22
Q:

# A, B, C, D and E play a game of cards. A says to B, "If you give me 3 cards, you will have as many as I have at this moment while if D takes 5 cards from you, he will have as many as E has." A and C together have twice as many cards as E has. B and D together also have the same number of cards as A and C taken together. If together they have 150 cards, how many cards has C got ?

 A) 28 B) 29 C) 31 D) 35

Explanation:

Clearly, we have :

A = B - 3 ...(i)

D + 5 = E ...(ii)

A+C = 2E ...(iii)

B + D = A+C = 2E ...(iv)

A+B + C + D + E=150 ...(v)

From (iii), (iv) and (v), we get: 5E = 150 or E = 30.

Putting E = 30 in (ii), we get: D = 25.

Putting E = 30 and D = 25 in (iv), we get: B = 35.

Putting B = 35 in (i), we get: A = 32.

Putting A = 32 and E = 30 in (iii), we get: C = 28.

Q:

O and O' are the center of the two circles with radii 7cm and 9cm respectively. The distance between the centers is 20cm. If PQ be the transverse common tangent to the circles, which cuts OO' at X, what is the length of O'X in cm ?

 A) 10 B) 6 C) 35/4 D) 45/4

Explanation:

As clear from the figure itself, triangle OQX and triangle O'PX are similar.

So,

OQ/O'P = OX/O'X

=> OX = (7/9)xO'X

And since, OX + O'X = 20

=> (7/9)xO'X + O'X = 20

=> O'X = 45/4

1 84
Q:

A kiddy bank of a child contains all Rs. 1, 50 paise and 25 paise coins. What is the total amount in the bank ?
(1) Total number of coins are 20
(2) The number of 50 paise and 25 paise coins are in the ratio of 7:3.

 A) only 1 is needed B) only 1 is needed C) Both and 2 are needed D) None

Explanation:

Given that 50 paise and 25 paise coins are in the ratio of 7:3, they can be either 7 & 3 or 14 & 6.
As total 20 coins also includes Rs. 1 coins, so 50 paise & 25 paise coins are 7 & 3 respectively & Rs. 1 coins are 10.
Total amount = (100x10) + (50x7) + (25x3)=1425 paise or Rs.14.25.

1 88
Q:

A cow was standing on a bridge, 5m away from the middle of the bridge. A train was coming towards the bridge from the ends nearest to the cow. Seeing this cow ran towards the the train and managed to escape when the train was 2m away from bridge. If it had run in the opposite direction, it would hit by the train 2m before the end of the bridge. What is the lenght of the bridge in meters assuming the speed of the train is 4 times that of cow ?

 A) 22 mts B) 28 mts C) 32 mts D) 34 mts

Explanation:

let 'b' be the Length of bridge from cow to the near end of the bridge and 'a' be the distance of the train from the bridge.
'x' be speed of cow => '4x' speed of train
Then the total length of the bridge 2b + 10.
(a-2)/4x = b/x
=> a-2 = 4b........(1)
Now if it had run in opposite direction
(a+2b+10-2)/4x = (b+10-2)/x
=> a - 2b = 24......(2)
Solving (1) and (2)
b = 11 ,
Therefore length of the bridge is 2 x 11 + 10 = 32mts.

1 84
Q:

There are 6561 balls are there out of them 1 is heavy. Find the minimum number of times the balls have to be weighted for finding out the heavy ball ?

 A) 2414 B) 204 C) 87 D) 8

Explanation:

Suppose there are 9 balls

Let us give name to each ball B1 B2 B3 B4 B5 B6 B7 B8 B9

Now we will divide all the balls into 3 groups.

Group1 - B1 B2 B3

Group2 - B4 B5 B6

Group3 - B7 B8 B9

Step1 - Now weigh any two groups. Let's assume we choose Group1 on left side of the scale and Group2 on the right side.

So now when we weigh these two groups we can get 3 outcomes.

Weighing scale tilts on left - Group1 has a heavy ball.
Weighing scale tilts on right - Group2 has a heavy ball.
Weighing scale remains balanced - Group3 has a heavy ball.
Lets assume we got the outcome as 3. i.e Group 3 has a heavy ball.

Step2 - Now weigh any two balls from Group3. Lets assume we keep B7 on left side of the scale and B8 on right side.

So now when we weigh these two balls we can get 3 outcomes.

Weighing scale tilts on left - B7 is the heavy ball.
Weighing scale tilts on right - B8 is the heavy ball.
Weighing scale remains balanced - B9 is the heavy ball.
The conclusion we get from this Problem is that each time weigh. We element 2/3 of the balls.

As we came to conclusion that Group3 has the heavy ball from Step1, we remove 6 balls from the equation i.e (2/3) of 9.

Simillarly we do the ame thing for the Step2.

Now going with this conclusion. We have 6561 balls.

Step - 1

Divided into 3 groups

Group1 - 2187Balls

Group2 - 2187Balls

Group3 - 2187Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 2

Divided into 3 groups

Group1 - 729Balls

Group2 - 729Balls

Group3 - 729Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 3

Divided into 3 groups

Group1 - 243Balls

Group2 - 243Balls

Group3 - 243Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 4

Divided into 3 groups

Group1 - 81Balls

Group2 - 81Balls

Group3 - 81Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 5

Divided into 3 groups

Group1 - 27Balls

Group2 - 27Balls

Group3 - 27Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 6

Divided into 3 groups

Group1 - 9Balls

Group2 - 9Balls

Group3 - 9Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 7

Divided into 3 groups

Group1 - 3Balls

Group2 - 3Balls

Group3 - 3Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 8

So now when we weigh 2 balls out of 3 we can get 3 outcomes.

Weighing scale tilts on left - left side placed is the heavy ball.
Weighing scale tilts on right - right side placed is the heavy ball.
Weighing scale remains balanced - remaining ball is the heavy ball.
So the general answer to this question is, it is always multiple of 3 steps.

For 9 balls  $\inline&space;\fn_jvn&space;\small&space;3^{2}$= 9. therefore 2 steps

For 6561 balls $\inline&space;\fn_jvn&space;\small&space;3^{8}$ = 6561 therefore 8 steps

2 95
Q:

Five farmers have 7,9,11,13 & 14 apple trees respectively in their orchards. Last year each of them discovered that every tree in their own orchard yields exactly the same number of apples. Further, if the 3rd farmer gives one apple to the 1st and the 5th gives 3 to each of the 2nd & d 4th, they would all exactly have the same number of apples, what were the yields per tree in the orchards of the 3rd & 4th farmers ?

 A) 17 & 9 B) 9 & 11 C) 9 & 9 D) 11 & 9

Explanation:

Let the number of apples in each tree of the 5 farmers be a, b, c, d,e respectively. Therefore total no of apples are 7a, 9b, 11c, 13d and 14e respectively.

Given,
7a+1 = 9b+3 = 11c-1 = 13d+3 = 14e-6 = x

9b+3 = 13d+3
===> 9b = 13d
so take b = 13 and d = 9

9b+3 = 9*13+3 = 120
ie, x=120

substituting
7a+1 = 120
7a = 119
a = 17

11c-1 = 120
11c = 121
c=11

14e-6 = 120
14e = 126
e = 9

Yields per tree in the orchards of d 3rd & 4th farmers= 11,9