SubQuestion No: 3
What is the total distance travelled by all the trucks together on Saturday?
To find the total distance travelled by all trucks together on Saturday, we need to locate the data for Saturday for each truck and sum up the distances. We will assume the question refers to a table or chart where this data is presented.
\[ \text{Total distance} = 1632 \text{ km} \]
D) 1632 km is the correct answer, as it is the sum of the distances travelled by all trucks on Saturday.
The question asks about the gas used to prevent rancidity in potato chips. Rancidity is caused by the oxidation of fats and oils. To prevent this, an inert gas is typically used to displace oxygen.
C) Nitrogen is an inert gas that is commonly used to flush food packaging, like chip bags, to remove oxygen and prevent the oxidation of fats and oils, thus preventing rancidity.
Mr. Piyush sold a bus for Rs. 3,400 with a loss of 15%. At what price should the bus be sold to get a profit of 15%?
To solve this problem, we first need to find the cost price (CP) of the bus. We are given the selling price (SP) and the percentage loss. Once we have the cost price, we can calculate the new selling price required to achieve a 15% profit.
Calculate the Cost Price (CP):
Given:
The formula for selling price with a loss is:
\[ \text{SP} = \text{CP} \times \left(1 - \frac{\text{Loss}\%}{100}\right) \]Substitute the given values:
\[ 3400 = \text{CP} \times \left(1 - \frac{15}{100}\right) \] \[ 3400 = \text{CP} \times \left(\frac{100 - 15}{100}\right) \] \[ 3400 = \text{CP} \times \frac{85}{100} \]Now, solve for CP:
\[ \text{CP} = \frac{3400 \times 100}{85} \] \[ \text{CP} = \frac{340000}{85} \] \[ \text{CP} = 4000 \]So, the Cost Price of the bus is Rs. 4,000.
Calculate the New Selling Price for 15% Profit:
We want to achieve a profit of 15% on the Cost Price (CP = Rs. 4,000).
The formula for selling price with a profit is:
\[ \text{New SP} = \text{CP} \times \left(1 + \frac{\text{Profit}\%}{100}\right) \]Substitute the values:
\[ \text{New SP} = 4000 \times \left(1 + \frac{15}{100}\right) \] \[ \text{New SP} = 4000 \times \left(\frac{100 + 15}{100}\right) \] \[ \text{New SP} = 4000 \times \frac{115}{100} \] \[ \text{New SP} = 40 \times 115 \] \[ \text{New SP} = 4600 \]Therefore, the bus should be sold for Rs. 4,600 to get a profit of 15%.
Correct Option: D) 4,600
This is a coding-decoding question where a specific pattern or rule is applied to transform words into numbers. We need to identify this rule by analyzing the given examples (TRACTOR = 14, TROLLEY = 14) and then apply it to find the value for SCOOTER.
B) 14. The pattern is to multiply the number of letters in the word by 2. SCOOTER has 7 letters, so \(7 \times 2 = 14\).
The question asks to find a pair that shares the same relationship as the first pair: LUNG : MAN. This is an analogy question where we need to identify the functional relationship between the first two words and apply it to the second pair.
Correct Option: A) FISH is the correct choice because gills are the respiratory organs for fish, just as lungs are for man.
To divide a total amount in a given ratio, first find the sum of the ratio parts. Then, divide the total amount by this sum to find the value of one ratio unit. Finally, multiply this unit value by each part of the ratio to find the respective shares.
The given ratio is 1:2:4:5.
\[ \text{Sum of ratio parts} = 1 + 2 + 4 + 5 = 12 \]The total amount to be divided is Rs. 156.
\[ \text{Value of one ratio unit} = \frac{\text{Total amount}}{\text{Sum of ratio parts}} = \frac{156}{12} = 13 \]Thus, the amounts in the respective ratios are Rs. 13, Rs. 26, Rs. 52, and Rs. 65.
Correct Option: C) 13, 26, 52 & 65
Answer the next 3 Questions.
Comprehension:
Read the data given below and answer the questions based on it.
There were 3 sections namely A, B, and C in a test. Out of three sections, 33 students passed in Section A, 34 students passed in Section B and 32 passed in Section C. 10 students passed in Section A and Section B, 9 passed in Section B and Section C, 8 passed in Section A and Section C. The number of students who passed each section alone was equal and was 21 for each section.
SubQuestion No: 1
The ratio of the number of students passed in one or more of the sections to the number of students who passed in Section A alone is:
\( \frac{78}{21} \)
3
\( \frac{73}{21} \)
\( \frac{75}{21} \)
To solve this problem, we need to find the number of students who passed in one or more sections (union of A, B, and C) and the number of students who passed in Section A alone. Then we will calculate the ratio of these two values.
Identify given data:
Find the number of students who passed in all three sections, \(n(A \cap B \cap C)\):
We know that the number of students who passed in Section A alone is given by:
\[ n(A \text{ alone}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C) \]Substitute the given values:
\[ 21 = 33 - 10 - 8 + n(A \cap B \cap C) \] \[ 21 = 33 - 18 + n(A \cap B \cap C) \] \[ 21 = 15 + n(A \cap B \cap C) \] \[ n(A \cap B \cap C) = 21 - 15 = 6 \]So, 6 students passed in all three sections.
Calculate the number of students who passed in one or more sections, \(n(A \cup B \cup C)\):
Using the Principle of Inclusion-Exclusion:
\[ n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \]Substitute the values:
\[ n(A \cup B \cup C) = 33 + 34 + 32 - 10 - 9 - 8 + 6 \] \[ n(A \cup B \cup C) = 99 - 27 + 6 \] \[ n(A \cup B \cup C) = 72 + 6 \] \[ n(A \cup B \cup C) = 78 \]Alternatively, we can sum the number of students in each distinct region of the Venn diagram:
Total passed in one or more sections:
\[ 21 + 21 + 21 + 4 + 3 + 2 + 6 = 78 \]Determine the number of students who passed in Section A alone:
This is directly given as 21.
Calculate the ratio:
\[ \text{Ratio} = \frac{\text{Number of students passed in one or more sections}}{\text{Number of students passed in Section A alone}} \] \[ \text{Ratio} = \frac{78}{21} \]Correct Option: A) \( \frac{78}{21} \)
SubQuestion No: 2
How many passed all the three sections?
The problem asks us to find the number of candidates who passed all three sections based on a Venn diagram. We need to identify the region in the Venn diagram that represents the intersection of all three sets.
B) 6 is the number located in the central intersection of all three circles in the Venn diagram, which signifies the candidates who passed all three sections.
SubQuestion No: 3
How many passed only one of the three sections?
To find the number of students who passed only one of the three sections, we need to identify the regions in the Venn diagram that represent students passing exactly one section. These are the non-overlapping parts of each individual circle.
Therefore, 63 students passed only one of the three sections.
Correct Option: C) 63
The question asks to identify a computer storage device that has become obsolete. We need to evaluate each option based on its current usage and relevance in modern computing.
Based on this analysis, the floppy disk is the storage device that has become largely obsolete.
Correct Option: A) Floppy