Numerical Reasoning

Problem on Numbers

Arithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d Sum of n terms of A.P S = n/2 *[2a+(n-1)d)] Geometrical Progression: Tn = arn – 1.

Sn = a(rn – 1) / (r-1)

Problems on Numbers

Basic Formulae

1. ( a+b)2 = a2 + b2 + 2ab

2. (a-b)2 = a2 +b2 -2ab

3. ( a+b)2 - (a – b)2 = 4ab

4. (a+b)2 + (a – b)2 = 2 (a2 +b2)

5. (a2 – b2) = (a+b) (a-b)

6. (a+b+c)2 =a2 +b2 +c2 + 2(ab +bc+ca)

7. (a3 +b3) = ( a+b) (a2 –ab +b2)

8. (a3 –b3) = (a-b) (a2 +ab + b2)

Problem - 1

A 2 digit number is 3 times the sum of its digits

if 45 is added to the number. Its digits are

interchanged. The sum of digits of the number

is?

Solution

The number is 3 times the sum of its digits

45 is added = 4 +5 = 9

So, common numbers in 3 and 9th table.

9, 18, 27, 36, 45….

27 + 45 = 72

2 + 7 = 9 or 4 + 5 = 9

Problem - 2

A number when divided by 119 leaves a

remainder of 19. If it is divided by 17. It will

leave a remainder of?

Solution

= 19/17 = 2 remainder

Problem - 3 A boy was asked to find the value of 3/8 of sum of

money instead of multiplying the sum by 3/8 he

divided it by 8/3 and then his answer by Rs.55.

Find the correct answer?

Solution

8/3 – 3/8 = 55/24

= 55/55/24

= 24

Problem - 4

A man spends 2/5rd of his earning. 1/4th of the

expenditure goes to food, 1/5th on rent, 2/5th on

travel and rest on donations. If his total earning

is Rs.5000, find his expenditure on donations?

Solution

5000*2/5 = 2000

Remaining amount has given as donation

2000* (1/5 + 2/5 + ¼)

Total amount = 200*17/20 = 1700

2000 – 1700 = 300

Problem - 5

From a group of boys and girls 15 girls leave.

There are then left, 2 boys for each girl. After

this 45 boys leave, there are then left 5 girls for

each boy, find the number of girls in the

beginning?

Solution 15 girls leave = 2 boys for each girl

45 boys leave = 5 girls fro 1 boy

Let the boys be x; Girls = x/2 +15

After the boys have left,

No.of boys = x – 45 and girls = 5(x-45)

x/2 = 5(x-45)

X = 2(5x-22)

X = 10x – 450

X =50

50/2 +15 =40

Problem - 6

An organization purchased 80 chairs fro

Rs.9700. For chairs of better quality they paid

Rs.140 each and for each of the lower grade

chair they paid Rs.50 less. How many better

quality chairs did the organization buy?

Solution Better quality chairs = x;

Lower quality = 80 –x

Price of better quality = Rs.140, Lower quality =

140-50 = 90

140*x + 90(80-x) = 9700

140x + 7200 – 90x = 9700

50x = 9700 – 7200;

50x = 2500

X = 50

Problem - 7

A labour is engaged for 30 days, on the condition

that Rs.50 will be paid for everyday he works

and Rs.15 will be deducted from his wages for

everyday he is absent from work. At the end of

30 days he received Rs.850 in all. For how many

days did he wanted?

Solution

Total wages = 30*50 = 1500 (without Absent)

Wages received in 30 days = 850 (with Absent)

Let the labourer work for x days

Absent = 30 – x

50x – (30-x)15 = 850

50x -450 +15x = 850

65x = 1300

X = 1300/65 = 20 days

Problem - 8

The rent is charged at Rs.50 per day for first 3

days Rs.100 per day next 5 days, and 300 per

day thereafter. Registration fee is 50 at the

beginning. If a person had paid Rs.1300 for his

stay how many days did he stay?

Solution

3 days = 150 + 50 = 200

5 days = 100*5 = 500

= 200 + 500 = 700

1300 – 700 = 600

2 days = 300*2 = 600

= 5 + 3 + 2 = 10 days

Problem - 9

In a school 20% of students are under the age of

8 years. The number of girls above the age of 8

years is 2/3 of the number of boys above the age

of 8 years and amount to 48. What is the total

number of students in the school?

Solution Girls above 8 yrs = 48

Boys above 8 yrs = 48 / 2/3

80% of students above 8 yrs = 48 + 72 = 120

80 120

20 x

80x = 120*20

X = 120*20/80 = 30

Total No.of students = 120+30 = 150

Ratio and Proportion

Ratio and Proportion

Ratio : Relationship between two variables.

= a : b

Proportion : Relationship between two ratios.

= a : b : : c : d

Proportion Calculation = a*d : b*c

Problem - 1

The ratio of number of boys to that of girls in a

school is 3:2. If 20% boys and 25% of girls are

scholarship holders, find the percentage of the

school students who are not scholarship holders?

Solution

Let the total number of students be 100

Boys = 100*3/5 = 60

Girls = 100*2/5 = 40

S. holders = 60*20/100 = 12, non S. holders = 60 -12 = 48

Girls s. holders = 40*25/100 = 10,

Non s. holders = 40 – 10 = 30

Students who do not have scholarship = 48 + 30 = 78

78/100*100 = 78%

Problem - 2

The cost of diamond varies as the square of its

weight. A diamond weighing 10 decigrams costs

Rs. 32000. Find the loss incurred when it breaks

into two pieces whose weights are in the ratio

2:3?

Solution

1st piece = 10*2/5 = 4

2nd piece = 10*3/5 = 6

Cost of the diamond varies as square of its weight

42 : 62 102 = 100k

16k : 36k

100k – 52 k = 48k(loss)

100k = 32000; k = 320

48*320 = 15360

Problem - 3

The ratio of the first and second class fares

between two railway stations 4 : 1 and the ratio

of the number of passengers traveling by first

and second class is 1:40. If the total of Rs.1100

is collected as fare from passengers of both

classes what was the amount collected from first

class passengers?

Solution

Fare = 4 : 1

Passengers traveling = 1 : 40

Amount = No. pas * fare = 4*1 :10*1 = 4 : 40

= 1:10

Total amount = 1100.

First class passengers’ amount = 1*1100/11

= 100

Problem - 4

A vessel contains a mixture of water and milk in

the ratio 1:2 and another vessel contains the

mixture in the ratio 3:4. Taking 1 kg each from

both mixtures a new mixture is prepared. What

will be the ratio of water and milk in the new

mixture?

Solution

1st vessel = water = 1/3 , milk = 2/3

2nd vessel = water = 3/7, milk = 4/7

Water = 1/3 + 3/7 = 16/21

Milk = 2/3 + 4/7 = 26/21

16 : 26 = 8:13

Problem - 5

Ratio of the income of A, B, C last year 3 : 4 : 5.

The ratio of their individual incomes of last year

and this year are 4:5, 2:3 and 3:4 respectively. If

the sum of their present income is Rs.78,800.

Find the present individual income of A, B and

C.

Solution A’s Present Income = 5/4*3x = 15x/4

B’s Present Income = 3/2*4x = 12x/2

C’s Present Income = 4/7*5x = 20x/7

15x/4 + 6x+20x/3 = 78,800

197x/12 = 78,800

X = 945600/197

X = 4,800

A’s Present income = 15x/4 = 15*4800/4 = 18,000

B’s Present income = 6*x = 6*4,800 = 28,800

C’s Present income = 20x/3 = 20*4800/3 = 32,000

Problem - 6

Of the three numbers, the ratio of the first and

the second is 8:9 and that of the second and third

is 3:4. If the product of the first and third

numbers is 2,400, then find the second number?

Solution

a : b = 8 : 9

b : c = 3 : 4

b : c = 3*3 : 4*3 = 9 : 12

a : b : c = 8 : 9 : 12

Product of first and third = 8k * 12k = 2400

96k2 = 2400; k2 = 2400/96 = 25

k = 5

Second number = 9 * 5 = 45

Problem - 7

Annual income of A and B are in the ratio of 4 : 3

and their annual expenses are in the ratio 3 : 2. If

each of them saves Rs.600 at the end of the year,

what is the annual income of A?

Solution

Income = 4 : 3, Expenses = 3 : 2

Savings 600 each

A’s income = 4x, expenses = 3x,

savings = x i.e 600

Income = 4*600 : 3*600

A : B = 2400 : 1800

A income = 2400

Problem - 8

The property of a man was divided among his

wife, son and daughter according to his will as

follows. Wife’s hare is equal to 6/7th of son’s

share and daughter share is equal of 4/7th of

Son’s. If the son and daughter together receives

Rs.1,02,300. How much does his wife get?

Solution

Let the Son’s share be x.

Daughter’s share = x*4/7 = 4x/7

Wife’s share = x* 6/7 = 6x/7

X + 4x/7 = 1,02,300

7x + 4x = 1,02,300

X = 1,02,300 /11 = 65,100

Wife Share = 65,100 *6/7 = Rs. 55, 800

Problem - 9

A pot containing 81 litres of pure milk of the

milk 1/3 is replaced by the same amount of

water. Again 1/3 of the mixture is replaced by

the same amount of water. Find the ratio of milk

to water in the new mixture?

Solution

Milk : Water

Initial = 81 : 0

1/3 removed = 54 : 27

1/3 mixture = 36 : 45

Ratio of Milk and Water = 4 : 5

Problem - 10

729 ml of mixture contains milk and water are in

the ratio 7 : 2. How much more water is to be

added to get a new mixture containing milk and

water in the ratio of 7 : 3.

Solution

Water = 729 * 2/9 = 162

Ratio Water

2 162

3 x

2x = 3*162/2 = 243

243 – 161 = 81 ml water is to be added

Problem - 11

Price of a scooter and a television set are in the

ratio 3 : 2. If the scooter costs Rs.600 more than

the television set, then find the price of

television?

Solution

Diff. in ratio = 3 – 2 = 1

1 ratio is 600 means, the television cost is 2 ratio

so, cost of television = 1200

Problem - 12

The annual income and expenditure of man and

his wife are in the ratio of 5:3 and 3:1

respectively, if they decide to save equally and

find their balance is 4000. Find their income at

the end of the year?

Solution

Man and Wife income = 5 : 3 = 2 (diff)

Man and Wife Expenses = 3 : 1 = 2 (diff)

so, both of them are saving ratio of 2

Total saving of Man and Women = 4000,

individual saving 2000

So, Man income = 5000 and Women income =

3000

Problem - 13

In a class room, ¾ of the boys are above 160 cm

in height and they are in 18 number. Also out of

the total strength, the boys are only 2/3 and the

rest are girls. Find the total number of girls in a

class?

Solution

¾ of the boys in 18 numbers means, ¼ of the boys = 6

Total number of boys = 18+6 = 24

Ratio Number

2/3 24

1/3 x

2/3*x = 24*1/3

x = 24/2 = 12 Girls

Problem - 14

Rs. 770 was divided among A, B and C such that

A receives 2/ 9th of what B and C together

receive. Find A’s share?

Solution

A = 2/9 (B+C)

B+C =9A/2

A+B+C = 770

A + 9A/2 = 770

11A = 770*2

A = 140

Problem - 15 A sporting goods store ordered an equal

number of white and yellow balls. The tennis

ball company delivered 45 extra white balls

making the ratio of white balls to yellow balls

1/5 : 1/6. How many white tennis balls did the

store originally order for?

Solution

Let the number of yellow balls be x

(x + 45) : x = 1/5 : 1/6

Solving the above equation,

The number of white balls originally ordered

would be = 225 balls

Alligation and Mixture

Alligation and Mixture

Alligation : It is the rule that enables us to find the ratio

in which two or more ingredients at the given price

must be mixed to produce a mixture of a desired

price.

(Quantity of cheaper / Quantity of costlier)

(C.P. of costlier) – (Mean price)

= --------------------------------------

(Mean price) – (C.P. of cheaper)

Alligation or Mixture

Cost of Cheaper Cost of costlier

c d

Cost of Mixture

m

d-m m-c

(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)

Problem -1

Three glasses of size 3 lit, 4 lit and 5 lit contain

mixture of milk and water in the ratio of 2:3, 3:7

and 4:11 respectively. The content of all the

three glasses are poured into a single vessel.

Find the ratio of milk and water in the resulting

mixture.

Solution 1st Vessel

= Milk = 3*2/5 = 6/5

= Water = 3*3/5 = 9/5

2nd Vessel:

= Milk = 4*3/10 = 12/10

= Water = 4*7/10 = 28/10

3rd Vessel:

= Milk = 5*4/15 = 20/15

= Water = 5*11/15 = 55/15

Milk : Water = 6/5 +12/10 + 20/15 : 9/5 + 28/10 + 55/15

= 18/15 + 18/15 +20/15 : 27/15 + 42/15+55/15

= 56 : 124 (or) 14:31

Problem - 2

How many kg of tea worth Rs. 25 per kg must be

blended with 30 kg tea worth Rs. 30 per kg, so

that by selling the blended variety at Rs.30 per

kg there should be a gain of 10%?

Solution

30*100/110 = 300/11

25 30

300/11

30/11 25/11

30 : 25

6 : 5

36 : 30kg

Problem - 3

A man buys cows for Rs. 1350 and sells one so

as to lose 6% and the other so as to gain 7.5%

and on the whole he neither gains nor loses. How

much does each cow cost?

Solution

6 7.5

0

7.5 6

15 12

5 : 4

1350*5/9 = 750

1350 *4/9 = 600

Problem - 4

There are 65 students in a class, 39 rupees are

distributed among them so that each boy gets

80p and each girl gets 30p. Find the number of

boys and girls in a class.

Solution Girls Boys

30 80

60

20 30

2 : 3

65*2/5 = 26

65*3/5 = 39

Problem - 5

A person covers a distance 100 kms in 10 hr

Partly by walking at 7 km per hour and rest by

running at 12 km per hour. Find the distance

covered in each part.

Solution

Speed = Distance / Time = 100 / 10 = 10

7 12

10

2 : 3

Time taken in 7 km/hr = 10 * 2/5 = 4

4*7 = 28 km

Time taken in 12 km/hours = 10*3/5 = 6

12*6 = 72 km

A merchant has 100 kg of salt, part of which

he sells at 7% profit and the rest at 17% profit.

He gains 10% on the whole. Find the quantity

sold at 17% profit?

Problem - 6

7 17

10

(17-10) (10-7)

7 : 3

The quantity of 2nd kind = 3/10 of 100kg

= 30kg

Solution

In what ratio two varieties of tea one costing

Rs. 27 per kg and the other costing Rs. 32 per

kg should be blended to produce a blended

variety of tea worth Rs. 30 per kg. How much

should be the quantity of second variety of tea,

if the first variety is 60 kg?

Problem - 7

27 32

30

2 3

Quantity of cheaper tea = 2

Quantity of superior tea 3

Quantity of cheaper tea =2*x/5 = 60 , x=150

Quantity of superior tea = 3 * 150/5 = 90 kg

Solution

A 3-gallon mixture contains one part of S and

two parts of R. In order to change it to mixture

containing 25% S how much R should be

added?

Problem - 8

R : S

2 : 1

75% : 25%

3 : 1

1 gallon of R should be added.

Solution

Three types of tea A,B,C costs Rs. 95/kg, Rs.

100/kg. and Rs 70/kg respectively. How many

kg of each should be blended to produce 100 kg

of mixture worth Rs.90/kg given that the

quantities of B and C are equal?

Problem - 9

B+C/2 A

85 95

90

5 5

Ratio is 1:1 so A = 50 , B + C = 50

The quantity would be 50 : 25 : 25

Solution

In what proportion water must be added to

spirit to gain 20% by selling it at the cost price?

Problem - 10

Profit%=20%

Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3

0 10

25/3

5/3 25/3

The ratio is 1: 5

Solution

In an examination out of 480 students 85% of

the girls and 70% of the boys passed. How

many boys appeared in the examination if total

pass percentage was 75%

Problem - 11

Solution

Solution:

70 85

75

10 5

Number of Boys = 480 * 10/15

Number of Boys = 320

Problem - 12

A painter mixes blue paint with white paint so

that the mixture contains 10% blue paint. In a

mixture of 40 litre paint how many litre of blue

paint should be added, so that the mixture

contains 20% of blue paint?

Solution

Quantity of blue paint in the mixture = 10% of

40

40*10/100 = 4

40 – 4 = 36 litre

Let x litre blur paint can be mixed

4+x/30 = 20/80 = 4+x = 9

x = 5

Problem - 13

From a 100 litre mixture containing water and

milk equal proportion, 10 litres of mixture is

replaced by 10 litres of water in succession

twice. At the end, what is the ratio of milk and

water?

Solution

Milk Water

10 lit(1st) 50 : 50

45 : 45

45 : 55

2nd 10 lit 40.5 : 49.5

Add Water 40.5 : 59.5

81 : 119

Problem - 14

In a mixture of 400 gms, 80% is copper, sliver is

20%. How much copper is to be added, so that

the new mixture has 84% copper?

Solution 400*80/100 = 320 Copper

400*20/100 = 80 Sliver

Percen Mixture

80 320

84 x

= 320*84/80 = 336

(320+x) = (400+x) 84/100

320+x = 400+84/100 + 84x/100

16x/100 = 336 – 320; 16x/100 = 16; x = 100

Problem - 15

A jar full of whisky contains 50% alcohol. A part

of this whisky is replaced by another containing

30% alcohol and now the percentage of alcohol

was found to be 35%. Find the quantity of

whisky replaced?

Solution

50 30

35

5 : 15

5 : 15 = 1 : 3

Replaced = 3/4

Partnership

Type - 1

A invest = 10000

B invest = 15000

Profit = 5000

Find their Individual Share ?

A : B = 10000 : 15000 = 2 : 3

A’s Share = 5000*2/5 = 2000

B’s Share = 5000*3/5 = 3000

This is a first and basic step for any Partnership Problem.

Type - 2

A invest = 5000,

After 3 months B joined A, with an investment

of 3000

Profit at the end of the year = 3500

Find their Share ?

Any thing happen after a month, like a person

joining a business, or withdraw from

business or withdraw some amount means

given amount is for month.

Cont…

Type - 2

A : B = 5000 : 3000 = 5*12 : 3*9 = 60 : 27 = 20 : 9

A’s share = 3500*20/29 = 2413.7

B’s Share = 3500*9/29 = 1086.3

Type - 3

A invest 5000

B invest 6000

After 3 months A withdraw amount 1000, after 5 months a withdraw amount 1000 again.

Profit at the end of the Year = 5000

Find their Share ?

A = 5*3 + 4*5 + 3*4 = 15 +20 + 12 = 47

B = 6*12 = 72

Type - 3

A’s share = 5000* 47/119 = 1974.8

B’s share = 5000*72/119 = 3025.2

Type - 4

A invest twice as much as B, B invest 1/3rd of C. At the end of the year their Profit is 6000. Find their Share?

A = 2B

B = 1/3C

C = x

A : B : C = 2x/3 : x/3 : x

A : B : C = 2x/3 : x/3 : 3x /3

A : B : C = 3 : 2 : 6

A’s Share = 6000*3/11 = 1636

B’s Share = 6000*2/11 = 1091

C’s Share = 6000*6/11 = 3273

Problem - 1

A, B and C started a business in partnership by

investing Rs.12000 each. After 6 months, C left

and after 4 months D joined with his capital of

Rs.24,000. At the end of a year, a profit of

Rs.8,500 shared among all the partners. Find B’s

share?

These are all the basic types remaining

we will see when we solve problems.

Solution

A : B : C : D

12000 : 12000 : 12000 : 24000

1 : 1 : 1 : 2

1*12 : 1*12 : 1*6 : 2*2

12 : 12: 6 : 4

6 : 6 : 3 : 2

B’s share = 6/17*8500 = 3000

Problem - 2

A, B and C enter into partnership. A contributes

one third of the capital while B contributes as

much as A and C together contributed. If the

profit at the end of the year amounted to Rs.840.

What would be B’s share?

Solution A’s share = 1/3 of the capital

A’s share = 1/3*840 = 280

B’s share = A + C = 280 + x

A + B + C = 840

280 + 280 + x + x = 840

560 + 2x = 840

2x = 840 – 560

X = 140

B’s share = 280+140 = 420

Problem - 3

Akilesh and Jaga enter into a partnership.

Akilesh contributing Rs.8000 and Jaga

contributing Rs.10000. At the end of 6 months

they introduce Prakash, who contributes

Rs.6000. After the lapse of 3 years, they find that

he firm has made a profit of Rs.9660. Find

Prakash’s share?

Solution

Akilesh : Jaga : Prakash

8 : 10 : 6

4 : 5 : 3

4*36 : 5836 : 3*30

144 : 180 : 90

8 : 10 : 5

Prakash’s share = 9660*5/23 = 2100

Problem - 4

Priya and Vijay enter into partnership. Priya

supplies whole of the capital amounting to

Rs.45,000 with the conditions that the profit are

to be equally divided and that Vijay pays Priya

interest on half of the capital of 10% p.a. but

receives Rs.120 per month for carrying on the

concern. Find their total yearly profit. When

Vijay’s income is one half of Priya’s income?

Solution 45,000 *1/2 = 22,500

22,500 *10//100 = 2250 (interest p.a)

Vijay receives Rs.120 per month = 120*12 = 1440

Total profit be x

Ratio of Profit sharing = 1 : 1

Priya’s income = x/2+2250

Vijay’s income = x/2 – 2250 +1440

1 Priya = ½ Vijay

Priya Income = Twice of Vijay income

x/2 + 2250 = 2(x/2 – 2250 +1440)

X+4500/2 = 2(x/2 – 810)

X+4500/2 = x – 1620 = x +4500 = 2x – 3240

X = 7740

Total Profit of the year = 7740+1440 = 9,180

Problem - 5

Revathy and Shiva are partners sharing profits in

the ratio of 2:1. They admit Pooja into

partnership giving her 1/5th share in profits

which she acquires from Revathy and Shiva in

the ratio of 1:2. Calculate the new profit sharing

ratio?

Solution

Pooja gets her share of 1/5th of total share of Profit from Revathy and Shiva in the ratio 1 : 2

From Revathy = 1/3*1/5 = 1/15

From Shiva = 2/3*1/5 = 2/15

Total Pooja share = 1/15+2/15 = 3/15 = 1/15

Revathy share = 2/3 – 1/15 = 9/15

Shiva share = 1/3 – 2/15 = 3/15

Shares = Revathy : Shiva : Pooja = 3 : 1 : 1

Problem - 6

A and B started a partnership business investing

some amount in the ratio of 3 : 5. C joined them

after six months with an amount equal to that of

B. In what proportion should the profit at the end

of 1 year be distributed among A, B and C?

Solution

Let the investment,

3 : 5 : 5

3*12: 5*12 : 5*6

36 : 60 : 30

6 : 10 : 5

Problem - 7

If 4(A’s capital) = 6(B’s capital) = 10 (C’s

capital) then out of a profit of rs.4650. Find C’s

share?

Solution

Let the unknown value be x

x/4 : x/6 : x/10

15x/60 : 10x/60 :6x/60

15 : 10 : 6

C’s share = 6/31*4650 = Rs. 900

Problem - 8

A, B, C subscribe Rs.50,000 fro business. A

subscribes Rs.4000 more than B and B Rs.5000

more than C. Out of total profit of Rs.35,000.

Find A’s share?

Solution C = x,

B = x + 5000

A = x+5000+4000 = x + 9000

x +x+5000 +x+9000 = 50000

3x+14000 = 50000

3x = 50000 – 14000

3x = 36000,

x = 12000

C : B : A

12000 : 17000 : 21000

A = 35000*21/50 = 14,700

Problem - 9

A and B are partners in a business, A contributes

¼ of he capital for 15 months and B received 2/3

of the profit. For how long B’s money was used?

Solution

B = 2/3

A = 1/3

A : B = 1/3 : 2/3 = 1 : 2

Investment

1/4x+15 : 3/4x*y

15x/4 : 3xy/4

15x/4 : 3xy/4 : : 1 : 2

30x/4 = 3xy/4

Y = 30x/4 * 4/3x = 10 months

Problem - 10

A, B and C invests Rs.4,000, Rs.5,000 and

Rs.6,000 respectively in a business and A gets

25% of profit for managing the business and the

rest of the profit is divided by A, B and C in

proportion to their investment. If in a year, A

gets Rs.200 less than B and C together, what was

the total profit for the year?

Solution Total Profit = 100

25% for managing the business = 100 – 25 = 75%

A : B : C

4000 : 5000 : 6000

4 : 5 : 6

4x : 5x : 6x = 25x

100*15x/75 = 20x

A gets 4x + 25% of 20x

= 4x + 20x *25/100 = 9x

B = 5x, C = 6x

(5x + 6x) – 9x = 200

11x – 9x = 200

2x = 200; x = 100

Total Profit 20x = 20*100 = 2000

Problem - 11

A and B entered into partnership with capitals in

the ratio of 4 : 5. After 3 months, A withdraw ¼

of his capital and B withdraw 1/5 of his capital.

The gain at the end of 10 months was Rs.760.

Find the share of B?

Solution A : B

4 : 5

4000 : 5000

A’s share = 4000*1/4 = 4000 – 1000 = 3000

B’s share = 5000*1/5 = 5000 – 1000 = 4000

A : B

3*4+3*7 : 5*3 +4*7

12 + 21 : 15+28

33 : 43

60*43/76 = 430

Problem - 12

Rs. 1290 is divided between A, B and C. So, that

A’s share is 1 ½ times B’s and B’s share is 1 ¾

times C. What is C’s share?

Solution

A : B = 1 ½ : 1 = 3/2 : 1 = 3 : 2

B : C = 1 ¾ : 1 = 7/4 : 1 = 7 : 4

A :B :C =3*7(A) : 2*7(B) : 7*2(B) : 4*2(C)

= 21 : 14 : 8

B = 1290*8/43 = Rs.240

Problem - 13

A man starts a business with a capital of

Rs.90000 and employs an assistant. From the

yearly profit he keeps an amount equal to 4 ½ of

his capital and pay 35% of the remainder of the

profits. Find how much the assistant receives in a

year, in which profit is Rs.30,000.

Solution

Investment = 90,000

4 ½ of investment = 9/2/100*90000 = Rs.4050

Profit = 30,000 – 4050 = 25,950

35/100*25,950 =9082.50

Problem - 14

A and B invest in a business in the ratio 3 : 2. If

5% of the total profit goes to charity and A’s

share is Rs. 855, what is the total profit %?

Solution

Let the total profit be Rs. 100

After paying charity A’s share = 3/5 *95 = 57

If A’s share is Rs. 57, the total profit is 100

If A’s share is Rs. 855, the total profit is

100 * 855/57 = Rs. 1500

The total profit = Rs. 1500

Problem - 15

A,B,C entered into a partnership by making an

investment in the ratio of 3 : 5 : 7. After a year

C invested another Rs. 337600 while A withdrew

Rs. 45600. The ratio of investments then

changed into 24: 59 : 167. How much did A

invest initially?

Solution

Solution:

Let the investments of A, B, and C be 3x, 5x, 7x

(3x – 45600) : 5x : (7x + 337600) = 24 : 59 : 167

(3x – 45600)/5x = 24/59

x = 47200

Initial investment of A = 47200 * 3 = Rs. 141600

Problems on Age

Problem - 1

The age of the Father is 4 times the age of his

Son. If 5 years ago, Father’s age was 7 times the

age of his Son, what is the Father’s present age?

Solution

F = 4S

F - 5 = 7(S - 5)

4S – 5 = 7S – 35

3S = 30

S = 10

Father’s age = 4* 10 = 40 years

Problem - 2

The age of Mr. Gupta is four times the age of his

Son. After Ten years, the age of Mr. Gupta will

be only twice the age of his Son. Find the present

age of Mr. Gupta’s Son.

Solution

G = 4S

G + 10 = 2 ( S + 10)

4S + 10 = 2S + 20

2S = 10

S = 5

Son’s Age = 5 years

Problem - 3

10 years ago Anu’s mother was 4 times older

than her daughter. After 10 years, the mother

will be twice as old as her daughter. Find the

present age of Anu.

Solution Ten years before:

M – 10 = 4(A – 10 )

M – 10 = 4A – 40

M = 4A – 40 + 10

M = 4A – 30

Ten Years After:

M + 10 = 2(A + 10)

M + 10 = 2A + 20

M = 2A + 20 – 10

M = 2A + 10

4A – 30 = 2A + 10

2A = 10 + 30

2A = 40: Anu’s Age = 20

Problem – 4

The sum of the ages of A and B is 42 years. 3

years back, the ages of A was 5 times the age of

B. Find the difference between the present ages

of A and B?

Solution A + B = 42

A = 42 – B

A – 3 = 5 ( B – 3)

A – 3 = 5B – 15

42 – B – 3 = 5B – 15

42 – 3 + 15 = 5B + B

54 = 6B

B = 54 /6 = 9

A = 42 – B; A = 42 – 9 = 33

Difference in their ages = 33 – 9 = 24 Years

Problem - 5

The sum of the ages of a son and father is 56

years. After 4 years, the age of the father will be

3 times that of the son. Find their respective

ages?

Solution

F + S= 56

S = 56 – F

F + 4 = 3 (S + 4)

F + 4 = 3 (56 – F + 4)

F + 4 = 168 – 3F + 12

4F = 168 + 12 – 4

4F = 176 ; F = 44

S = 56 – F ; S = 56 – 44 = 12

Father Age = 44; Son Age = 12

Problem – 6

The ratio of the ages of father and son at present

is 6:1. After 5 years, the ratio will become 7:2.

Find the Present age of the son.

Solution

6x + 5/x + 5 = 7/2

12x + 10 = 7x + 35

12x – 7x = 35 – 10

5x = 25

x = 25 / 5

x = 5 years

Son age = 1* 5 = 5 years

Problem - 7

The ages of Ram and Shyam differ by 16 years.

Six years ago, Shyam’s age was thrice as that of

Ram’s. Find their present ages?

Solution

S = R + 16

S – 6 = 3(R – 6)

S – 6 = 3R – 18

R + 16 – 6 = 3R – 18

R + 10 = 3R – 18

2R = 28 ; R = 14

Shyam’s Age = 14 + 16 = 30.

Problem - 8

A man’s age is 125% of what it was 10 years

ago, 83 1/3% of what it will be after 10 years.

What is his present age?

Solution

Let the age be x

125% of (x – 10) = 83 1/3 % of (x +10)

125/100 * x – 10 = 250/ 300 * x +10

5/4 x – 10 = 5/6 x – 10

5x / 4 – 5x / 6 = 50/6 + 50/4

5x /12 = 250/12

5x = 250 ; x = 50 years

Problem - 9

3 years ago, the average age of a family of 5

members was 17. A baby having born, the

average age of the family is the same today.

What is the age of the child?

Solution

Average age of 5 members = 17

Total age of 5 members = 17*5 = 85

3 years later, the age of 5 members will be

= 85 + 15 = 100

100 + x / 6 = 17

100 + x = 17*6

100 + x = 102

x = 102 – 100 = 2 years

Problem - 10

The sum of the age of father and his son is 100

years now. 5 years ago their ages were in the

ratio of 2 : 1. The ratio of the ages of father and

his son after 10 years will be?

Solution

F + S = 100

5 years ago 2 : 1

5 years ago

F + S = 100 – 10 = 90

90*2/3 = 60 : 30

Present age = 65 : 35

10 years ago = 75 : 45

= 5 : 3

Problem - 11

Six years ago, Sushil’s age was triple the age of

Snehal. Six years later, Sushil’s age will be 5/3

of the age of Snehal. What is the present age of

Snehal?

Solution

Six years ago,

Snehal = x; Sushil = 3x

Six years later,

3x + 6+6 = 5/3(x+6+6)

9x +36 = 5x+60

4x = 60 – 36

X = 6

Present Age of Snehal = 6+6 = 12 years

Problem - 12

Susan got married 6 years ago. Today her age is

1¼ times that at the time of her marriage. Her

son is 1/6 as old as she today. What is the age of

her son?

Solution

6 years ago Susan got married.

So her son’s age will be less than 6 years.

Let as consider, her son’s age is 5 years.

Susan’s Age is 5*6 = 30 yrs, since the son is 1/6th of

Susan’s age.

6 years ago her age must have been 24 yrs

24*1 ¼ = 24*5/4 = 30 yrs

As it satisfies the conditions her son’s age is 5 years

Problem - 13

My brother is 3 years elder to me. My father was

28 years of age when my sister was born, while

my mother was 26 years of age, when I was

born. If my sister was 4 years of age when my

brother was born, then, what was the age of my

father and mother respectively when my brother

was born?

Solution

My brother was born 3 years before I was born

and 4 years after my sister was born

Father’s age when brother was born

= 28 + 4 = 32 years

Mother’s age when brother was born

= 26 – 3 = 23 years

Problem - 14

If 6 years are subtracted from the present age of

Gagan and the reminder is divided by 18, then

the present age of his grandson Aunp is obtained.

If Anup is 2 years younger to Madan whose age

is 5 years, then what is Gagan’s present age?

Solution

Anup’s age = 5 – 2 = 3 years

Let Gagan’s age be x

= x – 6 / 18 = 3

x – 6 = 3*18 ; x – 6 = 54

x = 54 + 6

Gagan’s age = 60

Problem - 15

Ramu’s grandfather says, “ Ram, I am now 30

years older than your father. 15 years ago, I was

2½ times as old as your father”. How old is the

grandfather now?

Solution

Let the father’s age be x.

Grandfather’s age will be 30 + x

15 years ago,

X + 30 – 15 = 5/2 (x – 15)

X + 15 = 5/2 (x – 15)

2x + 30 = 5x – 75

105 = 3x

X = 105 / 3 = 35

Grandfather’s age = 35 + 30 = 65

Average

Average

Average = Sum of Quantities

Number of Quantities

Sum of quantities

= Average*Number of Quantities.

Number of quantities

= Sum of Quantities

Average

Problem - 1

The average age of a class of 22 students is 21

years. The average increases by 1 when the

teacher’s age is also included. What is the age of

the teacher?

Solution

Total age of the students be x

x/22 = 21; x = 21*22= 462

Teacher’s age is also included

x/23 = 22; x = 22*23 = 506

Total age of 23 people – Total age of 22 people

will be the age of teacher

506 – 462 = 44 years

The age of teacher = 44

Problem - 2

The average of 7 numbers is 25. The average of

first three of them is 20 while the last three is 28.

What must be the remaining number?

Solution

Average of 7 numbers = 25,

Sum of 7 numbers = 25* 7 = 175

Avg. of first three numbers = 20, 20* 3 = 60

Avg. of last three numbers = 28, 28*3 = 84

The 4th number = 175 – (60+84) = 175 – 144

= 31

Problem - 3

The average age of a team of 10 people remains

the same as it was 3 years ago, when a young

person replaces one of the member. How much

younger was he than the person whose place he

took?

Solution

Let Average be x

10 members’ Average = 10x

Average of 10 members (including new one) is

same as it was 3 yrs ago.

Now 10*3 = 30 years have increased, so a person

of 30 years should have replaced to keep the

average as same.

Problem - 4

The average age of a couple was 26 years at that

time of their marriage. After 11 years of marriage

the average age of the family with 3 children

become 19 years. What is the average age of the

Children?

Solution

Average of parents ages is 26, sum= 26*2 = 52

Parents age after 11 years = 52 +22 = 74

Average age of Family = 19, Sum = 19*5 = 95

Sum of family’s age – Sum of parents’ age

= 95 – 74 = 21

Sum of the ages of 3 children = 21,

Average Age = 21/3 = 7 yrs

Problem - 5

9 members went to a hotel for taking meals.

Eight of them spent Rs. 12 each on their meals

and the ninth person spent Rs. 8 more than the

average expenditure of all the nine. What was

the total money spent by them?

Solution

Average = x/9

Amount Spent by 8 members = 12 * 8 = 96

96 + x/9 + 8 = x

104 = x – x/9

104 = 8x/9

8x = 104 *9 = 936

x = 936/8 = 117

Problem - 7

A batsman makes a score of 87 runs in the 17th

inning and thus increases his average by 3. Find

his average after 17th innings?

Solution

17th innings avg. = x, Runs = 17x

16th innings avg. = x -3, Runs = 16 (x -3)

16 (x-3) + 87 = 17x

16x – 48 +87 = 17x

X = 39

Problem - 7

There are 24 students in a class. One of them,

who was 18 yrs old left the class and his place

was filled up by the newcomer. If the average of

the class thereby was lowered by one month,

what is the age of the newcomer?

Solution

Average reduced by 1 month,

24 * 1 = 2 years

So, the newcomer’s age is 18 -2 = 16 years

Problem - 8

The average of marks in mathematics for 5

students was found to be 50. Later, it was

discovered that in the case of one student the

mark 48 was misread as 84. What is the correct

average?

Solution

Difference = 84 – 48 = 36

36 /5 = 7.2 (Increased)

The corrected average = 50 – 7.2 = 42.8

Problem - 9

The average salary of all the workers in a factory

is Rs. 8000. The average salary of 7 technicians

is Rs. 12000 and the average salary of the rest is

Rs. 6000. What is the total number of workers in

the factory?

Solution

Members Avg.

7 12000

X 6000

6x = 7*12

X = 7812/6 = 14

Total no. of workers = 7 + 14 = 21

Problem - 10

Average salary of all the 50 employees including

5 officers of the company is Rs. 850. If the

average salary of the officers is 2500, find the

average salary of the remaining staff of the

company.

Solution

x/50 = 850; x = 42,500

5 officers’ salary = 2500*5 = 12500

50 – 5 members = 42500 – 12500

45 members = 30000

Avg. salary of 45 members = 30000/45

= 667(App)

Problem - 11

Find the average of 8 consecutive odd numbers

21,23,25,27,29,31,33,35

Solution

1st number + last Number /2

= 21 + 35 /2 = 28

Problem - 12

A train covers 50% of the journey at 30 km/hr,

25% of the journey at 25 km/hr, and the

remaining at 20 km/hr. Find the average speed of

the train during entire journey.

Solution

Total Journey = 100 km

S = Distance / Time = 100 / 5/3 + 1/1 + 5/4

= 100 * 12 /20+12+15

= 1200/47 = 25 25/47 km/hr

Problem - 13

The average of 10 numbers is 7. What will be the

new average if each number is multiplied by 8?

Solution

If numbers are multiplied by 8,

Average also to be multiplied by 8

= 7*8 = 56

{or}

x/10 = 7

x = 10*7 = 70

= 70* 5 = 560 /10 = 56

Problem - 14

The mean marks of 10 boys in a class is 70%

whereas the mean marks of 15 girls is 60%.

What is the mean marks of all 25 students?

Solution

Boys = x/10 = 70 = 700

Girls = x/15 = 60 = 900

10 + 15 = 700 + 900

25 = 1600

1600/25 = 64%

Problem - 15

Of the three numbers the first is twice the second

and the second is thrice the third. If the average

of the three numbers is 10, what are the

numbers?

Solution

A = 2x

B = x

C = x/3

2x + x + x/3/3 = 10

6x + 3x + x /9 = 10

6x + 3x + x = 90

10x = 90 ; x = 9.

A = 18, B = 9, C = 3

Percentage

Percentage

• By a certain Percent, we mean that many

hundredths.

• Thus, x Percent means x hundredths, written

as x%

•Finding out of Hundred.

If Length is increased by X% and Breadth is

decreased by Y% What is the percentage

Increase or Decrease in Area of the rectangle?

Formula: X+Y+ XY/100 %

Decrease 20% means -20

Percentage

Problem -1

When 75% of the Number is added to 75%, the

result is the same number. What is the number?

Solution

Percentage Number

75 x+75

100 x

100x + 7500 = 75x

25x = 7500

x = 300

Problem - 2

A tank is full of milk. Half of the milk is sold

and the tank is filled with water. Again half of

the mixture is sold and the tank is filled with

water. This operation is repeated thrice. Find the

percentage of milk in the tank after the third

operation?

Solution

Milk Water

100 0

50 50(1st)

25 75 (2nd)

12.5 87.5 (3rd)

After 3 operation Milk 12.5%

Problem 3

A large water-melon weighs 20kg with 96% of

its weight being water. It is allowed to stand in

the sun and some of the water evaporates so that

now, only 95% of its weight is water. What will

be its reduced weight?

Solution

20 *96/100=19.2kg of water

Let the evaporated water be x

19.2-x=95%(20-x)

19.2-x=95(20-x)/100

1920-100x=1900-95x

5x=20 ;x=4

20-4=16kg.

Problem 4

The population of a city is 155625. For

every1000 men, there are 1075 women. If 40%

of men and 24% of women be literate, then what

is the percentage of literate people in the city?

Solution Ratio of men and women=1000:1075=40:43

Number of men=40*155625/83=75000

Number of women=155625-7500=80625

Number of literate men=75000*40/100=3000

Number of literate women

=80625*24/100=19350

Literate people =30000+19350=49350

Percentage of literate people

=49350/155625*100=2632/83=31 59/83%

Problem 5

300 grams of sugar solution has 40% sugar in it.

How much sugar should be added to make it

50% in the solution?

Solution

Grams Sugar

300 40%

X 50%

50x = 40*300

x = 40*300/50 = 240

300 – 240 =60 Kg

Problem - 6

A man lost 12½% of his money and after

spending 70% of the remainder, he has Rs. 210

left. How much did the man have at first?

Solution Let the amount be 100

Then, 100.00 – 12.50 = 87.50

70% of 87.50 = 87.50 *70/100 =61.25

The remaining amount will be Rs. 26.25

Initial Final

100 26.25

X 210

26.25x = 21000; x = 21000/26.25 = 800

Problem - 7

During one year the population of a town

increases by 10% and during next year it

diminished by 10%. If at the end of the second

year, the population was 89,100, what was the

Population at the beginning of first year?

Solution

Let the population be 100

1st Year = 100 + 10 = 110

2nd Year = 110 * 10/100 = 110 -11 = 99

Percentage Population

99 89100

100 x

99x = 89100*100;

x = 8910000/99 = 90000

Problem - 8

When a number is first increased by 20% and

then again 20% by what percent should the

increase number be reduced to get back the

original number?

Solution

Let the number be 100

20% increase = 100*20/100 = 20

New Value = 120

Again increase by 20% = 120*20/100 = 24

New value = 144

Increased amount = 44/144*100 = 30 5/9%

Problem - 9

The number of students studying Arts,

Commerce and Science in an institute were in

the ratio 6 : 5 : 3 respectively. If the number of

students in Arts, Commerce and science were

increased by 10%, 30% and 15% respectively,

what was the new ratio between number of

students in the three streams?

Solution

A : C : S

6 : 5 : 3

6x : 5x : 3x

6x*110/100 : 5x*130/100 : 3x*115/100

6x*110 : 5x*130 : 3x*115

660 : 650 : 345

132 : 130 : 69

Problem - 10

In measuring the sides of rectangle errors of 5%

and 3% in excess are made. What is the error

percent in the calculated area?

Solution

Area = xy

X = 5% Excess = 100* 5/100 = 105

Y = 3% Excess = 100*3/100 = 103

103*105/100 = 10815/100 = 108.15

Error – Actual = 108.15 – 100 = 8.15% Excess

Problem - 11

In a certain examination there were 2500

candidates. Of them 20% of them were girls and

rest were boys. If 5% of boys and 40% of girls

failed, what was the Percentage of candidates

passed?

Solution

Girls = 2500*20/100 = 500

Boys = 2500*80/100 = 2000

Students who failed were

Boys = 2000*5/100 = 100

Girls = 500*40/100 = 200

Total Failed Students = 300

Total Pass students = 2500 – 300 = 2200

Pass Percentage = 2200/2500*100 = 88%

Problem - 12

A person saves every year 20% of his income. If

his income increases every year by 10% then his

saving increases by?

Solution

Every year saving, if the income is Rs. 100

= 100 *20/100 =Rs. 20

Salary increases = 110*20/100 = 22

Percentage increase (Savings) = 2/20*100 = 10%

Problem - 13

On a test containing 150 questions carrying 1

mark each, meena answered 80% of the first

answers correctly. What percent of the other 75

questions does she need to answer correctly to

score 60% on the entire exam?

Solution

Required correct answer = 150*60/100 = 90

Questions need to be correct.

80% of 75 questions = 60 q answered correctly.

Remaining 30 questions need be correct out of 75

= 30/75*100 = 40

Problem - 14

A boy after giving away 80% of his pocket

money to one companion and 6% of the

remainder to another has 47 paise left with him.

How much pocket money did the boy have in the

beginning?

Solution Let the amount be 100

To the first companion = 100*80/100 = 80

Remaining = 100 – 80 = 20

To the 2nd person = 20*6/100 = 1.20

The remaining = Rs.18.80 or 1880 paise

Initial Final

100 1880

X 47

1880x = 47*100

x = 4700/1880 = 2.5

Problem - 15

The length of a rectangle is increased by 10%

and breath decreased by 10%. Then the area of

the new rectangle?

Solution

I – D – I*D /100

10 -10 – 10*10/100

0 – 1 = -1

Decrease by 1%

Profit and Loss

• Gain =(S.P.)-(C.P.)

• Loss =(C.P.)-(S.P.)

• Loss or gain is always reckoned on C.P.

• Gain% = [(Gain*100)/C.P.]

• Loss% = [(Loss*100)/C.P.]

• S.P. = ((100 + Gain%)/100)C.P.

• S.P. = ((100 – Loss%)/100)C.P.

Profit and Loss

Problem - 1

A trade man allows two successive discount of

20% and 10%. If he gets Rs.108 for an article.

What was its marked price?

Solution

I1 + I2 – I1*I2/100

20 + 10 – 20*10 /100

= 28%

Discount = 28%, 72 Percent Cost is 108

Then 100percent cost = 72 108

100 x

100*108/72 = 150

Problem - 2

A trade man bought 500 metres of electric wire

at 75 paise per metre. He sold 60% of it at profit

of 8%. At what gain percent should he sell the

remainder so tas to gain 12% on the whole

Solution

500* 60/100 = 300

8 X

12

300 200

300 : 200 = 6 : 4

8 18

12

6 4

Remainder at 18% Profit

Problem - 3

A man purchased a box full of pencils at the rate

of 7 for Rs. 9 and sold all of them at the rate of 8

for Rs. 11. in this bargains he gains Rs. 10. How

many pencils did the box contains.

Solution

LCM = 7 and 8 = 56

56 pencil cost price = 8*9 = 72

56 Pencil selling price = 7*11

Profit = 77 – 72 = Rs. 5 for 56 pencil

Rs. 5 for 56 pencil means , for Rs. 10 the pencils

are 112

Problem - 4

A cloth merchant decides to sell his material at

the cost price, but measures 80cm for a metre.

His gain % is?

Solution

100 – 80 = 20 cm difference

Actual = 80

20/80*100 = 25% Gain

Problem - 5

Sales of a book decrease by 2.5% when its price

is hiked by 5%. What is the effect on sales?

Solution

Let the sales be 100 – 2.5 = 97.5

Profit = 100+5 = 105

Sales Profit

97.5 105

100 X

100x = 97.5*105

x = 97.5*105/100 = 102.375

100 – 102.375 = 2.375 = 2.4 profit (app)

Problem - 6

A dealer buys a table listed at Rs.1500 and gets

successive discount of 20% and 10%. He spends

Rs. 20 on transportation and sells it at a profit of

10%. Find the selling price of the table.

Solution

Discount = 20+10 – 20*10/100 = 28%

Actual price = 100 – 28 = 72

100 1500

72 x

72*1500/100 = 1080

Transport = 1080 +20 = 1100

100 1100

110 x

1100*110/100 = 1210

Problem - 7

A fridge is listed at Rs. 4000. due to the off

season, a shopkeeper announces a discount of

5%. What is the S.P?

Solution

= 4000*95/100 = 3800

Problem - 8

If the cost price of 9 pens is equal to the S.P of

11 pens. What is the gain or loss?

Solution

= 11 – 9 = 2

= 2/11*100 = 18 2/11% loss

Problem - 9

A machine is sold for Rs.5060 at a gain of 10%

what would have been the gain or loss percent if

it had been sold Rs.4370?

Solution

S.P = Rs.5060 = Gain = 10%

C.P = 100/110*5060 = 4600

IF S.P = Rs.4370 and C.P = Rs.4600

Loss = 230

Loss % = 230/4600 * 100 = 5% loss

Problem - 10

A person purchased two washing machines each

for Rs.9000. he sold one at a loss of 10% and

other at a gain of 10%. What is his gain or loss?

Solution

Each Rs.9000. one is 10% profit and other is

10% loss. So No profit and No loss

Problem - 11

Four percent more is gained by selling an article

for Rs.180, then by selling if for Rs.175. then its

C.P is?

Solution

Let the cost price = Rs. X

4% of x = 180 – 175 = 4x/100 = 5

4x = 500; x = 500/4 = 125

Problem - 12

An article is sold at a profit of 20%. If it had

been sold at a profit of 25%. It would have

fetched Rs.35% more. The Cost Price of the

article is?

Solution

Let C.P = Rs. X

125% of x – 120% of x = 35

5% of x =Rs.35 = x = 35*100/5 = 700

C.P = Rs. 700

Problem - 13

A reduction of 20% in the price of orange

enables a man to buy 5 oranges more for Rs. 10.

The price of an orange before reduction was,

Solution

20% Rs. 10 = Rs.2

Reduced price of 5 oranges = Rs. 2

Reduced price of 1 oranges = 40 p

Original price = 40/ 1- 0.20 = 400/8 = 50 Paise

Problem - 14

A man sells two horses for Rs.1475. The cost

price of the first is equal to the S.P of the second.

If the first is sold at 20% loss and the second at

25% gain. What is his total gain or loss? ( in

rupees)

Solution

Let cost price of 1st horse = S.P of 2nd = x

C.P of 2nd = S.P of 2nd * 100/125 = x*100/125 =

4x/5

S.P of 1st = C.P of 1st *80/100 = x*80/100 = 4x/5

Neither loss nor gain

Problem - 15

Rekha sold a watch at a profit of 15%. Had he

bought it at 10% less and sold it for Rs. 28 less,

he would have gained 20%. Find the C.P of the

Watch.

Solution

C.P be Rs. X

First S.P = 115% of x = 23x/20 and second C.P = 90% x = 9x/10

Second S.P = 120% of 9x/10 = 120/100 * 9x/10 = 27x/25

Given 23x/20 – 27x/25 = 28 = 115x – 108x/100 = 28

7x/100 = 28 = x = 28*100/7 = 400

C.P = Rs.400

Probability

Probability

• Probability:

P(є) = n(є) / n(s)

• (Addition theorem on probability:

n(AUB) = n(A) + n(B) - n(AB)

• Mutually Exclusive:

P(AUB) = P(A) + P(B)

• Independent Events:

P(AB) = P(A) * P(B)

Problem - 1

Four cards are drawn at random from a pack of

52 playing cards. Find the probability of getting

all face cards?

Solution

n(E) = 52C4

n(S) = 12C4 = 12C4/52C4

Problem - 2

Four persons are to be chosen at random from a

group of 3 men, 2 women and 4 children. Find

the probability of selecting 1 man, 1 woman or 2

children?

Solution

Total 3 M + 2 W + 4 C = 9 C 4 = 126

n (E) = 3C1 * 2C1 * 4C2 = 36

36/126 = 2/7

Problem - 3

A word consists of 9 letters, 5 consonants and 4

vowels. Three letters are chosen at random.

What is the probability that more than one

vowels will be selected?

Solution

n(E) = 9C3 = 84

More than one Vowels. So,

2V +1C or 3 V

4C2 *5C1 + 4C3 = 34

= 34/84 = 17/42

Problem - 4

A bag contains 10 mangoes out of which 4 are

rotten. Two mangoes are taken out together. If

one of them was found to be good, then what is

the probability that the other one is also good?

Solution

10 mangoes – 4 are rotten = 6 good mangoes

Getting good mangoes = 6C1/10C1 = 6/10

Getting second mango to be good = 5/9

1st and 2nd mangoes

6/10 *5/9 = 1/3

Problem - 5

Out of 13 applicants for a job there are 5 women

and 8 men. It is desired to select 2 persons for

the job. What is the probability that at least one

of the selected person will be a woman?

Solution

n(E) = 13C2 = 78

n(S) = 1m and 1 w or 2 w

= 8C1*5C1 + 5C2 = 50

= 50/78 = 25/39

Problem - 6

Two cards are drawn at random from a pack of

52 cards. What is the probability that either both

are black or both are queen?

Solution

P(A) = Both are Black

P(B) = Both are Queen

P(AnB) = Both are queen and Black

P(A) = 26C2/52C2 = 325/1326

P(B) = 4C2 /52C2 = 6/1326

P(AnB) = 2C2 /52C2 = 1/1326

325/1326 + 6/1326 - 1/ 1326 = 55/221

Problem -7

A man and his wife appear in an interview for

two vacancies in the same post. The probability

of husband’s selection is 1/7 and the probability

of wife’s selection is 1/5. Find the probability

that only one of them is selected?

Solution

Husband’s Selection = 1/7;

Not getting selected = 1 – 1/7 = 6/7

Wife’s selection = 1/5;

Not getting selected = 1 – 1/5 = 4/5

Only one of them is selected =

(Husband’s Selection + Wife Not getting selected) or

(Wife’s selection + Husband’s Not getting selected)

= (1/7*4/5) + 1/5*6/7) = 2/7

Problem - 8

Four persons are chosen at random from a group

of 3 men, 2 women and 4 children. What is the

chance that exactly 2 of them are children?

Solution

3 + 2 + 4 = 9C4 = 126

4 members 2(M and W) + 2(boy)

5C2 + 4C2 = 60

= 60 / 126 = 10/21

Problem - 9

Prakash can hit a target 3 times in 6 shots, Priya

can hit the target 2 times in 6 shots and Akhilesh

can hit the target 4 times in 4 shots. What is the

probability that at least 2 shots hit the target?

Solution

Prakash hitting = 3/6; not hitting = 3/6

Priya hitting = 2/6; not hitting = 4/6

Akilesh = 4/4 = 1

At least 2 shots hit target

= 3/6*4/6 + 3/6*2/6 = ½

Problem - 10

There are two boxes A and B. A contains 3 white

balls and 5 black balls and Box B contains 4

white balls and 6 black balls. One box is taken at

random and what is the probability that the ball

picked up may be a white one?

Solution

(Box A is selected and a ball is picked up ) or

(Box B is selected and a ball is picked up)

½*3/8 + ½*4/10 = 31/80

Problem - 11

A bag contains 6 white balls and 4 black balls.

Four balls are successively drawn without

replacement. What is the probability that they are

alternately of different colour?

Solution

Suppose the balls drawn are in the order white,

black, white, black…

= 6/10 *4/9*5/8*3/7 = 360/5040

Suppose the balls drawn are in the order black,

white, black, white…

= 4/10*6/9*3/8*5/7 = 360/5040

360/5040 +360/5040 = 1/7

Problem - 12

A problem in statistics is given to four students

A, B, C and D. Their chances of solving it are

1/3, ¼, 1/5 and 1/6 respectively. What is the

probability that the problem will be solved?

Solution

A is not solving problem = 2/3,

B is not solving problem = ¾

C not solving problem = 4/5

D not solving problem = 5/6

2/3*3/4*4/5*5/6 = 1/3

All together the probability of solving the

problem = 1 -1 /3 = 2/3

Problem - 13

There are 8 questions in an examination each

having only 2 answers choices ‘Yes’ or ‘No’. All

the questions carry equal marks. If a student

marks his answer randomly, what is the

probability of scoring exacting 50%?

Solution

Each questions having 2 ways of answering,

1 question = 2!........ 8 question = 2!

= 2!*2!*2!*2!*2!*2!*2!*2! = 256

To get 50%, 4 questions need to be correct,

8c4 = 8*7*6*5/1*2*3*4 = 70

= 70/256 = 35/128

A group consists of equal number of men and

women. Of them 10% of men and 45% of

women are unemployed. If a person is randomly

selected from the group find the probability for

the selected person to be an employee.

Problem - 14

Let the number of men is 100 and women be 100

Employed men and women = (100-10)+(100-45)

= 145

Probability = 145 / 200 = 29 / 40

Solution

Problem - 15

The probability of an event A occurring is 0.5

and that of B is 0.3. If A and B are mutually

exclusive events. Find the probability that

neither A nor B occurs?

Solution

It is Mutually exclusive events P(A n B)=0

Probability = 1 – ( P(A) + P (B) – P(A n B) )

= 1 – (0.5 + 0.3 – 0)

= 0.2

Permutation and Combination

Permutation and Combination

Permutation means Arrangement

Combination means Selection

Permutation and Combination • Permutations:

Each of the arrangements which can be made by taking some (or) all of a number of items is called permutations.

npr = n(n-1)(n-2)…(n-r+1)=n!/(n-r)!

• Combinations:

Each of the groups or selections which can be made by taking some or all of a number of items is called a combination.

nCr = n!/(r!)(n-r)!

Types

1. How many ways of Arrangement possible by

using word SOFTWARE?

SOFTWARE = 8!

2. How many ways of arrangement Possible by

using word SOFTWARE, vowels should come

together.

SFTWR (OAE) = 6! * 3!

Types

3. How many ways of Arrangement Possible by

using word SOFTWARE, vowels should not

come together?

SFTWR ( ARE)

Not together

= Total arrangement – Vowels together

= 8! – (5! * 3!)

Types 4. How many ways of arrangement possible by using

word MACHINE, so that vowels occupy only ODD

places.

- - - - - - - (7 places)

MCHN (AIE) 4 Consonant and 3 vowels.

7 places = 4 ODD places, 3 EVEN places

Vowels = 4P3 = 4!

Consonant = 4P4 = 4!

Total Number of arrangement = 4!*4!

Types

5. How many ways of arrangement possible by

using word ARRANGEMENT

Letter’s Repetition = 2(A) 2(R) 2 (E) 2 (N)

= 11!/2!*2!*2!*2!

In a given problem, any letter is repeated more

than once that should be divided with total

number.

Problem - 1

A committee of 5 is to be formed out of 6 gents

and 4 ladies. In how many ways this can be

done, when at least 2 ladies are included?

Solution

a. 2 ladies * 3 Gents

4C2 * 6 C3 = 120

b. 3 ladies * 2 Gents

4C3 * 6C2 = 60

c. 4 ladies * 1 Gent

4C4 *6C1 = 1*6 = 6

Total ways = 120 +60 +6 = 186

Problem - 2

It is required to seat 5 men and 4 women in a

row so that the women occupy the even places.

How many such arrangements are possible?

Solution

Total places = 9

Odd places = 5

Even places = 4

4 even places occupied by 4 women

= 4P4 = 4! = 24

5 odd places occupied by 5 men

= 5P4 = 5! = 120

Total ways = 120*24 = 2880 ways

Problem - 3

A set of 7 parallel lines is intersected by another

set of 5 parallel lines. How many parallelograms

are formed by this process?

Solution

Two parallel lines from the first set and any two

from the second set will from a parallelogram.

7C2 *5C2 = 21 * 10 = 210

Problem - 4

There are n teams participating in a football

championship. Every two teams played one

match with each other. There were 171 matches

on the whole. What is the value of n?

Solution

Total number of matches played = nC2

nC2 = 171

n(n-1)/2= 171

n2 – n – 342 = 0

(n+18) (n-19) = 0

n = 19

Problem - 5

In an examination, a candidate has to pass in

each of the 6 subjects. In how many ways can he

fail?

Solution

6C1 + 6C2 + 6C3 + 6C4+6C5+6C6

1 + 6 + 15 + 20 + 15 + 6 = 63 ways

Problem - 6

In how many ways can a pack of 52 cards be

distributed to 4 players, 17 cards to each of 3 and

one card to the fourth player?

Solution

17 cards can be given to 1st player = 52 C17

2nd player = 35C17

3rd player = 18C17

4th player = 1

= 52C17*35C17*18C17

= 52!/17!35! * 35!/17!*18! * 18!/17!*1!

= 52!/(17!)3

A foot race will be held on Saturday. How many

different arrangements of medal winners are

possible if medals will be for first, second and

third place, if there are 10 runners in the race …

Problem - 7

n = 10

r = 3

n P r = n!/(n-r)!

= 10! / (10-3)!

= 10! / 7!

= 8*9*10

= 720

Number of ways is 720.

Solution

To fill a number of vacancies, an employer must

hire 3 programmers from among 6 applicants,

and two managers from 4 applicants. What is

total number of ways in which she can make her

selection ?

Problem - 8

It is selection so use combination formula

Programmers and managers = 6C3 * 4C2

= 20 * 6 = 120

Total number of ways = 120 ways.

Solution

Problem - 9

A man has 7 friends. In how many ways can

he invite one or more of them to a party?

Solution

In this problem, the person is going to select his

friends for party, he can select one or more

person, so addition

= 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7

= 127

Number of ways is 127

Problem - 9

Find the number of different 8 letter words

formed from the letters of the word EQUATION

if each word is to start with a vowel

Solution

For the words beginning with a vowel, the first

letter can be any one of the 5 vowels, the

remaining 7 places can be filled by

7P7 = 5040

The number of words = 5 * 5040 = 25200

Problem - 10

In how many different ways can the letters of the

word TRAINER be arranged so that the vowels

always come together?

Solution

A,I,E can be arranged in 3! Ways

(5! * 3!) / 2! = 360 ways

Problem - 11

In how many different ways can the letters of the

Word DETAIL be arranged so that the vowels

may occupy only the odd positions?

Solution

___ ___ ___ ___ ___ ___

3P3 = 3! = 1*2*3 = 6

3P3 = 3! = 1*2*3 = 6

= 6*6 = 36

Problem - 12

There are 5 red, 4 white and 3 blue marbles in a

bag. They are taken out one by one and arranged

in a row. Assuming that all the 12 marbles are

drawn, find the number of different

arrangements?

Solution

Total number of balls = 12

Of these 5 balls are of 1st type (red), 4 balls are

the 2nd type and 3 balls are the 3rd type.

Required number of arrangements =

12!/5!*4!*3!

= 27720

Problem - 13

5 men and 5 women sit around a circular table,

the en and women alternatively. In how many

different ways can the seating arrangements be

made?

Solution

5 men can be arranged in a circular table in 4

ways = 24 ways

There are 5 seats available for 5 women they can

be arranged in 5 ways

No. of ways = 5!*4! = 2880 ways

Problem - 14

In a chess board there are 9 vertical and 9

horizontal lines. Find the number of rectangles

formed in the chess board.

Solution

Solution:

9C2 * 9C2 = 1296

Problem - 15

In how many ways can a cricket team of 11

players be selected out of 16 players, If one

particular player is to be excluded?

Solution

Solution:

If one particular player is to be excluded, then

selection is to be made of 11 players out of 15.

15C11= 15!/( 11!*4!)=1365 ways

Area and Volume

Area and Volume Cube:

• Let each edge of the cube be of length a. then,

• Volume = a3cubic units

• Surface area= 6a2 sq.units.

• Diagonal = √3 a units.

Cylinder:

• Let each of base = r and height ( or length) = h.

• Volume = πr2h

• Surface area = 2 πr h sq. units

• Total Surface Area = 2 πr ( h+ r) units.

Area and Volume

Cone:

• Let radius of base = r and height=h, then

• Slant height, l = √h2 +r2 units

• Volume = 1/3 πr2h cubic units

• Curved surface area = πr l sq.units

• Total surface area = πr (l +r)

Area and Volume

Sphere:

• Let the radius of the sphere be r. then,

• Volume = 4/3 πr3

• Surface area = 4 π r2sq.units

Area and Volume

Circle: A= π r 2

Circumference = 2 π r

Square: A= a 2

Perimeter = 4a

Rectangle: A= l x b

Perimeter= 2( l + b)

Area and Volume

Triangle:

A = 1/2*base*height

Equilateral = √3/4*(side)2

Area of the Scalene Triangle

S = (a+b+c)/ 2

A = √ s*(s-a) * (s-b)* (s-c)

Area and Volume

Problem - 1

A rectangular sheet of size 88 cm * 35 cm is bent

to form a cylindrical shape with height 35 cm.

What is the area of the base of the cylindrical

shape?

Solution

The circumference of the circular region = 88 cm

2r = 88

r = 88*7/22*2 = 14 cm

Area of the base = r2 = 22/7*14*14 v= 616 cm2

Problem - 2

The radius of the base of a conical tent is 7

metres. If the slant height of the tent is 15

metres, what is the area of the canvas required to

make the tent?

Solution

R = 7 m

L = 15 m

Area of Canvas required = Curved Surface Area

of cone

rl = 22/7*7*15 = 330 sq.m

Problem - 3

Three spherical balls of radius 1 cm, 2 cm and 3

cm are melted to form a single spherical ball. In

the process, the material loss was 25%. What

would be the radius of the new ball?

Solution

Vol. of sphere = 4/3 r3

Vol. of 3 small spherical balls = 4/3 ( 13+23+33)

= 4/3 (1+8+27) = 4/3 (36) = 48

Material loss = 25%

Vol. of the single spherical ball = 48*75/100

= 48 * ¾ = 36

V = 4/3r3 = 36

r3 = 36*3/4 = 27

r = 3 cm

Problem - 4

A rectangular room of size 5m(l)*4m(w)*3m(h)

is to be painted. If the unit of painting is Rs. 10

per sq.m, what is the total cost of painting?

Solution

Area of 4 walls = 2h(l+b)

The area to be painted includes the 4 walls and

the top ceiling.

Area to be painted = 2h (l+b) +lb

= 2*3 (5+4) + 5*4

= 54+20 = 74 sq.m.

Total cost of painting = 74*10 = Rs.740

Problem - 5

The radius of a sphere is r units. Each of the

radius of the base and the height of a right

circular cylinder is also r units. What is the ratio

of the volume of the sphere to that of the

cylinder?

Solution

Vol. of sphere = 4/3r3 and Vol. of Cylinder =

r2h = r3

Required Ratio = 4/3 r3: r3 = 4/3 : 1

= 4 : 3

Problem - 6

A measuring jar of internal diameter 10 cm is

partially filled with water. Four equal spherical

balls of diameter 2 cm each, are dropped in it

and they sink down in the water completely.

What will be the increase in the level of water in

the jar?

Solution

Radius of each ball = 1 cm

Vol. of 4 balls = 4* 4/3 (r)3 = 16/3 cm3

Vol. of water raised in the Jar = Vol. of 4 balls

Let h be the rise in water level, then

Area of the base *h = 16/3

*5*5*h = 16/3

H = 16/3*25 = 16/75 cm

What is the cost of planting the field in the form

of the triangle whose base is 2.8 m and height

3.2 m at the rate of Rs.100 / m2

Problem - 7

Area of triangular field = ½ * 3.2 * 2.8 m2

= 4.48 m2

Cost = Rs.100 * 4.48

= Rs.448..

Solution

Problem - 8

Find the length of the longest pole that can be

placed in a room 14 m long, 12 m broad, and 8 m

high.

Solution

Length of the longest pole = Length of the

diagonal of the room

= √(142 + 122 + 82)

= √ 404 = 20.09 m

Area of a rhombus is 850 cm2. If one of its

diagonal is 34 cm. Find the length of the other

diagonal.

Problem - 9

850 = ½ * d1 * d2

= ½ * 34 * d2

= 17 d2

d2 = 850 / 17

= 50 cm

Second diagonal = 50cm

Solution

A grocer is storing small cereal boxes in large

cartons that measure 25 inches by 42 inches by 60

inches. If the measurement of each small cereal

box is 7 inches by 6 inches by 5 inches then what is

maximum number of small cereal boxes that can be

placed in each large carton ?

Problem - 10

No. of Boxes = 25*42*60 / 7*6*5 = 300

300 boxes of cereal box can be placed.

Solution

Problem - 11

If the radius of a circle is diminished by 10%,

what is the change in area in percentage?

Solution

= x + y + xy/100

= -10 - 10 + 10*10/100

= -19%

Diminished area = 19%.

Problem - 12

A circular wire of radius 42 cm is bent in the

form of a rectangle whose sides are in the ratio

of 6:5. Find the smaller side of the rectangle?

Solution

length of wire = 2 πr = (22/7*14*14)cm

= 264cm

Perimeter of Rectangle = 2(6x+5x) cm

= 22xcm

22x =264 x = 12 cm

Smaller side = (5*12) cm = 60 cm

Problem - 13

A beam 9m long, 40cm wide and 20cm deep is

made up of iron which weights 50 kg per cubic

metre. Find the weight of the Beam.

Solution

Vol. of the Beam = lbh = 9*40/100*10/100

= 72 m3

Weight of the iron beam is given as lm3 = 50 kg

72/100 m3 = 72/100*50 = 36 kg

Problem - 14

If the length of a rectangle is reduced by 20%

and breadth is increased by 20%. What is the

percentage change in the area?

Solution

x + y + (xy/100)%

= - 20 + 20 – 400/100

= -4

The area would decrease by 4%

Problem - 15

Find the number of bricks measuring 25 cm in

length, 5 cm is breadth and 10 cm in height for a

wall 40 m long, 75 cm broad and 5 metres in

height?

Solution

Vol. of the wall = 40*72/100*5 = 150 m3

Vol. of 1 bricks = 25/100*5/100*10/100

= 1/80 m3

Number of bricks required = 150/1/800

= 150*800

= 120000

Calendar

Calendar Odd days:

0 = Sunday

1 = Monday

2 = Tuesday

3 = Wednesday

4 = Thursday

5 = Friday

6 = Saturday

Calendar Month code: Ordinary year

J = 0 F = 3

M = 3 A = 6

M = 1 J = 4

J = 6 A = 2

S = 5 O = 0

N = 3 D = 5

Month code for leap year after Feb. add 1.

Calendar

Ordinary year = (A + B + C + D )-2

-----------------------take remainder

7

Leap year = (A + B + C + D) – 3

------------------------- take remainder

7

Problem - 1

11th January 1997 was a Sunday. What day of

the week on 7th January 2000?

Solution

11th Jan 1997 = Sunday

11th Jan 1998 = Monday

11th Jan 1999 = Tuesday

11th Jan 2000 = Wednesday

7th Jan 2000 is on Saturday

Problem - 2

What day of the week was on 5th June 1999?

Solution

A+B+C+D – 2 / 7

A = 1999/7 = 4

B = 1999/4 = 499/7 = 2

C = June = 4

D = 5/7 = 5

= 4+2+4+5 – 2/7 = 13/7 = 5 = Saturday

Problem - 3

On what dates of August 1988 did Friday fall?

Solution

A = 1988 / 7 = 0

B = 1988/4 = 497/7 = 0

C = 3

D = x

0+0+3+x+3/7 = x/7 = 5(Friday)

Friday falls on = 5,12,19,26

Problem - 4

India got independence on 15 August 1947.

What was the day of the week?

Solution

A = 1947/7 = 1

B = 1947/4 = 486/7 = 3

C = 15/7 = 1

D = 2

1+3+1+2 – 2 /7 = 5/7 = Friday

Problem - 5

7th January 1992 was Tuesday. Find the day of

the week on the same date after 5 years. i.e on

7th January 1997.

Solution

7th January 1992 = Tuesday

7th January 1993 = Thursday (Leap)

7th January 1994 = Friday

7th January 1995 = Saturday

7th January 1996 = Monday ( Leap)

7th January 1997 = Tuesday

Problem - 6

The first Republic day of India was celebrated on

26th January 1950. What was the day of the

week on that date?

Solution

A = 1950/7 = 4

B = 1950/4 = 487/7 = 4

C = 0

D = 26/7 = 5

4+4+0+5 – 2/7 = 11/7 = 4 = Thursday

Problem - 7

Find the Number of times 29th day of the month

occurs in 400 consecutive year?

Solution

1 year = 1 (Ordinary Year)

1 year = 12 (Leap Year)

400 years = 97 leap year

97 * 12 = 1164

303*11 = 3333

= 1164+3333 = 4497 times

Problem - 8

If 2nd March 1994 was on Wednesday, 25 Jan

1994 was on,

Solution

A = 1994/7 = 6

B = 1994/4 = 498/7 = 1

C = 0

D = 25/7 = 4

= 6 + 1 + 0 + 4 – 2 / 7 = 3 = Tuesday

Problem - 9

Calendar for 2000 will serve also?

Solution

= 2000 + 2001 + 2002 + 2003 + 2004

= 2 + 1 + 1 + 1 + 2 = 7 (Complete Week)

2005

Problem - 10

If Pinky’s 1st birthday fell in Jan 1988 on one of

the Monday’s, the day on which are was born is,

Solution

Jan = 1988 = Monday

Jan = 1987 = Sunday

Problem - 11

Akshaya celebrated her 60th birthday on Feb 24,

2000. What was the day?

Solution

A = 2000 /7 = 7

B = 2000/4 = 500/7 = 3

C = 3

D = 24/7 = 0

= 7+3+3+0-3/7 = 10/7 = 3 = Wednesday

Problem - 12

On what dates of April 2008 did Sunday Fall?

Solution

Calculate for 1st April 2008

A = 2008/7 = 6

B = 2008/4 = 502/7 = 5

C = 1/7 = 1

D = 0

= 6+5+1+0 – 3/ 7 = 2 = Tuesday

1st April on Tuesday, then 1st Sunday fall on 6.

Sunday falls on 6, 13, 20, 27.

Problem - 13

Today is Friday. After 62 days it will be,

Solution

62 / 7 = 6 days after Friday then it will be

Tuesday

Problem - 14

What will be the day of the week on 1st Jan

2010?

Solution

A = 1

B = 5

C = 0

D = 1

= 1+5+0+1 – 2/ 7 = 5/7 = 5 = Friday

What is the day of the week on 30/09/2007?

Problem - 15

Solution:

A = 2007 / 7 = 5

B = 2007 / 4 = 501 / 7 = 4

C = 30 / 7 = 2

D = 5

( A + B + C + D )-2

= -----------------------

7

= ( 5 + 4 + 2 + 5) -2

----------------------- = 14/7 = 0 = Sunday

7

Calendar

Clock

Clocks

Clock:

Angle between hour hand and minute hand

= (11m/2) – 30h

Angle between minute hand and hour hand

=30h – (11m/2)

Problem - 1

What is the angle between the minute hand and

hour hand when the time is 2.15?

Solution

= 11 m/2 – 30(h)

= 11 15/2 – 30(2)

= 11(7.5) – 60

= 82.5 – 60 = 22 1/2

Problem - 2

At what time between 5 and 6 o’clock the hands

of a clock coincide?

Solution

Coinciding Angle = 0

Min. hand to hour hand = 25 min apart

60/55*25 = 12/11 * 25 = 300/11

= 27 3/11min past 5

Problem - 3

At what time between 12 and 1 o’clock both the

hands will be at right angles?

Solution

Right angle = 90 degrees

= 30(h) – 11 m/2

90 = 30(12) – 11 m/2

180 = 360 – 11m

11m = 360 – 180

M = 180/11

16 4/11 past 12

Problem - 4

Find at what time between 7 and 8 o’clock will

the hands of a clock be in the same straight line

but not together?

Solution

Minute hand to hour hand = 35 min apart

Straight line not together = 30 min apart

Difference = 35 – 30 = 5 min

= 60/55*5 = 12/11*5 = 60/11

= 55 5 / 11 past 7

Problem - 5

At what time between 5 and 6 are the hands of

the clock 7 minutes apart?

Solution

7 min space behind the hour hand:

25 min – 7 min = 18 min

60/55 *18 = 216/11 = 19 7/11 min past 5

7 Min space ahead the hour hand

25 min + 7 min = 32 min

60/55*32 = 12/11*32 = 384/11

= 34 10/11 min past 5

Problem - 6

A clock strikes 4 and takes 9 seconds. In order to

strike 12 at the same rate what will be the time

taken?

Solution

Strike Sec

3 (interval) 9

11 x

3x = 11*9

X = 11*9/3 = 33 Sec