Ysr navasakam
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A.D.
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how to 9999*9997 in 5 secs with Base method
Procedure Do the tricks
- Find the near base
- divide into two sections
- left side is cross sum and right side multiplication of two values
- for right side we need to take result number should be equal to the number of zeros in base
For better knowing we will take simple Example
1) Solve 7*8? It is easy for this but in the point of large numbers it is useful
Ans) step by step
- Finding the nearer base
7 near base is 10
8 near base is 10
- Finding the Deviation from Base
7 -3( it comes 10-7= -3)
8 -2 ( it comes 10-8= -2)
- Next step DO the the third step
7 -3
8 -2
-----------------------------
Either (7+(-2) or (8+(-3) | multliply -3*-2( take result euqal to num of zeros in base)
5 | 6
So ans is 7*8 = 56
2. Solve 999*996?
Ans)
- Near Base is 1000
- Finding the Deviation
999 -1
996 -4
- FInd the ans by cross sum and multiplication
999 -1
996 -4
--------------------------------
995 | 4( is the right ans but not because num of zeros in base is 3 so we will take 004)
=> 9997 -3
9999 -1
--------------------------------
9996 | 3(
So the final answer is 999*996 = 995004
3.Find 9997*9999?
Ans)
- Near Base is 10000
- Finding the Deviation
9997 -3
9999 -1
- FInd the ans by cross sum and multiplication
=> 9997 -3
9999 -1
--------------------------------
9996 | 3( is the right answer but not because num of zeros in the base is 3 so we will take 004)
=> 9997 -3
9999 -1
--------------------------------
9996 | 003
So the final ans is 999*996 = 9996003
DRDO-CEPTAM has extended date to apply for 21 educational categeries jobs
Name of the Exam : DRDO-CEPTAM
Vacancy details As per the categeory
| Post Code | Subject | SC | ST | OBC | UR | Total | Code of posting location |
|---|---|---|---|---|---|---|---|
| 0101 | Agriculture | 0 | 0 | 1 | 3 | 4 | H1,L1 |
| 0102 | Automobile Eng | 0 | 1 | 2 | 3 | 6 | A2,B2,C2 |
| 0103 | Botany | 0 | 0 | 2 | 1 | 3 | D2,H1 |
| 0104 | Chemical Eng | 1 | 1 | 1 | 10 | 13 | A3,H2,J1,N1,P2, V1 |
| 0105 | Chemistry | 3 | 2 | 9 | 10 | 24 | A3,D2,G1,H1,H2, J1,J2,K1,K2,N1, P2 |
| 0106 | Civil Enginnering | 0 | 0 | 1 | 3 | 4 | A1,D2,H2 |
| 0107 | Computer Science | 9 | 6 | 24 | 40 | 79 | A1,A2,A3,B1,B2, B3,D2,H2,J2,K2, P1,P2 |
| 0108 | Electrical & Electronics Engineering | 2 | 1 | 4 | 9 | 16 | B2,B3,C2,H2 |
| 0109 | Electrical Engineering | 5 | 2 | 8 | 20 | 35 | A2,B1,B2,H2,K2, P1,P2,V1 |
| 0110 | Electronics & Instrumentation | 1 | 0 | 2 | 4 | 7 | A1,C1,H2 |
| 0111 | Electronics or Electronics & Communication or Electronics & Telecommunication Engineering | 10 | 8 | 26 | 56 | 100 | B1,B2,C2,D1,D2, H2,J2,K2,M1,P2 |
| 0112 | Geology | 0 | 0 | 1 | 2 | 3 | D2 |
| 0113 | Instrumentation | 1 | 0 | 1 | 3 | 5 | B2,J1,N1,P2 |
| 0114 | Library Science | 0 | 1 | 3 | 7 | 11 | A1,C2,D2,H2,N1, P2 |
| 0115 | Mathematics | 3 | 0 | 2 | 3 | 8 | D2,M2 |
| 0116 | Mechanical Engineering | 30 | 9 | 35 | 66 | 140 | A1,A2,A3,B1,B2, C2,D1,D2,H2,J1, K1,K2,N1,P1,P2, V1 |
Age Limit: A) The candidate must be within prescribed age limit on the crucial date of eligibility. However, upper age limit is relaxable for SC, ST, OBC, PWD, ESM, Central Govt. Civilian Employees, Disabled Defence Services Personnel, Persons domiciled in Jammu & Kashmir state during the period from 01-01-1980 to 31-12-1989 as per the rule prescribed by the Government of India. No age relaxation is allowed to SC/ST/OBC candidates applying against unreserved (UR) vacancies. Candidates belonging to PWD, ESM, Central Govt. Civilian Employees, Disabled Defence Services Personnel, Persons domiciled in Jammu & Kashmir state during the period from 01-01- 1980 to 31-12-1989 categories, who are applying against UR vacancies will get age relaxation benefit only for their respective category as above and no additional relaxation will be given for belonging to SC/ST/OBC category (Refer FAQ for further details). B) For relaxation, proforma for certificates may be downloaded (if requireFee required: For Gen: Rs:100/- remaining categeries : nil
REMUNERATION AND SERVICE CONDITION: Pay, as per 7th CPC Pay matrix, at level 6 and other benefits include dearness allowance, house rent allowance, transport allowance, children education allowance, leave travel concession, medical facilities, CSD facility and other allowances/advances. Recruited candidates will be covered under national pension system (NPS) of the government unless provided otherwise as per Govt. of India rules. DRDO has beautiful well laid out green campuses with residential quarters, general amenities & sports facilities at most of the laboratories/ establishments. The recruited candidates will be governed by the central government rules. The personnel policies in DRDO are well laid down. The selected candidates will be appointed on probation and are liable to serve anywhere within limits of union of India including field locations / remote areas, as and when required, as per Govt. of India rules. These posts are covered under merit based limited flex
RESERVATION/RELAXATION BENEFITS: A) Reservation/ relaxation benefits regarding age, minimum qualifying criteria etc. are applicable to the SC/ST/OBC/PWD etc. candidates applying against vacancies earmarked for them, in accordance with the instructions/orders / circulars, as per extant Govt. of India orders. All candidates applying against unreserved (UR) vacancies will be treated as general candidates. B) Candidates seeking reservation/relaxation benefits must support their claim with duly selfattested copies of relevant certificates issued by Govt./notified competent authority, at the time of submitting online applications and document verification or whenever required by DRDO, else their claim for any relaxation/concession etc. will not be considered and their application will be treated under unreserved category. C) A candidate seeking reservation/ relaxation benefits of OBC must ensure that he/she possess a genuine caste/ community certificate in central Govt. format and does not fall in creamy layer on the crucial date of eligibility. D) Candidates with physical disability of 40% and more only would be considered as person with disability (PWD) and entitled to reservation for PWD. E) It may be noted that, candidature will remain provisional till the veracity of the concerned documents are verified/reverified by the appointing authority.
3.1 HOW TO APPLY: Candidates must satisfy themselves, before applying, about their eligibility for the post. Candidates are, therefore, urged to carefully read the advertisement and complete the application form and submit the same as per the instructions to avoid rejection later. A) All candidates must apply online through the link available on CEPTAM notice board (https://www.drdo.gov.in/drdo/ceptam/ceptamnoticeboard.html). Applications received by any other mode will be summarily rejected. B) The online application portal will be opened on 04 August 2018 and closed on 13 Sep 2018 (05:00 PM). C) Firstly, the candidate must register online by filling up the basic details. After registration, the candidate will get a user Id & password (note it down & keep safely), which will be used to login for filling of application. D) Candidates are advised not to wait until last date to submit their applications. CEPTAM will not be responsible, if candidates are not able to submit their applications on time due to last minute heavy rush, network congestion etc. E) The following documents and their scanned copies must be kept ready before filling the application: (i) 10th class or equivalent certificate for age proof. (ii) EQR certificates e.g. Graduation, Diploma etc. (as applicable). (iii) Photograph (Use only recent colour photograph taken within last 30 days. Keep 7 copies of the same photograph for future use.) (iv) Clear Left thumb impression taken with stamp pad on plain white paper. (v) Signature on plain white paper. (vi) Caste certificate in Central Govt. format (wherever applicable). (vii) PWD certificate (wherever applicable). (viii) Identity proof (ID) (e.g. Aadhaar, Passport, Voter ID, PAN, Official ID, etc.) which must be carried during the examination & document verification. F) Candidates must fill their name, date of birth, father’s name & mother’s name as given in matriculation/secondary examination certificate, otherwise their candidature may be cancelled. G) Candidates should have their own mobile number and valid & active personal email id. H) Contact details such as e-mail, mobile number, communication & permanent address etc. must be correct & active during the recruitment cycle as all communication will be done through them. I) Submission of Application: (i) Candidates should read the instructions carefully before making any entry or selecting options. The detailed instructions for filling-up of online application are available on our website. (ii) Candidates should save and review his/her application before submission. (iii) Candidates are required to make online payment of application fee (wherever applicable). (iv) After final submission, request for change/correction in the particulars given in the application form, shall not be entertained under any circumstances. CEPTAM will not be responsible for any consequences arising out of non-acceptance of any correction/deletion in any particular given by candidates in application form. Hence candidates are advised to fill the application form carefully. J) Candidates are advised to take a printout of application and keep safely, bring it, at the time of document verification (if shortlisted). No printed copy of application is required to be sent to CEPTAM. However, candidates have to produce duly signed printed copy of application at the time of document verification, if provisionally shortlisted.
- 3.2 APPLICATION FEE, EXEMPTION FROM PAYMENT OF FEE AND MODE OF PAYMENT: A) APPLICATION FEE: Non-refundable application fee of Rs. 100/- (Rupees one hundred only) is to be paid by the candidate. The fee should be paid separately for each post code applied.
- B) EXEMPTION FROM PAYMENT OF FEE: All women and SC/ST/PWD/ESM candidates are exempted from payment of application fee, as per Govt. of India rules.
- C) MODE OF PAYMENT: Fee is to be paid online through credit card/debit card/net banking. Fee once paid will not be refunded under any circumstances.
- D) Ex-servicemen, who have already secured employment in civil side under Central Government on regular basis after availing of the benefits of reservation given to ex-servicemen for their reemployment, are NOT eligible for fee concession.
3.3 A) EXAMINATION CITIES FOR TIER-I: Candidates are advised to choose any three different cities from the following list in order of preference for Tier-I examination. The option/preference once given by the candidate will be treated as final and irreversible. No request for change of examination city will be entertained. CEPTAM reserves the right to add/delete any examination city and allot the candidates to any examination city other than chosen by candidate depending upon the operational constraints. 01 AGRA 17 IMPHAL 33 MYSORE 02 AHMEDABAD 18 INDORE 34 NAGPUR 03 BALASORE 19 ITANAGAR 35 NASIK 04 BENGALURU 20 JABALPUR 36 PANAJI 05 BHOPAL 21 JAIPUR 37 PATNA 06 BHUBANESWAR 22 JALANDHAR 38 PORT BLAIR 07 BIKANER 23 JAMSHEDPUR 39 PUNE 08 CHANDIGARH 24 JAMMU 40 RAIPUR 09 CHENNAI 25 JODHPUR 41 RAJKOT 10 DEHRADUN 26 KANPUR 42 RANCHI 11 DELHI NCR 27 KOCHI 43 SILIGURI 12 GORAKHPUR 28 KOLKATA 44 THIRUVANANTHAPURAM 13 GUWAHATI 29 LUCKNOW 45 VARANASI 14 GWALIOR 30 MADURAI 46 VIJAYWADA 15 HAMIRPUR 31 MANGALORE 47 VISAKHAPATNAM 16 HYDERABAD 32 MUMBAI B) EXAMINATION CITIES FOR TIER-II: No choice for city is required to be given by the candidates for Tier-II examination. Examination cities for the same will be decided by CEPTAM based on the operational requirement.
3.4 REJECTION CRITERIA: The rejection of applications will be based on following grounds:
- A) Not meeting EQR.
- B) Incomplete or partially filled Application.
- C) Applications without Fees (wherever applicable).
- D) Applications not received through proper mode/channel.
- E) Applications having blurred/irrelevant photo, signature, thumb impression or other documents.
- F) Underage or overage as on crucial date of eligibility.
- G) Higher qualification viz. M.Sc. or B.Tech. or B.E. or Ph.D. degree etc. as on crucial date of eligibility. .
- H) If a candidate submits more than one application successfully for same post code, then only the latest application with application fee (if applicable) will be considered and other applications will be rejected.
Basics of Profit and Loss-Part1
Welcome to www.job-updates.com
First of all welcome to www.job-updates.com and we are providing all job updates, jobs information, shortcuts or tricks on Aptitude(Number series, simplification, time and work, time and distance, time and speed), reasoning, General Knowledge, as well as current affairs.
First of all welcome to www.job-updates.com and we are providing all job updates, jobs information, shortcuts or tricks on Aptitude(Number series, simplification, time and work, time and distance, time and speed), reasoning, General Knowledge, as well as current affairs.
Profit and Loss
Basic Definitions and Formulas
- Cost price (C.P.): Buying price.
- Selling price (S.P.): sales Price.
- Profit/Gain: If the selling price is more than the cost price, the difference between them is the profit occurred.
Formula: Profit/ Gain = S.P. – C.P.
- Loss: If the selling price is less than the cost price, the difference between them is the loss occurred.
Formula: Loss = Cost price (C.P.) – Selling Price (S.P.)
- Profit or Loss is always calculated on the cost price.
- Marked price: This is the price marked as the selling price on an article, also known as the listed price.
- Discount or Rebate: This is the reduction in price offered on the marked or listed price.
Below is the list of some basic formulas used in solving questions on profit and loss:
- Gain % = (Gain / CP) * 100
- Loss % = (Loss / CP) * 100
- SP = [(100 + Gain%) / 100] * CP
- SP = [(100 – Loss %) / 100]*CP
The above two formulas can be stated as,
If an article is sold at a gain of 10%, then SP = 110% of CP.
If an article is sold at a loss of 10%, then SP = 90% of CP.
- CP = [100 / (100 + Gain%)] * SP
- CP = [100 / (100 – Loss%)] * SP
Profit and Loss: Solved Examples
Question 1: An article is purchased for Rs. 450 and sold for Rs. 500. Find the gain percent.
Solution:
Gain = SP – CP = 500 – 450 = 50.
Gain% = (50/450)*100 = 100/9 %
Question 2: A man sold a fan for Rs. 465. Find the cost price if he incurred a loss of 7%.
Solution:
CP = [100 / (100 – Loss %)] * SP
Therefore, the cost price of the fan = (100/93)*465 = Rs. 500
Question 3: In a transaction, the profit percentage is 80% of the cost. If the cost further increases by 20% but the selling price remains the same, how much is the decrease in profit percentage?
Solution:
Let us assume CP = Rs. 100.
Then Profit = Rs. 80 and selling price = Rs. 180.
The cost increases by 20% → New CP = Rs. 120, SP = Rs. 180.
Profit % = 60/120 * 100 = 50%.
Therefore, Profit decreases by 30%.
Question 4: A man bought some toys at the rate of 10 for Rs. 40 and sold them at 8 for Rs. 35. Find his gain or loss percent.
Solution:
Cost price of 10 toys = Rs. 40 → CP of 1 toy = Rs. 4.
Selling price of 8 toys = Rs. 35 → SP of 1 toy = Rs. 35/8
Therefore, Gain = 35/8 – 4 = 3/8.
Gain percent = (3/8)/4 * 100 = 9.375%
Question 5: The cost price of 10 pens is the same as the selling price of n pens. If there is a loss of 40%, approximately what is the value of n?
Solution:
Let the price of each pen be Re. 1.
Then the cost price of n pens is Rs. n and
the selling price of n pens is Rs. 10.
Loss = n-10.
Loss of 40% → (loss/CP)*100 = 40
Therefore, [(n-10)/n]*100 = 40 → n = 17 (approx)
basics of simplification for SSC UPSC IBPS
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Now, Try It Yourself:
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Simplification
Step 1: Parts of the expression enclosed in ‘Brackets’ must be solved first. Inside the brackets, once again BODMAS rules apply!
Step 2: The mathematical operators ‘of’ and ‘order’ must be solved next. ‘Of’ means part of and is solved by substituting with a multiplication sign. ‘Order’ is the same as exponent. Powers are solved after brackets. Powers also include roots.
Step 3: Next, the parts of the equation that contain ‘Division’ and ‘Multiplication’ are calculated.
Step 4: Last but not least, the parts of the equation that contain ‘Addition’ and ‘Subtraction’ should be calculated.
Here is an example, to help you understand the BODMAS rule concept.
EXAMPLE:
What will come in place of question mark (?) in the following question?
240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (11 × 24 ÷ 8 × 3 – 69)2} = ?
(1) 4111
(2) 3441
(3) 4441
(4) 3841
Ans: (4)
Solution:
Given expression is,
240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (11 × 24 ÷ 8 × 3 – 69)2} = ?
According to the BODMAS, first we need to solve the Brackets.
Here in this equation first we will solve curly bracket and inside the bracket, we will again follow BODMAS Rule. In the curly bracket once again another bracket appears. So we solve that bracket first. Within these smaller brackets, there is no ‘of’ or ‘order’, so we proceed with the following steps in BODMAS.
We see multiplication, division and subtraction signs. Since multiplication and division are of the same rank, we go left to right to solve this. First we multiply, then we divide. Then we perform the second multiplication. After that, we proceed to the subtraction.
Here are the steps to illuminate this process.
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (264 ÷ 8 × 3 – 69)2} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (33 × 3 – 69)2} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (99 – 69)2} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ 302} = ?
Now we have reached the end of the small brackets. We encounter our first exponent. So we need to solve this ahead of division.
Now solving curly bracket,
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ 900} = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ of 2 = ?
So far we have cleared all the brackets. Now we move to the next step. We come across our first ‘of’. Here, we simply treat ‘of’ as product i.e. multiplication. But note that, ‘of’ will be solved before a regular multiplication.
⇒ 240 ÷ 8 × 512 ÷ 4 + ½ × 2 = ?
⇒ 240 ÷ 8 × 512 ÷ 4 + 1 = ?
Now, we have cleared our expression of all brackets, exponents and ‘of’s. We now move to the 3rd step. Here we perform all the divisions and multiplications. Note that these two operations are the same rank. So we perform either in the order we come across them in, from left to right.
⇒ 30 × 512 ÷ 4 + 1 = ?
⇒ 15360 ÷ 4 + 1 = ?
⇒ 3840 + 1 = ?
Now we have cleared the expression of all multiplication and division operations as well. All we are left with is addition and subtraction.
⇒ ? = 3841
Hence, the required answer is 3841.
Now, Try It Yourself:
Que. 1
What will come in place of question mark (?) in the following question?
24 + 13 – 5 × ½ of 10 - {45 ÷ (17 – 2)} =?
Solution
Answer:
Follow BODMAS rule to solve this question:
Given expression is,
24 + 13 – 5 × ½ of 10 – {45 ÷ (17 – 2)} =?
⇒ ? = 24 + 13 - 5 × ½ of 10 – {45 ÷ (15)}
⇒ ? = 24 + 13 – 5 × ½ of 10 – {3}
⇒ ? = 24 + 13 – 5 × (½ × 10) – 3 (of means ‘×’)
⇒ ? = 24 + 13 – 5 × 5 – 3
⇒ ? = 24 + 13 – 25 – 3
⇒ ? = 9
Hence, the required answer is 9.
how to solve five digit Square IBPS
how to solve five digit Square in less time
First of all welcome to www.job-updates.com and we are providing all job updates, jobs information, shortcuts or tricks on Aptitude(Number series, simplification, time and work, time and distance, time and speed), reasoning, General Knowledge, as well as current affairs.
Now we will go to the Topic How to square a number easily
Similer topic how-to-cube-number-in-just-5secs
First we are starting this topic we just learn three basic formula's for this topic
- (A+B)^2 =(A^2)+(B^2)+(2*A*B)
- (A+B)^2 =(A^2)+(B^2)-(2*A*B)
- (A^2)-(B^2) =(A+B)(A-B)
Now we will practice some problems on this
Based on Formala -1
Ex:- (102)^2=?
Ans) According to Formula we will devide 102 into 100+2
(102)^2 = (100+2)^2
= (100)^2+(2)^2+(2*100*2)
= 10000+4+400
(102)^2 = 10404
Like that we can solve 5-disit number also
Based on Formala -2
Ex:- (98)^2=?
Ans) According to Formula we will devide 98 into 100-2
(98)^2 = (100-2)^2
= (100)^2+(2)^2-(2*100*2)
= 10000+4-400
(98)^2 = 9604
Like that we can solve 5-disit number also
Based on Formala -3
Ex:- (95)^2-(5)^2=?
Ans) According to Formula we will write it as (95)^2-(5)^2 = (95+5)(95-5)
(95)^2-(5)^2 = (95+5)(95-5)
= 100*90
= 9000
Viral Aptitude Questions in japan & 60% failed to answer You?
Viral maths Basics rules of Aptitude
Do you know the Basic fundmentals of Aptitude ?
The question Which is viral in Japan is
Q) 9-3/(1/3)+1=?
Ans) Before going to the answer we just know about basic fundamentals of aptitude
We need to apply BODMAS rule to solve the equation.. which is
B - Bracket
O - of
D - Division
M - Multiplication
A - Addition
S - Substraction
we need to follow the above Priority to solve
9-3(1/3)+1 => 9-9+1 (Assume 3/(1/3) => (3/1)/(1/3) => 9/1 =>9
=0+1
=1
Example 2:
2. 5+63/9+2-12/3=?
Ans ) first we need to apply BODMAS rule
= 5+63/9+2-12/3
IN this problem no brackets so we go for solving divisions first
= 5+7+2-4
Next No multiplications available so we go for addition
= 14-4
Next we go for substraction
= 10
Ans 5+63/9+2-12/3 = 10
Do you know the Basic fundmentals of Aptitude ?
The question Which is viral in Japan is
Q) 9-3/(1/3)+1=?
Ans) Before going to the answer we just know about basic fundamentals of aptitude
We need to apply BODMAS rule to solve the equation.. which is
B - Bracket
O - of
D - Division
M - Multiplication
A - Addition
S - Substraction
we need to follow the above Priority to solve
9-3(1/3)+1 => 9-9+1 (Assume 3/(1/3) => (3/1)/(1/3) => 9/1 =>9
=0+1
=1
Example 2:
2. 5+63/9+2-12/3=?
Ans ) first we need to apply BODMAS rule
= 5+63/9+2-12/3
IN this problem no brackets so we go for solving divisions first
= 5+7+2-4
Next No multiplications available so we go for addition
= 14-4
Next we go for substraction
= 10
Ans 5+63/9+2-12/3 = 10
how to find cube and cube root of a number
Topic:1 How to find the cube root of a number
First of all welcome to job-updates.comBefore we are going to the topic of cube root of number faster we need to learn a simple table for easy shortcuts
Number Cube Last digit
----------------------------------------------------
1 1 1
2 8 8
3 27 7
4 64 4
5 125 5
6 216 6
7 343 3
8 512 2
9 729 9
10 1000 0
1 1 1
2 8 8
3 27 7
4 64 4
5 125 5
6 216 6
7 343 3
8 512 2
9 729 9
10 1000 0
The simple trick to remember the last digit of this table because this table is very important to find the cube root
For numbers 1,4,5,6,9,0 is same the last digit occurs
Number 2 flips 8 and vice versa
Number 3 flips 7 and vice versa
Now we will learn the basic procedure to solve the cube root of a number
- Take the unit digit same to the result.
- Leave the last three digits of the number.
- The last step is to find the nearest cube and put the left side of the unit digit.
- Now you will get the result of the cube root.
Now we will solve the examples of the cube root
1.find the cube root of 39,304?
Ans) From the above procedure
- first finding the unit digit so it is 4
- Leaving the last three numbers so leave304 we get remaining 39
- find the nearest cube for 39 which is 27= 33
- Finally, club the answer to get the cube root of 39,304
which is the cube root of 39,304= 34
1.find the cube root of 636,056?
Ans) From the above procedure
- first finding the unit digit so it is 6
- Leaving the last three numbers so leave 056 we get remaining 39
- find the nearest cube for 39 which is 512= 83
- Finally, club the answer to get the cube root of 636,056
which is the cube root of 636,056= 34
Topic2: How to cube a number in just 5sec
Now we are learning about cube of a two digit number
Now we divide into Four categories which is belowThe procedure is just 1, x1,x2,x3 Here X is the Second digit of a Number
and in a second line leave the first and last digits and then double the remaining numbers and Add the now then we get result
For Example 123= 1 2 4 8
+ 4 8
-----------------------------
1 7 2 8
Now 123= 1728
2.Number Ends With "1"
The procedure is reverse of above
x3,x2,x1,1 like this here x is the first value
and in a second line leave the first and last digits and then double the remaining numbers and Add the now then we get result
For Example 313 = 27 9 3 1
+ 18 6
----------------------------------
29 7 9 1
Ans 313= 29791
3.Same number repeated:
The Procedure for this forall four we just enter as x3,x3,x3,x3 same here X is same so no issue andin a second line leave the first and last digits and then double the remaining numbers and Add the now then we get result
For Example 223 = 8 8 8 8
+ 16 16
---------------------------
10 6 4 8
Ans 223 = 10648
4.Different number comes:
Here is different from all
Here we consider 1st digit as X and 2nd Digit as Y
The procedure x3,(x2*y),(y2*x), y3
and a second line leave the first and last digits and then double the remaining numbers and Add the now then we get result
For Example 233 = 8 12 18 27
+ 24 36
-----------------------------
12 1 6 7
Ans 223= 12167
Average shorcuts part-1
Average short cuts for all exams
Welcome to job-updates.comAverage shortcuts- part-1
Time,speed and Distance shortcuts
Basics
- 1) There is a relationship between speed, distance and time:
Speed = Distance / Time OR
Distance = Speed* Time
- 2) Average Speed = 2xy / x+y
where x km/hr is a speed for certain distance and y km/hr is a speed at for same distance covered.
**** Remember that average speed is not just an average of two speeds i.e. x+y/2. It is equal to 2xy / x+y
- 3) Always remember that during solving questions units must be same. Units can be km/hr, m/sec etc.
**** Conversion of km/ hr to m/ sec and m/ sec to km/ hr
x km/ hr = (x* 5/18) m/sec i.e. u just need to multiply 5/18
Similarly, x m/sec = (x*18/5) km/sec
- 4) As we know, Speed = Distance/ Time. Now, if in questions Distance is constant then speed will be inversely proportional to time i.e. if speed increases ,time taken will decrease and vice versa.
For Time and Work shortcuts Click here
Examples of Average on Time and Distance
Problem 1: A man covers a distance of 600m in 2min 30sec. What will be the speed in km/hr?
Solution: Speed =Distance / Time
⇒ Distance covered = 600m, Time taken = 2min 30sec = 150sec
Therefore, Speed= 600 / 150 = 4 m/sec
⇒ 4m/sec = (4*18/5) km/hr = 14.4 km/ hr.
Problem 2: A boy travelling from his home to school at 25 km/hr and came back at 4 km/hr. If whole journey took 5 hours 48 min. Find the distance of home and school?
Solution: In this question, distance for both speed is constant.
⇒ Average speed = (2xy/ x+y) km/hr, where x and y are speeds
⇒ Average speed = (2*25*4)/ 25+4 =200/29 km/hr
Time = 5hours 48min = 29/5 hours
Now, Distance travelled = Average speed * Time
⇒ Distance Travelled = (200/29)*(29/5) = 40 km
Therefore distance of school from home = 40/2 = 20km.
Speed∝Distance
⇒ Ratio of distance covered by both men = 60:40 = 3:2
⇒ Therefore, Ratio of speeds of both men = 3:2Using basic formula, Time = Total Distance / Average Speed
x/200 + x/400 + x/600 + x/800 = 4x/y ⇒ 25x/ 2400 = 4x/ y⇒
y= 384
⇒ Average speed = 384 km/hr