Mettl Equation Quiz 4

Let their speeds be x km/hr and y km/hr.
When they move in same direction, let them meet at M. Then, AM - BM = AB



Linear Equations Aptitude Questions with Short-tricks
∴ 6x - 6y = 60 or x - y = 10 ...(i)

When they move in the opposite direction, let them meet at N.
Then, AN + BN = AB



2x + 2y = 60 or x + y = 30 ...(ii)
Solving (i) and (ii) we get x = 20, y = 10.

Let first class fare be Rs. x and reservation charges be Rs. y. Then
x + y = 362 & (x + y) + {(1 / 2) x + y} = 554.
∴ x + y = 362 & 3x + 4y = 1108.
Solving these equations, we get: y = 22.

Given equations,
3x + 4y = 5 ........ (i)
x + 2y = 2 .........(ii)
Solving (i) and (ii), we get
y = (1 / 2) and x = 1. so x + y = 3 / 2

According to the given equations:

I. x² – 13x + 40 = 0
x² – 8x – 5x + 40 = 0
x (x – 8) – 5 (x – 8) = 0
(x – 5) (x – 8) = 0
x = 5, 8
II. y² – 21y + 110 = 0
y2 – 11y – 10y + 110 = 0
y (y – 11) – 10 (y – 11) = 0
(y – 10) (y – 11) = 0
y = 10, 11

After comparison of both equations, the conclusion is x < y

According to the given equations:
I. x = (208 – 14²) – 32

x = (208 – 196) – 9
x = 12 – 9
x = 3

II. y = 8³ – (21² ÷ 3) – 360
y = 512 – (441 ÷ 3) – 360
y = 512 – 147 – 360
y = 5

After comparison of both equations, the conclusion is x < y

I. x² = 30 – x
x² + x – 30 = 0
x² + 6x – 5x – 30 = 0
x (x + 6) – 5 (x + 6) = 0
(x – 5) (x + 6) = 0
x = 5, – 6

II. y² – 13y + 40 = 0
y² – 5y – 8y + 40 = 0
y (y – 5) – 8 (y – 5) = 0
(y – 8) (y – 5) = 0
y = 5 ,8

After comparison of both equations, the conclusion is x ≤ y or no relation

According to the given equations:
I. 35x² – 39x + 10 = 0
35x² – 25x – 14x + 10 = 0
5x (7x – 5) – 2 (7x – 5) = 0
(5x – 2) (7x – 5) = 0
x = \frac{2}{5}, \frac{5}{7}

II. 30y² + 2 = 17y
30y² – 17y + 2 = 0
30y² – 12y – 5y + 2 = 0
6y (5y – 2) – 1 (5y – 2) = 0
(6y – 1) (5y – 2) = 0
y =\frac{1}{6}, \frac{2}{5}

After comparison of both equations, the conclusion is x ≥ y

According to the given equations:
I. 18x² – 39x + 20 = 0
18x² – 15x – 24x + 20 = 0
3x (6x – 5) – 4 (6x – 5) = 0
(6x – 5) (3x – 4) = 0
x = \frac{5}{6}, \frac{4}{3}

II. 9y² – 51y + 52 = 0
9y² – 12y – 39y + 52 = 0
3y (y – 4) – 13 (y – 4) = 0
(3y – 4) (3y – 13) = 0
y = \frac{4}{3}, \frac{13}{3}

After comparison of both equations, the conclusion is x ≤ y