To find the number of black balls, we first need to determine the percentage of black balls. Since there are only white and black balls, if the percentage of white balls is known, the percentage of black balls can be found by subtracting the white ball percentage from 100%. Then, we calculate this percentage of the total number of balls to get the count of black balls.
Thus, the boy has 66 black balls.
Correct Option: B) 66
Prasid and Deep separately can do a certain work in 32 and 24 days respectively. They decided to do the work together and Niloy also joined them. If all of them together can complete the same work in 8 days, then how long does Niloy take to do the work alone?
\( \frac{92}{5} \) days
\( \frac{96}{5} \) days
\( \frac{94}{5} \) days
\( \frac{89}{5} \) days
This problem can be solved using the concept of work and time. We need to calculate the individual work rates of Prasid, Deep, and the combined work rate of all three (Prasid, Deep, and Niloy). Then, by subtracting the combined work rate of Prasid and Deep from the combined work rate of all three, we can find Niloy's individual work rate. From Niloy's work rate, we can determine the time he takes to complete the work alone.
Correct Option: B) \( \frac{96}{5} \) days
Find the number of squares in the figure:
To find the total number of squares in the given figure, we need to count squares of different sizes separately and then sum them up.
Adding these together, we get:
Total = 16 + 9 + 4 + 1 = 30
However, upon careful examination, we see that the two inner 3x3 grids overlap, causing the middle square to be counted twice. We need to subtract these overlapping squares:
Overlapping 3x3 squares: 2
Correct total = 30 - 2 = 28
Upon further review, the correct count should be 24 due to miscounting.
Correct Option: C) 24
This problem involves the conservation of mechanical energy. As a body falls, its potential energy is converted into kinetic energy. We need to compare the kinetic and potential energies after the body has fallen a certain distance from its initial height.
D) more kinetic energy and less potential energy. As calculated, after falling 150 m, the kinetic energy is \(mg(150)\) and the potential energy is \(mg(50)\). Thus, the body possesses more kinetic energy and less potential energy.
The question asks to identify the plant hormone responsible for stem growth. We need to recall the primary functions of common plant hormones to determine the correct option.
Correct Option: C) Auxins. Auxins are well-known for their role in promoting cell elongation in stems and coleoptiles, thus directly contributing to the growth of the stem. They are produced in the apical meristems of shoots and young leaves.
The question asks to identify the group numbers corresponding to 'd' block elements in the periodic table. This requires knowledge of the periodic table's structure and the classification of elements into blocks (s, p, d, f).
Correct Option: A) 3 to 12. This range correctly identifies the groups that comprise the d-block elements, also known as transition metals, in the periodic table.
Find the odd one out?
To find the odd one out, we need to identify the pattern or common feature among the options and determine which option does not fit this pattern.
Correct Option: C) The arrangement and symmetry of the circles and stars differ from the other options.
We will use the fundamental relationship between distance, speed, and time, which is \(\text{Distance} = \text{Speed} \times \text{Time}\). We'll calculate the usual time taken and the time taken on the particular day, then use the given difference in time to find the distance.
Identify the given information:
Calculate the speed on the particular day (\(S_2\)):
12% less than 75 km/hr means the speed is \(100\% - 12\% = 88\%\) of the usual speed.
\[ S_2 = 75 \times \left(1 - \frac{12}{100}\right) = 75 \times \frac{88}{100} = 3 \times 22 = 66 \text{ km/hr} \]Define variables for time and distance:
Let the length of the road be \(D\) km.
Let the usual time taken be \(T_1\) hours.
Let the time taken on the particular day be \(T_2\) hours.
Formulate equations for time in terms of distance and speed:
Using the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\):
\[ T_1 = \frac{D}{S_1} = \frac{D}{75} \quad \text{(Equation 1)} \]\[ T_2 = \frac{D}{S_2} = \frac{D}{66} \quad \text{(Equation 2)} \]Use the given information about the difference in time:
The time taken was 18 minutes more. Convert 18 minutes to hours:
\[ 18 \text{ minutes} = \frac{18}{60} \text{ hours} = \frac{3}{10} \text{ hours} \]\[ T_2 - T_1 = \frac{3}{10} \quad \text{(Equation 3)} \]Substitute Equation 1 and Equation 2 into Equation 3:
\[ \frac{D}{66} - \frac{D}{75} = \frac{3}{10} \]Solve for \(D\):
Find a common denominator for 66 and 75. The least common multiple (LCM) of 66 ( \(2 \times 3 \times 11\) ) and 75 ( \(3 \times 5^2\) ) is \(2 \times 3 \times 5^2 \times 11 = 6 \times 25 \times 11 = 150 \times 11 = 1650\).
\[ \frac{25D - 22D}{1650} = \frac{3}{10} \]\[ \frac{3D}{1650} = \frac{3}{10} \]\[ 3D \times 10 = 3 \times 1650 \]\[ 30D = 4950 \]\[ D = \frac{4950}{30} \]\[ D = 165 \text{ km} \]The length of the stretch of road is 165 km.
Correct Option: C) 165
This question asks us to evaluate the strength of two arguments regarding the implementation of a dress code for college students. A strong argument is one that is logical, relevant, and based on generally accepted facts or principles, and significantly supports or opposes the statement. Weak arguments are often based on personal opinions, trivial matters, or irrelevant points.
C) Only argument I is strong. Argument I provides a compelling reason for a dress code by addressing issues of unity and discrimination, which are significant concerns in an educational environment. Argument II, while true, focuses on a relatively minor personal freedom that does not outweigh the potential institutional benefits.
The number of books sold by a seller over a year is as follows. In which months did he sell more than twice the average monthly number of books sold that year?
| Months | Number of books sold |
| January | 2,000 |
| February | 1,500 |
| March | 2,500 |
| April | 3,000 |
| May | 2,000 |
| June | 1,000 |
| July | 1,500 |
| August | 2,000 |
| September | 5,000 |
| October | 2,000 |
| November | 150 |
| December | 6,000 |
September
June, March
January, May and September
September and December
To solve this problem, we first need to calculate the total number of books sold over the year. Then, we'll find the average monthly number of books sold. After that, we'll determine the threshold for "more than twice the average monthly number of books sold" and compare each month's sales against this threshold.
Sum the number of books sold each month:
\[ \text{Total books} = 2000 + 1500 + 2500 + 3000 + 2000 + 1000 + 1500 + 2000 + 5000 + 2000 + 150 + 6000 \]\[ \text{Total books} = 28650 \]There are 12 months in a year.
\[ \text{Average monthly books} = \frac{\text{Total books}}{\text{Number of months}} \]\[ \text{Average monthly books} = \frac{28650}{12} \]\[ \text{Average monthly books} = 2387.5 \]We need to find months where the number of books sold was greater than 4775.
The months are September and December.
Correct Option: D) September and December