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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
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:
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:
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:
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:
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.