**Answer : B**

**Solution :**

#### Product of two numbers = product of their HCF and LCM

Product of HCF and LCM = 13 x 273

Since HCF is 13, the number may be 13a and 13b where a and b are co-primes.

Then 13a x 13b = 13 x 273

ab = 13 x 273 / 13 x 13 = 21

#### The co-primes with product 21 are (1,21) and (7,3)

So, the pair of numbers are (13 x 1, 13 x 21) and (13 x 7 and 13 x 3).

Since one number lies between 60 and 140, the suitable pair (13 x 7 and 13 x 3) = (91, 39)

Hence the required number is 91.

**Answer : A**

**Solution :**

#### Greatest 5 digit number = 99999

The least number which is exactly divisible by 14, 17 and 21 is the LCM of 14, 17 and 21.

Now, the LCM of 14,17 and 21 is 714

We have to find the greatest number which leaves 13 when divided by 714

#### On dividing 99999 by 714 we have

99999 = 140 x 714 + 39.

Then the greatest 5 digit number exactly divisible by 714 is 99999 – 39 = 99960.

And the greatest 5 digit number which leaves 13 when divided by 714 is 99960 + 13 = 99973.

Hence the required answer is 99973.

**Answer : B**

**Solution :**

#### From the question, the distance between A and the destination is 100 meters and speed of the taxi is 50mph.

Then the time taken to reach the destination = Time = distance / speed = 100/50 hr = 2 hours.

Since the taxi started at 7.00 am, it should had reached the destination at 9.00am.

It is given that the taxi stayed at destination for 30 minutes. Therefore, it should have started from destination at 9.30am.

On return journey its speed had increased by 20%. This means, its new speed = Original speed + 20% of Original speed = 50 + (20/100)50 = 60mph.

Then the time taken to reach A = 100/60 = 1 + 2/3 hours = 1 hour and 40 minutes.

So it should had reached A at 11.10am.

Hence the answer is 11.10am.

**Answer : B**

**Solution :**

#### Total number of outcomes in the simultaneous throw of 3 dice is 6 x 6 x 6 = 216.

That is, n(S) = 216.

Let E be the event to get a product of even number with 1st die always an odd number.

Note, odd x odd = odd, odd x even = even.

#### Since 1st die is an odd number, the product of 2nd and 3rd must be an even number.

Possibilities of the product of 2nd and 3rd to be an even are

{(1,2),(1,4),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,2),(3,6),(3,4),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,2),(5,6),(5,4),

(6,1),(6,2),(6,3)(6,4),(6,5),(6,6)}

and the possibilities of 1st die are 1, 3 and 5.

#### Then E = 1 with above events + 3 with above events + 5 with above events = 27 + 27 + 27

Therefore n(E) = 3 x 27

And p(E) = 3 x 27 / 216 = 3/8.

hence the required answer is 3/8.

**Answer : B**

**Solution :**

#### Conclusion I cannot be true as only some watches could be wall clocks. However, conclusion II is true always as some wall clocks have to be hens.

**Answer : B**

**Solution :**

#### Given is a scalene triangle (no two sides are equal in length).

#### Draw the scalene triangle with base AC 21 cm. The apex of the triangle, point B, is 20 and 13 cm length from the points A and C respectively.

Drop a vertical line from B to the base and mark as D, thus forms a right angle with the line segment AC.

#### Given that, BD = 12 cm and BC = 13 cm, we have to find CD.

Now, from the right angled triangle BCD, apply the Pythagoras theorem as follows

BD^{2} + CD^{2} = BC^{2}

CD^{2} = 13^{2} – 12^{2}

CD^{2} = 25

CD = 5

Hence the answer is 5 cm.

**Answer : 16.5km/hr.**

**Solution :**

#### As we have speed and travel time of horse, we can get distance traveled by it.

#### Hence d = 22*6 = 132km,

#### Exactly this 132km was travelled by dog in 8 hours (as it started two hours earlier).

#### Hence speed of dog = 132/8 = 16.5km/hr

**Answer : B**

**Solution :**

#### A can write 3 notebooks in 48 days.

Then, he can write 1 notebook in 48/3 = 16 days.

And, A’s 1 day work = 1/16.

#### Similarly, B can write 4 notebooks in 48 days.

Then, he can write 1 notebook in 48/4 = 12 days.

And, B’s 1 day work = 1/12.

#### Now, (A+B)’s 1 day work = 1/16 + 1/12 … (1)

#### With the help of C, A and B can complete 5 note books in 20 days.

i.e., they complete 1 notebook in 20/5 = 4 days.

(A + B + C)’s 1 day work = 1/4.

And, C’s 1 day work = 1/4 – (A + B)’s 1 day work

= 1/4 – (1/16 + 1/12) = 5/48

#### Therefore, C can complete 1 notebook in 48/5 days.

C can write 5 note books in 48/5 x 5 = 48 days.

Hence, the answer is 48 days.

**Answer : B**

**Solution :**

#### Let the number of 6 litre cans of pure milk required be X.

Total quantity of pure milk required = 6 x X litres = 6X litres.

Since 1 can of pure milk is mixed with 3 cans of water then number of 6 litre cans of water required is 3X.

And, the total quantity of water required = 3X x 6 litres = 18X litres.

#### Note that, there are 100 servings of 3 litres.

Therefore, total quantity of coffee required for serving = 100 x 3 litres = 300 litres.

We have 6X + 18X = 300

24X = 300

X = 12.5

i.e., twelve and half cans are required.

Hence, the answer is 13.

**Answer : D**

**Solution :**

#### Suppose Q alone takes X hours to fill the tank.

Then P takes = 1/2 of 3X = 3X/2 hours.

Now, Q’s 1 hours work = 1/X and P’s 1 hours work = 2/3X.

#### Given that, (P+Q) takes = 18 hours.

Then (P+Q)’s 1 hour’s work = 1/18.

Therefore, 1/18 = 1/X + 2/3X.

5/3X = 1/18

X = 30.

Hence the answer is 30 hours.

**Answer : C**

**Solution :**

#### Given that Perimeters of square and rectangle are the same.

#### Perimeter of rectangle of size 12 cms and 10 cms is 2(Length + Breadth) = 2(12+10) = 44

#### Perimeter of square = 44. (4 x side of square)

#### Length of each side of the square = 44/4 =11

#### Area of the square = Side^{2} = 11 x 11 = 121 sq. meter

#### Area of the rectangle = Length X Breadth = 12 x 10 = 120 sq. meter

#### Percentage Increase of area of the square to that of the rectangle = (Area of Square – Area of Rectangle) / Area of Rectangle X 100 % = (121 – 120)/120 X 100 % = 5/6 %

**Answer : C**

**Solution :**

**Reason : **

#### Man’s son’s sister = Man’s daughter. Man’s daughter is the mother of Lakshmi’s husband.

**Answer : C**

**Solution :**

#### The general pattern of the given question is a + b + c = d where, a x c = bd. That is,

9 x 8 = 72 which gives 2 as the answer.

8 x 7 = 56 which gives 6 as the answer.

7 x 6 = 42 which gives 2 as the answer.

Then, 5 x 5 = 25 which gives 5 as the answer.

**Answer : A**

**Solution :**

**There are 7 letters in the word STATION.
out of the letters in the word STATION two letters T are alike.
Therefore, number of permutations = 7!/2! = 7!/2.**

**Let, S = The letters of the word STATION be arranged.
Then, n(S) = 7!/2 ….(1)**

**Let, E = the letters of the word STATION be arranged in such a way that two ‘T’s do not come together.**

**Now, we have to find the number of words in which two ‘T’ are together.
Considering the two T’s in the word STATION as one letter, we can arrange 6 letters in 6! ways.**

**Number of words in which two ‘T’s are never together = total number of words – number of words in which two ‘T’ are together
= (7!/2) – 6! = (7! – 2 x 6!)/2 = 6!(7 – 2)/2 = 6! x 5/2.**

**Therefore, n(E) = 6! x 5/2 ….(2)**

**Required probability = p(S) = n(E)/n(S) = (6! x 5/2)/(7!/2) = 6! x 5 / 7! = 5/7.**