UPSC CBI DSP LDCE

1. A sum of money invested at compound interest amounts to ₹ 400 in 2 yea

A sum of money invested at compound interest amounts to ₹ 400 in 2 years and to ₹ 420 in 3 years. The rate of interest per annum is

2.5%
5.5%
4%
5%
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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Let the principal amount be P and the annual rate of interest be r (as a decimal).
The formula for the amount A after t years at compound interest is $A = P(1+r)^t$.
Given:
Amount after 2 years ($A_2$) = ₹ 400
Amount after 3 years ($A_3$) = ₹ 420

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So, we have two equations:
1) $400 = P(1+r)^2$
2) $420 = P(1+r)^3$

To find the rate r, divide equation (2) by equation (1):
$\frac{420}{400} = \frac{P(1+r)^3}{P(1+r)^2}$
$\frac{42}{40} = (1+r)^{3-2}$
$\frac{21}{20} = 1+r$

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Now, solve for r:
$r = \frac{21}{20} – 1$
$r = \frac{21 – 20}{20}$
$r = \frac{1}{20}$

To express the rate as a percentage, multiply by 100:
Rate = $\frac{1}{20} \times 100\% = 5\%$.

In compound interest, the amount for any year becomes the principal for the next year. The ratio of the amount at the end of year (t+1) to the amount at the end of year (t) is equal to $(1+r)$, where r is the annual interest rate.
$\frac{A_{t+1}}{A_t} = 1+r$
In this problem, the amount after 2 years (₹400) acts as the principal for the 3rd year. The interest earned during the 3rd year is $420 – 400 = ₹20$.
The interest rate can be calculated as the interest earned in the 3rd year divided by the principal at the beginning of the 3rd year (which is the amount after 2 years).
Interest rate = $\frac{\text{Interest earned in 3rd year}}{\text{Amount after 2 years}} = \frac{20}{400} = \frac{1}{20}$.
As a percentage, this is $\frac{1}{20} \times 100\% = 5\%$. This method is valid only because the time difference is exactly one year.

2. Kumar completes a work in 8 days and Raj completes the same work in 16

Kumar completes a work in 8 days and Raj completes the same work in 16 days. In how many days can Kumar and Raj together complete the work?

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$3 rac{1}{3}$ days
$5 rac{1}{3}$ days
$1 rac{1}{3}$ days
$ rac{1}{3}$ day
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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Kumar completes the work in 8 days, so his work rate is $\frac{1}{8}$ of the work per day.
Raj completes the work in 16 days, so his work rate is $\frac{1}{16}$ of the work per day.
When working together, their work rates add up.
Combined work rate = Work rate of Kumar + Work rate of Raj
Combined work rate = $\frac{1}{8} + \frac{1}{16}$ per day.
To add the fractions, find a common denominator, which is 16.
Combined work rate = $\frac{2}{16} + \frac{1}{16} = \frac{3}{16}$ of the work per day.
The time taken to complete the work together is the reciprocal of the combined work rate.
Time taken together = $\frac{1}{\text{Combined work rate}} = \frac{1}{\frac{3}{16}} = \frac{16}{3}$ days.
Converting the improper fraction to a mixed number: $\frac{16}{3} = 5$ with a remainder of 1, so $5\frac{1}{3}$ days.
If a person completes a work in $n$ days, their work rate is $\frac{1}{n}$ of the work per day. When multiple people work together, their individual work rates add up to find the combined work rate. The total time taken is the reciprocal of the combined work rate.
This type of problem is a classic “work and time” problem. The formula for two people is $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$, where $T_1$ and $T_2$ are the times taken by individuals and $T$ is the time taken together.
$\frac{1}{T} = \frac{1}{8} + \frac{1}{16} = \frac{2+1}{16} = \frac{3}{16}$.
$T = \frac{16}{3} = 5\frac{1}{3}$ days.

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3. Which one of the following fractions is greater than $\frac{2}{7}$ and

Which one of the following fractions is greater than $\frac{2}{7}$ and less than $\frac{7}{9}$?

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$ rac{15}{19}$
$ rac{11}{17}$
$ rac{11}{14}$
$ rac{17}{21}$
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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To find the fraction greater than $\frac{2}{7}$ and less than $\frac{7}{9}$, we can compare each option with the given bounds. We can convert the fractions to decimals or find a common denominator, but cross-multiplication is usually efficient for pair-wise comparison.
The lower bound is $\frac{2}{7} \approx 0.2857$.
The upper bound is $\frac{7}{9} \approx 0.7778$.

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Let’s test option B, $\frac{11}{17} \approx 0.6471$.
Is $\frac{11}{17} > \frac{2}{7}$? Compare $11 \times 7$ with $17 \times 2$. $77 > 34$. Yes, $\frac{11}{17} > \frac{2}{7}$.
Is $\frac{11}{17} < \frac{7}{9}$? Compare $11 \times 9$ with $17 \times 7$. $99 < 119$. Yes, $\frac{11}{17} < \frac{7}{9}$. Since $\frac{11}{17}$ satisfies both conditions, it is the correct answer. Let's quickly check other options: A) $\frac{15}{19} \approx 0.7895$. Is $\frac{15}{19} < \frac{7}{9}$? Compare $15 \times 9 = 135$ with $19 \times 7 = 133$. $135 > 133$, so $\frac{15}{19} > \frac{7}{9}$. Incorrect.
C) $\frac{11}{14} \approx 0.7857$. Is $\frac{11}{14} < \frac{7}{9}$? Compare $11 \times 9 = 99$ with $14 \times 7 = 98$. $99 > 98$, so $\frac{11}{14} > \frac{7}{9}$. Incorrect.
D) $\frac{17}{21} \approx 0.8095$. Is $\frac{17}{21} < \frac{7}{9}$? Compare $17 \times 9 = 153$ with $21 \times 7 = 147$. $153 > 147$, so $\frac{17}{21} > \frac{7}{9}$. Incorrect.

To compare two fractions $\frac{a}{b}$ and $\frac{c}{d}$ (with $b, d > 0$), compare $ad$ and $bc$. If $ad > bc$, then $\frac{a}{b} > \frac{c}{d}$. If $ad < bc$, then $\frac{a}{b} < \frac{c}{d}$.
Estimating decimal values of fractions can help quickly eliminate options, especially in objective tests, but precise comparison using cross-multiplication is more accurate.
$\frac{2}{7} \approx 0.28$
$\frac{7}{9} \approx 0.78$
A) $\frac{15}{19} \approx \frac{15}{20} = 0.75$. Closer check needed. (Actual: 0.789)
B) $\frac{11}{17} \approx \frac{11}{16} = 0.6875$ or $\frac{10}{16} = 0.625$. Looks promising. (Actual: 0.647)
C) $\frac{11}{14} \approx \frac{11}{15} \approx 0.73$. Closer check needed. (Actual: 0.786)
D) $\frac{17}{21} \approx \frac{17}{20} = 0.85$. Too high. (Actual: 0.810)

4. What is the decimal equivalent of the octal number 325.412?

What is the decimal equivalent of the octal number 325.412?

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5195.213
339.5195
5195.339
213.5195
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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To convert an octal number (base 8) to a decimal number (base 10), we multiply each digit by the corresponding power of 8 and sum the results.
For the integer part (325):
3 * 8^2 + 2 * 8^1 + 5 * 8^0
= 3 * 64 + 2 * 8 + 5 * 1
= 192 + 16 + 5
= 213

For the fractional part (.412):
4 * 8^-1 + 1 * 8^-2 + 2 * 8^-3
= 4/8 + 1/64 + 2/512
= 0.5 + 0.015625 + 0.00390625
= 0.51953125

Combining the parts: 213.51953125…
Looking at the options, 213.5195 is the closest representation of the decimal equivalent.

Octal to decimal conversion involves summing the products of each digit and the base (8) raised to the power of its position (positive for integer part, negative for fractional part).
The position of a digit to the left of the radix point starts at 0 and increases by 1 for each position further left. The position of a digit to the right of the radix point starts at -1 and decreases by 1 for each position further right.

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5. Which one of the following logic gates produces true output when both

Which one of the following logic gates produces true output when both the values of its input are same?

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NAND
NOR
XNOR
XOR
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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Let’s look at the truth tables for the gates with inputs A and B and output Q:
– NAND: Q = NOT (A AND B). Output is true unless both A and B are true.
– NOR: Q = NOT (A OR B). Output is true only when both A and B are false.
– XNOR: Q = NOT (A XOR B). Output is true when A and B are the same (both 0 or both 1).
– XOR: Q = A XOR B. Output is true when A and B are different (one 0, one 1).

The question asks for the gate that produces true output when both inputs are the same. This matches the definition and truth table of the XNOR gate.

Understanding the basic logic operations and their truth tables is crucial for this question. XNOR (Exclusive NOR) outputs a true (1) when its inputs are identical (0,0 or 1,1).
XNOR is also known as the equivalence gate. It is often used in digital circuits for checking if two bits are equal. The XOR gate is the opposite of XNOR, producing a true output when inputs are different.

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6. The files and programs on which the user is currently working are stor

The files and programs on which the user is currently working are stored on which one of the following memories?

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Secondary memory
RAM
ROM
Register
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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RAM (Random Access Memory) is the primary volatile memory in a computer. When a user opens a program or a file, it is loaded from secondary storage (like an SSD or hard drive) into RAM. The CPU then accesses the instructions and data from RAM to execute the program or work with the file.
RAM holds the data and programs that are actively being used by the computer’s processor. Secondary memory is for long-term storage, ROM holds boot firmware, and registers are temporary storage within the CPU itself for immediate operations, not for the entire working files/programs.
RAM is volatile, meaning its contents are lost when the computer is turned off. Its speed allows the CPU to access data quickly, which is essential for smooth performance. The amount of RAM often determines how many programs can be run simultaneously and how large the files can be efficiently handled.

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7. Which of the following softwares can handle and manage bulk stored

Which of the following softwares can handle and manage bulk stored data?

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Database management system
Database
Compiler
Operating system
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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A Database Management System (DBMS) is a software application specifically designed to create, manage, and interact with databases. Databases are structured collections of data, and a DBMS provides the tools needed to store, retrieve, update, and manage this data efficiently, especially when dealing with large volumes (bulk stored data).
DBMS software is specialized for organizing and managing large datasets. A database is the data itself, not the software. A compiler translates code. An operating system manages system resources but not necessarily bulk structured data in the way a DBMS does.
Examples of DBMS include MySQL, PostgreSQL, Oracle Database, SQL Server, MongoDB, etc. They provide features like data integrity, security, concurrency control, and data recovery, which are essential for managing bulk data effectively.

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8. Consider the security cards of hotel rooms to gain access to rooms. Wh

Consider the security cards of hotel rooms to gain access to rooms. Which one of the following input/output devices (I/O devices) is used for this purpose?

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Optical mark reader
Scanner
Magnetic stripe reader
Cathode ray tube
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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Hotel security cards typically use a magnetic stripe on the back which stores information like room number, check-in/check-out dates, etc. The device used to read this information and grant access is a magnetic stripe reader.
Different I/O devices are designed for specific types of data input/output. Magnetic stripe readers are used to read data encoded on magnetic stripes found on cards.
Optical Mark Readers (OMR) read marks on paper. Scanners digitize images/text. Cathode Ray Tubes (CRT) are display devices (output). While some modern hotel cards use RFID or smart chip technology, magnetic stripe cards are still common, and the question specifically points to a magnetic stripe reader as the relevant I/O device for accessing rooms using such cards.

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9. One gigabyte is equal to

One gigabyte is equal to

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2<sup>30</sup> bytes
1024 bytes
2<sup>20</sup> bytes
2<sup>40</sup> bytes
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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In computing, prefixes like kilo, mega, giga, etc., are often based on powers of 2 (specifically 1024, which is 2^10) rather than powers of 1000, especially when referring to memory capacities.
1 Kilobyte (KB) = 1024 bytes = 2^10 bytes
1 Megabyte (MB) = 1024 KB = 1024 * 1024 bytes = 2^10 * 2^10 bytes = 2^20 bytes
1 Gigabyte (GB) = 1024 MB = 1024 * 1024 * 1024 bytes = 2^10 * 2^10 * 2^10 bytes = 2^30 bytes
Understanding the powers of 2 used in computer memory measurements is key. While 1 GB can also be defined as 10^9 bytes (decimal prefix), the 2^30 definition is standard for memory and is often used in computer science contexts.
The International Electrotechnical Commission (IEC) introduced binary prefixes (kibi, mebi, gibi, etc.) to distinguish from decimal prefixes (kilo, mega, giga). So, 1 Gibibyte (GiB) = 2^30 bytes, and 1 Gigabyte (GB) = 10^9 bytes. However, in common usage, especially referring to RAM or file sizes, ‘gigabyte’ often implicitly refers to 2^30 bytes. Given the options, 2^30 bytes is the intended answer.

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10. Which one of the following is *not* the characteristic feature of ‘El

Which one of the following is *not* the characteristic feature of ‘El Ni\u00f1o’ phenomenon impacting the Indian monsoon?

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It replaces the cold Peruvian current
Appearance of cold current in East Pacific region
Results in reduction of planktons
Results in drought or weak monsoon in India
This question was previously asked in
UPSC CBI DSP LDCE – 2023
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El Niño is characterized by the warming of the surface waters in the central and eastern tropical Pacific Ocean. This *replaces* the usual cold Peruvian current, not leads to the appearance of a cold current. Therefore, the statement “Appearance of cold current in East Pacific region” is incorrect.
El Niño is a phenomenon of warm sea surface temperatures in the eastern equatorial Pacific. It is generally associated with weak monsoon in India and reduction in plankton production due to suppressed upwelling of nutrient-rich cold water.
El Niño is part of the El Niño–Southern Oscillation (ENSO) cycle. Its opposite phase, La Niña, is characterized by colder than average sea surface temperatures in the central and eastern tropical Pacific and is generally associated with a stronger monsoon in India.

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