Answer : N=11 , x=2

Solution :

let N(N-1) = t;

(N+x)(N+x-1) = t+46;

trail and error method x=2 and N=11

Answer : A

Solution :

Total number of outcomes in the simultaneous throw of 3 dice is 6 x 6 x 6 = 216.
That is, n(S) = 216.
Then E = {(1,1,1),(1,1,3),(1,1,5),(2,2,1),(2,2,3),(2,2,5),(3,3,1),(3,3,3),(3,3,5),(4,4,1),(4,4,3),(4,4,5),(5,5,1),(5,5,3),(5,5,5),(6,6,1),(6,6,3),(6,6,5)}
Therefore, n(E) = 18
Required p(E) = 18 / 216 = 1 / 12.

Answer : C

Solution :

below. (We recommend you to try with other options, to prove that they are wrong.)
On day 1: Medals awarded = (1 + 24/6) = 5 : Remaining 20 medals
On day 2: Medals awarded = (2 + 18/6) = 5 : Remaining 15 medals
On day 3: Medals awarded = (3 + 12/6) = 5 : Remaining 10 medals
On day 4: Medals awarded = (4 + 6/6) = 5 : Remaining 5 medals
On day 5: Medals awarded = 5

Answer : B

Solution :

Jeya Celebrated her 18th birthday on 28 February 2009, Saturday.

Then her 19th birthday will be on 28 February 2010, Sunday. (This is because 2010 is not a leap year and hence contains 365 days. 365 divided by 7 gives remainder 1 which means any date on 2009 corresponds to the same date on 2010 advanced by one day.)

Her 20th birthday will be on 28 February 2011, Monday.

Her 21st birthday will be on 28 February 2012, Tuesday.

Her 22nd birthday will be on 28 February 2013, Thursday. (Since, 2012 is a leap year hence contains 366 days. 366 divided by 7 gives remainder 2 which means any date on 2013 corresponds to the same date on 2012 advanced by two days.)

Her 23rd birthday will be on 28 February 2014, Friday.

Her 24th birthday will be on 28 February 2015, Saturday.

Her 25th birthday will be on 28 February 2016, Sunday.

So, her 25th birthday will be on Sunday.

Answer : D

Solution :

Inspecting how RELATED was written as EFUBKDQ, we can easily find that,
RELATED – is first written as DETALER
DETALER becomes EFUBKDQ (+1,+1,+1, +1,-1,-1,-1 – alphabetically)
Applying the same logic to RETAINS we get,
RETAINS is first written as SNIATER
SNIATER becomes TOJBSDQ (applying the above rule)

Answer : B

Solution :

Though all animals are reptiles from the two statements, there is no indication that all reptiles are monkeys (and in turn animals). Hence option I cannot be true always. But second conclusion is always true which is straightforward to interpret.

Answer : A

Solution :

Let the two unequal numbers be a and b.
We have to find the product of numbers = ab
Sum of numbers = a + b
Difference of numbers = a – b.

The squares of sum and difference of the two unequal numbers are (a + b)2 and (a – b)2
Then the difference between the squares of sum and difference of two unequal numbers = (a + b)2 – (a – b)2

Given that, (a + b)2 – (a – b)2= 4928.
Remember the formulae (x – y)2 = x2 + y2 – 2xy and (x + y)2 = x2 + y2 + 2xy
Therefore, we have
(a + b)2 – (a – b)2 = 4928 = a2 + b2 + 2ab – (a2 + b2 – 2ab)
4928 = a2 + b2 + 2ab – a2 – b2 + 2ab
4928 = 4ab
ab = 1232.
Hence the required product is 1232.

Answer : A

Solution :

10

Let PQ be the height of the lighthouse.
And Given that, angle(PAQ) = a, angle(PBQ) = 2a and angle(PCQ) = 3a

We know that, “The size of an exterior angle at a vertex of a triangle equals the sum of the sizes of the interior angles at the other two vertices of the triangle”.
Here, angle(PBQ) = angle(A) + angle(APB)
2a = a + angle(APB)
angle(APB) = a

Since the opposite sides with equal angles are equal,
AB = PB
Therefore, PB = x

From triangle PQB,
PQ / PB = sin(2a)
PQ = xsin2a.
Hence the height of the lighthouse is xsin2a.

Answer : A

Solution :

Given is an isosceles triangle. (two sides are equal in length)11

Draw a vertical line from q to the straight line PR and mark it as D.
Now the base is divided into two segment, which form 2 right angled triangle with base of 30 cm each.
Now we have to find the length of QD.
By applying the Pythagoras theorem,
302 + QD2 = 502
900 + QD2 = 2500
QD2 = 2500 – 900 = 1600 = 40
Hence the answer is 40 cm

Answer : C

Solution :

6x+5x=55, so x=5, put first ratio after substitution is (6*5)/(5*5) and second ratio is 40/35 So difference in numerators 40-30=10 years

Answer : D

Solution :

Horse takes 3 hours to cover the distance Pony takes 3+2 =5 hours to cover the same distance, Velocity=distance/time, distance travelled by them is equal it is 81*3=243km, speed of pony=243/5=48.6

Answer : C

Solution :

side of cube = 5 cm
its thichness = 1 cm
so volume of outer cube = 5*5*5
volume of inner cube = 3*3*3
volume of the hollow cube = 5*5*5 – 3*3*3 = 98
so total no of small cubes of the size 1 cm = 98/1*1*1 = 98
we know a cube has 6 faces so total no of face = 98*6 = 588
one surface of outer cube contains a total of 25 surface of smaller cube , so when 4 surface of outer cube is painted total no of surface of small cubes i.e supposed to be painted is 4*25 = 100
so the total no of surfaces of small cube that will be remained unpainted is 588-100 = 488

Answer : A

Solution :

Mr.X’s salary for first three years = Rs.10,000 (per annum)

So, the total salary for first three years = Rs.30,000 ……(1)

Given that, his salary increased by Rs.1500(per annum) for next 10 years.

So, salary for the next 10 years are 11500, 13000, 14500,…
Total salary for the next 10 years is 11500+13000+14500,…

Note that, above sequence is an A.P with first term a = 11500, common difference = d = 1500, total number of terms = n = 10.
We know that, sum of first n terms in A.P = (n/2)[2a+(n-1)d] and nth term = a+(n-1)d
Therefore, 10th term = 11500+(10-1)1500 = 25000…..(2)

And the sum of 10 terms = 11500+13000+14500,…+25000 = (10/2)[2×11500 + 9×1500] = (5)(36500) = 182500……(3)

From (1) and (3), the total salary of first 13 years = Rs.30,000 + Rs.182500 = Rs. 212500…(4)

Form (2), we have, the salary of the 13th year = Rs.25000

That is, till his retirement, his salary = Rs.25000 (per annum)……(5)

Let Mr.X works for X more years.
Then, the total salary for this X years = Rs.25000 x X

And his experience = 13+X years ……(5)

Given that, the average salary for 13+X years = Rs.20312.5

That is, Rs.Rs.20312.5 = [Rs. 212500 + RS.25000 x X]/(13+X)
=> (13+X)20312.5 = [Rs. 212500 + RS.25000 x X]/(13+X)
=> (13+X)203.125 = [2125+250X]
Take the approximate value for 203.125 as 203 …(6)
=> (13+X)203 = [2125+250X]
=> 2639+ 203X = [2125+250X]
=> 514 = 47X => X = 10.93

If we hadn’t applied the approximation as in (6), X would be a whole number.
Therefore, we take it as X = 11.

Therefore, from eqn 5, Mr.X’s experience = 13+11 = 24 years.

Since, he was hired at his 30th age, his retirement age = 30+24 = 54.

Therefore, his retirement year is 1960+54 = 2014

Hence, the answer is 2014.

Answer : A

Solution :

Let W,X,Y and Z denote the quantities in respective containers.

Since, W contains 40% more than X, the quantity in W = 140% of X = 140/100 x X = 7X/5.

X contains 40% more than Y, then the quantity in X = 140% of Y = 140/100 x Y = 7Y/5.

Y contains 30% less than Z, then the quantity in Y = (100 – 30=)70% of Z = 70/100 x Z = 7Z/10.

From the above three, we have,

X = 5/7 W …(1),

Y = 5/7 X ….(2),

Z = 10/7 Y….(3)

Given that, W has 102.9 kg of contents, i.e., W = 102.9.

Substituting W = 102.9 in (1), we have, X = 5/7 x 102.9 = 73.5

Substituting above value of X = 73.5 in (2), we have, Y = 5/7 x 73.5 = 52.5

Substituting above value of Y = 52.5 in (3), we have, Z = 10/7 x 52.5 = 75.

Given that, the maximum quantity of container is 200 kg.

Percentage of maximum quantity contained in Z = (quantity in Z / maximum quantity) x 100 = 75/200 x 100 = 37.5 %

Answer : B

Solution :

Area of circle πr2 where r is the radius

Assume initially the radius was 10. Then area becomes 22/7x10x10 = 314.28 –> eq 1

Now if the radius is increased by 12%, it will become 10+(12/100)10 = 10+1.2=11.2

New area = 22/7 x 11.2x 11.2 = 394.24 –> eq 2

Increase in area in our case = eq 2 – eq 1 = 79.96

Percentage increase in area when radius is increased by 12% = Increase in Area / Original Area X 100% = 79.96 / 314.28 X 100 =25.44%

Answer : A

Solution :

Palani’s daughter in law is Cauvery. Cauvery’s daughter is Bimala.  Anand is uncle of Bimala. Cauvery and Anand are siblings.  Therfore Anand is son of Palani.

Answer : A

Solution :

The given sequence is of the form n3-n where n starts from 8. Hence the blank should be filled by 93-9 which is 720.

Answer : B

Solution :

The ten’s place of answers in the given question are obtained as follows:
85 + 17 = 85 / 17 = 5,
76 + 19 = 76 / 19 = 4,
91 + 13 = 91 / 13 = 7,
So, 120 + 15 = 120 / 15 = 8
Observing the given question, the unit digit of the answers are 1, 2 and 3 respectively.
So, the unit digit of required answer is 4.
Hence, the required answer is 84.

Justin was always prepared. His motto was “Never throw anything out, you never know when it might come in handy.” His bedroom was so full of flat bicycle tires, bent tennis rackets, deflated basketballs, and games with missing pieces that you could barely get in the door. His parents pleaded with him to clean out his room.


“What use is a fish tank with a hole in the bottom?” his father asked. But Justin simply smiled and repeated his motto, “Never throw anything out, you never know when it might come in handy.”  When Justin was away from home, he always carried his blue backpack. He liked to think of it as a smaller version of his bedroom—a place to store the many objects that he collected. It was so worn and stretched that it hardly resembled a backpack anymore. It was full of the kind of things that seemed unimportant, but when used with a little imagination, might come in handy. Justin had earned a reputation for figuring things out and getting people out of otherwise hopeless situations.

Many of his classmates and neighbors sought him out when they needed help with a problem. On the first day of school, his friend Kenny, came looking for Justin. “Do you think you have something in your bag that could help me remember my locker combination?” he asked. “I lost the scrap of paper it was written on. I have science class in two minutes and if I’m late on the first day it’ll make me look bad for the rest of the year.” Kenny looked genuinely worried. “Relax,” Justin said, taking his backpack off and unzipping the top. “Remember how you borrowed my notebook in homeroom to write the combination down? Well, I know how we can recover what you wrote.” He took the notebook and a soft lead pencil out of his bag. The page that Kenny had written on had left faint indentations on another page in the notebook. Justin held the pencil on its side and rubbed it lightly over the indentations.

Slowly but surely the numbers of the locker combination appeared in white, set off by the gray pencil rubbings. “That’s amazing!” Kenny said. “I owe you one.” And he dashed off to open his locker. During science class, Mr. Tran was lecturing on the structure of the solar system using a model. He made a sudden gesture and the model fell apart. Planets and rings and connector rods went everywhere, rolling and clattering and disappearing under desks.

The students scrambled around on the floor for ten minutes and were finally able to recover every piece except one—a connector rod that was lodged in a crack between two lab stations. “If we had a magnet,” said Mr. Tran, “we could easily coax it out that way. But I loaned all of the magnet kits to the elementary school yesterday.” Justin was already searching through his backpack. “I have some materials that will work just as well, I think,” he told Mr. Tran.

He pulled out a battery, an iron nail, and some electrical wire and tape, while Mr. Tran and the other students looked on in amazement. “Why do you have all of that stuff?” Louise Baxter asked. Justin just smiled and repeated his motto. “Never throw anything out, you never know when it might come in handy.” By wrapping the wire around the nail and taping each end to a battery terminal, he was able to make a magnet strong enough to lift the rod out of the crack. “Bravo!” said Mr. Tran. “No problem,” said Justin. 


After school, Justin rode the bus to the mall where he worked at a music store. His boss, Gail, was taking inventory of all of the CDs and tapes in the classical music section. As he helped a customer at the register, Justin heard her exclaim, “Oh, no! I forgot my glasses! There’s no way I can read this list without them.” Justin sighed, picked up his backpack, and walked over to Gail. 


“I think I can help you out,” he said, unzipping the bag. While Gail watched in surprise, he pulled out a jar of petroleum jelly, a washer, a glass slide, and a small bottle of water. He put the jelly on the bottom of the washer, placed it securely, jelly-side down, on the glass slide, and then put a drop of water in the center of the washer. He put the contraption on top of the inventory list and said to his boss, “See what happens when you look through the water droplet.” Gail looked and her eyes widened with delight. 
“Wow!” she cried. “It enlarges the print that I’m looking at, just like a magnifying glass!” She patted Justin on the back. “I’m all set now,” she said. “Thanks.” 


Justin smiled. “No problem,” he said, returning to the register. It was just another day in the life of the boy whose motto was “Never throw anything out, you never know when it might come in handy.”

Answer : B

Logic :

for(i=1;i {
for(j=1;j<=i;j++) { printf("%d,"j); } printf("\n"); } 
for(i=n;i>0;i–)
{
for(j=1;j<=i;j++) { printf("%d,"j); } printf("\n"); }