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We are going to have either a ice cream or a cake.
Ice cream can be selected from 3 flavors and cake from two flavors. Both the events cannot occur simultaneously selecting ice cream and cake.

∴ Number of possible ways = 3 + 2

= 5 ways

ENGLISH

V M T O R H S

To solve, we look up the corresponding letter from table to replace in code to get the actual word.

For V, it is E
for M, it is N
for T, it is G
for O, it is L
for R, it is I
for H, it is S
for S, it is H

Therefore we get E N G L I S H = ENGLISH

TAMIL

Now to solve, we look up the corresponding letter from the table to replace in code to get the actual word.

So, for G Z N R O, from table,

for G, it is T
for Z, It is A
for N, it is M
for R, it is I
for O, it is L

So , the actual word is TAMIL.

We have a letter followed by 3 digits in the roll number.

The letter is selected from A, B, C, D, E.
For these 5 letters we have to select a 3 digit number using the digits 0 to 9.

Once place can be formed using any one of the 10 number 0 to 9 in 10 ways.

Tens place can be formed in 10 ways.

∴ A two digit number can be formed in 10 × 10 = 100 ways.

Thousands place can be formed in 10 ways

∴ A 3 digit number can be formed in 10 × 10 × 10 = 1000 ways.

∴ 5 letters can be attached in 5 × 1000 = 5000 ways.

∴ The roll number can be formed in 5000 ways.

(i) (d) Sequence: 2,4,6,8,...

General from: y = 2n

(ii) (a) Sequence: 5,9,13,17,...

General from: y = 4n + 1

(iii) (e) Sequence: 4,16,36,64,...

General from: y = 4n²

(iv) (c) Sequence: 2,4,6,8,...

General from: y = 2n

(v) (b) Sequence: 1,4,9,16,...

General from: y = n²

1. 55 

2. 10 

3. 1

(iv) 1,5,10,10,5,1

Answer is (iv) y = -3x

Answer is (i) y = 4x

A tetromino is a shape obtained by 4 squares together.

(i) All possible routes from A to D are : 

(a) A ➝ G ➝ F ➝ E ➝ D 

(b) A ➝ G ➝ D 

(c) A ➝ B ➝ C ➝ D 

(d) A ➝ B ➝ D

(ii) Distance between E and C are 

(a) Route 1: E ➝ D ➝ C 

Distance: 120 m + 200 m = 320 m. 

(b) Route 2: E ➝ D ➝ B ➝ C 

Distance = 120 + 100 m + 120 m = 340 m. 

∴ Shortest distance is 320 m.

(iii) All possible routes between B and F are : 

(a) Route 1: B ➝ A ➝ G ➝ F 

Distance = 250 m + 100 m + 150 m = 600 m. 

(b) Route 2: 2 ➝ D ➝ E ➝ F 

Distance = 100 m + 120 m + 300 m = 520 m. 

(c) Route 3: B ➝ D ➝ G ➝ F 

Distance = 100 m + 200 m + 150 m = 450 m.

(d) Route 4: B ➝ C ➝ D ➝ E ➝ F 

Distance = 120 m + 200 m + 120 m + 300 m = 740 m. 

We find that Route 3 is shortest. 

i.e. B ➝ D ➝ G ➝ F is the shortest route.

Answer is (iii) y = x + 6

Route 1 : School ➝ Swimming pool ➝ canteen ➝ playground 

Distance 9km + 2km + 10km = 21 km. 

Route 2 : School ➝ canteen ➝ playground 

Distance : 10 km + 10 km = 20 km. 

Route 3 : School ➝ garden ➝ playground 

Distance : 2 km + 20 km = 22 km. 

Route 4 : School ➝ garden ➝ teashop ➝ playground 

Distance : 2km + 17km + 2km = 21 km.

Route 5 : School ➝ Park ➝ teashop ➝ playground 

Distance : 12 km + 12 km + 2 km = 24 km. 

∴ Route 2 is the shortest distance.

Possible routes from Mandapam to Vivekandar Memorial are 

Route 1:

(a) Mandapam ➝ Pullivasal Island ➝ Krusadai Island ➝ Vivekanandar Memorial Hall. 

Distance = 6 Km + 2 Km + 1.5 Km = 9.5 Km

Route 2: 

(b) Mandapam ➝ Krusadai Island ➝ Vivekanandar Memorial Hall. 

Distance = 7 Km + 1.5 Km = 8.5 Km 

8.5 km < 9.5 km 

∴ Shortest route : Mandapam ➝ Krusadai Island ➝ Vivekanandar Memorial Hall.

The relationship between x and y is y = 3x + 2

The relationship between x and y is y = -5x

Let x denote the number of steps and y denote the area. 

In the first shape let x = 1 and the area be 1 cm²

When x = 2 : Area = 2² = 4 cm² 

When x = 3 : Area = 3² = 9 cm² and so on. 

Tabulating the values of x and y

From the table: 

x = 1 ⇒ y = 1² 

x = 2 ⇒ y = 2² 

x = 4 ⇒ y = 4² 

x = 5 ⇒ y = 5²

Hence the relationship between x and y is y = x²

Let x denote the number of steps and y denote the number of matchsticks used. 

In step 1, x = 1 ⇒ y = number of mathsticks used is 1 

In step 2, x = 2 ⇒ y = number of mathsticks used is 4 

In step 3, x = 3 ⇒ y = number of mathsticks used is 7 and so on. 

The values of v and y are tabulated as 

x = 1 ⇒ y = 1 = 3(1) – 2 

x = 2 ⇒ y = 4 = 3(2) – 2 

x = 3 ⇒ y = 7 = 3(3) – 2 

x = 4 ⇒ y = 10 = 3(4) – 2 

From the table y = 3x – 2

Soil When x = – 2y 

= 2(-2) 

= -4 

When x = -1y 

= 2(-1) 

= -2 

When x = 0

y = 2(0) 

= 0 

When x = 1

 y = 2(1) 

= 2 

When x = 2 

y = 2 (2) 

= 4

When x = 8 

y = 2(8) 

= 16. 

Also y = 2x is the relation between x and y.

(i) Let the number of steps be x and the number of shapes be y.

Tabulating the values of x and y

From the table 

x = 1 ⇒ y = 1 = 12 

x = 2 ⇒ y = 4 = 22

x = 3 ⇒ y = 9 = 32 

x = 4 ⇒ y = 16 = 42 

Hence the relationship between x and y is y = x².

(ii) Let the number of steps be x and the number of shapes be y. 

Tabulating the values of x and y

From the table 

x = 1 ⇒ y = 1 = 1 

x = 2 ⇒ y = 2 + 1 = 3 

x = 3 ⇒ y = 3 + 2 = 5 

x = 4 ⇒ y = 4 + 3 = 7 

x = 5 ⇒ y = 5 + 4 = 9

Hence the relationship between x and y is y = 2x - 1.

Answer is (iii) 256

There are 4 stages

1. Choosing a Sandwich
2. Choosing a side
3. Choosing a dessert
4. Choosing a drink

There are 4 different types of sandwich, 3 different types of side two different type of desserts and five different types of drink.

∴ The number of meal combos possible is = 4 × 3 × 2 × 5 = 120

The correct option is (B) 8.

(10 × 5) + (9 × 4)

(a) 2 x 5

Prime factors of 30 are 2 x 3 x 5

Prime factors of 250 are 5 x 5 x 5 x 2

∴ Common prime factors are 2 x 5.

(d) 3 x 2 x 2

Prime factors of 36 are 2 x 2 x 3 x 3

Prime factors of 60 are 2 x 2 x 3 x 5

Prime factors of 72 are 2 x 2 x 2 x 3 x 3

∴ Common prime factors are 2 x 2 x 3

1. 56 and 12

56 & 12

Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 = 8 = 4
m – n = 8 – 44. now m = n

HCF of 56 & 12 is 4

2. 320, 120 and 95

Let us take 320 & 120 first m = 320, n = 120

m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = w = 40 → HCF of 320, 120

Now let us find HCF of 40 & 95

m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5

HCF of 40 & 95 is 5 10 – 5 = 5

HCF of 320, 120 & 95 is 5

Let m = 420, n = 280

m – n = 420 – 280 = 140

now m = 280, n = 140

m – n = 280 – 140 = 140

now m = n = 140

∴ HCF is 140

Let m = 1014, n = 654

m – n = 1014 – 654 = 360

now m = 654, n = 360

m – n = 654 – 360 = 294

now m = 360, n = 294

m – n = 360 – 294 = 66

now m = 294, n = 66

m – n = 294 – 66 = 228

now m = 66, n = 228

n – m = 228 – 66 = 162

now m = 162, n = 66

∴ m – n = 162 – 66 = 96

n – m = 96 – 66 = 30

Similarly 66 – 30 = 36

36 – 30 = 6

30 – 6 = 24

24 – 6 = 18

18 – 6 = 12

12 – 6 = 6

now m = n

∴ HCF of 1014 and 654 is 6

36 and 80

36 and 80 m = 80, n = 36

80 – 36 = 44, now n = 44, m = 36

Since n > m, we should do n m

44 – 36 = 8, now n = 8, m = 36

36 – 8 = 28

Similarly, processing, proceeding, we do repeated subtraction till m = n

28 – 8 = 20

20 – 8 = 12

12 – 8 = 4

8 – 4 = 4 now m = n = 4

∴ HCF is 4

Let m = 98, n = 56

m – n = 98 – 56 = 42 ; now m = 56 & n = 42
m – n = 56 – 42 = 14 ; now m = 42 & n = 14
m – n = 42 – 14 = 28 ; now m = 14 & n = 28
m – n = 28 – 14 = 14 ; now = 14 & n = 14

∴ HCF of 98 & 56 is 14.

(c) 6^th

Every 6^th number of the Fibonacci sequence is a multiple of 8.

The correct option is (c) 27.

The next term in the sequence 15, 17, 20, 22, 25, is 27.

The correct option is (b) 2.

If the Highest Common Factor of 26 and 54 is 2, then HCF of 54 and 28 is 2.

By Euclid’s game HCF (a, b) = HCF (a, a – b) if a > b.

Here HCF (188, 230) = HCF (230, – 188) because 230 > 188

= HCF (188, 42) = HCF (146, 42)

= HCF (104, 42) = HCF (62, 42)

= HCF (42, 20) = HCF (22, 20)

= HCF (20,2) = HCF (18, 2) = 2

∴ HCF (230, 188) = 2

(a) 199

The 11^th term in the Lucas sequence 1, 3, 4, 7,… is 199.

The correct option is (d) 3.

The difference between 6th term and 5th term in the Fibonacci sequence is 3.

The correct option is (a) B.

Write A, B, C with increasing number of A, B, and C.

Shanthi his 5 chudithar sets and 4 frocks. 

She wear either chudithar or a frock. 

∴ Total possible ways = 5 + 4 = 9 ways

The student either select any one of science group in 3 ways or any of the arts group in 3 ways or any of the vocational group in 2 ways. 

∴ Total possible ways = 3 + 3 + 2 = 8 ways

There are 5 stages

Question – 1
Question – 2
Question – 3
Question – 4
Question – 5

There are 2 choices for each question (Yes/No)

∴ Total number of possible ways to answer
= 2 × 2 × 2 × 2 × 2 = 32 ways.

Number of track events ⇒ (100m running, 4 × 100 m Relay) 2. 

Number of field events ⇒ (Long jump, High jump, Javelin Throw) 3. 

Students can take part in the given events in 2 × 3 = 6 ways.

18 different ways are there to organize your schedule.

(b) compare the choices before buying

1. True

2. False

3. True

4. False

(d) all the above