**Tata Motors Interview Series Questions And Answers**

**Series Questions**

**Exercise 1**

For the following, find the next term in the series

1. 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)

2. 1, 5, 13, 25

Answer : 41

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

3. 0, 5, 8, 17

Answer : 24

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

4. 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

5. 8,24,12,36,18,54

Answer : 27

6. 71,76,69,74,67,72

Answer : 67

7. 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

8. 1,2,4,10,16,40,64 (Successive terms are related)

Answer : 200

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

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

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

**Exercise 2**

Find the odd man out.

1. 3,5,7,12,13,17,19

Answer : 12

Explanation : All but 12 are odd numbers

2. 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;

3. 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;

**Exercise 3**

Solve the following.

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

Answer : 127

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. 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.

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

Answer : 4

5. 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)

6. 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

7. 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

8. 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.

9. A cylindrical container has a radius of eight inches with a height of three inches. Compute how many inches should be added to either the radius or height to give the same increase in volume?

Answer : 16/3 inches

Explanation : Let x be the amount of increase. The volume will increase by the same amount if the radius increased or the height is increased.

So, the effect on increasing height is equal to the effect on increasing the radius.

i.e., (22/7)*8*8*(3+x) = (22/7)*(8+x)*(8+x)*3

Solving the quadratic equation we get the x = 0 or 16/3. The possible increase would be by 16/3 inches.

10. With just six weights and a balance scale, you can weigh any unit number of kgs from 1 to 364. What could be the six weights?

Answer : 1, 3, 9, 27, 81, 243 (All powers of 3)

11. Diophantus passed one sixth of his life in childhood, one twelfth in youth, and one seventh more as a bachelor; five years after his marriage a son was born who died four years before his father at half his final age. How old is Diophantus?

Answer : 84 years

Explanation : x/6 + x/12 + x/7 + 5 + x/2 + 4 = x

12 . If time at this moment is 9 P.M., what will be the time 23999999992 hours later?

Answer : 1 P.M.

Explanation : 24 billion hours later, it would be 9 P.M. and 8 hours before that it would be 1 P.M.

13. How big will an angle of one and a half degree look through a glass that magnifies things three times?

Answer : 1 1/2 degrees

Explanation : The magnifying glass cannot increase the magnitude of an angle.

14. Divide 45 into four parts such that when 2 is added to the first part, 2 is subtracted from the second part, 2 is multiplied by the third part and the fourth part is divided by two, all result in the same number.

Answer: 8, 12, 5, 20

Explanation: a + b + c + d =45; a+2 = b-2 = 2c = d/2; a=b-4; c = (b-2)/2; d = 2(b-2); b-4 + b + (b-2)/2 + 2(b-2) = 45;

15. I drove 60 km at 30 kmph and then an additional 60 km at 50 kmph. Compute my average speed over my 120 km.

Answer : 37 1/2

Explanation : Time reqd for the first 60 km = 120 min.; Time reqd for the second 60 km = 72 min.; Total time reqd = 192 min

Avg speed = (60*120)/192 = 37 1/2

Questions 16 and 17 are based on the following :

Five executives of European Corporation hold a Conference in Rome

Mr. A converses in Spanish & Italian

Mr. B, a spaniard, knows English also

Mr. C knows English and belongs to Italy

Mr. D converses in French and Spanish

Mr. E , a native of Italy knows French

16. Which of the following can act as interpreter if Mr. C & Mr. D wish to converse

a) only Mr. A b) Only Mr. B c) Mr. A & Mr. B d) Any of the other three

Answer : d) Any of the other three.

Explanation : From the data given, we can infer the following.

A knows Spanish, Italian

B knows Spanish, English

C knows Italian, English

D knows Spanish, French

E knows Italian, French

To act as an interpreter between C and D, a person has to know one of the combinationsItalian&Spanish, Italian&French, English&Spanish, English&French

A, B, and E know atleast one of the combinations.

17. If a 6th executive is brought in, to be understood by maximum number of original five he should be fluent in

a) English & French b) Italian & Spanish c) English & French d) French & Italian

Answer : b) Italian & Spanish

Explanation : No of executives who know

i) English is 2

ii) Spanish is 3

iii) Italian is 3

iv) French is 2

Italian & Spanish are spoken by the maximum no of executives. So, if the 6th executive is fluent in Italian & Spanish, he can communicate with all the original five because everybody knows either Spanish or Italian.

What is the sum of the first 25 natural odd numbers?

Answer : 625

Explanation : The sum of the first n natural odd nos is square(n).

1+3 = 4 = square(2) 1+3+5 = 9 = square(3)

The sum of any seven consecutive numbers is divisible by

a) 2 b) 7 c) 3 d) 11

**Exercise 4**

1. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

A. 45 B. 60C. 75 D. 9

Answer & Explanation

Answer: Option D

Explanation:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:

A. 20 B. 80 C. 100 D.200

Answer & Explanation

Answer: Option C

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10 x - y = 20 .... (i)

and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

3. If a - b = 3 and a2 + b2 = 29, find the value of ab.

A. 10 B. 12 C. 15 D. 18

Answer & Explanation

Answer: Option A

Explanation:

2ab = (a2 + b2) - (a - b)2

= 29 - 9 = 20

ab = 10.

4. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?

A. 160 B. 175 C. 180 D. 195

Answer & Explanation

Answer: Option B

Explanation:

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

160 x 2.40 + x x 3.20 = 432

3.20x = 432 - 384 = 48

x = 15.

Hence, total hours of work = (160 + 15) = 175.

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