inequation why please elucidate x1*x2+x3*x4<=n its maximum when x1=x2, x3=x4

inequation why please elucidate x1*x2+x3*x4<=n its maximum when x1=x2, x3=x4

How can I start prepaing for Quants in shortdays??

Tushar

2 years, 6 months ago

Hi Prakash ,

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How many of the solutions to a+b+c+d=20 have one or two of the `digits' equal to 10?

Akhilesh Singh

3 years ago

Hi Utkarsh

The question is not clear. How can 1 digit be equal to 10. Please check the question once

In an examination it is required to get 1034 of the aggregate marks to pass.A student gets 940 marks and is declared failed by 5% marks.what are the maximum aggregate marks a student can get?

Atish Pradhan

3 years ago

let the max. aggregate marks be x,

more marks required to pass for the student =a= 1034-940=94,

now, a is the 5% of the total marks,

so, 94=5% of x

hence, x = 1880

Suri Gangula

2 years, 1 month ago

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A, B & C can independently do a job in 15,20 and 30 days. theywork together for some time after which C leave .a total of 18000/- is paid for the work & B gets 6000 more than C. for how many days did A work ?

Shivankan Gupta

2 years, 6 months ago

Hi,

Let the total work to be done be 60 units.

So, A must do 4 units/day, B 3 units/day and C 2 units/day.

A and B must have done work on equal number of days. So, their salaries must be in the ratio 3:4.

Lets suppose B gets 3x salary. So, A gets 4x salary and C 3x-6000.

=> 3x-6000+3x+4x=18000

=> x = 2400

So, C's total salary will be 3 X 2400 - 6000 = 1200

So in the days C worked, B and A must get 1800 and 2400 salaries i.e. a total of Rs 5400.

=> Work done in these days is 5400/18000 = 54/180 = 3/10 or 18 units of work.

18 units of work can be completed by them in 2 days together.

The rest of the 42 units of the work can be completed by A and B in 6 days.

So, A worked for a total of 6+2 = 8 days

if a+b+c=6 and ab+bc+ca= 1 than find ab(b+a)+bc(c+b)+ac(a+c)+3abc

D MOUNICA

3 years ago

ab(b+a)+bc(b+c)+ac(a+c)=ab(6-c)+bc(6-a)+ac(6-b)+3abc

=6ab-abc+6bc-abc+6ca-abc+3abc

=6(ab+bc+ca)-3abc+3abc

=6(1)=6

Suri Gangula

2 years, 4 months ago

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Just now only I started the VARC section... Is it possible to complete? If yes, then how?

Akhilesh Singh

3 years ago

Yeah, you should focus on areas which give you high returns like the RC's and PJ's

Two friends A and B run on a 550 m circular track such that A (who is faster) overtakes B in the middle of A's 6th round for the first time.If they were to run on a stright track of length 10000 m, find approximately how much headstart A can give to B such that they finish the race at the same time?

A 1818m B 1837m C 1762m D 1265m

[SOLUTION] Since A overtakes B in the middle of the 6th round,B must be in the middle of the 5th round. So A has covered 5.5 times the track length while B has covered 4.5 times the track length. For every 5.5m covered by A,B covers 4.5 m. 11m covered by A=9 m covered by B Headstart in 11m race=2m Headstart in 10000m race=(2*10000)/11=1818m

In the solution, it is said B must be in middle of the 5th round. How did we reach to this conclusion?

Muthu Vignesh

3 years ago

Because, it given that A is overtaking B for the FIRST TIME => A must be ONE lap ahead of B