Mettl Test Papers and Mettl Placement Papers

Let ‘m’ be the number of males, and let ‘x’ be the fraction of males refusing the bonus. Then the total bonus paid T= 8.15(350-m)+10(1-x)m
= 2852.5+m(1.85-10x),which will be independent of m only if x=0.185; so T=2852.5
Now, number of males refusing bonus= 0.185m, which has to be a +ve integer
Therefore 185m/1000 is a +ve integer or 185/(1000/m) is a +ve integer. 1000/m =37 or 5. 37 not possible; so 1000/m=5, which gives m=200. Thus females=150. Bonus paid to females= 8.15*150=1222.5

Approach :a flowers are for b rupees; and a+10 flowers are for 2 rupees
Therefore, b < 2 and since b is an integer, so b=1
(100/a) – (200/a+10)= 80/12
a = 5

Approach :52231-49966=2265
2265 should be exactly divisible by the three digit number (if x=kn+m, y=ln+m, the remainder being m when x and y are divided by n, then x-y= n(k-l))
Now 2265 = 453*5
Thus n= 453

Let the numbers be x and y
Then 4641*21 = x * y
Numbers can also be expressed as multiples of the HCF. Thus, x= 21a and y = 21b
Now 4641 * 21 = 21a * 21b
Or 13*17*21 *21= 21a *21b
Comparing LHS and RHS, a and b are 13*21 (=273) and 17*21(=357) in any order.
The answer is 357

Let the present age of the son = x
Then, present age of the father =(60 − x)
Six years ago father's age was 5 times the age of the son
⇒(60−x)−6=5(x−6)
⇒84=6x
⇒x=14
Son's age after 6 years
=x + 6=14 + 6=20

Total number of outcomes possible when a dice is rolled = 6
Hence, total number of outcomes possible when two dice are rolled = 6*6 = 36
To get a sum of 7, following are the favourable cases.
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
Therefore, number of ways in which we get a sum of 7 = 6
P(a sum of 7) =6/36

To get a sum of 11,following are the favourable cases(5,6),(6,5)
Therefore, number of ways in which we get a sum of 11=2
P(a sum of 11) =2/36

Therefore, 6/36 + 2/36 = 8/36 =2/9

We have a total of 10 books.

Given that a particular pair of books should always be together. Hence, just tie these two books together and consider as a single book.

Hence we can take total number of books as 9. These 9 books can be arranged in ^{9}P_{9}= 9! ways.

We had tied two books together. These books can be arranged among themselves in ^{2}P_{2} =2! ways.

Hence, required number of ways=9!×2!

Pipe A can fill 1/8 of the cistern in 1 hour.

Pipe B can empty 1/12 of the cistern in 1 hour

Both Pipe A and B together can effectively fill 1/8-1/12=1/24 of the cistern in 1 hour

i.e, the cistern will be full in 24 hrs.

Let the speed of the water in still water = x
Given that speed of the stream = 3 kmph

Speed downstream =(x+3)kmph
Speed upstream =(x−3)kmph

He travels a certain distance downstream in 4 hour and come back in 6 hour.
ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour

since distance = speed × time, we have
(x+3)4=(x−3)6
⇒(x+3)2=(x−3)3
⇒2x+6=3x−9
⇒x=6+9=15kmph

Distance travelled in 1st hour =35 km
Speed of the bus increases by 2 kmph after every one hour. Hence,distance travelled in 2nd hour =37 km
distance travelled in 3rd hour =39 km and so on

Total distance travelled =[35+37+39+...(12 terms)]
=12/2[2×35+(12−1)2]
=6(70+22)
=6×92
=552

Hence, total distance travelled =552 km

Only son of woman’s grandfather – Woman’s father;
Man’s brother father – Man’s father.
So, the woman is man’s sister.

C has two children A and B, if A is his son, B must be his daughter.

In the code, the first letter is the second alphabet, the second letter is the third alphabet, the third letter is the fourth alphabet and so on after the corresponding letter in the word.

Each letter of the word is one step ahead of the corresponding letter of the code

Based on the input given in the question, it is clear that in the final step, the numbers are arranged in ascending order. So, the answer to the input “74 32 45 90 21 77 80 82” will be as follows:

Step 1: 21 74 32 45 90 77 80 82

Step 2: 21 31 74 45 90 77 80 82

Step 3: 21 31 45 74 90 77 80 82

Step 4: 21 31 45 74 77 90 80 82

Step 5: 21 31 45 74 77 80 90 82

Step 6: 21 31 45 74 77 80 82 90

Step 6, being the final step.

If we analyse the input “Herb 30 Shrub 10 Grass 20 40 Tree Planter 50” and then each step carefully, we will observe that the change in order is happening both from the left corner and from the right corner.

In step 1, the smallest number “10” is placed at the 1st position from the left and “Grass” which begins with “G” and alphabetically comes first in comparison to the first alphabet of other words moved to the last.

Then, in step 2, the second smallest number, “20”, follows 10 and the alphabetical “H”, i.e. “Herb” is placed after Grass and so on.

In the final step, all the numbers are arranged in ascending order, followed by the words which are arranged in the Dictionary-order, i.e., alphabetically.

So, the output for “11 55 Green Blue Red 33 Orange 22 44 Pink” is as follows:

Step 1: 11 22 55 Green Red 33 Orange 44 Pink Blue

Step 2: 11 22 33 55 Red Orange 44 Pink Blue Green

Step 3: 11 22 33 44 55 Red Pink Blue Green Orange

Step 4: 11 22 33 44 55 Red Blue Green Orange Pink

Step 5: 11 22 33 44 55 Blue Green Orange Pink Red

Step 5, being the final step.

It is a simple division series in which each number is one-half of the previous number. We can also say that each number is divided by 2 to arrive at the next number;

On dividing 48 by 2, we get 24

On dividing 24 by 2, we get 12

So, on dividing 12 by 2, we will get 6

In this series, the difference between the consecutive numbers increases by 2;

7 - 4 = 3
12 - 7 = 5
19 - 12 = 7

Therefore, the next number would be 19 + 9 = 28