Answer : A

Solution :-

The first integer, after 60 which is divisible by 17 is 68
And the 1st integer, before 600 which is divisible by 17 is 595.
Therefore the sequence of integers between 60 and 600 which are divisible by 17 is
68, 85, 102,…,595
We have to find 68 + 85 + 102 + … + 595

It is an A.P with a = 68 and d = 17 and t(n) = 595
The formula to find nth term in A.P is, t(n) = a + (n – 1)d
Here, a +(n – 1)d = 595
68 + (n – 1)17 = 595
(n – 1)17 = 527
n – 1 = 31
n = 32
i.e., the 32-nd term of sequence is 595

We know that the sum of first n terms of the A.P series = s(n) = (n/2)[2a + (n – 1)d]
Here, s(32) = (32/2)[2(68) + (32 – 1)17]
16[136 + 31(17)]
16[136 + 527]
16[663] = 10608
Hence the required sum is 10608.

Answer : C

Solution :-

Average Points of juniors = 150
and the number of juniors = 20.
Total score obtained by juniors = 150 x 20

Average score of (juniors + seniors) = 152.
Total number of players = 32
Total score of (juniors + seniors) = 152 x 32

Total score of 12 seniors = 152 x 32 – 150 x 20
required average = (152 x 32 – 150 x 20)/12 = (152 x 8 – 150 x 5)/3 = (1216 – 750)/3 = 466/3 = 155.3.
Hence, the answer is 155.3 points.

Answer : B

Solution :

Let, 5 pens + 7 pencils + 4 erasers = x rupees
so 10 pens + 14 pencils + 8 erasers = 2*x rupees
also mentioned, 6 pens + 14 pencils + 8 erasers = 1.5*x rupees
so (10-6) = 4 pens = (2-1.5)x rupees
so 4 pens = 0.5x rupees => 8 pens = x rupees
so 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid by
amal) i.e 5/8 = 500/8% = 62.5% is the answer

Answer : A

Solution :

Let the speed of upstream be X km/hr.

Then, speed in downstream = 2X km/hr (since boat takes twice as much as time to cover up than as to cover down the same distance in running water).

Speed in still water = (2X+X)/2 km/hr. (formula 3)
= 3X/2 km/hr.

Given that, boat covers 10 1/3 km in 1 hour in still water.

Therefore, 3X/2 = 10 1/3
X = 62/9

So, speed in upstream = 62/9 km/hr.
And, speed in downstream = 2 x 62/9 = 124/9 km/hr

Hence, speed of the current = [(124/9 – 62/9)]/2 km/hr
= 62/9×2 = 34/9 = 3 4/9 km/hr.

Answer : B

Solution :

We know the property that, the unit’s digit of a perfect square cannot be 2, 3, 7 or 8.

Since, the numbers 1324802, 344578 and 868623 are ending with 2, 8 and 3. So, we can discard them.

Hence, the perfect square of given numbers is 127449.

Answer : C

Solution :

Time taken to rectify 0.128 errors is 1 second.
Time taken to rectify 25 errors,

Errors           Time

0.128            1

25               ?

Here the number of errors has been increased, so time also increases. Hence error is directly proportional to time.

? = (25 / 0.128)1
? = 25 x 1000 / 128
? = 195.31

Therefore, time taken to rectify 25 errors is 195 sec approximately.

Answer : C

Solution :

Given that, pq=28 and p, q are integers.

The factors of 28 are 1, 2, 4, 7, 14 and 28.

Then the possibilities of pq = 28 are (p,q) = (1, 28), (2, 14), (4, 7), (7,4), (14, 2) and (28,1)

We would have p > q since the given options are positive.

(That is, if p < q then (p2 – q2) will be negative).

Therefore, the possibilities of (p, q) reduced to (7,4), (14, 2) and (28,1).

If (p,q) = (28,1) then (p2 – q2) = 282 – 12 = 784 – 1 = 783.
If (p,q) = (14, 2) then (p2 – q2) = 142 – 22 = 196 – 4 = 192
If (p,q) = (7,4) = (p2 – q2) = 72 – 42 = 49 – 16 = 33

From the given options, required answer is option c.

Answer : C

Solution :

Total number of outcomes in the simultaneous throw of 2 dice is 6 x 6 = 36.
That is, n(S) = 36.
Let E be the event that 1st is a divisor of 2nd.
Then E = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,4),(2,2),(2,6),(3,3),(3,6),(4,4),(5,5),(6,6)}
N(E) = 14.
Required probability = p(E) = N(E) / n(S) = 14/36 = 7/18.

Answer : B

Solution :

The seating arrangement will be

9

From the above diagram, we can see that the second half arrangement is :
M M M M M M M W M M M M M M M M W M M M M M M M M M W M
Here, there are 25 men in the above arrangement.

Answer : D

Solution :

Ravi went to hospital on Monday.
Then, three days before Monday is Friday and the two days after Friday is Sunday.
And the next day of Sunday is Monday and the day after tomorrow of Monday is Wednesday.
So, Ravi will go to hospital on Wednesday.

Answer : B

Solution :

For calendar to repeat exactly, the dates and days have to match perfectly.
Consider 2014 :
Any date on 2014 will correspond to same date on 2013 advanced by one day. (same logic used in first question.) For example if Jan 1 is Tuesday on 2013, then Jan 1 will be Wednesday on 2014.

Year             2014     2015     2016 (leap year)    2017   2018    2019

Advanced  days   1       1       2                     1       1       1

In 2019 the total number of advancements will be 1 + 1 + 1 + 2 + 1 + 1 = 7 . Any week has seven days. Hence advancement of 7 days also means the days are going to be the same for any dates. That is if 1st Jan on 2013 is Tuesday, then 1st Jan on 2019 will also be Tuesday.
Hence the calendar for the year 2013 and 2019 is the same.

Answer : C

Solution :

Inspecting the encoding of AEGI :
Pairing two-two letters we have AE GI.
The middle letter between A and E is C and the middle letter between G and I is H.

Similarly applying the same logic to ACDF we get
Pairing two-two letters, we have CE DH
The middle letter between C and E is D while the middle letter between D and H is F
Hence the answer is DF.

Answer : A

Solution :

Required figure is:

13

Here, squares with side of 2 coins are 1-2-5-4, 2-3-6-5, 5-6-9-8, 8-7-4-5 and 2-4-8-6.
Required 4 squares of equal size are 1-2-5-4, 2-3-6-5, 5-6-9-8 and 8-7-4-5.
The coins that are shared is 2, 5, 8, 6 and 4.
Hence the required answer is 5.

Answer : D

Solution :

From the question, we can infer that the triangle would be a scalene triangle.

14

From the above diagram, we have to find AD and CD.
By applying the Pythagoras theorem for the triangle BAD,
BD2 + AD2 = AB2
AD2 = 132 – 122
AD = 5
Similarly for triangle BDC,
BD2 + CD2 = BC2
122 + CD2 = 372
CD2 = 1369 – 144 = 1225 = 352
CD = 35
Or alternately, CD = AC – AD = 40 – 5 = 35
Now, the required ratio is AD : CD = 5 : 35 = 1 : 7
Hence, the answer is 1:7

Answer : A

Solution :

15

Louis Pasteur started from his place and went 20 km towards north. Then he turned left and walked 40 km. At this stage he turned left and walked 20 km. Finally he moved 20 km after turning left. It may be seen that his present position is 20 km from the starting point. The starting point is to his right side now.

Answer : A

Solution :

General form is add the given two numbers and divide the results by 2, 3, 4 and 5 respectively, to get the answer.

That is,

51 + 23 = 74 = 74 / 2 = 37
62 + 34 = 96 = 96 / 3 = 32
59 + 49 = 108 = 108 / 4 = 27
And, 84 + 56 = 140 = 140 / 5 = 28.
Hence, the required answer is 28.

Answer : C

Solution :

A takes 5 hours to complete 1/8 of work.
He will take 5 x 8 = 40 hours to complete whole work.

B takes 3 1/3 days to complete 80% of the work;
i.e., he takes 10/3 days = 10/3 x 24 hours = 80 hours to complete 80% work.
Therefore, whole work will be done by B in 100 hours.

C takes 1 1/12 days = 13/12 days to complete 2/3 of work.
i.e., he takes 13/12 x 24 hours = 26 hours for 2/3 of work.

Therefore, whole work will be done by C in (26 x 3/2) = 39 hours.
And he takes minimum time to do the work.
Hence, C will complete the work first.

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 that sum of the three is greater than 12.
Then E = {(6,1,6), (6,2,6), (6,3,6), (6,4,6), (6,5,6),(6,6,6), (5,2,6), (5,3,6), (5,4,6), (5,5,6),(5,6,6), (4,3,6), (4,4,6), (4,5,6),(4,6,6), (3,4,6), (3,5,6), (3,6,6), (2,5,6), (2,6,6), (1,6,6)}
Therefore n(E) = 6 + 5 + 4 + 3 + 2 + 1 = 21
Hence p(E) = 21 / 216 = 7 / 72.

Answer : C

Solution :

From statement I, Total customers to three showrooms = (15 x 3) = 45.

From statement II, Given that A = C + 2 and A = B + 1

By sub. A = B + 1 in A = C + 2 we get,

B + 1 = C + 2

B = C + 1

Total Number of Customers = 45

i.e., A + B + C = 45

Sub. values for A and B,

C + 2 + C + 1 + C = 45

3C = 42

C = 14

Hence, I and II together gives the answer.

main()
{
int x=10,y=5,p,q;
p=x>9;
q=x>3&&y!=3;
printf(“p=%d q=%d”,p,q);
}

Answer : 1 , 1 

main()
{
int i=+1;
while(~i)
printf(“vicious circles”)
}

Answer : continuous loop 

Few areas of neuron behavioural research seemed more promising is the early sixties than that investigating the relationship between protein synthesis and learning. The conceptual framework for the research was derived directly from molecular biology, which had shown that genetic information is stored in nucleic acids and expressed in proteins why not acquired information as well.

The first step towards establishing a connection between protein synthesis and learning seemed to be to block memory (cause adhesion) by interrupting the production of proteins. We were fortunate in finding a non lethal dosage of puromycin that could, it first appealed, thoroughly inhibit brain protein synthesis as well as reliability produce amnesia.

Before the actual connection between protein synthesis and learning could be established however we began to have douche about whether inhibition of protein synthesis was in fact the method by which puromycin produced amnesia. First, other drugs, glutavimides themselves potent protein synthesis inhibitors either failed to cause amnesia in some situations where it could easily be induced by puromycin or produced an amnesia with a different time course from that of puromycin. Second, puromycin was found to inhabit protein synthesis by breaking certain amino acid chaim, and the resulting fragments were suspected of being the actual cause of amnesia is some eases. Third, puromycin was reported to cause abnormalities in the train, including seizures. Thus, not only were decreased protein synthesis and amnesia dissociated, but alternative mechanism for the amnestic action of puromycin were readily suggested.

So, puromycin turned out to be a disappointment. It came to be regarded as a poor agent for amnesia studies, although, of course, it was poor only in the context of our original paradigm of protein synthesis inhibition. In our frustration, our initial response was simply to change dregs rather than our conceptual orientation. After many such disappointments, however, it now appears unlikely, that we will make a firm connection between protein synthesis and learning merely by pursuing the approaches of the past our experience with drugs has shown that all the amnestic agents, often interfere with memory in ways that seem unrelated to their inhibition of protein synthesis. More importantly, the notion that the interruption or intensification of protein production in the train can be related in cause and affect fashion to learning non seems simplistic and unproductive. Remove the battery from a car and the car will not go Drive the car a long distance at high speed and the battery will become more highly charged. Neither of these facts proves that the battery power the car, only knowledge of the overall automotive system will reveal it mechanism of locomotion and the role of the battery with in the system.