The question asks to identify the type of organs that share a common basic structural design and origin but perform different functions. This is a fundamental concept in evolutionary biology related to homologous and analogous structures.
Correct Option: D) Homologous organs are those organs which have the same basic structural design and origin but have different functions.
A certain sum of money at simple interest amounts to ₹1,512 in 3 years and to ₹1,668 in 4.5 years. The rate of interest on this sum of money is:
\( \frac{26}{3} \) %
\( \frac{24}{3} \) %
\( \frac{28}{3} \) %
10%
We are given the amount of a sum of money at simple interest after two different time periods. We can use this information to find the simple interest accrued in the difference of these time periods. Once we have the simple interest for a certain duration, we can find the simple interest for one year and then calculate the principal amount. Finally, we can determine the rate of interest using the simple interest formula.
The difference in time is \(4.5 - 3 = 1.5\) years.
The difference in amounts is the simple interest earned during this \(1.5\) year period.
\[ \text{Interest for 1.5 years} = A_2 - A_1 = \text{₹}1,668 - \text{₹}1,512 = \text{₹}156 \]The amount after 3 years is the Principal plus the interest for 3 years.
\[ A_1 = P + \text{Interest for 3 years} \]\[ \text{₹}1,512 = P + \text{₹}312 \]\[ P = \text{₹}1,512 - \text{₹}312 = \text{₹}1,200 \]The formula for simple interest is:
\[ \text{Simple Interest} = \frac{P \times R \times T}{100} \]\[ \text{₹}312 = \frac{\text{₹}1,200 \times R \times 3}{100} \]\[ 312 = 12 \times R \times 3 \]\[ 312 = 36 \times R \]\[ R = \frac{312}{36} \]\[ R = \frac{78}{9} \]\[ R = \frac{26}{3} \% \]A) \( \frac{26}{3} \) % — The calculated rate of interest is \( \frac{26}{3} \% \).
A father can complete a task in 8 days, while the son can do it in 7 days. If they work on alternate days, with the father starting, then in how many days will the task be completed?
This problem involves calculating the time taken to complete a task when two individuals work on alternate days. We need to determine their individual work rates, combine them for a two-day cycle, and then calculate the total days required.
If the father can complete a task in 8 days, his one day's work is:
\[ \text{Father's 1 day work} = \frac{1}{8} \]If the son can complete a task in 7 days, his one day's work is:
\[ \text{Son's 1 day work} = \frac{1}{7} \]They work on alternate days, with the father starting. So, on Day 1, the father works, and on Day 2, the son works. A cycle consists of 2 days.
Work done in 2 days (1 cycle) = Father's 1 day work + Son's 1 day work
\[ \text{Work in 1 cycle} = \frac{1}{8} + \frac{1}{7} = \frac{7+8}{56} = \frac{15}{56} \]We need to find out how many full 2-day cycles are required to complete most of the work without exceeding the total work (which is 1).
Let \( N \) be the number of cycles.
\[ N \times \frac{15}{56} \le 1 \]\[ N \le \frac{56}{15} \]\[ N \le 3.73... \]So, they will complete 3 full cycles. This means \( 3 \times 2 = 6 \) days have passed.
Work done in 3 cycles = \[3 \times \frac{15}{56} = \frac{45}{56}\]
Remaining work = Total work - Work done in 3 cycles
\[ \text{Remaining work} = 1 - \frac{45}{56} = \frac{56-45}{56} = \frac{11}{56} \]After 3 full cycles (6 days), it's the father's turn to work (since the father starts the first day of each cycle).
The father's 1 day work is \( \frac{1}{8} \).
Since \[\frac{1}{8} = \frac{7}{56}\], and the remaining work is \( \frac{11}{56} \), the father can complete his full day's work and still some work will be left.
On Day 7 (father's turn), the father does \( \frac{1}{8} \) of the work.
Work remaining after Day 7 = \[\frac{11}{56} - \frac{1}{8} = \frac{11}{56} - \frac{7}{56} = \frac{4}{56}\]
Now it's the son's turn. The son's 1 day work is \( \frac{1}{7} \).
Time taken by son to complete the remaining \( \frac{4}{56} \) work = \[\frac{\text{Remaining work}}{\text{Son's 1 day work}} = \frac{4/56}{1/7} = \frac{4}{56} \times 7 = \frac{28}{56} = \frac{1}{2} \text{ day}\]
Total days = 6 days (for 3 cycles) + 1 day (father's turn) + \( \frac{1}{2} \) day (son's turn)
\[ \text{Total days} = 6 + 1 + \frac{1}{2} = 7\frac{1}{2} \text{ days} \]B) 7 \( \frac{1}{2} \)
The question asks to identify the specific decade designated by the United Nations as the "International Decade for Drinking Water and Sanitation." This is a factual question related to UN initiatives and requires knowledge of historical designations.
The United Nations General Assembly, in its resolution 35/18 of 10 November 1980, proclaimed the period 1981-1990 as the International Drinking Water Supply and Sanitation Decade (IDWSSD). The primary goal of this decade was to provide clean drinking water and adequate sanitation for all by 1990.
Correct Option: B) 1981-1990
This problem requires the application of a fundamental algebraic identity relating the sum of cubes to the sum of the numbers, the sum of their products taken two at a time, and their product. The identity is: \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)\). We are given the values for \(a+b+c\), \(abc\), and \(ab+bc+ca\). We need to find \(a^2 + b^2 + c^2\) first, which can be derived from the square of the sum \((a+b+c)^2\).
Recall the identity for the sum of three numbers squared:
We know that \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)\).
Calculate \(a^2 + b^2 + c^2\):
Given \(a+b+c = 17\) and \(ab+bc+ca = 94\).
\[ (17)^2 = a^2 + b^2 + c^2 + 2(94) \]\[ 289 = a^2 + b^2 + c^2 + 188 \]\[ a^2 + b^2 + c^2 = 289 - 188 \]\[ a^2 + b^2 + c^2 = 101 \]Recall the identity for the sum of cubes:
The identity is \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)\).
Substitute the known values into the identity:
We have \(a+b+c = 17\), \(abc = 168\), \(ab+bc+ca = 94\), and we just found \(a^2 + b^2 + c^2 = 101\).
\[ a^3 + b^3 + c^3 - 3(168) = (17)(101 - 94) \]\[ a^3 + b^3 + c^3 - 504 = (17)(7) \]\[ a^3 + b^3 + c^3 - 504 = 119 \]Solve for \(a^3 + b^3 + c^3\):
\[ a^3 + b^3 + c^3 = 119 + 504 \]\[ a^3 + b^3 + c^3 = 623 \]Correct Option: C) 623
This problem can be solved using the fundamental relationship between the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of two numbers. The product of two numbers is always equal to the product of their HCF and LCM.
Correct Option: D) 210
The question asks to identify the Prime Minister of India who signed the Assam Accord in 1985 with the leaders of the All Assam Students' Union (AASU).
Correct Option: B) Rajiv Gandhi
The question asks to identify the region where 'Viticulture' is a speciality. Viticulture refers to the cultivation of grapevines, primarily for wine production. We need to recall the geographical regions known for extensive grape cultivation and wine-making.
Correct Option: A) The Mediterranean region is renowned for its ideal climate (hot, dry summers and mild, wet winters) which is perfectly suited for grape cultivation. Countries like Italy, France, Spain, and Greece, all part of this region, have a long and rich history of viticulture and wine production, making it a defining speciality of the area.
The question asks to identify the true statements regarding medians and centroids of a triangle. We need to recall the definitions of a median and a centroid, and their properties.
B) Both statements are true. Both definitions and properties mentioned are standard in Euclidean geometry.
Two integers are selected at random from the first 11 natural numbers. If the sum of the integers is even, then the probability that both the numbers are odd is:
\( \frac{13}{121} \)
\( \frac{3}{5} \)
\( \frac{4}{9} \)
\( \frac{5}{11} \)
This is a problem of conditional probability. We need to find the probability that both numbers are odd, given that their sum is even. We will first list the natural numbers, identify odd and even numbers, then calculate the total number of ways to select two numbers such that their sum is even, and finally, the number of ways both selected numbers are odd.
Correct Option: B) \( \frac{3}{5} \)