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) 1
(b) 2
(c) 0
(d) –2
The correct answers to the above question in:
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
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
The maximum marks in a Test are converted from 250 to 50 for the purpose of an Internal Assessment. The highest marks scored were 170 and lowest marks were 70. What is the difference between the maximum and minimum marks scored in the Internal Assessment ?
a) 17
b) 20
c) 15
d) 24
Answer »Answer: (b)
${170 - 70}/{250/50} = {100}/5$ = 20
Question : 3
A number consists of two digits whose sum is 8. If 18 is added to the number, the digits are reversed. The number is equal to
a) 35
b) 53
c) 26
d) 62
Answer »Answer: (a)
Let tens digit and unit digits are x and y, respectively. Sum of digits = 8
x + y = 8 ... (i)
According to question,
(10x + y) + 18 = (10y + x)
⇒ 9y – 9x = 18
⇒ y – x = 2 ... (ii)
On solving equations (i) and (ii), we get
∴ y = 5 and x = 3
∴ Required number = 10x + y
= 10(3) + 5 = 30 + 5 = 35
Question : 4
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 : 5
Under what condition do the equations kx – y = 2 and 6x – 2y = 3 have a unique solution?
a) k ≠ 3
b) k = 0
c) k = 3
d) k ≠ 0
Answer »Answer: (a)
The equations kx – y = 2 and 6x – 2y = 3 have a unique solution. Then,
∴ $k/6 ≠ 1/2$ ⇒ k ≠ 3
Question : 6
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.
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