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# Test: Linear Equations- 1

## 10 Questions MCQ Test Quantitative Aptitude (Quant) | Test: Linear Equations- 1

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This mock test of Test: Linear Equations- 1 for Banking Exams helps you for every Banking Exams entrance exam. This contains 10 Multiple Choice Questions for Banking Exams Test: Linear Equations- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Linear Equations- 1 quiz give you a good mix of easy questions and tough questions. Banking Exams students definitely take this Test: Linear Equations- 1 exercise for a better result in the exam. You can find other Test: Linear Equations- 1 extra questions, long questions & short questions for Banking Exams on EduRev as well by searching above.
QUESTION: 1

### Mahesh visited his cousin Akash during the summer vacation. In the mornings, they both would go for swimming. In the evenings, they would play tennis. They would engage in at most one activity per day, i.e. either they went swimming or played tennis each day. There were days when they took rest and stayed home all day long. There were 32 mornings when they did nothing, 18 evenings when they stayed at home, and a total of 28 days when they swam or played tennis. What duration of the summer vacation did Mahesh stay with Akash?

Solution:

Let, the duration of Mahesh's vacation be 'n' days
According to the question, he was free on 32 mornings and on 18 evenings and on total 28 days they either went for swimming or tennis together.
Thus, they were together on (n − 32) mornings and (n − 18) evenings.

⇒ (n − 32) + (n − 18) = 28.
⇒ n = 39

Hence, the required number of days is 39.

QUESTION: 2

### In a factory, each day the expected number of accidents is related to the number of overtime hours by a linear equation. Suppose that on one day there were 1000 overtime hours logged and 8 accidents reported and on another day there were 400 overtime hours logged and 5 accidents. What is the expected number of accidents when no overtime hours are logged?

Solution:

► Let, the number of overtime hours be 'x' and the expected number of accidents be 'y'.
We will use the equation y = mx + c
where m and c are constant.

Now, according to the question:

If x = 1000, then y = 8
⇒ 8 = 1000x + c      ...(i)

And, if x = 400, then y = 5
⇒ 5 = 400x + c        ...(ii)

On solving these equations, we get:

m = 1/200 and c = 3

Hence, in case, no overtime is there i.e. x = 0, then the number of the expected accidents will be 3.

QUESTION: 3

### What is the least possible value of b/a, if it is given that |2a - 9| = 7 and |b - 4| = 5 ?

Solution:
• |2a - 9| = 7
⇒ -7 ≤ 2a-9 ≤ 7
⇒ 2 ≤ 2a ≤ 16
⇒ 1 ≤ a ≤ 8
• |b - 4| = 5
⇒ -5 ≤ b - 4 ≤ 5
⇒ -1 ≤ b ≤ 9

'b/a' is minimum when b is minimum and a is maximum (when b/a has a positive result)
OR
b is minimum and a is minimum (when b/a has negative result)

In our case, 'b' has a negative value (-1) and so b/a will be minimum when a is also minimum (1).
So, minimum value of b/a = -1

QUESTION: 4

Moli buys 3 ribbons,7 clips and 1 soap for Rs.120 exactly. At the same place it would cost Rs.164 for 4 ribbons, 10 clips and one soap. How much would it cost for one ribbon, one clip and one soap?

Solution:

Let us consider, Ribbon is denoted by r, Clip is denoted by c, Soap is denoted by s.
Now according to the question:

⇒ 3r + 7c + s = 120      ...(1)
⇒ 4r + 10c + s = 164    ...(2)

Equation (1) - (2):

⇒ (3r - 4r) + (7c - 10c) + (s - s) = 120 - 164
⇒ -r - 3c = - 44

Multiply the equation by -1, it becomes:

⇒ r + 3c = 44           ...(3)

Now Perform this operation:

2 * (1st equation) - (2nd equation)
⇒ 2r + 4c + s = 76          ...(4)

Finally Equation (4) - (3):

⇒ r + c + s = 32

Hence, the cost of 1 ribbon,1 clip and 1 soap would be Rs 32.

QUESTION: 5

In a pet shop, there are 120 eyes and 172 legs. How many birds and puppies are included in these?

Solution:

Let there are x birds and y puppies.
Each bird and puppy has 2 eyes. Each bird has 2 legs and each puppy has 4 legs.
Then:

2x + 2y = 120 ⇒ x + y = 60  ...(1)
2x + 4y = 172 ⇒ x + 2y = 86  ...(2)

Solving (1) and (2) we get:

y = 26 & x = 34

So birds = 34 and puppies = 26.

QUESTION: 6

The price of 10 mango is equal to that of 4 apples. The price of 15 mango and 2 apples together is Rs. 4000. The total price of 12 mango and 3 apples is:

Solution:

Let the cost of a apple and that of a mango be Rs. x and Rs. y respectively.
Given:

10x = 4y
y = 5x / 2

According to question:

15x + 2y = 4000
15x + 2 * (5x / 2) = 4000
20x = 4000
x = 200
y = (5/2) * 200
y = 500

Hence, the cost of 12 mangoes and 3 apples is:

= 12x + 3y
= Rs. (2400 + 1500)
= Rs. 3900

Hence (B) is the correct answer.

QUESTION: 7

Find (7x + 4y ) / (x - 2y) if x / 2y = 3/2?

Solution:

If  x / 2y = 3 / 2 , then:

⇒ x = (3 * 2y) / 2
⇒ x = 6y / 2
⇒ x = 3y

Putting value of x in our equation:

⇒ (7 * 3y + 4y) / (3y - 2y)
⇒ (21y+4y) /y
⇒ 25y / y
25

QUESTION: 8

Total Rs.700 are divided among 3 persons (A, B, and C). A gets 1/2 of B and B gets 1/2 of C. How much money does C has?

Solution:

⇒ A = B/2
⇒ B = C/2
⇒ A + B + C = 700
⇒ (B/2) + B + (2B) = 700
⇒ B = 200
⇒ A = B/2 = 200/2 = 100
⇒ C = 2B = 2*200 = 400
So, C has 400 Rs.

QUESTION: 9

Value of ‘x’ in 6x - 4 = 3x + 8 should be:

Solution:

⇒ 6x - 4 = 3x + 8
⇒ 6x - 3x = 12
⇒ 3x = 12
x = 4
So option D is correct.

QUESTION: 10

If the list price of a book is reduced by Rs 5 then a person can buy 2 more books for Rs 300. What is the original cost of the book?

Solution:

Let the original list price of the book be = Rs x
The original number of books purchased = 300/x​
When the price of the book is reduced by 5, the new number of books purchased will be = 300 / (x - 5)​
New number of books purchased -Original number of books purchased = 2

⇒ 300 / (x - 5)​ − 300 / x ​= 2
⇒ (300x - 300x + 1500) / (x - 5)x = 2
⇒ 2x2 - 10x = 1500

On solving this equation we get:

x = -25 or x = 30

x cant be negative. So the original cost of book is Rs 30
Hence (D) is the correct answer.