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Two trains are moving in opposite direction having speed in the ratio 5:7. First train crosses a pole in 12 second and second train crosses the same pole n 15 second. Find the time in which they can cross each other completely.

A.

59/4 sec

B.

57/4 sec

C.

55/4 sec

D.

53/4 sec

Answer with explanation

Answer: Option CExplanation

Let the length of first train and second train be a and b meter. Then

a = 5x*12 = 60x and b = 7x*15 = 105x

They are moving in opposite direction, 165x = (12x)*T

T = 165/12 = 55/4 sec

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A train travels a certain distance by taking 3 stops of 20min each. Considering the period of stoppage, the overall speed of the train comes to 40 km/h. While without consideration of stoppage, it is 45 km/h. How much distance must the train have travelled?

A.

360km

B.

550km

C.

400km

D.

360km

Answer with explanation

Answer: Option AExplanation

45×y=40(y+1)

distance travelled= 45×8=360

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A train is moving and crossing a man who is running on a platform of 100 m at a speed of 8 kmph in the direction of the train, in 9 sec. If the speed of the train is 74 kmph. Find the length of the train ?

A.

156 mts

B.

165 mts

C.

188 mts

D.

202 mts

Answer with explanation

Answer: Option AExplanation

Let the length of the train = L mts

Relative speed of train and man = 74 – 8 = 66 kmph = 66 x 5/18 m/s

=> 66 x 5/18 = L/9

=> L = 165 mts.

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A question paper consists of three sections 4,5 and 6 questions respectively. Attempting one question from each section is compulsory but a candidate need not attempt all the questions. In how many ways can a candidate attempt the questions?

A.

119

B.

29295

C.

209

D.

(4!-1) × (5!-1) × (6!-1)

Answer with explanation

Answer: Option BExplanation

At least 1 question from each section is compulsory, so from the 1^{st} section the candidate can attempt 1 or 2 or 3 or 4 questions.

In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it.

So far 4 question there are 2 × 2 × 2 × 2 ways to attempt.

As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 2^{4} – 1

Similarly for the 2^{nd} section there are 2^{5} – 1 ways in which he can attempt and for the 3^{rd} section there are 2^{6} – 1 ways.

The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections.

Thus, total number of ways in which he can attempt questions in that paper:

= (2^{4} – 1)(2^{5} – 1)(2^{6} – 1)

= 15 × 31 × 63

= 29295

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If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?

A.

B.

30

C.

32

D.

31

Answer with explanation

Answer: Option CExplanation

The 5 letter word can be rearranged in 5! = 120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then, the 25th word will start will CA _ _ _ .

The remaining 3 letters can be rearranged in 3! = 6 Ways.

i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.

The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24 + 6 + 2 = 32

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There are two identical red, two identical black, and two identical white balls. In how many different ways can the balls be placed in the cells (Each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutive cells?

A.

15

B.

18

C.

24

D.

30

Answer with explanation

Answer: Option CExplanation

Case I : 2 balls of the same colour and two balls are a different colour are arranged.

Two balls of the same colour and two balls of different colours can be arranged together in which two balls of the same colour are adjacent =4!/2!x2! = 6 ways

Therefore, Total number of arrangements = 6×3 =18 ways

Case II : Two colours out of 3 can be selected in = 3C1 = 3ways

Now 2 balls of each colour can be arranged alternatively in 2 ways

Thus 4 balls can be arranged(two of each colours)

= 3×2 = 6ways

Hence total number of arrangements = 18+6 =24 ways

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There are 8 men and 7 women. In how many ways a group of 5 people can be made such that at least 3 men are there in the group?

A.

1844

B.

1626

C.

1722

D.

1545

Answer with explanation

Answer: Option CExplanation

Case 1: 3 men and 2 women

^{8}C_{3}*^{7}C_{2} = 1176

Case 2: 4 men and 1 women

^{8}C_{4}*^{7}C_{1} = 490

Case 3: all 5 men

^{8}C_{5} = 56

Add all the cases.

Workspace

A.

70

B.

35

C.

105

D.

210

Answer with explanation

Answer: Option AExplanation

All the boxes contain distinct number of chocolates.

For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.

The number of ways of giving 4 boxes to the 4 person is,

^{8}C_{4} = 70

Workspace

Of the four numbers, whose average is 60, the first is one-fourth of the sum of the last three. The second number is one-third of the sum of other three, and the third is half of the other three. Find the fourth number.

A.

60

B.

80

C.

48

D.

52

Answer with explanation

Answer: Option DExplanation

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If the average income of a family of ‘x’ members is Rs. 25000 while average income of another family of 6 members is Rs. 32000, then find the value of ‘x’ if the average income of both families is Rs. 28000

A.

6

B.

7

C.

8

D.

9

Answer with explanation

Answer: Option CExplanation

Avg income of 1st family=25

Avg income of 2nd family=32

Both the families=28

1st family=28-25=3

2nd family=32-28=4

Requied ratio=4:3

Ratio of number of family members of 1^{st}to 2^{nd}is 4 : 3. In the second family, there are 6 members, hence in the 1^{st}family there are 8 members, only then the ratio = 8 : 6 or 4 : 3 hence x = 8.

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The average age of a group of 10 students was 18. The average age increased by 1 years when two new students joined the group. What is the average age of the two new students who joined the group?

A.

23.5

B.

23

C.

24

D.

24.5

Answer with explanation

Answer: Option CExplanation

The average age of a group of 10 students is 18.

Therefore, the sum of the ages of all 10 of them = 10 * 18 = 180

When two students joins the group, the average increase by 1. New Average = 19

Now there are 12 students Therefore, sum of all the ages of 12 students = 12 X 19 = 228

Therefore, the sum of the ages of two students who joined = 228 – 180 = 48

And the average age of these two students = 24

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Marks of a student were wrongly entered in computer as 83, actual marks of that student were 63. Due to this mistake average marks of whole class got increased by half (1/2). Find the total number of students in that class?

A.

40

B.

45

C.

35

D.

30

Answer with explanation

Answer: Option BExplanation

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