crt for jnu jaipur student

1. BASIC CALCULATION
(i) Square nearest to 50 (+) or 50 (-).
For numbernearestto50 we will take the base as 25 and forcomparisonwe will compare
the numberwith50.
50
50 – 18 =32
(32)
(a) (32)² = 25 - 18/(18)² = 7 /324.
We needonlytwodigitnumberinthe secondhalf.Sowe will addthe remainingdigitinto
the firsthalf.
So,7 + 3 /24 =1024.
(b) (47)²
Compare the numberwith50.Since itis 3 lessthan 50 .sowe will subtract3 fromitsbase
and take the square of 3 in the secondhalf.
25 – 3 / (3)²= 22/9.
We getonlyone digitinsecondhalf,we have toget a two digitnumbersowe make 9 as 09.
22/09 = 2209.
(ii) Square nearest to 100 (+) or 100 (-).
For numbernearestto100 we will take the base asnumberonlyandfor comparisonwe will
compare the numberwith100.
100
100 + 7 =107
107
(a) (107)² = 107 +7 /(7)²
=114 /49
=11449

2. (b) (97)² = 97 -3 /(3)²= 94 /9
We getonlyone digitinsecondhalf,we have toget a two digitnumbersowe make 9 as 09.
94/09 = 9409.
(iii) Square nearest to 200 (+) or 200 (-)
For numbernearestto200 we will take the base asnumberonlyandfor comparisonwe
will compare the numberwith200.
200
200 + 8 =208
208
(a) (208)² = 208 + 8 /(8)²
= 216/64 (Incase of numbernearestto200 ,we have to multiply2infirsthalf)
= 2 x 216 / 64
=43264.
200
(b) (197)² = 197 – 3 /(-3)² 200 – 3 =197
= 194 / 9 197
=2 x 194/09 (make secondhalf twodigitnumber)
= 38809.
(iv) Square nearest to 300 (+) or 300 (-)
For numbernearestto300 we will take the base asnumberonlyandfor comparisonwe
will compare the numberwith300.
(a) (309)² = 309 +9/ (9)²
=318/81 (Incase of numbernearestto300, we have to multiply3infirsthalf)
= 3 x 318 /81
=95481
(v) Square nearest to 400 (+) or 400 (-)
For numbernearestto400 we will take the base asnumberonlyandfor comparisonwe
will compare the numberwith400.
(a) (404)² = 404 +4/ (4)²
=416/16 (Incase of numbernearestto400, we have to multiply4infirsthalf)
= 4 x 416 /16

3. = 163216
(vi) Square nearest to 1000 (+) or 1000 (-)
For numbernearestto1000 we will take the base as numberonlyand forcomparisonwe
will compare the numberwith1000.
(a) (1004)² = 1004 +4/ (4)²
= 1008/16 (Incase of numbernearestto1000, we have to multiply10in first
half)
= 10 x 1008 /16 = 1008016
(vii) Square nearest to 1100 (+) or 1100 (-)
For numbernearestto1000 we will take the base as numberonlyandforcomparisonwe
will compare the numberwith1000.
(a) (1108)² = 1108 +8 / (8)²
= 1116/64 (Incase of numbernearestto1100, we have to multiply11in first
part)
= 11 x 1116 /16 = 1227664
(viii) ENDING WITH 5 (SQUARE)
(i) (35)² = 35 x 35 = 3 x (3 + 1 ) / 5 x 5
= 3 x 4 / 25
= 1225
(ii) (105)² = 105 x 105 = 10 x (10 + 1) / 5 x 5
= 10 x 11 / 25
11025.
(ix) UNIT DIGIT SUM = 10
In thiscase the othernumbernumbershouldbe same,andthe sumof digitsatunit
place mustbe = 10
(i) 76 x 74 = 7 x (7 +1) / 6 x 4
= 7 x 8 /24
=5624
(ii) 23 x 27 = 2 x (2 + 1) / 3 x 7
= 2 x 3 /21
= 621.
(x) SQUARE ROOT :-
NUMBERS LAST DIGIT
(i) 1, 9 1
(ii) 2, 8 4

4. (iii) 3, 7 9
(iv) 4, 6 6
(v) 5 5
(a) Findthe square root of 6084.
60 / 84 (We divide the numberintotwohalves,startingfromthe lefthandside)
60 / 84
7 / 2 7 / 8
7 x 8 = 56 ( lessthan60 ) ,so the square root is78.
(xi) CUBE ROOT :-
NUMBERS LAST DIGIT
(i) 1 1
(ii) 2 8
(iii) 3 7
(iv) 4 4
(v) 5 5
(vi) 6 6
(vii) 7 3
(viii) 8 2
(ix) 9 9
(a) Findthe Cube root of 912673
Steps:-
i) We divide the numberintotwohalves.
ii) From the lefthandside take the pairof three numbers,whichmakesone half
and remainingnumberformsthe other.

5. 912 / 673
9 (Cube of the number(9) is nearestto912) / 7.
Cube root is97
To multiply any two digit number by 11:
• For this example we will use 54.
• Separate the two digits in your mind (5 4).
• Add the 5 and the 4 together (5+4=9)
• Put the resulting 9 in the gap 594.
11 x 54=594
11 x 57 ... 57 ... 5+7=12 ... put the 2 in the hole and add the 1 from the 12 to the 5 in to get 6 for a result
of 627

6. NUMBER SYSTEM
Natural numbers: - The numberswhichwe use tocount anynumberof things.
Example:-1,2,3, 4, 5 ---- etc.
Whole number:- Whenzeroisincludedinnatural number,itbecomeswholenumber.
Example:-0,1,2, 3, 4, 5 ---- etc.
Integers:- Integersare whole numbers,whichare eitherpositiveornegative.
Example:-0,-1, -2,3, 4 --- etc.
Prime number: - All the numberswhichare divisibleby1 anditself onlyare prime numbers.
Example:-2,3,5, 7, 11 ---- etc.
Composite Number:- It isthe product of two or more than twodistinctor same prime number.
Example:- 4, 6, 8 ------ etc.
Even Number:- Anyintegerthatisa multipleof 2 isEven.
Example:-2,4,6, 8, 10 --- etc.
Odd Number:- Anyintegerthatis nota multipleof 2 isOdd.
Example:- 1, 3, 5, 7, 9 ----- etc.
L.C.M (Least CommonMultiple):- Itisa multiple of twoormore than twonumber.
H.C.F (HighestCommonFactor):- It is the highestvalue whichcandivide the givennumber.
For any 2 givennumbers HCFx LCM = Product of the 2 numbers.
HCF of fractions = HCF of numerators of all the givenfractions/LCM ofthe denominatorsof all the
fractions.
LCM of fractions = LCM ofnumerators of all the fractions/HCF ofthe denominatorsof all the fractions.
NumberSystem (Sumof differentnatural numbers):-
(i) Sum of first n natural number:-
1+2+3+4+5+------+n = n (n+1)/2.

7. (ii) Sum of square of first n natural number:-
1²+2²+3²+4²+5²+ ------ +n²= n (n+1)(2n+1)/6.
(iii) Sum of cube offirst n natural number:-
1³+2³+3³+4³+5³+ ------ +n³= {n(n+1)/2}².
(iv) Sum of first n evennatural number:-
2+4+6+8+10+ ----- +n = n(n+1).
(v) Sum of first n odd natural number:-
1+3+5+7+9+ ----- +n= (n) ².
DIVISIBILITY RULES:-
1) For 2:- Unit digitof any numberis0, 2, 4, 6, 8.
2) For 3:- Sum of all digitof anynumberisdivisibleby3.
3) For 4(2²):- If last twodigitof a numberisdivisible by4,thenthat numberisdivisibleby
4.
4) For 5:- If lastdigitof the numberis5 or 0, thennumberisdivisible by5.
5) For 6(2x3):- Last digitof any numberisdivisible by2 andsum total of all digitof a
numberisdivisible by3.
6) For 7:- Multiplyunitdigitby2, and subtract the numberwiththem. If itisdivisible
Example:- 795 is divisible by7or not check?
Unit digit5x2 = 10.
Remaining79 – 10 = 69.
Nowdivide 69 by7, resultisnotdivisibleby7.
7) For 8(2³):- Checkthe last three digit,whetheritisdivisibleby8 or not.
8) For 9:- Sum of all digitsmustbe divisibleby9.
9) For 10(2x5):- Checkthe divisibilityof 2and 5 or unitdigitmusthave zeroat the end.

8. 10) For 11:- A numberisdivisible by11,if difference betweensum of evenplace andthe
sumof digitof odd placesisdivisible by11.(Zeroisdivisibleby11).
11) For 12(3x4):- Checkdivisibilityof 3& 4.
12) For 13:- Multiplyunitdigitby4 and addthe value tothe remainingdigit.
Example:- Check1404 isdivisible by13 or not?
Unit digit4 x 4 =16
Now,addi.e.140+16 =156 /13. Thisnumberisdivisibleby13.
13) For 14:- Checkdivisibilityof 2& 7.
14) For 17:-Multiplyunitdigitby5 andsubtract withthe remainingdigit,checkif nowthe
numberisdivisible.
15) For 19:- Multiplyunitdigitby2 and addwiththe remainingdigit,checkif now the
numberisdivisible.
16) For 23:- Multiplyunitdigitby7 and addwiththe remainingdigit,checkif now the
numberisdivisible.
17) For 29:- Multiplyunitdigitby3 and addwiththe remainingdigit,checkif now the
numberisdivisible.
18) For 31:- Multiplyunitdigitby3 and subtractwiththe remainingdigit,checkif nowthe
numberisdivisible.
19) For 39:- Multiplyunitdigitby4 and addwith the remainingdigit,checkif now the
numberisdivisible.
NUMBER OF DIVISORS:-
N = aᴾ x bᴿx cᵀ
Numberof divisors:- (P+1)(R+1)(T+1).

9. N = aᴾ x bᴿx cᵀ
Sum of divisor:-
How to determine if a number is Prime?
The easiest& simplestmethodistodivide the numberuptothe closetsquare rootof thatnumber.
Example. Let’sconsider53. Numberclose to53 havinga perfectsquare is64 and itssquare root is8.
Nowstart dividing53from 2 to 8. There isno such numberbetween2to 8 whichdivides53so 53 isa
prime number.
Recurring decimal intoFractionConversion:
Convertthe decimal
intoa fraction.
Start withthe equation:
There are twodigitsinthe repeatingblock,somultiplybothsidesby102
= 100.
Nowsubtract N from bothsides.Notice thatthe repeatingpartcancelsout.
Divide bothsidesby99, multiplynumeratoranddenominatorbya powerof 10 to get ridof the decimal
point,andsimplify.
CYCLICITYMETHOD: - It isa methoduse tofindoutthe remainderof anyexpression.
Example:Findthe remainderwhen 4⁹⁶ isdividedby 6.
(aᴾ⁺¹-1)(bᴿ⁺¹-1)(cᵀ⁺¹ - 1) /(a-1)(b-1)(c-1).

10. Number/6 4¹ 4² 4³ 4⁴ 4⁵ 4⁶ 4⁷ 4⁸
Remainder 4 4 4 4 4 4 4 4.
Remainderinall casesis4, so final remainderwill be 4.
REMAINDER THEOREM:- Product of any twoor more thantwo natural numberhasthe same numberas
the product of theirremainder.
FERMAT’S REMAINDER THEOREM:-
Let P be a prime numberandN be a numbernotdivisible byP.ThenremainderobtainedwhenAᴾ⁻¹/P
=1, If H.C.F (A,P) is1.
(i) (A+1)ᴿwill alwaysgive 1as remainderforall valuesof A and N.
(ii) (A) ᴿwheneven,remainderis1 andwhenN isodd,remainderisA itself.
(iii) (aⁿ+ bⁿ) is divisiblebya+ b, if n isodd.
(iv) (aⁿ - bⁿ) is divisible bya+ b , if n is even.
(vi) (aⁿ - bⁿ) isdivisible bya- b , if n iseven.
Number of a’s (a prime factor) in N!
Number of a’s in N! = [N/a] + [N/a²] + [N/aᶟ] + …….
Q1) For example howmany2’sare in12!
Ans:It‘s [12/2] + [12/4] + [12/8] + [12/16]
= 6 + 3 + 1 + 0 = 10
Q2) 45! Ends withhowmanyzeros?
Numberzerosdependuponnumberof 5’s and2’s.
Numberof 5’s: 9 + 1 = 10
Numberof 2’s: 22 + 11 + 5 + 2 + 1 = 41
So numberof zero’s= 10

11. MIXTURES AND ALLIGATION
Alligation:Itisthe rule thatenablesustofindthe ratio inwhichtwoor more ingredientsatthe
givenprice mustbe mixedtoproduce a mixture of desiredprice.
Mean Price: The cost of a unitquantityof the mixture iscalledthe meanprice.
1. Rule of Alligation:If twoingredientsare mixed,then Quantityof cheaper = C.P.of dearer -
Mean Price
Quantity of dearerMean price - C.P. of cheaper
C.P.of a unitquantityof cheaper(c) C.P.of a unitquantityof dearer(d)
Mean Price (m)
(d- m) (m - c)
(Cheaperquantity) : (Dearer quantity) = (d - m) : (m - c).
2. Suppose acontainercontainsx of liquidfromwhichyunitsare takenoutand replacedbywater.
Afternoperations,the quantityof pure liquid=x{1 – (y/x)}ⁿ units.

12. PERCENTAGE
PERCENTAGE CHART:
1 2 3 4
1 100% 200 300 400
2 50 100 150 200
3 33.33 66.66 100 133.33
4 25 50 75 100
5 20 40 60 80
6 16.66 33.33 50 66.66
7 14.24 28.56 42.85 57.14
8 12.5 25 37.5 50
9 11.11 22.22 33.33 44.44
10 10 20 30 40
PERCENTAGE BASICS:-
1) Percentage increase = increase x 100 / Base Value. 100 120
2) Percentage decrease = decrease x 100 / Base Value. 120 100
3) Percentage = Value attainedx 100/ Total value.
Q 1) Salary of Ram increasesby30% in monthof January .If hismonthlysalary isRs 7000/-.Findthe
salaryof Ram inthe respective month.
Solution:- Assume Original salary=100
Salaryafterincrease = 130
100 7000
130 13 x 70 = 9100
Q2) A salaryis 20% lessthanB, thenby how much % B salarymore than A?
Solution: - Use Direct Formula 20 x 100/80 = 25%.
R x 100 /100 – R.

13. Q3) Price of Sugar increasesby 40%, familyreducesitsconsumptionsothattheirexpense remainsame.
Findthe reductioninpercentage of sugar.
Solution:- Use DirectFormula
40 x 100/140 = 28.56%
Q4) In a Companyprice of penincreasesby10% anddue to thisthe sale reducesby20%. What wouldbe
the neteffectdue tothis change?
Solution:- Use DirectFormula
A = increase by10% = +10 % {increases positive}
B = decrease by20% = - 20% {decrease negative}
10 – 20 + (-10 x 20/100) = 10 – 20 – 2 = -12 = 12 % decrease.
Q5) If side of square increase by10%, whatisthe netchange inthe area of the square?
Solution:- Use DirectFormula here,x = + 10% (increase)
2 x 10 + (10 x 10 /100) = 21%.
Q6) Ram donate 40% of hissalaryto charity organization,fromthe remaininghe put50% to hisbank
account ,fromthe remainingbalance he make aninvestmentof 30% ina companyand now he isleft
withRs 1470/-.FindRam’s monthlysalary.
Solution:- Use DirectFormula
Were p1, p2 andp3 are three percentagesgiven.
p1 = 40% p2 = 50% p3 =30%
R x 100 /100 + R.
A + B + (AB/100)
2x + (x²/100)
Salary = Saving x 100 x 100 x 100 /(100-p1)(100-p2)(100-p3)

14. Another Approach: - Total Salary(Assume) = 100
Donated(40%) = Rs 40
Left amount= 60
Bankaccount (50%) =Rs 50.
Leftamount=30
Investment(30%) =Rs30
LeftAmount=21
Here LeftAmount21 1470 (21 x 70)
AssumedSalary100 100 x 70 = Rs 7000/-

15. PROFITAND LOSS
C.P = COST PRICE OF AN ARTICLE
S.P = SELLING PRICE OF ANARTICLE
M.P = MARKED PRICE OF AN ARTICLE
L.P (LIST PRICE) = MARKED PRICE
1) PROFIT = S.P – C.P
(a) PROFIT % = PROFIT x 100 /C.P.
2) LOSS = C.P – S.P
(b) LOSS % = LOSS x 100 /C.P.
3) GENERAL FORMULA: C.P = S.P X 100 /100 + P%. (P% = PROFIT%)
Or
C.P = S.P X100/100 – L%. (L% = LOSS%)
4) SUCCESSIVEDISCOUNT: - It is a discountgivenonalreadydiscountedproduct.
A + B – (AB/100) where A and B both are discounts.
5) MARKED PRICE (M.P) – DISCOUNT = SELLING PRICE.
6) DISHONEST SHOPKEEPER (CASE ):-
PROFIT % = ERROR VALUE x 100/ORIGINAL – ERROR VALUE.
Q1) A man buysan article forRs 700/- and sellsitata gainof 20% .Findthe sellingprice of the
article.
Solution:C.P = 700, Gain % = 20
C.P=S.P x 100/100 + P%.
S.P= C.Px (100 +P %) /100 = 700 x 120/100 =Rs 840
Quick Approach: Letus assume C.P= 100, for 20% gain S.P=120
C.P:S.P=100:120
100 700 (100 x 7)

16. 120 840 (120 x 7).
Q2) Ram soldan article forRs 1500/- at a profitof 25%. Atwhat cost will she have tosell itto
geta profitof 30% ?
Solution:Assume C.P=100
At 25% profitS.P= 125
At 30% profitS.P= 130
125 1500 (125 x 12)
130 130 x12 = 1560.
Q3) The costof 12 pensisequal to sellingprice of 15 pens.Whatis the profitor loss% incurred?
Solution:C.P OF12 PEN = S.POF 15 PEN
C.P/S.P = 15/12 =5/4.
LOSS OF1 PEN (LOSS = C.P – S.P)
LOSS %= LOSS x 100/C.P = 1 x 100/5 =20%.
Q4) By selling35Orangesa vendorlosesthe sellingprice of 5 Oranges.Hisloss% is:
Solution:LOSS = C.P of 35 Orange – S.P of 35 Orange
LOSS = S.P of 5 Orange
S.P of 5 Orange = C.P of 35 Orange – S.P OF35 Orange
S.Pof 40 Orange = C.Pof 35 Orange
C.P/S.P= 40/35 = 8/7 LOSS of I Orange
LOSS %= 1 x 100/8 = 12.5%
Q5) A solda watch to B at a gainof 20% and B soldit to C at a lossof 10% .If C broughtthe
watch forRs 216,at whatprice didA buy?
Solution:In thisCase A B C
Use formula:
C.P = S.P x 100 x 100/ (100 – P %)( 100 – L %)
Q6) Successive discountof 10%,20% and 15% is:
Solution:Use formula:
A + B – (AB/100) =10 + 20 - (10 x 20/100) = 28%
Againuse same formula= 28 + 15 - (28 x 15/100) = 38.8%
Q7) A shopkeepersoldtwoitemsof Rs400 each. Onone he gain10% & on otherhe losses10%
.Findhisgain/losspercent.
Solution:Since S.Pis same forboth the items.
GAIN%=LOSS%= 10%. = x.
LOSS% = (x/10)²
= (10/10)² = 1% loss.

17. RATIO AND PROPORTION
1) A : B = 1 : 2 Here 1 = Antecedent, 2 = consequent
2) (a) A : B =1 : 2 DUPLICATE is A² : B² = 1 : 4
(b) A:B =1: 2 SUB- DUPLICATE is (A) ⅟₂: (B) ⅟₂ = 1: (2) ⅟₂
(c) A: B =1: 2 TRIPLICATE is Aᶟ: Bᶟ = 1: 8
(d) A:B =1: 2 SUB- TRIPLICATE is (A) ⅓: (B) ⅓ = 1: (2) ⅓
3) A: B:: C: D Here B and C are Means (mean values) and A and D are Extremes(extreme values).
4) MEAN PROPORTION: - Here b = Mean proportion
5) CONTINUED PROPORTION: -If a, b,c, d are in continuedproportionthen,
Q1) A:B = 1 : 3 , B:C = 5 : 7 and C:D = 9 : 7 thenA : B : C : D ?
Solution: A : B : C : D
1 3 ③ ③ fill vacantspace by puttingtheirnearestnumberi.e.(3)
⑤ 5 7 ⑦ fill vacantspace byputtingtheirnearestnumberi.e.(5,7)
⑨ ⑨ 9 7 fill vacantspace byputtingtheirnearestnumberi.e.(9)
_______________________________________________
A = 1 x 5 X 9 = 45
B = 3 x 5 x9 = 135
A/B = C/D (Basic Proportion)
b² = a x c
a/b = b/c =c/d

18. C = 3 x 7 x 9 = 108
D = 3 x 7 x 7 = 147
Q2) The monthlyincome of A & B are inratio 2: 3 and theirmonthlyexpenseare inratio5: 9.If each of
themsavesRs 600 per monththentheirmonthlyincomesare:
Solution:Let the income be intheirratio2x: 3x
Expense ratioof A & B is5y:9y
Income – Expense = Saving
2x -5y = 600 ------①
3x – 9y =600 ------②solvingboth,we getx =800.
Hence A = 2x = 2 x 800 = 1600 and B = 3x = 3 x 800 = 2400.
Q3) FindThirdproportionof 6, 24.
Solution:6: 24:: 24: x
6/24 = 24/x 6x = 24 x 24 => x = 24 x 24 /6 => 96.
Q4) FindFourthproportion8, 24, and36.
Solution:8:24:: 36: x.
8/24 =36/x 8x = 36 x 24 => x = 108.
A: B: C: D = 15: 45: 63: 49.

19. TIME AND DISTANCE
(i) Speed= Distance / time
 Distance ismeasuredinkilometer,meterormiles.
 Time ismeasuredinHour,minutesorseconds.
 SpeedismeasuredinKilometer/hour,meter/sec,miles/hour.
(ii) CONVERSION:-
 1 km= 1000 meter
 1 hour= 3600 seconds
54 km/h= 54 x 1000/3600 = 15 m/s.(km/hto m/s multiplyby5/18).
 1 meter= 1/1000 meter.
 1 second= 1/3600 hour.
15 m/s = 15 x 3600/1000 = 54 km/h (m/stokm/h multiplyby18/5)
(iii) Average Speed=Total distance travelled/Total time taken.
o For twovariables= 2ab/a+b ( a & b are speed )
o For three variables=3abc/ab +bc +ca (a, b & c are speed)
(iv) Time takenafter meeting(same distance):- If 2 personsstartsjourneyfromA & B and move
towards each otherandmeetat C, aftermeetingtheyreachopposite pointsinx & y hrs
respectively.
Then speedbefore theymeetis √y : √x
TRAINS CONCEPT:-
1) Same Direction:- Whentwotrains are goingin same directionandone traincrossesthe other.
Sa, La (TrainA) Sa = Speedof trainA,La = Lengthof trainA.
Sb, Lb (TrainB) Sb = Speedof trainB, Lb = lengthof trainB.
Relative Speed= Sa - Sb, Total distance = La + Lb

20. So, Sa – Sb = La + Lb / T.
2) Train andplatformcase :-
Sa, La (TrainA)
PLATFORM ( (P) Sa = La + P/ Time.
3) Opposite Direction:-
Sa, La (TrainA)
Sb,Lb (TrainB)
Relative Speed=Sa + Sb.
Total distance =La + Lb
Sa + Sb = La + Lb / Time
Boats and streams Concept :
Let the speedof boatupstream= x km/h (upstream=againstthe currentor flow of water)
Let the speedof boatdownstream= y km/h (downstream=withthe current or flow of water)
(i) Speedof boatin Still Water= ½ (x + y) km/h.
(ii) Speedof Stream(current) = ½ (y – x) km/h.
Let the Speedof boatin Still Water= a km/h
Let the Speedof Stream(current) = b km/h
(i) Speedof boatupstream= (a –b) km/h.
(ii) Speedof boatdownstream= (a + b) km/h.

21. TIME AND WORK
TIME AND WORKEQUIVALENCE:- Essence liesinthe facthasit exhibitsthe mostfundamental
relationshipbetweenthe three factorsi.e.work,time andthe agentcompletingthe work.
Work done = Number of Days x Number of Men (W = D x M)
Condition1:- Work(w) is constant.(Mx D = Constant)
Mα1/D (If work done isconstant,thenthe numberof personsisinverselyproportional to
the numberof days.)
Condition2:- D isconstant.
WαM (More workwill be done if we employmore men&vice versa.)
Condition3:-Mis constant
WαD (More workis done if we have more numberof days.)
In general,we summarizethat
M₁D₁/W₁ =M₂D₂/W₂
Individual Work & Individual Efficiency:-
Individual Work: - If A can do a certainworkin 10 days,thenhe will finish1/10th of
the workinone day.
Individual Efficiency:- Efficiencyisalsoknownasworkrate.
General expressioncorrelatingtime taken& efficiency:-
 If efficiencyof A isx percent more than efficiencyof Band B takes‘B’ daysto complete the
work,thenA will take (Bx 100/100+X ) daysto complete same work.
 If efficiencyof A isx percent lessthanefficiencyof Band B takes‘B’ daysto complete the work,
thenA will take (Bx 100 /100-X) daysto complete same work.

22.  If A doesthe work inx daysand B doesthe same work iny days,
Work done by A = 1/x
Work done by B = 1/y
Work done by A and B together= (1/x)+(1/y)
Days to finishthe worktogether=1/(workdone together) =(x*y)/(x+y)
 If A and B togethercan do a piece of workinx days,B and C togethercan do itin y daysand C
and A togethercando itin `z` days, thennumberof days requiredtodothe same work:If A,B,
and C workingtogether=(2xyz) / (xy+ yz + zx)
 A,B, C can do a piece of workinx, y,z days respectively.The ratioinwhichthe amountearned
shouldbe sharedis1/x : 1/y : 1/z = yz:zx:xy
Extension of the concept of time & work:-
1. Pipesand Cistern:- It isan applicationof conceptof time andwork.We see positive workbeing
done innormal casesof time andwork,in case of pipesandcisternnegative workisalso
possible.
Pipe A can fill a tankin 20 hours Positive workisdone.
Pipe B can emptya tankin 25 hours Negative workisdone.
2. Variable Work: - Thisconceptcomesfrom the possibilitythatthe rate of workingcan be
different,canbe dependentuponsome external agent.
Work done α external factor.
3. Alternate Work: -Thisconceptisanalogoustothe conceptof man-days.Aswe have seenin
the conceptof man-hourthatif 20 mencan do a workin 10 days,thenthisworkis equivalentto
200 man-days.

23. SIMPLE AND COMPOUND INTEREST
1. Principal:The moneyborrowedorlentoutfora certainperiodiscalledthe principal orthe sum.
2. Interest:Extramoneypaidforusingother'smoneyis calledinterest.
3. Simple Interest(S.I.): If the interestonasum borrowedforcertainperiodisreckoneduniformly,
thenit iscalled simple interest.
4. Amount(A) = Principal +Simple Interest(S.I)
 Let Principal =P, Rate = R% perannum(p.a.) andTime = T years.Then
 Simple Interest=P x T x R / 100.
ADDITIONAL FORMULA:
(i) AmountA₁ becomes afterTime T₁ year andAmountA₂ becomesafterTime T₂years.
(a) Principal =A₁T₂ – A₂T₁ /T₂ - T₁
(b) Rate of Interest=> R/100 = A₂ – A₁ / A₁T₂ – A₂T₁
(ii) A sumof moneybecomesntimes afterT years,Thento findrate of interestper
annum
R/100 = n – 1 /T
(iii) The annual installmentthatwill dischargeadebtof D due inT years at R % simple
interestperannum= 100 x D / 100T + {RT(T - 1) / 2}
Note:Simple interestshouldbe same foreveryyear.
COMPOUND INTEREST
Principal:The moneyborrowedorlentoutfor a certainperiodiscalledthe principal orthe sum.
Interest:Extra moneypaidforusingother'smoneyiscalled interest.
Let Principal =P, Rate = R% perannum, Time = n years.
1. Wheninterestiscompound Annually:

24. Amount = P { 1 + R /100}ⁿ
2. Wheninterestiscompounded Half-yearly:
Amount = p { 1 + (R/2)/100}²ⁿ
3. Wheninterestiscompounded Quarterly:
Amount = { 1 + (R/4)/100}⁴ⁿ
4. Wheninterestiscompounded Annuallybuttime isinfraction,say3 years.
Amount = p {1 + R/100}ᶟ x {1 + 2/5 x R/100}
5. WhenRatesare differentfordifferentyears,sayR1%,R2%,R3% for1st
, 2nd
and 3rd
year
respectively.
ADDITIONAL FORMULA:
(I) Difference betweenCompoundinterestandSimple interestfortime =2 yearsat R % rate of
interestis…….C.I – S.I = P x (R/100)²
(II) Difference between CompoundinterestandSimple interestfortime =3 yearsat R % rate of
interestis…….C.I – S.I = P(R/100)²(R/100 + 3)
(III) If a loanof Rs D at R% compoundinterestperannumistobe repaidinn equal yearly
installments,thenthe valueof eachinstallmentcanbe givenby
D / (100/100 + R) +(100/100 + R)² + (100/100 + R)ᶟ + ….
Then, Amount = P 1 +
R1
1 +
R2
1 +
R3
.
100 100 100

25. PERMUTATION AND COMBINATION
PERMUTATION:- Implies“arrangement”where“orderof the things”isimportant.
Permutationof nthingstakingr at a time isdenotedby ⁿPᵣ.
ⁿPᵣ = n! / (n-r)! Note 0! =1.
COMBINATION:- Implies“selection”where “orderof the things”isnotimportant.
Combinationof nthingstakingr at a time isdenotedby ⁿCᵣ.
ⁿCᵣ = n! / (n-r)! x (r)!
SOME IMPORTANT DERIVATIONS:-
1) Numberof arrangementsof nthingsof whichp are of one type,qare of secondtype and rest
are distinct …….. n! /p! x q! x r!
2) Numberof arrangementsof nthingstakingr at a time wheneachof the thingsmay be repeated
once,twice--- uptortimesinanyarrangementis…….nʳ.
CIRCULAR PERMUTATION:
1) The Numberof CircularPermutation(arrangement) of ndifferentarticles= (n-1)! .
2) The Numberof Circulararrangementsof n differentarticleswhenclockwise and
anticlockwisearrangementsare notdifferent.
I.e.whenobservationcanbe made fromboth the sides= (n-1)!/2.
COMBINATION:-
1) Out of n things,the numberof waysof selectingone ormore things: - 2ⁿ - 1, where n=
numberof things.
2) Distributingthe giventhings (r+n)intwogroupswhere one groupishaving r thingsand
otherone n things.
ʳ⁺ⁿCᵣ = (r + n)!/r! x n! .
If r=n , then²ʳCᵣ = (2r)!/r! x r!

26. PERMUTATION AND COMBINATION IN GEOMETRY:-
 Numberof diagonals inann sidedregularpolygon=n(n-3)/2.
 Maximumnumberof pointof intersectionamongnstraightlines= ⁿC₂ .
 Maximumnumberof pointof intersectionamongncircles= ⁿP₂ .

27. PROBABILITY
Probability:-Probabilityisanexpectationof the happeningof some events.Itisthe measure of
uncertainty.
Sample Space: - It isthe total numberof all the possible outcomes.
CompoundEvents: - Eventsobtainedbycombiningtwoormore elementaryevents.A compoundevent
issaid to occur if one of the elementaryeventsassociated withitoccurs.
Exhaustive Events: - The total numberof possible outcomesof arandomexperimentinatrial.
MutuallyExclusive Events: - If the occurrence of anyone of the two or more eventspreventsthe
occurrence of all othersi.e.if notwoor more eventsoccursimultaneously.
Pointsto Remember:-
(i)0≤P(E)≤1
(ii) P(E)+P(E)’=1 where P(E)’=Probabilityof nothappeningof eventE.
Odds in favour & Odds against:-
Conditional Probability: -LetA and B be two eventsassociatedwitharandomexperiment.Then
probabilityof occurrence of A underthe conditionthatB has alreadyoccurredandP (B) is not equal to
0.
Thus P (A/B) = Probabilityof occurrence of A giventhat B has happened.
P (B/A) = Probabilityof occurrence of B giventhat A has happened.
Probabilityof an Event (E) = Number offavorable outcomesof E / Total numberof possible outcomes.
Oddsin Favour:- Numberof Favourable Cases/Numberof unfavourableCases.
Oddsin Against:- Numberof unfavourable Cases/Numberof favourable Cases.

28. Addition Theorem:-
1. P(AUB) = P(A)+P(B) –P(AПB)
Let A={2,4,6} and B={3,6}.
P(AUB)={2,3,4,6} Means any one of the outcomesof A or B.
P(AПB) = {6} Means occurrence of both A and B.
2. If events are mutually exclusive then
P(AUB)=P(A)+P(B)

29. DATE INTERPRETATION
Data interpretationisthe act of transformingdatawiththe objectiveof extractinguseful information&
facilitatingconclusionsonthe basicof the givendata.
Data is a meansto representfacts,conceptsorinstructionsinaformalizedmannersuitablefor
communicationinterpretationorprocessingbyhumansorotherautomaticmodes.
Differentwaysof Datarepresentation:-
(i) Narration based: - Questioninvolvestoriesthatdefine asituationandgive detailsof various
parametersinvolved.
(ii) Pictorial: - In such a formdata is presentedinvariouspictorial formssuchasline graphs,bar
diagrams,line chartsetc.
(iii) Table: - Tabularmethodisthe mostfundamental wayof representingdata.Mostof
differentkindsof datapresentationformatlike barcharts,line chartsetc originate fromthe
table.
(iv) Pie Chart: - Pie Chart istypical type of data representationwhere dataisrepresentedasa
part of a circle.The circle representsthe total value (100%) anddifferentpartrepresents
certainproportionsof the total.
There are twoapproachesconstructinga pie chart fromany givendata:-
(A) Degree Approach: - In thiscentral angle ina circle represents360◦. So anypart or a segmentin
pie-chartiscalculatedasa proportionof 360◦.
(B) Percentage Approach: - In thiscase,any part or segmentina pie-chartiscalculatedasa part of
100%.
RADAR DIAGRAM:-
In thisdiagrameveryvalue isrepresentedwithrespecttoacentral point.All the changesinthe values
are expressedinthe formof distance fromthiscenterpoint.

30. CLOCKS
A clock is an example of circular motion where length of track is equal to 60km.
Assume (1 minute = 1 km)
Now on this track, two runner i.e. hour hand and minute hand are running with a speed 5 km/h
and 60 km/h. Since their direction is same, so relative velocity (speed) = 60-5 =55 km/h.
Speed = Distance / Speed.
Or
Time = Distance / Speed. Time = 60 /55 = 12/11 hour.
DEGREE CONCEPT:-
 Total Angle subtended at the centre of a clock = 360˚.
 Angle made by hour hand at the centre per hour =30˚ (per minute = 0.5˚).
 Angle made by minute hand at the centre per hour =360˚. (per minute = 6˚).
IMPORTANT DERIVATION:-
 The number of times hand of a clock are in straight line (either at 0˚ or at 180˚) in 24
hour = 44.
 The number of times hand of a clock are at right angle (90˚) in 24 hour = 44.
 Both the hands of a clock are together after every 65 5/11 minutes.

31. CUBES
A cube is a three dimensional figure of square in which all the sides are equal.
Number of Faces in the cube= 6
Number of Vertices in the cube= 8
Number of Edges in the cube= 12
General cases for Cubes:-
A cube was cut into 64 cubelets. (n×n×n)
So, n³= 64, n =4.
(i) Cubelets with 1 face painted = 6(n-2)².
(ii) Cubelets with 2 face painted = 12(n-2).
(iii) Cubelets with 3 face painted = 8 .
(iv) Cubelets with no face painted = (n-2)³.
Total n³=6(n-2)²+ 12(n-2) + 8+ (n-2)³.

32. SERIES
TYPE 1:- Number Series
(i) Prime number Series:- Prime numbershave onlytwodistinctfactorslike 2,3, 5, 7, 11, 13,
17….. Theiroccurrence doesnotfollow anyparticularrule.
(a) Examples:-
(i) 3, 5, 7, 11, 13, 17, ____
 Rule --- Seriescontainaprime numberserieswill have the nextnumberas 19.
(ii) 5, 7, 10, 15, 22, _____
 Rule --- Seriescontaindifference betweentwosuccessivenumberis2,3,5,7and so
on,whichare prime numbers.
Nextnumber=(Last number)+(Nextprimenumber) =33.
(iii) 4,9,25,49,121,_____
 Rule --- Seriesisthe square of prime numbers.Hence the nextnumberwillbe (169)
whichis(13)².
(ii) The Difference Series:- Inthistype ,patternwouldbe obtainedbyfindingthe difference
betweenthe terms.Differencesare categorizedas1st
orderdifference,2ndorder
difference, and3rdorder difference andsoon.
Order of difference:-
 1st
orderdifference isdifference betweentwoconsecutivetermsof series.
Example:- 2 5 10 17 26 37
3 5 7 9 11
 2nd
orderdifferenceisdifference betweenthe consecutive termsobtainedfromthe
1st
orderdifference.
Example:- 2 5 10 17 26 37
3 5 7 9 11
2 2 2 2
(iii) The product Series:- A seriesiscalledaproduct serieswhenthe termisobtainedby
multiplyingbyaconstantor anyothernumber.
Example:- 3 6 18 90 630 (630 x9).
x 2 x 3 x 5 x 7 x 9

33. (iv) The MixedSeries:- The mixedserieshasthree typesof formsinvolved,whichcanbe
classifiedas
(a) A mixedseriesmaybe formedbymixingtwodifferentseries.
Example:-3,6,24, 30, 63, 72, ______, 132.
ELEMENTARY IDEAS OF PROGRESSION:-
(1) ARITHMETIC PROGRESSION (A.P):- The progressionof the forma,a+d,a+2d---- isknownasA.Pwith
first term =a & common difference =d.
Example:-3,6,9, 12, ____ is an A.Pwitha =3 and d= 6-3= 3.
InA.P,we have nth term= a+ (n-1) d.
(2) GEOMETRIC PROGRESSION (G.P):- The progressionof the forma,ar, ar², ar³------ isknownasG.P
first term =a & common ratio =r.
Example:-1,5,25, 125, ____ is a G.P witha =3 and r= 5/1= 5.
InG.P, we have nth term= arⁿ⁻¹.

34. SYLLOGISM
Syllogismmeans‘Interference’or‘deduction’.
Types ofpropositionsin syllogism:-
(i) Positive Universal: ---Example:- All catsare dogs.(ALL)
Particular:---Example:- Some catsare dogs.(SOME)
(ii) Negative Universal:--Example:- Nocatsare dogs.(NO)
Particular:--Example:-Some catsare not dogs.(SOMENOT)
Euler’scircle or Venndiagram:-
SOME CATS ARE DOGS ( SOME) ALL CATSARE DOGS (ALL)
NO CATSARE DOGS (NO) SOME CATS ARE NOTDOGS (SOME NOT)
Standard Deductions:-
Statements DefinitelyTrue
1) All cats are dogs(ALL) Some dogsare cats.(SOME)
Or
Some cats are dogs.(SOME)
2) Some cats are dogs(SOME) Some dogsare cats.(SOME)

35. 3) No cats are dogs (NO) No dogsare cats.(NO)
Some dogsare not cats.(SOME NOT)
Or
Some cats are notdogs.(SOMENOT)
Deduction of two or more Statements:-
First Statement SecondStatement DefinitelyTrue Conclusion.
ALL
ALL SOME SOME
ALL ALL ALL/SOME
ALL NO SOME NOT/NO
SOME
SOME ALL SOME
SOME NO SOME NOT
NO
NO ALL NO/SOME NOT
NO SOME SOME NOT
NO NO NO/SOME NOT

36. BLOOD RELATION
RELATIONSHIP TABLE:-
Example1:- Amitintroduce Gautam as the son of the onlybrotherof hisfather’swife.How isGautamas
the son of the onlybrotherof hisfather’swife .How isGautam relatedtoAmit?
Solution:- Startingfrombackend.
Father’swife Mother
Onlybrotherof hisfather’swife Onlybrotherof hismother.
Maternal Uncle.
Sonof the onlybrotherof hisfather’swife Sonof maternal uncle cousin.
Gautam andAmitare Cousins.
Mother’sor father’sson Brother
Mother’sor father’sdaughter Sister
Mother’sor father’sbrother Uncle
Mother’sor father’ssister Aunt
Mother’sor father’sfather Grandfather
Mother’sor father’smother Grandmother
Son’swife Daughterinlaw
Daughter’shusband Son inlaw
Husbandor wife’ssister Sisterinlaw
Husband’sor wife’sbrother Brotherin law
Brother‘s son Nephew
Brother’sdaughter Niece
Uncle or aunt’s sonor daughter Cousin
Sister’shusband Brotherin law
Brother’swife Sisterinlaw
Grandson’sor Granddaughter’sdaughter Great Granddaughter

37. CALENDERS
ODD DAYS :- Numberof daysmore than the complete weeksare calledodddaysina givenperiod.
LEAP YEAR: - A leapyearhas 366 days.In a leapyear,the monthof February has29 days.
(a) Everyyear divisiblebyisa leapyear,if itis not a century.
(b) Every4th
centuryisa leapyearand noothercenturyis a leapyear.
ORDINARY YEAR: - The yearwhichisnot a leapyearis an ordinaryyear.Anordinaryyearhas 365 days.
COUNTING ODD DAYS AND CALCULATING THE DAY OF ANY PARTICULAR DATE
1) 1 Ordinaryyear= 365 days = (52 weeks+1 day)
Hence numberof Odd daysin1 Ordinaryyear= 1
2) 1 Leapyear = 366 = (52 weeks+ 2 days)
Hence numberof Odd daysin1 leapyear= 2
3) 100 years= (76 Ordinaryyear+ 24 leapyear)
= (76 x 1 + 24 x 2) odd days
=124 Odddays
= (17 weeks+ 5 days)
=5 Odd days.
4) Numberof Odddays in200 Years= (5 x2) =10 days
= 1 week+ 3 days
= 3 odd days.
5) Numberof Odddays in4th
centuries(400,800,1600,2400….etc) = 0

38. DATA SUFFICIENCY
Directions: The questionsbelowconsistof aquestionfollowedbytwostatementslabeledas(1) and(2).
You have to decide if these statementsare sufficienttoconclusivelyanswerthe question.
(A) If statement(1) alone issufficienttoanswerthe questionbutthe statement(2) alone isnot
sufficienttoanswerthe question.
(B) If statement(2) alone issufficienttoanswerthe questionbutthe statement(1) alone isnot
sufficienttoanswerthe question.
(C) If you getthe answerfrom(1) and (2) togetheralthoughneitherstatementbyitself suffices.
(D) If statement(1) alone issufficientandstatement(2) alone issufficient.
(E) If you cannot getthe answerfromstatements(1) and(2) togetherbutstill more dataare
needed.
Example:1) Whatis the relationof Xand Y?
(i) Y is motherof Z.
(ii) Z is brotherof A.
STEPS :-
Yes No
Yes No
No Yes
Yes No
Firstcheck statement1.Is itsufficient?
EitherB or C or E is the correct answer.EitherA or D is correct answer.
Now,checkstatement(2).Isit sufficient?
AnswerisD AnswerisA
AnswerisB
Try both the statementtogether.Are the statementtogethersufficient?
AnswerisEAnswerisC

39. SOME TYPICAL CASES:-
These casesare dividedintofollowingcategories:-
(i) Relationship
(ii) Dates
(iii) Comparison
(iv) Age
(v) Critical Analysis
(vi) Miscellaneous
Relationships:-
Checkthe genderof the personinvolved&followsthe previoussteps.
Dates:-
Check (A) The dayor the date of some earlierincidentismentioned.
(B) The numberof daysbetweenthatincidentandthe requiredday is given.
Critical Analysis:-
Revise general backgroundof argumentationtechniques,assumptionsof argumentsandinference
making.