Ratio And Proportion Important Formulas Concepts Essay

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Ratio And Proportion Important Formulas Concepts Essay

Four numbers are in a proportion if the product of extremes = product of means.

E.g. if a : b :: c : d, then ad = bc

If the ratio of first and second equals to the ratio of second and third (with three or more quantities), they are said to be in continued proportion.

E.g. if a/b = b/c, then a, b and c are in a continued proportion.

If a number, say Z, is divided into three parts, whose ratio is a:b:c, then:

First part =

Second part =

Third part =

Illustration 1: The ratio of number of ladies to gents in a party was 1:2, but when 2 ladies and 2 gents left, the ratio became 1:3. How many people were originally present at the party? Ratio And Proportion Important Formulas Concepts Essay.

12

Solution:Let the number of ladies be x& the number of gents be 2x

3x-6 =2x -2

Total number of persons at the party = 3x =3 × 4 = 12

Illustration 2: The sum of the present ages of A, B and C is 30 years. Six year ago, their ages were in the ratio 1: 2: 3. What is the present age of C?

42

Solution: Sum of the ages of A, B, C six years ago = 90 – 3

= 72 years

Now,

Six years ago, the ages of A, B & C were in ratio 1:2:3

Age of C six years ago =

= 36 years

Present age of C = 36 +6

= 42 years

Illustration 3: If bc : ac: ab = 1:2:3, find :

1 :3

9 :1

4 :9

4 :1

Solution: : / =

Also bc : ac = 1:2

=

: = = 2 =

Illustration 4: The sum of the squares of three numbers is 532 and the ratio of the first to the second as also of the second to the third is 3:2. What is the second number? Ratio And Proportion Important Formulas Concepts Essay.

12 Solution: = , =

A =B & C = B

Now,

A2 + B2 + C2 = 532

B2 + B2 + = 532

= 532

133B2 = 532 × 36

B =12

Illustration 5: A person distributes his pens among four friends A, B, C and D in the ratio 1/3: 1/4: 1/5: 1/6. What is the minimum number of pens that the person should have to perform this exercise?Ratio And Proportion Important Formulas Concepts Essay.

57

Solution: LCM of 3, 4, 5, 6 = 60

So, the pens are distributed among A, B, C & D in the ratio

15 : 12 : 10

Total number of pens = 20x + 15x + 12x + 10x

To gauge the minimum number of pens, the value of x should be taken as 1

=20 + 15 + 12 + 10 = 57

Practice Exercise

If half of one number is equal to 0.07 of another, the ratio of the numbers is:

50:7

5:7

7:50

1:14

Solution: x =0.07 y

=0.07 × 2 = × 2 = 7/50

If = 2:3 then =?

4/9

4/3

16/9

4/27

Solution: = {given}

=2 × =

A man divides his property so that ratio of his son’s share to his wife’s share and the ratio of the wife’s share to his daughter’s share are both 3:1. If the daughter gets Rs. 10,000 less than the son, then the total worth of his property is: Ratio And Proportion Important Formulas Concepts Essay.

25,000

16,250

65,000

27,350

Solution:Son : Wife = 3 : 1 & Wife : Daughter = 3 : 1

Son : Wife : Daughter = 9 : 3 : 1

Now,

Son’s share – Daughter’s share = 10, 000

9x – x =10, 000

8x = 10, 000

x = 1250

Total property = 13x

=13 × 1250

=16,250

The incomes of A & B is are in the ratio 3:2 and their expenditures are in the ratio 5:3. If each one of them saves Rs. 1000, then, A’s income can be:

Rs. 3000

Rs. 4000

Rs. 6000

Rs. 9000

Solution: Let the income of A & B be 3x and 2x

& the total expenditures of A & B be 5y & 3y

3x – 5y = 1000…… (I)

2x – 3y = 1000………….. (II)

Solving (I) & (II)

x = 2000

y = 1000

A’s income = 3x = 3 × 2000 = 6000

If Rs. 58 is divided among 150 children such that each girl and each boy gets 25 p and 50 p respectively, then how many girls are there in total? Ratio And Proportion Important Formulas Concepts Essay.

68

110

Solution: Let the number of boys & girls be x & y respectively

x + y =150 …….. (I)

x + y = 58

2x + y =232 ……. (II)

Solving (I) & (II)

y = 68

60 kg of an alloy A is mixed with 100 kg of alloy B. If alloy A has lead and tin in the ratio 3:2 and alloy B has tin and copper in the ratio 1:4, then the amount of tin in the new alloy is: Ratio And Proportion Important Formulas Concepts Essay.

36 kg

44 kg

53 kg

80 kg

Solution: Tin in Alloy ‘A’ = = 24 kg

Tin in Alloy ‘B’ = =20 kg

Tin in new Alloy C =24 + 20

= 44 kg

15 liters of mixture contains 20% alcohol and the rest water. If 3 liters of water is mixed with it, the percentage of alcohol in the new mixture would be:

15%

17 %

Solution: Amount of alcohol in mixture =20 % × 15

= = 3 liters

Percentage of alcohol after adding 3 liters water

=

= %

The speeds of the three cars are in the ratio 5:4:6. The ratio between the time taken by them to travel the same distance is:

5:4:6

6:4:5

10:12:15

12:15:10

Solution: Speeds are in ratio of 5:4:6

For same distance time will be in ratio: :

LCM (5, 4, 6) = 60

Required ratio = : : : 60

=12 : 15 : 10

Seats for Mechanical, Civil and Electrical branch in an engineering college are in the ratio of 5:7:8. Ratio And Proportion Important Formulas Concepts Essay. There is a proposal to increase these seats by 40%, 50% & 75% respectively. What will be the ratio of increased seats?

2:3:4

6:7:8

6:8:9

None of these

Given ratio 5:7: 8

After increment 5×140: 7×150: 8×175

700:1050:1400

= 2 : 3 : 4

Hence, the answer is (a)

The sides of the triangles are in the ratio of : : and its perimeter is 104 cm. The length of the longest side is:

52 cm

48 cm

32 cm

26 cm

Solution: Let the sides be , and

+ + = 104

6 x + 4x + 3x =104 × 12

13x =104 × 12

x =

x = 96

Longest side = ½ x= x × 96 = 48

Simple Interest/Compound Interest

Important Formulas/Concepts

Simple Interest =

From the above-mentioned formula, the following formulas can be derived:

P =

R =

T =

Let P = Principal, R = R% per annum, Time = T years and number of times the interest is compounded in a fixed interval = N, then (Compound Interest):

Amount = P×T

To find number of years it will take for money to double (in case of compound interest):

Number of years =

E.g. If rate = 18%, then number of years for money to double = = 4 years

Illustration 1: A man investsRs.3000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum?Ratio And Proportion Important Formulas Concepts Essay.

Rs.1200

Rs.1300

Rs. 1500

Rs. 2000

Solution: 30,000×+ (3000+ x) ×

15000+8x =18000+6x

Thus, x=1500

Illustration 2: A sum of money lent out at simple interest amounts to Rs.720 after 2years and to Rs.900 after a further period of 5 years. The sum and the rate % are:

500,10%

600,10%

500,15%

600,15%

Solution: A-P =

720-P = …..(i)

Similarly,

900 –p =………. (ii)

Dividing (i) by (ii)

=

5(720-P) =2(900-P)

3600 -5P =1800 -2P

P =600

Now,

Rate% 720 -600 =

10% =R

Illustration 3: Out of a certain sum, 1/3rd is invested at 3%, 1/6th at 6% and the rest at 8%. If the simple interest for 2 years from all these investments amount to Rs. 600, then the original sum is: Ratio And Proportion Important Formulas Concepts Essay.

Rs. 3000

Rs. 4500

Rs. 5000

Rs. 6000

Solution: Let the initial amount be x

Now,

+

12x = 60,000

Thus, x = 5000

Illustration 4: If Rs. 85 amounts to Rs.95 in 3years, what Rs.102 will amount to in 5 years at the same rate %?

117

122

132

142

Solution: A – P =

95-85 =…… (I)

Similarly,

A -102 =…… (II)

Dividing equation (I) by (II)

=85/102 3/5

Thus, A =122

Illustration 5: If simple interest on a certain sum is 16 over 25 of the sum, then the rate percent and time, if both are equal, are: Ratio And Proportion Important Formulas Concepts Essay.

8

Solution: Simple Interest =

Given, Simple Interest =16/25P& R =T

16/25P =

64 =R2

R =T =8

R = 8

Practice Exercise

If the difference between the Compound Interest and Simple Interest on a certain sum of money is Rs.72 at 12 percent per annum for 2 years, then the amount is:

Rs. 6000

Rs. 5000

Rs. 6500

Rs. 5500

Solution: Compound Interest – Simple Interest = 72

P – P – = 72

P=72

P =72

Thus, P =5000

A sum of money placed at a compound interest doubles itself in 3 years. Ratio And Proportion Important Formulas Concepts Essay. In how many years will it amount to 8 times of itself?

9 years

8 years

27 years

7 years

Solution: Given A = 2P

A = Pn

Or 2 = 3……………..(i)

Similarly 8P= P n

8 = n

23 = n …………(ii)

From (i) & (ii)

[()3]3= n

Thus, n = 9

If x is the simple interest on y and y is the simple interest on z, what is the relation between x, y and z, the rate % and the time being the same in both cases? Ratio And Proportion Important Formulas Concepts Essay.

x2 = y z

y2 = x z

x y z =1

x =2y+z

Solution: x ==…..(I)

y…..(II)

Dividing (I) & (II)

=

xz=y2

A sum of money invested triples itself in 8 years at simple interest. Find in how many years will at become 8 times itself at the same rate?

24 years

28 years

30 years

21 years

Solution: 2P =…….(i)

7P = ()………..(ii)

Dividing (I) by (II)

Thus, T =28 years

Find the compound interest at the rate of 10% for 3 years on the principle which in 3 years at the rate of 10% per annum gives Rs. 300 as simple interest. Ratio And Proportion Important Formulas Concepts Essay.

331

310

330

333

Solution: Simple Interest =

300=

Thus, P =1000

Now, Compound Interest = P [()n- 1]

=1000[()3- 1]

= 1000

Thus, Compound Interest =331

A man wants to invest Rs. 1, 50,000among two schemes at the rate of 5% and 10% to earn interest at Rs. 10,000 in 1 year. Thus, the amount invested in 5% and 10% schemes respectively is: Ratio And Proportion Important Formulas Concepts Essay.

1,00,000 & 50,000

50,000 & 1,00,000

75,000 & 75,000

1,25,000 &25,000

Solution: Let amount invested in 5% scheme be x

Thus, +=10000

5x +1500000 -10x =1000000

500000=5x

10000 =x

If Rs. 1100 is obtained after lending out same amount at 5% per annum for 2 years and Rs.1800 is obtained after lending the remaining amount at 10% per annum for 2 years, then the total amount lent out is:

2500

3000

2000

2200

Solution: A1 = P1 + I1

I1 = P1 ×

1100 =

P1 =1000

Similarly,

1800-P2

P2 =1500

Total principle P =P1+P2

=1000+1500

=2500

Shamit invested Rs 6000 in a company at a compound interest compounded semi-annually. He receives Rs 7986 after 18 months from the company.Find the rate of interest per annum. Ratio And Proportion Important Formulas Concepts Essay.

10%

20%

12.5%

Solution: 7986 = 6000 (1 + )3/2×2

R = 10%

Yearly rate = 2 × 10% = 20%

A sum was invested at a simple interest at a certain interest for 2 years. It would have fetched Rs 60 more had it been invested at 2% higher rate. What was the sum?

1500

1300

2500

1000

Solution: Condition…… (I)

I = P×R×2/100

Condition …… (II)

I+60 =

1500 = P

A sum of Rs.3,800 is lent out in two parts in such a way that the interest on one part at 8% for 5 years is equal to that on the another part at ½% for 15 years.Find the sum lent out at 8%. Ratio And Proportion Important Formulas Concepts Essay.

500

400

600

700

Solution: Let the amount invested at 8% be x

= ×

16x =11400-3x

19x =11400

x =600

Time and Work / Pipes and Cisterns

Important Formulas/Concepts

Work = Working units Time

If A can do a piece of work in b days, then his one day’s work =

Illustration 1: Varun can do a piece of work in 20 days. Tarun is 25% more efficient than Varun. The number of days taken by Tarun to do the same piece of work is:

16

Solution: Amount of work done by Varun in one day=

Amount of work done by Tarun in one day =125% × = =

Tarun will complete the same piece of work in 16 days

Illustration 2: Sam can do a piece of work in 40 days. He works on it for 8 days and then John finished it in 16 days. How long will they take to complete the work together? Ratio And Proportion Important Formulas Concepts Essay.

days

15 days

20 days

56 days

Solution: Amount of work completed by Sam in 8 days 8 × =

Work left to be completed by John =1 – =

Let the amount of work completed by John in one day =

Using work equation

Amount of work they can complete in one day = =

Required time = days

Illustration 3: Mukesh, Anil and Sumit are three civil engineers. Mukesh can design a plan alone of a multistoried apartment in 5 hours and Anil in 4 hours. They together can do it in 2 hours. In what time can Sumit do it alone? Ratio And Proportion Important Formulas Concepts Essay.

7 hours

15 hours

20 hours

18 hours

Solution: Using work equation,

= –

= =

Sumit can design the plan alone in 20 hours.

Illustration 4: If 4/7th of a piece of work is completed in 7/4th days, in how many days can rest of the work be completed?

7/3

3/7

21/16

16/21

Solution: of a work is completed in days

Total work is completed in = days

Work remaining to be completed = 1- =

Time taken to complete work = = days

Illustration 5: One man can paint a house in ‘r’ days and another man can do it in‘t’ hours. If they can together do it in ‘d’ days, then ‘d’ is given by:

Solution: Work completed by a man in one day= 1/r

Work completed by another man in one day = 24/t

Using work equation, Ratio And Proportion Important Formulas Concepts Essay.

=

d =

Practice Exercise

Mary has ‘m’ minutes of home work in each of her ‘s’ subjects. In one hour, she completes what part of her homework?

1/ms

60/ms

m/s

m/60s

Solution: Mary completes home work in each of the subjects in ‘m’ minutes

And there are in total‘s’ subjects

Total homework of Mary is completed in ms minutes

i.e. ms/60 hours

Hence, in one hour, Mary completed 60/ms work

Federer can do a job in 40 days. He worked on it for 5 days and then Paes finished it in 21 days. In how many days Federer and Paes can finish the work together? Ratio And Proportion Important Formulas Concepts Essay.

15

Solution: Work completed by Federer in 5 days = =

Work remaining to be completed = 1- =

Paes completed the remaining work in 21 days.

Let the work completed by Paes in one day be x

Using work equation,

Amount of work completed by both of them together in one day =

Thus, together they will take 15 days.

Rohit can copy 80 pages in 20 hours. Rohit and Rahul can copy 135 pages in 27 hours. Ratio And Proportion Important Formulas Concepts Essay. How many pages Rahul can copy in 20 hours?

20

Solution: Pages copied by Rohit in 1 hour = = 4 pages

Pages copied by Rohit and Rahul in 1 hour = = 5 pages

Number of pages Rahul can copy in an hour = 5 – 4 =1 page

Hence, pages completed by Rahul in 20 hours = 20 × 1 = 20 pages

A and B can do a piece of work in12 days, B and C in 15 days, C and A in 20 days. How long would each take separately to do the same amount of work?

30, 60, 20

30, 20, 60

20, 15, 30

15, 20, 35

Solution: ………………………… (I)

…………………………….. (II)

……………………………..(III

Adding (I), (II) & (III)

= ……….(IV)

Solving (I), (II), (III) & (IV)

Time taken by A = 30 days

Time taken by B =20 days

Time taken by C =60 days

Pipes P & Q can fill up a tank in 4 hours & 8 hours respectively and pipe R can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in: Ratio And Proportion Important Formulas Concepts Essay.

10 hours

7 hours

16/7 hours

24/7 hours

Solution: Using time and work equation:

Thus, T = 24/7 hours

Three taps A,B & C can fill a tank in 12,15& 20 hours respectively. If A is open all the time, B and C are open for one hour each alternately, the tank will be full in:

6 hours

6hours

.5 hours

7 hours

Solution: Work completed by A & B in 1 hour =

Work completed by A & C in 1 hour =

Work completed in 2 consecutive hours =

Work completed in 6 hours =

Remaining work = 1-

Now A & B can complete the remaining work in 1 hour

Total work is completed in 6 + 1 = 7 hours

12 buckets of water fill a tank when the capacity of each bucket is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each tank is 9 litres? Ratio And Proportion Important Formulas Concepts Essay.

18

Solution: Using concept of variation:

12 × 13.5 = x × 9

x = 18

A cistern is filled in 9 hours but it takes 10 hours when there is a leak in its bottom. If the cistern is full, in what time shall the leak empty it?

90 hours

94 hours

92 hours

91 hours

Solution: Let the time take by leak to empty the tank be x hours.

Using work equation

x =90 hours

From a leaking tap ‘a’ drop comes out in ‘b’ minutes. If there are ‘c’ drops in a litre, then in how many hours one litre of water will be wasted? Ratio And Proportion Important Formulas Concepts Essay.

Solution: Given ‘a’ drop falls in b minutes or hours

1 drop falls in hours

Hence, C drops will fall in hours

If a tanker is normally filled by a tap in 8 hours, but suddenly a leak develops and it empties the full tanker in 24 hours, the cistern will be filled (with a leak) in:

10 hours

12 hours

15 hours

18 hours

Solution: Using time equation,

=

t = 12 hours

Speed and Distance

Important Formulas/Concepts

Speed =

Time =

Distance = Speed × Time

If ratio of speeds of A and B is x : y, then ratio of time taken by them to cover the same distance = :

Illustration 1: Walking at 5/7th of his usual rate, a boy reaches his school 6 minutes late. Find his usual time to reach the school. Ratio And Proportion Important Formulas Concepts Essay.

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6 minutes

8 minutes

12 minutes

15 minutes

Solution: Let the usual speed of the boy be x, then his slower speed= x,

Also let the school be at a distance of D

Now using SDT equation,

= 6

D = 15x

Usual time to reach the school =15 minutes

Illustration 2: A train completes a journey without stopping in 8 hours. If it had travelled 5 km an hour faster, it would have completed the journey in 6 hours 40 minutes. What is its slower speed? Ratio And Proportion Important Formulas Concepts Essay.

23 km/hr

30 km/hr

25 km/hr

42 km/hr

Solution: Using distance relation for equal travel

Slow speed × More time= Fast speed × Less time

x × 8= (x × 5)

Thus x=25 km/hr.

Illustration 3: Without stoppages, a train travels certain distance with an average speed of 80 km/hr, and with stoppages; it covers the same distance with an average speed of 60 km/hr. How many minutes per hour does the train stop? Ratio And Proportion Important Formulas Concepts Essay.

20 minutes/hours

15 minutes/hours

12 minutes/hours

8 minutes/hours

Solution: Using the formula,

Stoppage time/hours =

hours

= hours

=15 minutes

Illustration 4: If a car moves from A to B at a speed of 60 km/hr, and comes back from B to A at a speed of 40 km/hr, then its average speed is:

46 km/hr

50 km/hr

56 km/hr

48 km/hr

Solution: Average speed =

= 48 km/hr

Illustration 5: A car during its journey travels 30 minutes at a speed of 40 km/hr, and another 45 minutes at a speed of 60 km/hr, and another 2 hours at a speed of 70 km/hr. Find its average speed.

65 km/hr

56 km/hr

48 km/hr

63 km/hr

Solution: Average speed =

=

= 63 km/hr

Practice Exercise

Two friends X and Y walk from A to B, a distance of 39 km, at 3 km/hour and 3 ½ km/hour respectively. Y reaches B, returns immediately and meets X at C. Find the distance from A to C. Ratio And Proportion Important Formulas Concepts Essay.

35 km

36 km

28 km

32 km

Solution: Let the distance of C from B be x

For same time, equation will be:2………………..(i)

On solving equation (i)

x = 3

Thus, distance A to C = 39-x = 39 – 3 = 36km

A long distance runner runs 9 laps of a 400 meters track every day. His timings for four consecutive days are 88, 96, 89 and 87 respectively. On an average how many meters/minutes does the runner cover?

36 meters/minute

12 meters/ minute

40 meters/minute

48 meters/minute

Solution: = 40 meters/minute

Average speed = = 40 m/min

A thief runs away with a car at a speed of 40 km/hr. The theft was discovered after half an hour and the owner sets off in another car at 50 km/hr. When will the owner overtake the thief from the time of theft? Ratio And Proportion Important Formulas Concepts Essay.

2 hours

hours

3 hours

1 ½ hours

Solution: Let the time taken by owner be t

Both thief and owner cover equal distances

50 t = (t + ) 40

Thus, T = 2 hours

He will be a caught after 2 ½ hours

The speed of the boat in still water is 12 km/hr and the speed of the stream is 3 km/hr. A distance of 27 km, going upstream, will be covered in:

3 hours

5 hours

4/3 hours

3/2 hours

Solution: Time taken to row upstream =

= = 3 hours

A boat rows 20 km upstream and 30 km downstream in 5 hours each. The velocity of the stream is: Ratio And Proportion Important Formulas Concepts Essay.

3km/hr

2 km/hr

1 km/hr

4 km/hr

Solution: Let the velocity of the stream be y & velocity of the boat be x

For upstream = 5, = 5

x – y = 4 ….(I)

x + y = 6…..(II)

Solving (I) & (II)

y =1 km/hr.

A 270 meter long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 seconds. What is the length of the other train?

230 meters

240 meters

260 meters

320 meters

Solution: Relative speed of train = (120 + 80)km/hr

200km/hr = 200 ×

500/9 m/s

Length of the train

Thus, x =230

A train of length 150 meters takes 40.5 seconds to cross a tunnel of length 300 meters. What is the speed of the train in km/hr.? Ratio And Proportion Important Formulas Concepts Essay.

13.33

26.67

40

66.67

Solution: Speed=

= 40 km/hours

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at the rate of 9 km/hr. The distance travelled on foot is:

14 km

15 km

16 km

17 km

Solution: Let the distance travelled on foot be ‘x’ km

Using time equation,

+ = 9

9x + 4(61 – x) = 9 × 36

9x + 244 – 4x = 324

5x = 80

x =16 km

Thus, distance travelled by farmer on foot is 16km

Two cars P & Q start at the same time from A & B which are 120 km apart. If the two cars travel in opposite directions, they meet after 1 hour and if they travel in the same direction, then P meets Q after 6 hours. What is the speed of the car P? Ratio And Proportion Important Formulas Concepts Essay.

60 km/hr

70 km/hr

100 km/hr

120 km/hr

Solution: Let the speed of car P be ‘x’ & speed of car Q be ‘y’

Then, the two equations will be:

x + y =120….(I)

6x – 6y = 120… (II)

Solving (I) & (II)

x =70 km/hr.

A bus while moving at an average speed of 40 km/hr reaches its destination on time. When its average speed becomes 35 km/hr, then it reaches 15 minutes late. Find the length of Journey.

30 km

40 km

70 km

80 km

Solution: Using time equation,

x =

x = 70 km

Number System

Important Formulas/Concepts

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

a2 + b2 = (a + b)2 – 2ab

(a – b)2 = a2 – 2ab + b2

a2 + b2 = (a – b)2 + 2ab

a2 – b2 = (a + b)(a – b)

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

(a + b)3 = a3 + b3 + 3ab(a + b)

(a – b)3 = a3 – b3 -3ab (a-b)

a3+ b3 = (a + b)3 – 3ab(a + b)

a3 + b3 = (a + b)(a2 – ab + b2)

a3 – b3 = (a – b)3 + 3ab (a – b)

a3 – b3 = (a – b)(a2 + ab + b2)

A number is divisible by

2

Its unit digit is any of 0,2,4,6,8

3

The sum of its digits is divisible by 3

4

The sum of its last two digits is divisible by 4

5

Its unit digit is either 5 or 0

6

It is divisible by both 2 and 3

8

The number formed by its last 3 digits is divisible by 8

9

The sum of its digits is divisible by 9

10

Its unit digit is 0

11

The difference of the sum of its digits at odd places and that at even places is either 0 or a number divisible by 11.

Illustration 1: The sum of two numbers is 25 and their product is 114. What is the sum of the reciprocals of these numbers?

5/19

19/5

114/25

25/114

Solution: Given- x + y =25&xy =114

Required answer = =

Illustration 2: Which of the following is not the perfect square?

1, 99, 809

1, 41,376

6, 59,344

17,82,362

Solution: Since the square of natural number never ends with 2, 3, 7, 8

1782362 is not a perfect square, as it ends with 2. Ratio And Proportion Important Formulas Concepts Essay.

Illustration 3: If the number 812×74 is completely divisible by 11, then smallest whole number in place of x will be:

1

Solution: Divisibility of 11 = (Sum of number at odd place) – (Sum of number at even place)

Thus, it should be either O or a number divisible by 11

= (8+2+7) – (1+x+4)

= (12-x)

Hence, minimum value of x should be 1 so as to make the number divisible by 11.

Illustration 4: The difference between place values of both 6s in the numerical 376982604 is:

4404

59,99,400

5,99,400

Solution: Place values of two 6s are 6000000 & 600

Difference = 6,00,0000 – 600 = 5999400.

Illustration 5: If the number78xis divisible by 6, then the value of x is:

6

Solution: For any number to be divisible by 6, it must be divisible by both 2 and 3

6 is the correct answer.

Since, 786 is divisible by:

2(the last digit is even)

3(7+8+6 =21 which is divisible by 3)

Practice Exercise

How many prime numbers are less than 30?

10

Solution: Prime numbers below 30 are ‘2, 3, 5, 7, 11, 13, 17, 19, 23, 29’. Thus, 10 is the correct answer.

=

105

10 -55

10 -12

Solution:

106+5-7

= 104 = 105

The value of 102×98 is:

996

986

9996

9986

Solution: (100+2) (100-2) =1002 -22

= 10000 – 4

=9996

=

Solution: = =

If f(x) = x2-6x – 5, then f(2) –f(-2)=

-24

-48

Solution: f (x) =x2-6x – 5

f (2) =22 -6(2) -5 = 4 -12 -5 =-13

f (-2) = (-2)2 -6(-2) -5 =4 +12 -5 = 11

f(2) – f (-2) = -13 – 11 =-24

Which of the following fractions has a non-terminating repeating decimal? Ratio And Proportion Important Formulas Concepts Essay.

33/50

17/625

171/800

17/90

Solution: To get a non-terminating repeating decimal expansion:

Denominator 2m 5n

50 =21×52,625 = 20× 54,800 = 25 ×52, 90 =21×32×5

Therefore, the answer is

Rational form of 0. is:

2/3

3/5

33/50

333/500

Solution: Let x = 0.……..(i)

x = 0.6…….(ii)

Multiplying equation (ii) by 10

10x = 6.…..(iii)

Subtracting (iii) from (i)

9x = 6

x = 6/9 = 2/3

is:

An integer

A rational number

An irrational number

None of these

Solution: = =

Here, 2 is a rational number& 3 isirrational.

Now, we know that Rational ×Irrational = Irrational

Hence,is an irrational number.

Rational form of 2.4 is:

12/5

22/9

11/7

6/5

Solution: Let, x =2.….. (i)

x = 2.4…… (ii)

On multiplying equation (ii) by 10

10x = 24.…… (iii)

Now, subtract (i) from (iii)

9x = 22

x =

The decimal expansion of the number 27683/625 will terminate after: Ratio And Proportion Important Formulas Concepts Essay.

One decimal places

Two decimal places

Three decimal places

Four decimal places

Solution: =

=

The correct answer is (d) as the number ends after four decimal expressions.

Average

Important Formulas/Concepts

Average =

Let average age of existing A people be x, and the number of new people be n, and increase in average age be y years. Ratio And Proportion Important Formulas Concepts Essay.

Then,

If n new people join,

If n people join and the average age increases, the average age of n people = x +

If n people join and the average age decreases, the average age of n people = x –

If n people leave,

If n people leave and average age increases, then average age of n people = x +

If n people leave and average age decreases, then average age of n people = x –

Illustration 1: If the mean of x, x + 3, x + 5, x + 7 & x + 10 is 9, then the mean of last three observations is:

11

Solution: Average of 5 numbers = 9

= 9

Thus, x = 4

Hence, the last 3 numbers are (4+5), (4+7) and (4+10)

Average (9, 11, 14) =

=

= 11 .

Illustration 2: If the mean of five observations is 4, then the mean of same observations after increasing 5 is:

9

Solution: Original sum of five observations = (x1+x2+x3+x4+x5) = 5×4 =20

Increase in sum after adding 5 to each observation

= (x1+5) + (x2+5)+(x3+5)+(x4+5)+(x5+5)

= (x1+x2+x3+x4+x5) + (5×5)

= 20 + 25

= 45

New mean = = 9

Illustration 3: The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. The corrected mean is: Ratio And Proportion Important Formulas Concepts Essay.

38.6

39.4

39.8

39.2

Solution: Wrong sum of 50 observations =50× wrong mean

= 50×39

=1950

Corrected sum =wrong sum –wrong reading +correct reading

=1950 – 23+ 43 =1970

Corrected mean = =39.4

Illustration 4: The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is:

38

Solution: Sum of five numbers =5×mean

= 5×30

=150

Sum of remaining four numbers =4× New mean

= 4× 28

= 112

The excluded number = sum of five observations – sum of four observations

= 150-112

= 38

Illustration 5: The mean of 31 results is 60. If the mean of first 16 results is 58 and that of the last 16 results is 62, the 16th result is:

60

Solution: Sum of first sixteen results = 16 × mean

= 16 × 58

= 928

Sum of last sixteen results = 16 × mean

= 16 × 62 = 992

Sum of all thirty one results = 31 × mean

= 31 × 60

= 1860

Sixteenth observation = (928 + 992 – 1860)

= 60

Practice Exercise

The average cost of 5 jeans & 7 shirts is Rs. 600 and the average cost of 5 jeans is Rs. 740 , then the average cost of 7 shirts is:

500

560

450

480

Solution: Total cost of 5 jeans & 7 shirts =Average (5J, 7S) ×12

= 600×12

= 7200

Total cost of 5 jeans = Average (5J) ×5

=740×5

=3700

Total cost of 7 shirts =7200-3700

= 3500

Thus, average cost of 7 shirts = = 500

At a shop ‘book A’ was priced thrice as ‘book B’ and ‘book B’ was priced 9 more than twice the price of ‘book C’.Ratio And Proportion Important Formulas Concepts Essay. If average price of the three books is Rs. 360, the price of ‘book C’ is:

310

312

256

116

Solution: Let the prices of books A, B and C be x,y and z respectively

x = 3y

y = 9 + 2z

According to given condition

= 360

y = 241

z =

Z = (y-9)/2 = (241-9) /2 = 116

Thus, the cost of book C = 116

Three years ago, the average age of Marsh & Mathew was 16. With Steve joining them now, the average age becomes 20. The present age of Steve is:

22

Solution: Sum of ages of Marsh and Mathew presently =16×2+3×2 =38

Sum of ages of Marsh, Mathew and Steve =20×3 =60

Therefore, Steve’s age =60-38 =22 years

A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was:

85

Solution: Let the number of wickets till the last match be ‘x’

New average after 5 more wickets for 26 runs is decreases by 0.4

= 12

The average of 5 subjects increased by 5, when marks in English were increased in rechecking. Ratio And Proportion Important Formulas Concepts Essay. If the marks in English before rechecking were 45, then the marks after rechecking in English are:

70

Solution: Total increase in marks after rechecking =5×5 =25

Therefore, corrected marks in English = 45+25=70

If the average of three numbers a, b and c is A, then the average of a, b, c and A is:

A/2

A

Solution: Given,

Thus,a+b+c = 3A

Average of a, b, c & A =

=

= = A

The total production of 10 tea estates is 550 tonnes. By opening two new tea estates the average increases by 3 tonnes. The average cash production of these two new tea estates (in tonnes)is:

573

570

568

564

Solution: Total production of 10 tea estates =550×10

= 5500 tonnes

Total production of 12 tea estates =553×12

= 6636 tonnes

Production by 2 new estates = 6636 -5500

= 1136

Average production by each new estate == 568

Find the average increase rate if increase in the population in the first year is 30% and that in the second year is 40%.

35%

38%

40%

41%

Solution: Population initially =100%

Population after 1 year =130%

Population after 2 years =130%+40% (of 130%)=182%

Thus, increase in population in 2 years =82%

Average increase = 82%/2 =41%

A person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively. What was his average speed during the whole journey?

xyz / (xy+yz+zx)

(xy+yz+zx) / xyz

3xyz / (xy+yz+zx)

None of these

Solution: Let, the equal distance be ‘D’ kms

Now,

Average speed =

=

The average salary of the entire staff in an office is Rs. 3200 per month. The average salary of officers is Rs.6800 and that of non-officers is Rs.2000. If the number of officers is 5, then find the number of non-officers in the office. Ratio And Proportion Important Formulas Concepts Essay.

15

Solution: Let the number of non-officers in the office be x

Now, average salary, 3200=

3200(x+5) =3400+2000x

Thus, x=15

HCF/LCM

Important Formulas/Concepts

Product of 2 numbers = HCF of 2 numbers × LCM of 2 (same) numbers

HCF of fractions =

LCM of fractions =

Illustration 1: If the HCF of 27 and 63 is of the form (63 – 27m), then value of m is:

2

Solution: HCF [27, 63] = 9

Also,HCF [27, 63] =63-27m (given)

63-27 m = 9

Thus, m = 2

Illustration 2: What is the largest number that divides 70, 97,125, leaving remainder 5, 6, & 8 respectively?

13

Solution: H C F [70-5, 97-6,125-8]

H C F [65, 91,117] = 13

Illustration 3: If r & s are positive integer such that r = a3 b2 & s = a2 b3 ,then their LCM is:

a2 b2

a3 b3

a6 b6

a b

Solution: L C M [r, s] = L C M [a3b2,a2b3]

=a3 b3

Illustration 4: Which among the following is a pair of co-primes?

(14, 21)

(18, 25)

(31, 62)

(32,62)

Solution: We know that, co-primes have their H C F=1

(18, 25) is pair of co-prime becauseits H C F=1

Illustration 5: What is the least number that is divisible by all prime numbers from 1 to 6? Ratio And Proportion Important Formulas Concepts Essay.

60

120

Solution: LCM [1, 2, 3, 4, 5, 6]

LCM [1, 2, 3, 2 = 1 = 60

Practice Exercise

Three measuring rods are 30, 45 & 60 cm in length. Find the least length of cloth that can be measured an exact number of times, using any one of the rods.

180

240

360

540

Solution: L C M [30, 45, 60] is the least length of cloth that can be measured exact number of times

L C M [30, 45, 60] =[2×3×5,3×3×5,2×2×3×5]

=2×3×5×3×2

=180cm

The HCF & LCM of two numbers is 16 & 2304. If one number is 256 then other number is:

169

144

125

Solution: We know that:

Product of numbers =H C F ×L C M

Thus, 256× x =16×2304

Thus, x=144

By what number should 162 be divided to get 10 as a quotient and 12 as a remainder?

15

Solution: Let the divisor be ‘x’

It is known that:

Dividend = Divisor × Quotient + Remainder

162 = x×10+ 12

Thus, x = 15

The least number of four digits which is exactly divisible by 12, 15 & 18 is:

1120

1060

1080

1180

Solution: L C M [12, 15, 18] = [2×2×3, 3×5, 2×3×3]

2×2×3×3×5

=180

Now, least four digits number is 1000, when divided by 180 gives remainder as 100

Least four digit number divisible by12, 15, 18 is (1000 +180-100)

=1080

What is the least number which when doubled will be exactly divisible by 6, 9, 12 and 15?

90

Solution: Let the number be x

L C M [6, 9, 12, 15] = [2×3,3×3,2×2×3, 3×5]

=2×2×3×3×5

=180

Now 2x =180

x =90

The HCF of (23×33×52),(22×33×52) & (24×3×53×7) is:

300

Solution: HCF of (23×33×52), (22×33×52) & (24×3×53×7)

=22×3×52

= 4×3×25

=12×25

=300

Richard takes 18 minutes to complete a circular track while John takes 12 minutes to complete the same track. If both start together in same direction, they both will meet again after: Ratio And Proportion Important Formulas Concepts Essay.

12 minutes

18 minutes

24 minutes

36 minutes

Solution: Richard and John will meet at the L C M (18, 12) minutes

L C M (18, 12) = 2×2×3×3

= 36minutes

The numbers nearest to 10,000 but greater than 10,000 which is exactly divisible by 5, 6, & 8 is:

10020

1040

10080

1025

Solution: We know that:

L C M (5, 6, 8) = 2×2×2×3×5

=120

Now, to obtain the number nearest but greater than 10,000 divisible exactly by 5, 6, & 8,

We will divide 10,000 by L C M (5, 6, 8)

= 10000/120 =

Required number =120 × 84

=10,080

The smallest fraction, which each of 2/7, 3/5, 4/21 will exactly divide is:

12/7

6/14

36/7

24/14

Solution: To obtain a number that is divisible by 2/7, 3/14 &4/21:

=12/7

The maximum number of students among whom 960 books and 720 copies can be distributed under “Sarv Shiksha Abhiyan” government scheme in such a way that each student gets the same number of books and copies is:

120

240

360

420

Solution: For distribution of equal number of books and copies,

We will have to take out the H C F of 960 and 720.

= 240

Percentage

Important Formulas/Concepts

Percentage equivalents of oft-appearing fractions:

1/10

10%

1/9

11.11%

1/8

12.5%

1/7

14.28%

1/6

16.66%

1/5

20%

1/4

25%

1/3

33.33%

1/2

50%

2/3

66.66%

¾

75%

If P is the current population of a town, and it increases by R% per annum, then population after n years:

P n

If P is the current population of a town, and it increases by R% per annum, then population n years ago:

If A is R% more than B, then B is less than A by:

If A is R% less than B, then B is more than A by:

Illustration 1: 30% of a number subtracted from 91, gives the number itself.Find the number. Ratio And Proportion Important Formulas Concepts Essay.

60 kg

65 kg

70 kg

75 kg

Solution: Let x be the number

According to given condition,

91 – 30% x = x

91 – x = x

91 =

x = 70

Illustration 2: Stuti spends 20% of her monthly income on her household expenditures, 15% of the rest on books, 30% of the rest on clothes and saves the rest. On counting, she came to know that she finally saved Rs. 9520. Find her monthly income.

10,000

25,000

20,000

12,000

Solution: Let her total income be x

x -20% x – = 9520

– -9520

= 9520

Illustration 3: 30% of x% of y is 150% of y% of z. Which of the following is z?

0.25 x

1.2 x

5/6 y

0.2 x

Solution: 30% × x% × y = 150% × y% × z

= z

0.20x = z

Illustration 4: Two years ago, Ravi used to purchase 2 mangoes more than today which he can afford at Rs. 40. If the price is raised by 20%, then what is the cost of a dozen mangoes today?

32

Solution: Let the cost of a mango 2 years ago be ‘x’

Then, the cost of a mango presently after 20%:

Price rice will be

According to given condition, Ratio And Proportion Important Formulas Concepts Essay.

– =3

x = Rs.

Cost of a mango today will be

Cost of a dozen mangos =12 × =32

Illustration 5: Two numbers are less than the third number by 25% and 40% respectively. How much percent is the second number less than the first?

15%

20%

25%

40%

Solution: Let the third number be ‘x’, then the first number = 75% x

Second number =60% x

Difference in the first & second number =15% x

Required percentage = ×100 = 20%

Practice Exercise

5% of income of A is equal to 15% income of B and 10% of income of B is equal to 20% of income of C. if C’s income is Rs. 2000, then the total income of A, B &C is:

18,000

16,000

10,000

8000

Solution: 5 % A = 15% B….. (I)

10% B = 20% C….. (II)

From (II)

B = × 20000

B = 4000

From (I)

A = × B

A = 3 × 4000 = 12,000

Therefore total income of A, B, C =12000 + 4000 + 2000 = 18,000

A building worth Rs. 12, 16, 700 is constructed on a land worth Rs. 4, 91, 300. After how many years will the values of both will be the same if the land appreciates at 15% per annum and the building depreciates at 15% per annum? Ratio And Proportion Important Formulas Concepts Essay.

1 year

1 ½ years

2 ½ years

3 years

Solution: According to the given condition,

4, 91, 300 × n = 12,16,700 × n

n ×

n =3

If the price of petrol is increased by 20%, by how much percent a car owner must reduce his consumption in order to maintain the same budget?

16 %

33 %

40%

23%

Solution: We know that:

Decrease in consumption = 1 ×100

=

37 ½% of the candidates in examination were girls, 75% of the boys and 62 ½% of the girls passed and 342 girls failed.The total number of boys failed were:

350

360

370

380

Solution: Let the total population of the class be x,

Number of girls failed:

37.5% × 37.5% × x = 342

(375/100)2 x = 342

x = 2432

Number of boys who failed:

= 380

If the price of a cricket bat is first decreased by 15% and then increased by 25%, then the net change in price of bat will be:

6.25%

7.5%

10%

Solution: Let the original price of the bat be Rs 1000

Price after decrement of 15% = 85% × 1000 = Rs 850

Further, price after increment of 25%:

= 125% × 850 = 1062.5

Net change in price =

The difference between one fifth of 1000 and one fifth percent of 1000 is: Ratio And Proportion Important Formulas Concepts Essay.

998

198

800

= 198

If price of sugar is increased by 7%, then by how much percent should a housewife reduce her consumption of sugar so as to incur no extra expenditure?

93%

6

12%

Solution: Decrease in consumption:

%=

The daily wage is increased by 15% and a person now gets Rs. 23 per day. What was his daily wage before the increase?

20

Solution: Original daily wage= ×100

=Rs. 20

The ratio of salary of a worker in July to that in June was. By what % was the salary of July more than salary of June?

10%

6 %

Solution: ×100

× 100

= 11

The population of a village is 1,00,000. The rate of increase is 10% per annum. Find the population increase in the third year?

12,100

10,000

33,100

21,000

Solution: Population after 2nd year =100000 × 2

=100000 ×

=121000

Population after 3rd year =100000 × 3

100000 × = 133100

Increase in population:

= 133100 – 121000

= 12,100

Profit, Loss and Discount

Important Formulas/Concepts

Profit = Selling Price – Cost Price

Loss = Cost Price – Selling Price

Profit % =

Loss % =

Selling Price = Cost Price

Cost Price = Selling Price

Discount = Marked Price – Selling Price

Rate of discount =

Illustration 1: A man purchased two watches for Rs. 560. He sold one at a 15% profit and the other at a 10% loss, and thus he neither gains nor loses. Find the cost price of each watch. Ratio And Proportion Important Formulas Concepts Essay.

320, 240

240, 320

224, 336

336, 224

Solution:CP1 ×0.15 =CP2 × 0.10

CP1/CP2 = 2/3…….(I)

CP1 + CP2 = 560………..(II)

On comparing (I) & (II)

CP1=224

CP2= 336

Illustration 2: A man buys 10 pencils for Rs. 3 and sold 8 of them for Rs 3. His gain percent is:

20%

25%

30%

27%

Solution: CP of 1 pencil = 3/10

SP of 1 pencil = 3/8

Gain% = × 100

= × 100

= 25%

Illustration 3: A radio costing Rs. 500 is available on 10% discount on cash purchase.The shopkeeper gives sequential discounts. Asha paid Rs 427.50 for the radio. Find the rate of the discount on cash price.

20%

15%

10%

5%

Solution: Price after 1st discount of 10% = 90%

= 450

If final cash price is Rs 427.50

Discount % =

= 5%

Illustration 4: A person marks his goods 20% higher than cost price and allows a discount of 5%. His percentage profit is:

15%

20%

14%

Solution: Let the cost price be 100,

Marked Price = 120% Cost Price

= × 100

= 120

Selling Price = 95% of Marked Price

=

= 19

= 114

=

= 14%

Illustration 5: A fruit seller has 24 kg of apples. He sells a part of these at a gain of 20% and the remaining at a loss of 5%. If on the whole he earns a profit of 10%, the amount of apples sold at a loss is:

4.6 kgs

6 kgs

9.6 kgs

11.4 kgs

Solution: Let the number of apples sold at a loss be x and the cost price of an apple because be Rs 1.

1× (24-x) ×120% – 1×x×95% =1×24×110%

(24-x)×6/5 – x×19/20 =24× 11/10

Thus, x =9.6 kgs

Practice Exercise

A retailer buys 30 articles from a wholesaler at the price of 27. If he sells them at their marked price the gain percent in the transactions is:

11

Solution: Let the marked price of one article be Rs.1

C P of 30 articles = 27

& S P/ M P of 30 articles =30

Profit %= 3/27×100=100/9

=%

A dishonest shopkeeper professes to sell potatoes at the cost price, but he weighs 875 grams instead of one kg. What is his percentage of profit?

6.5%

12.5%

18.75%

15.25%

Solution:Profit% =

= 12.5%

A man sold his book for Rs. 891, thereby gaining 1/10th of its costprice.The cost price is:

850

810

851

840

Solution: Selling Price = Cost Price + 1/10th Cost Price

891 = Cost Price

Cost Price =

= 810

A shopkeeper increased the price of a product by 50% and later on reduced the price by 50%.The shopkeeper’s loss was:

2.5%

25%

10%

Solution: Let the actual price of the article be Rs. 100

Price after increasing 50% =150% × 100

= 150

Price after reducing 50% = 50% × 150

= 75

Loss % = × 100 = 25%

A CD music system when sold at a certain price gives a gain of 20%. If sold for thrice that price, the gain percent will be:

260%

200%

360%

300%

Solution: Let the selling price of CD music system be x

Cost Price = =

If the new Selling Price is thrice that of Cost Price, then = 3

=

Profit % =× 100

× × 100

= 200%

A shopkeeper earns a profit of 15% on selling a book at 10% discount on the printed price. The ratio of the cost price and the printed price of the book is.

23:18

18:23

3:2

2:3

Solution: Let the Selling Price be x

Then CP =%

& Marked Price = %

= =

The cost price of a shirt and a pair of trousers is Rs. 371. If the shirt costs 12% more than the trousers, than the cost price of the trousers is:

Rs. 125

Rs. 150

Rs. 175

Rs. 200

Solution: Let the Cost Price of the trousers be x

Then,

x + 112% x = 371

212 x = 37,100

x = 175

The cost of manufacturing an article is made up of material, labour and overheads in the ratio 4: 3: 2. If the cost of laborers is Rs. 45, find the profit percent if the article is sold for Rs. 180.

50%

33.33%

25%

20%

Solution: Let the total cost of manufacturing be x

Then labour cost, x = 45

x = 135

Selling Price = 180 (given)

Gain% =

× 100

= 25%

A trader purchases apples at Rs. 60 per hundred. He spends 15% on the transportation. What should be the selling price per dozen to earn a profit of 20%?

Rs. 8.21

Rs. 9.936

Rs. 10.2

Rs. 3.362

Solution: Total Cost Price of 100 Apples = 60 = 69

of 100 apples earning profit 20% =

Selling Price of 1 dozen apples = = 9.936

A shopkeeper purchases 10 kg of rice at Rs.600 and sells at a loss as much as the selling price of 2 kg of rice. Find the sale of rice purchases/kg.

Rs.5

Rs.10

Rs.12.5

Rs.15

Solution: Let the Selling Price be Rs. x / kg

Loss = Cost Price –Selling Price

2x = 600 – 10 x

x = Rs. 5/ kg

Square Roots and Cube Roots

Illustration 1: If = than x=

17

Solution: =

On squaring both sides – 1+

-1

Thus, x = 17

Illustration 2: If x+ 1/x = 5, then the value of x2 + 1/x2 =

23

Solution: x += 5

On squaring both sides, we get:

(x+) 2 = 52

x2+ + 2 = 25

x2 + = 23

Illustration 3: 1+

1.732

1.414

2

None of these

(Take = 1.414, = 1.732, = 2)

Solution: 1 + + +

1 – (1 –2) – (2 – 3) – ( 3 – 4)

1 – 1 + 2 – 2+ 3 – 3 + 4

= 4

= 2

Illustration 4: If (x-7)2 + ( y–5 )2 +( z – 3)2 = 0,then value of x, y & z are …………… respectively.

√7,√ 5,√3

49,25,9

7, 5, 3

Can’t be determined

Solution: Since, x – 7 = 0 = x = 7

Also, y – 5 = 0 = y =5

z – 3 = 0 = z = 3

Illustration 5:

– 0.4472

+ 0.4472

– 0.5773

+0.5773

(Take = 1.732, = 2.236)

Solution: +

=

= = -2.236 / 5 = – 0.4472

Practice Exercise

The greatest four digit perfect square is:

9962

9921

9801

9726

Solution: 99

9 9999

9 81

181 1899

1701

198

9999 – 198 = 9801

The smallest three digit perfect square is:

100

120

125

144

Solution: 100 = 102

100 is the smallest perfect square

If x then values of x are:

2,3

-2,3

-5,1

5,1

Solution: x =

x =

x2= 6 + x

x2– x- 6 =0

x2– x – 6

x2– 3x + 2x -6

x (x – 3) +2 (x – 3)

(x+2) (x-3)

x = -2, 3

The cube root of 0.000000729 is:

0.0027

0.027

0.0009

0.009 Solution:

=

= 0.009

If x = 5 – 2√6 , then x2 – 1/x2 =

36 √6

– 27 √6

– 40 √6

72 √6

Solution: x= 5 – 2 √6

x2 = 25 + 24 – 20 √6 = 49 – 20 √6

== 5 + 2 √6

= 25 + 24 + 20 √6 = 49 + 20√6

x2 – = 49 – 20 √6 -49-20

= – 40√6

The value of is not equal to:

6√36

3√144

18 √12

Solution: = = 36

= 6

= 3

What is the least number which must be multiplied in 4851 to form a perfect square?

11

Solution: Factor of 4851 = 3×3×7×7×11

So, 11 must be multiplied in 4851 to get a perfect square.

What is the least number that must be divided in864 to form a perfect square?

6

Solution: Factor of 864 = 2×2×2×2×2×3×3×3

So, 6 must be divided in 864 to get a perfect square.

:

Solution:

If a = 13 , b = 5 , find value of 3

1896

1728

343

216

Solution: 3

= 3

= 1728

Partnership

Illustration 1: A, B and C enter into partnership with a total of Rs. 8200. A’s capital is Rs. 1000 more man B’s capital and Rs. 2000 less man C’s capital. What is B’s share of the year’s profit of Rs. 2460?

560

920

820

420

Solution: A + B + C = 8200…..(i)

A = 1000+B => B = A – 1000

A = C – 2000 => C = A + 2000

Substituting in equation (i)

A + A – 1000 + A + 2000 = 8200

3 A = 7200

A = 2400

B = 1400

B’s share =

= 420

Illustration 2: What amount of money is divided between Atul, Mradul, &Ishaan, if Mradul&Ishaan together get Rs. 3000 and Atul gets five times as much as Mradul while Ishaan and Atul together get Rs 6000?

8750

8000

7500

6750

Solution: M + I = 3000 ………(i)

A = 5 M …………….(ii)

I + A = 6000………..(iii)

Substitute equation (ii) in (iii)

5M + I = 6000 ……….(iv)

Solving (i) & (iv)

M = 750

I = 2250

A = 3750

Hence, total money shared between the three is (750 + 2250 + 3750) = 6750

Practice Exercise

A, B & C enter into a partnership. A contributes Rs. 320 for 4 months, B contributes Rs. 510 for 3 months, and C contributes Rs. 270 for 5 months. If the total profit is Rs. 208, find the profit share of the partners.

78, 98, 32

64, 76.50, 67.50

90, 48, 70

84,45,79

Solution: 320 × 4 : 510 × 3 : 270 × 5

128 : 153 : 135

A’s share × 208 = 64

B’s share × 208 =76.50

C’s share × 208 = 67.50

A began a business with Rs. 1,250 and was joined afterwards by B with Rs. 3,750. When did B join if the profits at the end of the year are divided equally?

After 6 months

After 8 months

After 4 months

After 7 months

Solution: Say B joined the business after ‘x’ months

If the profits are shared equally,

1250 × 12 = 3750 ×(12 – x)

= 3(12 – x) x = 8

The cost of a music system and an LCD TV are in the ratio of 3:8 and total price of both is Rs. 20,900. The difference in their price is:

7, 600

6, 050

9, 500

8, 100

Solution: The ratio of the price is 3:8

Let the cost of the music system be 3x and the cost of LCD TV be 8x

Difference in their prices = × 20,900

× 20,900

= 9500

Four milkmen rented a pasture. ‘A’ grazed 24 cows for 3 months; ‘B’ grazed 10 cows for 5 months; C grazed 35 cows for 4 months and D grazed 21 cows for 3 months. If A’s share of rent is Rs. 720, find the total rent of the field.

3250

2670

4500

2750

Solution: Let the total rent be ‘x’

Ratio of rent shared by four milkmen is

A B C D

24×3 : 10×5 : 35×4 : 21×3

72 : 50 : 140 : 63

A’ share of rent = Rs. 720 (given)

× x = 720

Thus, x = 3250

Hrithik and Ravish entered into partnership with capitals in the ratio 4:5. After 3 months, Hrithik withdrew ¼ of his capital and Ravish withdrew 1/5th of his capital. If the gain at the end of 10 months was Rs. 760, Ravish’s share in this profit is:

Rs. 330

Rs. 360

Rs. 380

Rs. 430

Solution: Let initially the sum invested by Hrithik and Ravish be 4x & 5x

Ratio of their profit share is

4x × 3 + 4x ×7 : 5x × 3+ × 5x × 7

12x +21x: 15x +28x

33x : 43x

Ravish share in profit× 760 = 430

A workman earned Rs. 180 in a certain number of days. If his daily wages had been Rs. 2. he would take one more day to earn the same amount, find how many days he worked at the higher rate.

18 days

9 days

6 days

12 days

Let the no of days initially be x

Using many equation,

= 2

x2+ x – 90 = 0

x2+ 10x – 9x – 90 = 0

x (x + 10) – 9 (x + 10) = 0

x = 9 days

Ayush, Sujoy&Kartik hired a taxi for Rs. 780 and used it for 17, 8 & 14 hours respectively. The charges paid by Karthik are:

320

280

460

300

Solution: Kartik’s share = = Rs. 280

In a partnership A invested 1/6th of the capital for 1/6th of the time, B invest 1/3rd of the capital for 1/3rd of the time and C, the rest of the capital for the whole time. Out of a profit of Rs. 4600, B’s share is:

650

800

960

1000

Solution: Let x be the total amount & y be the total time period investment

Ratio of the profit for A, B, & C is

× x × 1 y

1 : 4 : 18

Profit share of B = 4600 ×

= 800

Adil & Akram invest in a grocery business in the ratio 4:3. If 9% of the total profit goes to certain taxes and Akram’s share is Rs. 35,100, the total profit is:

1,15, 000

1,09, 000

90,000

68, 000

Solution: Net profit after paying all taxes = 91% x

Given ratio 4:3

× 91% x =35,100

x =

x = 90, 000

A and B are partners in a business. A contributes ¼ of the capital for 15 months and B received 2/3rd of the profit. For how long B’ money was used?

ORDER A PLAGIARISM -FREE PAPER NOW

6 months

9 months

10 months

1 year

Solution: The invested ratio is 1: 3

& the profit ratio is 1:2

Equally born,

=

y = 10 months

Probability

Important Formulas/Concepts

Probability =

Illustration 1: If two dices are thrown at random, probability of getting a sum more than 9 is:

1/6

5/6

2/3

½

Solution: Favorable outcomes – (5, 5) (4, 6) (6, 4) (5, 6) (6, 5) (6, 6)

Also, total outcomes = 36

Probability = 6/ 36 = 1/6

Illustration 2: Probability of getting a black face card, when a card is drawn randomly from a 52 card deck is:

3/52

3/26

1/13

2/13

Solution: Favorable outcomes = 6

Total number of outcomes = 52

Probability =

Illustration 3: Find the probability of getting a prime number on throwing a dice.

¼

1/3

½

Solution: Favorable outcomes = 3; {2, 3, 5}

Total number of outcomes = 6

Probability = 3/6 = 1/2

Illustration 4: Find the probability of getting a composite number on throwing a dice.

1/3

½

2/3

Solution: Favorable outcomes = 2 ;{ 4, 6}

Total number of outcomes = 6

Probability = 2/6 = 1/3

Illustration 5: The probability of getting atleast 2 heads, on tossing a coin three times is:

½

3/8

¾

¼

Solution: The various possibilities (of getting at least 2 heads) are:

H HH

H H T

H T H

H T T

T H H

T H T

T T H

T TT

Thus, total number of favorable outcomes = 4

Total number of outcomes = 8

Thus, probability = 4/8 = ½

Practice Exercise

The probability of getting a doublet on throwing a pair of dice is:

1/2

1/3

1/6

5/6

Solution: Favorable outcomes = 6; {(1,1),( 2,2),(3,3),(4,4),(5,5),(6,6)}

Total number of outcomes =36

Probability = =

The probability of getting 53 Sundays in a leap year is:

1/7

2/7

3/7

4/7

Solution: Total complete weeks in a leap year =52

Also, remaining days in a leap year =2 days

Now, favorable outcomes = {(sat, sun), (sun, mon)}

Thus, 2/7 is the correct answer.

The probability of getting 53 Mondays in a non-leap year is:

1/7

2/7

3/7

4/7

Solution: Total number of complete weeks in a non-leap year = 52

Also, remaining days in a leap year =1

Number of favorable outcomes = 1

Total number of outcomes = 7

Hence, probability =

Among a group of 9 males & 6 females, a President & a Vice-President is to be appointed.Find the probability of occupying either both places by a female or one each by a male and a female. Ratio And Proportion Important Formulas Concepts Essay.

12/105

25/105

37/105

54/105

Solution:

=

A class consists of 12 boys and 8 girls. If a panel of 3 students is to be made, then the probability of choosing two girls and one boy is:

28/95

7/20

11/75

12/19

Solution: Number of ways of choosing 2 girls and 1 boys = 12c1×8c2

Total number of ways to choose = 20c3

To select 3 such students:

Required probability =

=

A lady wardrobe consists of five black pants, three brown pants and six white tops and four red tops. The probability of choosing a black pant with a red top for a party is:

4/5

1/5

1/4

2/5

Solution: Number of ways of choosing a black pant = 5c1

Number of ways of choosing a red top = 4c1

Total number of ways of choosing a pant and a top = 8c1×10c1

Required probability =

=

If two dices are thrown at random, find the probability of getting an odd number on both faces.

1/2

1/3

1/4

1/6

Solution: Number of favorable cases = {(1, 1) (3, 3) (5, 5), (1, 3) (3, 1)(5, 3), (1, 5)(3, 5)(5, 1)}

= 9

Total number of cases = 36

Required probability = =

If three coins are tossed simultaneously, find the probability of getting at most one tail.

½

3/8

¼

5/8

Solution: Number of favorable cases =4 ;{(H,H,H,)(H,T,H)(H,H,T)(T,H,H)}

Total number of outcomes = 8

Required probability =4/8 =1/2

From the deck of 52 cards, if a card is drawn at random, find the probability of getting either a Black jack or a Red king.

1/26

3/26

1/13

2/13

Solution: Number of ways of getting a black jack= 2c1

Number of ways of getting a red king =2c1

Total number of ways of getting a card =52c1

Required probability = =

If out of a sample of 40 pens, 12 are defective, find the probability of choosing a defective and a perfect pen.

12/30

84/195

2/19

7/40

Solution: Number of ways to select a defective pen=12c1

Number of ways to select a correct/perfect pen =28c1

Total number of ways of selecting two pens =40c2

Required probability =

=84/195

Ratio And Proportion Important Formulas Concepts Essay