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.

Solution :

Label the given 6 coins as A, B, C, D, E and F respectively.

Required figure is,

6

And, the equal triangles are ADE, ABE, BEF and BCF.

Answer : B

Solution :

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

9

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
BD2 + CD2 = BC2
CD2 = 132 – 122
CD2 = 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 = Side2 = 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.

Answer : B

Solution :

Let the number of steps in escalator be n.
Given that, he covers n – 16 steps in 13 seconds and n – 36 steps in 8 seconds.
Then the required speed is either (n – 16)/13 or (n – 36)/8.
Now, we have (n – 16)/13 = (n – 36)/8
8(n – 16) = 13(n – 36)
13n – 8n = 468 – 128
5n = 340
n = 68 steps
Therefore speed, n – 16/13 steps per second = 68 – 16 / 13 = 4
Hence, the answer is 4 steps per second.

Answer : C

Solution :

Given ratio of girls and boys = 4:6 = 2:3
Let boys = 3X and girls = 2X.
Number of girls who are non-degree holders = (100 – 50)% of 2X = 50% of 2X
Number of boys who are non-degree holders = (100 – 40)% of 2X = 60% of 3X.
Total non-degree holders = 50% of 2X + 60% of 3X.
= X + 9X/5 = 14X/5
Required percentage = [(14X/5) / 5X] x 100 = 14X/5 x 1/5X x 100 = 14 X 4 = 56%.

Answer : D

Solution :

Here the scientist goes two metres towards east from his house.(starting point). Then he turned right and walked 25 metres towards south. At this stage he turned left and travelled towards east 15 metres. He stopped there and turned left and travelled 18 metres and then he turned right and walked towards east 7 metres. He stopped at that place and got the black bear.

18

When the movement of the scientist is drawn his present position can be found out using Pythagoras theorem.
On the one side it will be 7 metres and on the other side it will be 24 m (2 + 15 + 7).
Applying Pythagoras theorem — √ 72 + 242
= 25 m.

Note: In this problem you have to draw the diagram carefully and also identify the correct distances – Some may be misled to take one side as 2 m and other side as 24 m. The first sentence in the problem guides some people to mark the figure wrongly.

Answer : B

Solution :

Speed of boat – 15 kmph —- B
Speed of current(river) —3 kmph – C
Effective downward speed D (downward ) = Speed of boat + Speed of river = 15 +3 = 18 kmph
Effective upward speed U (upward ) = Speed of boat – Speed of river = 15 – 3 = 12 kmph
Let the place of interest be at a distance of x Km from start.
Time taken to row to x and come back = x/D + x/U = 40mins = 2/3 hours
Or, x/18 + x/12 = 2/3 (40 minutes)
x = 4 4/5 km

Answer : C

Solution :

Average of 12 boys = 41.5
Total weight of 12 boys = 12 x 41.5
Average of 4 of them = 36 then their total weight = 4 x 36
Average of other 4 = 42.5 then their total weight = 4 x 42.5
Total of remaining 4 = 12 x 41.5 – (4 x 36 + 4 x 42.5)
= 4 [ 3 x 41.5 – (36 + 42.5) ]
= 4 [ 124.5 – 78.5 ]
= 4 x 46
Average of remaining 4 = 4 x 46/4 = 46.

The Mayan Indians lived in Mexico for thousands of years before the Spanish arrived in the 1500s. The Mayan were an intelligent, culturally rich people whose achievements were many. They had farms, beautiful palaces, and cities with many buildings. The Mayan people knew a lot about nature and the world around them. This knowledge helped them to live a better life than most people of that time, because they could use it to make their lives more comfortable and rewarding.

Knowledge about tools and farming, for instance, made their work easier and more productive. In ancient Mexico there were many small clearings in the forest. In each clearing was a village with fields of corn, beans, and other crops around it. To clear the land for farms, the Mayan cut down trees with stone axes. They planted seeds by digging holes in the ground with pointed sticks. A farmer was able to grow crops that produced food for several people. But not every Mayan had to be a farmer. Some were cloth makers, builders, or priests. The Mayan believed in many gods, including rain gods, sun gods, and corn gods. The people built large temples to honor the Mayan gods. Skillful workers built cities around these temples. I

t was difficult for them to construct these cities, because they had no horses to carry the heavy stone they used to build with. Workers had to carry all of the building materials themselves. Today, many of these ancient Mayan cities and temples are still standing. Although the cities that the Mayan built were beautiful, and the people worked hard to build them, very few of the people lived in them. Usually, only the priests lived in the cities.The other people lived in small villages in the forests. Their houses were much simpler than the elaborate structures in the cities. They lived in small huts with no windows.

The walls were made of poles covered with dried mud, and the roof was made of grass or leaves. Most Maya lived a simple life close to nature. Measuring time was important to the Maya, so they developed a system for measuring it accurately. Farmers needed to know when to plant and harvest their crops. Mayan priests made a system to keep track of time.

They wrote numbers as dots (…) and bars (-). A dot was one and a bar was five. The Mayan priests studied the Sun, Moon, stars, and planets. They made a calendar from what they learned. The year was divided into 18 months of 20 days each with five days left over.

The Mayan calendar was far more accurate than the European calendars of the time. Around the year 800, the Maya left their villages and beautiful cities, never to return. No one knows why this happened.

They may have died from an infectious disease. They may have left because the soil could no longer grow crops. Archaeologists are still trying to find the lost secrets of the Maya. They are still one of our greatest mysteries. 

main()
{
char *src = “Hello World”;
char dst[100];
strcpy(src,dst);
printf(“%s”,dst);
}strcpy(char *dst,char *src)
{while(*src) *dst++ = *src++;
}

A) “Hello World”
B) “Hello”
C) “World”
D) NULL
E) unidentified

Answer : D

main()
{
int x=20;
int y=10;
swap(x,y);
printf(“%d %d”,y,x+2);
}
swap(int x,int y)
{
int temp;
temp =x;
x=y;
y=temp;
}

A)10,20

B) 20,12

C) 22,10

D)10,22

E) None

Answer : D

Answer : GCC , LCC and Visual C++ do support ANSI C standards.