APTITUDE MATHEMATICS BY YOGESH KUMAR YADUVANSHI

TIME AND DISTANCE -> IMPORTANT FACTS AND FORMULAE

1. Speed = [Distance/Time],

Time=[Distance/Speed],

Distance = (Speed*Time)

2. x km/hr = [x*5/18] m/sec.

3. If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is 1/a : 1/b or b:a.

4. x m/sec = [x*18/5] km/hr.

5. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. then, the average speed during the whole journey is [2xy/x+y] km/hr.

PROFIT AND LOSS -> IMPORTANT FACTS AND FORMULAE

Cost Price : The price at which an article is purchased, is called its cost price, abbreviated as C.P.

Selling Price : The price at which an article is purchased, is called its cost price, abbreviated as C.P.

Profit or Gain : The price at which an article is purchased, is called its cost price, abbreviated as C.P.

Loss : If S.Pis less than C.P., the seller is said to have incurred a loss.

1. Gain = (S.P.) - (C.P.)

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

3. gain% = [Gain*100/C.P.]

4. Loss = (C.P.) - (S.P.)

5. Loss% = [Loss*100/C.P.]

6. S.P. = (100+Gain%)/100 * C.P.

7. S.P. = (100-Loss%)/100 * C.P.

8. C.P. = 100/(100+Gain%) * S.P.

9. C.P. = 100/(100-Loss%) * S.P.

10. If an article is sold at a gain of say, 35%, then S.P. = 135% of CP

11. If an article is sold at a loss of say, 35%, then S.P. = 65% of C.P.

VOLUME AND SURFACE AREA -> IMPORTANT FACTS AND FORMULAE

I. CUBIOD

Let length = l, breadth = b and height = h units. Then,

1. Volume = (l x b x h) cubic units.

2. Surface area = 2 (lb + bh + lh)

II. CUBE

Let each edge of a cube be of length a. Then, 1. Volume = a³ cubic units.

2. Surface area = 6a² sq. units.

3. Diagonal = √3 a units.

III. CYLINDER

Let radius of base = r and Height (or length) = h Then,

1. Volume = (Πr²h) cubic units.

2. Curved surface area = (2Πrh) sq. units.

3. Total surface area = (2Πrh + 2Πr² sq. units)

= 2Πr (h + r) sq. units.

IV. CONE

Let radius of base = r and Height = h. Then,

1. Slant height, l = √h² + r ² units.

2. Volume = [1/3 Πr²h] cubic units.

3. Total surface area = (Πrl + Πr²) sq.units.

V. SPHERE

Let the radius of the sphere be r. Then,

1. Volume = [4/3 Πr3] cubic units.

2. Surface area = (4Πr²) sq. units.

VI. HEMISPHERE

Let the radius of a hemisphere be r. Then,

1. Volume = [2/3 Πr3] cubic units.

2. Curved surface area = (3Πr²) sq. units.

3. Total surface area = (3Πr²) sq. units.

Remember : 1 litre = 1000 cm³.

BOATS AND STREAMS -> IMPORTANT FACTS AND FORMULAE

I. In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream.

II. If the speed of a boat in still water is u km/ht and the speed of the stream is v km/hr, then :

Speed downstream = (u + v) km/hr

Speed upstream (u - v) km/hr.

III. If the speed downstream is a km/hr and the speed upstream is b km/hr, then :

Speed in strill water = 1/2 (a + b) km/hr

Rate of stream = 1/2 (a - b) km/hr

PARTNERSHIP -> IMPORTANT FACTS AND FORMULAE

I. Partnership : When two or more than two persons run a business jointly, they are called partners and the deal is known as partnership.

II. Ratio of Division of Gains :

(i) When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments.

Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at the end of the year :

(A’s share of profit) : (B’s share of profit) = x : y.

(ii) When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital * number of units of time). Now, gain or loss is divided in the ratio of these capitals.

Suppose A invests Rs. x for p months and B invests Rs. y for q months, then (A’s share of profit) : (B’s share of profit) = xp : yq.

III. Working and Sleeping Partners : A partner who manages the business is known as working partner and the one who simply invests the money is a sleeping partner.

BANKERS DISCOUNT -> IMPORTANT CONCEPTS

Bankers’ Discount : Suppose a merchant A buys googds worth, say Rs. 10,000 from another merchant B at a credit of say 5 months. Then, B prepares a bill, called the bill of exchange. A signs this bill and allows B to withdraw the amount from his bank account after exactly 5 months.

The date exactly after 5 months is called nominally due date. Three days (known as grace days) are added to it to get a date, known as legally due date.

Suppose B wants to have the money before the legally due date. Then he can have the money from the banker or a broker, who deducts S.I. on the face value (i.e., Rs. 10,000 in this case) for the period from the date on which the bill was discounted (i.e., paid by the banker) and the legally due date. This amount is known as Banker’s Dicount (B.D.)

Thus, B.D. is the S.I. on the face value for the period from the date on which the bill was discounted and the legally due date.

Banker’s Gain (B.G.) = (B.D.) - (T.D.) for the unexpired time.

Note : When the date of the bill is not given, grace days are not to be added.

BANKERS DISCOUNT -> IMPORTANT FORMULAE

I. B.D. = S.I. on bill for unexpired time.

II. B.G. = (B.D.) - (T.D.) = S.I. on T.D. = (T.D.)² / R.W.

III. T.D. = √P.W. * B.G.

IV. B.D. = [Amount * Rate * Time / 100]

V. T.D. = [Amount * Rate * Time / 100 + (Rate * Time)]

VI. Amount = [B.D. * T.D. / B.D. - T.D.]

VII. T.D. = [B.G. * 100 / Rate * Time]

CLOCKS -> IMPORTANT FORMULAE

The face or dial of a watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.

A clock has two hands, the smaller one is called the hour hand or short hand while the larger one is called the minute hand or long hand.

I. In 60 minutes, the minute hand gains 55 minutes on the hour hand.

II. In every hour, both the hands coincide onece.

III. The hands are in the same straight line when they are coincident or opposite to each other.

IV. When the two hands are at right angles, they are 15 minute spaces apart.

V. When the hands are in opposite directions, they are are 30 minute spaces apart.

VI. Angle traced by hour hand in 12 hrs = 360°.

VII. Angle traced by munute hand in 60 min. = 360°.

Too Fast and Too Slow : If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15 minutes too fast.

On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be 15 minutes too slow.

TRUE DISCOUNT -> IMPORTANT CONCEPTS

Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to Rs. 156 in 4 years. So, the payment of Rs. 100 now will clear off the debt of Rs. 156 due 4 years hence. We say that :

Sum due = Rs. 156 due 4 years hence;

Present worth (P.W.) = Rs.100;

True Discount (T.D.) = Rs. (156 - 100) = Rs. 56 = (Sum due) - (P.W.).

We define : T.D. = Interest on P.W.

Amount = (P.W.) + (T.D.).

Interest is reckoned on P.W. and true discount is reckoned on the amount.

TRUE DISCOUNT -> IMPORTANT FORMULAE

Let rate = R% per annum and Time = T years. Then,

I. P.W. = 100 * Amount / 100 + (R*T) = 100 * T.D. / R * T

II. T.D. = (P.W.)* R * T / 100 = Amount * R * T / 100 + (R * T)

III. Sum = (S.I.) * (T.D.) / (S.I.) - (T.D.)

IV. (S.I.) - (T.D.) = S.I on T.D.

V. When the sum is put at compound interest, then P.W. = Amount / [1+R/100]T;

PROBLEMS ON TRAINS -> IMPORTANT FORMULAE

1. a km/hr = [a * 5/18]m/s.

2. a m/s = [a * 18/5] km/hr.

3. Time taken by a trian of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.

4. Time taken by a train of length l metres to pass a stationary object of length b metres is the time taken by the train to cover (l + b) metres.

5. Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relatives speed = (u - v) m/s.

6. Suppose two trains or two bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s

7. If two trains of length a metres and b metres are moving in opposite directions at u

8. If two trains of length a metres and b metres are moving in the same direciton at u m/s and v m/s, then the time taken by the faster train to cross the slower train = (a + b)/(u - v) sec.

9. If tow trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then

(A’s speed) : (B’s speed) = (√b : √a).

SIMPLE INTEREST -> IMPORTANT FORMULAE

1. Principal : The money borrowed or lent out for a certain priod is called the principal of he sum.

2. Interest : Extra money paid for using other’s money is called interest.

3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years, Then,

(i) S.I. = [P * R * T / 100]

(ii) P = [100 * S.I. / R * T]

R = [100 * S.I / P * T] and T = [100 * S.I. / P * R]

PROBLEMS ON NUMBERS -> DESCRIPTION

In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyse the given conditions, assume the unknown the numbers and form equations accordingly, which on solving yield the unknown numbers.

AVERAGE -> IMPORTANT FACTS AND FORMULAE

I. Average = [Sum of observations / Number of observations]

II. Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is [2xy / x + y] kmph.

Numbers -> IMPORTANT FACTS AND FORMULAE

1. Natural Numbers :

Counting numbers 1, 2, 3, 4, 5, .. are called natural numbers.

II. Whole Numbers :

All counting numbers together with zero form the set of whole numbers. Thus,

I. 0 is the only whole number which is not a natural number.

II. Every natural number is a whole number.

III.Some Important Formulae :

I. ( 1 + 2 + 3 + .....+ n) = n (n + 1 ) / 2

II. (1 2 + 22 + 32 + ..... + n²) = n ( n + 1 ) (2n + 1) / 6

III. (1 3 + 23 + 33 + ..... + n³) = n² (n + 1)² / 4

SURDS ADN INDICES -> IMPORTANT FACTS AND FORMULAE

1. LAWS OF INDICES :

(i) am * an = am + n

(ii) am / an = am - n

(iii) (am)n = amn

(iv) (ab)n = anbn

(v) (a/b)n = an/ bn

(vi) a0 = 1

2. SURDS : Let a be rational number and n be a positive integer such

that a(1/n) = n√a

3 LAWS OF SURDS :

(i)               n√a = a (1/n)

(ii)            n√ab = n√a x n√b

(iii)          n√a/b = n√a / n√b

(iv)          (n√a)n = a

FOR GK NOTES

https://www.facebook.com/groups/135852876565706/

GK TRICKS BY YOGESH KUMAR YADUVANSHI

ए दोस्त जब जाना ही था

ऐ दोस्त जब जाना ही था,
तो पलकों के आँसू बन कर क्यों रह गये?
हर पल हम तुझे ही सोचते रहे,
दुआ में सिर्फ तेरा ही नाम लेते रहे,
दुआ कबूल करके खुदा भी प्यार से कहते गये,
कि ‘वो सब तो आगे चले गये,
फिर तुम क्यों यूँ ही भटकते रह गये,
ऐ दोस्त जब जाना ही था,
तो दिल की आश बन कर क्यों रह गये?
तुझे पाने की ख्वाहिश में हम खुद को भूल गये,
तुझसे बिछड़ने के गम में हम जीना ही भूल गये,
आज खुदसे इतने रुठे कि,
खुदा से मौत की भीख मांग बैठे
पता नहीं था मरने की होगी इतना तमन्ना
न जाने क्यों आज तक यूँ ही जीते रह गये,
ऐ दोस्त जब जाना ही था,
तो दु:ख में मेरा साथ छोड़ तुम कहाँ गये?
हम बेवक़ूफ़ हर दिन ये सोचते रहे कि,
कल मेरे दोस्त से बाते होंगी,
और दुनिया की हर खुशी मेरे पास होगी,
पर ना जाने अब वो दिन सिर्फ
सपनों में दिल की आशा बन कर रह गये,
ऐ दोस्त जब जाना ही था
तो मेरी सांसे बन कर क्यों रह गये?
ऐ दोस्त  जब जाना ही था
तो मुझे जिंदा छोड़ कर क्यों चले गये?

मुहब्बत

जिसने भी की मुहब्बत, रोया जरूर होगा।
वो याद में किसी के खोया जरूर होगा।

दिवार के सहारे, घुटनों में सिर छिपाकर ,
वो ख्याल में किसी के खोया जरुर होगा।

आँखों में आंसुओ के, आने के बाद उसने,
धीरे से उसको उसने, पोंछा जरुर होगा।

जिसने भी की मुहब्बत, रोया जरूर होगा।

अपने ही पीछे छोड़ जाते हैं

खवाहिश नही मुझे मशहुर होने की।
आप मुझे पहचानते हो बस इतना ही काफी है।
अच्छे ने अच्छा और बुरे ने बुरा जाना मुझे।
क्यों की जीसकी जीतनी जरुरत थी उसने उतना ही पहचाना मुझे।
ज़िन्दगी का फ़लसफ़ा भी कितना अजीब है,
शामें कटती नहीं, और साल गुज़रते चले जा रहे हैं....!!
एक अजीब सी दौड़ है ये ज़िन्दगी,
जीत जाओ तो कई अपने पीछे छूट जाते हैं,
और हार जाओ तो अपने ही पीछे छोड़ जाते हैं।

NUMBERS by yogesh kumar yaduvanshi

       

IMPORTANT FACTS AND FORMULAE

I..Numeral : In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number.

 A group of digits, denoting a number is called a numeral.

 We represent a number, say 689745132 as shown below :

Ten Crores (108)	Crores(107)	Ten Lacs (Millions) (106)	Lacs(105)	Ten Thousands (104)	Thousands (103)	Hundreds (102)	Tens(101)	Units(100)
6	8	9	7	4	5	1	3	2

We read it as : 'Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred and thirty-two'.

II  Place Value or Local Value of a Digit in a Numeral :

In the above numeral :

Place value of 2 is (2 x 1) = 2; Place value of 3 is (3 x 10) = 30;

Place value of 1 is (1 x 100) = 100 and so on.

Place value of 6 is 6 x 10⁸ = 600000000          

III.Face Value : The face value of a digit in a numeral is the value of the   digit itself at whatever place it may be. In the above numeral, the face value of 2 is 2; the face value of 3 is 3 and so on.

IV.TYPES OF NUMBERS

1.Natural Numbers : Counting numbers 1, 2, 3, 4, 5,..... are called natural

numbers.

2.Whole Numbers : All counting numbers together with zero form the set of whole
numbers. Thus,

        (i) 0 is the only whole number which is not a natural number.

        (ii) Every natural number is a whole number.

3.Integers : All natural numbers, 0 and negatives of counting numbers i.e.,
{…, -3,-2,-1, 0, 1, 2, 3,…..} together form the set of integers.

(i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers.

(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers.

(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor

negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while

{0, -1,-2,-3,…..} represents the set of non-positive integers.

4.  Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, etc.

5.  Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc.

6.  Prime Numbers : A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself.

Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,

47,  53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Prime numbers Greater than 100 : Let p be a given number greater than 100. To find out whether it is prime or not, we use the following method :

Find a whole number nearly greater than the square root of p. Let k > *jp. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime.

e.g,,We have to find whether 191 is a prime number or not. Now, 14 > V191.

Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.

191 is not divisible by any of them. So, 191 is a prime number.

7.Composite Numbers : Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12.

Note :    (i) 1 is neither prime nor composite.

(ii) 2 is the only even number which is prime.

(iii) There are 25 prime numbers between 1 and 100.

8.  Co-primes : Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes,

V.TESTS OF DIVISIBILITY

1.  Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.

Ex. 84932 is divisible by 2, while 65935 is not.

2.  Divisibility By 3 : A number is divisible by 3, if the sum of its digits is divisible by 3.

Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3.

But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3.

3.  Divisibility By 4 : A number is divisible by 4, if the number formed by the last two digits is divisible by 4.

Ex. 892648 is divisible by 4, since the number formed by the last two digits is

48,  which is divisible by 4.

But, 749282 is not divisible by 4, since the number formed by the last tv/o digits is 82, which is not divisible by 4.

4.  Divisibility By 5 : A number is divisible by 5, if its unit's digit is either 0 or 5. Thus, 20820 and 50345 are divisible by 5, while 30934 and 40946 are not.

5.  Divisibility By 6 : A number is divisible by 6, if it is divisible by both 2 and 3. Ex. The number 35256 is clearly divisible by 2.

Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.

6.   Divisibility By 8 : A number is divisible by 8, if the number formed by the last

three digits of the given number is divisible by 8.

Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8.

But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.

7.   Divisibility By 9 : A number is divisible by 9, if the sum of its digits is divisible

by 9.

Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9.

But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9.

8.   Divisibility By 10 : A number is divisible by 10, if it ends with 0.

Ex. 96410, 10480 are divisible by 10, while 96375 is not.

9.   Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.

Ex. The number 4832718 is divisible by 11, since :

(sum of digits at odd places) - (sum of digits at even places)

(8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.

10.  Divisibility By 12 ; A number is divisible by 12, if it is divisible by both 4 and

3.

Ex. Consider the number 34632.

(i) The number formed by last two digits is 32, which is divisible by 4,

(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is divisible by 4 as well as 3. Hence, 34632 is divisible by 12.

11.   Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.

12.  Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.

13.  Divisibility By 16 : A number is divisible by 16, if the number formed by the last4  digits is divisible by 16.

Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16.

14.  Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3  and 8.

15.  Divisibility By 40 : A given number is divisible by 40, if it is divisible by both

5  and 8.

16.  Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.

Note : If a number is divisible by p as well as q, where p and q are co-primes, then the given number is divisible by pq.

If p arid q are not co-primes, then the given number need not be divisible by pq,

even when it is divisible by both p and q.

Ex. 36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since

4  and 6 are not co-primes.

VI   MULTIPLICATION BY SHORT CUT METHODS

1.  Multiplication By Distributive Law :

(i) a x (b + c) = a x b + a x c    (ii) ax(b-c) = a x b-a x c.

Ex.   (i) 567958 x 99999 = 567958 x (100000 - 1)

= 567958 x 100000 - 567958 x 1 = (56795800000 - 567958) = 56795232042. (ii) 978 x 184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.

2.  Multiplication of a Number By 5ⁿ :  Put n zeros to the right of the multiplicand and divide the number so formed by 2ⁿ

Ex. 975436 x 625 = 975436 x 5⁴= 9754360000 =   609647600

                                                            16

VII.   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)

9. (a³ + b³ + c³ -3abc) = (a + b + c) (a² + b² + c² - ab - bc - ca)

10. If a + b + c = 0, then a³ + b³ + c³ = 3abc.

VIII.  DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM

If we divide a given number by another number, then :

Dividend = (Divisor x Quotient) + Remainder

IX.     {i) (xⁿ - aⁿ ) is divisible by (x - a) for all values of n.

           (ii) (xⁿ - aⁿ) is divisible by (x + a) for all even values of n.

           (iii) (xⁿ + aⁿ) is divisible by (x + a) for all odd values of n.

X. PROGRESSION

A succession of numbers formed and arranged in a definite order according to certain definite rule, is called a progression.

1. Arithmetic Progression (A.P.) : If each term of a progression differs from its preceding term by a constant, then such a progression is called an arithmetical progression. This constant difference is called the common difference of the A.P.

An A.P. with first term a and common difference d is given by a, (a + d), (a + 2d),(a + 3d),.....

The nth term of this A.P. is given by T_n =a (n - 1) d.

The sum of n terms of this A.P.

S_n = n/2 [2a + (n - 1) d] = n/2   (first term + last term).

SOME IMPORTANT RESULTS :

 (i) (1 + 2 + 3 +…. + n) =n(n+1)/2

(ii) (l² + 2² + 3² + ... + n²) = n (n+1)(2n+1)/6

(iii)  (1³ + 2³ + 3³ + ... + n³) =n²(n+1)²

2.   Geometrical Progression (G.P.) : A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression.

The constant ratio is called the common ratio of the G.P. A G.P. with first term a and common ratio r is :

a, ar, ar²,

In this G.P. T_n = ar^n-1

sum of the n terms, Sn=   a(1-rⁿ)

                                          (1-r)

SOLVED EXAMPLES

Ex. 1. Simplify :   (i) 8888 + 888 + 88 + 8   

                              (ii) 11992 - 7823 - 456  

Sol.   i )  8888                       ii) 11992 - 7823 - 456 = 11992 - (7823 + 456)

                888                                                            = 11992 - 8279 = 3713-

                  88                                  7823                         11992

         +         8                               +   456                      -   8279

               9872                                  8279                          3713

Ex. 2, What value will replace the question mark in each of the following equations ?

(i) ? - 1936248 = 1635773            (ii) 8597 - ? = 7429 - 4358

Sol.  (i) Let x  - 1936248=1635773.Then, x = 1635773 + 1936248=3572021.                 (ii) Let 8597 - x = 7429 - 4358.

Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.

 Ex. 3. What could be the maximum value of Q in the following equation?          5P9 + 3R7 + 2Q8 = 1114                     

Sol. We may analyse the given equation as shown :                     1   2     

Clearly, 2 + P + R + Q = ll.                                                           5   P  9

So, the maximum value of Q can be                                                3   R  7

(11 - 2) i.e., 9 (when P = 0, R = 0);                                               2   Q  8

                                                                                                      11  1   4

Ex. 4. Simplify : (i) 5793405 x 9999  (ii) 839478 x 625

Sol.

i)5793405x9999=5793405(10000-1)=57934050000-5793405=57928256595.b

ii) 839478 x 625 = 839478 x 5⁴ = 8394780000 = 524673750.

                                                            16

Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863    (ii) 983 x 207 - 983 x 107

Sol.

(i) 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.

(ii) 983 x 207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300.

Ex. 6. Simplify : (i) 1605 x 1605    ii) 1398 x 1398

Sol.

i) 1605 x 1605 = (1605)² = (1600 + 5)² = (1600)² + (5)² + 2 x 1600 x 5

                                                               = 2560000 + 25 + 16000 = 2576025.

(ii) 1398 x 1398 - (1398)² = (1400 - 2)²= (1400)² + (2)² - 2 x 1400 x 2

                                                         =1960000 + 4 - 5600 = 1954404.

Ex. 7. Evaluate : (313 x 313 + 287 x 287).

Sol.

 (a² + b²) = 1/2 [(a + b)² + (a- b)²]

(313)² + (287)² = 1/2 [(313 + 287)² + (313 - 287)²] = ½[(600)² + (26)²]

= 1/2 (360000 + 676) = 180338.

Ex. 8. Which of the following are prime numbers ?

(i) 241           (ii) 337         (Hi) 391           (iv) 571

Sol.                             

(i)            Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.

             241 is not divisible by any one of them.

         241 is a prime number.

                                               

(ii)           Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.          

             337 is not divisible by any one of them.

          337 is a prime number.

(iii)          Clearly, 20 > Ö39l". Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.

           We find that 391 is divisible by 17.

          391 is not prime.

(iv)          Clearly, 24 > Ö57T. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.

            571 is not divisible by any one of them.

         571 is a prime number.

 Ex. 9. Find the unit's digit in the product (2467)163 x (341)72.

Sol. Clearly, unit's digit in the given product = unit's digit in 7¹⁵³ x 1⁷².                                 

       Now, 74 gives unit digit 1.

       7¹⁵²  gives unit digit 1,

  \ 7¹⁵³  gives unit digit (l x 7) = 7. Also, 1⁷² gives unit digit 1.

       Hence, unit's digit in the product = (7 x 1) = 7.

Ex. 10. Find the unit's digit in (264)¹⁰² + (264)¹⁰³

Sol. Required unit's digit = unit's digit in (4)¹⁰² + (4)^103.

       Now, 4²  gives unit digit 6.

   \(4)¹⁰² gives unjt digit 6.

   \(4)103 gives unit digit of the product (6 x 4) i.e., 4.

       Hence, unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.

 Ex. 11. Find the total number of prime factors in the expression (4)¹¹ x (7)⁵ x (11)².

Sol. (4)¹¹x (7)⁵ x (11)² = (2 x 2)¹¹ x (7)⁵ x (11)² = 2¹¹ x 2¹¹ x7⁵x 11² = 2²² x 7⁵ x11²

       Total number of prime factors = (22 + 5 + 2) = 29.

Ex.12. Simplify :    (i) 896 x 896 - 204 x 204

                                (ii) 387 x 387 + 114 x 114 + 2 x 387 x 114

                                (iii) 81 X 81 + 68 X 68-2 x 81 X 68.

Sol.

(i)  Given exp  =  (896)² - (204)² = (896 + 204) (896 - 204) = 1100 x 692 = 761200.

(ii) Given exp  = (387)²+ (114)²+ (2 x 387x 114)

                        = a² + b² + 2ab,  where a = 387,b=114

                            = (a+b)² = (387 + 114 )² = (501)² = 251001.

(iii) Given exp = (81)² + (68)² – 2x 81 x 68 = a² + b² – 2ab,Where a =81,b=68

                        =  (a-b)² = (81 –68)² = (13)^{2 = 169.}

Ex.13. Which of the following numbers is divisible by 3 ?

(i) 541326                                 (ii) 5967013

Sol.

(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.

Hence, 541326 is divisible by 3.

(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.

Hence, 5967013 is not divisible by 3.

Ex.14.What least value must be assigned to * so that the number 197*5462 is r 9 ?

Sol.

Let the missing digit be x.

Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +»2) = (34 + x).

For (34 + x) to be divisible by 9, x must be replaced by 2 .

Hence, the digit in place of * must be 2.

Ex. 15. Which of the following numbers is divisible by 4 ?

(i) 67920594                    (ii) 618703572

Sol.     

(i) The number formed by the last two digits in the given number is 94, which is not divisible by 4.

Hence, 67920594 is not divisible by 4.

(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4.

Hence, 618703572 is divisible by 4.

Ex. 16. Which digits should come in place of * and $ if the number 62684*$ is divisible by both 8 and 5 ?

Sol.     

Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $.

Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4.

Hence, digits in place of * and $ are 4 and 0 respectively.

Ex. 17. Show that 4832718 is divisible by 11.

Sol.    (Sum of digits at odd places) - (Sum of digits at even places)

                                          = (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.

            Hence, 4832718 is divisible by 11.

Ex. 18. Is 52563744 divisible by 24 ?

Sol.  24 = 3 x 8, where 3 and 8 are co-primes.           

        The sum of the digits in the given number is 36, which is divisible by 3. So, the                       given number is divisible by 3.

         The number formed by the last 3 digits of the given number is 744, which is  divisible by 8. So, the given number is divisible by 8.

         Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.

So, it is divisible by 3 x 8, i.e., 24.

Ex. 19. What least number must be added to 3000 to obtain a number exactly divisible by 19 ?

Sol. On dividing 3000 by 19, we get 17 as remainder.

    \Number to be added = (19 - 17) = 2.

Ex. 20. What least number must be subtracted from 2000 to get a number exactly divisible by 17 ?

Sol. On dividing 2000 by 17, we get 11 as remainder.

     \Required number to be subtracted = 11.

Ex. 21. Find the number which is nearest to 3105 and is exactly divisible by 21.

Sol. On dividing 3105 by 21, we get 18 as remainder.

     \Number to be added to 3105 = (21 - 18) - 3.

        Hence, required number = 3105 + 3 = 3108.

Ex. 22. Find the smallest number of 6 digits which is exactly divisible by 111.

Sol. Smallest number of 6 digits is 100000.

       On dividing 100000 by 111, we get 100 as remainder.

    \Number to be added = (111 - 100) - 11.

        Hence, required number = 100011.-

Ex. 23. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor.

                            Dividend - Remainder      15968-37     

Sol.     Divisor = -------------------------- = ------------- = 179.

                                       .Quotient                    89

Ex. 24. A number when divided by 342 gives a remainder 47. When the same number ift divided by 19, what would be the remainder ?

Sol.   On dividing the given number by 342, let k be the quotient and 47 as remainder.

          Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.

       \The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Ex. 25. A number being successively divided by 3, 5 and 8 leaves remainders 1, 4

and 7 respectively. Find the respective remainders if the order of divisors be reversed,

Sol.

3	X
5	y	- 1
8	z	- 4
	1	- 7

\z = (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 + 1) = 238.

Now,

8	238
5	29	- 6
3	5	- 4
	1	- 9,

\Respective remainders are 6, 4, 2.

Ex. 26. Find the remainder when 2³¹ is divided by 5.

 Sol.    2¹⁰ = 1024. Unit digit of 2¹⁰ x 2¹⁰ x 2¹⁰ is 4 [as 4 x 4 x 4 gives unit digit 4].

        \Unit digit of 231 is 8.

            Now, 8 when divided by 5, gives 3 as remainder.

            Hence, 231 when divided by 5, gives 3 as remainder.

Ex. 27. How many numbers between 11 and 90 are divisible by 7 ?

Sol.  The required numbers are 14, 21, 28, 35, .... 77, 84.

        This is an A.P. with a = 14 and d = (21 - 14) = 7.

        Let it contain n terms.

        Then, T_n = 84   =>  a + (n - 1) d = 84

                                  =>   14 + (n - 1) x 7 = 84   or n = 11.

     \Required number of terms = 11.

Ex. 28. Find the sum of all odd numbers upto 100.

Sol. The given numbers are 1, 3, 5, 7, ..., 99.

        This is an A.P. with a = 1 and d = 2.

        Let it contain n terms. Then,

        1 + (n - 1) x 2 = 99 or n = 50.

     \Required sum = n (first term + last term)

                                   2

                                = 50 (1 + 99) = 2500.

                                     2

Ex. 29. Find the sum of all 2 digit numbers divisible by 3.

 Sol. All 2 digit numbers divisible by 3 are :

        12, 51, 18, 21, ..., 99.

        This is an A.P. with a = 12 and d = 3.

        Let it contain n terms. Then,

        12 + (n - 1) x 3 = 99 or n = 30.

     \Required sum = 30 x (12+99) = 1665.

                                    2

Ex.30.How many terms are there in 2,4,8,16……1024?

Sol.Clearly 2,4,8,16……..1024 form a GP. With a=2 and r = 4/2 =2.

       Let the number of terms be n  . Then

       2 x 2^n-1 =1024 or 2^{n-1 =512 = 2}9.

    \n-1=9 or n=10.

Ex. 31. 2 + 2² + 2³ + ... + 2⁸ = ?

 Sol.    Given series is a G.P. with a = 2, r = 2 and n = 8.

        \sum =  a(rⁿ-1) = 2 x (2⁸ –1) = (2 x 255) =510

                         (r-1)        (2-1)