Logical Reasoning1
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Question 1 of 6
1. Question
Sheela is a typist and she always types the number 1 as one, 2 as two and so on.
And she always counts the letters she types e.g. if she types one two three four the number of letters she types is 3 + 3 + 5 + 4 = 15.
She charges 0.5 paisa per letter for typing.Note: Do not count spaces. For example, 197 (one hundred and ninety seven) contains 24 letters. And take 100 as hundred.
How many letters did she type, if she typed from 1 to 99?
Correct
One two three four five six seven eight nine = (3+3+5+4+4+3+5+5+4) = 36
Ten Eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen = (3+6+6+8+8+7+7+9+8+8) = 70
Twenty Thirty Forty Fifty Sixty Seventy Eighty Ninety =(6+6+5+5+5+7+6+6) = 46
Total words from 1 to 99 = 36*9 + 70 + 46*10 = 854Incorrect
One two three four five six seven eight nine = (3+3+5+4+4+3+5+5+4) = 36
Ten Eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen = (3+6+6+8+8+7+7+9+8+8) = 70
Twenty Thirty Forty Fifty Sixty Seventy Eighty Ninety =(6+6+5+5+5+7+6+6) = 46
Total words from 1 to 99 = 36*9 + 70 + 46*10 = 854 
Question 2 of 6
2. Question
Sheela is a typist and she always types the number 1 as one, 2 as two and so on.
And she always counts the letters she types e.g. if she types one two three four the number of letters she types is 3 + 3 + 5 + 4 = 15.
She charges 0.5 paisa per letter for typing.Note: Do not count spaces. For example, 197 (one hundred and ninety seven) contains 24 letters. And take 100 as hundred.
What will be the cost of typing (in Rs), if she types from 60 to 147?
Correct
Sixty Seventy Eighty Ninety = (5+7+6+6) = 24
Words till 99 = 24*10 + 36*4 = 384
Hundred = 7, and = 3
Words from 101 to 139 = 39*(3+7+3) + 36*3 + 70 = 685
Words from 140 to 147 = 8*(3+7+3) + (27) = 131
Total words from 60 to 147 = 384 + 7 +6*10+6*10+5*7+ 685 + 131 = 1331
Cost = 1331*0.5 = Rs.665.5Incorrect
Sixty Seventy Eighty Ninety = (5+7+6+6) = 24
Words till 99 = 24*10 + 36*4 = 384
Hundred = 7, and = 3
Words from 101 to 139 = 39*(3+7+3) + 36*3 + 70 = 685
Words from 140 to 147 = 8*(3+7+3) + (27) = 131
Total words from 60 to 147 = 384 + 7 +6*10+6*10+5*7+ 685 + 131 = 1331
Cost = 1331*0.5 = Rs.665.5 
Question 3 of 6
3. Question
Sheela is a typist and she always types the number 1 as one, 2 as two and so on.
And she always counts the letters she types e.g. if she types one two three four the number of letters she types is 3 + 3 + 5 + 4 = 15.
She charges 0.5 paisa per letter for typing.Note: Do not count spaces. For example, 197 (one hundred and ninety seven) contains 24 letters. And take 100 as hundred.
What will be the difference in the cost of typing (in Rs), if she starts charging 25 paisa for typing a single space and she types from 47 to 120?
Correct
We just need to count total number of spaces in order to calculate the difference.
47 – 99 = 1 space each, hence total spaces = 53 – 5(fifty, sixty….) = 48
101 – 120 = 3 spaces each, hence total = 20*3 = 60
Total space from 47 to 120 = 48 + 60 = 108
Cost to type spaces = 108*0.25 = Rs.27Incorrect
We just need to count total number of spaces in order to calculate the difference.
47 – 99 = 1 space each, hence total spaces = 53 – 5(fifty, sixty….) = 48
101 – 120 = 3 spaces each, hence total = 20*3 = 60
Total space from 47 to 120 = 48 + 60 = 108
Cost to type spaces = 108*0.25 = Rs.27 
Question 4 of 6
4. Question
Collatz conjecture in mathematics, named after Lothar Collatz, can be summarized as follows:
n→n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
According to Collatz, the numbers will end at 1.Let n be the starting number. Now, if n > 8 then what is the sum of the last 5 terms of the series?
Correct
No, matter what the starting number is, series will always end as 16 → 8 → 4 → 2 → 1
So, the sum will be 31.Incorrect
No, matter what the starting number is, series will always end as 16 → 8 → 4 → 2 → 1
So, the sum will be 31. 
Question 5 of 6
5. Question
Collatz conjecture in mathematics, named after Lothar Collatz, can be summarized as follows:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
According to Collatz, the numbers will end at 1.If the starting number is an odd number greater than 5 and less than 20, then for which number will the number of terms be maximum in the series?
Correct
Lets start with 5,
5 → 16 → 8 → 4 → 2 → 1
For 7,
7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
For 9,
9 → 28 → 14 → 7→ … then it will the series of 7
11, 13 and 17 are already part of series of 7. So lets skip to 15
15 → →46→23→70→35→106→53→160→80→40→ … then it will follow series of 40 as shown in series of 7
19→58→29→88→44→22→11→ … then it will follow series of 11 as shown in series of 7
So, the series of 19 has highest number of terms.Incorrect
Lets start with 5,
5 → 16 → 8 → 4 → 2 → 1
For 7,
7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
For 9,
9 → 28 → 14 → 7→ … then it will the series of 7
11, 13 and 17 are already part of series of 7. So lets skip to 15
15 → →46→23→70→35→106→53→160→80→40→ … then it will follow series of 40 as shown in series of 7
19→58→29→88→44→22→11→ … then it will follow series of 11 as shown in series of 7
So, the series of 19 has highest number of terms. 
Question 6 of 6
6. Question
Collatz conjecture in mathematics, named after Lothar Collatz, can be summarized as follows:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
According to Collatz, the numbers will end at 1.For how many such series only 7 terms are possible?
Correct
Last 5 terms will 16→8→4→2→1 , 16 can come from 32 and 5 both So, 32→16→8→4→2→1 and 5→16→8→4→2→1
Now, 32 and 5 can only come from 64 and 10 as both are not of the form 3n+1.
64→32→16→8→4→2→1 and 10→5→16→8→4→2→1
So, only two series are possible.Incorrect
Last 5 terms will 16→8→4→2→1 , 16 can come from 32 and 5 both So, 32→16→8→4→2→1 and 5→16→8→4→2→1
Now, 32 and 5 can only come from 64 and 10 as both are not of the form 3n+1.
64→32→16→8→4→2→1 and 10→5→16→8→4→2→1
So, only two series are possible.