# Reasoning: Machine Input-Output Questions — Set 42

Directions (1-5): To answer the questions, study the following example based on the corresponding conditions given below it.
1. If the sum of two digits in triangle 1 is less than 10, then increase each digits by 2 and then take the sum else decrease the digits by 2 and then take the product.
2. If the number of alphabets in English alphabet series between the 2 letters in triangle 2 is odd, then each alphabet is changed to their immediate successive alphabet in the alphabetical series else each alphabet is changed to their immediate preceding alphabet in the alphabet series.
3. If the product of two digits in triangle 3 is greater than 15, then increase digits by 1 and take the product else take the sum of two digits.
4. If the difference in the position of alphabets(in English alphabet series) of the two alphabets in triangle 4 is less than 10, then the alphabet exchange their positions to their respective immediate succeeding alphabets in the alphabetical series else alphabets exchange their positions in the triangle.

1. Question figure:

Options:

Cannot be determined
Option C
Solution:

In triangle 1: Sum of digits = 6 + 5 = 11.
Sum of digits > 10 it implies decrease each digit by 2 and take the product.
6 – 2 = 4, 5 – 2 = 3 → Product of 4 × 3 = 12.
In triangle 2: Number of letters between ‘h’ and ‘p’ = 7. Number of letters is odd. It implies change each alphabet to their immediate succeeding alphabet. ‘h’ → ‘i’ and ‘p’ → ‘q’
In triangle 3: Product of 3 and 6 = 3 × 6 = 18 Product of two digits > 15 it implies increase each digit by 1 and take the product. 3 + 1 = 4, 6 + 1 = 7 → Product of 4 × 7 = 28
In triangle 4: Difference in position of alphabets ‘l’(12) and ‘t’(20) is 20 – 12 = 8
Difference in position of alphabets < 10 it implies exchange the position of alphabets to their respective immediate succeeding alphabets. ‘l’ → ‘m’ and ‘t’ → ‘u’

2. What is the sum of the digits of triangle (1) and triangle (3) in the following figure?

40
53
41
43
60
Option D
In triangle 1: Sum of digits = 9 + 0 = 9.
Sum of digits < 10 it implies increase each digit by 2 and take the sum. 9 + 2 = 11, 0 + 2 = 2 → Sum of 11 + 2 = 13 In triangle 2: Number of letters between ‘d’ and ‘m’ = 8. Number of letters is even. It implies change each alphabet to their immediate preceding alphabet. ‘d’ → ‘c’ and ‘m’ → ‘l’ In triangle 3: Product of 4 and 5 = 4 × 5 = 20 Product of two digits > 15 it implies increase each digit by 1 and take the product.
4 + 1 = 5, 5 + 1 = 6 → Product of 5 × 6 = 30
In triangle 4: Difference in position of alphabets ‘f’(6) and ‘o’(15) is 15 – 6 = 9
Difference in position of alphabets < 10 it implies exchange the position of alphabets to their respective immediate succeeding alphabets. ‘f’ → ‘g’ and ‘o’ → ‘p’ Sum of the digits of triangle (1) and (3) = 13 + 30 = 43

3. How many meaningful words can be formed with the alphabets of triangle (2) and triangle (4)?

One
Two
Three
Four
None
Option E

In triangle 1: Sum of digits = 4 + 8 = 12.
Sum of digits > 10 it implies decrease each digit by 2 and take the product.
4 – 2 = 2, 8 – 2 = 6 → Product of 2 × 6 = 12
In triangle 2: Number of letters between ‘c’ and ‘j’ = 6. Number of letters is even.
It implies change each alphabet to their immediate preceding alphabet. ‘c’ → ‘b’ and ‘j’ → ‘i’
In triangle 3: Product of 7 and 2 = 7 × 2 = 14 Product of two digits < 15 it implies take the sum of two digits. 7 + 2 = 09. In triangle 4: Difference in position of alphabets ‘b’(2) and ‘h’(8) is 8 – 2 = 6 Difference in position of alphabets < 10 it implies exchange the position of alphabets to their respective immediate succeeding alphabets. ‘b’ → ‘c’ and ‘h’ → ‘i’ No word is formed by alphabet ‘b’, ‘i’, ‘c’, ‘i’

4. Find the sum of positions of alphabets in English alphabet series of triangle (2) and triangle (4)?

50
55
45
60
65
Option D

In triangle 1: Sum of digits = 5 + 2 = 7. Sum of digits < 10 it implies increase each digit by 2 and take the sum. 5 + 2 = 7, 2 + 2 = 4 → Sum of 7 + 4 = 11 In triangle 2: Number of letters between ‘a’ and ‘j’ = 8. Number of letters is even. It implies change each alphabet to their immediate preceding alphabet. ‘a’ → ‘z’ and ‘j’ → ‘i’ Position of ‘z’ in English alphabet series = 26 Position of ‘i’ in English alphabet series = 9 In triangle 3: Product of 6 and 1 = 6 × 1 = 6 Product of two digits < 15 it implies take the sum of two digits. 6 + 1 = 07. In triangle 4: Difference in position of alphabets ‘g’(7) and ‘p’(16) is 16 – 7 = 9 Difference in position of alphabets < 10 it implies exchange the position of alphabets to their respective immediate succeeding alphabets. ‘g’ → ‘h’ and ‘p’ → ‘q’ Position of ‘h’ in English alphabet series = 8 Position of ‘q’ in English alphabet series = 17 Sum of position of alphabets = 26 + 9 + 8 + 17 = 60

5. Find the product of numbers in triangle (1) and triangle (3)?

410
340
440
240
310
Option C

In triangle 1: Sum of digits = 2 + 5 = 7.
Sum of digits < 10 it implies increase each digit by 2 and take the sum. 2 + 2 = 4, 5 + 2 = 7 → Sum of 4 + 7 = 11 In triangle 2: Number of letters between ‘k’ and ‘v’ = 10. Number of letters is even. It implies change each alphabet to their immediate preceding alphabet. ‘k’ → ‘j’ and ‘v’ → ‘u’ In triangle 3: Product of 3 and 9 = 3 × 9 = 27 Product of two digits > 15 it implies increase each digit by 1 and take the product.
3 + 1 = 4, 9 + 1 = 10 → Product of 4 and 10 = 4 × 10 = 40
In triangle 4: Difference in position of alphabets ‘d’(4) and ‘l’(12) is 12 – 4 = 8
Difference in position of alphabets < 10 it implies exchange the position of alphabets to their respective immediate succeeding alphabets. ‘d’ → ‘e’ and ‘l’ → ‘m’ Product of numbers of triangle (1) and triangle (3) = 11 × 40 = 440

6. Directions (6-10): Study the given information carefully and answer the given questions: An input-output is given in different steps. Some mathematical operations are done in each step. No mathematical operation is repeated in next step.

7. What is the sum of all the numbers in step 2?
40
42
39
44
45
Option C

In first step: 5 × 3 = 15, 5 × 8 = 40 and 15 + 40 = 55 4 × 3 = 12, 2 × 6 = 12 and 12 + 12 = 24 1 × 8 = 8, 6 × 1 = 6 and 8 + 6 = 14
In second step: Multiply the digits of first box: 5 × 5 = 25 and multiply the answer with first digit of second box = 25 × 2 = 50 Multiply the digits of third box: 4 × 1 = 4 and multiply the answer with second digit of second box = 4 × 4 = 16
In third step: Adding both the digits of first box = 5 + 0 = 5 and square of 5 = 25 Adding both the digits of second box = 6 + 1 = 7 and square of 7 = 49 In fourth step: Adding both the digits = 25 + 49 = 74

For given question:
In first step: 6 × 4 = 24, 1 × 7 = 7 and 24 + 7 = 31 2 × 2 = 4, 3 × 3 = 9 and 4 + 9 = 13 7 × 2 = 14, 4 × 3 = 12 and 14 + 12 = 26
In second step: Multiply the digits of first box: 3 × 1 = 3 and multiply the answer with first digit of second box = 3 × 1 = 3
Multiply the digits of third box: 2 × 6 = 12 and multiply the answer with second digit of second box = 12 × 3 = 36
In third step: Adding both the digits of first box = 0 + 3 = 3 and square of 3 = 9
Adding both the digits of second box = 3 + 6 = 9 and square of 9 = 81
In fourth step: Adding both the digits = 9 + 81 = 90

8. What is the difference between the numbers obtained in step 3?
63
90
81
73
None of these
Option E

9. What is the minimum value must be added in number of step 4 to make it a perfect square?
9
8
7
6
10
Option E

Hence, 10 must be added to make 90 a perfect square.

10. What is the smallest number obtained in all the steps together?
9
3
13
10
5
Option B

Hence, 3 is the smallest number obtained in all the steps together.

11. What is the result if numbers of step 3 divide with each other?
8
8.5
9
7.4
None of these
Option C

Hence, 9 is the answer.