Latest Capgemini Aptitude Question SOLUTION: 7. if x-y + z = 19 , y + z =20 , x-z=3 , find d value of x+4y-5z 22/38/17/none

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(#M40157551) CAPGEMINI Numerical Ability question Keep an EYE

7. if x-y + z = 19 , y + z =20 , x-z=3 , find d value of x+4y-5z 22/38/17/none

Asked In Capgemini G.Sushmitha naidu (7 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

gaurav kumar  (7 years ago)

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
ans- none

Is this solution Helpfull? Yes (27) | No (3)  

Shrestha Chowdhury  (7 years ago)

x-y+z=19....(1)
y+z=20...(2)
x-z=3...(3)
by adding all these 3 equations we get,
x-y+z+y+z+x-z=19+20+3
=>2x+z=42...... (4)
by comparing euation (3) & (4) we get, 3x=45. So, x=15
so z=x-3=12 and y=20-z=8
given equation, x+4y-5z
=15+(4*8)-(5*12)
=47-60
=-13 (Ans)

Is this solution Helpfull? Yes (3) | No (1)  

sahil khumman  (7 years ago)

x-y=z=19------(1)
y+z=20---------(2)
Add both the equations (1) and (2)
x+2z=39-------(3)
x-z=3----------(4)
solve 3 and 4
you will get x=15,y=8 and z=12
Put it in the equation
15+32-60=-13
So, the ans is none

Is this solution Helpfull? Yes (2) | No   

siva  (7 years ago)

x - y + z = 19 ---- (1)
y + z = 20 ---- (2)
x - z = 3 ------ (3)

From (3)
x = z + 3 --- (4)

Place (4) in (1)

z + 3 - y + z = 19
2z - y = 16 --- (5)

Add (1) and (5) we get

z = 12
y = 8
x = 15

Substitute x, y, z values in x + 4y - 5z

15 + 4 * 8 - 5 * 12 = -13

So there is no option -13 the answer is none

Is this solution Helpfull? Yes (2) | No   

mahaveer kumar dhaker  (7 years ago)

None and
X=15;y=8;z=15
-13 answer

Is this solution Helpfull? Yes (1) | No   

Pratiksha Pandey  (7 years ago)

none

x-y+z=19
x-20=19 => x=39

x-z=3
39-z=3 => z=36

Is this solution Helpfull? Yes (1) | No (1)  

Dheeraj Gaur  (7 years ago)

X-Y+Z=19 --------------(1)
X-Z=3 ----------------(2)
Y+Z=20 ----------------(3)
Now put the value of (2) in (1)
X-20=19
=>X=39

Put the value of X in (2)
=> 39-Z=3
=> Z=36
Now,
Put the value of X and Z in (1)
=>39-Y+36=19
=>Y=56
or
put the value of Z in (2)
=>Y+36=20
=> Y= -16

Hence X=39 Y=56 or -16 Z=36
How can value of Y be different ?????????
So I think Question is wrong

Is this solution Helpfull? Yes (1) | No (3)  

vinay kumar  (7 years ago)

from second equation y=20-z
from 3rd equation x=3+z
put these value of x and y in 1st equation and solve it we get z=12
now put this value of z in 2nd and 3rd equation and solve for y and x we get y=8 and x=15
now put these three values of x,y,z in last equation then we get
x+4y-5z=-13

Is this solution Helpfull? Yes (1) | No (1)  

Harshit  (7 years ago)

x-y+z=19, y+z=20, x-z=3
3 equations 3 unknowns Thus we get x=15 y=8, z=12
Hence x+4y-5z = 15+32-60= -13

Is this solution Helpfull? Yes (1) | No (1)  

ABHISHEK KUMAR  (7 years ago)

x/6=4
at x=10
10/6 give remainder 4
y/6=5
11/6= 5
(10+11)/6=3

Is this solution Helpfull? Yes | No (3)  

amit anand  (7 years ago)

x=15 and y=8 and z=12
so x+4y-5z=>15+32-60=-13

Is this solution Helpfull? Yes | No   

AASHRAY AJAY SHARMA  (5 years ago)

Answer is -13 (None)
On computing 1st & 3rd,
2x-y=22
y=2x-22
Putting this in 2nd,
2x-22+z=20 => 2x+z=42
x-z=3
3x=45
x=15,z=12,y=8
15+4(8)-5(12)= -13.

Is this solution Helpfull? Yes | No   

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