quantitive aptitude

QUANTITATIVEAPTITUDE

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TechRel Technologies Pvt. Ltd., Pune. www.facebook.com/techrelindia

QUANTITATIVE APTITUDE

TechRel Technologies Pvt. Ltd.

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INDEX

1. Percentages, Profit And Loss…………………………………..03

2. Ratio, Proportion & Variation……………………………………..09

3. Averages And Mixtures …………………………….……………..13

4. Time And Work …………………………………………………..18

5. Time, Speed And Distance …………………………….…………..23

6. Number Theory …………………………………………………..29

7. Simple And Compound Interest …………….……………………..41

8. Quadratic And Higher Order Equations ……..……………………46

9. Set Theory ………………………………………………………..51

10. Plain Geometry ………………….………………...……………..54

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CHAPTER 1

PERCENTAGES, PROFIT AND LOSS 1.1 Percentage: The word percentage means per hundred, for example ,if a person saves of his income it means he saves 30 part for every 100 part he earns. Ex: quantity of water in milk constitutes 5 parts every 15 parts of the mixture what is the percentage of water in the mixture?

Solution:

Percentage of Water = No. of parts of water/No. of parts of mixture

= 5/15 x100=33.33%

Percentage increase or decrease is calculated with respect to the base value unless mentioned otherwise.

Percentage increase or decrease = increase or decrease/base value x 100

Ex: if A’s income is 25% more than B’s income then by what percentage is B’s income less than A’s?

Solution:

Let B’s income =100

A’s income =100+25% of 100=125

B”s income is less than A’s income by 25

Percentage decrease= decrease/base value x100

= 25/125x100=20%

Note here b’s income is compared with A’s income, so A’s income should be taken as base value.

If a quantity is increased by x% then the final quantity is obtained by multiplying original quantity with (100+x/100)

If a quantity is reduced by r%, then the final quantity is obtained by multiplying original quantity with (100+r/100)

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Ex: if the price of petrol is increased by 20% and subsequently by 40% , what is the final price if the original price is rs. 25?

Solution:

Final price= original price x (100+a/100)x(100+b/100)

Therefore Final price = 25x(100+25/100)x (100+40/100)= 42 Rs.

Note : use negative sign in case of price reduction.

Successive increase or decrease of percentage:

Let ‘a’ be the first percentage change and ‘b’ be the second , the net change is given by [a+b+axb/100]%

1.2 Profit and Loss:

Ex: Find the profit % , if an article worth 300% is sold for 312 ?

Solution:

Profit= selling price-cost price=312-300=12

Profit %= profit/cost price x100

= 12/300 x 100= 4%

Percentage reduction in consumption

Profit= selling price-cost price

Loss= cost price- selling price

Profit %= profit/cost pricex100

Loss % = loss/cost price x 100

Selling price = [1+ gain%/100] x cost price

Selling price = [1+ loss%/100] x cost price

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If the rate of a commodity is increased , then the consumption should be reduced to maintain the same expenditure.

Percentage reduction in consumption =( % change/100% + change) x100

Ex: If the rate of rice is increased by 25% then what should be the consumption to maintain the same expenditure?

Solution:

=(25/100+25)x100=20%

So the quantity of rice consumption should be reduced by 20% to maintain the same expenditure.

Successive Discount:

If two successive discounts of a% and b% are offered then the net discount offered is [a+b-axb/100]%

Ex: Find the net discount offered on two successive discounts of 20% and 30%?

Solution:

Net discount = [a+b – a x b/100]

= [20+30 – 20 x 30/100]

= 44 %

Discount is the reduction of price . if a% discount is offered then the article would be sold for (100-a)% of the cost price.

Selling price = (100-a)/100 x Cost price

1.3 False Weight:

In case a false weight which is less than the actual weight is used then the transaction ends in a profit.

% profit = [error/ true value- error x 100]

Ex: If a shop keeper sells 800 gms at cost price claiming it to be 1 kg, find the gain %?

Solution:

% profit = [ error/true value-error x 100]

= 200/1000-200 x 100=25%

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In case where selling price of two articles is the same and one is sold at a loss of x% then after at a profit of x% then this transaction always leads to a loss of x^2/100

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EXERCISE 1

1. By selling 11 apples for a dollar a man loses 10%, how many apples for a dollar should he sell to gain 10%?

(A) 9 (B) 10 (c) 11 (D) 12

2. Mr smith credits 20% of his salary into a fixed deposit account and spends 35% of the remaining on groceries , if cash in hand is rs.2600 what is his salary?

(A) 4000 (B) 4500 (C) 5000 (D) 5500

3. The price of an article has been reduced by 30% , in order to restore the original price the current price must be increased by?

(A) 30% (B) 42 6/7% (C) 23 1/3% (D) 42 1/9%

4. If the denominator is increased by 20% and the numerator is diminished by 10% the value of fraction is 21/6, the original fraction is ?

(A) 5/4 (B) 7/4 (C) 4/7 (D) 14/3

5. If the side of a square is increased by 20% its area is increased by?

(A) 4% (B) 20% (C) 44% (D) 40%

6. The price per Kg of rice increases by 20% by what percentage should the consumption be decreased such that expenditure remains the same?

(A) 20% (B) 16.67% (C) 25% (D) 16.33%

7. Ram has to travel from Hyderabad to Chennai which is a certain distance apart. 23% of the distance was travelled by a bus ,50% of the remaining by train and the rest of distance 231 km by taxi, find the distance between Hyderabad and Chennai in KM?

(A) 600 (B) 462 (C) 231 (D) 856

8. The daily wages of a company worker is increased by 25% and the decreased by 25% , what is the net change in his wage?

(A) 6.25% increase (B) 6.25% Decrease (C) No change (D) None of these

9. Adding 30% of 50% of n to n is multiplying of n by what?

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(A) 1.3 (B) 0.5 (C) 1.5 (D) 1.15

10. The price of a toy is first increased by 30% and then decreased by 20%. If the price of the toy is now Rs.312 what was the original price?

(A) 350 (B) 250 (C) 300 (D) 400

11. The population of a town decreases at the rate of 5% per year if after two years the population of town is 902500, then what is the population of the town before two years?

(A) 100000 (B) 814507 (C) 1000000 (D) None of these

12. If the cost price of 12 articles is equal to the selling price of 10 articles find the gain or loss percentage?

(A) 20% loss (B) 20% profit (C) 16.67% loss (D) 16.67% Profit

13. In a direct election between two contestants, 4% of the votes cast are declared to be illegal. One contestant secures 55% of the valid votes and wins with a majority of 240 Votes. What is the total number of votes cast?

(A) 455 (B) 2400 (C) 2500 (D) None of these

14. A dishonest business man professes to sell his article at cost price but he uses false weight with which he cheats by 10% while buying and 10% while selling find his profit percentage?

(A) 20% (B) 21% (C) 22.22% (D) 25%

15. A dishonest shopkeeper profeses to sell 1 kg wheat at Rs.18 but he bought 1 Kg at Rs.16 , and

uses faulty weight of 0.9 kg for 1 Kg, find his gain %?

(A) 25% (B) 20% (C) 22.22% (D) 12.5%

16. A and B start a business with a capital of Rs.2000 and Rs.3000 respectively and at the end of the

year their profit is Rs.2500 what is the profit of A?

(A) 500 (B) 1000 (C) 250 (D) 1250

17. A,B and C start a business . A invest 2 times as much as B invests and B invests 3 times as much

as C invests . Then the ratio of capital of A,B and C is?

(A) 1:3:6 (B) 6:3:1 (C) 1:2:3 (D) 3:2:1

18. Three partners A,B, C invest Rs.1000, Rs.2000 and Rs.4000 respectively in a Business out of a

profit of 1400 B’s share is ?

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(A) 200 (B) 700 (C) 400 (D) 800

19. Sunil started a business with an investment of Rs.1200, 4 months later Suresh joined him

investing 1800, if they made a profit of Rs.1300, then Suresh’s profit is?

(A) 650 (B) 260 (C) 520 (D) 780

20. The value of a machine depreciates at the rate of 20% every year. It was purchased 2 years ago. If

its present value is 6400, its purchase price was?

(A) 9240 (B) 7920 (C) 6400 (D) 10000

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CHAPTER 2

RATIO , PROPORTION & VARIATION 2.1 Ratio: A ratio is a comparison of two numbers by division. The ratio of a to b is expressed as A:b= a/b Here a is called antecendent B is called consequent 2.2 Proportion: Proportion is the equality of two ratios.

Eg. 4/20= 1/5 is a proportion

If a,b,c,d are in proportion

Then a/b=c/d

Ad = cb

Product of extremes = product of means

2.3 Properties of Proportion:

If a/b = c/d then

(a) b/a = d/c (b) a/c=b/d (c) a+b/a=c+d/d

(d) a-b/a=c-d/d (e) a+b/a-b=c+d/c-d

Here a is the 1st proportional

B is the 2nd proportional

C is the 3rd proportional

D is the third proportional

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2.4 Continued Proportion:

Three quantities a,b,c are said to be in continued proportion

If a=b=b:c=b=√ac

Here , b is called the mean proportional af a nad c

Eg. The mean proportional of 0.32 and 0.02 is

X=√0.32 x0.02= 0.08

Third proportional of a and b is b²/a

Eg. Find the third proportional of 16 and 24

X=24²/16=36

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EXERCISE 2

1. On a scale of a map,1.2 mm represents 20.5 miles, if the distance between the points of the map is 6 mm the actual distance between these points is

a. 102.5 mm b. 102.5 miles c.10.25 mm d. 10.25 miles

2. If 28 men can construct a wall in 40 days , then how many days will 70 men do it?

a. 20 b. 16 c. 100 d. none

3. A square and an equilateral triangle all equal sides of length . the ratio between area of triangle and square is

a. √3 :3 b. 3: √3 c. 4: √3 d.√3 :4

4. The sum of three numbers which are in ration 5:7:13 is 1250. The difference between the greatest number and the least number is

a. 50 b. 100 c. 400 d. 300

5. If amit can 35 words in 7 sec then how many word can he read in an hour?

a. 3600 b. 18000 c.245 d. none

6. The weight of a 17m long carbon rod is 30.6 Kgs the the weight of 11m long carbon rod is ?

a. 198 kgs b.19.8 kgs c. 18.8 kgs d. 20.8 kgs

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7. If the sum of two whole numbers is 24, which of the following cannot be their ratio?

a. 1:2 b. 1:3 c. 3:5 d. 2:5

8. If 10, 20 ,16 and x are in proportion , then x=

a. 8 b. 24 c. 26 d. 32

9. The mean proportion of 1/6 and ½.16 is

a. 36 b.3.6 c.1/3.6 d.1/36

10. If half of A one third of B and one seventh of C are equal then A:B:C is

a. ½:1/3:1/7 b. 7:3:2 c. 21:14:6 d. 2:3:7

11. The third proportion of 0.3 and 0.9 is

a. 2.7 b. 27 c. 0.027 d.0.27

12. If x is added to each term of ratio 7:11 and the ratio becomes 4;5 then x is

a. 3 b. 6 c. 9 d. cannot be determined

13. If thrice of A , twice of B and 7 times of C are all equal, then the ratio of B to A & B to C is

a. 1:6 b. 6:1 c. 7:15 d. 15:7

14. Rs 170 is generated using a combination of 10ps,25 ps and 50ps coins , if the ratio of 10ps ,25 ps and 50 ps coins is 5:10:11, then the total number of coins is

a. 100 b. 200 c. 220 d. 520

15. The side of square is in ratio 2;3 then the ratio of their areas is

a. 2:3 b. √2:√3 c. 4:9 d. 8:27

16. A,B and can do a work in 7,21, and 14 days respectively, if they undertook to finish the work together for 1210, the share of A exceeds the share of B by

a. 660 b. 440 c. 330 d. 220

17. Acertain no. of person can do a piece of work in 36 day , three times the number4 of such person will do twice the original work in how many days?

a. 36 b. 12 c. 18 d. 24

18. If 10 pigieons make 10 nests in 10 days , 1 pigeons can make 1 nest in how many days?

a. 10 b. 1 c. 100 d. 1/10

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19. The proportion of milk and water in two samples is 5:2 and 7:5 if a mixture comprising of equal quantities of two samples is made, the proportion of milk and water in the mixture is

a. 12:7 b. 7:12 c. 109:59 d. 59:109

20. The annual salary of Mr. John , Mr.Adam and Mr.Joe is in the ratio 2;3:5 if the salary of Mr. Joe is 90,000 more than that of Mr.John, then the monthly salary of Mr.Adam is

a. 7,500 b.75,000 c.90,000 d.none.

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CHAPTER 3 AVERAGES AND MIXTURES

3.1 Average: Average is the mean value of a set of numbers of Values

Therefore Average = (x1+x2+x3…….xn)/n= i ∑ x1 N=1 i Ex: if the ages of four (4) students are 20,22,18 and 24 years respectively, find the average age.

Average = sum of ages = 20+22+18+24 = 84 = 21 No. of students 4 4

3.2 Weighted Average :

Weighted average is the average of two or more groups whose individual averages are known

W.A = n1.a1 + n2.a2 n1+n2

Group Average

N1 a1

N2 a2

3.3 Average Speed:

Average speed is the ratio of Total distance to the time taken

Average Speed = Total Distance Total Time

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3.4 Age and Average:

If the average age of n persons is decreased by ‘a’ years then the total age decreases by ( n x a years and vice versa)

Ex:The average weight of six (6) men decreases by 3 kgs when a man whose weight is 60 kg is replaced by another man. Find the weight of the man newly included.

Solution:

Decrease in total weight = 6 x 3 = 18 Kg so , the weight of the newly included man is 60-18 i.e is 42 kgs.

Note:- if a value which is less than the actual value is entered, then the average would reduce.

3.5 Rules of Allegation:

If two quantity are mixed in a ratio , then

Quantity of cheaper = cp of dearer – Kean price Quantity of Dearer Mean price- Cp of Cheaper

Cp of cheaper(c) cp of dearer(d)

Mean Price(m)

(d-m) (m-c)

Ratio = x1 : x2

Hence , cheaper quantity : dearer quantity is (d-m): (m-c)

Container originally contains ‘v’ units of liquid and ‘K’ units of liquid is taken out of this operation is repeated ‘n’ times , then the final quantity of liquid in container is

V x [1-k/v] n units

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Ex: for a container with 120 lts of milk, 40 lts is taken out and replaced with water , if this process is repeated twice, find the quantity of milk in the container now?

Solution:

Quantity of milk in container

= 120 [1-40/120]²

= 120[1-1/3]²

=120 x 4/9

=53.33 lts

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Exercise 3

1. There are two groups A and B consisting of 30 and 70 students respectively, if the average weight of Group A is 30 KGs and that of group B is 70 KGs find the average weight of all the students of Group A and B

a. 58 b. 50 c. 40 d. 42

2. The average weight of x, y and z is 50 kgs , if the average weight of x and y is 38 kgs and that of Y and z is 52 kgs , find the weight of Z.

a. 44 b. 34 c.30 d. 12

3. The average of the four digit numbers which remain the same when the digits inter change their position is

a. 4444 b. 3333 c. 5555 d. 6666

4. The average temperature on Friday , Saturday and Sunday was 38°c, the average temperature on Saturday, Sunday and Monday was 40°c.Then the difference between on Monday and Friday was

a. 6°c b. 36°c c. 39°c d. None of these

5. The average cost of 4 apples and 7 bananas is Rs.16, The average cost of 7 apples and 4 bananas is Rs.24. The cost of 1 apple and 1 banana is Rs.

a. 11 b. 40 c. 30 d. 8

6. The average of 50 numbers was 36, later it was found that one number was misread as 67 instead of 76, find the correct average

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a. 36.18 b. 35.88 c. 45 d. 27

7. The average age of three women is 30. If their ages are in ratio 2:5:3 the age of youngest woman is

a. 9 b. 18 c. 27 d. 4.5

8. The average score of Shikhar Dhawan for 10 matches is 52.9 runs, if the average for the first six matches is 50 runs the average for the last four matches is

a. 47.25 b.57.25 c.57.75 d. 67.25

9. The average of 10 numbers is 15. If each number is multiplied by 3 and added to 5, the average of new set of 10 numbers is

a. 15 b. 45 c. 40 d. 50

10. The average age of P,Q,R and S three years ago was 42 years, by including A,the present average age of all the five is 47 years. The present age of A is?

a. 45 b. 50 c. 55 d. 60

11. A van covers four sides of a square at speeds of 5, 10, 15 and 20 Kms per hour respectively. What is the average speed of the van in its journey?

a. 12.5 Kmph b. 9.6 kmph c. 2.4 Kmph d. none

12. A person covers equal distance by Car, bus, and train with speed 20, 30 and 40 kmph respectively . The average speed of the total journey is

a. 30 kmph b. 22 9/13 kmph c. 27 9/13 kmph d.90 kmph

13. On a journey across Hyderabad , a car averages 25 kmph for 50% of the distance , 30 kmph for 30% of the distance and 20 kmph for the remaining distance. The average speed of the whole journey is

a. 20 kmph b. 25 kmph c. 30 kmph d. none

14. A container contains 1000 litres of Milk. From this container 100 litres of milk was taken out and replaced by water .If this process was repeated another three times , how much milk is now contained by the container?

a. 729 lts b. 700 lts c. 656.1 lts d. 343.9 lts

15. In what proportion must wheat at rs 11:00 per Kg be mixed with wheat at Rs. 16:00 per Kg so that the mixture is worth Rs.13 per Kg?

a. 2:3 b. 3:2 c. 11:16 d. 16:11

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16. If 3 Kg of metal , which is one third silver and rest alumunium is mixed with 7 Kg of another metal, which is two-seventh silver and rest alumunium. What is the ratio of silver is to alumunium in the mixture

a. 3:7 b. 7:3 c. 1:7 d. 2:7

17. In what proportion must water be mixed with milk in order to gain 25% by selling it at cost price

a. 1:4 b. 4:1 c. 4:5 d. 5:4

18. 10 litres of a mixture contains 20% water and rest milk, if 2 litres of water is added the ratio of milk to water in the new mixture is

a. 4:1 b. 1:4 c. 2:1 d. 1:2

19. In a mixture of 60 litres , the ratio of milk and water is 3:1 if the ratio of the milk and water is to be further added to the mixture

a. 52.5 lts b. 62.5 lts c. 67.5 lts d. 45 lts

20. A shopkeeper mixes 13 kgs of sugar at rs.13 per kg with 17 kg of sugar at Rs.17 per Kg. find the cost per Kg. of mixture

a. Rs.14 b. Rs 15 c Rs.15.25 d. Rs.16

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CHAPTER 4 TIME AND WORK

4.1 Work: Quantity of work is directly proportional to the time taken i.e W∞T Ex: If a man can reap 100 coins in 4 hours find the time taken by him to reap 5 coins?

W∞T

Therefore w/t is constant

100/4 = 5/9 ? = 20/100 (or) 1/5 hours

If ‘A” can do a piece of work in ‘n’ days then in one day he can finish 1/n part of work in 1 day.

Ex: If a person can finish a work in 3 days then he can finish 1/3rd of work in 1 day.

If A can do a piece of work in ‘N’ days and ‘B’ in ‘M’ , then both of them can finish the work in

N xM M+N Days Ex: If A can do a piece of work in 4 days and B in 6 days , find the time taken by both of them to finish the work Solution: Time taken to finish the work = n xm / m + n days = 6 x4 / 6 + 4 = 2.4 days If A & B can do a work in ‘n’ days , B and C in ‘y’ days , C and A in ‘z’ days then A,B,C working together can finish the work in 2x ( X x Y x Z) / X x Y + Y x Z + Z x X days A can do it in y x d/ y-d days B can do it in Z x d / Z- d days C Can do it in X x d/ X- d days D = (2 x X x Y x Z) / X x Y + Y x Z + Z x X

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4.2 Efficiency: Efficiency is the rate of doing work

1. Efficiency (E) is Inversely proportional to time E ∞ 1/t where E x T is constant

2. Efficiency E is directly proportional to Wages/ Works E∞ W where E/W = constant

Ex: If A can finish a piece of work in 60 days find the time taken by B to finish the work given that B is 50% more efficient than A? Solution : Let the efficiency of A be 100% Time taken by A is 60 days Efficiency of B is 100+50= 150% Efficiency ∞ 1/time Where EaTa = EbTb Therefore 100 x 60= 150 x Tb Where Tb= 40 days. If a pipe can fill a tank in ‘n’ hours and another pipe empties the same tank in ‘m’ hours (n<m), then the tank is filled in m x n / m-n hours. If m < n then the tank is emptied in m x n / n-m hours Ex: Pipe A can fill a tank in 4 hours and B empties it in 6 Hours find the time taken to fill the tank, if both are opened at the same time? Solution: Time taken to fill the tank = m x n/ m-n = 6 x 4/ 6-4 = 12 hours ----------------------------------------------------------------------------------------------------

EXERCISE 4

1. Ramu can do a piece of work in 15 days ,Ravi can do the same work in 30 days. If both of them are doing the same work together in how many days will they complete half the work? a. 10 days b. 15 days c. 22 ½ days d. 11 ¼ days

2. Ajay can do a piece of work in 24 days, sunil can do the same work in 8 days . If they work at it on alternative days with Ajay beginning the work then in how many days the work will be completed?

a. 6 days b. 16 days c. 12 days d. none

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3. Sanjay can do ¼ of work in 5 days and Shyam can do 4/5 of the work in 64 days in how many days both together can do the work?


4. A,B and C working in shoe manufacturing company can do assignments in 14, 28, and 56 days respectively working together, the assignment will be completed in how many days?

a. 1/8th day b. 49 days c. 8 days d. 28 days

5. Mr. X can complete a job in 18 days , Mr.Y in 20 days and Mr. Z in 30 days , Mr. Y and Mr.Z start the work and are forced to leave after 4 days . The time taken to complete the remaining work by mr.X is

a. 12 days b. 18 days c. 20 days d. 26 days

6. Sharma and shyam can do a piece of work in 24 days , shyam and shastri in 30 days , shastri and Sharma in 40 days in how many days will Sharma finish it alone?

a. 20 days b. 60 days c. 40 days 120 days

7. If 3 women working for 6 hours a day earns Rs.650 in 10 days then how much will 18 women working 9 hours a day earn in 10 days?

a. Rs.6500 b. Rs. 5850 c. Rs.925 d. none of these

8. 3 men and 5 women can do a job in 12 days , how long will 6 men and 5 women take to finish the same job?

a. 4 days b. 6 days c. 8 days d. none of these

9. Henry and maggy can complete a work in 50 and 75 days respectively. Henry starts the work and after 25 days maggy also joins him. How many days will the work get completed?


10. A,B and C together earn Rs.2700 in 9 days .A and B together earn Rs.1050 in 6 days . B and C together earn Rs.1400 in 7 days . Find earning of B per day

a. Rs.300 b. Rs. 200 c. Rs.175 d. Rs.75

11. Ramesh and Roy contract to do a work together for $ 600. Ramesh alone can do it in 2 days and roy alone in 3 days . But with the help of Joe they finish it in a day, how much is joe’s share?

a. $100 b. $ 200 c. $ 300 d. $ 400

12. Mr. Ricky can do a piece of work in 60 days Mr.peter is 50% more efficient thanMr.Ricky. the number of days taken by Mr.Peter to do the same work is

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13. Working together peter and sue can do a work together in 12 days. Sue and rose can do it in 20 days while peter and rose can do it in 10 days. How long will it take if all of them work together to complete the job?

a. 4 2/9 days b. 3 2/9 days c. 7 2/4 days d. 8 4/7 days

14. Pipe A can fill a tank in 140 hours while pipe B alone can fill it in 210 hours and pipe c can empty the full tank in 280 hours. If all the pipes are opened together how much time will be needed to fill the tank?

a. 100 hours b. 110 hours c. 70 hours d. 120 hours

15. Two pipes A and B can fill a tank in ! hour 12 minutes and 1 hour 30 minutes respectively. Pipe C can empty the tank in 1 hour . Initially Pipes A and B are opened and after 14 minutes C is also opened. In how much time will the tank be full?

a. 1 hr b. 80 min c. 84 min d. 1 hr 32 min

16. A cistern can be filled by two pipes X and Y in 2 hours and 3 hours respectively. While full tank can be emptied by third pipe Z in 4 hours. If all the taps be turned on at the same time the cistern will be full in

a. 1 hour 39 minutes b. 1 hour 43 minutes c. 1 hour 51 minutes d.1 hour 54 minutes

17. A tank is filled by pipe A in 16 minutes and Pipe B in 18 minutes. When full it can be emptied by a pipe C in 10 minutes. If all the three pipes are opened simultaneously two tanks of same volume can be filled in

a. 55 5/13 min b. 50 10/13 min c. 1 hr. 50 10/13 min d. 1 hr 55 5/13 min

18. A can do a piece of work in 120 days He works at it for 24 days and then B finishes it in 48 days. How long will they take to complete the work if they do it together?


19. There is a leak in the bottom of the cistern, when the cistern had no leak it was filled in 2.5 hours, it now takes half hour longer. If the cistern is full of water , how long will it take in leaking itself empty, in case the water leaks out at double the rate after half the cistern becomes empty?

a. 15 hrs b. 11 hrs 15 min c. 11 hrs 25 min d. 7.5 hrs

20. If two pipes function simultaneously, a reservoir will be filled in 6 hrs . one pipe fills the reservoir 5 hrs faster than the other. How many hours does the faster one take to fill the reservoir.

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a. 10 hours b. 12 hours c. 15 hours d. none of these.

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CHAPTER 5 TIME, SPEED AND DISTANCE

5.1 Speed: Speed = Distance/ Time or Distance = Speed x Time

1. Speed is directly proportional to distance

2. Distance is directly proportional to time

3. Speed and time are inversely proportional

5.2 Conversion Factors:

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1 Km/Hr = 5/18 m/s

1 M/S = 18/5 Km/Hr

Ex: A bowler has a run up of 150 m. if the speed of the bowler is 54 kmph, how much time would he take to complete the run up?

Solution:

54 km/Hr = 54 x 5/18 m/s = 15 m/s

Time = distance /speed =150/15 = 10 sec.

In travelling equal distance with speeds of X and Y , the average speed is given by

2x XxY/X+Y

Ex: Find the average speed of a car that covers 1st 50% of distance at 40 Km/Hr and the 2nd 50% of the distance at 60 Km/ Hr.

Solution:

Average speed = 2 x X x Y / X+Y

= 2 x 60 x 40/ 60 +40

=4800/100=48 km/hr

5.3 Relative Speed :

The speed of a body with the respect to another moving body is defined as relative speed.

If two bodies are moving in same direction , the relative speed is given by the difference of speeds.

X

y

Relative speed = (X-Y)

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If two bodies are moving in the opposite direction, then the relative speed is given by the sum of the speeds.

X

Y

Relative speed = (X+Y)

Ex: If two trains of length 200m and 250m are having speed 18 km/hr and 36 km/ hr , find the time taken to cross each other

1. If they move in the same direction

2. If they move in opposite direction

Solution:

Speed of first train = 18 x 5/18= 5m/s

Speed of second train = 36 x 5/18= 10 m/s

Distance they have to cross to cross each other is the sum of lengths of train = 200 + 250 =450m

If they move in same direction, relative speed = 10-5 = 5 m/s

Time taken = distance /Speed= 450/5 = 90 sec

If they move in opposite direction, relative speed = 10 + 5=15 m/s

Time taken = distance / speed = 450 /15 = 30 sec.

5.4 Boats and Streams:

If the speed of boat in still water is X km /hr and speed of stream is Y km/hr then

Speed down stream =X + Y

Speed Up Stream = X-Y

Speed of boat in still water = Speed up stream+speed Down stream/2

Speed of Stream = Speed down stream- speed upstream/2

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Ex: If the speed down stream is 12 Km/hr and speed upstream is 8 km/hr , find the speed of boat in still water and speed of stream.

Solution:

Speed of boat in still water = speed upstream + speed down stream/2

= 12+8/2= 10 km/hr

Speed of stream = Speed down stream – speed upstream/2

=12 -8/2= 2 km/hr

EXERCISE 5

1. John covers half of a certain distance by bus at 40 Km/hrand the remaining by by train at 60 km/hr. John return to starting point riding a scooter at 52 Km/hr. Find his average speed for the whole Journey.

a. 48 km b. 49.52 km c. 50.66 km d.51 km

2. A train travelling with uniform speed crosses a platform of length 750 meters in 15 seconds and crosses a bridge of length 1.8 km in 30 sec. what is the length of the train.

a. 1 km b. 1.3 km c. 0.3 km d.0.03 km

3. The average speed of a train with stoppages is 75 kmph and without stoppage is 90 kmph. How many minutes per hour does the train stop.

a. 18 b. 16 c. 12 d. 10

4. If 601 signal poles are arranged such that the distance betweentwo consecutive signal poles is 0.1 km and a train 240 m long crosses them completely in 45 minutes. What is the speed of the train.

a. 80.45 kmph b. 80.40 kmph c.80.32 kmph d. none

5. Ram and shyam start driving cars in same direction at 36 kmph and 15 m/s respectively. In how many hours they will 100 kms apart.

a. 5 5/9 b. 1 1/9 c. 6 2/3 d 4 2/3

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6. A train running at a uniform speed crosses two women coming in opposite direction at 2.4 and 6 kmph in 45 seconds and 36 seconds respectively. Find the speed of the train.

a. 10.8 km/hr b.12 km/hr c. 13.2 km/hr d. 14.4 km/hr

7. A person standing on railway platform which is 200 meters long . He finds that a train with a uniform speed crosses the platform in 12 seconds and himself in 4 seconds ,what is the length of the train?

a. 100 meters b. 110 meters c. 140 meters d. 180 meters

8. A car after travelling for 25 km in a city , increases its speed by ¼ of its original speed and reaches its destination 40 minutes early. Had it increased the speed 10 km before , it would have reached its destination 5 minutes earlier. What is the total distance travelled by the car?

a. 25 Km b. 80 Km c. 100 Km d. 105 Km

9. The distance between two station A and B is 600 Km. One train leaves station A towards station B at avg speed of 54 km/hr, after an hour another train left station b towards station A at avg speed of 66 km/hr The distance from station A where the two meets?

a. 272.7 Km b. 299.7 km c. 333.3 Km d. 387.3 km

10. Smith covers a total distance of 90 km in 45 minutes by travelling at a speed of ‘p’ km for 9 minutes ,’p/3’ km for the next 18 minutes and rest of time with speed of ‘p/4’. Find the value of P.

a. 77 1/3 m/s b. 76 1/3 m/s c. 76 12/13 m/s d. 77 12/13 m/s

11. A man can row 3 kmph in still water . When the river is running at 0.6 kmph ,it takes him half an hour to row to a place and back. How far is the place?

a. 0.7 km b. 0.3 km c.280 mts d. 720 mts

12. A man can row 15 km upstream and 22 km downstream in 5 hours. Also he can row 20 km upstream and 27.5 km down stream in 6 ½ hours. Find the rate of current and speed of the man in still water respectively.

a. 8 km/ 3km b. 3 km/8 km c. 5km/ 11 km d. 11 km/ 5km

13. A runs 1 2/3 times as fast as B if A gives start of 80 meters to B, how far must the wining post be for the race to end in atie?

a. 120 m b. 200 m c. 160 m d. 180 m

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14. In a 100 meters race , A runs at 6 kmph, if A gives B a start of 4 meters and still beats him by 12 seconds , what is the speed of B/

a. 4.8 km/h b. 1.33 m/s c. 5 km/hr d. 2 m/s

15. If A and B start at the same time from the same point and are running around a circular track of length 100 mtrs with speed 25 m/s and 20 m/s respectively. If they are running in the same direction, how much distance will the faster man have covered when they meet again for the first time?

a. 100 mts b. 200 mts c. 400 mts d. 500 mts

16. If a,b,c start at the same point at the same time with respective speeds of 20, 25 and 30 kmph. Find the time taken for them to meet at the starting point for the first time in circular tract of length 1 Km.

a. 5 mts b.12 mts c. 7 mts d. none

17. If A,B and C start at the same point , at the same time and in the same direction with speeds of 17,25 and 33 Kmph, then find time taken for all of them to meet for the second time, given circumference of the track is 100 Km.

a. 7 ½ hr b. 15 hours c. 6 ½ hours d. 13 hours

18. Two men A and B walk round a circular track of length 1.5 km , A walks at the rate 200 m per minute and B at the rate of 140 m per minute . If they both start at the same time , from same point and walk in the same direction , when will they be together for the first time, and when will they be together again for the first time in starting point?

a. 25 mts/75 mts b. 75 mts/25 mts c. 25 mts/25 mts d. 75 mts /75 mts

19. Two men , A and B are going round a circle of circumference 50 km. A’s speed is 54 km/hr and B’s speed is 36 Km/hr. If they both start in opposite direction , when will they be together again for the first time and when will they be together again at the starting point for the first time?

a. 33.33 min/33.33 min b. 33.33 min/133.33 min

C . 33.33 min/ 166.67 min d. 33.33 min / 333.33 min

20. Rajesh started walking along the boundaries of a square field from corner point A after half an hour, he reached corner point C diagonally opposite to A if his speed was 8 km/hr what is the area of the field in sq km?

a. 16 b. 4 c. 8 d. none

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Chapter 6 Number Theory

6.1 Types of Numbers:

The number theory or number systems happens to be the back bone for CAT preparation. Number systems not only form the basis of most calculations and other systems in mathematics, but also it forms a major percentage of the CAT quantitative section. The reason for that is the ability of examiner to formulate tough conceptual questions and puzzles from this section. In number systems there are hundreds of concepts and variations, along with various logics attached to them, which makes this seemingly easy looking topic most complex in preparation for the CAT examination. The students while going through these topics should be careful in capturing the concept correctly, as it’s not the speed but the concept that will solve the question here. The correct understanding of concept is the only way to solve complex questions based on this section.

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6.1.1 Real numbers: The numbers that can represent physical quantities in a complete manner. All real numbers can be measured and can be represented on a number line. They are of two types:

6.1.2 Rational numbers: A number that can be represented in the form p/q where p and q are integers and q is not zero. Example: 2/3, 1/10, 8/3 etc. They can be finite decimal numbers, whole numbers, integers, fractions.

6.1.3 Irrational numbers: A number that cannot be represented in the form p/q where p and q are integers and q is not zero. An infinite non recurring decimal is an irrational number. Example: √2, √5 , √7 and Π(pie)=3.1416.

The rational numbers are classified into Integers and fractions

6.1.4 Integers: The set of numbers on the number line, with the natural numbers, zero and the negative numbers are called integers, I = {…..-3, -2, -1, 0, 1, 2, 3…….}

6.1.5 Fractions: A fraction denotes part or parts of an integer. For example 1/6, which can represent 1/6th part of the whole, the type of fractions are:

1. Common fractions: The fractions where the denominator is not 10 or a multiple of it. Example: 2/3, 4/5 etc. 2. Decimal fractions: The fractions where the denominator is 10 or a multiple of 10. Example 7/10, 9/100 etc.

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3. Proper fractions: The fractions where the numerator is less than the denominator. Example ¾, 2/5 etc. its value is always less than 1. 4. Improper fractions: The fractions where the numerator is greater than or equal to the denominator. Example 4/3, 5/3 etc. Its value is always greater than or equal to 1. 5. Compound fraction: A fraction of a fraction is called a compound fraction Ex: 3/5 of 7/9 = 3/5 x 7/9 = 21/45 6. Complex fractions: The combination of fractions is called a complex fraction. Ex: (3/5)/ (2/9) 7. Mixed fractions: A fraction which consists of two parts, an integer and a fraction. Example 3 ½, 6 ¾

Ex: Express 27/8 as a mixed fraction Ans. Divide the numerator by denominator; note the multiplier, whatever remainder is left divide it with the original denominator. For 27/8, 24/8 = 3, and remainder left is 3, therefore 3 3/8 is the mixed fraction

Ex: Express 35 7/17as an improper fraction. Ans. Here we need to multiply the denominator with the non-fraction part and add it to numerator and using same denominator. For 35 7/17= = 602/17

The integers are classified into negative numbers and whole numbers

6.1.6 Negative numbers: All the negative numbers on the number line, {…..-3, -2, -1}

6.1.7 Whole numbers: The set of all positive numbers and 0 are called whole numbers, W = {0, 1, 2, 3, 4…….}.

6.1.8 Natural numbers: The counting numbers 1, 2, 3, 4, 5……. are known as natural numbers, N = {1, 2, 3, 4, 5…..}. The natural numbers along with zero make the set of the whole numbers. 6.1.9 Even numbers: The numbers divisible by 2 are even numbers. e.g., 2, 4, 6,8,10 etc. Even numbers can be expressed in the form 2n where n is an integer other than 0.

6.1.10 Odd numbers: The numbers not divisible by 2 are odd numbers. e.g. 1, 3, 5, 7, 9 etc. Odd numbers are expressible in the form (2n + 1) where n is an integer other than 0.

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6.1.11 Composite numbers: A composite number has other factors besides itself and unity .e.g. 8, 72, 39 etc. A real natural number that is not a prime number is a composite number.

6.1.12 Prime numbers: The numbers that has no other factors besides itself and unity is a prime number. Example: 2, 23,5,7,11,13 etc. Here are some properties of prime numbers:

• The only even prime number is 2 • 1 is neither a prime nor a composite number • If p is a prime number then for any whole number a, ap – a is divisible by p. • 2,3,5,7,11,13,17,19,23,29 are first ten prime numbers (should be remembered) • Two numbers are supposed to be co-prime of their HCF is 1, e.g. 3 & 5, 14 & 29 etc. • A number is divisible by ab only when that number is divisible by each one of a and b, where a and b are co prime. • To find a prime number, check the rough square root of the given number and divide the number by all the prime number lower than the estimated square root • All prime numbers can be expressed in the form 6n-1 or 6n+1, but all numbers that can be expressed in this form are not prime Ex: If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is: (a)1 (b) 2 (c) 3 (d) more than 3 (a) a, a + 2, a + 4 are prime numbers. The number fits is 1, 3, 5 and 3, 5, 7 but post this nothing will fit. Now 1, 3, 5 are not prime numbers as 1 is not prime number. So, only one possibility is there 3, 5, 7 for a = 3.

6.2 Prime Factors: The composite numbers express in factors, wherein all the factors are prime. To get prime factors we divide number by prime numbers till the remainder is a prime number. All composite numbers can be expressed as prime factors, for example prime factors of 150 are 2,3,5,5. A composite number can be uniquely expressed as a product of prime factors. Ex: 12 = 2 x 6 = 2 x 2 x 3 = 22 x 31 20 = 4 x 5 = 2 x 2 x 5 = 22 x 51 etc

Note : The number of divisors of a given number N ( including one and the number itself ) where N = am x bn x cp ……. Where a, b, c are prime numbers are = ( 1 + m ) ( 1 + n ) ( 1 + p ) ………….. Ex: 90 = 2 x 3 x 3 x 3 x 5 = 21 x 32 x 51

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Hence here a = 2 b = 3 c = 5 m = 1 n = 2 p = 1 then the number of divisors are = ( 1 + m ) ( 1 + n ) ( 1 + p ) = 2 x 3 x 2 = 12 the factors of 90 = 1 , 2 , 3 , 5 , 6 , 9 , 10 , 15 , 18 , 30 , 45 , 90 = 12

the sum of divisors of given number N is ( am+1 – 1 ) ( bn+1 - 1 ) ( cp+1 – 1 ) …….. ___________________________________ ( a – 1 ) ( b – 1 ) ( c – 1 ) ……….

Perfect number: If the sum of the divisor of N excluding N itself is equal to N , then N is called a perfect number. e.g. 6, 28, 496

6.3 Finding a perfect number through Euclid’s method

Euclid's method makes use of the powers of 2, which are numbers obtained by multiplying by 2 by itself over and over again, which are 1, 2, 4, 8, 16, 32, 64, 128….

Note that the sum of the two numbers in this series (in ascending order) is equal to the third number minus 1:

1+2 = 3 = 4 - 1, 1+2+4 = 7 = 8 - 1,

STARTING FROM THE NUMBER 1, IF YOU ADD THE POWERS OF 2 AND IF THE SUM IS A PRIME NUMBER, THEN YOU GET A PERFECT NUMBER BY MULTIPLYING THIS SUM TO THE LAST POWER OF 2.

If you add 1+2, the sum is 3, which is a prime number. Therefore 3 x 2 = 6 is a perfect number.

If you add 1+2+4, the sum is 7, a prime number. Therefore 7 x 4 = 28 is a perfect number.

If you add 1+2+4+8, the sum 15 is not a prime number, so you can't use Euclid's method here.

If you add 1+2+4+8+16, the sum is 31, a prime number. Therefore 31 x 16 = 496 is another perfect number.

6.4 Absolute value of a number: The absolute value of a number a is | a | and is always positive.

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6.5 Fibonacci numbers: The Fibonacci numbers is a sequence where X (n+2) = X (n+1) + X (n), X (1) = 1, X (2) = 1 Example1,1,2,3,5,8,13,21,34,55,89,144.., it can be clearly seen that any number in the series is the addition of the last two numbers, other than the first two numbers

Ex: The price of pens has increased over the years. Each year for the last 7 years the price has increased, and the new price is the sum of the prices for the two previous years. Last year a pen cost 60 rupees. How much does a pen cost today? How much did a pen cost 7 years ago?

Solution:

Let this year price be x. Last year it was 60, so the previous year it must have been x-60, continuing this process backwards gives us a sequence of expressions:

x, 60, x-60, 120-x, 2x-180, 300-3x, 5x-480, 780-8x, 13x-1260

All of these increases must be positive as every year prize has gone up. That gives us a sequence of inequalities, each of which can be solved to find a range for x:

x>0 60>0 x-60>0x>60 120- >0 x < 120 2x-180>0x>90 300-3x>0x<100 5x-480>0>96 780-8>0x<97.5 13x-1260 > 0 x > 96.92

Looking at this, we can say

96.92 < x < 97.5

The whole number value x can have is 97, with which we get

x = 97 60 = 60 x-60 = 37 120-x = 23 2x-180 = 14 300-3x = 9 5x-480 = 5 780-8x = 4 13x-1260 = 1

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Seven years ago, the price was 4 rupees

In the MCQ format, where you have the four answers, you can check it by working forward and seeing if the results are correct. You can try putting the given answers for original price and see which one fits in the equation.

6.6 Golden ratio: The golden ratio is a special number approximately equal to: 1.6180339887498948482... Golden ratio = (1 + √ 5)/2

To find the golden ratio, we define the golden ratio as the ratio between x and y if

x y --- = ----- y x+y

Let's say x is 1. Then we have 1/y = y/(y+1). If we solve this equation to find y, we'll find that it is the value given above, about 1.618 A golden rectangle is a rectangle in which the ratio of the length to the width is the golden ratio.

The concepts like Fibonacci and golden ratio are reference concepts, students are advised not to cram them but just understand the concepts as they are.

6.7 Number System

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36


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EXERCISE 6

1. If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of 'a'?

a.121 b.3267 c.363 d. 33

2. Find the largest five digit number that is exactly divisible by 7, 10, 15, 21 and 28.

a 99840 b.99900 c. 99960 d.99990 3. Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

a.540 b.1440 c.1520 d.1590 4. Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?

a (x - z)2 y is even b. (x - z) y2 is odd c. (x - z) y is odd d. (x - y)2 z is even

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6. When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

a.10 b. 7 c.19 d.192

6. The sum of the first 100 natural numbers, 1 to 100 is divisible by

a.2, 4 and 8 b. 2 and 4 c. 2 d.100

7.How many different factors does 48 have, excluding 1 and 48?

a. 12 b.4 c.8 d.10 8. 1025-7 is divisible by

1. 2 2. 3 3. 9 4. Both (2) and (3)

9. Find the G.C.D of 12x2y3z2, 18x3y2z4, and 24xy4z3

a. 6xy2z2 b.6x3y4z3 c.24xy2z2 d.24xy3z2

10. What is the value of M and N respectively? If M39048458N is divisible by 8 & 11; Where M & N are single digit integers?

a. 7, 8 b. 8, 6 c. 6, 4 d.5, 4

11.48 students have to be seated such that each row has the same number of students as the others. If at least 3 students are to be seated per row and at least 2 rows have to be there, how many arrangements are possible?

a. 4 b. 10 c.8 d. 7

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12. If two fractions, each of which has a value between 0 and 1, are multiplied together, the product will be:

a.always greater than either of the original fractions b.always less than either of the original fractions c. sometimes greater and sometimes less than either of the original fractions d.remains the same 13.Three Vice Presidents (VP) regularly visit the plant on different days. Due to labour unrest, VP (HR) regularly visits the plant after a gap of 2 days. VP (Operations) regularly visits the plant after a gap of 3 days. VP (sales) regularly visits the plant after a gap of 5 days. The VPs do not deviate from their individual schedules. CEO of the company meets the VPs when all the three VPs come to the plant together. CEO is on leave from January 5th to January 28th, 2012. Last time CEO met the VPs on January 3, 2012. When is the next time CEO will meet all the VPs?

a.February 6, 2012 b.February 7, 2012 c.February 8, 2012 d. February 9, 2012

14. How many divisors does 7200 have?

1. 20 2. 4 3. 54 4. 32

15. What is the highest power of 7 that can divide 5000! without leaving a remainder? (5000! Means factorial 5000)

1. 4998 2. 714 3. 832 4. 816

17.'a' and 'b' are the lengths of the base and height of a right angled triangle whose hypotenuse is 'h'. If the values of 'a' and 'b' are positive integers, which of the following cannot be a value of the square of the hypotenuse?

1. 13 2. 23 3. 37

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18. If x, y, z are chosen from the three numbers, -3, and 2, what is the largest possible value of

the expression z2?

a. b. 16 c. 24 d. 36

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Chapter 7 Simple and Compound Interest

The lending and borrowing of money has been happening since thousands of years. Any sum of money, borrowed for a certain period, will invite an extra cost to be paid on the money borrowed; this extra cost at a fixed rate is called interest. The money borrowed is called principal. The sum of interest and principal is called the amount. The time for which money is borrowed is called period.

Amount = Principal + Interest

The interest paid per hundred (or percent) for a year is called the rate percent per annum. The rate of interest is almost always taken as per annum, in calculations we will always consider it per annum unless indicated.

The interest is of two types, one is simple, the other is compound:

7.1 Simple interest

It is the interest paid as it falls due, at the end of decided period (yearly, half yearly or quarterly), the principal is said to be lent or borrowed at simple interest.

Simple Interest, SI = PRT / 100

Here P = principal, R = rate per annum, T = time in years Therefore Amount, A = P + PRT/100 = P [1 + ( RT / 100 )]

If T is given in months, since rate is per annum, the time has to be converted in years, so the period in months has to be divided by 12. if T = 2 months = 2/12 years)

Ex 1: Find the amount on S.I. when Rs 4000 is lent at 5 % p.a. for 5 years. By the formula, A = P (1 + RT/100) = 4000( 1 + 5 x 5/100 ) = Rs 5000

7.2 Compound Interest

The compound interest is essentially interest over interest. The interest due is added to the principal and that becomes the new principal for the interest to be levied. This method of interest calculation is called compound interest, this can be for any period (yearly, half yearly or quarterly) and will be called “Period compounded” like Yearly compounded or quarterly compounded and so on.

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First period’s principal + first period’s interest = second period’s principal Compound interest = principal {1 + Rate/100}time - Principal CI = P { 1 + R/100 }T – P Here Amount = principal {1 + Rate/100 }time

Ex2: Find the compound interest on Rs 4500 for 3 years at 6 % per annum Using the formula, A = P (1 + R/100)T = 4500(1 + 6/100)3 = 4500 (1.06)3 = 5360 Compound interest = 5360 – 4500 = Rs 860

7.3 THE RULE OF 72 The rule of 72 is a quick way to show how long it will take to double your money under

The equation for the rule of 72 is: Number of years for money to double = (72/Annual Interest Rate) interest rate At 8% interest, it will take 72/8 = 9 years for your money to double. Here are more examples: At 6%, it will take 12 years ( 72/6 = 12) At 12%, it will take 6 years ( 72/12 = 6)

The rule of 72 is a short cut to estimate the magic of compound interest that makes your money grow.

• Remember that the rule of 72 is an approximation and its accuracy reduces as the interest rate becomes high.

Important notes

1. In case interest is paid half yearly, then the interest is divided by 2, and used as (R/2) in the formula and the time is multiplied by 2, and used as 2T in the formula, given by A = P [ 1 + ( R / 200 ) ]2T

Ex3: Find the compound interest on Rs 5000 for 3 years at 6 % per annum compounded half yearly. Using the formula, A = P [ 1 + ( R / 200 ) ]2T = 5000(1 + 6/200)3x2 = 5000 (1.03)6 = 5971 Compound interest = 5971 – 5000 = Rs 971

2. In case interest is paid quarterly, then the interest is divided by 4, and used as (R/4) in the formula and the time is multiplied by 4, and used as 4T in the formula, given by A = P [ 1 + ( R / 400 ) ]4T payable quarterly (rate = R/4, time = 4T)

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Ex4: Find the compound interest on Rs 5000 for 3 years at 6 % per annum compounded quarterly. Using the formula, A = P [ 1 + ( R / 200 ) ]2T = 5000(1 + 6/400)3x4 = 5000 (1.015)12 = 5978 Compound interest = 5978 – 5000 = Rs 978

3. In case the rates are different(R1, R2, R3….) for different years, the amount is given by P{1 + R1/100}{1 + R2/100}{1 + R3/100}

Ex5: Find the compound interest on Rs 5000 for 3 years at 6 % per annum for first year, 7% for the second year and 8% for the third year Using the formula, P{1 + R1/100}{1 + R2/100}{1 + R3/100} = 5000(1 + 6/100) (1 + 8/100) (1 + 9/100) = 6125 Compound interest = 6125 – 5000 = Rs 1125

4. For population increase the formula to be used is P {1 + R/100 }T, and for decrease P { 1 - R/100 }T. It can also be used for depreciation factor.

Ex6: The death rate of a town with population of 100000 is 5 %, considering there are no new births, what is the population of town in next three years? Using the formula, P { 1 - R/100 }T = 100000(1-5/100)3 = 100000(0.857) = 85738

5. In case the period is a fraction like 3 and ½ years, or a and b/c years, then the amount should be calculated by this formula A = P { 1 + R/100 }3{1+(1/2 x R)/100} Or A = P { 1 + R/100 }a{1+(b/c x R)/100}

Ex7: The birth rate of a town with population of 100000 is 5 %, considering there are no deaths in the town, what is the population of town in next three years and fours months? Three years and four months mean 3 1/4 Using the formula, A = P { 1 + R/100 }a{1+(b/c x R)/100} = 100000(1+5/100)3(1+ ¼ x 5/100) = 100000(1.157) (1.012) = 117210 will be the population

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6. The SI and CI earned during the first period remains the same.

Ex8: The compound interest on a certain sum of money in 2 years is 210 and the simple interest on the same amount is 200, what are the principle and the rate of interest Since SI and CI for first year is the same, and SI for each year is the same, so SI for the first year = 200/2 = 100, CI for year I = 100, that means CI for the year II = 210 – 100 = 110. Here the excess of interest over year I = 10. Since the excess of interest in CI is interest over first years interest, assuming I is the interest, I/100 x 100 = 10, so I = 10, and the principal is obviously 1000.(calculate yourself) Ex9: A sum of money placed at Compound Interest doubles in every 5 years, then in how many years it will become 16 times? Now, it is given that the principle gets doubled in every 5 years. So, if we start from initial amount P, then in first 5 years it will become 2P. In the next 5 years 2P will become 4P, next 5 years 4P will become 8P and finally in next 5 years 8P will become 16P. So, it will take (5+5+5+5) = 20 years.

7.4 Net present value (NPV)

Money received or paid today is not the same as money received or paid after a period. This is because the money has an opportunity cost of interest in the same period. What it simply means is that you can earn interest on money if you have it now, and if you get the money later, you loose the opportunity to make interest on that. For example, if the going interest rate in the market is 10%, and someone has to pay me Rs. 1000, and he pays after an year, so he should pay, 1100 (100 has the interest), Here 1100 is called the future value and 1000 is called the present value.

Here the Future value (FV) = Present value (PV) {1 + Rate/100 }time, which is the basic formula for amount in the case for compound interest, this is the formula to be used for calculating present value. From here

PV = FV / {1 + Rate/100 }time

This is the same formula as of the compound interest; herein we are calculating principal from the amount, which’s it!

Practical applications of the NPV

1. Installment schemes

Today there are all kinds of loans and financing of various products right from two wheelers to houses. When a loan is taken, customer generally pays a monthly installment, his dues reduced by that amount and the interest is charged only on the balance amount which is known as

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reducing balance. Also there are many other concepts like floating rates etc, but they are out of purview of CAT. Here is the monthly installment formula for a fixed rate of interest (fixed means which does not change over time, floating means which changes with market conditions):

Monthly Installment, M = [A/(1-B)] x P

Here A = R/1200 (where R is the rate of interest) B = [1/(1+A)]T (Where T is time in months) And P is the principal amount that is the amount of loan taken

The installment can be calculated with this formula by using concept of NPV also. This formula is derived from there only. You can find this formula in Microsoft excel also under PMT in the formula section. But these annuity formula questions will not be asked in CAT.

Ex10: If Ram takes a home loan of 500000 for 3 years at the rate of 7.5%, what will be his monthly installment?

T = 12 x 3 = 36 months R = 7.5% P = 500000 Using the formula, M = [A/(1-B)] x P A = 7.5/1200 = 0.00625 B = [1/(1+A)]T = [1/(1+0.00625)]36 M = [0.00625 / {1 - [1/(1+0.00625)]36}] x 500000 M = 15553

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CHAPTER 8 QUADRATIC AND HIGHER ORDER EQUATIONS

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CHAPTER 9 SET THEORY

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Chapter 10 Plain Geometry

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quantitive aptitude

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