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Q:

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) 32 B) 24
C) 72 D) 36
 
Answer & Explanation Answer: A) 32

Explanation:

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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Q:

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

A) 27 and 23 B) 24 and 23
C) 25 and 23 D) 22 and 23
 
Answer & Explanation Answer: A) 27 and 23

Explanation:

Let the two parallel sides of the trapezium be a cm and b cm. 

Then, a - b = 4 

And, 12×a+b×19=475=>a+b=50 

Solving (i) and (ii), we get: a = 27, b = 23. 

So, the two parallel sides are 27 cm and 23 cm.

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Q:

How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never beings separated?

A) 1440 B) 720
C) 360 D) 640
 
Answer & Explanation Answer: A) 1440

Explanation:

There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.

 

But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.

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Q:

In how many ways can 4 sit at a round table for a group discussions?

A) 9 B) 3
C) 6 D) 12
 
Answer & Explanation Answer: C) 6

Explanation:

We get this using the formula of circular permutations i.e., (4 - 1)! = 3! =6

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Q:

When Dr. Ram arrives in his dispensary, he finds 12 patients waiting to see him. If he can see only one patients at a time,find the number of ways, he can schedule his patients if 3 leave in disgust before Dr. Ram gets around to seeing them.

A) 479001600 B) 79833600
C) 34879012 D) 67800983
 
Answer & Explanation Answer: B) 79833600

Explanation:

There are 12 - 3 = 9 patients.They can be seen in 12P9 = 79833600 ways

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Q:

When John arrives in New York,he has eight stops to see, but he has time only to visit six of them.In how many different ways can he arrange his schedule in New York?

A) 20610 B) 24000
C) 20160 D) 21000
 
Answer & Explanation Answer: C) 20160

Explanation:

He can arrange his schedule in 8P6 = 20160 ways

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Q:

Find the length of the altitude of an equilateral triangle of side  33cm.

A) 4.5 B) 3.5
C) 2.5 D) 6.5
 
Answer & Explanation Answer: A) 4.5

Explanation:

Let the length of the altitude be h.
Then, h = a32   = 33 × 32 = 4.5

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Q:

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

A) 24 B) 48
C) 60 D) 72
 
Answer & Explanation Answer: B) 48

Explanation:

Let ABC be the isosceles triangle and AD be the altitude 

Let AB = AC = x. Then, BC = (32 - 2x). 

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC,AC2=AD2+DC2 

x2=82+16-x2x=10 

BC = (32- 2x) = (32 - 20) cm = 12 cm. 

Hence, required area =  12*BC*AD=12*12*8=48cm2

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