ad

In this section, you can find numerous aptitude questions with answers and explanation. The quantitative aptitude questions with answers mentioned above covers various categories and extremely helpful for competitive exams. All the answers are explained in detail with very detailed answer descriptions.

The quantitative aptitude questions mentioned above also contain aptitude questions asked for various placement exams and competitive exams. These will help students who are preparing for any type of competitive examinations.

Quantities aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT),MAT, GMAT, IBPS Exam, CSAT, CLAT , Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams , Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.

Titas

2015-07-28 21:30:00

$4.$ Since we do not know at what time Shyam actually sets on his journey, we will consider any one of the time limits provided (either $12$ p.m or $2$ p.m) to be taking $x$ hours from the time he started, that is we have to consider either duration to be $x$ (or $y$ or $z$ or whatever you wish to take) hours. Now, if we consider this for $2$ p.m, then by time = distance/speed, we have

$x=\dfrac{d}{10}~~\cdots(1)$

Now, $(x-2)=\dfrac{d}{15}$

$\Rightarrow 15x-30=d\\

\Rightarrow 15x=30+d\\

\Rightarrow x=\dfrac{d+30}{15}~~\cdots(2)$

From $(1)$ and $(2)$, we have

$\dfrac{d+30}{15}=\dfrac{d}{10}\\

\Rightarrow 10d+300=15d\\

\Rightarrow 5d=300\\

\Rightarrow d=60\text{ km}$

Now,

$x-1=\dfrac{d}{10}-1\\

x-1=\dfrac{60}{10}-1=5$

[We are talking about $1$ p.m and we have chosen the time duration of up to to $2$ p.m to be $x$, so up to $1$ p.m,it is $(x-1)$ hours]

So,the time taken up to $1$ p.m is $5$ hours.

Therefore, speed = distance/time $=\dfrac{60}{5}$ km/hr

$=12$ km/hr

$x=\dfrac{d}{10}~~\cdots(1)$

Now, $(x-2)=\dfrac{d}{15}$

$\Rightarrow 15x-30=d\\

\Rightarrow 15x=30+d\\

\Rightarrow x=\dfrac{d+30}{15}~~\cdots(2)$

From $(1)$ and $(2)$, we have

$\dfrac{d+30}{15}=\dfrac{d}{10}\\

\Rightarrow 10d+300=15d\\

\Rightarrow 5d=300\\

\Rightarrow d=60\text{ km}$

Now,

$x-1=\dfrac{d}{10}-1\\

x-1=\dfrac{60}{10}-1=5$

[We are talking about $1$ p.m and we have chosen the time duration of up to to $2$ p.m to be $x$, so up to $1$ p.m,it is $(x-1)$ hours]

So,the time taken up to $1$ p.m is $5$ hours.

Therefore, speed = distance/time $=\dfrac{60}{5}$ km/hr

$=12$ km/hr

0
0

dinesh

2015-07-27 17:17:08

As many discussed on 3rd problem I would like to give my answer that it is exactly 47.5 days.

Try it. Work done by 1 person in one day is work/2400.

After every 10 days, 5 will go off. So for 1st 10 days there will be 60, then 55 and so on...

now, it follows as

10*(60+55+50+45)/2400 +40*x/2400=1. (here x is no of more days required and 1 is the total work)

solving this we get

2100/2400+40x/2400=1

40x=300

x=7.5

so the total no of days required to do the given work as per the condition is 40+x=40+7.5=47.5.....

Try it. Work done by 1 person in one day is work/2400.

After every 10 days, 5 will go off. So for 1st 10 days there will be 60, then 55 and so on...

now, it follows as

10*(60+55+50+45)/2400 +40*x/2400=1. (here x is no of more days required and 1 is the total work)

solving this we get

2100/2400+40x/2400=1

40x=300

x=7.5

so the total no of days required to do the given work as per the condition is 40+x=40+7.5=47.5.....

0
0

Pushpalatha.N

2015-12-29 08:06:22

Hi

Dear Madam / Sir,

Please tell me how to solve the problem because i am try lot difference is came

50,45,40 how to came this numbers

I am attending the exam in ESIC so please send answer

waiting for your reply

Dear Madam / Sir,

Please tell me how to solve the problem because i am try lot difference is came

50,45,40 how to came this numbers

I am attending the exam in ESIC so please send answer

waiting for your reply

1
0

sam

2016-01-07 12:16:29

60 men can build a wall in 40 days

assume 1 man does 1 unit of work in 1 day

Then the total work is 60×40=2400 units

60 men work in the fist 10 days and completes 60×10 = 600 units of work

55 men work in the next 10 days and completes 55×10=550 units of work

50 men work in the next 10 days and completes 50×10=500 units of work

45 men work in the next 10 days and completes 45×10=450 units of work

So far 600 + 550 + 500 + 450 = 2100 units of work is completed and remaining work is 2400 - 2100 = 300 units

40 men work in the next 10 days. In each day, they does 40 units of work. Therefore, additional days required = 300/40 = 7.5

Thus, total 10+10+10+10+7.5 = 47.5 days required.

assume 1 man does 1 unit of work in 1 day

Then the total work is 60×40=2400 units

60 men work in the fist 10 days and completes 60×10 = 600 units of work

55 men work in the next 10 days and completes 55×10=550 units of work

50 men work in the next 10 days and completes 50×10=500 units of work

45 men work in the next 10 days and completes 45×10=450 units of work

So far 600 + 550 + 500 + 450 = 2100 units of work is completed and remaining work is 2400 - 2100 = 300 units

40 men work in the next 10 days. In each day, they does 40 units of work. Therefore, additional days required = 300/40 = 7.5

Thus, total 10+10+10+10+7.5 = 47.5 days required.

1
0

Anusree

2015-07-13 17:50:54

Que.3

60 men can do work in 40 days

Thus for one man completes 1/2400 in one day

Every 10 days 5 men quit,thus

On the first 10 days,60 men were working

Next 10 days,55 men were working

Next 10 days, 50 men were working,

Next, 45 men and so on

ie:

60 men * 10days + 55men * 10days + 50 men * 10 days + 45men * 10days+ ... + 5men*10 days

= 600 man days + 550 man days + 500 man days + ... + 50 man-days

This follows an arithmetic progression, A.P formula to find sum of n numbers is

Sn=n(a1+aL)/2 where a1 is the first term and aL is the last or the n th term,n is the no. of terms.

Using this formula

12(600+50)/2

=650*6

=3900 man-days

The total amount of work done before there are zero workers is 3900/2400.

60 men can do work in 40 days

Thus for one man completes 1/2400 in one day

Every 10 days 5 men quit,thus

On the first 10 days,60 men were working

Next 10 days,55 men were working

Next 10 days, 50 men were working,

Next, 45 men and so on

ie:

60 men * 10days + 55men * 10days + 50 men * 10 days + 45men * 10days+ ... + 5men*10 days

= 600 man days + 550 man days + 500 man days + ... + 50 man-days

This follows an arithmetic progression, A.P formula to find sum of n numbers is

Sn=n(a1+aL)/2 where a1 is the first term and aL is the last or the n th term,n is the no. of terms.

Using this formula

12(600+50)/2

=650*6

=3900 man-days

The total amount of work done before there are zero workers is 3900/2400.

0
0

Yash

2015-07-25 12:33:14

3. If 60 people would have worked for 40 days, it will complete the work after applying 2400 man-days (i.e. 60*40)

But here we have, 60 people for 10 days = 60*10 = 600 MD ;(MD refers man-days, Shortened for time ease)

55 people for next 10 days = 55*10 = 550 MD ;

50 people for other 10 days = 50*10 = 500 MD ;

45 people for more 10 days = 45*10 = 450 MD ; (10+10+10+10=40 days, considering our initial time budget)

Adding our calculated database for MD, we get a total of 2100MD ; (i.e. (600+550+500+450)MD)

Now, we can conclude that the manpower must work for 2400 MD to complete the work,instead, we are getting it as 2100 days as according to the question's statement, means it is causing a lack of 300 days, which must be replenished an excessive time period of work and that can be calculated as follows:

we can solve that, 2400 is 1.14 times the 2100

equating the manpower, the time must also have a leap of 1.14 times, thats 40*1.14=45.6 days.

clearly causing an increase of 5.6 days, certainly let to the sign out of approximately 2.8 people.

(ie 5 people leave after 10 days, 2.5 people in 5 days, 0.25 people in 0.5 days, giving an rough idea for so)

it will then also leap an lack of MD of 2.8 * 5.6=15.68~15.5 MD, which will cause a minor decrease in MD and the time period to increase as well by 0.00625 days. hence, it will require about 45.6+0.00625= 45.606 days for the so work to complete.

__Thanking careerbless as platform and greeting the readers__

I do regard Ms. Anusreeji for her contribution, but it can be noticed that she got out the mandays and man work until the no. of labours become 0, but not time which was required by the question (as i identify her solution), to the above extent. I am a grade 9 student and so not know the progressions, this is my answer for the question for those having same as me. Any modifications are kindly welcomed.

But here we have, 60 people for 10 days = 60*10 = 600 MD ;(MD refers man-days, Shortened for time ease)

55 people for next 10 days = 55*10 = 550 MD ;

50 people for other 10 days = 50*10 = 500 MD ;

45 people for more 10 days = 45*10 = 450 MD ; (10+10+10+10=40 days, considering our initial time budget)

Adding our calculated database for MD, we get a total of 2100MD ; (i.e. (600+550+500+450)MD)

Now, we can conclude that the manpower must work for 2400 MD to complete the work,instead, we are getting it as 2100 days as according to the question's statement, means it is causing a lack of 300 days, which must be replenished an excessive time period of work and that can be calculated as follows:

we can solve that, 2400 is 1.14 times the 2100

equating the manpower, the time must also have a leap of 1.14 times, thats 40*1.14=45.6 days.

clearly causing an increase of 5.6 days, certainly let to the sign out of approximately 2.8 people.

(ie 5 people leave after 10 days, 2.5 people in 5 days, 0.25 people in 0.5 days, giving an rough idea for so)

it will then also leap an lack of MD of 2.8 * 5.6=15.68~15.5 MD, which will cause a minor decrease in MD and the time period to increase as well by 0.00625 days. hence, it will require about 45.6+0.00625= 45.606 days for the so work to complete.

I do regard Ms. Anusreeji for her contribution, but it can be noticed that she got out the mandays and man work until the no. of labours become 0, but not time which was required by the question (as i identify her solution), to the above extent. I am a grade 9 student and so not know the progressions, this is my answer for the question for those having same as me. Any modifications are kindly welcomed.

0
0

sony

2015-07-29 12:22:02

didnt understand 1.14 times of 2100 is 2400

0
0

Karan

2015-08-02 10:49:01

@yash, 1.14×2100= 2394. But I guess, you are taking the approximation.

0
0

yash

2015-07-30 13:51:55

2400/2100=1.14

or, 2100*1.14=2400

or, 2100*1.14=2400

0
0

rugmani

2015-06-04 17:15:41

A water tank is hemispherical below and cylindrical at the top. If the radius is 12m and capacity is 3312*pi cubic metre the height of the cylindrical portion in metres is:

Solution

Let h be the height.

Volume 3312 pi = 12*12 pi h + 2/3*pi* 12*12*12

Dividing this by 12*12 pi

23 = h + 2/3*12

23- 8 = h

ie., h = 15

Is it not the correct answer? whether this answer has been given ?

Solution

Let h be the height.

Volume 3312 pi = 12*12 pi h + 2/3*pi* 12*12*12

Dividing this by 12*12 pi

23 = h + 2/3*12

23- 8 = h

ie., h = 15

Is it not the correct answer? whether this answer has been given ?

0
0

preview