The following table gives population and total income of a city for four years: 
Which one of the following statements is correct in respect of the above data?
Population increased by 5% or more every year.
Income increased by 10% or more every year.
Per capita income was always above ₹ 5, 000.
per capita income was highest in 1994.
To determine the correct statement, we need to calculate the per capita income for each year and compare it to ₹ 5,000. We also need to verify the percentage increase in population and income to evaluate the other options.
Correct Option: C) Per capita income was always above ₹ 5,000.
Consider the table given below in which the numbers bear certain relationship among themselves along the rows: 
Which one of the following numbers is the missing number indicated above by X ?
To find the missing number indicated by X in the table, we need to identify the pattern or relationship among the numbers in each row.
The correct missing number should be 8 to maintain consistency with the pattern across all rows.
Correct Option: D) 8
Consider the following matrix with one empty block in the lower extreme corner: 
Which of the following figures could fit in the empty block and thus complete the matrix?
The matrix consists of a pattern of circles and triangles. To solve the problem, we need to identify the pattern and determine which option completes the matrix correctly.
Correct Option: A) The block with two triangles fits the pattern of increasing triangles.
With reference to the figure given below, the number of different routes from S to T without retracing from U and/or V, is
To find the number of different routes from S to T without retracing from U and/or V, we need to analyze the possible paths using the diagram's structure.
Correct Option: D) 18
Consider the following figures: 
Change in the positions of beads in the four figures above follows a sequence. Following the same sequence, which of the figures below should appear as the fifth figure above?
This question involves identifying a pattern in the sequence of diagrams and predicting the next figure based on that pattern.
Correct Option: B) The shading has rotated correctly to the bottom-left bead, consistent with the sequence.
To find out when the three bells will ring together again, we need to find the least common multiple (LCM) of the time intervals at which each bell rings. The LCM will give us the minimum time after which all three events (bells ringing) will coincide again.
Correct Option: B) 12:48 hrs
The bells will ring together again at 12:48 PM.
The question describes a scenario where the prices of all goods and services double. It then asks what the implication of this universal price increase would be on values. The core concept to understand here is the relationship between price, value, and money.
D) the value of money only would be halved. If all prices double, it means that the purchasing power of money has decreased by half. You need two units of currency to buy what one unit could buy before. The relative value of goods to each other remains unchanged because all prices have increased proportionally.
The problem requires calculating the total bus fare for two persons, determining how much money A has initially, how much he has after receiving money from B, and then comparing this total with the required fare to find the remaining amount or deficit.
A has ₹ 10, which is 80% of the bus fare for two persons.
Let the total bus fare for two persons be \(F\).
\[ 80\% \text{ of } F = ₹ 10 \] \[ 0.80 \times F = 10 \] \[ F = \frac{10}{0.80} \] \[ F = \frac{100}{8} \] \[ F = ₹ 12.50 \]So, the total bus fare for two persons is ₹ 12.50.
Initially, A has ₹ 10.
B has ₹ 3 and hands it over to A.
Total money A has now \( = ₹ 10 + ₹ 3 = ₹ 13 \).
Money A has \( = ₹ 13 \).
Bus fare for two persons \( = ₹ 12.50 \).
Since A has ₹ 13 and the fare is ₹ 12.50, A has more than enough money.
Money left with A \( = \text{Total money A has} - \text{Bus fare} \)
\[ = ₹ 13.00 - ₹ 12.50 \] \[ = ₹ 0.50 \]₹ 0.50 is equivalent to 50 paise.
C) After buying the two tickets A will be left with 50 paise.
As calculated, A will have ₹ 0.50 or 50 paise remaining after purchasing the tickets.
This problem can be solved by setting up a proportion. We are given the number of instalments paid and the percentage of the loan refunded. We need to find the total number of instalments required to refund 100% of the loan.
Let \(T\) be the total number of instalments in the agreement.
We are given that after paying 18 instalments, 60 percent of the loan was refunded. This can be written as a ratio:
\[ \frac{\text{Instalments paid}}{\text{Percentage of loan refunded}} = \frac{18}{60\%} \]We want to find the total number of instalments \(T\) when 100 percent of the loan is refunded. So, we can set up a proportion:
\[ \frac{18}{60} = \frac{T}{100} \]Now, we solve for \(T\):
\[ T = \frac{18 \times 100}{60} \]Simplify the expression:
\[ T = \frac{1800}{60} \] \[ T = 30 \]Therefore, there were 30 instalments in the agreement.
Correct Option: C) 30
This problem involves calculating the distance given different speeds and corresponding time differences. We can use the fundamental relationship between distance, speed, and time, and set up equations based on the given conditions. Let the distance be \(D\) and the usual time taken be \(T\). We will express the time taken in each scenario and relate them to the usual time.
Define variables:
Let the distance from the house to the factory be \(D\) km.
Let the usual time taken to reach the factory be \(T\) hours.
Formulate equations based on the first condition:
If the speed is 5 km/hr, the worker is 3 minutes late. This means the time taken is \(T + 3 \text{ minutes}\).
Convert minutes to hours: \[3 \text{ minutes} = \frac{3}{60} \text{ hours} = \frac{1}{20} \text{ hours}\].
So, the time taken is \(T_1 = T + \frac{1}{20}\) hours.
Using the formula \(\text{Distance} = \text{Speed} \times \text{Time}\):
\[ D = 5 \times \left(T + \frac{1}{20}\right) \quad \ldots (1) \]
Formulate equations based on the second condition:
If the speed is 6 km/hr, the worker is 7 minutes early. This means the time taken is \(T - 7 \text{ minutes}\).
Convert minutes to hours: \(7 \text{ minutes} = \frac{7}{60} \text{ hours}\).
So, the time taken is \(T_2 = T - \frac{7}{60}\) hours.
Using the formula \(\text{Distance} = \text{Speed} \times \text{Time}\):
\[ D = 6 \times \left(T - \frac{7}{60}\right) \quad \ldots (2) \]
Solve the equations:
Since the distance \(D\) is the same in both cases, we can equate equations (1) and (2):
\[ 5 \times \left(T + \frac{1}{20}\right) = 6 \times \left(T - \frac{7}{60}\right) \]
Distribute the speeds:
\[ 5T + \frac{5}{20} = 6T - \frac{42}{60} \]
Simplify the fractions:
\[ 5T + \frac{1}{4} = 6T - \frac{7}{10} \]
Rearrange the terms to solve for \(T\):
\[ \frac{1}{4} + \frac{7}{10} = 6T - 5T \]
\[ T = \frac{1}{4} + \frac{7}{10} \]
Find a common denominator (20) for the fractions:
\[ T = \frac{5}{20} + \frac{14}{20} \]
\[ T = \frac{19}{20} \text{ hours} \]
Calculate the distance \(D\):
Substitute the value of \(T\) into either equation (1) or (2). Using equation (1):
\[ D = 5 \times \left(T + \frac{1}{20}\right) \]
\[ D = 5 \times \left(\frac{19}{20} + \frac{1}{20}\right) \]
\[ D = 5 \times \left(\frac{20}{20}\right) \]
\[ D = 5 \times 1 \]
\[ D = 5 \text{ km} \]
Correct Option: C) 5 km
