Basic Equations - Online Quiz


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Following quiz provides Multiple Choice Questions (MCQs) related to Basic Equations. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Questions and Answers

Q 1 - The arrangement of x/2+ y/9 = 11 and x/3 + y/6 =9 are:

A - x=36, y=9

B - x=9, y=9

C - x=18, y=18

D - x=18, y=9

Answer : C

Explanation

The given equations are
9x+ 2y = 198...(i)
2x+y =54...(iii)
On multiplying (ii) by 2 and subtracting it from (i), we get: 5x= 90 ⇒x= 18
Putting x=18 in (ii), we get: 36+y = 54 ⇒y = 18
∴x = 18, y= 18

Q 2 - On solving 4/x+5y=7 and 3/x+4y =5 we, get:

A - x=1/3, y=1

B - x=- 1/3, y=-1

C - x=1/3, y=-1

D - x=- 1/3, y=1

Answer : C

Explanation

Given equations are 4/x+5 y= 7 ...(i)
3/x+4y = 5  ...(ii)
On multiplying (i) by 3, (ii) by 4 and subtracting, we get -y =1 ⇒y= -1
Putting y= -1 in (i), we get 4/x-5 = 7 ⇒4/x= 12 ⇒12x= 4 ⇒x= 1/3
∴x= 1/3, y= -1

Q 3 - If 3x+7y= 75 and 5x-5y= 25, then what is the estimation of x+y?

A - 14

B - 15

C - 16

D - 17

Answer : D

Explanation

Given 3x+7y = 75  ...(i)
5x-5 y= 25 ⇒x-y = 5...(ii)
Multiplying (ii) by 7 and adding to (i), we get:
10 x = 110   ⇒x = 11
Putting x = 11 in (ii), we get: y=(11-5) = 6
∴ x+y = (11+6) = 17

Answer : B

Explanation

Given equations are p/x+q/y = m...(i),   q/x+ p/y = n ...(ii)
On multiplying (i) by q, (ii) by p and subtracting, we get:
q2/y- p2/y = mq-np
⇒y (mp-np) = (q2- p2)
⇒y = (q2-p2)/(mq- np)
= (p2- q2)/(np-mq)
On multiplying (i) by p, (ii) by q and subtracting, we get:
p2/x - q2/x = mp- nq
⇒ (p2- q2) = x (mp- nq)
⇒x = (p2- q2)/ (mp-nq)
∴ x= (p2-q2)/(mp-nq) , y = (p2-q2)/(np- mq)

Q 5 - The arrangement of 3x-y+1/3 = 2x+y+2/5 = 3x+2y+1/6 are given by:

A - x=1, y=2

B - x=-1, y=-1

C - x=1, y=1

D - x=2 , y=1

Answer : C

Explanation

We have (3x-y+1)/3= (2x+y+2)/5
⇒5 (3x-y+1) =3(2x+y+2)
⇒15x-5y+5 = 6x+3y+6
⇒9x-8y -1 = 0 ⇒9x-8y = 1 ...(i)
And (2x+y+2)/5 = (3x+2y+1)/6
⇒ 6 (2x+y+2) = 5(3x+2y+1)
⇒12x+6y +12=   15x+10 y+5
⇒3x+ 4y= 7...(ii)
Multiplying (ii) by 2 and adding (i) to it, we get: 15x= 15 ⇒x= 1
Putting x =1 in .., (ii), we get 3*1+4y= 7 ⇒4y = 4 ⇒y = 1
∴ x= 1, y = 1

Q 6 - On the off chance that x+1/y= 5, 2x+ 3/y= 13, then (2x-3y) =?

A - 1

B - 2

C - 3

D - 5

Answer : C

Explanation

X+1/y = 5...(i),     2x+3/y =13 ...(ii)
On multiplying (i) by 3 and subtracting (ii) from it, we get: x=2
Putting x= 2 in (i), we get 1/y =3 ⇒3y= 1 ⇒y = 1/3
∴ (2x-3y) = (2*2-3*1/3) = (4-1) = 3

Q 7 - The arrangement of x/2+y/3 =4 and x+y = 10 are given by:

A - x=6, y=-4

B - x=-6, y=4

C - x=4, y=6

D - x=6, y=4

Answer : C

Explanation

Given equation are 3x+2y = 24 ...(i), x+y =10 ...(ii)
On multiplying (ii) by 2 and subtracting from (i), we get: x=4
Putting x= 4 in (ii), we get: y = (10-4) = 6

Q 8 - The arrangement of mathematical statements 2x+ℏy= 11 and 5x-7y = 5 have no arrangement when:

A - ℏ= 13/5

B - ℏ=-13/5

C - ℏ=-14/5

D - ℏ=-16/5

Answer : C

Explanation

For no solution , we have a₁/a₂ = b₁/b₂ ≠c₁/c₂
i.e. 2/5 = ℏ/-7 ≠11/5 ⇒ℏ= -14/5 

Q 9 - The arrangement of comparisons 4x+7y= 10 and 10x+ky= 25 have boundless number of arrangements, when:

A - ℏ=17/2

B - ℏ=5

C - ℏ=27/2

D - ℏ=35/2

Answer : D

Explanation

For infinite number of solutions, we have a₁/a₂ = b₁/b₂ =c₁/c₂
∴ 4/10 = 7/ℏ= 10/25 ⇒7/ℏ= 2/5 ⇒ℏ= 35/2

Q 10 - In the event that 3 seats and 1 table expense Rs. 900 and 5 seats and 3 tables cost Rs. 2100, then the expense of 4 seats and 1 table is:

A - Rs. 1000

B - Rs. 1050

C - Rs. 1100

D - Rs. 1150

Answer : B

Explanation

Let cost of 1 chair =Rs x and cost of 1 table =Rs y. Then,
3x+y=900   ...(i)             5x+3y=2100      ...(ii)
On multiplying (i) by 3 and subtracting (ii) from it, we get: 4x=600⇒x=150.
Putting x=150 in (i), we get: 3*150+y=900⇒y= (900-450) =450.
Cost of (4 chairs+1 table) =Rs (4*150+1*450) =Rs (600+450) =Rs1050.

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