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Basic Equations - Online Quiz
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.
Q 1 - The arrangement of x/2+ y/9 = 11 and x/3 + y/6 =9 are:
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:
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
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
Q 4 - On solving p/x+q/y = m, q/x+p/y = n, we get:
A - x=(q2-p2)/(mp-nq) , y = (q2-p2)/(np-mq)
B - x=(p2-q2)/(mp-nq), y=(p2-q2)/(np-mq)
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:
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 4x+6y =32 and 4x-2y= 4, then 8y =?
Answer : D
Explanation
4x+6y = 32...(i) 4x-2y = 4...(ii) On subtracting (ii) from (i), we get: 8 y= 28
Q 7 - The arrangement of 2x+3y=2 and 3x+2y =2 can be spoken to by a point in the direction plane in:
Answer : A
Explanation
2x+3y = 2...(i) , 3x+2y= 2...(ii) Multiplying (i) by 2 and (ii) by 3 and subtracting, we get: -5x= -2 ⇒x= 2/5 Putting x= 2/5 in (i), we get 4/5+3y= 2 ⇒3y = (2-4/5) = 6/5 ⇒y = 6/5*1/3 =2/5 ∴ the solution can be represented by a point (2/5, 2/5) which lies in 1st quadrant.
Q 8 - The System of mathematical statements ℏx-y= 2 and 6x-2y =3 have an exceptional arrangement when:
Answer : D
Explanation
For a unique solution , we must have ℏ/6 ≠-1/-2 ⇒ ℏ≠(6*1/2 )= 3
Q 9 - The arrangement of comparisons x+2y = 3 and 2x+ 4y = 3 have:
Answer : B
Explanation
Here a₁/a₂= 1/2, b₁/b₂=2/4=1/2 and c₁/c₂=3/3=1. ∴ a₁/a₂=b₁/b₂≠c₁/c₂. ∴Give system has no solution.
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:
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.