20
Q:

# In a class of 60 students, the number of boys and girls participating in the annual sports is in the ratio 3 : 2 respectively. The number of girls not participating in the sports is 5 more than the number of boys not participating in the sports. If the number of boys participating in the sports is 15, then how many girls are there in the class ?

 A) 20 B) 25 C) 30 D) Data inadequate

Explanation:

Let the number of boys and girls participating in sports be 3x and 2x respectively.

Then, 3x = 15 or x = 5.

So, number of girls participating in sports = 2x = 10.

Number of students not participating in sports = 60 - (15 + 10) = 35.

Let number of boys not participating in sports be y.

Then, number of girls not participating in sports = (35 -y).

Therefore (35 - y) = y + 5 ${\color{Blue}&space;\Leftrightarrow&space;}$ 2y ${\color{Blue}&space;\Leftrightarrow&space;}$ 30 ${\color{Blue}&space;\Leftrightarrow&space;}$ y = 15.

So, number of girls not participating in sports = (35 - 15) = 20.

Hence, total number of girls in the class = (10 + 20) = 30.

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

1 10
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

1 4
Q:

Virus doubles every 3 mins. It is 'K' in 1 hour, when was it K/4 ?

 A) 29 min B) 25 min C) 54 min D) 57 min

Explanation:

If virus is K in 1hr i.e 60 min
then in 57 min it was K/2 as it doubles every 3 minutes
=> In 54 min it was K/4.
At 54 minutes it was K/4.

2 19
Q:

If in an A.P series 18th term is 29 and 29th term is 18. Then find the 49th term ?

 A) -3 B) -2 C) 0 D) 1

Explanation:

We know that in A.P series
18th term = a + 17d
29th term = a + 28d
But given
a + 17d = 29..........(i)
a + 28d = 18......... (ii)
Solving equation (i) and (ii), we get
d = -1
put d = -1 in any of the above equations and we get,
a = 46
Now, we know 49th term can be written as, a + 48d
putting the value of a and d in the above equation,
a + 48d = 46 + 48(-1)
= 46 - 48
= -2

Hence, the 49th term is -2.

1 7
Q:

What is the time required to punch 1500 cards of 45 column each at the rate of 9,000 punches per hour ?

 A) 8 hrs B) 7 hrs 15 min C) 7 hrs 30 min D) 8 hrs 20 min