is:

$\dfrac { 21\times 8 }{ 7 } =\dfrac { 168 }{ 7 } =24$

And, number of chairs made by

$C$ is:

$\dfrac { 21\times 8 }{ 8 } =\dfrac { 168 }{ 8 } =21$

thus, the total number of chairs made in

$21$ days are

$28+24+21=73$

Hence,

$73$ chairs made in

$21$ days.

The mean of prime numbers between 20 and 30 is :

0%
21
0%
26
0%
25
0%
27

Explanation

The prime numbers between

$20$ and

$30$ are

$23,29$

Mean of the data set is the average of values in the data set.

Therefore the mean of the prime numbers is

$\dfrac{23+29}{2}=\dfrac{52}{2}=26$

There are

$50$ numbers. Each number is subtracted from

$53$ and the mean of the numbers so obtained is found to be

$3.5$ . The mean of the given numbers is:

0%
$46.5$
0%
$49.5$
0%
$53.5$
0%
$56.5$

Explanation

Total numbers

$= 50$
Mean of numbers after subtracting

$53$ from each

$= 3.5$
Sum of numbers after subtracting

$53$ from each

$=3.5\times50 = 175$
Sum of the original numbers

$=175 + 53\times50 = 2825$

Mean of the original numbers

$=\dfrac{2825}{50} = 56.5$

If the volume of a prism is

$1920$

$\sqrt{3} cm^3$ and the side of the equilateral base is

$16$

$cm$ , then the height (in cm) of the prism is?

0%
$19$
0%
$20$
0%
$30$
0%
$40$

Explanation

Volume of prism

$=$ Area of equilateral triangle

$\times$ height

Now, Area of triangle

$= \dfrac{\sqrt{3}}{4}a^2$

$\dfrac{\sqrt{3}}{4} 16^2$ =

$64 \sqrt{3}$

$1920 \sqrt{3} = 64\sqrt{3}\times$ height

$\therefore$ height

$=$

$\displaystyle \dfrac{1920}{64}$

$= 30 \ cm$

$15$ labourers can fill

$35$ boxes in

$7$ days. How many labourers can fill

$65$ boxes in

$5$ days ?

0%
$13$
0%
$39$
0%
$45$
0%
$55$

Ten pipes of the same type can fill up a tank in

$16$ minutes. If two pipes go out of order, how long will the remaining pipes take to fill the tank ?

0%
$18$ mins
0%
$20$ mins
0%
$22$ mins
0%
$24$ mins

Explanation

$x$ is the required number of hour then,
In order to solve this problem, first, we need to find out whether the proportionality is direct or inverse.
Now, here we note that:
For more number of pipes, the time required to fill is less.
For less number of pipes, the time required to fill is more.
Hence, there is an inverse proportionality.

$10\times 16 = (10-2)\times x$
where, x is the time required.

$\Rightarrow 10\times 16 = 8\times x$

$\Rightarrow x = \cfrac{10\times 16}{8}$

$= 10\times 2$

$=20$ mins.

At what time are the hands of a clock together between 5 and 6?

0%
$33\displaystyle \frac{3}{11}$ min. past $5$
0%
$28\displaystyle \frac{3}{11}$ min. past $5$
0%
$30$ min. past $5$
0%
$27\displaystyle \frac{3}{11}$ min. past $5$

Explanation

Minute hand moves

$6^0$ in one minute.

Hour hand moves

$\dfrac12^0$ in one minute.

Total degree moved by hour hand in

$x$ minutes past

$5\>$ O'clock is

$150^0 + \dfrac x2^0$

Total degree moved by minute hand in

$x$ minute is

$6x^0$

Now both have to coincide, that only happens when,

$150^0+\dfrac x2^0 = 6x^0$

$\Rightarrow x = \dfrac {300}{11} = 27\dfrac3{11}$ min. past

$5\>$ O'clock

Hence, option D is the correct answer.

Kate was born on Saturday, 22nd MarchOn what day of the week was she 14 years 7 month and 8 days of age ?

0%
Sunday
0%
Tuesday
0%
Wednesday
0%
Monday

Explanation

One normal year has

$365$ Days i.e

$52$ weeks and

$1$ day extra.

So, if we go

$14$ years ahead from 1982, we will reach to 1996. In this duration, we will have

$4$ leaps years and a leap year gives us

$2$ Extra days.

So, calculating the extra days:

$10 \times 1 + 4 \times 2 = 18$ and

$18$ mod

$7$

$= 4$

So, on 22nd March 1986, we will get a (Saturday + 4 days) Wednesday,

Plus additional

$7$ months will constitute:

22 nd April

$\rightarrow$

$31$ days

22 nd May

$\rightarrow$

$30$ days

22 nd June

$\rightarrow$

$31$ days

22 nd July

$\rightarrow$

$30$ days

22 nd August

$\rightarrow$

$31$ days

22 nd September

$\rightarrow$

$31$ days

22 nd October

$\rightarrow$

$30$ days

And 8 days will result in

$222$ days,

So,

$222$ mod

$7$

$= 5$

which is

$31$ weeks and

$5$ Days (Wednesday +5 ) will result in

$Monday$

The attendance of a class of

$45$ boys for

$10$ days is given as

$40, 42, 30, 35, 45, 44, 41, 38, 44$ and

$41$ , then the mean attendance of a class is

0%
$39$
0%
$40$
0%
$41$
0%
$43$

Explanation

Mean attendance of the class for 10 days

$= \cfrac{\text{Total attendance}}{\text{Number of days}}$

$=\cfrac{40 + 42 + 30 + 35 + 45 + 44+ 41 + 38+ 44 +41}{10}$

$=\cfrac{400}{10}$

$= 10$

If a regular square pyramid has a base of side 8 cm and height of 30 cm, then its volume is

0%
120 c.c.
0%
240 c.c.
0%
640 c.c.
0%
900 c.c.

Explanation

Side of square base

$=$ 8 cm.
Height of pyramid

$=$ 30 cm

$\therefore$ Volume of square pyramid

$= \frac{1}{3}\times$ area of base

$\times$ height

$= \frac{1}{3}\times 82\times 30 = 640$ c.c.

In U.P. on 17th October, 1996 President's rule was declared. Find the day of week on that date.

0%
Tuesday
0%
Friday
0%
Wednesday
0%
Thursday

Explanation

We use the concept of odd days, for such type of questions.

The number of days more than the complete weeks are called odd days.

$1$ ordinary year

$=365$ days

$= ( 52$ weeks

$+ 1$ odd day

$)$

$1$ leap year

$= 366$ days

$= (52$ weeks

$+ 2$ odd days

$)$

$100$ years

$= 76$ ordinary years

$+ 24$ leap years

$= 76(1)+24(2) = 124$ odd days

$= 17$ weeks

$+ 5$ odd days

$= 5$ odd days

Number of odd days in

$200$ years

$=5 \times 2=3$ odd days

Number of odd days in

$300$ years

$= 5 \times 3 = 1$ odd day

Number of odd days in

$400$ years

$= 5 \times 4+1= 0$ odd days

If number of odd days come as

$0$ , that means given date is on a Sunday,

$1$ for Monday ,

$2$ for Tuesday,

$3$ for Wednesday,

$4$ for Thursday,

$5$ for Friday and

$6$ for Saturday.

$17$ oct.

$1996$

$=$

$1995$ years

$+ 9$ months

$+ 17$ days
No. of odd days in

$1600$ years

$= 0$
No. of odd days in

$300$ years

$= 1$
No. of odd days in

$95$ years

$= 6$

$\therefore$ Total no. of odd days upto

$31$ st Dec

$'95$

$= 7 = 0$ odd days

Number of odd days in

$1996$ till

$17$ Oct

$= 31+29+31+30+31+30+31+31+30+17=291$ days

$=41$ weeks

$+4$ odd days

$=4$ odd days

$\Rightarrow 17$ Oct

$1996$ is a Thursday.

$3$ persons weave

$168$ shawls in

$14$ days, how many shawls will 8 people weave in

$5$ days?

0%
$90$
0%
$105$
0%
$115$
0%
$160$

Explanation

Set the proportion for number of persons with the number of shawls. Both the quantities are in direct ratio.

Here, we are asked to find out the number of shawls, we compare each item with the number of shawls as shown below :
More number of persons

$\Rightarrow$ More shawls
Less number of persons

$\Rightarrow$ Less shawls
i.e. the relationship is direct.
Also, More number of days

$\Rightarrow$ More shawls
Less number of days

$\Rightarrow$ Less shawls
i.e. the relationship is direct.
Persons 3 : 8 : : 168 : x
Days 14 : 5

$\therefore \dfrac{3}{8}\times \dfrac{14}{5} = \dfrac{168}{x}$

$\therefore 3\times 14\times x = 168\times 8\times 5$

$\Rightarrow x = \dfrac{168\times 8\times 5}{3\times 14} = 160\ shawls$

if x men working x hours per day can do x units of a work in x days, then y men working y hours per day would be able to complete how many units of work in y days ?

0%
$\displaystyle \frac{x^{2}}{y^{3}}$
0%
$\displaystyle \frac{x^{3}}{y^{2}}$
0%
$\displaystyle \frac{y^{2}}{x^{3}}$
0%
$\displaystyle \frac{y^{3}}{x^{2}}$

Explanation

x men working x hours in x days = x units work Then 1 men working 1 hour in 1 day =

$\displaystyle \frac{x}{x^{3}}$ units
work y men working y hours in y days =

$\displaystyle \frac{xy^{3}}{x^{3}}=\frac{y^{3}}{x^{2}}$

Find the Arithmetic Mean of the numbers 7, 6, 5, 9, 8, 0, 7

0%
5
0%
6
0%
7
0%
14

Explanation

The Arithmetic Mean of numbers is calculated by using the below formula,

A.M of numbers

$=\dfrac{\text{sum of the numbers}}{ \text{Total numbers}}$

$=\dfrac{7+6+5+3+8+6+7}{7}$

$=\dfrac{42}{7}$

$=6$

Option

$[B]$ is correct.

If 1st October is a Sunday, then 1st November will be

0%
Monday
0%
Tuesday
0%
Wednesday
0%
Thursday

How many times in a day, the two hands of a clock coincide?

0%
$11$
0%
$12$
0%
$22$
0%
$24$

Find the exact time between

$7$ am and

$8$ am when the two hands of a watch meet?

0%
$7$ hrs $35$ min
0%
$7$ hrs $36.99$ min
0%
$7$ hrs $38.18$ min
0%
$7$ hrs $42.6$ min

Explanation

The minute hand covers

$6^o$ in one minute.

The hour hand covers

$(\dfrac{1}{2})^o$ in one minute

Suppose both hands meet after

$x$ minutes

so angle made by both with the

$12$ is same

$6x=\dfrac{x}{2}+210$

$\dfrac{11x}{2}=210$

$x=38.18 min.$

If the

${14}^{th}$ day after

$5^{th}$ March is a Wednesday, then what day of the week will fall on

${10}^{th}$ December of the same year?

0%
Friday
0%
Wednesday
0%
Thursday
0%
Tuesday

Explanation

On the

${14}^{th}$ day after

$5^{th}$ March, will be

$19^{th}$ March which will be Wednesday

Now, no. of days from

$19^{th}$ March to

$10^{th}$ December.

$= 12 + 30 + 31 + 30 + 31 + 31 + 30+ 31 + 30 + 10 = 266 = 266 / 7 = 38$

Since, it is divisible by

$7$ , so, it will be a Wednesday on

$10^{th}$ December.

At what time between 9 to 10 will the hands of a watch be together?

0%
$45$ minutes past $9$
0%
$50$ minutes past $9$
0%
$49\dfrac {1}{11}$ minutes past 9
0%
$48\dfrac {2}{11}$ minutes past 9

Explanation

To be together between 9 and 10, the minute hand has to gain 45 minute spaces.
Now, 55 min spaces are gained in 60 minutes.

$\therefore 45$ min spaces are gained in

$\left(\dfrac {60}{55}\times 45\right)$ min or

$49\dfrac {1}{11}$ min
So, the hands are together at

$49\dfrac {1}{11}$ minutes past

$9$

If the day before yesterday was Thursday, when will be Sunday?

0%
Today
0%
Two days after today
0%
Tomorrow
0%
Day after tomorrow

Explanation

$\textbf{Step-1: Apply day calculation according to the language of the problem.}$

$\text{We have,}$

$\text{Day before yesterday was Thursday}$

$\text{So, today is Saturday. and tomorrow will be Sunday.}$

$\textbf{Hence, option - C is the correct answer}$

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?

0%
Sunday
0%
Monday
0%
Wednesday
0%
Friday

If Thursday was the day after the day before yesterday five days ago, then what is the least number of days ago, when Sunday was three days before the day after tomorrow?

0%
Two
0%
Three
0%
Four
0%
Five

Explanation

${\textbf{Step1: Day after the day before yesterday is yesterday}}$

${\textbf{Step2: Now,five days ago, yesterday was thursday}}$

${\textbf{Step3: So, five days ago it was friday, therefore today is wednesday}}$

${\textbf{Step4: Now,three days before the day after tomorrow is yesterday}}$

${\textbf{Step5: Now, it is on monday that we say yesterday was sunday}}$

${\textbf{Hence, 2 days ago sunday was three days befor$$e the day after tomorrow}}$

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 4 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 4 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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