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Question 1 of 10
1. Question
1 pointsCategory: ReasoningDirection: Q(15)
Manoj had a calculator which had ten digits and all the basic mathematical operators. He decided to clean the calculator by removing all the keys and while reassembling it, he interchanged the locations of three pairs of keys. For example, if he placed the key ‘1’ in the place of the key ‘5’, he placed the key ‘5’ only in the place of the key ‘1’ and not any other key. All the six keys that were interchanged corresponded only to numbers and not to any mathematical operators. Later, he input some calculations and obtained outputs as provided below. The input is shown in terms of the keys that Manoj pressed, while the output is the output displayed by the calculator.
Output:
i)128+64=195 ii)912+43=942 iii)95+84=128 iv)178+19=192Key for which of the following is not interchanged?
Correct
Explanation:
From the first calculation we can see that 1 must be interchanged with either Zero or must not have been interchanged. If 1 was interchanged with a higher number i.e. 2,3, the output would have been greater than 200.
From the 4th calculation we can see that if 1 was with interchanged with 0,the input would have been a Sum Of one twodigit number and one Onedigit number and result could not have been greater than 108. Hence, 1 must not have been interchanged.
From the 2nd calculation,9 could have been interchanged with 8 or it might not have been interchanged. If 9 was interchanged with 8, the third calculation would be 8x + 9x (since 8 and 9 are interchanged) which must be around 180, But since the sum of these two was given as 128, 9 could not have been interchanged 8. Hence. 9 must also not have been interchanged.
Now 1 and 9 are not interchanged, hence In the 4th calculation
1xx + 19 = 192. 1xx must be 173. Therefore 7 must not have been interchanged and 8 must have been interchanged with 3
From the third calculation, 9x + 3x= 128 .5 and 4 in the units place must add up to 8. This means that 5 and 4 or any one of them must have been interchanged with (2 or 6) or (0 or 8). To get a sum of 8, the only possible way is to interchange both of them with 2 and 6 as 8 is already interchanged with 3. Therefore 5 and 4 must have been interchanged with 2 and 6 any order. This implies 0 was not interchanged.
From 2nd equation.considering only the units place 2+3= 2. 3 is replaced with 8. This implies that 2 must be replaced with 4. Hence, 5 must be replaced with 6.Incorrect
Explanation:
From the first calculation we can see that 1 must be interchanged with either Zero or must not have been interchanged. If 1 was interchanged with a higher number i.e. 2,3, the output would have been greater than 200.
From the 4th calculation we can see that if 1 was with interchanged with 0,the input would have been a Sum Of one twodigit number and one Onedigit number and result could not have been greater than 108. Hence, 1 must not have been interchanged.
From the 2nd calculation,9 could have been interchanged with 8 or it might not have been interchanged. If 9 was interchanged with 8, the third calculation would be 8x + 9x (since 8 and 9 are interchanged) which must be around 180, But since the sum of these two was given as 128, 9 could not have been interchanged 8. Hence. 9 must also not have been interchanged.
Now 1 and 9 are not interchanged, hence In the 4th calculation
1xx + 19 = 192. 1xx must be 173. Therefore 7 must not have been interchanged and 8 must have been interchanged with 3
From the third calculation, 9x + 3x= 128 .5 and 4 in the units place must add up to 8. This means that 5 and 4 or any one of them must have been interchanged with (2 or 6) or (0 or 8). To get a sum of 8, the only possible way is to interchange both of them with 2 and 6 as 8 is already interchanged with 3. Therefore 5 and 4 must have been interchanged with 2 and 6 any order. This implies 0 was not interchanged.
From 2nd equation.considering only the units place 2+3= 2. 3 is replaced with 8. This implies that 2 must be replaced with 4. Hence, 5 must be replaced with 6. 
Question 2 of 10
2. Question
1 pointsCategory: ReasoningDirection: Q(15)
Manoj had a calculator which had ten digits and all the basic mathematical operators. He decided to clean the calculator by removing all the keys and while reassembling it, he interchanged the locations of three pairs of keys. For example, if he placed the key ‘1’ in the place of the key ‘5’, he placed the key ‘5’ only in the place of the key ‘1’ and not any other key. All the six keys that were interchanged corresponded only to numbers and not to any mathematical operators. Later, he input some calculations and obtained outputs as provided below. The input is shown in terms of the keys that Manoj pressed, while the output is the output displayed by the calculator.
Output:
i)128+64=195 ii)912+43=942 iii)95+84=128 iv)178+19=192What will be output for the input “345+765”?
Correct
Incorrect

Question 3 of 10
3. Question
1 pointsCategory: ReasoningDirection: Q(15)
Manoj had a calculator which had ten digits and all the basic mathematical operators. He decided to clean the calculator by removing all the keys and while reassembling it, he interchanged the locations of three pairs of keys. For example, if he placed the key ‘1’ in the place of the key ‘5’, he placed the key ‘5’ only in the place of the key ‘1’ and not any other key. All the six keys that were interchanged corresponded only to numbers and not to any mathematical operators. Later, he input some calculations and obtained outputs as provided below. The input is shown in terms of the keys that Manoj pressed, while the output is the output displayed by the calculator.
Output:
i)128+64=195 ii)912+43=942 iii)95+84=128 iv)178+19=192Which of the following pair of digits is unchanged?
Correct
Incorrect

Question 4 of 10
4. Question
1 pointsCategory: ReasoningDirection: Q(15)
Manoj had a calculator which had ten digits and all the basic mathematical operators. He decided to clean the calculator by removing all the keys and while reassembling it, he interchanged the locations of three pairs of keys. For example, if he placed the key ‘1’ in the place of the key ‘5’, he placed the key ‘5’ only in the place of the key ‘1’ and not any other key. All the six keys that were interchanged corresponded only to numbers and not to any mathematical operators. Later, he input some calculations and obtained outputs as provided below. The input is shown in terms of the keys that Manoj pressed, while the output is the output displayed by the calculator.
Output:
i)128+64=195 ii)912+43=942 iii)95+84=128 iv)178+19=1925 is interchanged with which of the following?
Correct
Incorrect

Question 5 of 10
5. Question
1 pointsCategory: ReasoningDirection: Q(15)
Manoj had a calculator which had ten digits and all the basic mathematical operators. He decided to clean the calculator by removing all the keys and while reassembling it, he interchanged the locations of three pairs of keys. For example, if he placed the key ‘1’ in the place of the key ‘5’, he placed the key ‘5’ only in the place of the key ‘1’ and not any other key. All the six keys that were interchanged corresponded only to numbers and not to any mathematical operators. Later, he input some calculations and obtained outputs as provided below. The input is shown in terms of the keys that Manoj pressed, while the output is the output displayed by the calculator.
Output:
i)128+64=195 ii)912+43=942 iii)95+84=128 iv)178+19=192What will be output for the input “36+71”?
Correct
Incorrect

Question 6 of 10
6. Question
1 pointsCategory: ReasoningDirection: Q(68)
In code language, ‘0’ is represented as ‘$’ and ‘1’ is represented as ‘α’. For numbers greater than 1 only two symbols given above are to be used. The value of symbol for ‘1’ doubles itself every time it shifts one place to the left. For example:
0 is written as $, 1 is written as α, 2 is written as α$, 3 is written as αα.
Which of the following represents 27?
Correct
Explanation:
27 can be written in binary form as follows:
=(1* 2^{4)}+(1* 2^{3})+(0*2^{2}) +(1* 2^{1})+ (1×2^{0})
=αα$ααIncorrect
Explanation:
27 can be written in binary form as follows:
=(1* 2^{4)}+(1* 2^{3})+(0*2^{2}) +(1* 2^{1})+ (1×2^{0})
=αα$αα 
Question 7 of 10
7. Question
1 pointsCategory: ReasoningDirection: Q(68)
In code language, ‘0’ is represented as ‘$’ and ‘1’ is represented as ‘α’. For numbers greater than 1 only two symbols given above are to be used. The value of symbol for ‘1’ doubles itself every time it shifts one place to the left. For example:
0 is written as $, 1 is written as α, 2 is written as α$, 3 is written as αα.
Which of the following will represent the value of (α$α *α$)?
Correct
Explanation:
α$α= 2^{2} + (0 x 2^{1} ) + 2^{0}=5
α$= 2^{1}+(0x2^{0}) = 2
(α$α X α$) = (5 x 2) = 10
10 can be written in binary form as follows:
=2^{3 }+(0*2^{2})+2^{1}+(0*2^{0})
α$α$Incorrect
Explanation:
α$α= 2^{2} + (0 x 2^{1} ) + 2^{0}=5
α$= 2^{1}+(0x2^{0}) = 2
(α$α X α$) = (5 x 2) = 10
10 can be written in binary form as follows:
=2^{3 }+(0*2^{2})+2^{1}+(0*2^{0})
α$α$ 
Question 8 of 10
8. Question
1 pointsCategory: ReasoningDirection: Q(68)
In code language, ‘0’ is represented as ‘$’ and ‘1’ is represented as ‘α’. For numbers greater than 1 only two symbols given above are to be used. The value of symbol for ‘1’ doubles itself every time it shifts one place to the left. For example:
0 is written as $, 1 is written as α, 2 is written as α$, 3 is written as αα.
Which of the following number will be represented by α$α$αα?
Correct
Explanation:
α$α$αα
=2^{5} + 2^{3}+ 2^{1} +2^{0} = 43Incorrect
Explanation:
α$α$αα
=2^{5} + 2^{3}+ 2^{1} +2^{0} = 43 
Question 9 of 10
9. Question
1 pointsCategory: ReasoningConvert (1101001)_{2} to Decimal number system?
Correct
Explanation:
1101001
=2^{6}+2^{5}+2^{3}+2^{0}
=64+32+8+1
=105Incorrect
Explanation:
1101001
=2^{6}+2^{5}+2^{3}+2^{0}
=64+32+8+1
=105 
Question 10 of 10
10. Question
1 pointsCategory: ReasoningConvert binary number 11010011 to hexadecimal number
Correct
Explanation:
11010011_{2} = 1∙2^{7}+1∙2^{6}+0∙2^{5}+1∙2^{4}+0∙2^{3}+0∙2^{2}+1∙2^{1}+1∙2^{0} = 128+64+0+16+0+0+2+1 = 211_{10}
211 /16 = 13 Quotient and 3 remainder
A=10, B=11, C=12, D=13
Hence 211= D3Incorrect
Explanation:
11010011_{2} = 1∙2^{7}+1∙2^{6}+0∙2^{5}+1∙2^{4}+0∙2^{3}+0∙2^{2}+1∙2^{1}+1∙2^{0} = 128+64+0+16+0+0+2+1 = 211_{10}
211 /16 = 13 Quotient and 3 remainder
A=10, B=11, C=12, D=13
Hence 211= D3
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