Chapter 1 — Arithmetic I: Percentage, Profit, Interest, Ratio, Average — Non-Technical Part

Percentage · Profit, loss & discount · Simple & compound interest · Ratio · Partnership · Average · Ages

These topics give the largest share of bank-exam mathematics: in the MCQ part (15 marks) roughly half the questions come from here, and at least one or two of the four written problems (5 marks each) are usually percentage, profit or interest. Interest — especially compound interest vs simple interest — is the banks' favourite. No calculator is allowed, so learn the fraction equivalents and the shortcuts below.

How to earn full 5 marks in a written problem

Let — define the unknown clearly ("Let the cost price be Tk x").
Equation — write the condition from the question as an equation.
Solve — show every line; work in fractions (7/20, not 0.35).
Answer — write a clear final line with units ("∴ Cost price = Tk 6,000").
Check — a one-line check proves the answer and often catches a slip.

Steps carry marks even if the final number is wrong.

1.1Percentage and successive change

x% of y = xy/100 | % change = (change ÷ original) × 100 | new value = old × (1 ± r/100)

Fraction ↔ percentage table (memorise)

Fraction	%	Fraction	%	Fraction	%
1/2	50	1/7	14 2/7	1/12	8 1/3
1/3	33 1/3	1/8	12.5	1/15	6 2/3
1/4	25	1/9	11 1/9	1/16	6.25
1/5	20	1/10	10	1/20	5
1/6	16 2/3	1/11	9 1/11	3/8	37.5

Key rules

Rules for percentage change

Successive change of a% then b% ⇒ net change = a + b + ab/100 % (use minus sign for a decrease).
+a% followed by −a% ⇒ always a net decrease of a²/100 %.
If A is r% more than B, then B is r/(100 + r) × 100 % less than A.
If A is r% less than B, then B is r/(100 − r) × 100 % more than A.
Price rises r% ⇒ to keep expenditure the same, cut consumption by r/(100 + r) × 100 %. Price falls r% ⇒ consumption can rise by r/(100 − r) × 100 %.
Length +a%, width +b% ⇒ area changes by a + b + ab/100 % (same as successive change).

Example 1 — successive change The price of an item rises by 20% and then falls by 10%. Net change?
Net = 20 + (−10) + (20 × −10)/100 = 20 − 10 − 2 = 8% increase. Check: 100 → 120 → 108 ✔

Example 2 — consumption The price of sugar rises by 25%. By what percentage must a family cut consumption so that expenditure does not change?
Cut = 25/(100 + 25) × 100 = 25/125 × 100 = 20%. Check: 1.25 × 0.80 = 1 ✔

Example 3 — "more than" vs "less than" A's salary is 25% more than B's. By what percentage is B's salary less than A's?
= 25/125 × 100 = 20%. (Take B = 100, A = 125; difference 25 is 20% of 125.)

Example 4 — election In an election between two candidates, one gets 40% of the votes and loses by 2,000 votes. Total votes (all valid)?
Winner 60%, loser 40% ⇒ difference 20% = 2,000 ⇒ total = 2,000 × 100/20 = 10,000.

Example 5 — pass marks A student must get 33% to pass. He gets 125 marks and fails by 40 marks. Full marks?
Pass mark = 125 + 40 = 165 = 33% ⇒ full marks = 165 × 100/33 = 500.

Example 6 — pass/fail in two subjects 20% fail in English, 25% fail in Mathematics and 10% fail in both. What percentage pass in both?
Failed in at least one = 20 + 25 − 10 = 35% ⇒ passed in both = 100 − 35 = 65%.

Past question AP 17.01.2026 In a class, the number of boys is 35% more than the number of girls. Total students = 94. If 6 more boys are admitted, find the ratio of boys to girls.
Solution. Let girls = g. Then boys = g + 35% of g = 1.35g = 27g/20.
g + 27g/20 = 94 ⇒ 47g/20 = 94 ⇒ g = 40; boys = 94 − 40 = 54.
After admission, boys = 54 + 6 = 60 ⇒ boys : girls = 60 : 40 = 3 : 2.
Check: 54 is 35% more than 40 (40 × 1.35 = 54) ✔

Past question BB AP 2023 A basketball team has won 15 games and lost 9. If these games represent 16⅔% of the games to be played, how many more games must the team win to average 75% for the season?
(The percentage is printed unclearly in the collection; the known version of this question uses 16⅔%.)
Solution. Games played = 15 + 9 = 24. 16⅔% = 1/6 ⇒ total games = 24 × 6 = 144.
Wins needed for 75% = 3/4 × 144 = 108 ⇒ more wins needed = 108 − 15 = 93 games.

Traps

Percent vs percentage point: interest rate rising from 8% to 10% is a rise of 2 percentage points, but a 25% rise in the rate.
+20% then −20% is not "no change" — it is a 4% decrease.
Always divide the change by the original (base) value, not the new value.

1.2Profit, loss and discount

▶ Step-by-step animation: profit and loss as bars (Example 7)

Key frames from the animation: Profit and loss are percentages of the cost price

Profit % = (SP − CP)/CP × 100 | SP = CP × (100 ± p)/100 | CP = SP × 100/(100 ± p)

Discount is on Marked Price (MP): SP = MP × (100 − d)/100 | MP/CP = (100 + profit%)/(100 − discount%)

Rules and shortcuts

Profit and loss % are always on CP; discount % is always on MP.
Successive discounts d₁% and d₂% ⇒ single discount = d₁ + d₂ − d₁d₂/100 %.
Two items sold at the same SP, one at x% gain and one at x% loss ⇒ always an overall loss of x²/100 %.
Profit % at SP₁ equals loss % at SP₂ ⇒ CP = (SP₁ + SP₂)/2.
Dishonest dealer (sells at CP but uses a false weight): gain % = error/(true value − error) × 100.
CP of x articles = SP of y articles ⇒ profit % = (x − y)/y × 100.
Chain of sales (A → B → C): C's price = A's CP × (1 + p₁)(1 + p₂).

Example 7 — from loss to gain An article sold for Tk 1,140 gives a 5% loss. At what price should it be sold to gain 5%?
CP = 1,140 × 100/95 = Tk 1,200 ⇒ SP for 5% gain = 1,200 × 105/100 = Tk 1,260.
Shortcut: 1,140 × 105/95 = 1,260.

Example 8 — chain of sales A sells to B at 20% profit; B sells to C at 25% profit. C pays Tk 1,500. A's cost price?
A's CP × 6/5 × 5/4 = 1,500 ⇒ A's CP × 3/2 = 1,500 ⇒ Tk 1,000.

Example 9 — same SP, equal gain and loss Two phones are sold at Tk 990 each, one at 10% gain and the other at 10% loss. Overall?
Shortcut: loss = 10²/100 = 1% loss.
Check: CP₁ = 990 × 100/110 = 900, CP₂ = 990 × 100/90 = 1,100; total CP 2,000, total SP 1,980 ⇒ loss Tk 20 = 1% ✔

Example 10 — marked price and discount A shirt is marked Tk 2,500 and sold at a 12% discount. If the CP is Tk 2,000, find the profit %.
SP = 2,500 × 88/100 = Tk 2,200 ⇒ profit = 200 ⇒ profit % = 200/2,000 × 100 = 10%.

Example 11 — successive discounts Discounts of 20% and 10% on Tk 1,000. Equivalent single discount?
20 + 10 − (20 × 10)/100 = 28%. Check: 1,000 × 0.8 × 0.9 = 720 ⇒ 28% off ✔
Note: 20% + 10% is not 30%; the second discount is on the already reduced price.

Example 12 — dishonest dealer A shopkeeper claims to sell at cost price but uses a 900 g weight for 1 kg. Gain %?
Gain = 100/(1,000 − 100) × 100 = 100/900 × 100 = 11 1/9 %.
With an 800 g weight: 200/800 × 100 = 25%.

Example 13 — buy x for Tk a, sell y for Tk b Oranges bought at 12 for Tk 10 and sold at 10 for Tk 12. Gain %?
CP of 1 = 10/12 = 5/6; SP of 1 = 12/10 = 6/5 ⇒ SP/CP = (6/5) ÷ (5/6) = 36/25 = 1.44 ⇒ 44% gain.

Past question BB AME 2023 The percentage profit earned by selling an article for Tk 1,920 is equal to the percentage loss incurred by selling the same article for Tk 1,280. At what price should the article be sold to make 25% profit?
Solution. Let CP = Tk C. (1,920 − C)/C × 100 = (C − 1,280)/C × 100
⇒ 1,920 − C = C − 1,280 ⇒ 2C = 3,200 ⇒ C = Tk 1,600.
SP for 25% profit = 1,600 × 125/100 = Tk 2,000.
Check: profit at 1,920 = 320 = 20%; loss at 1,280 = 320 = 20% ✔

Past question (MCQ) Officer IT 2026 Even after allowing a 20% discount, a trader makes a 20% profit. By what percentage is the marked price above the cost price?
Solution. Let CP = 100 ⇒ SP = 120. SP = 80% of MP ⇒ MP = 120 × 100/80 = 150.
MP is 150 − 100 = 50% above CP. Formula check: MP/CP = 120/80 = 1.5 ✔

Traps

Profit % on the SP is a common wrong option — always divide by CP unless the question says "on selling price".
"Marked 40% above CP and 40% discount" is not break-even: 1.4 × 0.6 = 0.84 ⇒ 16% loss.

1.3Simple interest

I = P × r × t / 100 | A = P + I | P = 100I/(rt) | r = 100I/(Pt) | t = 100I/(Pr)

P = principal, r = rate % per annum, t = time in years, A = amount. If time is in months, t = months/12; if in days, t = days/365.

Shortcuts

A sum becomes k times in t years at SI ⇒ r = 100(k − 1)/t. Doubles (k = 2) ⇒ r = 100/t.
Amount A₁ after t₁ years and A₂ after t₂ years ⇒ yearly interest = (A₂ − A₁)/(t₂ − t₁); then P = A₁ − t₁ × yearly interest.
SI is the same every year (interest is always on the original P).

Example 14 — basic SI Find the SI and amount on Tk 12,000 at 7.5% p.a. for 4 years.
I = 12,000 × 7.5 × 4/100 = Tk 3,600; A = 12,000 + 3,600 = Tk 15,600.

Example 15 — interest for days SI on Tk 7,300 at 5% p.a. for 146 days?
I = 7,300 × 5/100 × 146/365 = 365 × 146/365 = Tk 146.

Example 16 — two amounts A sum amounts to Tk 5,800 in 2 years and Tk 7,000 in 5 years at simple interest. Find the sum and the rate.
Interest for 3 years = 7,000 − 5,800 = 1,200 ⇒ per year = 400.
P = 5,800 − 2 × 400 = Tk 5,000; r = 400/5,000 × 100 = 8%.

Example 17 — doubling A sum doubles in 8 years at SI. Rate? In how many years will it become 3 times?
r = 100/8 = 12.5%. To become 3 times, interest = 2P ⇒ t = 100 × 2/12.5 = 16 years.

1.4Compound interest

In compound interest, the interest of each period is added to the principal, and the next period's interest is calculated on this larger amount ("interest on interest"). Banks pay deposits (FDR, DPS) and charge loans this way.

A = P(1 + r/100)ⁿ | CI = A − P

Compounding	Rate per period	Number of periods	Amount
Yearly	r	n	P(1 + r/100)ⁿ
Half-yearly	r/2	2n	P(1 + r/200)²ⁿ
Quarterly	r/4	4n	P(1 + r/400)⁴ⁿ
Monthly	r/12	12n	P(1 + r/1200)¹²ⁿ
Different rates r₁, r₂, r₃ in successive years	—	—	P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100)

CI − SI shortcuts (asked very often)

2 years: CI − SI = P(r/100)²
3 years: CI − SI = P(r/100)²(3 + r/100) = P r²(300 + r)/100³
For 1 year (yearly compounding), CI = SI.
CI for 2 years − SI for 2 years = interest on the first year's interest.

Population growth and depreciation (same formula)

Population after n years = P(1 + r/100)ⁿ
Population n years ago = P ÷ (1 + r/100)ⁿ
Value of a machine after depreciation = V(1 − r/100)ⁿ; value n years ago = V ÷ (1 − r/100)ⁿ
Amount after n years A₁ and after n + 1 years A₂ ⇒ r = (A₂ − A₁)/A₁ × 100.
If a sum doubles in n years at CI, it becomes 4 times in 2n years and 8 times in 3n years.
Rule of 72: years to double ≈ 72 ÷ rate (e.g. at 8% ≈ 9 years) — good for checking options.

Powers to remember: 1.1² = 1.21, 1.1³ = 1.331 · 1.05² = 1.1025, 1.05³ = 1.157625 · 1.04² = 1.0816, 1.04³ = 1.124864 · 1.2² = 1.44, 1.2³ = 1.728 · 0.9² = 0.81, 0.8² = 0.64, 0.8³ = 0.512

Figure 1.1: Amount on Tk 10,000 at 10% p.a. — SI grows by a fixed Tk 1,000 each year; CI grows faster (the gap is Tk 100 after 2 years and Tk 310 after 3 years).

Example 18 — yearly CI CI on Tk 8,000 at 10% p.a. for 3 years?
A = 8,000 × 1.1³ = 8,000 × 1.331 = Tk 10,648 ⇒ CI = Tk 2,648.

Example 19 — half-yearly Tk 10,000 at 8% p.a. compounded half-yearly for 1 year.
Rate per half-year = 4%, periods = 2 ⇒ A = 10,000 × 1.04² = 10,000 × 1.0816 = Tk 10,816 ⇒ CI = Tk 816 (yearly compounding would give only Tk 800).

Example 20 — quarterly Tk 16,000 at 20% p.a. compounded quarterly for 9 months.
Rate per quarter = 5%, periods = 3 ⇒ A = 16,000 × 1.05³ = 16,000 × 1.157625 = Tk 18,522 ⇒ CI = Tk 2,522.

Example 21 — CI − SI for 2 years (find P) The difference between CI and SI on a sum for 2 years at 5% p.a. is Tk 25. Find the sum.
P × (5/100)² = 25 ⇒ P × 1/400 = 25 ⇒ P = Tk 10,000.

Example 22 — CI − SI for 3 years Find the difference between CI and SI on Tk 8,000 at 10% p.a. for 3 years.
= 8,000 × (1/10)² × (3 + 1/10) = 80 × 3.1 = Tk 248.
Check: CI = 2,648 (Example 18), SI = 8,000 × 10 × 3/100 = 2,400 ⇒ difference 248 ✔

Example 23 — find rate and principal from two amounts A sum amounts to Tk 9,680 in 2 years and Tk 10,648 in 3 years at CI. Find the rate and the sum.
3rd year's interest = 10,648 − 9,680 = 968 on 9,680 ⇒ r = 968/9,680 × 100 = 10%.
P = 9,680 ÷ 1.1² = 9,680 ÷ 1.21 = Tk 8,000.

Example 24 — different rates Tk 5,000 is invested at 10% in the first year and 20% in the second year (compounded yearly). Amount?
A = 5,000 × 1.1 × 1.2 = Tk 6,600.

Example 25 — population growth The population of a town is 50,000. It grows at 4% per year. Population after 2 years?
= 50,000 × 1.04² = 50,000 × 1.0816 = 54,080.

Example 26 — population in the past The present population is 1,32,300. It has grown at 5% per year. What was it 2 years ago?
= 1,32,300 ÷ 1.05² = 1,32,300 ÷ 1.1025 = 1,20,000.

Example 27 — depreciation A machine bought for Tk 1,25,000 depreciates at 20% per year. Its value after 3 years?
= 1,25,000 × 0.8³ = 1,25,000 × 0.512 = Tk 64,000.

Example 28 — effective annual rate (bank context) A bank offers 12% p.a. compounded quarterly. What is the effective annual rate?
(1 + 0.03)⁴ − 1 = 1.12550881 − 1 ≈ 12.55%. (Compounded monthly: (1.01)¹² − 1 ≈ 12.68%.) The more often interest is compounded, the higher the effective rate.

Past question BB AME 2023 A father divided Tk 25,000 between his sons A and B. A invests his share at 8% p.a. compound interest and B invests at 10% p.a. simple interest. After 2 years, B's interest is Tk 1,336 more than A's. Find A's share.
Solution. Let A's share = Tk a ⇒ B's share = Tk (25,000 − a).
A's CI = a[(1.08)² − 1] = a(1.1664 − 1) = 0.1664a
B's SI = (25,000 − a) × 10 × 2/100 = 5,000 − 0.2a
Condition: (5,000 − 0.2a) − 0.1664a = 1,336 ⇒ 0.3664a = 3,664 ⇒ a = Tk 10,000.
Check: A's CI = 1,664; B's share 15,000 ⇒ SI = 3,000; difference = 1,336 ✔

Traps

Half-yearly: halve the rate and double the time — students often do only one.
The question may ask for the amount or only the interest — read the last line again.
For depreciation use (1 − r/100), not (1 + r/100).

1.5Ratio and proportion

a : b = a/b | a : b = c : d ⇔ ad = bc (product of extremes = product of means)

Term	Meaning	Example
Compound ratio	(a:b) and (c:d) ⇒ ac : bd	2:3 and 4:5 ⇒ 8:15
Duplicate / triplicate ratio	a² : b² / a³ : b³	of 2:3 ⇒ 4:9 / 8:27
Sub-duplicate ratio	√a : √b	of 9:16 ⇒ 3:4
Inverse ratio	1/a : 1/b = b : a	of 3:5 ⇒ 5:3
Mean proportional of a, b	√(ab)	of 9, 16 ⇒ 12
Third proportional to a, b	b²/a	to 4, 6 ⇒ 9
Fourth proportional to a, b, c	bc/a	to 3, 5, 12 ⇒ 20

Working rules

Joining ratios A:B and B:C — make the common term (B) equal using its LCM.
Dividing a sum S in the ratio a:b:c ⇒ shares = S × a/(a + b + c), etc.
Ratio given as fractions (1/2 : 1/3 : 1/4) ⇒ multiply all by the LCM of denominators to get whole numbers.
Most ratio problems: take the quantities as ax and bx, then form one equation in x.

Example 29 — joining ratios A:B = 2:3 and B:C = 4:5. Find A:B:C.
LCM of 3 and 4 = 12 ⇒ A:B = 8:12, B:C = 12:15 ⇒ A:B:C = 8 : 12 : 15.

Example 30 — fractional ratio Divide Tk 7,800 among A, B, C in the ratio 1/2 : 1/3 : 1/4.
× 12 ⇒ 6 : 4 : 3, total 13 parts ⇒ 1 part = 7,800/13 = 600.
A = Tk 3,600, B = Tk 2,400, C = Tk 1,800.

Example 31 — adding to both terms What number must be added to each term of 5 : 9 to make it 3 : 4?
(5 + x)/(9 + x) = 3/4 ⇒ 20 + 4x = 27 + 3x ⇒ x = 7. Check: 12 : 16 = 3 : 4 ✔

Example 32 — coins A bag has Tk 1, 50-paisa and 25-paisa coins in the ratio 5 : 6 : 8. Total value is Tk 210. Number of each coin?
Value ratio = 5×1 : 6×½ : 8×¼ = 5 : 3 : 2 (10 parts) ⇒ 1 part = Tk 21.
Values: Tk 105, Tk 63, Tk 42 ⇒ coins: 105 one-taka, 63 ÷ 0.5 = 126 fifty-paisa, 42 ÷ 0.25 = 168 twenty-five-paisa. Check: 105 : 126 : 168 = 5 : 6 : 8 ✔

Example 33 — mixture 60 litres of a mixture has milk and water in the ratio 2 : 1. How much water must be added to make the ratio 1 : 2?
Milk = 40 L, water = 20 L. Milk stays the same: 40/(20 + x) = 1/2 ⇒ 20 + x = 80 ⇒ x = 60 litres.

MCQ speed tip In ratio questions, the answer must fit the ratio — test the options. If A:B = 3:5, A's share must be a multiple of 3 parts; options that do not divide correctly can be eliminated at once.

1.6Partnership

Profit share ∝ Capital × Time | A : B = C₁t₁ : C₂t₂

Rules

Same time for all ⇒ profit ratio = capital ratio.
Partner joins late or leaves early ⇒ multiply capital by the number of months it was invested.
Capital changed in the middle ⇒ add (capital × months) for each period.
Working partner: first pay the salary/commission out of the profit, then divide the rest in the capital ratio.

Example 34 — same time A, B and C invest Tk 30,000, Tk 45,000 and Tk 60,000. Profit Tk 36,000. Shares?
Ratio 30 : 45 : 60 = 2 : 3 : 4 (9 parts), 1 part = 4,000 ⇒ Tk 8,000, Tk 12,000, Tk 16,000.

Example 35 — late joining A starts a business with Tk 40,000. After 3 months B joins with Tk 60,000. Profit at the end of the year is Tk 34,000. Shares?
A : B = 40,000 × 12 : 60,000 × 9 = 4,80,000 : 5,40,000 = 8 : 9 (17 parts), 1 part = 2,000.
A = Tk 16,000, B = Tk 18,000.

Example 36 — withdrawal A invests Tk 20,000 and B Tk 30,000. After 6 months A withdraws half his capital. Year-end profit Tk 27,000. Shares?
A = 20,000 × 6 + 10,000 × 6 = 1,80,000; B = 30,000 × 12 = 3,60,000 ⇒ 1 : 2.
A = Tk 9,000, B = Tk 18,000.

Example 37 — working partner A and B invest in the ratio 3 : 2. A, the working partner, takes 10% of the profit as salary. Total profit Tk 50,000. A's total income?
Salary = Tk 5,000; remaining 45,000 in 3 : 2 ⇒ A 27,000, B 18,000.
A's total = 5,000 + 27,000 = Tk 32,000.

1.7Average

Average = Sum ÷ Number | Sum = Average × Number

Rules and shortcuts

Adding a member: new person's value = new sum − old sum = (n + 1) × new average − n × old average.
Removing a member: removed value = old sum − new sum.
Replacing a member: new person = old person + n × (increase in average). If the average falls, subtract.
Combined average of two groups = (n₁a₁ + n₂a₂)/(n₁ + n₂).
Consecutive numbers (or any arithmetic series): average = middle term = (first + last)/2.
First n natural numbers: average (n + 1)/2 · first n odd numbers: n · first n even numbers: n + 1.
If every value is increased by k, the average increases by k; if every value is multiplied by k, the average is multiplied by k.

Example 38 — adding a member The average age of 30 students is 15 years. When the teacher's age is included, the average becomes 16. Teacher's age?
= 31 × 16 − 30 × 15 = 496 − 450 = 46 years.

Example 39 — replacement The average weight of 8 persons increases by 2.5 kg when a new person replaces one weighing 65 kg. Weight of the new person?
= 65 + 8 × 2.5 = 85 kg.

Example 40 — overlapping groups The average of 11 results is 50. The average of the first 6 is 49 and of the last 6 is 52. Find the 6th result.
6th result is counted twice: 6 × 49 + 6 × 52 − 11 × 50 = 294 + 312 − 550 = 56.

Example 41 — raising an average A batsman's average in 10 innings is 32. How many runs must he score in the 11th innings to raise his average by 4?
Required = 11 × 36 − 10 × 32 = 396 − 320 = 76 runs.

Example 42 — combined average Section A has 40 students with average 60; section B has 60 students with average 70. Combined average?
= (40 × 60 + 60 × 70)/100 = (2,400 + 4,200)/100 = 66.

Example 43 — consecutive odd numbers The average of 5 consecutive odd numbers is 27. The largest?
Middle = 27 ⇒ 23, 25, 27, 29, 31 ⇒ largest = 31.

1.8Problems on ages

Method

Let the present ages be x and y (or ax and bx if a ratio is given).
"n years ago" ⇒ subtract n from every person; "after n years" ⇒ add n to every person.
The difference between two people's ages never changes.
In MCQs, test the options in both conditions — usually faster than algebra.

Example 44 — multiple A father is 4 times as old as his son. After 5 years he will be 3 times as old. Present ages?
Son = x, father = 4x ⇒ 4x + 5 = 3(x + 5) ⇒ x = 10 ⇒ son 10, father 40. Check: 45 = 3 × 15 ✔

Example 45 — ratio now and later The ratio of two people's ages is 3 : 5. After 6 years it will be 2 : 3. Present ages?
(3x + 6)/(5x + 6) = 2/3 ⇒ 9x + 18 = 10x + 12 ⇒ x = 6 ⇒ 18 and 30 years. Check: 24 : 36 = 2 : 3 ✔

Example 46 — sum and past ratio The sum of the ages of A and B is 50. Five years ago the ratio of their ages was 7 : 3. Present ages?
Five years ago the sum was 50 − 10 = 40 ⇒ ages then 28 and 12 ⇒ now A = 33, B = 17.

Example 47 — "half as old" Ten years ago A was half as old as B. The ratio of their present ages is 3 : 4. Present ages?
3x − 10 = ½(4x − 10) ⇒ 6x − 20 = 4x − 10 ⇒ x = 5 ⇒ A = 15, B = 20. Check: ten years ago 5 and 10 ✔

1.9Practice MCQ

Remember the −0.25 negative mark: answer only when you can eliminate at least two options.

1. 30% of 40% of 500 = ?

Answer: (b) — 0.3 × 0.4 × 500 = 60

2. A's income is 25% more than B's. By what percentage is B's income less than A's?

Answer: (b) — 25/125 × 100 = 20%

3. A price is increased by 10% and then decreased by 10%. The net effect is:

Answer: (c) — 10 − 10 − 100/100 = −1%

4. If 40% of a number is 120, then 75% of the number is:

Answer: (b) — number = 300; 0.75 × 300 = 225

5. Successive discounts of 25% and 20% are equal to a single discount of:

Answer: (b) — 25 + 20 − 500/100 = 40%

6. The cost price of 20 articles equals the selling price of 16 articles. Profit %?

Answer: (c) — (20 − 16)/16 × 100 = 25%

7. Selling an item at Tk 450 causes a 10% loss. To gain 20%, it must be sold at:

Answer: (c) — CP = 450/0.9 = 500; 500 × 1.2 = 600

8. A shopkeeper sells at cost price but uses a 950 g weight for 1 kg. His gain is about:

Answer: (b) — 50/950 × 100 = 5 5/19 % ≈ 5.26%

9. Marked price Tk 800, discount 15%. Selling price?

Answer: (a) — 800 × 0.85 = 680

10. Simple interest on Tk 6,000 at 8% p.a. for 3 years 6 months?

Answer: (c) — 6,000 × 8 × 3.5/100 = 1,680

11. A sum doubles itself in 10 years at simple interest. The rate is:

Answer: (c) — r = 100/10 = 10%

12. Compound interest on Tk 10,000 at 10% p.a. for 2 years (yearly)?

Answer: (b) — 10,000 × 1.21 − 10,000 = 2,100

13. Difference between CI and SI on Tk 20,000 at 5% p.a. for 2 years?

Answer: (b) — 20,000 × (5/100)² = 50

14. CI on Tk 40,000 at 10% p.a. compounded half-yearly for 1 year?

Answer: (b) — 40,000 × 1.05² = 44,100 ⇒ CI = 4,100

15. A town's population is 20,000 and increases by 10% every year. After 3 years it will be:

Answer: (b) — 20,000 × 1.331 = 26,620

16. A machine depreciates 10% every year. Its present value is Tk 81,000. Its value 2 years ago was:

Answer: (c) — 81,000 ÷ 0.9² = 81,000 ÷ 0.81 = 1,00,000

17. The effective annual rate of 12% p.a. compounded quarterly is about:

Answer: (c) — 1.03⁴ − 1 ≈ 0.1255 (12.36% is half-yearly; 12.68% is monthly)

18. A:B = 3:4 and B:C = 8:9. Then A:C = ?

Answer: (c) — A:B = 6:8, B:C = 8:9 ⇒ A:C = 6:9 = 2:3

19. The fourth proportional to 4, 9 and 12 is:

Answer: (c) — 9 × 12/4 = 27

20. The mean proportional between 4 and 25 is:

Answer: (a) — √(4 × 25) = 10

21. A invests Tk 20,000 for 12 months and B invests Tk 30,000 for 8 months. Out of a profit of Tk 12,000, A gets:

Answer: (b) — 2,40,000 : 2,40,000 = 1:1 ⇒ Tk 6,000 each

22. The average of 4 numbers is 25. When a fifth number is added, the average becomes 27. The fifth number is:

Answer: (d) — 5 × 27 − 4 × 25 = 135 − 100 = 35

23. The average of the first 10 natural numbers is:

Answer: (b) — (10 + 1)/2 = 5.5

24. The ages of A and B are in the ratio 4:5. After 5 years the ratio will be 5:6. B's present age is:

Answer: (b) — (4x + 5)/(5x + 5) = 5/6 ⇒ x = 5 ⇒ B = 25

1.10Written practice (5 marks each)

Write each answer in the "Let → Equation → Solve → Answer → Check" format shown below.

Problem 1 — Percentage (election) Q. In an election between two candidates, 20% of the voters did not vote and 500 of the votes cast were invalid. The winner got 55% of the valid votes and won by 750 votes. Find the total number of voters on the list.
Solution.
Let the number of valid votes = V.
Winner got 55% of V, the loser got 45% of V.
Margin = 55% − 45% = 10% of V = 750 ⇒ V = 750 × 100/10 = 7,500.
Votes cast = valid + invalid = 7,500 + 500 = 8,000.
Votes cast are 80% of the voters on the list (20% did not vote).
⇒ Total voters = 8,000 × 100/80 = 10,000.
∴ Total number of voters = 10,000.
Check: winner 4,125, loser 3,375 ⇒ margin 750 ✔

Problem 2 — Marked price and discount Q. A trader marks his goods 40% above the cost price and allows a discount of 15%. If he gains Tk 1,140 on an article, find its cost price, marked price and selling price.
Solution.
Let the cost price = Tk C.
Marked price = C + 40% of C = 1.4C = 7C/5.
Selling price = 85% of MP = 0.85 × 1.4C = 1.19C.
Gain = SP − CP = 1.19C − C = 0.19C = 1,140 ⇒ C = 1,140/0.19 = 6,000.
∴ CP = Tk 6,000; MP = 1.4 × 6,000 = Tk 8,400; SP = 0.85 × 8,400 = Tk 7,140.
Check: 7,140 − 6,000 = 1,140 ✔ (gain 19%)

Problem 3 — Bank deposit: CI half-yearly vs SI Q. A customer deposits Tk 50,000 in a bank at 8% p.a. interest compounded half-yearly. Find the amount and interest after 1½ years. How much more is this than simple interest at the same rate?
Solution.
Rate per half-year = 8/2 = 4%; number of half-years = 1.5 × 2 = 3.
A = P(1 + r/100)ⁿ = 50,000 × (1.04)³ = 50,000 × 1.124864 = Tk 56,243.20
CI = 56,243.20 − 50,000 = Tk 6,243.20
SI = 50,000 × 8 × 1.5/100 = Tk 6,000
Difference = 6,243.20 − 6,000 = Tk 243.20
∴ Amount = Tk 56,243.20; CI = Tk 6,243.20; it is Tk 243.20 more than SI.

Problem 4 — CI − SI for 3 years Q. The difference between compound interest and simple interest on a certain sum for 3 years at 10% p.a. is Tk 930. Find the sum and the compound interest.
Solution.
Let the sum = Tk P, r = 10% = 1/10.
For 3 years: CI − SI = P(r/100)²(3 + r/100) = P × (1/100) × (31/10) = 31P/1,000.
31P/1,000 = 930 ⇒ P = 930 × 1,000/31 = 30,000.
CI = 30,000 × [(1.1)³ − 1] = 30,000 × 0.331 = Tk 9,930.
∴ Sum = Tk 30,000; CI = Tk 9,930.
Check: SI = 30,000 × 10 × 3/100 = 9,000; 9,930 − 9,000 = 930 ✔

Problem 5 — Population (past value) Q. The present population of a town is 1,75,760. It has been increasing at 4% per annum for the last 3 years. What was the population 3 years ago?
Solution.
Let the population 3 years ago = P.
P(1 + 4/100)³ = 1,75,760 ⇒ P × (26/25)³ = 1,75,760
⇒ P = 1,75,760 × 15,625/17,576 = 10 × 15,625 = 1,56,250.
∴ Population 3 years ago = 1,56,250.
Note: 26³ = 17,576 and 25³ = 15,625, so the fraction method avoids long decimals.
Check: 1,56,250 × 1.124864 = 1,75,760 ✔

Problem 6 — Partnership Q. A starts a business with Tk 50,000. After 3 months B joins with Tk 60,000, and after 6 months C joins with Tk 80,000. At the end of the year the profit is Tk 54,000. Find each partner's share.
Solution.
Profit is divided in the ratio of capital × time (in months):
A = 50,000 × 12 = 6,00,000; B = 60,000 × 9 = 5,40,000; C = 80,000 × 6 = 4,80,000.
A : B : C = 600 : 540 : 480 = 10 : 9 : 8 (total 27 parts).
1 part = 54,000/27 = 2,000.
∴ A = Tk 20,000; B = Tk 18,000; C = Tk 16,000.
Check: 20,000 + 18,000 + 16,000 = 54,000 ✔

Problem 7 — Average (two groups) Q. The average monthly salary of all employees of a bank branch is Tk 32,000. The average salary of 8 officers is Tk 50,000 and that of the other employees is Tk 26,000. Find the total number of employees.
Solution.
Let the number of other employees = n.
Total salary: 8 × 50,000 + 26,000n = 32,000(8 + n)
⇒ 4,00,000 + 26,000n = 2,56,000 + 32,000n
⇒ 6,000n = 1,44,000 ⇒ n = 24.
∴ Total employees = 8 + 24 = 32.
Check: (4,00,000 + 6,24,000)/32 = 10,24,000/32 = 32,000 ✔

Problem 8 — Ages Q. The present ages of a father and his son are in the ratio 5 : 2. After 8 years the ratio will be 2 : 1. Find their present ages. How many years ago was the father 4 times as old as the son?
Solution.
Let the present ages be 5x and 2x years.
(5x + 8)/(2x + 8) = 2/1 ⇒ 5x + 8 = 4x + 16 ⇒ x = 8.
Father = 40 years, son = 16 years.
Let t years ago the father was 4 times as old: 40 − t = 4(16 − t) ⇒ 40 − t = 64 − 4t ⇒ 3t = 24 ⇒ t = 8.
∴ Father 40, son 16; the father was 4 times as old 8 years ago (32 and 8).
Check: after 8 years 48 : 24 = 2 : 1 ✔

Problem 9 — Ratio (income and expenditure) Q. The incomes of A and B are in the ratio 5 : 4 and their expenditures are in the ratio 3 : 2. If each saves Tk 1,600 per month, find their incomes.
Solution.
Let incomes = 5x and 4x, expenditures = 3y and 2y.
Saving = income − expenditure: 5x − 3y = 1,600 …(i) and 4x − 2y = 1,600 …(ii)
From (ii): y = 2x − 800. Put in (i): 5x − 3(2x − 800) = 1,600 ⇒ −x + 2,400 = 1,600 ⇒ x = 800.
∴ A's income = Tk 4,000; B's income = Tk 3,200.
Check: y = 800 ⇒ expenditures 2,400 and 1,600; savings 1,600 each ✔

Last-minute checklist for this chapter

Successive change a + b + ab/100 — also gives successive discounts (with minus signs) and area change.
Profit/loss on CP, discount on MP; MP/CP = (100 + p)/(100 − d).
CI − SI: 2 years P(r/100)²; 3 years P(r/100)²(3 + r/100).
Half-yearly: r/2 and 2n; quarterly: r/4 and 4n. Depreciation: (1 − r/100)ⁿ.
Partnership: capital × months. Average: work with sums, not averages.