AFCAT Numerical Ability · Percentage

Percentages in Interest, Work and Speed

Interest as a percentage of the principal, work done as a percentage of a job, and speed changed by a percentage, plus the speed and distance basics these questions lean on.

All Percentage notes

Why this matters

About a quarter of this page is HARD. The percentage part is usually one factor; what costs marks is the inverse step, since time goes down when speed or efficiency goes up, and by a smaller percentage. The last group of questions is plain speed, distance and time.

Concept 1 of 4: Simple and compound interest

Simple interest pays the same percentage of the principal every year. Compound interest pays it on the growing amount, the growth formula from the population page, so it pulls ahead a little more each year.

Definition

Simple interest: $SI = \dfrac{P \times R \times T}{100}$ ; amount $= P + SI$ .
A sum that doubles at simple interest has earned 100% of itself; one that becomes four times itself has earned 300%.
Compound interest: amount $= P\left(1 + \tfrac{R}{100}\right)^T$ ; $CI =$ amount $- P$ .
For 2 years, $CI - SI = P\left(\tfrac{R}{100}\right)^2$ : the interest earned on the first year's interest.

Simple and compound interest

SI = \dfrac{PRT}{100}, \qquad A = P\left(1 + \dfrac{R}{100}\right)^T

Worked example

Find the simple interest and the compound interest on ₹8,000 at 10% a year for 2 years.

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The same idea in a practice question:

Practice questionModerate

Example 1 · Percentage · Interest, Work and Speed

Simple interest on a sum for 2 years at

5\%

\text{Rs.}\,200

. What is the compound interest for the same period?

Amount is not interest

₹4,000 at 10% compound for 2 years amounts to ₹4,840, but the interest is ₹840. Read whether the question asks for the amount or the interest.

Drill 8 more on simple and compound interest

Concept 2 of 4: Work done as a percentage of a job

If A does 40% of a job in 10 days, A does 4% a day and needs 25 days for the whole job. Turn every percentage into a daily rate, then add or subtract rates.

Definition

$p\%$ of a job in $d$ days: the whole job takes $\dfrac{100d}{p}$ days; the daily rate is $\dfrac{p}{d}\%$ .
Working together, rates add: $\tfrac{1}{T} = \tfrac{1}{a} + \tfrac{1}{b}$ .
'A is 40% more efficient than B' means A's rate is 1.4 times B's, so A takes $\tfrac{1}{1.4}$ of B's time.
Efficiency and time are inversely proportional: efficiency $\times$ time is the job. So is men $\times$ days.

Rates add

\dfrac{1}{T} = \dfrac{1}{a} + \dfrac{1}{b}

Worked example

A does 40% of a job in 6 days. A and B together finish the whole job in 10 days. How long would B take alone?

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Practice questionModerate

Example 2 · Percentage · Interest, Work and Speed

A and B together can complete

60\%

of a work in

18

days. If A alone can complete the work in

50

days, in how many days can B alone complete the work?

More efficient does not mean that much less time

40% more efficient takes

\tfrac{1}{1.4}

of the time, about 29% less, not 40% less. Rates and times are inverse.

Drill 14 more on work done as a percentage of a job

Concept 3 of 4: Speed changed by a percentage

Over a fixed distance, time is distance over speed. Raise the speed by 20% and the time is divided by 1.2, so it falls by one sixth, not by 20%. It is the undo-a-change rule from the successive-change page.

Definition

Same distance: new time $=$ old time $\div$ speed factor.
Speed up $r\%$ : time falls by $\dfrac{r}{100 + r} \times 100\%$ . Speed down $r\%$ : time rises by $\dfrac{r}{100 - r} \times 100\%$ .
To finish in a set time, the speed factor is old time $\div$ new time.
The same inverse rule holds for fuel: if the kilometres per litre fall, the litres needed rise.

Time after a change in speed

t_{\text{new}} = \dfrac{t_{\text{old}}}{1 + r/100}

Worked example

A car takes 6 hours for a journey. Its speed is raised by 20%. How long does the journey take now, and by what percent has the time fallen?

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The same idea in a practice question:

Practice questionHard

Example 3 · Percentage · Interest, Work and Speed

A person increases his speed by

50

% and reduces his travel time by what percentage?

Time does not fall by the same percentage

20% faster saves

16\tfrac 23\%

of the time, not 20%. The time is divided by 1.2, not multiplied by 0.8.

Drill 14 more on speed changed by a percentage

Concept 4 of 4: Speed, distance and time: the basics used here

Some questions filed with percentages are plain speed problems. Three facts carry them: distance is speed times time, speeds add or subtract when two things move, and an average speed is total distance over total time.

Definition

$\text{distance} = \text{speed} \times \text{time}$ . To change km/h into m/s multiply by $\tfrac{5}{18}$ .
A train passing a pole covers its own length; passing a platform or bridge, its length plus the platform's.
Relative speed: add the speeds when moving towards each other, subtract when moving the same way.
A boat: downstream speed $= b + c$ , upstream $= b - c$ ; still-water speed is the average of the two.
Average speed is total distance over total time, never the average of the speeds.

Speed, distance and time

d = s \times t, \qquad \text{average speed} = \dfrac{\text{total distance}}{\text{total time}}

Worked example

A train 120 m long crosses a 230 m platform in 14 seconds. Find its speed in km/h.

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The same idea in a practice question:

Practice questionEasy

Example 4 · Percentage · Interest, Work and Speed

A train passes a telegraph post in

8

seconds and a platform

264

m long in

20

seconds. What is the length of the train?

Average speed is not the average of speeds

30 km at 10 km/h and 30 km at 15 km/h average 12 km/h, not 12.5. More time is spent at the slower speed.

Drill 17 more on speed, distance and time: the basics used here

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Simple and compound interest
Simple and compound interest
$SI = \dfrac{PRT}{100}, \qquad A = P\left(1 + \dfrac{R}{100}\right)^T$
Work done as a percentage of a job
Rates add
$\dfrac{1}{T} = \dfrac{1}{a} + \dfrac{1}{b}$
Speed changed by a percentage
Time after a change in speed
$t_{\text{new}} = \dfrac{t_{\text{old}}}{1 + r/100}$
Speed, distance and time: the basics used here
Speed, distance and time
$d = s \times t, \qquad \text{average speed} = \dfrac{\text{total distance}}{\text{total time}}$

Watch out for (4)

Amount is not interest
→ Simple and compound interest
More efficient does not mean that much less time
→ Work done as a percentage of a job
Time does not fall by the same percentage
→ Speed changed by a percentage
Average speed is not the average of speeds
→ Speed, distance and time: the basics used here

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