Mathematics Test 289

Result

Mathematics Test 289

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Weekly Quiz Competition

Question 1
4 / -1

A
x

B
−x

C
−y

D
y

Solution
Question 2
4 / -1

ax + by + c = 0 does not represent an equation of line when ____.

A
a = c = 0, b ≠ 0

B
b = c = 0, a ≠ 0

C
a = b = 0

D
a = b ≠ 0
Solution
Given: The general equation of a line is ax + by + c = 0.

Step 1: For this to be an equation of a line, at least one of the coefficients a or b must be non-zero.

If both a = 0 and b = 0, the equation becomes 0x + 0y + c = 0, or simply c = 0.
- If c ≠ 0, the equation c = 0 is a contradiction, and no points ( x,y ) satisfy it (it represents an empty set).
- If c = 0, the equation 0 = 0 is always true, so all points (x,y) in the plane satisfy it (it represents the entire plane).
Conclusion: In neither case (when a = b = 0) does the equation represent a line. Therefore, the condition for the equation not to represent a line is when both a and b are zero.
Question 3
4 / -1

A

B

C

D

Solution
Question 4
4 / -1

A five digit number divisible by 3 has to formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is

A
216

B
240

C
600

D
3125
Solution
Given: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. The available digits are 0,1,2,3,4,5. The sum of all these digits is 0 + 1 + 2 + 3 + 4 + 5 = 15, which is divisible by 3.

Solution:
- Case 1: Exclude the digit 0. The remaining digits are {1,2,3,4,5}. Their sum is 1 + 2 + 3 + 4 + 5 = 15, which is divisible by 3.
(using digits 1, 2, 3, 4, 5): All the digits are non-zero, so any arrangement will form a 5-digit number.

The number of arrangements is 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.
- Case 2: Exclude the digit 3. The remaining digits are {0,1,2,4,5}. Their sum is 0 + 1 + 2 + 4 + 5 = 12, which is divisible by 3.
(using digits 0, 1, 2, 4, 5): The number 0 cannot be in the first (ten thousands) place, otherwise it would be a 4-digit number.
- There are 4 choices for the first digit (1, 2, 4, or 5).
- The remaining 4 digits can be arranged in the remaining 4 places in 4! = 4 × 3 × 2 × 1 = 24 ways.
- So, the number of valid 5-digit numbers in this case is 4 × 4! = 4 × 24 = 96 ways.
Answer: Total waysTotal waysTotal ways = 120 + 96 = 216
Question 5
4 / -1

If all permutations of the letters of the word AGAIN are arranged as in dictionary, then fiftieth word is

A
NAAGI

B
NAGAI

C
NAAIG

D
NAIAG

Solution
Question 6
4 / -1

A
-4 and 2

B
2 and 4

C
-2 and -4

D
8 and -8

Solution

Find the value of a

Given:
Question 7
4 / -1

Condition that the pair of straight lines represented by the equation ax² + 2hxy + by² = 0 to be perpendicular is

A
ab = −1

B
a = −b

C
a = b

D
ab = 1

Solution
Question 8
4 / -1

The value of a for which the lines x = 1, y = 2 and a²x + 2y − 20 = 0 are concurrent, is

A
1

B
8

C
-4

D
-2

Solution
Question 9
4 / -1

A
A² + B²

B
A² − B²

C
A²

D
B²

Solution
Question 10
4 / -1

A

B

C

D

Solution