Andhra Bank Clerk Recruitment 2012 Notification - 850 Clerk posts

Andhra Bank Clerk Recruitment 2012 Notification - 850 Clerk posts - ibpsreg.sifyitest.com - bank job in andhra pradesh - 850 clerk job vacancy in andra pradesh - Andhra bank recruitment 2012 online application fom

Andhra Bank issued a latest recruitment notification 2012 for Filling up 850 Clerk Recruitment posts . All those candidates who have taken the Common Written Examination (CWE) For Clerk Examination Which examination conducted by IBPS In 2011-12 And Also have a valid score card then you are eligible. Eligible Candidates can apply easly via online mode before 10 November 2012 . More Details can see below in shortly About Andhra Bank Clerk Recruitment 2012 Notificaiton .

Andhra Bank Number of Recruitment Posts of Clerk - 850 posts

Post Name - Clerk

Age Details - 

Candidates age should be between 18 Years To 28 Years as on 01 August 2011 .

Educational Qualification - 

Candidates should have Graduation Degree or 12th passes in First Class with minimum 60% Marks and also have valid score card of IBPS .

Application Form Fee -

1. General ANd OBC Category Candidates need to pay Rs. 150/- As Application fee . 

Application form fee can be paid in the form of Challan Download through Andhra Bank official website at any branch of andhra bank or you can also be paid through NEFT at any Andhra bank branch .

2. SC , ST , PWD , EXSM Category Candidates need to pay Rs. 50/- As Application fee .

Selection process - 

Selected on basis of IBPS Score Card and Interview Performance base .

How To Apply -  

Eligible and Interested can apply via online mode through Andhra Bank official website before 10 November 2012 .After submit online application form take printout of application form and it to be submitted at the time ofInterview .

http://ibpsreg.sifyitest.com/ibpspo_andhraoct12/

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SSC FCI Answer key 2012 download - FCI Assistant Grade 3 Key answer - Required cutoff mark 2012-2013

SSC FCI Assistant Grade 3 Answer Key Download 2012

Staff Selection Committee(SSC) is conducting the written examination for the post of FCI assistant grade 3 (General,Accounts, technical  & Depot cadres) on 11th November 2012. Around 6545 job vacancies are available for which the exam is going to be conducted.


SSC FCI Exam Date : 11.11.2012 ( 11th November 2012 )

Food Corporation of India (FCI), one of the largest Public Sector Undertakings, dealing with food grain supply-chain management called for vacancies for Assistant Grade III in General, Depot, Technical and Account Cadres in its various depots and offices located at five zones  i.e. North, South, East, West and North East

SSC FCI Answer Key

SSC FCI Answer Key 2012

SSC FCI Exam has 3 papers (paper 1,2 and paper 3)
SSC FCI Paper I (Post Codes A, B, C, D), candidate will be required to solve multiple-choice objective type questions based on:
•    General Intelligence,
•    General Awareness,
•    Quantitative Aptitude (Basic Arithmetic Skill)
•    English Language (Basic Knowledge).

SSC FCI Paper II (for Post Codes A, B and D), the questions would be on
•    Quantitative Methods
•    English Language and Comprehension.

SSC FCI Paper III (Post Code C) will consist of questions in Botany, Zoology, Bio-technology, Bio-chemistry, Microbiology, Food Science and Technology, Agriculture and Agricultural Engineering

SSC FCI Question Paper Download

Written by Shankar Krish

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Time and work Quantitative aptitude questions and answers with general rules

 

As many exams in private sector and public sector Time and Work are important so here we listed  Time and Work  quantitative aptitude questions and answers with general rules

Time and Work

General Rules (or) short cuts
(i) If A can do a piece of work in n days, then A's 1 day's work = 1/n.
(ii) If A's 1 day's work = 1/n, then A can finish the work in n days.
(ii) If A is thrice as good a workman as B, then:
Ratio. of work done by A and B = 3 : 1,
Ratio of times taken by A & B to finish a work = 1 : 3.

Solved Problems
Ex. 1.A can do a piece of work in 10 days which B alone can do in 12
days. In how many days will they finish the work, both working together?
Sol. A's 1 day's work = 1/10, B's 1 day's work = 1/12
       (A + B)'s 1 day's work = (1/10 +1/12)= 11/60
       :. Both will finish the work in 60/11= 5 5/11days.
Ex. 2. Two persons A and B working together can dig a trench in 8 hours while A alone can dig it in 12 hours. In how many hours B alone can dig such a trench?
Sol. (A + B)'s 1 hour's work = 1/8, A's 1 hour's work = 1/12
        :. B's 1 hour's work=(1/8-1/12 )=1/24
         Hence B alone can dig the trench in 24 hours.
Ex.3. A and B can do a piece of work in 12 days; Band C can do it
in 15 days ; A and C can do it in 20 days. In how many days will A, Band' C finish it, working all together?Also ,find the number of days taken by each to finish it working alone.

Sol. (A+B)'s 1 day's work =1/12; (B+C)'s 1 day's work = 1/15
        and (A + C)'s 1 day's work = 1/20
Adding, we get: 2 (A + B + C)'s 1 day's work =(1/12 + 1/15 + 1/20)= 1/5
         :. (A + B + C)'s 1 day's work =1/10
Thus, A, Band C together can finish the work in l0 days.
          Now, A's 1 day's work
           = [(A + B + C)’s 1 day's work] - [(B + C)'s 1 day's work]
           =(1/10 – 1/15) = 1/30
:.   A alone can finish the work in 30 days.
Similarly, B's 1 day's work =(1/10 – 1/20)= 1/20
   :. B alone can finish the work in 20 days.
And, C's 1 day's work=(1/10 – 1/12)= 1/60 .
:.   C alone can finish the work in 60 days.

Ex. 4. 3 men can complete a piece of work in 6 days. Two days after
they started the work, 3 more men joined them. How many days will they take to complete the remaining work?

Sol. Work done by 3 men in 2 days = (1/6x 2 )=1/3.
       Remaining work = (1 – 1/3)= 2/3
       Now 3 men's 1 day's work =1/6
       1 man's 1 day's work = 1/18
       6 men's 1 day's work = (6 x 1/18=1/3
       Now, 1/3 work is done by them in 1 day.
       :. 2/3 work is done by them in (3 x2/3)= 2 days.

Ex. 5.2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1. boy do the work? . 
Sol. Let 1 man's 1 day's work = x &.1 boy's 1 day's work = y.
       Then, 2x+3y=1/10 and3x+2y= 1/8
       Solving, we get: x = 7/200 and y = 1/100
        :.' (2 men + 1 boy)'s 1 day's work = (2 x 7/200+ 1 x 1/100)= 16/200 =         2/25
       So, 2 men & 1 boy together can finish the work in 25/2 = 121/2 days.

Ex. 6. A is twice as good a workman as B and together they finish a
piece of work in 18 days. In how many days will A alone finish the work?
Sol. (A's 1 day's work) : (B's 1 day's work) = 2: 1.
        (A + B)'s 1 day's work =1/18.
        Divide 1/18 in the ratio 2 : 1.
        :. A's 1 day's work = [1/18 X 2/3] = 1/27
        Hence, A alone can finish the work in 27 days.
Ex.7. A and B undertake to do a piece of work for Rs. 600. A alone
can- do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. Find the share of each.
Sol. C’s 1 day s work = 1/3 – [1/6 + 1/8]= 1/24
        :. A: B : C = Ratio of their 1 day's work = 1/6 : 1/8 : 1/24 = 4:3:1
        :. A's share=Rs(600X4/8)=Rs.300,
        B's share = Rs. (600 x 3/8)= Rs. 225.
        C’s share = Rs. [600 - (300 + 225)] = Rs. 75.

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PROFIT & LOSS SHORTCUTS - QUANTITATIVE APTITUDE with solved question paper

 In QUANTITATIVE APTITUDE  they are various topic are present like  PROFIT & LOSS  .here we listed out  profit and loss short cut key  with solved question and answer

PROFIT & LOSS SHORTCUTS FOR QUANTITATIVE APTITUDE

1. Profit = Selling Price - Cost price


2. Selling Price = Cost Price + Profit


3. Cost Price = Selling Price - Profit


4. Loss = Cost Price - Selling Price
5. Selling Price = Cost Price - Loss


6.Cost price = Selling Price + Loss


7. Percentage profit / loss is always calculated on CP unless otherwise stated.


8. Profit Percentage = (Profit x 100) / CP


9. Loss Percentage = (Loss x CP) / CP


10. Selling Price = {[(100+ Gain %) x CP] / 100}


11. Selling Price = {[100- Loss %) x CP] /100}


12.Cost Price = {(100 x SP) / (100+ Gain %)}

solved question and answer  for QUANTITATIVE APTITUDE  PROFIT & LOSS

1. A trader uses a 800gm weight instead of 1 kg. Find his profit %.

Soln: (He is buying 800 gm but selling 1000 gm.

So, CP is for 800 gm and SP is for 1000 gm.)

SP/CP = 1000/800 = 1.25 => 25% profit.

1. A trader uses 1 kg weight for 800 gm and increases the price by 20%. Find his profit/loss %.

Soln: 1 kg weight for 800 gm => loss (decrease) => 800/1000 = 0.8

20% increase in price => profit (increase) => 1.2

So, net effect = (0.8) X (1.2) = 0.96 => 4% loss.

1. A milk vendor mixes water to milk such that he gains 25%. Find the percentage of water in the mixture.

Soln: To gain 25%, the volume has to be increased by 25%.

So, for 1 lt of milk, 0.25 lt of water is added => total volume = 1.25 lt

% of water = 0.25 / 1.25 X 100 = 20%.

1. A trader bought an item for Rs. 200. If he wants a profit of 22%, at what price must he sell it?

Soln: CP=200, Profit = 22%.

So, SP = 1.22CP = 1.22 X 200 = 244/-.

1. A person buys an item at Rs. 120 and sells to another at a profit of 25%. If the second person sells the item to another at Rs. 180, what is the profit % of the second person?

Soln: SP of 1^st person = CP of 2^nd person = 1.25 X 120 = 150.

SP of 2^nd person = 180.

Profit % = SP/CP = 180/150 = 1.2 => 20%.

1. A milk vendor mixes water to 20 lt of milk such that the ratio of milk and water is 4:3. He sold the mixture at Rs. 12 per liter but bought the milk at Rs. 10 per liter. Find the profit % of the vendor.

Soln: milk : water = 4:3 => he bought 4 parts (milk) but sold 7 parts (mixture)

CP = 10 and SP = 12.

So, profit % = (SP/CP) X (SP/CP) = (7/4) X (12/10) = 2.1 => 110% gain.

1. A trader buys some apples at a price of 10 apples for Rs. 8 and sold them at a price of 8 apples for Rs. 10. Find his profit or loss %.

Soln: He bought 10 apples for Rs. 8 and sold 8 apples for Rs. 10 => clearly got profit

ð     SP > CP => (SP/CP) X (SP/CP) = (10/8) X (10/8) = 100/64 = 1.5625 => 56.25 % gain.

1. A trader allows a discount of 25% on his articles but wants to gain 50% gain. How many times the CP should be marked on the items?

Soln: CP applied with profit = MP applied with discount = SP

ð     1.5CP = 0.75MP (since 50% gain and 25% discount) => MP = 2CP.

1. By selling an item at a price a trader gains 40%. What is the profit / loss % if the item is sold at half the price?

Soln: SP =1.4CP => (SP/2) = 0.7CP => 30% loss.

1. A trader gets a profit of 25% on an article. If he buys the article at 10% lesser price and sells it for Rs. 2 less, he still gets 25% profit. Find the actual CP of the article.

Soln: 25% gain => SP = 1.25CP…..1.

Now, CP is 10% less => 0.9CP and SP is Rs. 2 less => (SP-2).

Still, profit is 25% => (SP-2)=1.25(0.9CP) , where SP = 1.25CP (From  1)

ð     CP = Rs. 16.

1. A trader gets a discount of 20% from the dealer and marks it at 20% more price then the actual MP to the customer. Find his overall gain %.

Soln: Let MP be the price on the item.

Then, CP=0.8MP (20% discount) and SP = 1.2MP.

So, gain => SP/CP = 1.2/0.8 = 1.5 => 50%.

1. A trader allows a discount of 20% to the customer after marking the item up by 25%. Find his gain/loss% if he is given a commission of 20% of the MP by the dealer.

Soln: Trader’s SP = 0.8 X (1.25MP) = MP (since 20% discount on 25% raised price)

Trader’s CP = 0.8 MP (20% commission)

So, gain = SP/CP = MP/0.8MP = 1

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Ages - quantitative aptitude questions for problems on ages in different shortcut keys

In many interview problem on age is important one to face here we list out more than short cut key to understood clearly with question and answer

short cut key to understood problems on age

1. Odd Days:

   We are supposed to find the day of the week on a given date.

   For this, we use the concept of 'odd days'.

   In a given period, the number of days more than the complete weeks are called odd days.

2. Leap Year:

   (i). Every year divisible by 4 is a leap year, if it is not a century.

   (ii). Every 4^th century is a leap year and no other century is a leap year.

   Note: A leap year has 366 days.

   Examples:

   1. Each of the years 1948, 2004, 1676 etc. is a leap year.

   2. Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.

   3. None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

3. Ordinary Year:

   The year which is not a leap year is called an ordinary years. An ordinary year has 365 days.

4. Counting of Odd Days:

   1. 1 ordinary year = 365 days = (52 weeks + 1 day.)

      1 ordinary year has 1 odd day.

   2. 1 leap year = 366 days = (52 weeks + 2 days)

      1 leap year has 2 odd days.

   3. 100 years = 76 ordinary years + 24 leap years

        = (76 x 1 + 24 x 2) odd days = 124 odd days.

        = (17 weeks + days) 5 odd days.

      Number of odd days in 100 years = 5.

      Number of odd days in 200 years = (5 x 2) 3 odd days.

      Number of odd days in 300 years = (5 x 3) 1 odd day.

      Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.

      Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days

question and answer

1) A man divides Rs.8600 among 5 sons,4 daughters and 2 nephews. If each daughter receives four times as much as each nephew, and each son receives five times as much as each nephew, how much does each daughter receive?

  Sol : Let the share of each nephew be Rs.x

  Then, share of each daughter = Rs.(4x); share of each son = Rs.(5x).

  So, 5 * 5x + 4 * 4x + 2 * x = 8600

  => 25x + 16x + 2x = 8600 => 43x = 8600

=> x = 200. Therefore, Share of each daughter = Rs. (4 * 200) = Rs.800

2) Kumar spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%. Find the percentage increase in his savings.

  Sol : Let original income = Rs. 100. Then, expenditure = Rs.75 and savings = Rs. 25.

  New income = Rs.120 & New expenditure = Rs.[110/100 * 75] = Rs.165/2

  New savings = Rs.[120-165/2] = Rs.75/2

  Increase in savings = Rs.[75/2 - 25] = Rs. 25/2

  Therefore, Increase% = [25/2 * 1/25 * 100]% => 50%

3) A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost of the article.

  Sol: Let the C.P. be Rs,x.

  First S.P. = 125% of x = 125/100x = 5x/4;

  2nd S.P. = 80% of x = 80/100x = 4x/5

  2nd S.P. = 130% of 4x/5 = [130/100 * 4x/5] = 26x/25

  => 5x/4 - 26x/25 = 10.50

  => 21x/100 = 10.50 =>x=[10.50 * 100 / 21] => 50

  Hence, C.P. = Rs. 50

4) A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?

  Sol: (A's 1 day's work) : (B's 1 day's work) = 2:1

  (A+B)'s 1 day's work = 1/18.

  Therefore A's 1 day's work = [1/18 * 2/3] = 1/27.

  Hence, A alone can finish the work in 27 days.

5) Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. After 7 minutes, C is also opened. In how much time, the tank is full?

  Sol: Part filled in min. = 7[1/36 + 1/45] = 7/20

  Remaining part = [1-7/20] = 13/20

  Net part filled in 1 min. when A,B and C are opened = [1/36 + 1/45 - 1/30] = 1/60

  Now, 1/60 part is filled in 1 min.

  13/20 part is filled in [60 * 13/20] = 39 min.

  Therefore, Total time taken to fill the tank = (39+7)min. => 46 min.

6) From a group of boys and girls 15 girls leave. There are then left 2 boys for each girl. After this, 45 boys leave. There are then 5 girls for each boy. Find the number of girls in the beginning.

  Sol: Let at present there be x boys. Then number of girls at present = 5x.

  Before the boys had left : Number of boys = x+45 and number of girls = 5x.

  => x+45 = 2*5x =>9x = 45 => x=5.

  Hence, number of girls in the beginning = 5x+15 = 25+15 = 40.

7) A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, thereby increasing his average by 6 runs. His new average is :

  Sol: Let average for 10 innings be x. Then,

  10x + 108/11 = x+6

  => 11x+66 = 10x+108 => x=42.

  Therefore, New Average = (x+6) = 48 runs.

8) Present ages of Abi and Suji are in the ratio of 5:4 respectively. Three years hence, the ratio of their ages will become 11:9 respectively. What is Suji's present age in years?

  Sol: Let the present ages Abi and Suji be 5x years and 4x years respectively.

  Then, 5x+3 / 4x+3 = 11/9

  => 9(5x+3) = 11(4x+3) => x=6.

  Therefore, Suji's present age = 4x = 24 years.

9) The difference between the ages of two person is 10 years. Fifteen years ago, the elder one was twice as old as the younger one. The present age of the elder person is

  Sol: Let the ages be x years and (x+10) years respectively.

  Then, (x+10) - 15 = 2(x-15)

  => x-5 = 2x-30 => x=25.

  Therefore, present age of the elder person = (x+10) = 35 years.

10) Pranav spends 30% of his monthly income on food articles, 40% of the remaining on conveyance and clothes and saves 50% of the remaining. If his monthly salary is Rs.18400, how much money does he save every month?

  Sol: Saving = 50% of (100-40)% of (100-30)% of Rs.18400

  => Rs. 50/100 * 60/100 * 70/100 * 18400 => 3864

  Therefore, he saves Rs.3864.

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Problems on Trains quantitative aptitude questions and answer for ibps and competitive exams

quantitative aptitude contain many topics but Problems on Trains  is most important to attend the exam .Here we list out questions and answer with shortcuts key

APTITUDE – Trains – Formulas:shortcuts key

• a km/hr= (a* 5/18) m/s.

• a m / s = (a*18/5) km/hr.

• Time taken by a train of length 1 metres to pass a pole or a standing man or a signal  post is equal to the time taken by the train to cover 1 metres.

• Time taken by a train of length 1 metres to pass a stationary object of length  b metres is the time taken by the train to cover (1 + b) metres.

• 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.

 

A train 150 m long is running with a speed of 68 kmph. In what time will it pass a man who is running at 8 kmph in the same direction in which the train is going?(www.naukribuzz.com)

Sol: Speed of the train relative to man = (68 – 8 ) kmph

= (60* 5/18) m/sec = (50/3)m/sec

Time taken by the train to cross the man            I

= Time taken by It to cover 150 m at 50/3  m / sec = 150 *3/ 50  sec  = 9sec

Two trains 137 metres and 163 metres in length are running towards each other on parallel lines, one at the rate of 42 kmph and another at 48 kmpb. In what  time will they be clear of each other from the moment they meet?(www.naukribuzz.com)

Sol. Relative speed of the trains = (42 + 48) kmph = 90 kmph

=(90*5/18) m / sec = 25 m /sec.

Time taken by the  trains  to’pass each other

= Time taken to cover (137 + 163) m at 25 m /sec =(300/25) sec = 12 sec

A train  100 metres long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train. Find the speed of the train.?

Sol:Let the speed of the train be x kmph.

Speed of the train relative to man = (x + 5) kmph = (x + 5) *5/18 m/sec.

Therefore 100/((x+5)*5/18)=6 <=> 30 (x + 5) = 1800 <=> x = 55

Speed of the train is 55 kmph.

. A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.?

Sol: Relative speed = 280/9 m / sec = ((280/9)*(18/5)) kmph = 112 kmph.

Speed of goods train = (112 – 50) kmph = 62 kmph.

A train 100 m long is running at the speed of 30 km / hr. Find the time taken  by      it to pass a man standing near the railway line.

Sol. Speed of the train = (30 x 5/18_) m / sec = (25/3) m/ sec.

Distance moved in passing the standing man = 100 m.

Required time taken = 100/(25/3) = (100 *(3/25)) sec = 12 sec

A man is standing on a railway bridge which is 180 m long. He finds that  a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed?

Sol. Let the length of the train be x metres,

Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 sec

      x/8=(x+180)/20 ó 20x = 8 (x + 180)     <=>   x = 120.

Length of the train = 120 m.

Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 km

A train is moving at a speed of 132 km/br. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metres long?

Sol. Speed of train = 132 *(5/18) m/sec = 110/3  m/sec.

Distance covered in passing the platform = (110 + 165) m = 275 m.

Time taken =275 *(3/110)  sec =15/2 sec = 7  ½ sec

A train 220 m long is running with a speed of 59 kmph.. In what will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going?(www.naukribuzz.com)

sol. Speed of the train relative to man = (59 + 7) kmph

= 66 *5/18  m/sec = 55/3 m/sec.

Time taken by the train to cross the man = Time taken by it to cover 220 m at (55/3) m / sec = (220 *3/55)sec = 12 sec

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quantitative aptitude average shortcuts tricks with solved question paper

Average quantitative aptitude are basic and easy topic in aptitude so in many exam they asked two are more question so we provide shortcut and question and answer listed below

general rules or shortkeys

The sum of first n natural numbers = n (n+1)/2
The sum of squares of first n natural numbers is n (n+1)(2n+1)/6
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n^2

 

1.  The average age of a class of 39 students is 15 years .If the age of the teacher be included, then the average increases by 3 months .Find the age of the teacher.

A) 25 years  B) 27 years   C) 35 years   D) 28 years

2. The average of eight numbers is 14. The average of six of these numbers is 16.The average of the remaining two numbers is :

A)4     B) 8    C)16     D)none

3. The average of ten numbers is 7 . If each number is multiplied by 12 ,then the average of new set of numbers is :

A) 7    B)19    C) 82   D) 84

4.  The average of six numbers is 30.If the average of first four is 25 and that of last three is 35, the fourth number is :

A) 25   B) 30  C)35   D)40

5. The average age of 30 students is 9 years .If the age of their teacher is included ,it becomes 10 years . The age of the teacher (in years ) is :

A) 27   B) 31   C)  35   D) 40

6. After replacing an old member by a new member, it was found that the average age of five members of a club  is the same as it was 3 years ago.what is the difference between the ages of the replaced and the new member ?

A) 15 yrs   B) 8 yrs   C) 4 yrs   D) 2 yrs

7. Three years ago , the average age of A, B and C was 27 years and that of B and C, 5 years ago was 20 years. A’s present age is :

A) 30 yrs  B) 35 yrs  C) 40 yrs  D)48 yrs

8. three years ago, the average age of a family of 5 members was 17 years. A baby having been born , the average age of the family is the same today. The present age of the baby is :

A) 2 yrs B) 2.4 yrs C) 3 yrs D)1.5 yrs

9. The average weight of 6 men decreases by 3 kg when one of them weighing 80 kg is replaced by a new man . The weight of the new man is :

A) 56kg   B) 58 kg   C)62 kg  D) 76 kg

10. The average of ten numbers is 7 .If each number is multiplied by 12 , then the average of new set of numbers is :

a) 7 B) 19  C) 82 d)  84

11. A batsman makes  a score of  87 runs in the 17th inning and thus increases his average by 3 . Find his average after 17th inning.

A) 39  B) 38  C) 38.5  D) 39.5

12. Out of four numbers ,the average of first three is 16 and that of the last three is 15 .If the last number is 18,the first number is :

A) 20  B) 21  C) 23  D) 25

13. Narasimha, Madhu and pavan started a business by investing Rs.1, 20,000, Rs. 1,35,000 and Rs 1, 50,000   respectively. Find the share of Pavan, out of an annual profit of Rs 56,700.

A) Rs .16, 800   B) Rs.18, 900     C) Rs.21, 000   D) none

14. Anand and Deepak started a business investing Rs. 22,500 and Rs.35, 000 respectively. Out of a total profit of Rs. 13,800, Deepak’s share is_______.

A) Rs .5400   B) Rs. 7200     C) Rs. 84,000   D) 94,000

15. A starts business with Rs. 35,000 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2:3. What is B’s contribution in the capital?

A) Rs .7500   B) Rs. 8000     C) Rs. 8500  D) Rs. 9000

16. A and B are partners in a business. A contributes 1/4 of the capital for 15 months and B received 2/3 of the profit .For how long B’s money was used.

A) 6 months B) 9 months C) 10 months   D) 1 year

17. X and Y invested in a business. They earned the profit, which they divided in the ratio of 2:3. If X invested Rs. 40,000, the amount invested by Y is

A) Rs .60, 000   B) Rs. 8000     C) Rs. 50,000 D) Rs. 90,000

18. A, B and C enter into a partnership and their shares are in the ratio 1/2:1/3:1/4.After 2 months, A withdraws half of his capital and after 10 months, a profit of Rs. 378 is divided among them. What is B’s share?

A) Rs .129   B) Rs. 144     C) Rs. 156 D) Rs. 168

19. A, B, C enters into a partnership investing Rs 35000 Rs 45000 and Rs 55000 respectively. The respective shares of A, B and C in annual profit of Rs 40,500 are

A) Rs.10, 500 Rs.13, 500 Rs.16, 500         B) Rs.11, 500 Rs.13, 000Rs.16, 000

C) Rs.11, 000 Rs.14, 000 Rs.15, 500         D) Rs.11, 500 Rs.12, 500 Rs16,500

20.    2 is what percent of 50?

A)2%    B)10%   C)100%  D)4%

21.    5% of (25% of Rs1600) is

A)Rs. 5    B)Rs 17.50    C) Rs 20   D)Rs 25

22.  In an examination, 35% of the students passed and 455 failed. How many students appeared for the examination?

A) 490       B) 700    C) 845   D) 1300

23. At an election a candidate who gets 84% of the votes is elected by a majority of 476

Votes what is the total number of votes polled?

A) 672      B) 700     C) 749     D) 848

24.1100 boys and 700 girls are examined in a test; 42% of the boys and 30% of thegirlsPass.The percentage of the total who failed is:

A) 58%    B) 62 2/3%      C) 64%    D) 78%

25. Find S.P when C.P= Rs 56.25,gain=20%.

A) 68.34   B) 67.50   C) 61.25   D) 66.66

26. Find C.P when S.P =Rs 40.60, gain =16%

A) 25   B) 45   C) 35   D) 55

27. A man buys a cycle for Rs 1400 and sells it at loss of 15% what is the selling price of the cycle?

A) Rs 1090  B) 1160   C) 1202 D) 1190

28. A shopkeeper purchased 70 kg of potatoes for Rs. 420 and sold the whole lot at the rate of Rs 6.50 per kg .What will be his gain percent?

A) 4 1/6 %     B) 6 1/4 %   C) 8 1/3 %   D) 20%

29. A man buys an article for Rs 27.50 and sells it for Rs 28.60 .Find his gain %?

A) 4   B) 8  C) 2   D) none

30.How is my father’s wife’s grandfather’s only child’s sons is related to    me?

A)     Brother

B)      Uncle

C)      Cousin

D)      Nephew

31. A man said to a lady, “Your mother’s husband’s sister is my aunt.” How is the lady related to the man?

A) Mother     B) Aunt     C) Sister     D) Grandmother

32. If P + Q means P is the brother of Q; P – Q means P is the mother of Q and P * Q means P is the sister of Q. Which of the following means M is the maternal uncle of R, if you can assume a third person K to be involved in establishing the relationship?

A) M-K*P     B) M+K*R     C) M+K-R       D) M+K+R

33.  If P + Q means P is the mother of Q, P ¸ Q means P is the father of Q, P – q means P is the sister of Q. Then which of the     following relationship shows that M is the daughter of R:

A)     R – M ¸ N       B) R + N ¸ M         C) R ¸ M + N      D) None of these

34. Some forms are books. All books are made of paper.

A)     Some forms are made of paper

B)      Some forms are not made of paper

C)      No forms are made of paper

D)      None of the above

35. Some books are not forms. All books are made of paper.

A)     Some forms are made of paper.

B)      Some forms are not made of paper

C)      Some things made of paper are not forms

D)      None of the above

36.   No M are P.

Some M are S

A) Some S are P         B) Some P are S

C) No S are M             D) Some S are not P

37.    All M are P.

All S are M

A) All P are S         B) All M are S

C) All S are P        D) None of the above

38.    Some M are not P.

All M are S

A) No S are P                 B) All S are P

C) Some S are not P     D) No S are P

(Q39-Q40): A cube  painted yellow on all faces in cut in to 27 small cubes of equal sizes. Answer the questions that following

39.  How many cubes are painted on one face only?

A)1       B)6             C)8             D)12

40.  How many cubes  are not painted on any face?

A)1        B)4          C)6          D)8

(Q41-Q42):There was a cube painted in red. IF it’s cut into 512 equal pieces, how many faces will be with

41. How many cubes are 2 red color faces ?

A) 27    B) 1    c) 8     D) None

42. How many cubes are 3 red color faces ?

A) 27    B) 72    C)8     D) None

43.729ml of a mixture contains milk and water in the ratio 7:2. how much water is to be added to get a new mixture containing milk and water in the ratio 2:1?

A)112.5ml            B)5 ml         C)121.5 ml           D)215 ml

44. In an examination, 80% of the candidates pass in History and 75% in Geography. If 10% fail in both the subjects, what is the pass percentage?

A) 65                       B) 45                   C) 75                   D) 55

45. Find the percentage saving, in buying a dozen articles for Rs. 126 instead of Rs. 12 for one?
A) 12                       B) 12 1/2                   C) 15                 D) 16 2/3

46. By selling 300 apples a seller gains the selling price of 60 apples. The gain of the seller is …
A)Rs. 200      B)Rs. 20%          C)Rs 25%     D)Rs. 16 2/3%

47. A reduction of 20% in the price of sugar enables a consumer to obtain 2.5kg more for Rs.160. Find the reduced price per kg.of sugar.

A) Rs. 12.70        B) Rs. 12.75      C) Rs12.80     D) Rs. 12.85

48. Salary of Suresh is 25% less than the salary of Ramesh. By how much percent the salary of Ramesh is more than the salary of Suresh?

A)12 1/2 %    B)16 2/3%           C)33 1/3%        D) 26 2/3%

49. A man’s income was reduced by 10%. How much percent it must now be raised so that it  may be equal to the original amount?

A)11 %         B)12%                C)15%             D) 16 %

50. A man’s wages was reduced by 50%. Again the reduced wages was increased by 50%.  Find out the percentage effect on his salary.

A)Remains same                  B)Increases by 50%

C)Decreases by 50%            D)Decreases by 25%

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quantitative aptitude question and answer solved Problems on H.C.F and L.C.M lesson

 

H.C.F and L.C.M are basic topic in quantitative and aptitude so we gave simple method to follow it and auestion and answer:

H.C.F and L.C.M

1. If number a divided another number b exactly, we say that a is a factor of b.

2. In this case, b is called a multiple of a.

3. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.):

   The H.C.F. of two or more than two numbers is the greatest number that divided each of them exactly.

   There are two methods of finding the H.C.F. of a given set of numbers:

   1. Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.

   2. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

      Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.

      Similarly, the H.C.F. of more than three numbers may be obtained.

4. Least Common Multiple (L.C.M.):

   The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

   There are two methods of finding the L.C.M. of a given set of numbers:

   1. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

   2. Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

5. Product of two numbers = Product of their H.C.F. and L.C.M.

6. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.

7. H.C.F. and L.C.M. of Fractions:

       1. H.C.F. =
   H.C.F. of Numerators

   L.C.M. of Denominators

       2. L.C.M. =
   L.C.M. of Numerators

   H.C.F. of Denominators

8. H.C.F. and L.C.M. of Decimal Fractions:

   In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

9. Comparison of Fractions:

   Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

---

 

1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

   i.

       13

   ii.

       7

   iii.

       4

   iv.

       9

   Answer & Explanation

   Answer: 4

   Explanation

   Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4.

2. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

   i.

       299

   ii.

       276

   iii.

       345

   iv.

       322

   Answer & Explanation

   Answer: 322

   Explanation

   Clearly, the numbers are (23 x 13) and (23 x 14). Larger number = (23 x 14) = 322.

3. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

   i.

       15

   ii.

       10

   iii.

       16

   iv.

       4

   Answer & Explanation

   Answer: 16

   Explanation

   L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together(30/2) 1 = 16 times.

4. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

   i.

       4

   ii.

       5

   iii.

       6

   iv.

       7

   Answer & Explanation

   Answer: 4

   Explanation

   N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 1 2 0 ) = 4

5. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

   i.

       9000

   ii.

       9800

   iii.

       9600

   iv.

       9400

   Answer & Explanation

   Answer: 9600

   Explanation

   Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Required number (9999 - 399) = 9600.

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Quantitative aptitude question with answer permutation and combination solved

In  campus interview quantitative aptitude question and  answer permutation and combination  are important so we provide many question and answer here:

Combinations and permutation:shortcuts

The number of different groups of r objects that can be formed from a total of n objects when the order of selection is irrelevant is called the number of combinations of n objects taken r at a time.  
The number of these r combinations is denoted by ⁿC_rorC(n, r)and

Recall that ⁿP_r is the number of ways in which r objects can be selected from n objects, when the order of selections is relevant.
We can also choose an ordered selection of r object out of n objects in two steps.
1. Choose a subset consisting of r objects without paying any attention to the orderin which they are chosen. This can be done in ⁿC_r ways.
2. For a given choice of r objects, count the number of ways in which these object could be ordered. Any r objects can be permuted amongst them in r! ways.

Hence from the fundamental rule of counting we must have :

The second and the third equalities follows from the fact that:

(a) In a game of poker a player receives 5 cards from an ordinary deck of playing cards. These 5 cards are called a poker hand. How many poker hands are there?    
(b) A poker hand is said to be a flush if all the cards are of the same suit. In how many ways can one be dealt a flush?
Solution:

(a) The number of poker hands that a player can receive is

(b) There are four suits in a deck of cards. Out of these 4 suits, one suit has to be chosen and from this suit 5 cards have to be chosen (out of 13 cards in the four suits). The number of ways one can be dealt a flush is:

Example:

(a) A student has to answer 6 out of 10 questions in an examination. How many choices does she have?
(b) How many choices are there if she must answer at least 2 of the first 3 questions?

Solution:

(a) The student has to select 6 out of 10 of questions. There are

choices for her.

(b) If the student must answer at least 2 of the first 3 questions, it is convenient to divide her choices into tow mutually exclusive groups.
(i) She answers all three of the first three questions and answers any 3 of the remaining 7 questions. This can be done in

(ii) She chooses 2 out of the first 3 questions and answers any 3 of the remaining 7 questions. This can be done in

Hence the number of choices that are available to the student is

35 + 105 = 140

Example:

A man has 10 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is:
[1] 1024

[2] 10!

[3] 1023

[4] 10! - 1
Solution:
Suppose the man has to invite only one of his friends (out of 10) for dinner, then we know that he has to choose one amongst his 10 friends. This can be done in: 10_C1ways
The key word in the question is: at least, this means the friend can call 1 or 2 or 3 or…10 friends over for dinner. Mathematically, this can be shown as:

Therefore, the man can invite his 10 friends in (1024 – 1) or 1023 ways.
Example:
Assume the above problem is framed as follows:
A man has 10 friends whom he wants to invite for dinner. The number of ways in which he can invite at least three of them is:
(1) 1021                        

(2) 968

(3) 864                         

(4) None of these
Solution:

Suppose the man has to invite only three of his friends (out of 10) for dinner, then we know that he has to choose three amongst his 10 friends. This can be done in: 10_C3
The key word in the question still is: at least, this means the friend can call 3 or 4 or 5 or…10 friends over for dinner. Mathematically, this can be shown as:

Therefore, the man can invite his 10 friends in [(1024 – 1)– (10+45)]or 968 ways.
Example:
A question paper has 5 multiple choice questions and 5 true or false questions. Each multiple choice question is provided with 3 choices. Find the number of ways that an examinee can attempt one or more questions of the given 10 questions.
(1)  3¹⁰ (2) 6⁵-1
(3)  8⁵ -1    (4) 12⁵ -1

Solution:

Number of ways of dealing with each multiple choice question = 4(3 ways of attempting and one way of not attempting)
Number of ways of dealing with each true or false question = 3 (2 ways of attempting and one way of not attempting)
Therefore Number of ways of attempting one or more questions = (4⁵ × 3⁵) – 1 = 12⁵ – 1

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CHAIN RULE in quantitative aptitude question with answer in short methods

 

quantitative and aptitude in chain rule with simple steps and solved paper are provided here to learn chain rule:

CHAIN RULE:IMPORTANT FACTS AND FORMULAE

1. Direct Proportion: Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increases (or decreases) to the same

Ex. 1. Cost is directly proportional to the number of articles.
(More Articles, More Cost)
Ex. 2. Work done is directly proportional to the number of men working on it
(More Men, More Work)

2. Indirect Proportion: Two quantities are said to be indirectly proportional,if on the increase of the one, the other decreases to the same extent and vice-versa.

Ex. 1. The time taken by a car in covering a certain distance is inversely proportional to the speed of the car.
(More speed, Less is the time taken to cover a distance)
Ex. 2. Time taken to finish a work is inversely proportional to the num of persons working at it.
(More persons, Less is the time taken to finish a job)

Remark: In solving questions by chain rule, we compare every item with the term to be found out.

SOL VED EXAMPLES

Ex. 1. If 15 toys cost Rs, 234, what do 35 toys cost?
Sol. Let the required cost be Rs. x. Then,
More toys, More cost (Direct Proportion)
. 15 : 35 : : 234 : x ? (15 x x) = (35 x 234) ? x=(35 X 234)/15 =546
Hence, the cost of 35 toys is Rs. 546

.
Ex. 2. If 36 men can do a piece of work in 25 hours, in how many hours will 15 men
do it ?
Sol. Let the required number of hours be x. Then,
Less men, More hours (Indirect Proportion)
15 : 36 : : 25 : x ?(15 x x) = (36 x 25) ?(36 x 25)/15 = 60
Hence, 15 men can do it in 60 hours.

Ex. 3. If the wages of 6 men for 15 days be Rs.2100, then find the wages of
for 12 days.
Sol. Let the required wages be Rs. x.
More men, More wages (Direct Proportion)
Less days, Less wages (Direct Proportion)
Men 6: 9 : :2100:x
Days 15:12
Therefore (6 x 15 x x)=(9 x 12 x 2100) ? x=(9 x 12 x 2100)/(6 x 15)=2520
Hence the required wages are Rs. 2520.

Ex. 4. If 20 men can build a wall 66 metres long in 6 days, what length of a similar can be built by 86 men in 8 days?
Sol. Let the required length be x metres
More men, More length built (Direct Proportion)
Less days, Less length built (Direct Proportion)
Men 20: 35
Days 6: 3 : : 56 : x
Therefore (20 x 6 x x)=(35 x 3 x 56)?x=(35 x 3 x 56)/120=49
Hence, the required length is 49 m.

Ex. 5. If 15 men, working 9 hours a day, can reap a field in 16 days, in how many
days will 18 men reap the field, working 8 hours a day?
Sol. Let the required number of days be x.
More men, Less days (indirect proportion)
Less hours per day, More days (indirect proportion)
Men 18 : 15
Hours per day 8: 9 } : :16 : x
(18 x 8 x x)=(15 x 9 x 16)? x=(44 x 15)144 = 15
Hence, required number of days = 15.

Ex. 6. If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours
day, bow much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?
Sol. Let 3 engines of former type consume 1 unit in 1 hour.
Then, 4 engines of latter type consume 1 unit in 1 hour.
Therefore 1 engine of former type consumes(1/3) unit in 1 hour.
1 engine of latter type consumes(1/4) unit in 1 hour.
Let the required consumption of coal be x units.
Less engines, Less coal consumed (direct proportion)
More working hours, More coal consumed (direct proportion)
Less rate of consumption, Less coal consumed(direct prportion)
Number of engines 9: 8
Working hours 8 : 13 } :: 24 : x
Rate of consumption (1/3):(1/4)
[ 9 x 8 x (1/3) x x) = (8 x 13 x (1/4) x 24 ) ? 24x = 624 ? x = 26.
Hence, the required consumption of coal = 26 metric tonnes.

Ex. 7. A contract is to be completsd in 46 days sad 117 men were said to work 8 hours a day. After 33 days, (4/7) of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?
Sol. Remaining work = (1-(4/7) =(3/7)
Remaining period = (46 - 33) days = 13days
Let the total men working at it be x.
Less work, Less men (Direct Proportion)
Less days, More men (Indirect Proportion)
More Hours per Day, Less men (Indirect Proportion)
Work (4/7): (3/7)
Days 13:33 } : : 117: x
Hrs/day 9 : 8
Therefore (4/7) x 13 x 9 x x =(3/7) x 33 x 8 x 117 or x=(3 x 33 x 8 x 117)/(4 x 13 x 9)=198
Additional men to be employed = (198 - 117) = 81.

Ex. 8. A garrison of 3300 men had provisions for 32 days, when given at the rate of 860 gns per head. At the end of 7 days, a reinforcement arrives and it was for that the provisions wi1l last 17 days more, when given at the rate of 826 gms per head, What is the strength of the reinforcement?
Sol. The problem becomes:
3300 men taking 850 gms per head have provisions for (32 - 7) or 25 days,
How many men taking 825 gms each have provisions for 17 days?
Less ration per head, more men (Indirect Proportion)
Less days, More men (Indirect Proportion)
Ration 825 : 850
Days 17: 25 } : : 3300 : x
(825 x 17 x x) = 850 x 25 x 3300 or x = (850 x 25 x 3300)/(825 x 17)=5000
Strength of reinforcement = (5500 - 3300) = 1700.

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SBI Associates Clerical Exam current affairs question for oct 2012 – IBPS

SBi has been conducted clerk exam on October 2012 in that they asked different types of question including current affairs

current affairs question for oct 2012

India imports gold from which country-options brazil,Switzerland,France,Korea

1. The girl under 21 age,who run 15000 kms in London Olympics ?
2. The organization which is going to sponsor Bangladesh Cricket team?
3. FII full form ?
4. Name the raman megasaysay award winner ? (list given)
5. Name the state which we call as God's Own country ?
6. Name the first Chinese Woman Astronaut ?
7. Writer of God of Small Things book ?
8. Name the Gajal singer who died recently?
9. Which state is called as the state of god
10. Wisdon outstading award goes to ?
11. Full form of IPL ?
12. Rajiv gandhi national sadbavana award goes to ?
13. Chief election commissioner of india is?
14. Sustanaible development conference named as ?
15. Blood donor's day observed on?
16. world blood donation day is celebrated on
17. The Next Olympics are Going to held at?
18. The person who won medal in Air Rifle event from India?
19. Name the small credit card sized chip in Laptop ?
20. Name the attorny general of India ?
21. Name the communication satellite of India?
22. Name the book written by P. Chidambaram?
23. Which is not included in the classification of Computers ?
24. Which Computer works in a speed of TB? 
25. Meaning of NAVOTHSAV
26. Rajiv Gandhi Sadbhavana award?
27. Who is the chairman of the Rajya Sabha?
28. Who is the founder of Microsoft?
29. The Organizations holds bulk Quantity of Products and selling them in smaller quantities for Lower margin are called ?

Current Affairs Questions asked in SBI Associates Clerical Exam Held on 14th October 2012 (Evening Session)

These questions shared by Vinodhini Thana Sekharan T

1.IPL Full form

2.face book owner name

3.white tiger author

4.which is not game is olympics-golf

5.10m air

rifle bronze winner

6.largest trading partner of india

7. Next common wealth game venue

8.ICC brand ambassador to promote games

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IBPS Clerk, PO/MT, Specialist officer- model question paper with answer

 

 

 

IBPS conducting two times exam per year so the are various question paper and answer are available for aptitude  here for who are writing common written exam

1. In a division sum, the divisor is 10 times the quotient and 5 times the remainder. If the remainder is 46, the dividend is:

(1) 4236

(2) 4306

(3) 4336

(4) 5336

Check Answer--

2. If 1.5 x= 0.04 y, then the value of (y-x) (y+x) is:

(1) 730/77

(2) 73/77

(3) 7.3/77

(4) 703/77

Check Answer--

3. An employee may claim Rs. 7.00 for each km when he travels by taxi and Rs. 6.00 for each km if he drives his own car. If in one week he claimed Rs. 675 for traveling 90 km. how many kms did he travel by taxi?

(1) 135

(2) 155

(3) 162

(4) 170

Check Answer--

4. The square root of 3 + √5 is :

(1) √3 /2 + 1/√2

(2) √3 /2 - 1/√2

(3) √5 /2 - 1/√2

(4) √(5/2) + √(1/2)

5. The mean temperature of Monday to Wednesday was 370C and of Tuesday to Thursday was 340C, if the temperature on Thursday was 4/5th that of Monday, then what was the temperature on Thursday?

(1) 36.50C

(2) 360C

(3) 35.50C

(4) 340C

6. A certain number of two digits is three times the sum of its digits. If 45 be added to it, the digits are reversed. The number is:

(1) 72

(2) 32

(3) 27

(4) 23

7. Three years ago the average age of A and B was 18 years. While C joining them now, the average becomes 22 years. How old (in years) is C now?

(1) 24

(2) 27

(3) 28

(4) 30

8. If 2^(2x-1) = 8^(3-x), then the value of x is:

(1) -1

(2) -2

(3) 2

(4) 3

9. A man's basic pay for a 40 hours' week is Rs. 200. Overtimes is paid at 25% above the basic rate. In a certain week, he worked overtime and his total was Rs. 300. He therefore, worked for a total of (in hours):

(1) 52

(2) 56

(3) 58

(4) 62

10. On a Rs. 10, 000 payment order, a person has choice between 3 successive discounts of 10%, 10% and 30% and 3 successive discounts of 40%, 5% and 5%. By choosing the better one he can save (in Rupees):

(1) 200

(2) 255

(3) 400

(4) 433

11. Rs. 600 are divided among A, B, C so that Rs. 40 more than 2/5 th of A's share, Rs. 20 more that 2/7 th of B's share and Rs. 10 more than 9/17 th of C's may all be equal. What is A's share (in Rupees)?

(1) 150

(2) 170

(3) 200

(4) 280

 

12. A, B, C started a business with their investment in the ratio 1 : 3 : 5. After 4 months, A invested the same amount as before and B as well as C withdrew half of their investments. The ratio of their profits at the end of the year was:

(1) 5 : 6 : 10

(2) 6 : 5 :10

(3) 10 : 5 : 6

(4) 4 : 3 : 5

Check Answer--

13. If 9 men working 71/2 hours a day can finish a piece of work in 20 days, then how many days will be taken by 12 men, working 6 hours a day to finish the work? It is being given that 2 men of latter type work as much as 3 men of the former type?

(1) 91/2

(2) 11

(3) 121/2

(4) 13

14. Three pipes A, B and C can fill a cistern in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the cistern is:

(1) 12

(2) 14

(3) 16

(4) 18

15. A train B speeding with 120 kmph crosses another train C running in the same direction, in 2 minutes. If the lengths of the trains B and C be 100 m and 200 m respectively, what is the speed (in kmph) of the train C?

(1) 111

(2) 123

(3) 127

(4) 129

16. River is running at 2 kmph. It took a man twice as long to row up as to row down the river. The rate (in km/hr) of the man in still water is:

(1) 8

(2) 10

(3) 4

(4) 6

17. A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity (in kg.) sold at 18% profit is:

(1) 560

(2) 600

(3) 400

(4) 640

Check Answer--

18. A sum of money becomes Rs. 13, 380 after 3 years and Rs. 20, 070 after 6 years on compound interest. The sum (in Rupees) is:

(1) 8800

(2) 8890

(3) 8920

(4) 9040

19. A well with 14 m inside diameter is dug 10 m deep. Earth taken out of it, has been evenly spread all around it to a width of 21m to form an embankment. The height (in metres) of the embankment is:

(1) 1/2

(2) 2/3

(3) 3/4

(4) 3/5

20. A rectangular carpet has an area of 120 sq metres and a perimeter of 46 metres. The length of its diagonal (in metres) is:

(1) 11

(2) 13

(3) 15

(4) 17

 

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