# Latest Cognizant Fully Solved Placement Test Papers.

Latest Cognizant Fully Solved Placement Test Papers. Most Frequently Asked Cognizant Test Paper. Sample Cognizant Placement Papers. Cognizant Placement Papers.

1. Find a number which exceeds the aggregate of its 72, 81 and 108 parts by 8750/27.
a) 336 b) 128 c) 824 d) 538
Solution :
Let the number be X.
Then its respective 72, 81 and 108 parts are X/72, X/81 and X/108.
The aggregate of 72, 81 and 108 parts = X/72 + X/81 + X/108.
Given, X exceeds X/72 + X/81 + X/108 by 8750/27.
Then X - (X/72 + X/81 + X/108) = 8750/27
X - (9X + 8X + 6X)/648 = 8750/27
X - 23X / 648 = 8750/27
625X/648 = 8750/27
X/24 = 14
X = 24 x 14 = 336
Hence the required number is 336.

2. If a number is much greater than the sum of its 12th and 8th parts as is less than 258. Then what will be the number?
a) 258 b) 506 c) 144 d) 129
Solution :
Let the number be X.
Then its 12th and 8th parts are X/12 and X/8
Sum of X/12 and X/8 = X/12 + X/8 = 5X/24
From the given question, we have X - 5X/24 = 258-X
19X/24 = 258 - X
19X = 24 x 258 - 24X
19X + 24X = 6192
43X = 6192
X = 144
Hence the number is 144.

3. Sum of 4-times of a number and 5 times of the aggregate of its 12th, 3rd and 10th part is 387 more than thrice of the number. Then find the number.
a) 1244 b) 108 c) 144 d) 1098
Solution :
Let X be a number.
Aggregate of its 12th, 3rd and 10th part = X/12 + X/3 + X/10 = 31X/60
And 5 times of the aggregate of its 12th, 3rd and 10th parts = 5 x 31X/60 = 31X/12
Sum of 4-times of a number and 5 times of the aggregate of its 12th, 3rd and 10th parts (4X + 31X/12) which is 387 more than the thrice of X.
4X + 31X/12 = 387 + 3X
(48X + 31X)/12 = 387 + 3X
79X/12 - 3X = 387
43X/12 = 387
X = 108.
4. A sum of a number and the reciprocal of aggregation of 2nd, 3rd and 4th parts of the number is 1/26 more than the number. Find the number.
a) 26 b) 52 c) 72 d) 24
Solution :
Let the number be X.
Then aggregation of its 2nd, 3rd and 4th parts = X/2 + X/3 + X/4 = (12X + 8X + 6X)/24 = 26X/24
And the reciprocal of aggregation of 2nd, 3rd and 4th parts of X = 1/(26X/24) = 24/26X
From the given question, we have X + 24/26X = 1/26 + X
Cancel out X on both sides, we have
24/26 X = 1/26
24 / X = 1
X = 24
Hence the required number is 24.

5. Two trains are travelling at 25 kmph, starting at the same time from city 1 and city 2 respectively which are 75 km apart. There is a bee, which can fly at 50 kmph starts from city1 at the same time the train starts and hits the second train.After hitting the train, takes rest till the trains meet each other. Find the time taken by the fly to rest ?
a)15 minutes b)30 minutes c)45 minutes d)65 minutes
Solution :
Distance between the two cities = 75 km
Speed of the two trains = 25 km per hour.
Speed of the bee = 50 km per hour.
The bee rests until the two trains meet each other.
Time taken by the bee to rest = Time taken to hit the 2nd train - Time taken by the 1st train to meet the second train
We know that,"Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.
and the time taken by them to meet is = the distance travelled by two objects / their relative speed".
Relative speed of the bee and 2nd train = (50 + 25) = 75 km/hr
And the total distance travelled by them = 75 km (distance between two cities)
Then the required time = 75/(50 + 25) = 75/75 hr = 1 hour.
Relative speed of two trains = (25 + 25) = 50 km/hr
And the total distance travelled by them = 75 km (distance between two cities)
Then the required time = 75 /(25 + 25) = 75/50 hr = 3/2 hr = 1 and half hour.
Therefore the required difference = (3/2 - 1)hr = 1/2 hr = 30 minutes.
Hence the bee rests for 30 minutes.

6. Two buses are travelling at 36 kmph and are 120 km apart. There is a fly in one bus which starts to fly at 40 kmph when the bus is started. It flies between two buses until the buses meet each other. Find the distance travelled by the fly.
a)66.6 km b)76.8 km c)56.5 km d)87.5 km
Solution :
From the given question, we can know that the required distance (total distance travelled by the fly at 40kmph) is the distance travelled by the fly in the time period before the two buses meet.
Distance between two buses = 120 km
Speed of buses = 36 km/hr
Travelling towards each other, so the relative speed = (36 + 36) = 72 km/hr
Time to meet = 120 / 72 hours = 10 / 6 hours = 5 / 3 hours.
Given that the speed of the fly = 40 kmph
And the time period it flies = 5/3 hours.
Distance travelled by the fly = speed x time = 40 x 5/3 km = 200/3 km = 66.66 km
Hence the required distance is 66.6 km.

7. Two buses are travelling at 36kmph and are 120 km apart. There is a fly in one bus which starts to fly at 72kmph when the bus is started.It hits the second bus and starts flying between two buses until the buses meet each other.Find the distance travelled by the fly after hitting the second bus is.
a)120 km b)60 km c)40 km d)72 km
Solution :
Here, the distance travelled by the fly after hitting the second bus = the distance travelled by n hours (where n is the difference between the time taken by the fly to hit the 2nd bus and the time taken by the two buses to meet).
We have to find the value of n
Distance between two buses = 120 km
Speed of buses = 36 km/hr
Travelling towards each other, so the relative speed = (36 + 36) = 72 km/hr
Time to meet = 120/72 hours = 10/6 hours = 5/3 hours.
Given that the speed of the fly = 72 kmph
Speed of the 2nd bus = 36km/hr and the distance = 120 km
Time taken to hit the 2nd bus = 120/(72 + 36) = 120/108 = 30/27 = 10/9 hours
The required time difference(n) = 5/3 - 10/9 = 5/9 hour.
Now, the required distance = Speed x time = 72 x 5/9 = 40 km
Hence the required distance is 40 km

8. The digits 2,3,8,7 and 6 are arranged to form 5 digit numbers in such a way that the repetition of the digits are not allowed. Then find the exact sum of all such possibilities:
a) 6031214 b) 9031214 c) 6933264 d) 5933264
Solution :
We have 5 different values 2, 3, 6, 7 and 8.
The number of 5-digit numbers possible without repetition of any digit is equivalent to the number of unique ways to arrange our 5 different digits
That is, the number of 'permutations'.
The method to calculate these numbers:
For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we have 4 numbers to choose from, for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit.
The total unique arrangement is then: 5 x 4 x 3 x 2 x 1 = 5! = 120 different 5-digit numbers.
Out of these 120, our 5 possible values will each appear an equal number of times in the units place, tens place, hundreds place, etc.
Therefore,120/5 = 24, so there will be 24 numbers starting with 2, and 24 starting with 3 and so on.
So we know each of our 5 values will appear in each position 24 times.
This means that we can sum up them as follows:
Sum of ten thousands digits = 24 x 10,000 x (2 + 3 + 6 + 7 + 8)
+ Sum of thousands digits = 24 x 1,000 x (2 + 3 + 6 + 7 + 8)
+ Sum of hundreds digits = 24 x 100 x (2 + 3 + 6 + 7 + 8)
+ Sum of tens digits = 24 x 10 x (2 + 3 + 6 + 7 + 8)
+ Sum of units digits = 24 x 1 x (2 + 3 + 6 + 7 + 8)
= 24x10,000(26)
+   24x1000(26)
+    24x100(26)
+     24x10(26)
+        24(26)
= 6,933,264

9. The sum of 4 digit numbers that can be formed by using the digits 3, 2 , 4 and 1 such that the digits 2 and 1 always comes together is: (note that the repetition of digits are not allowed)
a) greater than 34000 b) less than 34000 c) none of these d) cannot be determined
Answer : a) greater than 34000
Solution :
Here , we have 4 different values 1,2,3 and 4.
Given that 2 and 1 always comes together in required 4-digit numbers.
And so, take 2 and 1 as a single digit,(21)
Now we have 3 values to arrange and they are (21), 3 and 4.
Here, total unique arrangements are 3! = 6
And 2 and 1 are arranged themselves in 2! times = 2
Therefore the total unique arrangements are 6 x 2 = 12
And 12/3 = 4, So there will be 4 numbers starting with 4 and 4 numbers starting with 3 and 2 numbers starting with 2 and 2 numbers starting with 1.
Now, the thousand place gives the values
Thousand value of the numbers starting with 4 = 4 x 1000 x 4 = 16,000
Thousand value of the numbers starting with 3 = 3 x 1000 x 4 = 12,000
Thousand value of the numbers starting with 2 = 2 x 1000 x 2 = 4,000
Thousand value of the numbers starting with 1 = 1 x 1000 x 2 = 2,000
Thus the total value = 16,000 + 12,000 + 4,000 + 2,000 = 34,000
Therefore, the required sum value must be greater than 34000.
Hence the answer is option a.

10. Find the sum of the 4 digit numbers which are the multiples of 5 and are formed by using the digits 5, 3, 2 and 6.
a) 24450 b) 22250 c) 23350 d) 26450
Solution :
Here, we have 4 different values 2,3,5 and 6.
A multiple of 5 will end with 5 or 0.
Here the numbers that end with 5 are 2365, 2635, 6325, 6235, 3625 and 3265.
Required sum = 2365 + 2635 + 6325 + 6235 + 3625 + 3265 = 24450

11. Find the product of distinct prime factors of 34476.
a) 1321 b) 1326 c)34476 d)2873.
Solution :
The method to find the prime factor is,
1. Find the lowest prime number that divides exactly.
2. Check if the result of division is prime.
3. If not, find the next lowest.
4. Repeat the step 2.
Here, we have to find the prime factors of 34476.
Apply the above method, we get 34476 = 2 x 2 x 3 x 13 x 13 x 17.
Required product = 17 x 13 x 3 x 2 = 1326.

12. If N is a prime number and 8N is less than 160 and greater than 72 then 12N-7 is
a) 197 b) 239 c) 97 d) 251
Solution :
Given that, N is a prime number we have to find 12N-7.
8N is less than 160 and greater than 72 can be expressed as 72 < 8N < 160.
The above inequality can be simplified to get the term 12-N as follows,
Dividing by 8, we get
9 < N < 20
Multiplying by 12 we get,
108 < 12N < 240.
Subtracting 7 from each term, we get
101 < 12N - 7 < 233
Therefore, 12N-7 lies between 101 and 233.
From the given options, 197 is less than 233 and greater than 101.

13. What is the first prime number that enters in the series 12, 21, 30,... ?
a) 67 b) 73 c) 83 d) cannot be determined.
Answer : d) cannot be determined.
Solution :
Given series is 12, 21, 30,...
Here the difference of successive terms is 9.
Therefore, the series is an A.P with a = 12 and d = 9.
The general notation of A.P series is a, a+d , a+2d,...
Here, 12, 12+9, 12+2(9), 12+3(9),...
We can rewrite the above series as,
12, 3(4+3), 3(4+2(3)), 3(4+3(3)),...
Therefore, every term of the series is a multiple of 3. i.e., composite number and none of these is a prime number.
Note that, if a and d are not co-prime then none of the terms in the A.P series is prime and if a and d are co-prime then there are infinite primes.
Hence, the prime number does not exist in the given series

14. A man saves Rs.16,500 in 10 years. If he increases savings amount every year by Rs.100 then what will be his savings amount during the first year ?
a) Rs.1200 b) Rs.500 c) Rs.1500 d) Rs.1700
Solution :
Let X be the amount saved in 1st year.
Then in next year he saves Rs.X + 100
In 3rd year, he saves Rs.X + 100 + 100
Therefore, we have X + X+100 + X+200 +...(10 terms) = Rs.16,500
Now, X, X+100, X+200,... which is an arithmetic progression with a = X, d = 100 and n = 100.
We know that, Sn = (n/2)(2a + (n-1)d)
Here Sn = Rs.16500 = (10/2)(2X + 900)
2X + 900 = 3300
X = 1200
Hence, he saves Rs.1200 in first year.

15. In every hour, a wall clock makes a beep as many as the time it shows. Find how many beeps the clock will make in a day ?
a) 78 b) 156 c) 180 d) 39
Solution :
The wall clock makes 1 beep when it shows time 1.
It makes 2 beeps when it shows time 2.
Similarly, it makes 12 beeps when the time is 12.
Then the total number of beeps = 1 + 2 + 3 + ... + 12
1,2,3,...12 is an A.P series with first term a = 1 and last term = l = 12 and n = 12
We know that, the sum of n-terms of an A.P = Sn = (n/2)(a + l) = (12/2)(1 + 12) = 78
That is, it makes 78 beeps in 12 hours.
Since a day has 24 hours, then the required number of beeps = 2 x 78 = 156.

16. Sunil parents gave him Rs.50 in the first month and asked him to save that. In the second month they repeated with Rs.55, then with Rs.60 in the third month and so on. This continued for 10 long years. Every year he gave certain amount from his savings to his parents for his expenses. The amount was increased by 10% every year. At the end of 5 years, only Rs.5069.40 was remaining. Find the amount he gave for his expenses during the first year.
a) Rs.5000 b) Rs.6826.40 c) Rs.6000 d) Rs.5295.20
Solution :
First, we have to find the total savings amount of 10 years. (120 months).
His parents gave Rs. 50 in first month and increase the amount by Rs.5 for successive months.
Then the total amount, 50 + 55 + 60 + 65 + .....(120 months)
The above series is an arithmetic progression.
we know that, the sum of first n-terms of an A.P is Sn = (n/2)(2a + (n-1)d)
Here, n = 120, a = 50 and d = 5
Then Sn = (120/2)(2(50)+(119)5) = 60(100 + 595) = 41700
Therefore, total savings amount is Rs.41700.
From Rs.41700, he gave amount for 5 years and the remaining amount is Rs.5069.40
Total amount given by him in 5 years = Rs.41700 - Rs.5069.40 = Rs.36630.60
Let X be the amount given by him in 1st year.
In the next year, he gave X + 10X/100 = 11X/10
And in 3rd year, he gave 11X/10 + 10/100 (11X/10) = 121X/100
That is, X, 11X/10, 121X/100,.. which is a geometric progression with first term X and r = 11/10 and n = 5.
Then X + 11X/10 + 121X/100 + ... (5 terms) = 36630.60
We know that, the sum of first n terms in G.P = Sn = a(rn - 1)/r - 1
Here, X((11/10)5 - 1)/(11/10 - 1) = 36630.60
X[(161051-100000)/100000]/(1/10) = 36630.60
61051X = 366306000
X = 6000.
Hence, he gives Rs.6000 for his expenses in first year.

17. Find the actual profit percentage of a milk vendor, if he sells milk using a faulty measure which reads 750 ml as 1000 ml.
a) 12% b) 25% c) 33% d) 40%
Solution :
The vendor sells 750 ml instead of 1000 ml (i.e., 1 litre)
Profit made by the vendor is 1000 - 750 = 250 ml.
250 ml = 1/3 litre
Profit = 1/3
Profit % = 1/3 x 100 = 33.33% = 33%(approx)

18. A seller sold 2 items at Rs.23998.80 each, such that one is sold at a profit of 40% and another one at a loss of 40%. Then what is the net loss?
a) Rs.9142.40 b) Rs.8142.40 c) Rs.7142.40 d) Rs.6142.40
Solution :
Selling price of 1st one = Rs.23998.80
Profit = 40%
Remember that, "if there is a profit of R% then Cost price = [SP /(100 + R)] x 100 and in case of loss of R%, CP = [SP / 100 - R] x 100
Here, Cost price of 1st item = 23998.80 x 100 / 140 = Rs.17,142
Selling price of 2nd item = Rs.23998.80
Here loss = 40%
Then cost price of 2nd item = Rs.23998.80/(100-40) x 100 = Rs.23998.80 x 100/60
= Rs.39,998
Total cost price = Rs.17,142 + Rs.39,998 = Rs.57140
Total Selling price = Rs.47997.60
Net loss = Rs.57140 - Rs.47997.60 = Rs.9142.40
19. Two salesmen sales their item for Rs.8000 each. One salesmen calculates his profit % on his Cost price and another calculates his Profit % on selling price. If both claim to have a profit of 30% then what is the difference in their profit amount?

a) Rs.564 b) Rs.554 c) Rs.654 d) Rs.614
Solution:
In 1st case,
Selling price = Rs.8000
Profit = 30% of Cost price
CP = [SP /(100 + R)] x 100 = 8000 / 1.3 = Rs.6153.84
Profit = SP - CP = Rs.8000 - Rs.6153.84 = Rs.1846.16
In case 2,
Selling price = Rs.8000
Profit = 30% of selling price
Profit = 30/100 x 8000 = Rs.2400
Therefore required difference = Rs.2400 - Rs.1846.16 = Rs.553.84

20. Four friends Mr.A, Mr.B, Mr.C and Mr.D are walking in a park. On the way, they have to cross a narrow bridge in which only two person can cross at a time. They walk at different speed and can cross at 12, 24, 21 and 5 minutes respectively. Find the minimum time taken by them to cross the narrow bridge.
a)24 minutes b)33 minutes c)36 minutes d)26 minutes
Solution:
Given that only two person can cross the bridge at a time.
We have to find the minimum time taken by them.
Let C and B be allowed to cross the bridge first.
Time taken by C and B to cross the bridge is 21 and 24 minutes.
At 21st minute C could reach the other side of the bridge and B has 3 minutes more to cross.
Next A is allowed to cross the bridge and the time taken by him to cross is 12 minutes.
In the fist 3 minutes of A, B would have reach the other side of the bridge and A has 9 minutes more to cross the bridge.
Meanwhile D is allowed to cross and time taken by him is 5 minutes.
In the 5th minute D would have reach the end but A has 4 minutes more to reach the end.
Total time taken by them to cross = 21 + 3 + 5 + 4 = 33 minutes.
Hence the minimum time taken by them to cross the bridge is 33 minutes.

21. Seven girls namely P, Q, R, S, T, U and V can cross a rope bridge in 22, 34, 14, 4, 13, 29 and 25 minutes respectively. If only three girls can cross the bridge at a time then the minimum time taken by them to cross the bridge is:
a)34 minutes b)29 minutes c)47 minutes d)40 minutes
Solution :
Given that, P, Q, R, S, T, U and V can cross a rope bridge in 22, 34, 14, 4, 13, 29 and 25 minutes respectively.
i.e., S, T, R, P, V, U and Q takes 4, 13, 14, 22, 25, 29 and 34 minutes respectively.
Only three girls can cross the bridge at a time.
Then allow the first 3 slowest girls to cross the bridge.
i.e., allow V, U and Q and they take 25, 29, and 34 minutes respectively.
Therefore at 25th minute V will reach the other side of the bridge ....(1)
And U, Q has 4 and 9 minutes more to reach the other side.
Now, allow the next slowest girl P and she takes 22 minutes.
At the 4th minute (after 25 minutes)of P, U would reach the other side of the bridge....(2)
Q and P has 5 and 19 minutes more to reach the other side.
Now the next slowest girl R is allowed and she takes 14 minutes.
In the 5th minute (after 29 minutes) of R, Q would reach the other side of the bridge....(3)
R and P respectively has 9 and 13 minutes more to reach the other side.
Now, the next slowest girl T is allowed and she takes 13 minutes.
In the 9th minute (after 34 minutes)of T, R have reached the other side of the bridge....(4)
T and P separately has 4 minutes more to reach the other side.
Finally, the last girl S is allowed to cross and she takes 4 minutes to cross the bridge.
Now, T, P and S takes 4 minutes to reach the other side.
At the 4th minute (after 43 minutes) all seven girls are in the other side of the bridge.
Therefore the required time = 25 + 4 + 5 + 9 + 4 = 47 minutes.
Hence the minimum time is 47 minutes.

22. Eight members P, Q, R, S, T, U, V and W can cross a bridge in 3, 13, 21, 17, 5, 9, 1 and 25 minutes respectively. If only two can cross the bridge at a time, then find which of the following group crosses in minimum time?
a)(P,Q,V,W) b)(R,S,T,U) c)(Q,S,U,W) d)(P,R,T,V)
Solution :
First let us find the minimum time taken by group a. (P,Q,V,W)
Time taken by P, Q, V and W is 3, 13, 1, 25 minutes respectively.
Here, we would allow the two slowest persons Q and W to cross a bridge at a time.
Then at 13th minute Q would reach the other side of the bridge.
And W has 12 minutes more to cross the bridge.
Now allow the third person P to cross the bridge and he takes 3 minutes.
In the 3rd minute P would reach the other side of the bridge.
And W has 9 more minutes to reach the other side.
Now, allow the 4th person V to cross the bridge and he takes 1 minute.
Then in the next minute (after the above 3 minutes) V will reach the other side.
and W has 8 minutes more and in the next 8th minute W would also reach the other side.
Therefore, the minimum time taken by the group a is = 13 + 3 + 1 + 8 = 25 minutes.
Similarly,
The minimum time taken by the group b (R,S,T,U) is = 17 + 4 + 5 = 26 minutes
And the minimum time taken by the group c (Q,S,U,W) is = 17 + 8 + 9 = 34 minutes
And the minimum time taken by the group d (P,R,T,V) is = 5 + 3 + 13 = 21 minutes
Therefore, group d takes the minimum time to cross the bridge.
Hence the required answer is option d.

23. The straight distance between X and Y is 1000 meters. A taxi starts travelling from X at morning 8 o'clock. There is a taxi for every 30 minutes interval. Each taxi travels at a speed of 25 km/hr. A man travels from Y at a speed of 20 km/hr. If the man starts his journey at 2 pm, find the number of taxi he encounters in his journey.

a)17 b)5 c)12 d)8

Solution :
Distance between X and Y = 1000 meters = 100km
The taxi starts from X at 8 am and travels at a speed of 25km/hr.
Time required for the taxi to reach Y = 4 hours (100/25 = 4)
The man travels from Y at a speed of 20km/hr.
Time required to reach X = 5 hours (100/20) and his arrival to X is at 7pm.
Between 8 am to 2 pm, 12 taxi's are started from X. (2 x 6 = 12, since every 30 mins a taxi is started (per hour 2 taxi))
Now the taxi started at 8 am reaches Y at 12 noon, since the time taken to reach Y is 4hrs.
Similarly, taxi starting at 8.30 am would reach Y at 12.30 pm, starting at 9am would reach at 1pm, and so on.
At 2 pm, the taxi started at 10 am would have reached Y.
By 2 pm, 5 taxi's had arrived to Y.
So, remaining(12 - 5) 7 taxi's are on their way to Y and will meet the man along the way.
Now, the man starts at 2 pm takes 5 hrs to reach X.
Within 5 hours (5 x 2) 10 more taxi's are started, which will also meet the man on his way.
so the man would meet (10 + 7) 17 taxis on his way.
(i.e., Excluding the taxi's that start at 10 and 7 the answer will be 17.
10.30,11,11.30,12,12.30,1,1.30,2,2.30,3,3.30,4,4.30,5,5.30,6,6.30...
totally 17.)

24. A taxi starts travel from P at morning 8 am to Q which is at a distance of 50 km from P. There is a taxi for every 30 minutes from P and each taxi travels at a speed of 25 km/hr. Two men A and B starts at 2 pm from Q at a speed of 12.5 km/hr and 10 km/hr respectively. During their journey,find the number of taxi's B encounter more than A.
a)1 b)4 c)2 d)3
Solution :
Total distance of journey = 50km
A travels from Q at a speed of 12.5km/hr.
Time required by A to reach P = 50/12.5 = 4 hours
Therefore the time of arrival to P is at 6 pm
Similarly, B travels from Q at a speed of 10km/hr.
Time required by B to reach P = 50/10 = 5 hours
Therefore the time of arrival to P is at 7pm
Given that there is a taxi for every 30 minutes.
And, B need 1 hour more than A. Within this 1 hour period B encounters 2 more taxi's than A.

25. The distance between two stations A and B is 75 km. A taxi starts from A at morning 7 o'clock. There is a taxi for every 45 minutes interval and each taxi travels at a speed of 25 km/hr. A man travels from B and reaches station A such that 15km at a speed of 15 km/hr and remaining distance by 30 km/hr respectively. If the man starts at 1 pm then find the number of taxi's he encounters during his journey.
a)7 b)8 c)6 d)9
Solution:
Distance between A and B = 75 km
A taxi starts from A at 7 am and each taxi travels at a speed of 25km/hr.
Time required for the taxi to reach Q = 3 hours (75/25 = 3)
A man travels from B at a speed of 15 km/hr for 15 km and 30 km/hr for remaining 50 km
Time required by him to reach A = (15/15) hours + (60/30) = 1 + 2 = 3 hours
So the actual time of arrival is at 4 pm.
Since there is a taxi for every 45 minutes interval
Between 7 am to 1 pm, 8 taxi's are started from A.
The taxi started at 7 am reaches B at 10 am, since time taken by taxi to reach B is 3 hrs.
Similarly, taxi starting at 7.45am would reach B at 10.45 am, starting at 8.30 am would reach at 11.30 am, and so on.
By the time 1 pm, the taxi starting at 10 am would gave reached B.
In this process a total of 5 taxi's had already arrived to B by 1 pm.
So remaining(8 - 5) 3 taxi's are on their way to B and will meet the man along the way.
Now the man starting at 1 pm takes 3 hrs to reach A.
Within this 3 hours, 4 more taxi's are released which will also meet the man on his way.
so the man would meet (3 + 4) 7 taxi's on his way.
(i.e., Excluding the taxi's that start at 10 am and 4 pm the answer will be 7
10.45,11.30,12.15,1,1.45,2.30 and 3.15
totally 7.)