Test: Linear Equations- 1


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


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QUESTION: 1

Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for "Linear Equations" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.

 

Q. 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. 
Acording 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.

=>x=39.

Hence, the require number of days is 39.

QUESTION: 2

In a factory, each day the expected number of accidents is related to the number of overtime hour by 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 over time hours be x and 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
i.e 8=1000x + c           ----(i)

And, if x=400, then y=5
i.e 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 expected accident 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 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

                          Clips is denoted by c

                          Soaps is denoted by s

Now according to the question,

=> 3r+7c+s = 120                     ------- (1)

=>  4r+10c+s  = 164                 ------- (2)

Equa (1) - (2)

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

   =>      -r - 3c = - 44

Multily the equation by -1 , it becomes

=>   r + 3c = 44                         ------- (3)

Now Perform this operation

2 * (1st equation) - (2nd equation)

=> 2r+4c+s=76                          ------ (4)

Finaly (4) - (3)

=> r + c + s=32

Hence, the cost of 1 ribbon ,1 clip and 1 saop would be 32 rs.     

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 (i) and (ii) we get 

y = 26 & x = 34

So birds are 34 and puppies are 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 --------(1)

  y=5x/2---------(2)

 15x + 2y = 4000

=> 15x + 2*(5x/2) = 4000

=>  20x = 4000

=>   x = 200

So,

    =>    y = (5/2) * 200

   = >    y = 500

Hence, the cost of 12 mangoes and 3 apples = 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 x 2y) / 2

       x = 6y/2

       x=3y
then:
putting value of x in our eq
==> (7 x 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 C have

Solution:

Here's the 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

Hence A is the correct answer.

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 ,

A =  list price

B =  No of book

According to the question :

=>                 AxB = 300

=>       (A-5)x(B+2) = 300
=>      2A-5B+AB-10=300

=>                 2A-5B=10
=>             (2A-10)A=300x5
=>              A = 30

Hence (D) is the correct answer.

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