Practice Model question set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The value of k, for which the system of equation 3x – ky – 20 = 0 and 6x – 10y + 40 = 0 has no solution, is
(a)
(b)
(c)
(d)
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.
Q-2) 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)
(b)
(c)
(d)
$x^2 – y^2$ = 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 = $√2$
Q-3) If 65x – 33y = 97 and 33x – 65y = 1, then what is xy equal to?
(a)
(b)
(c)
(d)
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
Q-4) The sum of digits of a two-digit number is 8 and the difference between the number and that formed by reversing the digits is 18. What is the difference between the digits of the number?
(a)
(b)
(c)
(d)
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.
Q-5) 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)
(b)
(c)
(d)
(e)
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.
Q-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)
(b)
(c)
(d)
(e)
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.
Q-7) If x + $1/x$ = 2, then what is value of x – $1/x$ ?
(a)
(b)
(c)
(d)
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
Q-8) If ${5x - 7y + 10}/1 = {3x + 2y + 1}/8 = {11x + 4y - 10}/9$ then what is x + y equal to?
(a)
(b)
(c)
(d)
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
Q-9) If 2x + 3y = 87 and 3x – 3y = 48, what is the value of x?
(a)
(b)
(c)
(d)
(e)
| 2x | + | 3y | = | 87 |
| 3x | - | 3y | = | 48 |
| 5x | = | 135 |
∴ x = ${135}/5$ = 27
Q-10) A positive number, when increased by 10 equals 200 times its reciprocal. What is number?
(a)
(b)
(c)
(d)
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.