Answer: Option A
Explanation:
M ≥ P T = M …… (ii)
Combining all statements, we get
V ≥ T = M ≥ P P is true.
Again, T ≥ H is not true.
2.
Statements: A ≥ B = C ≥ D, V ≥ G ≤ H = D
Conclusion:
I. C ≥ D
II. A ≥ H
III. B ≥ G
IV. C ≤ V
Answer: Option C
Explanation:
A ≥ B = C ≥ D …..(i)
V ≥ G ≤ H = D ……(ii)
Combining both statement, we get
A ≥ B = C ≥ D = H ≥ G ≤ V
Thus, C ≥ G is true.
A ≥ H is true. B ≥ G is true. C ≤ V is not true.
Hence, only I, II and III are true.
3.
Statements: M ≤ N T ≥ Q
Conclusions:
I. R ≥ L
II. T ≤ N
III. L ≥ M
IV. R ≥ M
Answer: Option B
Explanation:
M ≤ N T ≥ Q ……(ii)
Combining both the statements, we get
M ≤ N ≤ L ≥ Q ≤ T ≤ R
Thus, R ≥ L is not true.
T ≤ N is not true.
Again, M M is true.
R ≥ M is not true.
4.
Statement: M ≥ P T = M
Conclusions:
I. V ≥ P
II. T ≥ H
Answer: Option A
Explanation:
M ≥ P T = M …… (ii)
Combining all statements, we get
V ≥ T = M ≥ P P is true.
Again, T ≥ H is not true.
5.
Statements: E = G ≥ H = N, C ≥ F ≥ M = N
Conclusions:
I. F ≥ E
II. E ≥ M
III. C ≥ G
IV. C ≥ H
Answer: Option C
Explanation:
E = G ≥ H = N ……(i)
C ≥ F ≥ M = N …….(ii)
Combining both the statements, we get
E = G ≥ H = N = M ≤ F H or H ≤ C is true.
Answer: Option B
Explanation:
I. \(x^{5}\) – 41\(x^{3}\) + 400x = 0
x(\(x^{4}\) – 41\(x^{2}\) + 400) = 0
x = 0 or \(x^{4}\) – 41\(x^{2}\) + 400 = 0
Let \(x^{2}\) = a
\(a^{2}\) – 41a + 400 = 0 ∴ \(a^{2}\) – 25a − 16a + 400 = 0
(a – 16)(a – 25) = 0
a = 16 or a = 25
\(x^{2}\) = 16 or \(x^{2}\) = 25
x = ±4 or x = ±5
x = −5 or −4 or 0 or 4 or 5
II. \(y^{2}\) – 14y + 30 = −18
\(y^{2}\) – 14y + 48 = 0
\(y^{2}\) – 8y − 6y + 48 = 0
(y – 6)(y – 8) = 0
y = 6 or y = 8
For both values of y
2.
I. \(x^{2}\) – 3 = 2x
II. \(y^{2}\) + 5y + 6 = 0
Answer: Option B
Explanation:
I. \(x^{2}\) – 3 = 2x
\(x^{2}\) – 2x – 3 = 0
\(x^{2}\) – 3x + x – 3 = 0
(x + 1)(x – 3) = 0
x = 3 or x = −1
II. \(y^{2}\) + 5y + 6 = 0
\(y^{2}\) + 3y + 2y + 6 = 0
(y + 3)(y + 2) = 0
y = –3 or y = –2
For both values of x, x ≥ y
3.
I. 4\(x^{3}\) + 24\(x^{2}\) – 64x = 0
II. 3\(y^{2}\) + 39y + 126 = 0
Answer: Option E
Explanation:
I. 4\(x^{3}\) + 24\(x^{2}\) – 64x = 0
4x(\(x^{2}\) + 6x – 16) = 0
4x = 0 or \(x^{2}\) + 6x – 16 = 0 (Here 4x = 0 is nothing but x = 0)
\(x^{2}\) + 6x – 16 = 0
\(x^{2}\) + 8x − 2x – 16 = 0
(x − 2)(x + 8) = 0
x = 2 or x = −8
x = −8, 0 or 2
II. 3\(y^{2}\) + 39y + 126 = 0
\(y^{2}\) + 13y + 42 = 0
\(y^{2}\) + 7y + 6y + 42 = 0
(y + 7)(y + 6) = 0
y = –7 or y = –6 When x = −8, x y Hence, no relation can be established between x and y.
4.
I. \(x^{2}\) – 25x + 114 = 0
II. \(y^{2}\) – 10y + 24 = 0
Answer: Option C
Explanation:
I. \(x^{2}\) – 25x + 114 = 0
\(x^{2}\) – 19x – 6x + 114 = 0
(x – 6)(x – 19) = 0
x = 19 or x = 6
II. \(y^{2}\) – 10y + 24 = 0
\(y^{2}\) – 6y – 4y + 24 = 0
(y – 4)(y – 6) = 0
y = 6 or y = 4
When x = 19, x ≥ y
When x = 6, x ≥ y
Hence, x ≥ y
5.
I. 48\(x^{2}\) – 24x + 3 = 0
II. 55\(y^{2}\) + 53y + 12 = 0
Answer: Option A
Explanation:
48\(x^{2}\) – 24x + 3 = 0 is of the form a\(x^{2}\) – bx + c = 0
Both roots of this equation are positive i.e. x ≥ 0
55\(y^{2}\) + 53y + 12 = 0 is of the form a\(y^{2}\) + by + c = 0
Both roots of this equation are negative i.e. y
Answer: Option D
Explanation:
Let a two-digit number be (10x + y) and reversing number be (10y + x)
Required sum = 10x + y + 10y + x = 11x + 11y = 11(x + y)
Hence, it’s divisible by 11.
2. What is the sum of two numbers whose differences is 45 and the quotient of the greater number by the lesser number is 4?
Answer: Option D
Explanation:
Let the greater number be x and smaller number be y
x – y = 45 ….. (i)
x = 4y ….. (ii)
From eqs. (i) and (ii)
4y – y = 45
y = \(\frac{45}{3}\)= 15
On putting the value of y in Eq. (ii), we get
x = 4 × 15 = 60
Hence, required sum = x + y = 60 + 15 = 75.
3. If one-third of a two-digit number exceeds its one-fourth by 8, then what is the sum of the digits of the number?
Answer: Option C
Explanation:
Let the number be y,
\(\frac{y}{3}\)=\(\frac{y}{4}\) + 8
\(\frac{4y – 3y}{12}\)=8
y = 12 × 8 = 96
Sum of digits = 9 + 6 = 15.
4. A person bought 5 tickets from a station P to a station Q and 10 tickets from the station P to a station R. He paid Rs. 350. If the sum of a ticket from P to Q and a ticket from P to R is Rs. 42, then what is the fare from P to Q?
Answer: Option B
Explanation:
Let the fare from station P to station Q is ₹ x and the fare from station P to station R is ₹ y.
By given condition, x + y = 42 …..(i)
and 5x + 10y = 350 …. (ii)
On solving Eqs. (i) and (ii), we get
x = 14 and y = 28
Hence, fare from station P to station Q is Rs. 14.
5. If 6x – 5y = 13, 7x + 2y = 23 then 11x + 18y = ?
Answer: Option B
Explanation:
6x – 5y = 13 …… (i)
7x + 2y = 23 …… (ii)
By equation (i) * 2 & (ii) * 5,
12x – 10y = 26 ….. (iii)
35x + 10y = 115 …… (iv)
by solving these equation we get
x = 3, y = 1
11x + 18y = 11 × 3 + 18 × 1 = 33 + 18 = 51.
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