MCQ: Linear Equations - 3 - SSC CGL MCQ

# MCQ: Linear Equations - 3 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Linear Equations - 3

MCQ: Linear Equations - 3 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Linear Equations - 3 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Linear Equations - 3 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Linear Equations - 3 below.
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MCQ: Linear Equations - 3 - Question 1

### The system of equations 3x + y - 4 = 0 and 6x + 2y - 8 = 0 has

Detailed Solution for MCQ: Linear Equations - 3 - Question 1

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.

MCQ: Linear Equations - 3 - Question 2

### In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1 : 2, Based on the information; the total number of coins in the collection now becomes.

Detailed Solution for MCQ: Linear Equations - 3 - Question 2

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.

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MCQ: Linear Equations - 3 - Question 3

### If 6x - 10y = 10 and x / (x + y) = 5/7, then (x - y) = ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 3

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

MCQ: Linear Equations - 3 - Question 4

If 2a + 3b = 17 and 2a + 2 - 3b +1 = 5, then

Detailed Solution for MCQ: Linear Equations - 3 - Question 4

Given, 2a + 3b = 17
and 2a + 2 - 3b + 1 = 5
⇒ 22 x 22 - 3b x 31 = 5
⇒ 4.2a - 3.3b = 5
Let 2a = x amd 3b = 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, 2a = x
⇒ 2a = 8 (2)3
∴ 3b = y = 9
⇒ 3b = 32
∴ b = 2
Hence, a = 3 and b = 2

MCQ: Linear Equations - 3 - Question 5

The graph of ax + by = c, dx + ey = f will be
I. parallel, if the system has no solution.
II. coincident, if the system has finite number of solutions.
III. intersecting. if the system has only one solution.
Which of the above statements are correct ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 5

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.

MCQ: Linear Equations - 3 - Question 6

If (x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11 then find the values of x and y, respectively.

Detailed Solution for MCQ: Linear Equations - 3 - Question 6

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

MCQ: Linear Equations - 3 - Question 7

If the number obtained on the interchanging the digits of two-digits number is 18 more than the original number and the sum of the digits is 8, then what is the original number ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 7

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

MCQ: Linear Equations - 3 - Question 8

The solution of the equations (p/x) + (q/y) = m and (q/x) + (p/y) = n is

Detailed Solution for MCQ: Linear Equations - 3 - Question 8

(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)

MCQ: Linear Equations - 3 - Question 9

In a two-digit number, the digit in the unit's place is three times the digit in tenth's place. The sum of the digits is equal to 8. What is the number ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 9

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.

MCQ: Linear Equations - 3 - Question 10

If the digits of a two-digit number are interchanged , the number formed is greater than the the original number by 45. If the difference between the digits is 5, then what is the original number ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 10

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.

MCQ: Linear Equations - 3 - Question 11

The difference between a two- digit number and the number obtained by interchanging the position of its digits is 36.What is the difference between the two digits of that number ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 11

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

MCQ: Linear Equations - 3 - Question 12

A series of books was published at seven years interval. When the seventh book was issued, the sum of the publication year was 13,524. When was the first book published ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 12

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

MCQ: Linear Equations - 3 - Question 13

There are two numbers such that the sum of twice the first number and thrice the second number is 100 and the sum of thrice the first number and twice the second number is 120. Which is the larger number ?

Detailed Solution for MCQ: Linear Equations - 3 - Question 13

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 .

MCQ: Linear Equations - 3 - Question 14

In a two-digit number, the digit at the unit's place is 1 less than twice the digit at the ten's place. 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. The original number is

Detailed Solution for MCQ: Linear Equations - 3 - Question 14

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.

MCQ: Linear Equations - 3 - Question 15

A man ordered 4 pairs of black socks and some pairs of brown socks. The price of a black sock is double that of a brown pair. While preparing the bill the clerk interchanged the number of black and brown pairs by mistake which increased the bill by 50%. The ratio of the number of black and brown pairs of socks in the original order was :

Detailed Solution for MCQ: Linear Equations - 3 - Question 15

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

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests