Which amendment of the constitution enabled 27% reservation for Other Backward Classes in higher educational institutions, including private institutions?
91st amendment
92nd Amendment
93rd amendment
94th amendment
The question asks about the specific constitutional amendment that enabled 27% reservation for Other Backward Classes (OBCs) in higher educational institutions, including private ones. To answer this, we need to recall the key provisions of various constitutional amendments related to reservations and education.
Correct Option: C) The 93rd Amendment inserted Clause (5) into Article 15 of the Constitution, enabling the state to make special provisions for the advancement of socially and educationally backward classes (including OBCs) in admission to educational institutions, including private ones (excluding minority institutions). This amendment specifically allowed for the 27% reservation for OBCs in higher educational institutions.
Who developed the vaccine for smallpox?
Edward Jenner
Robert Koch
Pearl Kendrick
Alexander Fleming
The question asks to identify the scientist responsible for developing the smallpox vaccine. This requires knowledge of key historical figures in medicine and their major contributions.
Correct Option: A) Edward Jenner
Edward Jenner, an English physician, is known as the pioneer of the smallpox vaccine. In 1796, he observed that milkmaids who had contracted cowpox (a milder disease) seemed to be immune to smallpox. He famously inoculated a young boy, James Phipps, with material from cowpox lesions and later exposed him to smallpox, finding that the boy was immune. This groundbreaking work laid the foundation for modern vaccinology.
______ is an endangered Indian medicinal plant species.
The question asks to identify an endangered Indian medicinal plant species from the given options. This requires knowledge of plant conservation status and their medicinal uses, particularly in the Indian context.
Correct Option: B) Podophyllum (specifically Podophyllum hexandrum) is an endangered Indian medicinal plant species. It is critically endangered due to over-exploitation for its medicinal properties, particularly podophyllotoxin, which is used in anti-cancer drugs. It is native to the Himalayan regions of India.
Calculate: 3668 ÷ 524 – 9
To solve this problem, we need to follow the order of operations (BODMAS/PEMDAS), which dictates that division should be performed before subtraction. First, we will divide 3668 by 524, and then subtract 9 from the result.
We need to calculate \(3668 \div 524\).
Let's perform the division:
\[ \frac{3668}{524} \]We can estimate or perform long division. Notice that \(524 \times 7 = 3668\).
So, \(3668 \div 524 = 7\).
Now, substitute the result of the division back into the original expression:
\[ 7 - 9 \]Subtracting 9 from 7 gives:
\[ 7 - 9 = -2 \]Thus, the final answer is -2.
Correct Option: B) \(-2\)
To find the HCF (Highest Common Factor) of two numbers, we can use the Euclidean algorithm, which is an efficient method for computing the HCF of two integers. The principle is that the HCF of two numbers does not change if the larger number is replaced by its difference with the smaller number. This process is repeated until one of the numbers becomes zero, and the other number is the HCF. Alternatively, we can use the division method repeatedly.
C) 1108 is the HCF of 4432 and 3324, as determined by the Euclidean algorithm.
Read the statement and choose one conclusion from the given alternatives. :
Statement:
All countries are districts
Some villages are not districts
Conclusions:
None follows
Only 2 and 4 follow
Only 2 and 3 follow
Only 3 and either 2 or 4 follow
This is a syllogism problem. We will use Venn diagrams to represent the given statements and then evaluate each conclusion based on these diagrams. We need to consider all possible valid diagrams to ensure the conclusions hold true universally.
Let's represent the statements using Venn diagrams:
Statement 1: All countries are districts.
Statement 2: Some villages are not districts.
Now let's combine these and analyze the conclusions.
Possible Scenarios for combining Statement 1 and Statement 2:
Since "Countries" are entirely inside "Districts", and some "Villages" are outside "Districts", it implies that those "Villages" that are outside "Districts" are definitely also outside "Countries".
However, the relationship between "Countries" and the part of "Villages" that *is* inside "Districts" (or the entire set of "Villages") is not fully defined. We need to consider different overlaps.
Conclusion 1: All countries are villages.
Conclusion 2: Some countries are villages.
Conclusion 3: Some villages are not countries.
Conclusion 4: Some countries are not villages.
Let's re-evaluate Conclusion 2 and 4 with the possibility of "either/or" scenarios.
From Statement 1: All C are D.
From Statement 2: Some V are not D.
This means there is a part of V, let's call it \(V_1\), such that \(V_1 \cap D = \emptyset\). Since \(C \subset D\), it implies \(V_1 \cap C = \emptyset\). So, Conclusion 3 (Some villages are not countries) is definitely true.
Now consider the remaining part of V, let's call it \(V_2\), which might be inside D. The relationship between C and \(V_2\) is not explicitly defined. C is entirely within D.
We have two possibilities for the relationship between C and V:
It is also possible for "Some C are V" and "Some C are not V" to be true simultaneously (i.e., partial overlap).
However, the question asks for conclusions that follow. Conclusion 3 definitely follows. For conclusions 2 and 4, they are contradictory. If one is true, the other is false, and vice-versa, assuming there are countries. If there are countries, then either some countries are villages or some countries are not villages (or both, if there's partial overlap). This means that either Conclusion 2 or Conclusion 4 must be true. This is a complementary pair.
Therefore, Conclusion 3 definitely follows, and either Conclusion 2 or Conclusion 4 follows.
D) Only 3 and either 2 or 4 follow
Find the length of the diagonal of a rectangle of length 6 cm and width 2 cm.
2 \(\sqrt{10}\) cm
± 2 \(\sqrt{10}\) cm
4 \(\sqrt{2}\) cm
± 4 \(\sqrt{2}\) cm
To find the length of the diagonal of a rectangle, we can use the Pythagorean theorem. A rectangle's diagonal divides it into two right-angled triangles, where the length and width are the perpendicular sides, and the diagonal is the hypotenuse.
Correct Option: A) \(2 \sqrt{10}\) cm
Answer the questions based on the information given below.
If '+' is '×', '-' is '+', '×' is '÷' and '÷' is '-' 21 ÷ 8 + 2 - 12 × 3 = ?
This problem involves substituting new operational symbols into a given mathematical expression and then evaluating the expression using the BODMAS/PEMDAS rule.
B) 9 is the correct answer, as derived from the step-by-step evaluation.
This problem involves finding the Least Common Multiple (LCM) of two numbers given their ratio and their Highest Common Factor (HCF). We will use the fundamental relationship between two numbers, their HCF, and their LCM: the product of two numbers is equal to the product of their HCF and LCM.
A) 576 is the correct LCM calculated from the given ratio and HCF.
This problem involves calculating the total profit percentage when a shopkeeper cheats by using less weight during both buying and selling. We can assume a base weight (e.g., 100 units) and calculate the effective weight gained and lost due to cheating, then determine the overall profit.
Understand the cheating mechanism: The shopkeeper cheats by 12% in both buying and selling. This means:
Assume a base quantity: Let's assume the shopkeeper intends to buy 1000 grams (or any convenient unit) of fruit.
Calculate effective weight bought: When buying, he cheats by 12%. This means for every 100 units he pays for, he gets 112 units. So, if he pays for 1000 grams, he actually receives:
\[ \text{Weight received} = 1000 \times \frac{100 + 12}{100} = 1000 \times \frac{112}{100} = 1120 \text{ grams} \]Let the Cost Price (CP) of 1000 grams be \(P\). So, he gets 1120 grams for price \(P\).
Calculate effective weight sold for the same price: When selling, he cheats by 12%. This means for every 100 units he charges for, he gives only 88 units. If he sells all the 1120 grams he bought, he will charge for more than 1120 grams. Let's consider the profit in terms of weight.
Alternatively, consider the profit on a per-unit basis. Let the actual cost of 1 gram be \(C\).
Let the actual cost of 1 unit of weight be $1.
Let's consider the amount of goods. Suppose he intends to deal with 100 units of weight.
Let the true value of 1 unit of weight be $1.
When buying, he gets 112 units for the price of 100 units. So, his cost price (CP) for 112 units is $100.
When selling, he sells 88 units but charges the price of 100 units. So, his selling price (SP) for 88 units is $100.
To compare, let's find the CP and SP for a common quantity, or calculate profit per unit.
Let's use the formula for successive profit percentages due to cheating:
If a shopkeeper cheats by \(x\%\) at the time of buying and \(y\%\) at the time of selling, the total profit percentage is given by:
\[ \text{Total Profit Percentage} = \left( \frac{(100 + x)(100 + y)}{100 \times 100} - 1 \right) \times 100\% \]However, the cheating here is on weight. When cheating by 12% in buying, he gets 112 units for the price of 100 units. This is a profit of 12% on the cost. So, \(P_1 = 12\%\).
When cheating by 12% in selling, he gives 88 units instead of 100 units. He charges for 100 units but gives 88 units. The profit here is on the actual quantity delivered.
Profit on selling = \[\frac{\text{Weight saved}}{\text{Weight delivered}} \times 100 = \frac{100 - 88}{88} \times 100 = \frac{12}{88} \times 100\].
\[ P_2 = \frac{12}{88} \times 100 = \frac{3}{22} \times 100 = \frac{300}{22} = \frac{150}{11}\% \]Now, we can use the successive percentage formula for total profit:
\[ \text{Total Profit} = P_1 + P_2 + \frac{P_1 P_2}{100} \] \[ \text{Total Profit} = 12 + \frac{150}{11} + \frac{12 \times \frac{150}{11}}{100} \] \[ \text{Total Profit} = 12 + \frac{150}{11} + \frac{12 \times 150}{1100} \] \[ \text{Total Profit} = 12 + \frac{150}{11} + \frac{1800}{1100} \] \[ \text{Total Profit} = 12 + \frac{150}{11} + \frac{18}{11} \] \[ \text{Total Profit} = 12 + \frac{150 + 18}{11} \] \[ \text{Total Profit} = 12 + \frac{168}{11} \] \[ \text{Total Profit} = \frac{12 \times 11 + 168}{11} \] \[ \text{Total Profit} = \frac{132 + 168}{11} \] \[ \text{Total Profit} = \frac{300}{11} \] \[ \text{Total Profit} = 27.2727...\% \]Correct Option: B) 27.27