To find the average price per book, we need to calculate the total cost of all books and divide it by the total number of books. This involves calculating the cost from each shop and then summing them up.
Number of books from first shop = \(51\)
Price per book from first shop = ₹\(1,120\)
\[ \text{Cost}_1 = 51 \times 1120 = \text{₹}57,120 \]Number of books from second shop = \(49\)
Price per book from second shop = ₹\(900\)
\[ \text{Cost}_2 = 49 \times 900 = \text{₹}44,100 \]Correct Option: C) 1,012.2
Three numbers are in the ratio \( \frac{2}{7} \) : \( \frac{1}{2} \) : \( \frac{16}{15} \). The difference between the largest and the smallest number is 41. Find the largest number.
To solve this problem, we first need to convert the given ratio of fractions into a ratio of whole numbers. Then, we can represent the numbers using a common variable and use the given difference between the largest and smallest numbers to find the value of that variable. Finally, we can calculate the largest number.
Convert the ratio of fractions to a ratio of whole numbers:
The given ratio is \[\frac{2}{7} : \frac{1}{2} : \frac{16}{15}\].
To convert this to whole numbers, find the Least Common Multiple (LCM) of the denominators (7, 2, 15).
The LCM of 7, 2, and 15 is \( 7 \times 2 \times 15 = 210 \).
Multiply each fraction by the LCM:
Identify the smallest and largest parts of the ratio:
The ratio of the three numbers is \( 60 : 105 : 224 \).
Smallest part = 60
Largest part = 224
Represent the numbers using a common variable:
Let the three numbers be \( 60x \), \( 105x \), and \( 224x \).
Use the given difference to find the value of \(x\):
The difference between the largest and the smallest number is 41.
Largest number \( - \) Smallest number \( = 41 \)
\( 224x - 60x = 41 \)
\( 164x = 41 \)
\( x = \frac{41}{164} \)
\( x = \frac{1}{4} \)
Calculate the largest number:
The largest number is \( 224x \).
Substitute the value of \( x \):
\[\text{Largest number} = 224 \times \frac{1}{4}\]
\( \text{Largest number} = 56 \)
Correct Option: C) 56
The problem involves arranging seven people (P, Q, R, S, T, U, V) for exams on different days of a week, from Monday to Sunday. We need to use the given clues to deduce the exam schedule and then determine how many people have exams after Q.
List the days of the week:
Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Place V:
V has an exam on Thursday.
| Day | Person |
| Monday | |
| Tuesday | |
| Wednesday | |
| Thursday | V |
| Friday | |
| Saturday | |
| Sunday |
Analyze U and Q:
Exactly 4 people have their exams between U and Q. Neither U nor Q has an exam on Sunday. This means U and Q cannot be on Monday and Saturday (4 people between them, but Sunday is excluded for Q/U), or Tuesday and Sunday (Sunday excluded). Also, if U is on Monday, Q must be on Saturday. If U is on Tuesday, Q must be on Sunday (not possible). If U is on Wednesday, Q must be on Monday (not possible). If U is on Friday, Q must be on Tuesday. If U is on Saturday, Q must be on Monday. If U is on Sunday, Q must be on Tuesday (not possible).
Let's consider the possible pairs for (U, Q) with 4 people in between, excluding Sunday for U and Q:
So, the possible pairs are (U-Mon, Q-Sat) or (Q-Mon, U-Sat).
Analyze T and S:
T has an exam immediately before S. This means T and S must be consecutive days (TS block).
Analyze U and S:
U does not have an exam after S. This means U must have an exam before S.
Combine clues:
Case 1: U is on Monday, Q is on Saturday.
| Day | Person |
| Monday | U |
| Tuesday | |
| Wednesday | |
| Thursday | V |
| Friday | |
| Saturday | Q |
| Sunday |
Now, place the TS block. Since U is on Monday, and U must be before S, S cannot be on Tuesday (T-Mon, S-Tue, but U is Mon). S cannot be on Wednesday (T-Tue, S-Wed). S cannot be on Friday (T-Thu, S-Fri, but V is Thu). S cannot be on Sunday (T-Sat, S-Sun, but Q is Sat). So, the TS block can only be on Tuesday-Wednesday or Friday-Saturday (not possible as Q is Sat).
If T is on Tuesday and S is on Wednesday:
| Day | Person |
| Monday | U |
| Tuesday | T |
| Wednesday | S |
| Thursday | V |
| Friday | |
| Saturday | Q |
| Sunday |
Remaining people: P, R. Remaining days: Friday, Sunday. R does not have an exam on Friday. So, R must be on Sunday and P on Friday.
| Day | Person |
| Monday | U |
| Tuesday | T |
| Wednesday | S |
| Thursday | V |
| Friday | P |
| Saturday | Q |
| Sunday | R |
Let's check all conditions: V on Thursday (Yes). 4 people between U and Q (T, S, V, P - Yes). Neither U nor Q on Sunday (Yes). T immediately before S (Yes). R not on Friday (Yes). U not after S (U is Mon, S is Wed - Yes). All conditions are met.
Case 2: Q is on Monday, U is on Saturday.
| Day | Person |
| Monday | Q |
| Tuesday | |
| Wednesday | |
| Thursday | V |
| Friday | |
| Saturday | U |
| Sunday |
Place the TS block. U is on Saturday, and U must be before S. This is impossible as S must be after T, and U is already late in the week. If S is on Sunday, T must be on Saturday, but U is on Saturday. If S is on Friday, T is on Thursday, but V is on Thursday. This case is not possible because U must be before S.
Final Schedule (from Case 1):
| Day | Person |
| Monday | U |
| Tuesday | T |
| Wednesday | S |
| Thursday | V |
| Friday | P |
| Saturday | Q |
| Sunday | R |
Answer the question:
How many people have exams after Q? Q has an exam on Saturday. The only person after Q is R (on Sunday).
Therefore, 1 person has an exam after Q.
B) 1
This problem involves setting up a system of two linear equations with two variables based on the given information. We will define variables for the cost of one pen and one notebook, form the equations, solve them simultaneously to find the individual costs, and then calculate the final required cost.
Define variables:
Let \(P\) be the cost of one pen (in ₹).
Let \(N\) be the cost of one notebook (in ₹).
Formulate equations from the given statements:
From the first statement: "The cost of 3 pens and 12 notebooks is ₹195."
\[ 3P + 12N = 195 \quad \text{(Equation 1)} \]
From the second statement: "The cost of 7 notebooks exceeds the cost of 4 pens by ₹39." This means the cost of 7 notebooks is ₹39 more than the cost of 4 pens.
\[ 7N = 4P + 39 \quad \text{(Equation 2)} \]
Solve the system of equations:
We can simplify Equation 1 by dividing by 3:
\[ P + 4N = 65 \quad \text{(Equation 3)} \]
From Equation 3, express \(P\) in terms of \(N\):
\[ P = 65 - 4N \quad \text{(Equation 4)} \]
Substitute Equation 4 into Equation 2:
\[ 7N = 4(65 - 4N) + 39 \] \[ 7N = 260 - 16N + 39 \] \[ 7N + 16N = 260 + 39 \] \[ 23N = 299 \] \[ N = \frac{299}{23} \]
\[ N = 13 \]
Now substitute the value of \(N\) back into Equation 4 to find \(P\):
\[ P = 65 - 4(13) \]
\[ P = 65 - 52 \]
\[ P = 13 \]
So, the cost of one pen is ₹13 and the cost of one notebook is ₹13.
Calculate the required cost:
We need to find the cost of 12 pens and 6 notebooks.
\[ \text{Cost} = 12P + 6N \]
\[ \text{Cost} = 12(13) + 6(13) \]
\[ \text{Cost} = 156 + 78 \]
\[ \text{Cost} = 234 \]
The cost of 12 pens and 6 notebooks is ₹234.
Correct Option: C) ₹234
The problem asks us to identify the highest and lowest numbers from a given set of three-digit numbers, then extract specific digits from them, and finally perform a subtraction. We need to follow the steps of identifying the numbers, finding their second digits, and then calculating the difference.
The question asks about the primary purpose of the Eastern Peripheral Expressway (EPE). This requires knowledge of major infrastructure projects in India and their geographical impact.
Correct Option: A) The Eastern Peripheral Expressway (EPE) is a 135 km long, six-lane access-controlled expressway that forms a part of the larger Peripheral Expressway around Delhi. Its primary purpose is to divert non-Delhi bound traffic, especially commercial vehicles, from entering the capital city, thereby significantly reducing traffic congestion and pollution within Delhi. It is also known as the National Expressway 2 (NE-2).
The problem presents a series of letter-number clusters. We need to identify the pattern in the letters and numbers separately to find the next cluster in the series.
S (19) → Q (17) → O (15) → M (13)
The pattern for the first letter is a decrease of 2 in alphabetical position (19 - 2 = 17, 17 - 2 = 15, 15 - 2 = 13). Following this pattern, the next letter will be M (13) - 2 = K (11).
M (13) → K (11) → I (9) → G (7)
The pattern for the second letter is also a decrease of 2 in alphabetical position (13 - 2 = 11, 11 - 2 = 9, 9 - 2 = 7). Following this pattern, the next letter will be G (7) - 2 = E (5).
13 → 25 → 37 → 49
Let's find the difference between consecutive numbers:
The pattern for the number is an addition of 12 to the previous number. Following this pattern, the next number will be \(49 + 12 = 61\).
The next cluster will have the first letter K, the second letter E, and the number 61. So, the next cluster is KE61.
C) KE61 is the correct option because it follows all the identified patterns for the letters and the number.
The question asks for the number of Scheduled Tribes notified under Article 342 of the Indian Constitution as of December 2022, according to the Ministry of Tribal Affairs. This requires knowledge of the official statistics provided by the relevant government ministry.
D) Over 730. As per the Ministry of Tribal Affairs, there are 753 Scheduled Tribes notified under Article 342 of the Constitution of India as of December 2022. Therefore, "Over 730" is the correct representation of this number.
The question asks about the primary building material of the Jahangiri Mahal in Agra Fort. This requires knowledge of the architectural characteristics of Mughal-era constructions, specifically those found in the Agra Fort.
Correct Option: C) Red sandstone — The Jahangiri Mahal, located within the Agra Fort, was primarily constructed using red sandstone. This was a characteristic building material during the reign of Emperor Akbar, who initiated much of the construction at Agra Fort. The use of red sandstone gives the palace its distinctive appearance and reflects the architectural style of the period.
The question asks about a significant milestone achieved by INSV Tarini during the third leg of Navika Sagar Parikrama II in January 2025. This requires knowledge of the expedition's details and the specific events that occurred.
C) Crossed Point Nemo, the Oceanic Pole of Inaccessibility
INSV Tarini, with its all-women crew, successfully crossed Point Nemo, the Oceanic Pole of Inaccessibility, in January 2025 during the third leg of the Navika Sagar Parikrama II. This is a significant navigational and endurance achievement, marking the vessel's journey through one of the most remote parts of the world's oceans.