To solve this problem, we need to determine the ratio of milk to water in the initial mixture. Then, we will use this ratio to find out how much water is needed for a larger quantity of milk to maintain the same proportion.
A) 16 ℓ is the correct answer. By maintaining the same ratio of milk to water (\(9:4\)), if the milk quantity is 36 ℓ, the water quantity must be 16 ℓ.
The question asks to identify which of the given options is NOT considered biomedical waste. Biomedical waste refers to any waste generated during diagnosis, treatment, or immunization of human beings or animals, or in research activities pertaining thereto, or in the production or testing of biologicals. We need to evaluate each option based on this definition.
D) Food wastes are not considered biomedical wastes. They are typically general waste unless they come into contact with infectious materials in a healthcare setting, in which case they would be treated as contaminated waste, but not inherently biomedical waste themselves.
The question asks for the date on which Earth Day is observed. This is a factual question related to environmental awareness and international observances. Recalling the specific date associated with Earth Day will lead to the correct answer.
Correct Option: C) April 22
To find the volume of a cuboidal tank, we need to use the formula for the volume of a cuboid, which is the product of its length, width, and height. The given dimensions are 6 m, 5 m, and 4 m.
Identify the dimensions of the cuboidal tank:
Recall the formula for the volume of a cuboid:
\[ \text{Volume} = L \times W \times H \]Substitute the given dimensions into the formula:
\[ \text{Volume} = 6 \text{ m} \times 5 \text{ m} \times 4 \text{ m} \]Perform the multiplication:
\[ \text{Volume} = (6 \times 5) \times 4 \text{ m}^3 \] \[ \text{Volume} = 30 \times 4 \text{ m}^3 \] \[ \text{Volume} = 120 \text{ m}^3 \]Correct Option: A) 120
The volume of the cuboidal tank is \(120 \text{ m}^3\).
Refer to the following letter series and answer the question.
K D N C I C L O A S C S O P M C O D K C V N D C M M
How many consonants are there in the given series between the letter that is fifth from the right end and the letter that is sixth from the left end of the given series?
To solve this problem, we need to first identify the specified letters in the given series based on their positions from the left and right ends. Then, we will extract the segment of the series between these two letters. Finally, we will count the number of consonants within that segment.
K D N C I C L O A S C S O P M C O D K C V N D C M M
Counting from the right end:
M (1st), M (2nd), C (3rd), D (4th), N (5th)
So, the fifth letter from the right end is N.
Counting from the left end:
K (1st), D (2nd), N (3rd), C (4th), I (5th), C (6th)
So, the sixth letter from the left end is C.
The letters between 'C' (sixth from left) and 'N' (fifth from right) are:
L O A S C S O P M C O D K C V
Consonants are all letters of the alphabet except A, E, I, O, U.
Counting the consonants: L, S, C, S, P, M, C, D, K, C, V. There are 11 consonants.
A) Eleven. As calculated in the step-by-step explanation, there are 11 consonants between the specified letters.
To find the greatest number that divides two given numbers leaving specific remainders, we first subtract the remainders from the respective numbers. The resulting numbers will be perfectly divisible by the required greatest number. Therefore, the problem reduces to finding the Highest Common Factor (HCF) of these new numbers.
The first number is 2000, and the remainder is 22. So, the number perfectly divisible by the required greatest number is:
\[ 2000 - 22 = 1978 \]The second number is 2200, and the remainder is 38. So, the number perfectly divisible by the required greatest number is:
\[ 2200 - 38 = 2162 \]Now, we need to find the HCF of 1978 and 2162. We can use the prime factorization method or the Euclidean algorithm.
Using prime factorization:
The common prime factors are 2 and 23.
The HCF is the product of the common prime factors raised to their lowest powers:
\[ \text{HCF}(1978, 2162) = 2 \times 23 = 46 \]If we divide 2000 by 46:
\[ 2000 = 46 \times 43 + 22 \]The remainder is 22, which is correct.
If we divide 2200 by 46:
\[ 2200 = 46 \times 47 + 38 \]The remainder is 38, which is correct.
Correct Option: D) 46
This problem involves calculating the final displacement from the starting point using vector addition or by breaking down movements into North-South and East-West components. We will track the net displacement in the North-South direction and the East-West direction separately and then use the Pythagorean theorem to find the straight-line distance.
A) 61 km is the correct distance from Siddharth's home to his destination.
The question asks to identify the founder of Bank of Baroda from the given options. This requires knowledge of the history of major Indian banks and their founders.
Correct Option: C) Maharaja Sayajirao Gaekwad III
Maharaja Sayajirao Gaekwad III was a visionary ruler of the Baroda State. He played a crucial role in the industrial and economic development of his state, which included the establishment of the Bank of Baroda in 1908. His efforts led to the bank becoming one of the prominent financial institutions in India.
To find the rate of interest per annum, we will use the formula for simple interest. The formula relates principal amount, time, rate of interest, and the simple interest earned.
The formula for simple interest is:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]Where:
We need to find R, so we can rearrange the formula as:
\[ \text{R} = \frac{\text{SI} \times 100}{\text{P} \times \text{T}} \]So, the rate of interest is 6% per annum.
Correct Option: B) 6%
Select the option in which the numbers are related in the same way as are the numbers of the given sets.
NOTE: Operations should be performed on the whole numbers without breaking down the numbers into their constituent digits.
For example, for 13, operations such as adding/subtracting/multiplying, etc. can be performed on 13. Breaking down 13 into 1 and 3 and then performing mathematical operations on 1 and 3 is not allowed.
(9, 8, 15)
(4, 5, 11)