Capgemini
Company
Numerical Ability
Algebra
if x-y+z=19,y+z=20,x-z=3,find d value of x+4y-5z
Read Solution (Total 19)
-
- x-y+z=19 -----eqn1 y+z=20-----eqn2 x-z=3-------eqn3
Now substract eqn 2 from eqn3,we will get x-y-2z=-17-------eqn4.
Now substract eq4 and eqn1, we get,
z=12.
Using z=12 in eqn2,we get- y=8
Using z=12 in eqn3,we get- x=15
Put values in x+4y-5z=15+32-60=-13.
SO ans is -13.
- 8 years agoHelpfull: Yes(27) No(4)
- x-z=3
x=3+z,y+z=20,y=20-z
substitute this equation in x-y+z=19,ten we get z=12,x=15&y=8.finally we have to find x+4y-5z.then substitute x,y,z values in the above equation then we get -13. - 8 years agoHelpfull: Yes(9) No(3)
- x-y+z=19---------(1)
y+z=20
thus y=20-z---------(2)
x-z=3
thus x =3+z---------(3)
substitute x and y in equation (1)
you will get,
(3+z) - (20-z) + z = 19
3z = 36
z = 12
substitute z in (2) and (3)
thus x = 15
and y = 8
substitute value of x,y,z in x+4y-5z
answer is - 13 - 7 years agoHelpfull: Yes(5) No(0)
- x=3+z, y=20-z
substitute in eq: x-y+z=19 then find z=12
from this x=15 and y= 6 , put this value in x+4y-5z that gives -21
ANS IS -21 - 8 years agoHelpfull: Yes(1) No(12)
- y+z= 20 & x-z =3 on solving y+x =23
x-y+z =19 & y+x= 23 on solving 2x+z =23
x-z =3 & 2x+z =23 on solving x= 13 y =10
z =10
So x+4y-5z =3 - 7 years agoHelpfull: Yes(1) No(0)
- x-y+z=19---------->eq1
y+z=20------------->eq2
x-z=3---------------->eq3
we know the value of y+z=20
then
subtitute in eq1
x-y+z=19
x-20=19
x=19+20
x=39------------------------------------------------------------------------------------->1
now ew know the value of x=39
then
substitute in eq3
x-z=3
39-z=3
39-3=z
z=36------------------------------------------------------------------------------------>2
now we know the values of x and z
now substitute the x and z values in eq1
x-y+z=19
39-y+36=19
add 39 and 36
75-y=19
75-19=y
y=56----------------------------------------------------->3
now got the values of x,y and z
x=39
y=56
z=36
now given x+4y-5z
39+4(56)-5(36)
39+224-180
39+44
83 - 7 years agoHelpfull: Yes(1) No(10)
- x-z=3,x-y+z=19,y+z=20
substitute y+z value in x-y+z=19 we getx=39,y=-16,z=36
so value is -43 - 8 years agoHelpfull: Yes(0) No(12)
- x-(20)=19
x=39
39-z=3
z=39-3
z=36
y+36=20
y=-16
ans:
x+4y-5z=155 - 8 years agoHelpfull: Yes(0) No(6)
- y+z=20 =>y=20-z
x-(20-z)+z=19 => x+2z= 39 subtracting from x-z =3
we get
=> z=12 and y=8 and x= 15 using in equation
ans is -13 - 7 years agoHelpfull: Yes(0) No(1)
- from given,
y=20-z
x=3+z
3+z-20+z+z=19
on solving , z=11
x=14
y=17
14+4(9)-5(11)=-5 - 7 years agoHelpfull: Yes(0) No(3)
- x-y+z=19-----(1)
y+z=20,=>y=20-z--(2)
x-z=3,=>x=3+z--(3)
substitute eq (3),(2) in(1)
then we get z=12;x=15;y=8
then substitute the values of x,y,z in x+4y-5z
th ans is -13 - 7 years agoHelpfull: Yes(0) No(0)
- x-y+z=19
y=20-z
x=3+z
so
3+z-(20-z)+z=19
3z-17=19
3z=36
z=12
y+12=20
y=8
x-12=3
x=15
15+4*8-5*12=-13 - 7 years agoHelpfull: Yes(0) No(0)
- x-y+z=19
y+z=20
--------------------
x+2z=39
x+2z=39
x-z=3
-------------------
3z=36
z=12----**
now x-z=3
z=12
so, x=15---**
x-y+z=19
15-y+12=19
so y=8---**
now we know x=15
y=8
z=12
putting the values we will get the answer as [-13] - 7 years agoHelpfull: Yes(0) No(0)
- Substitute y+z in equ1 then get X value 39 and X value substitute in equ3 then get z value 36 then z value substitute in equ2 get y value -16 then get -205
- 7 years agoHelpfull: Yes(0) No(1)
- y=20-z.
x-20+z+z=19,x-20+2z=19,x-2z=19+20,
x-2z=39.so
z=-36
y-36=20,y=56.
x+36=3
x=-33
so
-33+4*56-39*5
-33+224-180
Ans 11 - 7 years agoHelpfull: Yes(0) No(0)
- 15+32-60=-13
- 7 years agoHelpfull: Yes(0) No(0)
- hello, are questions repeated in capgemini written test and can you share the questions pls, i have to crack the first round.
sunilmalviya23061996@gmail.com - 6 years agoHelpfull: Yes(0) No(1)
- x-y+z=19
x-20=19
x=39
x-z=3
39-z=3
z=39-3
z=36
y=56
x+4y-5z=83 - 6 years agoHelpfull: Yes(0) No(1)
- Can any one send the Capgemini previous papers with solutions plz to this mail divyakanthi.v@gmail.com
- 6 years agoHelpfull: Yes(0) No(0)
Capgemini Other Question