Linear Equations Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Linear Equations topic of quantitative aptitude
(a) k ≠ 3
(b) k = 0
(c) k = 3
(d) k ≠ 0
The correct answers to the above question in:
Answer: (a)
The equations kx – y = 2 and 6x – 2y = 3 have a unique solution. Then,
∴ $k/6 ≠ 1/2$ ⇒ k ≠ 3
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Question : 1
A player holds 13 cards of four suits, of which seven are black and six are red. There are twice as many diamonds as spades and twice as many hearts as diamonds. How many clubs does he hold ?
a) 5
b) 4
c) 6
d) 7
e) None of these
Answer »Answer: (c)
Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts.
Let the number of spades be x. Then, number of clubs = (7 – x).
Number of diamonds = 2 x number of spades = 2x;
Number of hearts = 2 x number of diamonds = 4x.
Total number of cards = x + 2x + 4x + 7 – x – 6x + 7.
Therefore 6x + 7 = 13⇔6x = 6⇔x – 1.
Hence, number of clubs = (7 – x) = 6.
Question : 2
If x + $1/x$ = 2, then what is value of x – $1/x$ ?
a) 1
b) 2
c) 0
d) –2
Answer »Answer: (c)
Given that x + $1/x$ = 2 ... (i)
Squaring both sides, we get
$(x + 1/x)^2$ = 4
⇒ $x^2 + 1/{x^2}$ + 2 = 4
⇒ $x^2 + 1/{x^2}$ = 2 ... (ii)
Now, $(x - 1/x)^2 = (x^2 + 1/{x^2})$ – 2
= 2 – 2 = 0 [from equation (ii)]
∴ x – $1/x$ = 0
Question : 3
If ${5x - 7y + 10}/1 = {3x + 2y + 1}/8 = {11x + 4y - 10}/9$ then what is x + y equal to?
a) 2
b) 3
c) 1
d) –3
Answer »Answer: (b)
Given,
${5x - 7y + 10}/1 = {3x + 2y + 1}/8 = {11x + 4y - 10}/9$
= ${(5x + 10 - 7y) + (3x + 2y + 1) - (11x + 4y - 10)}/{(1 + 8) - 9}$
= ${- 3x - 9y + 21}/0$
⇒x + 3y = 7 ... (i)
On taking first two terms,
8(5x – 7y + 10) = 3x + 2y + 1
37x – 58y + 79 = 0 ... (ii)
From equation (i), on putting the value of x in equation (ii), we get
37(7 – 3y) – 58y + 79 = 0
⇒259 – 111y – 58y + 79 = 0
⇒169y = 338
⇒y = 2
From equation (i),
x = 7 – 3(2) = 1
∴ x + y = 1 + 2 = 3
Question : 4
The solution of linear inequalities x + y ≥ 5 and x – y ≤ 3 lies
a) In the first and second quadrants
b) In the second and third quadrants
c) Only in the first quadrant
d) In the third and fourth quadrants
Answer »Answer: (a)
x+ y ≥ 5
let first draw graph of x + y = 5 ... (i)
put y = 0 in (i)
x = 5
put x = 0 in (i)
y = 5
Checking for (0, 0)
x + y ≥ 5
0 ≥ 5, which is false
Hence, origin does not lie in x + y ≥ 5
Now,
x – y ≤ 3
let first drawn graph of x – y = 3 ...(ii)
put x = 0 in (ii)
y = –3
put y = 0 in (ii)
x = 3'
Checking for (0, 0)
x – y ≤ 3
0 ≤ 3, which is true
Hence, origin lies in x – y ≤ 3
So, shaded region will be the solution of these linear inequalities lies in the first and second quadrant.
Question : 5
Assertion (A) : The equations 2x – 3y = 5 and 6y – 4x = 11 cannot be solved graphically.
Reason (R) :The equations given above represent parallel lines.
a) A and R are correct but R is not correct explanation of A
b) A is correct but R is wrong
c) A and R are correct and R is correct explanation of A
d) A is wrong but R is correct
Answer »Answer: (c)
Given that,
2x – 3y = 5 ... (i)
and –4x + 6y = 11 ... (ii)
Also, ${a_1}/{a_2} = 2/{-4} = {-1}/2$
${b_1}/{b_2} = {-3}/6 = {-1}/2$
⇒${a_1}/{a_2} ={b_1}/{b_2} = {-1}/2 ≠ {c_1}/{c_2}$
So, both equations are in parallel. So, it is not solved graphically.
Hence, A and R are individually true and R is correct explanation of A.
Question : 6
In a family, the father took 1/4 of the cake and he had 3 times as much as each of the other members had. The total number of family members is
a) 7
b) 3
c) 10
d) 12
e) None of these
Answer »Answer: (c)
Let there be (x + 1) members. Then,
Father's share = $1/4$ , share of each other member = $3/{4x}$.
∴ 3 $(3/{4x}) = 1/4$⇔4x = 36⇔x=9
Hence, total number of family member = 10.
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