You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
Not categorized0%
Click view Questions button to view Explanation
Post your marks in the comment section
Prepare well,Keep Working Hard and never Give Up !!
1
2
3
4
5
6
7
8
9
10
Answered
Review
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 : p2 + 23p + 130 = 0
II : q2 – 25 = 0
Correct
Here for equation I
p2 +23p + 130 = 0
⇒ (p + 13)(p + 10) = 0
⇒ p = -13, -10
for equation II
q2 – 25 = 0
⇒ q2 = 25
⇒ q = ± 5
So, ±5 > -13, -10
Incorrect
Here for equation I
p2 +23p + 130 = 0
⇒ (p + 13)(p + 10) = 0
⇒ p = -13, -10
for equation II
q2 – 25 = 0
⇒ q2 = 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.
a2– 60a + 899 = 0
b2– 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:
x2– 80 = 244
y + 122 = 136
Correct
x2– 80 = 244
or, x2 = 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
x2– 80 = 244
or, x2 = 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
28a2– a – 2 = 0
26b2– 27b + 7 = 0
Correct
28a2– a – 2 = 0
⇒ 28a2 – 8a + 7a – 2 = 0
⇒ 4a(7a – 2) + 1(7a – 2) = 0
⇒ (4a + 1) (7a – 2) = 0
Then, a = (-1/4) or a = 2/7
26b2– 27b + 7 = 0
⇒ 26b2 – 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
28a2– a – 2 = 0
⇒ 28a2 – 8a + 7a – 2 = 0
⇒ 4a(7a – 2) + 1(7a – 2) = 0
⇒ (4a + 1) (7a – 2) = 0
Then, a = (-1/4) or a = 2/7
26b2– 27b + 7 = 0
⇒ 26b2 – 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
