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Question 1 of 10

1. Question

For the given two equations solve them and establish the reaction between p and q and mark the option.
I : p^{2} + 23p + 130 = 0
II : q^{2} – 25 = 0

Correct

Here for equation I
p^{2} +23p + 130 = 0
⇒ (p + 13)(p + 10) = 0
⇒ p = -13, -10
for equation II
q^{2} – 25 = 0
⇒ q^{2} = 25
⇒ q = ± 5
So, ±5 > -13, -10

Incorrect

Here for equation I
p^{2} +23p + 130 = 0
⇒ (p + 13)(p + 10) = 0
⇒ p = -13, -10
for equation II
q^{2} – 25 = 0
⇒ q^{2} = 25
⇒ q = ± 5
So, ±5 > -13, -10

Question 2 of 10

2. Question

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer.
a^{2}– 60a + 899 = 0
b^{2}– 71b + 1240 = 0

⇒ b(b – 40) – 31(b – 40) = 0
⇒ (b – 31)(b – 40) = 0
Then, b = +31 or b = + 40
So, when b = +31, a = b for a = 31 and b > a for a = + 29
And when b = +40, b > a for a = +29 and b > a for a = +31
∴ So, we can observe that b ≥ a.

⇒ b(b – 40) – 31(b – 40) = 0
⇒ (b – 31)(b – 40) = 0
Then, b = +31 or b = + 40
So, when b = +31, a = b for a = 31 and b > a for a = + 29
And when b = +40, b > a for a = +29 and b > a for a = +31
∴ So, we can observe that b ≥ a.

Question 3 of 10

3. Question

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer:
7x – 3y = 13
5x + 4y = 40

Correct

I. 7x – 3y = 13
Multiplying on both sides by 4, we get,
⇒ 28x – 12y = 52 —-(1)
II. 5x + 4y = 40
Multiplying on both sides by 3, we get,
⇒ 15x + 12y = 120 —-(2)
Adding equations 1 and 2, we get,
⇒ 43x = 172
⇒ x = +4 —-(3)
Substituting equation 3 in equation 2, we get,
⇒ 60 + 12y = 120
⇒ 12y = 60
⇒ y = +5
∴ We can observe that x < y.

Incorrect

I. 7x – 3y = 13
Multiplying on both sides by 4, we get,
⇒ 28x – 12y = 52 —-(1)
II. 5x + 4y = 40
Multiplying on both sides by 3, we get,
⇒ 15x + 12y = 120 —-(2)
Adding equations 1 and 2, we get,
⇒ 43x = 172
⇒ x = +4 —-(3)
Substituting equation 3 in equation 2, we get,
⇒ 60 + 12y = 120
⇒ 12y = 60
⇒ y = +5
∴ We can observe that x < y.

Question 4 of 10

4. Question

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer:
x^{2}– 80 = 244
y + 122 = 136

Correct

x^{2}– 80 = 244
or, x^{2} = 324
⇒ x = ±18
y + 122 = 136
or, y = 14
Now, we can observe that →
+ 18 > 14, ∴ x > y-18 < 14, ∴ x < y
So, no relation cannot be established between x and y.

Incorrect

x^{2}– 80 = 244
or, x^{2} = 324
⇒ x = ±18
y + 122 = 136
or, y = 14
Now, we can observe that →
+ 18 > 14, ∴ x > y-18 < 14, ∴ x < y
So, no relation cannot be established between x and y.

Question 5 of 10

5. Question

In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer
28a^{2}– a – 2 = 0
26b^{2}– 27b + 7 = 0

Correct

28a^{2}– a – 2 = 0
⇒ 28a^{2} – 8a + 7a – 2 = 0
⇒ 4a(7a – 2) + 1(7a – 2) = 0
⇒ (4a + 1) (7a – 2) = 0
Then, a = (-1/4) or a = 2/7
26b^{2}– 27b + 7 = 0
⇒ 26b^{2} – 14b – 13b + 7 = 0
⇒ 2b(13b – 7) – 1(13b – 7) = 0
⇒ (2b – 1) (13b – 7) = 0
Then, b = (1/2) or b = (7/13)
So, when a = (-1/4), a < b for b = (1/2) and a < b for b = (7/13)
And when a = (2/7), a < b for b = (1/2) and a < b for b = (7/13)
∴ we can observe that a < b.

Incorrect

28a^{2}– a – 2 = 0
⇒ 28a^{2} – 8a + 7a – 2 = 0
⇒ 4a(7a – 2) + 1(7a – 2) = 0
⇒ (4a + 1) (7a – 2) = 0
Then, a = (-1/4) or a = 2/7
26b^{2}– 27b + 7 = 0
⇒ 26b^{2} – 14b – 13b + 7 = 0
⇒ 2b(13b – 7) – 1(13b – 7) = 0
⇒ (2b – 1) (13b – 7) = 0
Then, b = (1/2) or b = (7/13)
So, when a = (-1/4), a < b for b = (1/2) and a < b for b = (7/13)
And when a = (2/7), a < b for b = (1/2) and a < b for b = (7/13)
∴ we can observe that a < b.

Question 6 of 10

6. Question

Find the next term in the given series:
1335, 444, 147, 48, 15, ?

Correct

÷3-1, ÷3-1, ÷3-1, ÷3-1…

Incorrect

÷3-1, ÷3-1, ÷3-1, ÷3-1…

Question 7 of 10

7. Question

What will come in place of question mark ‘?’ in the given number series?
2.5, 3, 4.5, 5, 6.5, 7, 8.5, 9, ?

Correct

+0.5, +1.5, +0.5, +1.5, +0.5, +1.5

Incorrect

+0.5, +1.5, +0.5, +1.5, +0.5, +1.5

Question 8 of 10

8. Question

1, 1, 2, 3, 4, 7, ?

Correct

⇒ Original series: 1, 1, 2, 3, 4, 7
⇒ Sum of adjacent terms: 2, 3, 5, 7, 11 (Consecutive prime numbers)
∴ The next term = 13 – 7 = 6

Incorrect

⇒ Original series: 1, 1, 2, 3, 4, 7
⇒ Sum of adjacent terms: 2, 3, 5, 7, 11 (Consecutive prime numbers)
∴ The next term = 13 – 7 = 6

Question 9 of 10

9. Question

3, 4, 12, 39, 103, ?

Correct

The pattern of the series is as follows:
⇒ 3
⇒ 3 + 13 = 4
⇒ 4 + 23 = 12
⇒ 12 + 33 = 39
⇒ 39 + 43 = 103
⇒ 103 + 53 = ?
∴ ? = 228

Incorrect

The pattern of the series is as follows:
⇒ 3
⇒ 3 + 13 = 4
⇒ 4 + 23 = 12
⇒ 12 + 33 = 39
⇒ 39 + 43 = 103
⇒ 103 + 53 = ?
∴ ? = 228