What is Mathematics? **Mathematics** is the study of numbers, shapes and patterns. The **word** comes from the Greek **word** meaning “science, knowledge, or learning”, and is sometimes shortened to **maths **or **math.**

This page will continuously be updated with Mathematics Questions across all maths topics. Practice as many Maths Questions as possible.

**TABLE OF CONTENTS**

## Business Mathematics Questions

1. John sold a pig and an Infinix s4 at #30,000 if he gains 20% profit on Infinix s4 and loses 10% on Pig. If he gets 10% total profit, then in how much should he sold the car?

**Assumptions**:

So Let ”˜x’ be cost price of the Infinix s4 and ”˜y’ be cost price of the pig.

Let ”˜X’ be selling price of the Infinix s4 and ”˜Y’ be selling price of the pig.

we know that X + Y = 30000.

He gains 20% on the Infinix s4. Implies Selling price = 1.2 times cost price (100/100 + 20/100).

Implies X = 1.2*x .

Similarly Y = 0.9*y .

His total profit is 10% implies that he gets 10% profit on his total investment meaning

Total cost price of both = (total selling price) / 1.1

Let z be total cost price. i.e x + y = z. z = 30000/1.1 Now we have 2 equations. x + y = 27272.72 . - 1 1.2x + 0.9y = 30000 . - 2 (eq 1)*1.2 - (eq2) we get 1.2x + 1.2y - 1.2x - 0.9y = 32727 - 30000 implies 0.3y = 2727.27 implies y = 9090. we know x + y = 27272.72 means x = 27272.72 - 9090 = 18182.72 = 18182. therefore his selling price of scooter would be X = 1.2x=1.2 * (18182) = 21819. therefore his selling price of car would be X = 1.2x=1.2 * (18182) = 21819. Is it correct?

2. A man’s income increases by 10%, then decrease by 10%. Calculate the percentage change in his income?

Let initial man income be 100 then after increasing 10 per cent it would be 110 then after degrades it by 10 per cent (10/100 X110) then 99. so overall percentage changes in his income would be 1 per cent

3. A used car salesman sold two cars for $2,000. On the first car he made a 25% profit, and on the other he lost 25% of the cost. How much did he make or lose in the dual transaction?

4. Mercy buys a car for $50,000. Shee then sells the car for 20% more than she bought it, how much did she sell the car for?

So we now know that Mercy sold the car for one fifth more than the price she paid for it. In other words we want to know what is one-fifth of $50,000 ($10,000) and then add that to the price he paid.

$50,000+$10,000=$60,000

5. Abigail sells her car for $7,500.00. She buys a new car that is $4,800 less than five times the selling price of her old car. How much did Abigail spend on her new car?

As the selling price of “his” old car is not given in the question, the price of “her” car spent by Abigail cannot be determined.

In the event the question is incorrectly written and assumptions are made to correct, the answer is $32,700 and she likely paid too much.

6. By selling a car for #300,000, a dealer makes a profit of 25%. At what price did the dealer buy the car?

7. Two laptops were sold at #1599 each. First was sold at 25% profit and second at 20% loss. Find the overall profit or loss %?

8. A salesman sold 2 markers at #2,500 each. His profit on one was 50% and the loss on the other was 25%. How much was his profit or loss percentage in total?

9. If a store sold 25 pairs of socks, making a 25% profit of $150, how much did they charge for each pair?

10. I bought a car for $8,000. Then sold it for $10,000. I then bought the car back for $12,000, and I sold the same car again for $16,000. In the end, how much money did I make or lose?

12. A man buys a car for $15,000. He sells the car for $18,000. He then buys the car back for $21,000, and he sells the car again for $23,000. In the end, how much money did the man make or lose? Or did he break even?

**Solutions**

Let’s say he only has $15,000 cash, no debt, and no car.

He uses all his money to buy the car, so now he has $0 cash, no debt, and a car.

He sells the car and now has $18,000 cash, no debt, and no car.

He borrows $3,000. He now has $21,000 cash, $3,000 debt, and no car.

He buys the car for all his money. He now has $0 cash, $3,000 debt, and a car.

He sells the car and now has $23,000 cash, $3,000 debt, and no car.

He pays the debt and now has $20,000 cash, no debt, and no car.

He started with $15,000 cash, no debt, and no car.

$20,000 – $15,000 = $5,000.

He’s $5,000 richer than when he started.

**MORE LOADING…..**

## PERMUTATION AND COMBINATION

1. How many ways can a committee of 5 be formed from 6 lecturers and 4 instructors such that the committee contains at least 2 lecturers?

6C2*4C3+6C3*4C2+6C4*4C1+ 6C5

6C2= 6!/(2!*4!) = 15, 4C3= 4!/(3!*1!) = 4,

6C2*4C3= 15*4= 60.

6C3= 6!/(3!*3!)= 20, 4C2 = 6,

6C3*4C2= 120.

6C4 = 6!/(4!*2!)= 15, 4C1= 4,

6C4*4C1= 15*4=60.

6C5=6.

So, 60*2+6+120=120+6+ 120=246 ways.

2. A committee of 5 is to be formed from 8 women and 12 men. How many committees are possible if there must be 3 women and 2 men?

3. A committee of 12 is to be formed from 9 women and 8 men. In how many ways can this be done if at least five women have to be included in a committee? In how many of these committees are the men in a majority?

3. Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is included in each committee?

4. A committee of 5 men and 3 women is to be formed out of 7 men and 6 women. If two particular women are not to be together in the committee, what is the number of committees formed?

There is no restriction on selection of men so the no of ways to select 5 men out of 7= 7C5.

Now consider the following three cases for selection of the 3 women.

SOLUTION: None of the 2 women(ones who can’t be together in committee) are in the committee. So now we have to select 3 out of the remaining 4. No of ways=4C3.

One of the two are in the committee-the other is automatically out of committee. So now we have to select only 2 out of the remaining 4. Ways of selection= 4C2.

Now suppose the other one is in the committee. Again 4C2.

Hence,no of ways of selection of women= 4C3 + 4C2 + 4C2.

Total ways to form committee is equal to: (ways of selection of men)(ways of selection of women)= 7C5* (4C3 + 4C2 + 4C2)= 336 ways.

5. From a group of 5 women and 7 men, a committee of 4 is to be constituted. How many committees are possible if at least 2 men are to be considered?

6. A group consists of 6 women and 8 men. How many ways are there to form a committee of a size of 5, such that one man and one woman can’t be on the same committee?

7. From a group of 5 men and 5 women, how many committees of size 3 is possible, with 2 men and 1 woman, if a certain woman must be in a committee?

SOL:** From a group of 5 men and 5 women, how many committees of size 3 is possible, with 2 men and 1 woman, if a certain woman must be in a committee?**

We want to pick 2 men and 1 woman, but the woman is chosen for us. So we just need to pick the 2 men from the pool of 5 = (52)=10(52)=10.

8. How many committees can be formed of 5 members from 4 men and 6 women if at least 2 men always included and 1 particular woman excluded?

9. From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together?

Ways of choosing 2 women from 5

= 5C2 = 10.

Ways of choosing 3 men from 7

= 7C3 = 35.

Total ways = 10*35 = 350.

Now, 2 men refuse to serve on a committee together. Let’s find the number of possible combinations with these 2 men & subtract it from 350 to get the answer.

Choices for 3rd man = 5.

Choices for women = 5C2 = 10.

Possible combinations = 5*10 = 50.

Number of required combinations

= 350 – 50 = 300.

10. A committee of 5 people is to be formed from a selection pool of 12 people. If Carmen must be on the committee, how many committees can be formed?

SOL: The 12 is made up of 2 groups. There is a group of Carmen, and a group of not Carmen. The group of Carmen has 1 person in it, the group of not Carmen has 11 people in it.

**MORE LOADING…..**

## BEARING

1. A ship leaves port and travels 21km on a bearing of 032^{0}and then 45km on a bearing of 287^{0}.

(a) Calculate its distance from the port.

(b) Calculate the bearing of the port from the ship.

2. A surveyor leaves her base camp and drives 42km on a bearing of 032^{0}, she then drives 28km on a bearing of 154^{0}. How far is she then from her base camp and what is her bearing from it?

3. Two ships leave port at the same time and travels at 5km/hr on a bearing of 046^{0}. The other travels at 9km/hr on a bearing of 127^{0}. How far apart are the ships after 2 hours?

4. A triangular field has two sides 50m and 60m long, and the angle between them is 096^{0}. How long is the third side?

5. An airplane flies faster than a cargo plane at a speed of 40km per hour. The airplane flies 500km in the amount of time that the cargo plane flies 400km. What is the speed of each plane?

6. What’s the minimum speed for an aeroplane to stay airborne?

**Solution: **It depends on its aerodynamics and design of its wings. There is no ”˜slowest speed an aircraft can fly’ per sÃ©; the speed is always dependent of the design of the particular aircraft. The speed where an aircraft just and just stays aloft on level flight is called ”˜stall speed’.

7. An airplane is headed due south”‹ (bearing 180°”‹) with an airspeed of 200 mph. The wind is from the northeast with a bearing of 210° and a speed of 50 mph. What is the ground speed, drift angle, and course of the plane?

8. Why do planes fly at 30-50k feet in the air? Why don’t they fly any lower?

a. **Weather.** You get to fly above most of it.

b. **Air density**. The air is nice and thin up there, which decreases parastic drag and thus increases fuel c. economy.

c. **Engines.** For reasons 1 and 2, those turbofan engines were designed with an altitude of 30ish thousand feet in mind, so that is where they operate most efficiently.

d. **Safety.** In case those engines fail, the higher you are the farther you can go unpowered. A 737 90 miles out at 30k feet can make it to the runway just fine.

9. What will happen if a plane flies above 60,000 feet?

Solution: If a conventional airplane is “put” at 60,000 ft (a conventional aeroplane cannot reach this height by itself), its engines will not be working, because of lack of sufficient air. The aeroplane will glide until it reaches about 40,000 ft. Everybody onboard will be using oxygen masks.

10. A plane flies 90km on a bearing of 030 degrees and then flies 150km due east. How far east of the starting point is the plane? Tell your teacher this question only makes sense if you assume the plane in on a flat plain.

11. What causes an aeroplane to fly?

12. What is the mean distance if a plane flies 1,000 km in 73 minutes?

In this “question”, the plane only does one journey. The time it takes is irrelevant, and the “mean distance” is 1,000 / 1 or 1,000 km.

**MORE LOADING…..**

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