Aptitude Question And Answer For Fresher- Trains and Directions

1. From height of 8 mts a ball fell down and each time it bounces half the distance back. What will be the distance travelled

Ans.: 24

Sol. 8+4+4+2+2+1+1+0.5+0.5+ and etc .. =24

Hence ,the required distance is 6 km

2. Two trains 200mts and 150mts are running on the parallel rails at this rate of 40km/hr and 45km/hr. In how much time will they cross each other if they are running in the same direction?

Ans: 252sec

Sol: Relative speed=45-40=5km/hr=25/18 mt/sec

Total distance covered =sum of lengths of trains =350mts.

So, time taken =350*18/25=252sec.

3.Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance traveled by train was ?

Ans.800km

Sol: Let the total trip be x km.

Then 2x/5=1200

x=1200*5/2=3000km

Distance travelled by car =1/3*3000=1000km

Journey by train =[3000-(1200+1000)]=800km.

4. The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?

Ans. 250

Sol. Distance to be covered in 1 min=(66*1000)/60 m=1100m

Circumference of the wheel =(2*22/7*0.70)m=4.4m.

So, Number of revolutions per min=1100/4.4=250.

5. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?

Ans:50 min

Sol: New speed = 5/6 of usual speed

New time = 6/5 of usual time

Therefore, (6/5 of usual time) – usual time = 10min

Therefore Usual time = 50min

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

Ans: length of train=120m

Speed of train=54kmph

Sol: Let the length of the train be x meters

Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.

Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120

Therefore Length of the train = 120m

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

7. A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.

Ans. length of the train=160m

length of the platform=140 m.

Sol: Let the length of the train be x meters and length of the platform be y meters.

Speed of the train relative to man=(54-6) kmph =48 kmph.

=(48*5/18) m/sec =40/3 m/sec.

In passing a man, the train covers its own length with relative speed.

Therefore, length of the train=(Relative speed *Time)

=(40/3 * 12) m =160 m.

Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.

Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.

Therefore, Length of the platform=140 m.

8. If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

Ans:6km.

Sol: Let the required distance be x km.

Difference in the times taken at two speeds=12mins=1/5 hr.

Therefore x/5-x/6=1/5 or 6x-5x=6 or x=6km.

Submitted by subbaiah

Quatitative Aptitude Model Paper for freshers

1)A rectangular plate with length 8 inches, breadth 11 inches and thickness 2 inches is available. What is the length of the circular rod with diameter 8 inches and equal to the volume of the rectangular plate?

Answer : 3.5 inches

Explanation : Volume of the circular rod (cylinder) = Volume of the rectangular plate

(22/7)*4*4*h = 8*11*2

h = 7/2 = 3.5

2. A fast typist can type some matter in 2 hours and a slow typist can type the same in 3 hours. If both type combinely, in how much time will they finish?

Answer : 1 hr 12 min

Explanation : The fast typist's work done in 1 hr = 1/2

The slow typist's work done in 1 hr = 1/3

If they work combinely, work done in 1 hr = 1/2+1/3 = 5/6

So, the work will be completed in 6/5 hours. i.e., 1+1/5 hours = 1hr 12 min

3. What is the sum of all numbers between 100 and 1000 which are divisible by 14?

Answer : 35392

Explanation : The number closest to 100 which is greater than 100 and divisible by 14 is 112, which is the first term of the series which has to be summed. The number closest to 1000 which is less than 1000 and divisible by 14 is 994, which is the last term of the series.

112 + 126 + .... + 994 = 14(8+9+ ... + 71) = 35392

4. What can you conclude from the statement: All green are blue, all blue are red. ?

(i) some blue are green                                           (ii) some red are green

(iii) some green are not red                                    (iv) all red are blue

(a) i or ii but not both                                   (b) i & ii only

(c) iii or iv but not both                                 (d) iii & iv

Answer : (b)

5) Gavaskar's average in his first 50 innings was 50. After the 51st innings, his average was 51. How many runs did he score in his 51st innings.(supposing that he lost his wicket in his 51st innings)

Answer : 101

Explanation : Total score after 50 innings = 50*50 = 2500

Total score after 51 innings = 51*51 = 2601

So, runs made in the 51st innings = 2601-2500 = 101

If he had not lost his wicket in his 51st innings, he would have scored an unbeaten 50 in his 51st innings.

6. If s(a) denotes square root of a, find the value of s(12+s(12+s(12+ ...... upto infinity.

Answer : 4

Explanation : Let x = s(12+s(12+s(12+.....

We can write x = s(12+x). i.e., x^2 = 12 + x. Solving this quadratic equation, we get x = -3 or x=4. Sum cannot be -ve and hence sum = 4.

7. What is the number of zeros at the end of the product of the numbers from 1 to 100?

Answer : 127

8.Out of 80 coins, one is counterfeit. What is the minimum number of weighings needed to find out the counterfeit coin?

Answer : 4

Submitted by Subbaiah

Aptitude Question And Answer For Fresher- Number Series

1)1, 8, 9, 64, 25 (Hint: Every successive terms are related)

Answer: 216

Explanation: 1^2, 2^3, 3^2, 4^3, 5^2, 6^3

2) 71, 76, 69, 74, 67, 72

Answer: 67

3)1, 2, 4, 10, 16, 40, 64 (Successive terms are related)

Answer: 200

Explanation: The series is powers of 2 (2^0,2^1,..).

4) 8, 24, 12, 36, 18, 54

Answer: 27

5) 6, 24, 60,120, 210

a) 336 b) 366 c) 330 d) 660

Answer: a) 336

Explanation: The series is 1.2.3, 2.3.4, 3.4.5, 4.5.6, 5.6.7, ..... ( '.' means product)

6) 5, 9, 16, 29, 54

Answer: 103

Explanation: 5*2-1=9; 9*2-2=16; 16*2-3=29; 29*2-4=54; 54*2-5=103

7) 1, 5, 13, 25

Answer: 41

Explanation: The series is of the form 0^2+1^2, 1^2+2^2,...

8) 0, 5, 8, 17

Answer: 24

Explanation: 1^2-1, 2^2+1, 3^2-1, 4^2+1, 5^2-1

All digits are less than 8. Every second number is in octal number system.

128 should follow 64. 128 base 10 = 200 base 8.

9) 2, 5, 10, 17, 26, 37, 50, 64

Answer: 64

Explanation: 2+3=5; 5+5=10; 10+7=17; 17+9=26; 26+11=37; 37+13=50; 50+15=65;

10) 105, 85, 60, 30, 0, -45, -90

Answer: 0

Explanation: 105-20=85; 85-25=60; 60-30=30; 30-35=-5; -5-40=-45; -45-45=-90;

11) 3, 5, 7, 12, 13, 17, 19

Answer: 12

Explanation: All but 12 are odd numbers.

Submitted by subbaiah

Aptitude Material - With Solutions For fresher interview

1) The value of ¼ + 1/4.3 + 1/4.32 + 1/4.33 correct to four places of decimals is

     Sol:   ¼ + 1/12 + 1/36 + 1/108 = 27+9+3+1/108

          -- > 40/108   -- > 10/27

             =   0.3704

2) A boy was asked to find the value of 7/12 of a sum of money. Instead of multiplying the sum by 7/12 he divided it by 7/12 and thus his answer exceeded the correct answer By Rs.95. Find the correct?

Sol: Let sum = Rs. K

  : . 12/7 k – 7k/12 = 95

-- > 144k – 49k/84   = 95 -- > k = 84

  : .   7/12 k -- > 7/12 * 84 = Rs. 49

3) Simplify     3.7*3.7+2.3*2.3+2*3.7*2.3 / 4.6*4.6-3.4*3.4

  Sol:    -- >          (3.7 + 2.3)2 / (4.6 + 3.4) (4.6 – 3.4)

               -- >    36/8*1.2 -- > 36/9.6 = 360/96 = 3 ¾

  4) If 2/3rd of a number is subtracted from 7/3 of the number, the result is 2 more than the number itself. Find the number.

  Sol: Let the number be k

        : . 7/3 k – 2/3 k = k+2 -- > 5/3 k = k+2

            -- > 2/3 k = 2 -- > = 3

5) Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the Sum of the third and 52. The smallest number is?

 Sol: Let the numbers be x, y, z

                    : .   x/3 = y/4 = z/5 = k

                  : . x = 3k, y = 4k, z = 5k

                  : . 3k + 5k= 4k + 52

                  -- > k = 13

                 : . Smallest number = 39

6) The least number of five digits which is exactly divisible by 12, 15 and 18 is

 1)       10010

2)       10015

3)       10020

4)       10080

  Sol: Least number of 5 digits is 10000.

        L.C.M of 12, 15, 18 is 180

        On dividing 10000 by 180, the remainder is 100.

        : . Required number = 10000 + (180 – 100) = 10080

7) Of the three numbers, second is twice the first and also thrice the third. If the average of the three numbers is 44, the largest number is?

1)       24

2)       36

3)       72

4)       108

 Sol: Let the 3rd number be x, Then, second number = 3x

                 : . First number = 3x/2

                 : . x+3x + 3x/2 = (44*3)

                  -- > 11x/2 = 132 -- > x = 24

                  Largest number -- > 3x

                  -- > 3*24 = 72      

8) If a/2 = b/3 = c/5, the value of a+b+c/c is

1)       2

2)       5

3)       ½

4)       1/5

      Sol:   a/2 = b/3 = c/5 = k

        -- > a = 2k, b = 3k, c = 5k

       : .   a+b+c/c = 2k+3k+5k/5k = 10k/5k = 2     

9) If 1/3rdof a number subtracted from ½ of that number, then the difference is 10more than 1/7th of the same number. How much is that number?

 Sol: Let the number = x

      : . x/2 – x/3 = x/7+ 10

              -- > 3x-2x/6 = x/7+10

              -- > x/6 – x/7 = 10

              -- >7x-6x/42 = 10          -- > x = 420

10) Simplify: 2 ½ of ¾ * ½ /3/2 + ½ / 3/2 (2/3 – ½ of 2/3) to get

1)       1 5/8

2)       2 5/8

3)       1 8/11

4)       3 5/8

Sol: 1) 2 ½ of ¾ * ½ / 3/2 + ½ / 3/2 (2/3 – ½ of 2/3)

      -- > (5/2 * ¾) * ½ * 2/3 + ½ * 2/3 * (2/3 – ½ * 2/3)

      -- > 15/8 * 1/3 + 1/3 * [ ( 2/3 – 1/3 )]

     -- > 15/8 * 1/3 + 1/3 * [(2/3- 1/3) ]

      -- > 5/8 + 1/3 / 1/3

      -- > 5/8 + 1 = 1 5/8

11) Find the smallest number which when divided by 6,10 and 15 respectively leaves 5 as Remainder in each case?

 Sol: L.C.M of 6,10,15 -- > 30

                Required number =30 + 5 = 35

 12) If a = 16 and b = then a2+b2+ab/a3-b3 will be

 1)       11

2)       1/11

3)       121

4)       Non

 Sol: 1) 162+52+80 / 163-53 = 256+25+80/4096-125

                                              = 361/3971   = 1/11

13) Simplify 2/3 / 4/9 of 7 ½ + 999 494/495 * 99 

1)       99999

2)       10000

3)       99000

4)       11111

  Sol: 2/3 / 4/9 of 15/2 + (999 +494/495) * 99

         = 2/3 / 10/3 + 999*99 + 494/495 * 99

          = 2/3 * 3/10 + (1000 – 1) 99 + 494/495 * 99

       = 1/5 + 99000 – 99 + 494/5

       = 494/5 + 99000 – 99 = 99000 

14) If we multiply a fraction by itself and divide the product by its reciprocal; the fraction thus obtained is 15 5/8. The original fraction is?

   Sol: Let the original fraction be x/y

              X/y *x/y/y/x = 15 5/8

        -- > x/y * x/y * x/y =125/8

    -- > x3/y3 = 125/8

        > x/y =   3√125/8 = 5/2 = 2 1/2  

 15) Find the H.C.F. of 852, 1065 and 1491.

Sol:   852) 1065(1

                   852

                   213) 852 (4

                           852

                             0

          213) 1491(7

                  1491

                      0

            Hence 213 is H.C.F of given numbers.

 16) Find the greatest number which divides 3460 and 9380 leaving as remainder 9 and 13 respectively?

  Sol: Since on dividing 3460 remainder 9 is left, the required number must divide (3460-9) = 3451 exactly. Similarly it must divide 9380 – 13 = 9367 exactly. Hence H.C.F of 9367 and 3451 will be required number.

Submitted by M.U.Subbaiah