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2) If three metallic spheres of radii 6 cms, 8 cms and 10 cms are melted to from a single sphere, the diameter of the new sphere will be:

3) Sixty men can build a wall in 40 days, but though they begin the work together, 5 men quit every ten days. The time needed to build the wall is:

4) Shyam is travelling on his cycle and has calculated to reach point 'A' at 2 PM If he travels at 10 kmph. he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach point 'A' at 1 PM:

5) An aeroplane travelling at 700 km. p. h. in level flight drops a bomb from a height of 1000 metres on a target. The time taken from releasing the bomb to hitting the target is nearest to the figure:

6) 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

s = ut + 0.5at

^{2}

here s=1000

u=0(initial speed)

a=g=9.8 m/s

^{2}

So time can be calulated as 14.14sec

__Answer for Q2:__

If three metallic spheres melted down and formed a new sphere, volume of all three sphere will be equivalent to the volume of new sphere and final equation will be

(r1^3)+(r2^3)+(r3^3) = R^3

here,

(6^3)+(8^3)+(10^3) = R^3

=>216+512+1000 = R^3

=> R = (1728)^1/3

i.e. R = 12 cm.

Hence, diameter of the new sphere will be (2R) i.e 24cm.

S1*t1=s2*t2. (Speed 1=10 kmh, s2 is 15)

10*(14-t)=15*(12-t). (2 o'clock on 24 hrs clock is 14, while 12 is 12)

140-10t=180-15t. (t is the start time)

5t=40

t=8 hrs, means he had started journey at 8AM

D=s1*(14-t)

D=10*(14-8)

D=60 km

Time to reach destination=13-8=5 hrs

Speed=d\t=60/5=12kmh

x = d/10 ......Eqn 1.

Now,(x - 2) = d/15 or 15x - 30 = d or 15x = 30 + d or x = (d + 30)/15......Eqn 2.

From eqns 1 and 2,we have (d + 30)/15 = d/10 or 10d + 300 = 15d or 5d = 300 or d = 60 km.

Now,

x -1 = d/10 - 1 or x - 1 = 60/10 - 1 = 5 [We are talking about 1 p.m and we have chosen the time duration of upto to p.m to be x,so upto 1 p.m,it is (x -1) hours]

So,the time taken upto 1 p.m is 5 hours.Therefore,the speed = distance/time = 60/5 km/hr

= 12 km/hr

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.....

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

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.

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.

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.

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