CTS Aptitude Question paper

 If all the 6 are replaced by 9, then the algebraic sum of all the numbers from 1 to 100(both inclusive) varies by

the solution is as follows:
 1)sum of numbers from 1 to 100 is 5050

2)numbers having 6 are 6,16,..56,60,61,...69,76,86,96 whose sum is 1089.

3)remove the above sum from 5050 we get 3961.

4)replace 6 in numbers in step 2 with 9,which we get 9,19,..59,90,91,..99,79,89,99 whose sum is 1419.

5)add above sum to 3961 we get 5380.

6)so increase in sum is by 5380-5050=330.

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B is 50% faster than A. If A starts at 9 A.M. and B starts at 10 A.M. A travels at a speed of 50 km/hr. If A and B are 300 kms apart, The time when they meet when they travel in opposite direction is Ans:12 noon

speed of A=50 so speed of B=75.

let at T am they meet.

so distance travelled by A=50(T-9)
abd distance by B=75(T-10).

total=50(T-9)+75(T-10)=300 so T=12 noon.

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A,B,C, can do a work in 8,14,16 days respectively. A does the work for 2 days. B continues from it and finishes till 25% of the remaining work. C finishes the remaining work. How many days would have taken to complete the work Ans:

A compltes 1/4th in 2 days and remaining is 3/4th.
B completes 3/16 th work in 14*3/16 21/8 days
C completes 9/16 th work in 9 days. so total no of days =2+21/8+9=13 5/8 days.

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80 coins.

Ans: 6

1st time: 40 coins in each. weigh them.

Hence 20 coins remain. continue in this manner. until 5 coins remain.

then place 2 coins in each.

The problem may be solve in this way also .

A travel =50km/hr.

b travel=50 + 50*50/100.=75km/h.

Let x be the distance travel, after that they meet.

then x be the distance travel by A in time t1.

& the same x distance is travel by B in time t2.

then there time ratio is,

t1:t2=x/50:x/75=3:2.

Therefore if A travel x distance in time 3 hour & same distance travel by B
in 2 hour.

so, A start at 9AM+3=12 noon & B start at 10AM+2=12noon.

so they meet at 12 noon.

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shouldnt the answer fer coins b 7???????

srry 6 is rite!!!!

question no12 ans 9.00pm

12. A starts at 11:00AM and travels at a speed of 4km/hr. B starts at 1:00PM and travels at 1km/hr for the first 1hr and 2km/hr for the next hr and so on. At what time they will meet each other.

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22)this is true when B is negative

divya: Let Fathers age and sons age be 5x and 3x respectively
           then, Fathers age after 10 years= (5x+10)
                     Son's age after 10 years =(3x+10)
                    (5x+10)/(3x+10)=3/2
                      2(5x+10)=3(3x+10)
                     x=10
                  father's age is 50
                  son's age is 30

ans for question no 30 is 32

Answer: 4
Split the coins into 3 groups
27, 27, 26

1) measure 27 & 27..
if equal the coin is in the remaining 26,
else it must be in any one of the 27.

2) 9,9,9

3) 3,3,3

4) 1,1,1

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80 coins - One Counterfeit-  No of weighings required.

The answer is 4.

Step 1:  Divide 80 as : 27 -  27- 26 : If the first two weighing are the same select 26 coins later.

 Step 2:  Taking the worst case: 27
Divide into 3 groups : 9 -9-9 

Step 3: Divide 9 as : 3-3-3

Step 4: Divide 3 as 1-1-1

The logic is: On weighing 1 and 1 , if one one tilts you know, which coin is counterfeit, else the remaining one coin is counterfeit.

  

Question:
How will you find distance between Nagpur and Mumbai? I took one hour more when I travel at 80 km/hr than at 90 km/hr.

Distance = Speed x Time

Hence ,
Distance = 80 x (t +1 ) --(1), since it takes one hour more

Distance = 90 x t -- (2), where t is the time in hours, since speed (given) also has dimensions in hours.

Now, dividing one by the other. Since distance is the same , it will result to 1, hence you get the equation after solving:

t= 8

On solving for both the equations above, you get the distance as: 720 Kms.

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26. A car travels from B at a speed of 20 km/hr. The bus travel starts from A at a time of 6 A.M. There is a bus for every half an hour interval. The car starts at 12 noon. Each bus travels at a speed of 25 km/hr. Distance between A and B is 100 km. During its journey , The number of buses that the car encounter is

The Answer is 3.

The car encounters the first bus when it starts, or in some mins if car starts earlier :)

The car travels slower than any bus , as given in the question, hence it will not be able to catch up with the other buses which had left before.

Distance = Speed x time. The car takes 5 hours to reach the destination. Buses take 4 hours.

Hence the bus which will start 30 minutes or 1 hour later than the car will be able to catch up with the car.

Hence the total no. of buses encountered is: 3  

If all the 6 are replaced by 9, then the algebraic sum of all the numbers from 1 to 100(both inclusive) varies by Ans: 330


Another way of looking at this question is where all do we need to replace:

Exluding 60 - 69 , we are left with 9 places 6, 16,26,36,46,56,76,86,96 where adding 3 each will make the changes, hence partial sum now is 9 x 3 =27.

In the remaining part: excluding 66, 60 - 69 can to be replaced  by adding 30 , which comes as 30 x 9 =270 .

Finally , for 66 adding 33 will suffice.

Hence the difference will be:


27 + 270 +33 =   330

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There will be 51 multiples of 2 between 100 and 200 (both are inclusive)
There will be 33 mutiples of 3 between 100 and 200 (both are inclusive)
Number of mutiples in common is 17
(51+33)-17 = 67
Its just logical just think cool.

Answer for 24th Question
90 x n=80 x (n+1)
=> n=8
so distance =90 x 8=720km

Answer for 12th Question is 7
Method-Let it take n hours
when they meet the distance travelled by both shud be same.
so 4*n=n(n+1)/2n=7

Answer for 27th Question - Father's age will be-60 and son's age will be 40

Answer for 22nd Question - It can hold true for both condtions.he main  thing is that the value of C>B 

Answer 19th Question - They will take 40 seconds

Answer 15th Question - 13.55 days

Question 3: 

From a pack of cards Jack, Queen, King & ace are removed. Then the algebraic sum of rest of the cards is

The total no divisible by 2 b/w 100 and 200 is 50(102,104....)
The total no divisible by 3 b/w 100 and 200 is 29(120-150 is 10 then 151-180 is 10 then 181-200 is 6)
so total is 79.

Answer:

1. Amount of work completed by A in 2 days:1/4
2. B completes half of the work in:7 days
3. Remaining 25% ie 1/4 work is completed by C in 4 days

Answer 5 is rite.

Total = 80 coins

1 time - 40 + 40

2 time - 20 + 20

3 time - 10 + 10

4 time - 5 + 5

5 time - 2 + 2 here odd coin found no need to further weighing those coins.

There is a simpler way of doing it.

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13th questions answer is 5.

Step 1 : weigh 27,27 and 26 coins are left
Step 2 : if one 27 measure less, weigh 9, 9 and 9 is left
Step 3 : weigh 3, 3 and 3 is left
Step 4 : weigh 1, 1 and 1 is left.

but we have to check whether odd coin is in 26 coins which is left.
therefore...

Step 2 : weigh 9, 9 and 8 is left
Step 3 : weigh 3, 3 and 2 is left (if odd coin is in 8 coins which is left)
Step 4 : weigh 1, 1 and 2 is left(if odd coin is in 2 coins which is left)
Step 5 : weigh 1, 1