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) 2
(b) 3
(c) 1
(d) 4
The correct answers to the above question in:
Answer: (a)
Let x be the first digit and y be the second digit of two digit number.
According to question,
x + y = 8 ... (i)
(10x + y) – (10y + x) = 18
⇒9x – 9y = 18
⇒x – y = 2 ... (ii)
Adding (i) and (ii),
x + y = 8
x - y = 2
We get
2x = 10
⇒ x = 5
⇒ y = 8 – 5 = 3
⇒ x = 5 and y = 3
∴ Required difference of digits, x – y
= 5 – 3 = 2.
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If 2x + 3y = 87 and 3x – 3y = 48, what is the value of x?
a) 29
b) 11
c) 72
d) 28
e) None of these
Answer »Answer: (d)
2x | + | 3y | = | 87 |
3x | - | 3y | = | 48 |
5x | = | 135 |
∴ x = ${135}/5$ = 27
Question : 2
A positive number, when increased by 10 equals 200 times its reciprocal. What is number?
a) 10
b) 20
c) 100
d) 200
Answer »Answer: (a)
Let the positive number be x.
According to the question,
x + 10 = ${200}/x$
⇒ $x^2$ + 10x = 200
⇒ $x^2$ + 10x – 200 = 0
⇒ (x – 10) (x + 20) = 0
∴ x = 10, –20
But x ≠ –20, since x is a positive number
The required number is 10.
Question : 3
If a and b are negative real numbers and c is a positive real number, then which of the following is/are correct?
1. a – b < a – c
2. If a < b then $a/c < b/c$
$1/b < 1/c$
Select the Correct answer using the code given below.
a) 2 only
b) 3 only
c) 1
d) 2 and 3
Answer »Answer: (a)
Since a, b are negative numbers.
a < 0 and b < 0
C is a positive real number
⇒ c > 0
(1) a – b < a – c
⇒ – b < – c
⇒ b > c
It is not true as b < c
(2) if a < b ⇒ $a/c < b/c$ this is true.
(3) $1/b < 1/c$
Since c > b ⇒ $1/c < 1/b$ this is not true.
∴ option (a) is correct.
Question : 4
If 65x – 33y = 97 and 33x – 65y = 1, then what is xy equal to?
a) 3
b) –2
c) 2
d) –3
Answer »Answer: (c)
65x – 33y = 97 .... (i)
(given)
33x – 65y = 1 ....(ii)
(given)
From eq (i) + eq (ii)
98x – 98y = 98
98(x – y) = 98 ∴ x – y = 1 ....(iii)
Now
From eq (i) × 1 – eq (iii) × 33
65x - 33y = 97
- 33x + 33y = - 33
32x = 64
∴ $x = 64/32$ = 2
From eq (iii)
x – y = 1
2 – y = 1
∴ y = 2 – 1 = 1
Hence, xy = 2 × 1 = 2
Question : 5
It is given that the equations $x^2 – y^2$ = 0 and $(x – a)^2 + y^2$ = 1 have single positive solution. For this, the value of 'a' is
a) 2
b) - $√2$
c) $√2$
d) 1
Answer »Answer: (c)
$x^2 – y^2$ = 0
(Equation of y = |x|)
$(x – a)^2 + y^2$ = 1
[Equation of a circle with centre (a, 0) and radius 1]
These have single positive solution
⇒ y = |x| to be tangent to the circle
AB = radius of circle
∠AOB = 45°
∠OAB = 90°
[tangent to the circle]
∠ABO = 45°
Length of OB = $√{AB^2 + OA^2} = √{1^2 + 1^2}$
OB = $√2$
Question : 6
The value of k, for which the system of equation 3x – ky – 20 = 0 and 6x – 10y + 40 = 0 has no solution, is
a) 6
b) 5
c) 10
d) 3
Answer »Answer: (b)
Given that system of equations has no solution
⇒ |A| = 0
|A| = – 30 + 6 k
But |A| = 0
⇒ 6k – 30 = 0
⇒ k = 5
∴ Option (b) is correct.
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