MCQ: Linear Equations - 3 - SSC CGL MCQ

Given equations of system
3x + y = 4 ...(i)
x + 2y = 8 ...(ii)
Here, a₁ = 3 , b₂ = 2 and c₂ = B
∵ a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2
So, the system of equations has infinite solutions, because it represents a parallel line.

Let the number of gold coins initially be x and the number of non-gold coins be y.
According to the question,
3x = y

When 10 more gold coins, total number of gold coins becomes x + 10 and the number of non-gold coins remain the same at y.
Now, we have 2(x + 10) = y

Solving these two equations, we get
x = 20 and y = 60.
Total number of coins in the collection at the end is equal to
x + 10 + y = 20 + 10 + 60 = 90.

Given, 6x - 10y = 10 ..........(i)
and x/(x + y) = 5/7
⇒ 7x = 5x + 5y
⇒ 2x - 5y = 0 ...(ii)

On multiplying Eq. (ii) by 2 and subtracting from Ed.(i), we get
6x - 10y = 10
4x - 10y = 0
---------------------
2x = 10
∴ x = 5
Putting the value of x in Eq. (i), we get
30 - 10y = 10
⇒ 10y = 20
⇒ y = 2
∴ (x - y) = 5 - 2 = 3

Given, 2^a + 3^b = 17
and 2^{a + 2} - 3^{b + 1} = 5
⇒ 2² x 2² - 3^b x 3¹ = 5
⇒ 4.2^a - 3.3^b = 5
Let 2^a = x amd 3^b = y
Then, x + y = 17 ...(i)
4x - 3y = 5 ...(ii)
On multiplying Eq. (i) by 3 and adding to Eq (ii), we get
3x + 3y = 51
4x - 3y = 5
------------------
7x = 56
⇒ x = 8

On putting the value of x in Eq. (i), we get
8 + y = 17
∴ y = 9

Now, 2^a = x
⇒ 2^a = 8 (2)³
∴ 3^b = y = 9
⇒ 3^b = 3²
∴ b = 2
Hence, a = 3 and b = 2

ax + by = c and dx + ey = f
a₁/a₂ = a/d, b₁/b₂ = b/e, c₁/c₂ = c/f
∵ b₁/b₂ ≠ c₁/c₂
∴ b/e ≠ c/f
it represent a pair of parallel lines.
∵ a₁/a₂ ≠ b₁/b₂
∴ a/d ≠ b/e
Therefore, system has unique solutions and represents a pair of intersecting lines.

Given,
(x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11
⇒ (x + y - 8)/2 = (x + 2y - 14)/3
⇒ 3x + 3y - 24 = 2x + 4y - 28
⇒ 3x + 3y - 2x - 4y = -28 + 24
x - y = -4 ...(i)

Again, (x + 2y - 14)/3 = (3x + y - 12)/11
⇒ 11x + 22y - 154 = 9x + 3y - 36
⇒ 2x + 19y = 118 ..(ii)

On multiplying Eq. (i) by 2 and subtracting from Eq., we get
2x - 2y = -8
2x + 19y = 118
-------------------
-21y = -126
∴ y = 6

On putting the value of y in Eq. (i), we get
x - 6 = -4
∴ x = 2
∴ x = 2 and y = 6

Method 1 to solve the given question.
Lets assume the unit's digit is y and the ten's digit is x. then, the number is 10 x + y.
According to question, after interchanging the digits, the new number is 10y + x.
Then,
New Number - 18 = Original number
10y + x = 10x + y + 18
⇒ 9y - 9x = 18
⇒ y - x = 2 .....................(1)
Again according to question,
Sum of digits of Original Number = 8
x + y = 8 ..............................(2)
Add the equation (1) and (2) , we will get
y - x + x + y = 2 + 8
⇒ 2y = 10
⇒ y = 5
Put the value of Y in equation (2) , we will get
x + 5 = 8
⇒ x = 8 - 5 = 3
Then , the original number = 10x + y
Put the value of x and y and get the original number.
The original number = 10x + y = 10 x 3 + 5 = 30 + 5 = 35

Method 2 to solve the given question.
Given that sum of digits of original number is 8.
Let us assume that unit digit is x , then ten digit will be 8 - x. (since sum of digits is 8.)
Now Original number = 10( 8 - x) + x
After interchanging the digits , the number = 10x + (8 - x) = 9x + 8
According to question.
New number = original number + 18
⇒ 9x + 8 = 10(8 - x) + x + 18
⇒ 9x + 8 = 80 - 10x + x + 18
⇒ 9x + 8 = 98 - 9x
⇒ 9x + 9x = 98 - 8
⇒ 18x = 90
⇒ x = 90/18
⇒ x = 5
The original number = 10(8 - x) + x = 10(8 - 5) + 5 = 10(3) + 5 = 30 + 5 = 35

(p/x) + (q/y) = m ..(i)
(q/x) + (p/y) = n ...(ii)

On multiply Eq (i) by q and Eq. (ii) by p and subtracting, we get
(pq/x) + (q²)/y = mq
(pq/x) + (p²)/y = np
-----------------------------------------
(q²/y) - (p²/y) = mq - np
∴ (q² - p²) = y(mq - np)
∴ y = (q² - p²)/mp - np = (p² - q²)/np - mq

Again, on multiplying Eq. (i) by p and Eq. (ii) by q and subtracting, we get
(p²/x) + (pq/y) = mp
(q²/x) + (pq/y) = nq
-----------------------------------------
(p²/x) - (q²/x) = mp - nq
⇒ (p² - q²) = x (mp - nq)
⇒ x = (p² - q²)/(mp - nq)
∴ x = (p² - q²)/(mp - nq)
and y = (p² - q²) / (np - mq)

Let us assume the ten's digit of the number is n. Then the unit's digit will be 3n.
According to question,
The sum of the digits is equal to 8.
sum of the digit = 8
n + 3n = 8
⇒ 4n = 8
⇒ n = 2
So, ten's digit is 2 and unit's digit is 6.
So the number = 10n + 3n = 10 x 2 + 3 x 2 = 20 + 6 = 26.

Let us assume the two digits number of the number is a and b.
According to given question,
The original number is 10a + b.
After interchanging the digits , the Number is 10b + a.
The number formed is greater than the the original number by 45,
We will get,
New number = Original number + 45
⇒ 10b + a = 10a + b + 45
⇒ 10b + a - 10a - b = 45
⇒ 9b - 9a = 45
⇒ b - a = 5
According the question the difference between the digits is 5 which will not help us to solve this question because both condition given in question are same. So can't be determined the answer.

Now try Hit and Trail method
Try one by one all option. All 3 Options are true.
But we cannot choose all 3 option, but as per question we cannot find the answer.

Let the ten's digits be x and unit's digit be y.
Then,(10x + y) - (10y + x)=36 ⇔9(x-y)=36
⇔ x - y=4

Let us assume the first book published in year x.
According to question,
Books are published at seven years interval,
First edition book published in year = x
Second edition book published in year = x + 7
Third edition book published in year = x + 14
Four edition book published in year = x + 21
Five edition book published in year = x + 28
Six edition book published in year = x + 35
Seven edition book published in year = x + 42
When the seventh book was issued, the sum of the publication year was 13,524.
x + x + 7 + x +14 + x +21 + x + 28 + x + 35 + x + 42 =13524
147 + 7x = 13524
7x = 13524 -147
⇒ x = 13377/7 = 1911
The first book published in year x = 1911

Let us assume the first number is x and second number is y.
According to question,
sum of twice the first number and thrice the second number is 100.
2x + 3y = 100 ...................(1)
and sum of thrice the first number and twice the second number is 120.
3x + 2y = 120...................(2)
Subtract equation (1) from Equation (2) after multiply 2 with Equation (1) and 3 with Equation (2), we will get.
3(3x + 2y) - 2(2x + 3y) = 3 x 120 - 2 x 100
9x + 6y - 4x - 6y = 360 - 200
5x = 160
x = 32
put the value of x in equation (1)
2 x 32 +3y = 100
3y = 100 - 64
3y = 36
y = 12
The larger Number is 32 .

let us assume the ten's digit number be x and unit's place digit be y.
Then two digit Original Number = 10x + y
According to question → y = 2x - 1 ................(1)
When digits are interchanged then new number = 10y + x
If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20,
New Number - Original Number = Original number - 20
⇒ 10y + x - (10x + y) = 10x + y - 20
⇒ 10y + x - 10x - y - 10x - y = - 20
⇒ 8y - 19x = - 20
⇒ 19x - 8y = 20 ...................(2)
Put the value of y from equation (1) in above equation (2).
⇒ 19x - 8(2x - 1) = 20
⇒ 19x - 16x - 8 = 20
⇒ 7x = 20 + 8
⇒ 7x = 28
⇒ x = 28/7
⇒ x = 4
Put the value of y in equation (1)
From (i) y = 2 x 4 -1 = y = 7
∴ original number = 10x + y =10 x 4 + 7 = 47.

Let us assume the number of brown socks = B
Let us assume the price of brown socks = ₹ P per pair
Then price of black socks = ₹ 2P per pair
Before interchanging the shocks the amount = amount of black shocks + amount of brown shocks
⇒ Before interchanging the shocks the total amount = 4 x 2P + B x P
⇒ Before interchanging the shocks the total amount = 8P + BP
After interchanging the shocks the total amount = amount of black shocks + amount of brown shocks
⇒ After interchanging the shocks the total amount = 4 x P + B x 2P
⇒ After interchanging the shocks the total amount = 4P + 2BP
According to question,
After interchanging the shocks the total amount = Before interchanging the shocks the amount + Before interchanging the shocks the amount x 50 %
4P + 2BP = 8P + BP + (8P + BP ) x 50%
⇒ 4P + 2BP = 8P + BP + (8P + BP ) x 50/100
⇒ 4P + 2BP = 8P + BP + (8P + BP ) x 1/2
⇒ 4P + 2BP = ( 16P + 2BP + 8P + BP ) x 1/2
⇒ (4P + 2BP) x 2 = ( 16P + 2BP + 8P + BP )
⇒ 8P + 4BP = 24P + 3BP
⇒ 8P + 4BP = 24P + 3BP
⇒ 4BP - 3BP = 24P - 8P
⇒ BP = 16P
⇒ B = 16
The ratio of the number of black and brown pairs of socks = Number of block shocks / Number of Brown Shocks
⇒ The ratio of the number of black and brown pairs of socks in the original order = 4/16 = 1/4 = 1: 4