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Series: AP, GP and other formula:

Arithmetic Progression (AP): If a, a+d, a+2d, a+3d …………….are said to be in AP in which „a‟ is the first term and„d‟ is the common difference.Then,

  • nth term = a+(n-1)d
  • Sum of the n terms = [2a+(n-1)d]
  • Sum of the n terms = (a+l), where „l‟ is the last term of the series.

Geometric Progression (GP): If a,ar,ar 2 ,ar 3 ……………are said to be in GP in which „a‟ is the first termand „r‟ is the common ratio. Then,

  • nth term = \(ar^{n-1}\)
  • Sum of n terms =a\( \frac{(1-r^{n)}}{(1-r)}\) when r<1.
  • Sum of n terms = a  \(a\frac{(r^n-1)}{(r-1)}\), when r>1.

For natural number series,
(1+2+3+4+5+…………..+n) =\(\frac n2(n+1)\)
For natural square number series,
(1 2 +2 2 +3 2 +……………..+n 2 ) =\(\frac n6(n+1)(2n+1)\)
For natural cube number series,
(1 3 +2 3 +3 3 +………………+n 3 ) =\(\frac{n^2}4(n+1)^2\)

Simplification:

There is three main points to remember to solve simplification problem easily. Three points are,
  • BODMAS rule: This rule is about the correct sequence to be executed to find the value of thesimplification problem. Here  B means ‘Bracket’, O means of, D means ‘Division’, M means ‘Multiplication’, A means ‘Addition’ and S means “Subtraction”. For Bracketoperation we must follow the order of (),{} and [] in order to remove the Brackets. Then accordingly we must follow of, “Division”, “Multiplication”, “Addition” and “Subtraction”.
  • Modulus of a real number “a” is defined as,
    |a| = a, if a>1 and |a| = -a, if a<1.
  • Virnaculum or Bar: We an expression contain “Virnaculum” then before applying “BODMAS” rulewe must simplify the expression under “Virnaculum”. Example: (a+b), here we (a+b) is under the”Virnaculum” or “Bar”.
Examples:

Example: Simplify: – [{(2+3) +5}-6] + (6-3+2).

Answer: In the above example there is Virnaculum above the term ‘6-3’ so we need to simplify the expression under Virnaculum first before applying BODMAS.

= [{(2+3) +5}-6] + (3+2)
= [{5+5}-6] +5
= [10-6] +5
= 4+5
= 9, Final answer.

Example: Simplify:-[2×3÷3+ {(2-1) +6}]
Answer: In the above example we need to apply the BODMAS rule to find out the simplified value.
According to BODMAS rule bracket must remove first in order to (), {} and [].
= [2×3÷3+ {1+6}]
= [2×3÷3+ 7]
= [2×1+7]
= [2+7]
= 9, Final answer.

Profit and Loss:

To solve profit-loss numerical we need to go through some formulae,
  • Gain= \(Selling\;Price(SP)-Cost\;Price(CP)\)
  • Loss= \(Cost Price(CP)-Selling Price(SP)\)
  • Gain%= \(\frac{Gain8\times100}{CP}\)
  • Loss%= \(\frac{Loss\times100}{CP}\)
  • SP=\(\frac{(100+Gain\%)\times CP}{100}\)
  • SP=  \(\frac{(100-Gain\%)\times CP}{100}\)
  • CP=  \(\frac{100\times SP}{(100+Gain\%)}\)
  • CP=  \(\frac{100\times SP}{(100-Loss\%)} \)
  • If a person sell two similar items, one at gain of x% and other at loss of x% then seller always loss which is given by,Loss%=\(\left\{\frac{Common\;Loss\;and\;Gain\%}{10}\right\}^2=\left\{\frac x{10}\right\}^2 \)
  • If trader use wrong weight to sell his goods at cost price then ,Gain%= \(\left[\frac{Error\times100\%}{(True\;Value)-(Error)}\right]\)

Example 1: A man sold cloths at a loss of 10%. If the selling price had been increased by Rs. 150, there would have been a gain of 15%. What was the cost price of the article?

Solution: Let CP of the cloths be Rs. X.

At loss of 10 %,( applying unit system of solving)

For CP of Rs. 100 SP is Rs. 90

For CP of Rs. 1 SP is Rs.\(\frac{90}{100}\)

For CP of Rs. X SP is Rs.\(\frac{90\times x}{100} \)

Now SP is increased by Rs. 150, so new SP = \(\left(\frac{90\times x}{100}+150\right)\) Rs. = \(\left(\frac{90x+1500}{100}\right)\) Rs.
Since gain is 15% for the new SP, using CP formula for gaining we get,

$$X=\frac{100\times SP}{100+Gain\%}=\frac{100\times(9X+1500)}{(100+15)\times10}=\frac{90x+15000}{115}$$

115X=90X+15000

$$X=\frac{15000}{25}=600Rs.$$

So, CP of the cloths is Rs. 600. (Whole problem can be solved using formula only instead of unit system).

Example 2: After getting two successive discounts, a shirt with a list price of Rs. 150 is available at Rs.105. If second discount is 12.5%, find the first discount?

Solution: Let the first discount be X%.

Then, 87.5% of (100 – X) % of 150 = 105

$$\frac{87.5\times(100-X)\times150}{100\times100}=105\;<=>$$ 100-X <=> X =20.

So, first discount is 20%.

Time and Work:

To solve time and work problem we must use unit system. Because there is no specific formula for
solving these type of question. These types of questions can be understood with the help of some
examples.
Example 1: A can do a work in 5 days of 9 hours each and B can do it in 7 days of 7 hours each. How long will they take to do it, working together 8 hours a day?

Solution:
A can complete the work in (5×9) = 45 hours
B can complete the work in (7×7) = 49 hours

A‟s 1 hours work =\(\frac1{45}\)
B‟s 1 hours work =\(\frac1{49}\)
Both (A+B)‟s 1 hours work =\(\left(\frac1{45}+\frac1{49}\right)=\frac{94}{2205}\)

both will finish the work in\(\left(\frac{2205}{94}\right)\)hours.

Number of days of 8 hours each = \(\left(\frac{2205}{94\times8}\right)\) days = 2.93 = 3 days (approx.).

Example 2: A and B working separately can do a work in 8 and 12 days respectively. If they work for a day alternately, A beginning, in how many days the work will be completed?

Solution:
(A+B)‟s 2 days‟ work =\(\left(\frac18+\frac1{12}\right)=\left(\frac5{24}\right)\)

Work done in 4 pairs of days(8 days) =\(\left(\frac{5\times4}{24}\right)=\left(\frac{20}{24}\right)\)

Remaining work =\(\left(1-\frac{20}{24}\right)=\left(\frac16\right)\)

On 9 th day it is A‟s turn.

\(\left(\frac18\right)\)work is done by A in 9 th day. So now remaining work = \(\left(\frac16-\frac18\right)=\frac1{24}\)

So, now it is B‟s turn on 10 th day\(\frac1{12}\)work is done by B in 1 day.

So,\(\frac1{24}\)work is done by B in\left(\frac{12}{24}\right)day = 0.5 day.

So, total time taken = (9+0.5) = 9.5 days.

Example 3: 2 men and 5 boys can do a piece of work in 10 days while 3 men and 6 boys can do the same work in 8 days. In how many days can 3 men and 1 boy do the work?

Solution:
Let 1 man‟s 1 day‟s work = x
Let 1 boy‟s 1 day‟s work = y
Then according to the question we have two condition as below,

Solution:
Let 1 man‟s 1 day‟s work = x
Let 1 boy‟s 1 day‟s work = y
Then according to the question we have two condition as below,

\(2x+5y=\frac1{10}\) \(3x+6y=\frac18\)

Solving above two equations we get,

\(x=\frac1{120}\),\(y=\frac1{60}\)

(3 men + 1 boy)‟s 1 day‟s work =\(\left(\frac{3\times1}{120}+\frac1{60}\right)=\left(\frac1{40}+\frac1{60}\right)=\frac1{24}\)

So, 3 men and 1 boy can finish the work together in 24 days.

Relative speed, distance and problems on trains:

For this type of problems we need to remember some important formula, which are given below,

  • \(Time=\frac{Dis\tan ce}{Speed}\)
  • \(\frac{km}{hr\;}=\frac{5\;meter}{18\;second}\)
  • \(\frac{meter}{second}=\frac{18\;km}{5\;hr}\)

Some other important facts are given below,

  1. When a man covers a certain distance at x km/hr. and an equal distance at y km/hr. Then theaverage speed during the whole journey is\(\left(\frac{\left(2xy\right)}{\left(x+y\right)}\right)\).
  2. If two trains of length a meters and b meters are moving in the same direction at u m/s and v m/sthen time taken by the faster train to cross the slower train = \(\left(\frac{\left(a+\right)}{\left(u-v\right)}\right)sec\) sec.
  3. If two trains moving in opposite directions with same parameter =  \(\left(\frac{\left(a+b\right)}{\left(u+v\right)}\right)sec\)sec.

Example 1: While covering a distance of 24 km, a man noticed that after walking for 1 hour, the distancecovered by him was 5/6 of the remaining distance. What was his speed in meters per second?

Solution:
Let the speed be x km/hr.
Then distance covered in 1 hr. = x km
Remaining distance = (24 – x) km
So according to question,

\(x=\frac{5\times\left(24-x\right)}6;11x=120;x=10.9=11(approx).\)

Hence speed =\(=\frac{11\times5}{18}m/sec=3.06m/sec\)

Example 2: Walking at 5/6 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey?

Solution:
New speed = \(\frac56\)of usual speed.
New time taken = \(\frac65\)of usual time.
So, \(\left(\frac65of\;usual\;time\right)-\left(\;usual\;time\right)=10min\)
So, usual time = 50 min.

Example 3: A train 450 meter long is running with a speed of 60 kmph. In what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going?

Solution:
Speed of the train relative to the man = (60+7) = 67 kmph =\(\frac{67\times5}{18}\)m/sec.\(\frac{335}{18}\) =m/sec.
Now, time taken by the train to cross the man = Time taken by it to cover its own length of 450 meter at \(\frac{335}{18}\)m/sec. =\(\frac{450\times18}{335}\)=24.18sec

Example 4: A boy standing in a platform which is 200 meter long. He finds that a train crosses theplatform in 20 seconds but him in 10 seconds. Find the length of the train and speed?
Solution:
Let the length of the train be x meter.
Then, the train covers x meters in 10 seconds and (x+200) meters in 20 seconds.

so,\(\frac x{10}=\frac{x\times200}{20}\);x=200meters.

Speed of the train =\(rac{200}{10}\frac m{sec}\)=\(\left(20\times\frac{18}5\right)\)kmph.=72kmph.

Example 5: Two trains 160 meters and 185 meters in length are running towards each other on parallel lines, one at 45 kmph and another at 55 kmph. In what time will they be clear of each other from the moment they meet?

Solution:
Relative speed of the trains = (45+55) kmph = 100 kmph = \(\)m/sec= \(\left(\frac{250}9\right)\) m/sec.
Time taken by the train to pass each other = Time taken to cover (160+185) meter at \( \left(\frac{250}9\right)\)m/sec is  \( \left(\frac{345\times9}{250}\right)\)seconds=12.42seconds
Final answer = 12.42 seconds.

Boats and streams:

Important facts and formula,

  • In water the direction along the stream is called downstream and opposite the stream is upstream.
  • If speed of the boat in still water is u km/hr. and the speed of the stream is v km/hr. then,

speed doenstream=(u+v)km/hr

speed upenstream=(u-v)km/hr

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

speed in still water=\( \left(\frac{a+b}2\right)\)km/hr.

rate of stream=\(\left(\frac{a-b}2\right) \)Km/hr

Example 1: A man can row 22 kmph in still water. It takes him thrice as long to row up as to row down the river. Find the rate of the stream.

Solution:
Let man‟s rate upstream be x kmph.
Then his rate downstream = 3x kmph

So, rate in still water =\(\frac{\left(3x+x\right)}2\)=2xkmph

so,2x=22 or x=11.

So, rate upstream = 11 kmph, rate downstream = 33 kmph.

Hence rate of stream =\(\frac{\left(33-11\right)}2\)kmph = 11 kmph.

Example 2: A boat takes 12 hours for travelling downstream from a place A to place B and coming back to place C midway between A and B. If the velocity of the stream is 6 kmph and speed of the boat in still water is 10 kmph, what is the distance between A and B?

Solution:
Given that,
speed of the boat in still water = 10 kmph
rate of stream = 6 kmph

So,
speed downstream = (10 + 6) = 16 kmph
speed upstream = (10 – 6) = 4 kmph
Now, Let the distance between A and B be x kmph.

Then, Distance between A and C or B and C =\( \left(\frac x2\right)\) kmph.

So,

$$\frac x{10}+\frac{\displaystyle\frac x2}6=12\;\Leftrightarrow\frac x{10}+\frac x{12}=12\Leftrightarrow\frac{11x}{60}=12\Leftrightarrow x=\frac{12×60}{11}\Leftrightarrow x=65.45\;km$$

Distance between A and B = 65.45 km.

Alligation or Mixture:

First all, what is Alligation?
“Alligation is the rule with which we can find the ratio of two or more ingredient at the given price which
are mixed to produce a mixture of desired price”

Important Formula and facts

 

1. Rule of Alligation: If two ingredients are mixed together, then

 

\( \frac{quantity of cheaper}{quantity of dearer} \)=\( \frac{(CP\;ofdearer)-(Mean\;price)}{(Mean\;price)-(CP\;of\;cheaer)} \),where CP=Cost price

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Here C = Cost price of a cheaper unit, D = Cost price of a dearer unit, M = Mean Price

So,                       (Cheaper quantity) ( Dearer quantity) = (D-M)( M-C)

2. Suppose a vessel contains X unit of liquid from which Y units are taken out and replaced by water. After n operations, the quantity of pure liquid =\( \left[X\left(1-\frac yx\right)^n\right]\) units.

Example 1: How much water must be added to 60 liters of milk at 2 liters for Rs. 30 so as to have a mixture worth Rs. 12 a liter?

Solution:
Cost price of 1 liter of milk = Rs. 15. (As cost price of 2 liter of milk is Rs. 30)
Water is assumed to be price less in alligation type problems.

So cost price of 1 liter water = 0

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In the above flow diagram we find that,
The ratio of water and milk = (15-12):(12-0) = 3:12 = 1:4
So quantity of water to be added to 60 liters of milk =\(\left(\frac{1\times60}4\right) \)liters=15liters.

Example 2: A vessel contains 30 liter of alcohol. From this vessel 3 liters of alcohol was taken out and replaced by milk. This process was further repeated 3 times. How much alcohol now contained in the vessel?

Solution:
Apply the formula we get,

amount of alcohol after 4 operation=\( \left[30\left(1-\frac3{30}\right)^4\right]\)liters=\( 30\times\frac{\left(9\times9\times9\times9\right)}{\left(10\times10\times10\times10\right)}\)

Example 3: A container is filled with liquid, 4 parts of which are water and 6 parts are milk. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half milk?

Solution:
Suppose the container initially contains 10 liters of liquid (4part + 6 part)
Let x liters of this liquid be replaced with water.

Quantity of water in new mixture =\( \left(4-\frac{4x}{10}+x\right)\)liters

Quantity of milk in the new mixture =\( \left(6-\frac{6x}{10}\right)\)liters

Since in the new mixture amount of milk and water is same,

\( \left(4-\frac{4x}{10}+x\right)\)=\( \left(6-\frac{6x}{10}\right)\),=>6x+40=60-6x,12x=20,=>x=5/3

So the part of the mixture replaced =\(\frac{5\times1}{3\times10} \)=\(\frac16 \)

Simple and Compound interest:

Some important formula and facts are given below for simple interest in which interest is reckoned uniformly.

  •  S.I=\(\left(\frac{P\;x\;R\;x\;T}{100}\right)\),where S.I=Simple intrest,P=Principal,R=Rate of interest in %,T=Time in year.
  • From the above formula we get,P=\(\left(\frac{100\;x\;S.I}{R\;x\;T}\right)\; \),R=\(\left(\frac{100\;x\;S.I}{P\;x\;T}\right)\;\),T=\(\left(\frac{100\;x\;S.I}{P\;x\;R}\right)\;\)

In compound interest (C.I) the amount after the first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and this process is go on. Some formula related to C.I is given below (in the next page):-

  • When interest is compounded annually,

A=P\(\left(1+\frac R{100}\right)\;^n\),A=Amount,P=Principal,R=R% per annum,n=Times in years.

  • When interest is compounded half yearly,

A=P\(\left(1+\frac{\displaystyle\frac R2}{100}\right)\;^{2n} \),where A,P,R and n have usual meaning.

  • When interest is compounded quarterly,

A=P   \( \left(1+\frac{\displaystyle\frac R4}{100}\right)\;^{4n}\),where A,P,R and n have usual meaning.

  • Present worth of Rs. X due n years hence is given by,

Present Worth=\(\frac X{\left(1+\frac R{100}\right)\;^n}\),where R=R%per annum,n=Times in years.

Example 1: A certain sum of money amounts to Rs. 1000 in 4 years and Rs. 1200 in 5 years. Find the sum and the rate of interest.

Solution:
S.I for 1 years = Rs. (1200-1000) = Rs. 200
S.I for 4 years = Rs. (200×4) = Rs. 800
So, principal = Rs. (1000 – 800) = Rs. 200
Now P = 200, T= 4, S.I = 800

R=\(\frac{100×800}{200×4}\)%=100%

So, rate of interest is 100%

Example 2: What the annual installment will discharge a debt of Rs. 1200 due in 3 years at 15% simple interest?

Solution:
Let each installment be Rs. X, Then,\(\left(X+\frac{X\;x\;15\;x\;1}{100}\right)+\left(X+\frac{X\;x\;15\;x\;2}{100}\right)+X=1200\)

\(\frac{23X\;}{20}+\frac{13X}{10}\)+X=1200<=>(23X+26X+20X)=(1200*20)<=>X=347.83

So, each installment is Rs. 347.83.

Example 3: If the simple interest on sum money at 6% per annum for 2 years is Rs. 900, find the compound interest for same period with 5% per annum for the same sum?

Solution:
Principal, P = Rs.\(\frac{100×900}{2×6}\)=Rs.7500

Amount, A = Rs.\( \left[7500x\left(1+\frac5{100}\right)^2\right]\)=Rs.\(\left(7500x\frac{21×21}{20×20}\right)\)=Rs.768.75.

Example 4: In what time will Rs. 4000 becomes Rs. 4300 at 15% per annum compounded half yearly?

Solution:
Let the time be n years.
Applying formula we get,

\(\left[4000\;x\;\left(1+\frac{\displaystyle\frac{15}2}{100}\right)^{2n}\right]\;=4300\;Or\;\left(\frac{43}{40}\right)^{2n}\;=\frac{43}{40}\;Or\;n=\frac12years.\)

Required time is 6 month.

Example 5: What annual payment will discharge a debt of Rs. 7600 due in 2 years at 20% per annum compound interest?

Solution:
Let each installment be Rs. X. Then,
(Present worth of Rs. X due 1 years hence) + (Present worth of Rs. X due 2 years hence)= 7600

\(\Leftrightarrow\frac X{\left(1+{\displaystyle\frac{20}{100}}\right)^1}+\frac X{\left(1+{\displaystyle\frac{20}{100}}\right)^2}\)=7600

\(\Leftrightarrow\frac{5X}6+\frac{25X}{36}\)=7600

30X+25X=7600*36

\(\Leftrightarrow X=\frac{7600×36}{55}=4975\;\left(approx.\right)\)

So, amount of each installment = Rs. 4975

Permutation and Combination:

Some Definitions,

Factorial Notation: Let n be positive integer. Then, factorial n denoted by n! is defined by,
n! = n (n-1) (n-2)……..3.2.1

Permutation: The different arrangements of a given number of things by taking some or all at a time are called permutation. Permutation is defined as,

\( {}_r^nP=\frac{n!}{(n-r)!}\)

Combination: each of the different groups or selections which can be formed by taking some or all of a number of objects is called combination. It is defined as,

\({}_r^nC=\frac{n!}{r!(n-r)!} \)

Example 1: How many words can be formed by using all letters of the word “DELHI”?

Solution: There is 5 different letters in “DELHI”. Required number of words = 5! = (5×4×3×2×1) = 120.

Example 2: How many words can be formed from the letters of the word “ASSOM” so that vowels never come together?

Solution:
There is 5 letters in word „ASSOM‟
Now writing vowels together we get „AOSSM‟
If we consider AO as one letter the word like “(AO) SSM”.
These 4 letters can be arranged in 4! = 24 ways.
And two vowels AO can be arranged among them in 2! = 2 ways.
But „S‟ occurs twice

So number of words having vowel together =\(\frac{4!\times2!}{2!} \)=24

Total number of words formed by using all letter in the given word = 5! = 120.
Number words each having vowels never together = (120 – 24) = 96.

Example 3: How many ways can a football eleven team be chosen out from a batch of 15 players?

Solution:
Required number ways =\( {}_{11}^{15}C=\frac{15\times14\times\times1\times12\times11!}{11!\times4\times3\times2\times1}\)=15*13*7=1365

Example 4: A box contains 3 white cubes, 4 blue cubes and 2 red cubes. In how many ways can 3 cubes be drawn so that at least 1 white cube is to be included in the draw?

Solution:

200
For getting at least 1 white cube we may have,
(3 white cube) or (1 white and 2 non-white) or (2 white and 1 non-white)

So required number ways =\({}_3^3C+({}_1^3C+{}_2^6C)+({}_2^3C\times{}_1^6C) \)

= 1+ (3×15) + (3×6)

=1+45+18=64.

Probability:

“Probability is defined as the measure of the likeliness of an event occurrence. Uncertain event can be predicted with the idea of probability”

The idea of probability can be understand with the help of tossing a coin. When a coin is tossed, there are two possible outcomes:

  • Heads or H
  • Tails or T

So, we say that probability of coin landing H is 1⁄2 and probability of coin landing T is also 1⁄2.

So probability of occurrence of an Event if „S‟ be the sample space and „E‟ be the event,

P(E)=\(\frac{n(E)}{n(S)} \)

Some results on probability:

  1. P(S)=1
  2. P(0)=10
  3. \( \overset\rightharpoonup{P(A)}\)=1-P(A),where
    \(\overset\rightharpoonup A \) denotes (not A).

Example 1: Three unbiased coins are tossed. What is the probability of getting at least one tail?

Solution:
Here S = {HHH,HHT,HTH,THH,TTT,THT,TTH,HTT}
Let E = event of getting at least one Tail(T).
So, E = {HHT, HTH, THH, TTT, THT, TTH, HTT}

So,\(P(E)=\frac{n(E)}{n(S)}=\frac78 \)

Example 2: A speaks truth in 80% cases and B in 90% case. In what percentage of cases are they likely to contradict each other, narrating the same incident?

Solution:
Let A = Event that A speaks the truth.
Let B = Event that B speaks the truth.

Then,\( P(A)=\frac{80}{100}=\frac45\)and\(P(B)=\frac{90}{100}=\frac9{10} \)

And,\( P(\overline A)=\left(1-\frac45\right)=\frac15\)and\(P(\overline B)=\left(1-\frac9{10}\right)=\frac1{10} \)

Now, P (A and B contradict each other)

= P [(A speak the truth and B tells a lie) or (A tells a lie and B speaks the truth)]

=P[(Aand\( \overline B\)) or (\( \overline A\) and B)]

=P(Aand\( \overline B\)) or (\( \overline A\) and B)

=P(A).P(\( \overline A\) )+P(\( \overline A\)).P(B)

\(=\frac{(4\times1)}{(5\times10)}+\frac{(1\times9)}{(5\times10)}=\frac4{50}+\frac9{50}=\frac{13}{50}=\left(\frac{13}{50}\times100\right)\%=26\% \)

So, A and B contradict each other in 26% cases.

Example 3: In a single throw of a die, what is the probability of getting a number greater than 3?

Solution:
When a die is thrown we have S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number greater than 3 = {4, 5,6}

So,P(E)=\(=\frac{n(E)}{n(S)}=\frac36=\frac12 \)

 

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