GDS To MTS Question Paper 2024 Telangana Circle

Logistics Post provides business customers a cost-effective and efficient solution, managing the entire value chain from collection, storage, transmission, to distribution across the country.

Customers can send their consignments either in Full Truck Load (FTL) or Less than Truck Load (LTL), as single or multiple parcels, based on their requirements. It is flexible and convenient.

Post Offices in India are divided into three classes – Head Post Offices, Sub Post Offices (including E.D. Sub Offices), and E.D. Branch Post Offices.

A Camp Bag is used to enclose office files and other official papers, and is closed by the Secretariat or Headquarters offices of the High Officers mentioned in Clause 198 of Post Office Guide, Part-I, for the High Officer in Camp, and vice versa.

There are seven countries in BIMSTEC: Bangladesh, Bhutan, India, Myanmar, Nepal, Sri Lanka, and Thailand.

Member (PLI) is renamed as Member (Financial Services).

Any postal article found to contain goods transmitted/imported in contravention of any law for the time being in force will be returned to the sender.

If an article is suspected to contain such goods, it will be forwarded marked “Doubtful” to the office of destination, where it will be opened in the presence of the addressee.

Reinforcing efforts to increase the reach of Philately, the Department of Posts launched a Scholarship Scheme called Deen Dayal SPARSH Yojana for students of Classes VI to IX.

Under the Scholarship for Promotion of Aptitude & Research in Stamps as a Hobby (SPARSH Yojana), annual scholarships are awarded to students with a good academic record who also pursue Philately as a hobby.

The objective of the scholarship is to promote Philately among children at a young age in a sustainable manner, helping to reinforce and supplement the academic curriculum, while also providing a hobby for relaxation and de-stressing.

In order to meet the requirements of Bulk Customers for collection of amount for costlier goods, the Cash on Delivery (COD) facility is available, which provides collection of amount up to ₹50,000 at the time of delivery of consignments.

The Cash on Delivery (COD) facility is available to Express Parcel, Business Parcel, and Speed Post customers.

Let the required number of days be $D_2$. Using the formula $M_1 \times D_1 = M_2 \times D_2$, we have $5 \times 10 = 1 \times D_2$. Therefore, $D_2 = 50$ days.

The formula for Simple Interest is $SI = \frac{P \times R \times T}{100}$. Substituting the given values, we get $1540 = \frac{X \times 14 \times 5}{100}$. Simplifying, $1540 = \frac{70X}{100}$, which gives $1540 = 0.7X$. Solving for $X$, $X = \frac{1540}{0.7} = 2200$. The principal amount is Rs $2200$.

Simple Interest formula is $SI = \frac{P \times R \times T}{100}$. Here $P = 7200$, $R = 12\%$, and $T = 7$ years. $SI = \frac{7200 \times 12 \times 7}{100} = 72 \times 12 \times 7 = 72 \times 84 = 6048$. The simple interest is Rs $6048$.

Let the value of $Y$ be $100$. Then the value of $X$ is $100 + 25\% \text{ of } 100 = 125$. The difference between $X$ and $Y$ is $125 - 100 = 25$. To find out how much $Y$ is less than $X$ in percentage, we calculate: $\frac{\text{Difference}}{X} \times 100 = \frac{25}{125} \times 100 = \frac{1}{5} \times 100 = 20\%$.

The total distance covered by the train when it crosses a tunnel is the sum of the length of the train and the length of the tunnel. Total distance = Speed $\times$ Time = $100 \text{ m/s} \times 10 \text{ s} = 1000 \text{ m}$. Since the length of the train is $500 \text{ m}$, the length of the tunnel is $1000 \text{ m} - 500 \text{ m} = 500 \text{ m}$.

Using the BODMAS rule, first solve the brackets: $(22+6) = 28$. The expression becomes $2 + 4 \div 28 \times 2$. Next, perform division: $4 \div 28 = \frac{4}{28} = \frac{1}{7}$. Then, perform multiplication: $\frac{1}{7} \times 2 = \frac{2}{7}$. Finally, perform addition: $2 + \frac{2}{7} = \frac{14}{7} + \frac{2}{7} = \frac{16}{7}$.

According to the BODMAS rule, perform division first: $10 \div 10 = 1$. The expression becomes $10 + 10 \times 1$. Next, perform multiplication: $10 \times 1 = 10$. The expression becomes $10 + 10$. Finally, perform addition: $10 + 10 = 20$.

Let the old average after $9$ innings be $x$. Total runs after $9$ innings $= 9x$. In the $10$th innings, he scores $100$ runs, so the new total runs $= 9x + 100$. The new average becomes $x + 8$. Therefore, we can write the equation: $\frac{9x + 100}{10} = x + 8$. Solving this, we get $9x + 100 = 10x + 80$, which gives $x = 20$. His new average is $x + 8 = 20 + 8 = 28$.

Cloth required for $1$ shirt = $2 \text{ m } 47 \text{ cm}$. Total cloth required for $13$ shirts $= 13 \times (2 \text{ m } 47 \text{ cm})$. First, multiply meters: $13 \times 2 \text{ m} = 26 \text{ m}$. Then multiply centimeters: $13 \times 47 \text{ cm} = 611 \text{ cm}$. Since $100 \text{ cm} = 1 \text{ m}$, $611 \text{ cm} = 6 \text{ m } 11 \text{ cm}$. Adding them together gives $26 \text{ m} + 6 \text{ m } 11 \text{ cm} = 32 \text{ m } 11 \text{ cm}$.

Work done by A in $1$ day $= \frac{1}{10}$. Work done by B in $1$ day $= \frac{1}{40}$. Work done by A and B together in $1$ day $= \frac{1}{10} + \frac{1}{40} = \frac{4}{40} + \frac{1}{40} = \frac{5}{40} = \frac{1}{8}$. Since they complete $\frac{1}{8}$ of the work in one day, they will take $8$ days to complete the entire work together.