Given:

• Principal (P) = Rs 50,000
• Rate (R) = 10% per annum
• Time (T) = 2 years

Simple Interest (SI):

$SI = \frac{P \times R \times T}{100}$

$SI = \frac{50000 \times 10 \times 2}{100} = Rs\,10,000$

Compound Interest (CI):

$A = P\left(1+\frac{R}{100}\right)^T$

$A = 50000\left(1+\frac{10}{100}\right)^2 = 50000 \times (1.1)^2 = Rs\,60,500$

$CI = A - P = 60500 - 50000 = Rs\,10,500$

Difference:

$CI - SI = 10500 - 10000 = Rs\,500$

Therefore, the difference between compound interest and simple interest is Rs 500.

Let the present salaries of Dipesh and Sumit be $2x$ and $3x$ respectively.

After an increase of Rs. 4000 each, the new ratio is $40:57$.

$$\frac{2x+4000}{3x+4000}=\frac{40}{57}$$

Cross-multiplying:

$$57(2x+4000)=40(3x+4000)$$ $$114x+228000=120x+160000$$ $$228000-160000=120x-114x$$ $$68000=6x$$ $$x=\frac{68000}{6}=\frac{34000}{3}$$

Sumit’s present salary:

$$3x=3\times \frac{34000}{3}=\text{Rs. }34000$$

Therefore, Sumit’s present salary is Rs. 34,000.

$\frac58+\frac7{12}-\frac34$

LCM of 8, 12 and 4 is 24, so

$\frac{15}{24}+\frac{14}{24}-\frac{18}{24}=\frac{15+14-18}{24}=\frac{11}{24}$

In decimal, $\dfrac{11}{24}=0.4583\overline{3}\approx 0.458$.

Total work = (12 \times 15 = 180) worker-days.

Work done in first 5 days = (12 \times 5 = 60) worker-days.

Remaining work = (180 - 60 = 120) worker-days.

After 4 more workers join, total workers = (12 + 4 = 16).

Time required to finish remaining work:

[ \frac{120}{16} = 7.5 \text{ days} ]

So, additional days after the new workers join = 7.5 days.

Total days from the start:

[ 5 + 7.5 = 12.5 \text{ days} ]

Answer: 12.5 days total from the beginning, or 7.5 more days after the 4 workers join.

Given:

• Principal (P) = Rs 8,000
• Rate (R) = 5% per annum
• Time (T) = 3 years
• Compounded annually

Formula: $A = P\left(1+\frac{R}{100}\right)^T$

Substitution: $A = 8000\left(1+\frac{5}{100}\right)^3$ $A = 8000(1.05)^3$ $A = 8000 \times 1.157625$ $A = Rs\ 9,261$

Compound Interest: $CI = A - P$ $CI = 9261 - 8000 = Rs\ 1,261$

Therefore, the compound interest earned after 3 years is Rs 1,261.

Factorise each expression:

1. $2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2)$

2. $x^2 + 2x = x(x+2)$

3. $x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = (x-4)(x+2)$

So the factors are:

$2,\ x,\ (x+2),\ (x-2),\ (x-4)$

Taking the highest power of each distinct factor, the LCM is:

$\boxed{2x(x+2)(x-2)(x-4)}$

If expanded:

$2x(x+2)(x-2)(x-4)=2x(x^2-4)(x-4)=2x(x^3-4x^2-4x+16)$

$\boxed{2x^4-8x^3-8x^2+32x}$

(B) $2:3$

As printed, "$xy=4.5$" does not fix $x:z$. The intended reading is $x:y=4:5$ (the "$=4.5$" is a misprint of "$=4:5$"). Then

$x:y=4:5,\quad y:z=5:6 \Rightarrow x:y:z=4:5:6$

so $x:z=4:6=2:3$.

None of the printed options is correct. The correct LCM is $(x-3)(4x^2-25)$, which is closest to option (C), whose sign is misprinted as $+25$.

$4x^2-25=(2x-5)(2x+5)$ and $2x^2-x-15=(2x+5)(x-3)$.

$\text{LCM}=(x-3)(2x+5)(2x-5)=(x-3)(4x^2-25)$

Nepal Bank Assistant (Level 4) — Mathematics, Statistics and Information Technology

Model Answer / Revision Notes

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2.1 Mathematics

Key Concepts and Formulae

Integers: Whole numbers with sign ($\ldots,-2,-1,0,1,2,\ldots$). Follow the sign rules: $(+)\times(+)=+$, $(-)\times(-)=+$, $(+)\times(-)=-$. Always apply BODMAS — Bracket, Of, Division, Multiplication, Addition, Subtraction.

Unitary method: First find the value of one unit, then multiply by the required number of units. For inverse variation (more workers → fewer days), use $M_1D_1 = M_2D_2$.

Percentage: $\text{Percent} = \dfrac{\text{Part}}{\text{Whole}}\times 100$. Profit percent is always calculated on cost price; discount percent on marked price.

Ratio and proportion: If $a:b = c:d$ then $ad = bc$ (product of extremes = product of means). To divide a sum $S$ in the ratio $a:b$, shares are $\dfrac{a}{a+b}S$ and $\dfrac{b}{a+b}S$.

Profit and Loss: $\text{Profit} = SP - CP,\qquad \text{Profit\%} = \frac{SP-CP}{CP}\times 100$ $\text{Loss\%} = \frac{CP-SP}{CP}\times 100,\qquad SP = CP\left(1+\frac{P\%}{100}\right)$

Simple Interest: $SI = \dfrac{P\times T\times R}{100}$, Amount $= P + SI$.

Compound Interest: $A = P\left(1+\dfrac{R}{100}\right)^T$, $CI = A - P$. For half-yearly compounding, rate is halved and time doubled.

LCM and HCF: For two numbers, $\text{LCM}\times\text{HCF} = \text{Product of the numbers}$. HCF divides the difference of the numbers exactly.

Index (Indices): $a^m\times a^n = a^{m+n}$, $\dfrac{a^m}{a^n}=a^{m-n}$, $(a^m)^n = a^{mn}$, $a^0=1$, $a^{-n}=\dfrac{1}{a^n}$.

Solved Examples

1. Profit percent. A trader buys 25 kg of rice at Rs 92 per kg and sells it at Rs 105 per kg. Find the profit percent. $CP = 25\times 92 = \text{Rs }2300,\quad SP = 25\times 105 = \text{Rs }2625$ $\text{Profit\%} = \frac{2625-2300}{2300}\times 100 = \frac{325}{2300}\times100 = 14.13\%$

2. Simple interest. Find the SI and amount on Rs 25,000 at 9% per annum for 3 years. $SI = \frac{25000\times 3\times 9}{100} = \text{Rs }6750,\qquad A = 25000+6750 = \text{Rs }31{,}750$

3. Compound interest. Find the CI on Rs 40,000 at 10% p.a. for 2 years, compounded annually. $A = 40000\left(1+\frac{10}{100}\right)^2 = 40000\times 1.21 = \text{Rs }48{,}400$ $CI = 48400 - 40000 = \text{Rs }8400$

4. Unitary method (inverse). 12 workers complete a wall in 18 days. How long will 8 workers take? $12\times 18 = 8\times D \Rightarrow D = \frac{216}{8} = 27\text{ days}$

5. LCM and HCF. The LCM of two numbers is 180 and their HCF is 6. If one number is 36, find the other. $\text{Other} = \frac{\text{LCM}\times\text{HCF}}{\text{One number}} = \frac{180\times 6}{36} = 30$

6. Ratio. Divide Rs 45,000 between two persons in the ratio 4 : 5. $\text{First} = \frac{4}{9}\times45000 = \text{Rs }20{,}000,\qquad \text{Second} = \frac{5}{9}\times45000 = \text{Rs }25{,}000$

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2.2 Statistics

Tabulation and Graphic Methods

Tabulation means arranging collected data in rows and columns for easy comparison. A frequency distribution table shows class interval, tally marks, frequency ($f$), mid-value ($m$), cumulative frequency ($cf$) and relative frequency. Types of class intervals: inclusive (10–19) and exclusive (10–20).

Graphic methods:

• Histogram — adjoining rectangles on class intervals; area represents frequency.
• Frequency polygon — obtained by joining the mid-points of the histogram bars.
• Ogive (cumulative frequency curve) — "less than" or "more than" type; the median is read at $N/2$.
• Bar diagram and pie chart — used for qualitative or categorical data.

Measures of Central Tendency

Arithmetic Mean: $\bar x = \frac{\sum fx}{N}\quad\text{(direct method)},\qquad \bar x = A + \frac{\sum fd}{N}\quad\text{(assumed mean)}$

Median:

• Discrete series: $M = \left(\dfrac{N+1}{2}\right)^{th}$ item.
• Continuous series: $M = L + \dfrac{N/2 - cf}{f}\times h$, where $L$ = lower limit of median class, $cf$ = cumulative frequency before it, $f$ = frequency of median class, $h$ = class size.

Mode: $M_o = L + \frac{f_1-f_0}{2f_1-f_0-f_2}\times h$ where $f_1$ = frequency of modal class, $f_0$ and $f_2$ = frequencies of the classes before and after it.

Measures of Dispersion

Standard Deviation: $\sigma = \sqrt{\frac{\sum fx^2}{N} - \left(\frac{\sum fx}{N}\right)^2}$ Coefficient of Variation: $CV = \dfrac{\sigma}{\bar x}\times 100\%$ — used to compare consistency; smaller CV means more consistent data.

Solved Example (Mean and SD)

$\bar x = \frac{580}{20} = 29$ $\sigma = \sqrt{\frac{19200}{20} - (29)^2} = \sqrt{960-841} = \sqrt{119} = 10.91$

Solved Example (Median and Mode — Continuous Series)

$M = 20 + \frac{22.5-13}{15}\times10 = 20 + 6.33 = 26.33$ $M_o = 20 + \frac{15-8}{2(15)-8-12}\times10 = 20 + \frac{7}{10}\times10 = 27$

Simple Correlation and Regression

Karl Pearson's coefficient of correlation: $r = \frac{N\sum xy - \sum x\sum y}{\sqrt{\left[N\sum x^2-(\sum x)^2\right]\left[N\sum y^2-(\sum y)^2\right]}}$ $r$ lies between $-1$ and $+1$. $r=+1$ perfect positive, $r=-1$ perfect negative, $r=0$ no correlation.

Regression lines: $y - \bar y = b_{yx}(x-\bar x),\qquad b_{yx} = \frac{N\sum xy-\sum x\sum y}{N\sum x^2-(\sum x)^2}$ $x - \bar x = b_{xy}(y-\bar y),\qquad b_{xy} = \frac{N\sum xy-\sum x\sum y}{N\sum y^2-(\sum y)^2}$ Also $r = \sqrt{b_{yx}\times b_{xy}}$.

Solved example. For $x$: 1, 2, 3, 4, 5 and $y$: 2, 4, 5, 4, 5: $\sum x=15,\ \sum y=20,\ \sum xy=66,\ \sum x^2=55,\ \sum y^2=86,\ N=5$ $r = \frac{5(66)-15(20)}{\sqrt{[5(55)-225][5(86)-400]}} = \frac{330-300}{\sqrt{50\times30}} = \frac{30}{38.73} = 0.77$ $b_{yx} = \frac{30}{50} = 0.6,\qquad \bar x = 3,\ \bar y = 4$ $y - 4 = 0.6(x-3) \Rightarrow y = 0.6x + 2.2$ When $x = 6$, $y = 0.6(6)+2.2 = 5.8$.

Probability and Sampling

Probability: $P(A) = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$, $0\le P(A)\le 1$.

• Addition rule: $P(A\cup B) = P(A)+P(B)-P(A\cap B)$; for mutually exclusive events, $P(A\cup B)=P(A)+P(B)$.
• Multiplication rule: $P(A\cap B) = P(A)\times P(B|A)$; for independent events, $P(A\cap B)=P(A)\times P(B)$.

Solved example. A bag contains 5 red and 7 blue balls. Two balls are drawn without replacement. Find the probability that both are red. $P = \frac{5}{12}\times\frac{4}{11} = \frac{20}{132} = \frac{5}{33}$

Sampling: Selecting a part (sample) from the whole (population) to draw conclusions about the population. Types: random (lottery), systematic (every $k^{th}$ item), stratified (population divided into homogeneous groups), cluster (whole groups selected), and quota sampling. A good sample must be representative, adequate in size and free from bias. Difference between sample result and population value is sampling error.

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2.3 Information Technology

Computer Introduction

A computer is an electronic device that accepts data, processes it according to instructions and gives meaningful information. Its main parts are input unit, CPU (ALU + control unit + memory), output unit and storage. Characteristics: speed, accuracy, diligence, versatility, storage capacity and no intelligence of its own.

Operating System

An operating system is system software that manages hardware, files and programs and provides a user interface. Examples:

• DOS — single-user, character-based (command line) OS; commands like DIR, COPY, DEL, MD.
• Windows — GUI-based, multitasking, user-friendly; versions from Windows 95 to Windows 11.
• Linux — open-source, multi-user, multi-tasking, secure and free; distributions like Ubuntu, Red Hat, Fedora.

Functions: process management, memory management, file management, device management and security.

Office Package

• MS Word — word processing: typing, formatting, tables, mail merge.
• MS Excel — spreadsheet: formulas, functions (SUM, AVERAGE, IF), charts, what-if analysis.
• MS PowerPoint — presentation: slides, animation, slide show.
• MS Access — database: tables, queries, forms, reports.

Internet, Intranet and E-mail

• Internet — a global network of interconnected computers using TCP/IP; services include the World Wide Web, e-mail, FTP, chat and e-commerce.
• Intranet — a private network within one organisation using Internet technology, accessible only to its members; used for internal notices, circulars and records. An extranet extends limited access to outsiders such as suppliers.
• E-mail — electronic mail; sending and receiving messages and files. Parts of an address: username@domain (e.g. info@nepalbank.com.np). Advantages: fast, cheap, written record; disadvantages: virus, spam, needs Internet.

Database Management System (DBMS)

A DBMS is software for creating, storing, retrieving and managing data in a database (e.g. MS Access, Oracle, MySQL). Advantages: reduces data redundancy, maintains consistency, ensures security and integrity, allows sharing and quick query. Key terms: field, record, table, primary key, foreign key, query, report. SQL is the standard query language.

Management Information System (MIS)

MIS is a computer-based system that collects, processes, stores and supplies information to managers for planning, controlling and decision-making. It converts data into reports (daily, summary, exception, on-demand). Benefits: faster and accurate decisions, better coordination, reduced cost and timely reporting. In banking, MIS gives daily transaction summaries, deposit/loan positions, NPL status and branch-wise performance.

Role of Information Technology in Banking

1. Any Branch Banking (ABBS) — customers can deposit/withdraw at any branch.
2. ATM and Debit/Credit Cards — 24-hour cash withdrawal and cashless payment.
3. Mobile and Internet Banking — balance enquiry, fund transfer, bill payment, QR payment.
4. Core Banking System (CBS) — centralised, real-time posting of all transactions.
5. Electronic Clearing and RTGS/connectIPS — instant interbank transfer and settlement.
6. Cheque truncation, SMS/e-mail alerts, digital KYC and AML monitoring.
7. Reduced cost and errors, faster service, better record-keeping and fraud control.

Thus IT has made banking faster, safer, cheaper and more customer-friendly, and is the backbone of modern banking in Nepal.