Answer: Option B
Explanation:
I. \(x^{3}\) – 4913 = 0
or, \(x^{3}\) = 4913
x = 17
II. \(y^{2}\) = 361
or, y = ± 19
While comparing the values of x and y, one root value of y lies between the root values of x
2.
I. \(x^{2}\) = 361
II. \(y^{3}\) = 7269 + 731
Answer: Option A
Explanation:
I. \(x^{2}\) = 361
x = ± 19
II. \(y^{3}\) = 7269 + 731
\(y^{3}\) = 8000
y = 20
x ≤ y
3.
I. 15\(x^{2}\) + x – 6 = 0
II. 5\(y^{2}\) – 23y + 12 = 0
Answer: Option B
Explanation:
I. 15\(x^{2}\) + x – 6 = 0
15\(x^{2}\) + 10x – 9x – 6 = 0
5x (3x + 2) – 3 (3x + 2) = 0
(5x – 3) (3x + 2) = 0
x = \(\frac{3}{5}\), –\(\frac{2}{3}\)
II. 5\(y^{2}\) – 23y + 12 = 0
5\(y^{2}\) – 20y – 3y + 12 = 0
5y (y – 4) – 3 (y – 4) = 0
(y – 4) (5y – 3) = 0
y = 4, \(\frac{3}{5}\)
x ≤ y
4.
I. \(x^{3}\) – 2744 = 0
II. \(y^{2}\) – 256 = 0
Answer: Option E
Explanation:
I. \(x^{3}\) – 2744 = 0
\(x^{3}\) = 2744
x = 14
II. \(y^{2}\) – 256 = 0
\(y^{2}\) = 256
y = ± 16
While comparing the values of x and y, one root value of x lies between the root values of y.
5.
I. \(x^{2}\) – 8x – 20 = 0
II. 3\(y^{2}\) – 60y + 297 = 0
Answer: Option B
Explanation:
I. \(x^{2}\) – 8x – 20 = 0
\(x^{2}\) – 10x + 2x – 20 = 0
x (x – 10) + 2 (x – 10) = 0
(x – 10) (x + 2) = 0
Then, x = + 10 or x = – 2
II. 3\(y^{2}\) – 60y + 297 = 0
\(y^{2}\) – 20y + 99 = 0 [Dividing both sides by 3]
\(y^{2}\) – 11y – 9y + 99 = 0
y (y – 11) – 9 (y – 11) = 0
(y – 11) (y – 9) = 0
Then, y = + 11 or y = + 9
So, when x = + 10, x y for y = + 9
And when x = – 2, x ≤ y for y = + 11 and x ≤ y for y = + 9
So, we can observe that one root value of x lies between the root values of y. Therefore, the relation between x and y can't be determined.
Answer: Option B
Explanation:
These equation are consistent if \(\frac{a}{c}\) = \(\frac{b}{d}\), i.e ad = bc
2. Points A and B are 60km apart. A bus starts from A and another from B at the same time. If they go in the same direction they meet in 6 hours and if they go in opposite direction they meet in 2 hours. The speed of the bus with greater speed is:
Answer: Option B
Explanation:
Let their speeds be x km/hr and y km/hr.
When they move in same direction, let them meet at M. Then, AM – BM = AB
6x – 6y = 60 or x – y = 10 …(i)
When they move in opposite direction, let them meet at N. Then, AN + BN = AB
2x + 2y = 60 or x + y = 30 …(ii)
Solving (i) and (ii) we get x = 20, y = 10.
3. A railway half ticket costs half the full fare but the reservation charges are the same for half ticket as well as for full ticket. one reserved first class ticket for a journey between two stations is Rs. 362 and one full and one half reserved first class tickets cost Rs. 554. The reservation charges are:
Answer: Option A
Explanation:
Let first class fare be Rs. x and reservation charges be Rs. y. Then
x + y = 362 & (x + y) + (\(\frac{1}{2}\)x+y)= 554
x + y = 362 & 3x + 4y = 1108.
Solving these equations, we get: y = 22.
4. What is the value of x + y in the solution of the equations?
\(\frac{x}{4}\)+\(\frac{y}{3}\)=\(\frac{5}{12}\) and \(\frac{x}{2}\)+ y= 1 is
Answer: Option B
Explanation:
Given equations,
3x + 4y = 5 …….. (i)
x + 2y = 2 ………(ii)
Solving (i) and (ii), we get
y= \(\frac{1}{2}\) and x = 1. so x + y =\(\frac{3}{2}\)
5. The number of solutions of these equations x+\(\frac{1}{y}\)= 2 and 2xy – 3y = – 2 is:
Answer: Option A
Explanation:
First equation gives\(\frac{1}{y}\) = 2 – x or y =\(\frac{1}{2 – x}\)
Second equation is: y(2x – 3) = -2 or\(\frac{2x – 3}{2 – x}\) = – 2
2x – 3 = – 2(2 – x). This gives 1 = 0.
This is impossible. So there is no solution at all.
Answer: Option D
Explanation:
E = F ≥ G ≤ H = I
We can’t compare H and F because between H & F opposite symbol used. We know that the inequalities does not works between opposite symbol
2. L ≤ O ≥ V = E ≥ S Which of the following ones is correct?
1) L ≤ V
2) O = E
3) O ≥ S
4) S ≥ L
Answer: Option C
Explanation:
L ≤ O ≥ V = E ≥ S
We can compare O and S. which shows that the option 3rd is correct because the common symbol between O and S is ‘≥’.
3. B ≥ E ≤ A = T ≥ S Which of the following ones is correct?
1) B ≥ S
2) E = T
3) E ≤ T
4) E ≤ S
Answer: Option B
Explanation:
B ≥ E ≤ A = T ≥ S
We can compare E and T but either 2 or 3 equation is correct.
4. M = O ≤ N = K ≤ S Which of the following ones is correct?
1) M = S
2) O ≤ S
3) N ≥ S
4) O = K
Answer: Option E
Explanation:
M = O ≥ N = K ≤ S
We can compare O & S. which shows that the option 2 is correct because the common symbol between O & S is ‘≤’.
5. C ≥ H = A ≥ T ≥ S Which of the following ones is correct?
1) S ≥ C
2) T = C
3) H ≤ T
4) H ≤ S
Answer: Option A
Explanation:
C ≥ H = A ≥ T ≥ S
We can compare S & C. which shows that the option first one is correct because the common symbol between S & C is ‘≤’.
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