MCQ: Graphs of Linear Equations - 1 - SSC CGL MCQ

Let present age be x
x = – p% of (x – 20).
x = – q% of (x + 10).
p% of (x – 20) = q% of (x + 10)
px – 20p = qx + 10q
x (p – q) = 10q + 20 p = 10 (q + 2p)
(p – q)x = 10q + 20p = 10(q + 2p)

Hence, option C is correct.

Let the parts be A, B and C.
Acc. To question –

A + B + C = 243

2A + 3A + 4A = 486.
9A = 486; A = 54

C = 2A = 54 × 2 = 108.
Hence, option D is correct.

Let the number be xy = 10x + y
x + y = 8 .......(i)
(10x + y) – (10y + x) = 18
9x – 9y = 18
x – y = 2 .....(ii); Solving equation (i) & (ii)
x = 5, y = 3.
So, the number is 53.
Hence, option C is correct.

Multiplying equation (i) by 5 and equ. (ii) by 8, we get,

Subtracting eq. (b) from (a)

Putting the value of u – v in eq. (i) we get
u + v = 10 → [B]
Solving eq. [A] and [B]
u – v = 2; u + v = 10
2u = 12 ⇒ u = 6. So, v = 4.
Hence, option A is correct.

Dividing equation (i) by 2


Hence, option B is correct.

∴ 2x – 3 = – 2(2 – x). This gives 1 = 0.
This is impossible. So there is no solution at all.
Hence, option A is correct.

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

Hence, option B is correct.

Let first class fare be Rs. x and reservation charges be Rs. y. Then

∴ x + y = 362 & 3x + 4y = 1108.
Solving these equations, we get: y = 22.
Hence, option B is correct.

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

∴ 6x – 6y = 60 or x – y = 10 ...(i)
When they move in 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.
Hence, option B is correct.

These equation are consistent if a/c = b/d i.e. ad = bc.
Hence, option A is correct.

The given equations are 3x + y = 1 & 2(3x + y) = 2.
or 3x + y = 1 & 3x + y = 1
Thus, there is one equation in two variables.
Hence, option D is correct.

Let the larger number be x and the smaller be y.
Then, 2x – 20 = y and 2y – 5 = x
∴ 2x – y = 20 & x – 2y = – 5.
by solving these equation, we get: x = 15 & y = 10.
Hence, the larger number = 15.
Hence, option C is correct.

Let the number be x and y. then,
x + y = 80 ...... (i)
and x - 4y = 5. ....... (ii)
Solving these equation, we get:
y = 15.
Hence, option B is correct.

Let there be x hens and y cows. Then,
x + y = 48 and 2x + 4y = 140
Solving x + y = 48 and x + 2y = 70, we get: x = 26.
Hence, option A is correct.

Let ten's digit = x & unit digit = y. Then,
x + y = 10 & 10x + y - 18 = 10y + x
∴ x + y = 10 & x - y = 2. So, x = 6 and y = 4.
So, the number is 64.
Hence, option A is correct.