The post Aptitude: Problems on H.C.F and L.C.M appeared first on NewWalkin.
]]>1. Factors and Multiples: –If number a divided another number b exactly, we say that a is a factor of b. In this case, b is called a multiple of a.
2. 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 divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers:
I. 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.
II. 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.
3. 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:
I. 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.
II. Division Method (shortcut): 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.
4. Product of two numbers = Product of their H.C.F. and L.C.M.
5. Coprimes: Two numbers are said to be coprimes if their H.C.F. is 1.
6. 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.
QUESTIONS
1.Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
A.4
B.7
C.9
D.13
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:
A.276
B.299
C.322
D.345
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 ?
A.4
B.10
C.15
D.16
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:
A.4
B.5
C.6
D.8
5.The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
A.9000
B.9400
C.9600
D.9800
6.The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
A.101
B.107
C.111
D.185
7.Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
A.40
B.80
C.120
D.200
8.The G.C.D. of 1.08, 0.36 and 0.9 is:
A.0.03
B.0.9
C.0.18
D.0.108
9.The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A.1
B.2
C.3
D.4
10.The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
A.74
B.94
C.184
D.364
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]]>The post Aptitude: Problems on Numbers appeared first on NewWalkin.
]]>QUESTION
1.If onethird of onefourth of a number is 15, then threetenth of that number is:
A.35
B.36
C.45
D.54
2.Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
A.9
B.11
C.13
D.15
3.The difference between a twodigit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
A.3
B.4
C.9
D.Cannot be determined
E.None of these
4.The difference between a twodigit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
A.4
B.8
C.16
D.None of these
5.A twodigit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:
A.18
B.24
C.42
D.81
6.The sum of the digits of a twodigit number is 15 and the difference between the digits is 3. What is the twodigit number?
A.69
B.78
C.96
D.Cannot be determined
E.None of these
7.The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
A.20
B.30
C.40
D.None of these
8.A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:
A.3
B.5
C.9
D.11
9.In a twodigit, if it is known that its unit’s digit exceeds its ten’s digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:
A.24
B.26
C.42
D.46
10. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
A. 3
B. 10
C. 17
D. 20
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]]>Some Basic Formula
i. (a + b)(a – b) = (a2 – b2)
ii. (a + b)2 = (a2 + b2 + 2ab)
iii. (a – b)2 = (a2 + b2 – 2ab)
iv. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
v. (a3 + b3) = (a + b)(a2 – ab + b2)
vi. (a3 – b3) = (a – b)(a2 + ab + b2)
:
vii. (a3 + b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ac)
viii. When a + b + c = 0, then a3 + b3 + c3 = 3abc
QUESTION
1.Which one of the following is not a prime number?
A.31
B.61
C.71
D.91
2.(112 x 54) = ?
A.67000
B.70000
C.76500
D.77200
3.It is being given that (232 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?
A.(216 + 1)
B.(216 – 1)
C.(7 x 223)
D.(296 + 1)
4.What least number must be added to 1056, so that the sum is completely divisible by 23 ?
A.2
B.3
C.18
D.21
E.None of these
5.1397 x 1397 = ?
A.1951609
B.1981709
C.18362619
D.2031719
E.None of these
6.How many of the following numbers are divisible by 132 ? 264, 396, 462, 792, 968, 2178, 5184, 6336
A.4
B.5
C.6
D.7
7.(935421 x 625) = ?
A.575648125
B.584638125
C.584649125
D.85628125
8.The largest 4 digit number exactly divisible by 88 is:
A.9944
B.9768
C 9988
D.8888
E.None of these
9.Which of the following is a prime number ?
A.33
B.81
C.93
D.97
10.What is the unit digit in {(6374)1793 x (625)317 x (341491)}?
A.0
B.2
C.3
D.5
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]]>The post Aptitude :: Compound Interest appeared first on NewWalkin.
]]>Amount = P  1 +  R  n  
100 
Amount = P  1 +  (R/2)  2n  
100 
Amount = P  1 +  (R/4)  4n  
100 
Amount = P  1 +  R  3  x  1 +  R  
100  100 
Then, Amount = P  1 +  R_{1}  1 +  R_{2}  1 +  R_{3}  .  
100  100  100 
Present Worth =  x  .  

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]]>Extra money paid for using other’s money is called interest.
If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.
Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then
(i). Simple Intereest =  P x R x T  
100 
(ii). P =  100 x S.I.  ; R =  100 x S.I.  and T =  100 x S.I.  .  
R x T  P x T  P x R 
QUESTION
1.A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
2.Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
3.A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
4.How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
5.Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
6.A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?
7.utomobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:
8.lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
9.Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
10 an took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
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]]>The post TIME AND DISTENCE appeared first on NewWalkin.
]]>FORMULA
1. Speed, Time and Distance:
Speed =
Distance
,
Time =
Distance
,
Distance = (Speed x Time).
Time
Speed
2. km/hr to m/sec conversion:
x km/hr =
x x
5
m/sec.
18
3. m/sec to km/hr conversion:
x m/sec =
x x
18
km/hr.
5
4. If the ratio of the speeds of A and B is a : b, then the ratio of the
the times taken by then to cover the same distance is
1
:
1
or b : a.
a
b
5. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then,
the average speed during the whole journey is
2xy
km/hr.
x + y
(QUESTION)
1.A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A. 3.6
B. 7.2
C. 8.4
D. 10
2.An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 hours, it must travel at a speed of:
A. 300 kmph
B. 360 kmph
C. 600 kmph
D. 720 kmph
3.If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
A. 50 km
B. 56 km
C. 70 km
D. 80 km
View Answer Discuss in Forum Workspace Report
4.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
A. 100 kmph
B. 110 kmph
C. 120 kmph
D. 130 kmph
View Answer Discuss in Forum Workspace Report
5.Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
A. 9
B. 10
C. 12
D. 20
6.
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
A. 1 hour
B. 2 hours
C. 3 hours
D. 4 hours
7.A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A. 220 km
B. 224 km
C. 230 km
D. 234 km
8.The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
A. 70 km/hr
B. 75 km/hr
C. 84 km/hr
D. 87.5 km/hr
9.A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:
A. 35.55 km/hr
B. 36 km/hr
C. 71.11 km/hr
D. 71 km/hr
10.A car travelling with of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
A. 6 km/hr
B. 25 km/hr
C. 30 km/hr
D. 35 km/hr
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]]>The post Problems on Trains appeared first on NewWalkin.
]]>(Formula)
1. km/hr to m/s conversion:
a km/hr =a * 5/18 m/s.
2. m/s to km/hr conversion:
a m/s =a * 18/5 km/hr.
3. Formulas for finding Speed, Time and Distance
4. Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.
5. Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.
6. Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u – v) m/s.
7. Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.
8. If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:
The time taken by the trains to cross each other =(a + b)/(u + v)sec.
9. If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s,
then:The time taken by the faster train to cross the slower train =(a + b)/(u – v)sec.
10. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then: (A’s speed) : (B’s speed) = (b : a)
Question:
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]]>
Problems on Trains  Time and Distance  Height and Distance 
Time and Work  Simple Interest  Compound Interest 
Profit and Loss  Partnership  Percentage 
Problems on Ages  Calendar  Clock 
Average  Area  Volume and Surface Area 
Permutation and Combination  Numbers  Problems on Numbers 
Problems on H.C.F and L.C.M  Decimal Fraction  Simplification 
Square Root and Cube Root  Surds and Indices  Ratio and Proportion 
Chain Rule  Pipes and Cistern  Boats and Streams 
Allegation or Mixture  Logarithm  Races and Games 
Stocks and Shares  Probability  True Discount 
Banker’s Discount  Odd Man Out and Series 
]]>