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In a college cricket club of 22 players, captain and wicketkeeper are selected from seniors team. For remaining players they have to select from 8 seniors and 12 juniors. Now, for a new team of 11 players except captain and wicketkeeper, out of 9 they select 4 from juniors and 5 from seniors. Find the number of methods to select players of new team.
A12220
B27620
C15870
D27720
Answer: Option D
Explanation:New team of 11 players will be formed.
As per quesiton 5 seniors selected out of 8 in 8C5 ways = 8! / 5! (8 - 5)! = 8! / 5!3! = 56 ways
and 4 juniors selected out of 12 in 12C4 ways = 12! / 4! (12 - 4)! = 495 ways
nCr = n! / r!(n-r)! <=== point to remember
2 players [captain & wicketkeeper] already selected = 1 possible way.
Hence the total number of possible ways = 56 x 495 x 1 = 27720 ways.
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282 / 291
A train P starts from Delhi at 8 pm, reaches its destination UP at 1:00 am. Another train Q starts from UP at 8:00 pm and reaches Delhi at 2:00 am. The two trains P and Q will cross each other at :
A10:45 AM
B10:45 PM
C12.45 pm
D12.45 am
Answer: Option B
Explanation:Lets assume distance between UP and Delhi = X km.
So, Time taken by P to reach UP = 8 pm to 1 am = 5 hours.
=> Speed of P = X/5 km/hr.
Time taken by Q to reach Delhi = 8 pm to 2 am = 6 hours.
=> Speed of Q = X/6 km/hr.
Lets assume them meet Y hours after 8 pm.
Distance covered by P at Y hours at X/5 km/hr speed = Y * X/5 km
Distance covered by Q at Y hours at X/6 km/hr = Y * X/6 km.
As given train are running in opposit direction so
=> XY/5 + XY/6 = X
XY (1/5 + 1/6 ) = X
Y (1/5 + 1/6 ) = 1
Y (11/30) = 1
11*Y = 30.
Y = 30/11 = 2.72 hours
Therefore, they meet after 2.72 hours from 8 pm.
That is, 8 pm + 2 hours + 45 minutes (nearly).
Hence, the answer is 10.45 pm.
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283 / 291
There are two bus stands, namely X and Y. Buses leave from X for every 30 minutes and its first bus starts at 8:05 am. Every hour number of buses leaving from Y increases by 1 and its first bus starts at 7:00 am. From Y there is only 1 bus for the 1st hour. Any bus from either of the bus stations takes 15 minutes to reach a nearby bus stop. Suppose a person reaches the stop in between 12:15 pm and 1:15 pm. The probability that the person will get a bus from Y is:
A1/4
B1
C2/3
D3/4
Answer: Option D
Explanation:From bus stand X :
The first bus will leave by 8.05 am and reach the bus stop in 15 minutes, i.e. at 8.20 am
The second bus will leave after 30 minutes i.e. at 8.35 am and will reach the stop at 8.50 am
Therefore, buses will reach the stop at the following times: 8.20am, 8.50am, 9.20 am,.....,12.20 pm, 12.50 pm, 1.20 pm and so on.
Between 12.15 pm and 1.15 pm, two buses will reach the stop at 12.20 pm and 1.20 pm.
Therefore, the person will get 2 buses from X.
From bus stand Y :
The first bus will leave by 7 am and reach the bus stop in 15 minutes, i.e. at 7.15 am.
There is only one bus for first 1 hour.
i.e., the second bus will leave at 8 am.
Note that, the number of buses leaving from Y is increased by 1 per hour.
From 8 am to 9 am, two buses will leave from Y and reach the stop between 8.15 am to 9.15 am.
And from 9 am to 10 am, 3 buses will leave from Y and reach the stop between 9.15 am to 10.15 am.
Proceeding like this, we have,
From 12 pm to 1 pm, 6 buses will leave from Y and reach the stop between 12.15 pm to 1.15 pm.
Therefore, the person will get 6 buses from Y between 12.15 pm to 1.15 pm.
The probability of getting the bus from Y between 12.15 pm to 1.15 pm = Number buses from Y in between 12.15 pm to 1.15 pm / Total number of buses from X and Y in between 12.15 pm to 1.15pm = 6/(2+6) = 6/8 = 3/4.
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284 / 291
Jaya Towers and Karuna towers are two tall buildings standing 400 metres apart. Two fast flying birds were sitting on top of these two towers and were aiming at catching the grains put up in one place. Jaya Tower's height is 300 metres and the grains 160 metres apart from Jaya Towers. Both the birds were flying down at the same speed and reached the grains spot at the same time. What is the approximate height of Karuna Towers?
A200 m
B240 m
C160 m
D180 m
Answer: Option B
Explanation:Jaya Tower is 300 metre tall
Lets assume Karuna Tower's height be x
Distance of grains spot from Jaya Towers -- 160 metres.
Both the birds were flying down at the same speed and reached the grains spot at the same time
i.e Time taken for bird on Jaya tower to reach grain = Time taken for bird on Karuna tower to reach grain
Or Distance travelled by bird on Jaya tower / Speed of bird on Jaya tower = Distance travelled by bird on Karuna tower / Speed of bird on Karuna tower ->(1)
But as it is given that both the birds travelled with same speed, Speed of bird on Jaya tower = Speed of bird on Karuna tower. Applying this to equation (1) we get,
Distance travelled by bird on Jaya tower = Distance travelled by bird on Karuna tower -> (2)
Apply Pythagoras theorem, we know that (Distance travelled by bird on Jaya tower)2 = (300)2 + (160)2 -> (3)
Again by applying Pythagoras theorem to the other triangle. (Distance travelled by bird on Karuna tower)2 = (400-160)2 + x2 -> (4)
From equations 2,3 and 4 we get
(300)2 + (160)2 = (400-160)2 + x2
115600 = 57600 + x2
58000 = x2
240.83 m = x
Therefore x is approximately 240 m
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285 / 291
A can do a work in 10 days and B can do the same work in 15 days. So how many days they will take to finish the same work ?
A5 days
B7 days
C9 days
D6 days
Answer: Option D
Explanation:First find the 1 day work of both (A & B)
A 1 day's work = 1/10
and
B 1 day's work = 1/15
So (A + B) 1 day's work = (1/10+1/15)
= (3/30+2/30) = 5/30 = 1/6
So Both (A & B) together can finish work in 6 days
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286 / 291
Radha moves towards South-east a distance of 7 km, then she moves towards west and travels a distance of 14 m. From here , she moves towards North-West a distance of 7 m. and finally she moves a distance of 4 m towards East and stood at that point . How far is the staring point from where she stood?
A3 m
B4 m
C10 m
D11 m
Answer: Option C
Explanation:Here is no explanation for this answer
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287 / 291
Statements:
All mangoes are golden in colour.
No golden-coloured things are cheap.
Conclusions:
I. All mangoes are cheap.
II. Golden-coloured mangoes are not cheap.
AOnly conclusion I follows
BOnly conclusion II follows
CEither I or II follows
DNeither I nor II follows
EBoth I and II follow
Answer: Option B
Explanation:Clearly, the conclusion must be universal negative and should not contain the middle term. So, it follows that 'No mango is cheap'. Since all mangoes are golden in colour, we may substitute 'mangoes' with 'golden-coloured mangoes'. Thus, II follows.
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288 / 291
find the number that can be put in place of the question mark
5,6,?,87,412,2185.
A13
B14
C18
D20
Answer: Option D
Explanation:series in the form of 5*1+1^3=6,6*2+2^3=20,20*3+3^3=87,87*4+4^3=412,412*5+5^3=125.
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289 / 291
"p" and "q" are pointers to a node of the linked list, "head" points to the first node of the list, "next" points to the next node in the list, which of the following is true for the following piece of code:
for (p=head, q = head; p!=NULL; q = p)
{
p = p->next;
free(q);
}
ADeletes all nodes
BDeletes all but the last node
CDoes not delete any node
DThe program will crash
Answer: Option A
Explanation:Here is no explanation for this answer
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290 / 291
Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:
ARs. 175.50
BRs. 180
CRs. 170
DRs. 169.50
Answer: Option A
Explanation:Here is no explanation for this answer
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