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) 17
(b) 20
(c) 15
(d) 24
The correct answers to the above question in:
Answer: (b)
${170 - 70}/{250/50} = {100}/5$ = 20
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
If m and n are the roots of the equation $x^2$ + ax + b = 0 and $m^2, n^2$ are the roots of the equation $x^2$ - cx + d = 0, then which of the following is/are correct ?
1. 2b - $a^2$ = c
2. $b^2$ = d
Select the correct answer using the codes given below:
a) Both 1 and 2
b) Only 2
c) Neither 1 nor 2
d) Only 1
Answer »Answer: (b)
Here m and n are the roots of the equation
$x^2$ + ax + b = 0.
m + n = - a .......(i)
mn = b ......(ii)
Also m $m^2 and n^2$ are the roots of the equation of
$x^2$ - cx + d = 0.
$m^2 + n^2$ = c...(iii)
$m^2 n^2$ = d .....(iv)
by squaring Eq. (i) both sides, we get
$m^2 + n^2 + 2mn = a^2$ [from Eqs. (i) and (ii)]
⇒ c + 2b = $a^2 ⇒ c = a^2 - 2b$
⇒ 2b - $a^2$ = -c
Therefore, Statement 1 is incorrect.
From Eq. (ii)
$m^2 n^2 = b^2 ⇒ b^2$ = d
Therefore, Statement 2 is correct.
Question : 3
If the equations $x^2 + 5x + 6 = 0 and x^2$ + kx + 1 = 0 have a common root, then what is the value of k?
a) $5/2 or - {10}/3$
b) $5/2 or {10}/3$
c) $- 5/2 or {10}/3$
d) $- 5/2 or {10}/3$
Answer »Answer: (b)
We know that two equation
$a_1 x^2 + b_1 x + c_1$ = 0
$a_2 x^2 + b_2 + c_2$ = 0
have common root when
$(c_1 a_2 - a_1 c_2)^2 = (b_1 c_2 - c_1 b_2) (a_1 b_2 - b_1 a_2)$
So, for $x^2 + 5x + 6 = 0 and x^2 + kx = 1 = 0$
we have $(5)^2$ = (5 - 6x) (x - 5)
⇒ 25 = $-6x^2$ + 35x - 25
⇒ $6x^2$ - 35x + 50 = 0
⇒ x = $5/2 or {10}/3$
Question : 4
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 : 5
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 : 6
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.
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