TCS Mini Mock Test

1) 11, 23, 47, 83, 131, ___. What is the next number?

145

178

176

191

The sequence is formed by adding consecutive prime numbers starting from 11.
11 + 12 = 23
23 + 24 = 47
47 + 48 = 95
95 + 96 = 191
Therefore, the next number in the sequence is 191.

Explanation

The sequence is formed by adding consecutive prime numbers starting from 11.
11 + 12 = 23
23 + 24 = 47
47 + 48 = 95
95 + 96 = 191
Therefore, the next number in the sequence is 191.

2) An old man and a young man are working together in an office and staying together in a near by apartment. The old man takes 30 minutes and the young 20 minutes to walk from appartment to office. If one day the old man started at 10.00 AM and the young man at 10:05AM from the apartment to office, when will they meet?

10:00

10:15

10:45

10:30

The young man takes 20 minutes to walk from the apartment to the office, while the old man takes 30 minutes. Since the young man starts 5 minutes after the old man, he will be 5 minutes behind. The young man will catch up to the old man at a rate of 10 minutes per hour (30 minutes - 20 minutes). Therefore, it will take the young man 15 minutes (5 minutes behind + 10 minutes to catch up) to meet the old man. Hence, they will meet at 10:15 AM.

Explanation

The young man takes 20 minutes to walk from the apartment to the office, while the old man takes 30 minutes. Since the young man starts 5 minutes after the old man, he will be 5 minutes behind. The young man will catch up to the old man at a rate of 10 minutes per hour (30 minutes - 20 minutes). Therefore, it will take the young man 15 minutes (5 minutes behind + 10 minutes to catch up) to meet the old man. Hence, they will meet at 10:15 AM.

3) A and B are running around a circular track of length 120 meters with speeds 12 m/s and 6 m/s in the same direction. When will they meet for the first time?

After 20 seconds

After 10 seconds

After 25 seconds

They never meet

A and B are running in the same direction on a circular track. The relative speed between them is the difference of their individual speeds, which is 12 m/s - 6 m/s = 6 m/s. The time taken for them to meet can be calculated by dividing the circumference of the circular track (120 meters) by the relative speed (6 m/s). 120 meters / 6 m/s = 20 seconds. Therefore, A and B will meet for the first time after 20 seconds.

Explanation

A and B are running in the same direction on a circular track. The relative speed between them is the difference of their individual speeds, which is 12 m/s - 6 m/s = 6 m/s. The time taken for them to meet can be calculated by dividing the circumference of the circular track (120 meters) by the relative speed (6 m/s). 120 meters / 6 m/s = 20 seconds. Therefore, A and B will meet for the first time after 20 seconds.

4) How many 2's are there between the terms 112 to 375?

313

159

156

315

Between the terms 112 and 375, the numbers that contain the digit 2 are: 120, 121, 122, 123, ..., 220, 221, ..., 299, 302, 312, ..., 322, ..., 372, 382, 392, ..., 522, ..., 572, ..., 622, ..., 672, ..., 722, ..., 772, ..., 822, ..., 872, ..., 922, ..., 972, ..., 1002, ..., 1052, ..., 1102, ..., 1152, ..., 1202, ..., 1252, ..., 1302, ..., 1352, ..., 1402, ..., 1452, ..., 1502, ..., 1552, ..., 1602, ..., 1652, ..., 1702, ..., 1752, ..., 1802, ..., 1852, ..., 1902, ..., 1952, ..., 2002, ..., 2052, ..., 2102, ..., 2152, ..., 2202, ..., 2252, ..., 2302, ..., 2352, ..., 2402, ..., 2452, ..., 2502, ..., 2552, ..., 2602, ..., 2652, ..., 2702, ..., 2752, ..., 2802, ..., 2852, ..., 2902, ..., 2952, ..., 3002, ..., 3052, ..., 3102, ..., 3152, ..., 3202, ..., 3252, ..., 3302, ..., 3352, ..., 3402, ..., 3452, ..., 3502, ..., 3552, ..., 3602, ..., 3652, ..., 3702, ..., 3752. Thus, there are 156 numbers that contain the digit 2.

Explanation

Between the terms 112 and 375, the numbers that contain the digit 2 are: 120, 121, 122, 123, ..., 220, 221, ..., 299, 302, 312, ..., 322, ..., 372, 382, 392, ..., 522, ..., 572, ..., 622, ..., 672, ..., 722, ..., 772, ..., 822, ..., 872, ..., 922, ..., 972, ..., 1002, ..., 1052, ..., 1102, ..., 1152, ..., 1202, ..., 1252, ..., 1302, ..., 1352, ..., 1402, ..., 1452, ..., 1502, ..., 1552, ..., 1602, ..., 1652, ..., 1702, ..., 1752, ..., 1802, ..., 1852, ..., 1902, ..., 1952, ..., 2002, ..., 2052, ..., 2102, ..., 2152, ..., 2202, ..., 2252, ..., 2302, ..., 2352, ..., 2402, ..., 2452, ..., 2502, ..., 2552, ..., 2602, ..., 2652, ..., 2702, ..., 2752, ..., 2802, ..., 2852, ..., 2902, ..., 2952, ..., 3002, ..., 3052, ..., 3102, ..., 3152, ..., 3202, ..., 3252, ..., 3302, ..., 3352, ..., 3402, ..., 3452, ..., 3502, ..., 3552, ..., 3602, ..., 3652, ..., 3702, ..., 3752. Thus, there are 156 numbers that contain the digit 2.

5) A series of book was published at seven year intervals. When the seventh book was published the total sum of publication year was 13, 524. First book was published in?

1911

1910

2002

1932

The sum of the publication years of the books is given as 13,524. Since the books are published at seven-year intervals, we can assume that the difference between the publication years of consecutive books is 7. To find the publication year of the first book, we can subtract the sum of the publication years of the remaining six books (6 x 7 = 42) from the total sum. This gives us 13,524 - 42 = 13,482. Therefore, the first book was published in 1911.

Explanation

The sum of the publication years of the books is given as 13,524. Since the books are published at seven-year intervals, we can assume that the difference between the publication years of consecutive books is 7. To find the publication year of the first book, we can subtract the sum of the publication years of the remaining six books (6 x 7 = 42) from the total sum. This gives us 13,524 - 42 = 13,482. Therefore, the first book was published in 1911.

6) In a city there are few engineering, MBA and CA candidates. Sum of four times the engineering, three times the MBA and 5 times CA candidates is 3650. Also three times CA is equal to two times MBA and three times engineering is equal to two times CA. In total how many MBA candidates are there in the city?

200

300

400

450

not-available-via-ai

Explanation

not-available-via-ai

7) A owes B Rs.50. He agrees to pay B over a number of consecutive days starting on a Monday, paying single note of Rs.10 or Rs.20 on each day. In how many different ways can A repay B.

10

8

6

15

A can repay B in 8 different ways. The possible combinations of notes that A can use to repay B are as follows: 10, 10, 10, 10, 10; 10, 10, 10, 20; 10, 10, 20, 10; 10, 20, 10, 10; 20, 10, 10, 10; 10, 20, 20; 20, 10, 20; 20, 20, 10.

Explanation

A can repay B in 8 different ways. The possible combinations of notes that A can use to repay B are as follows: 10, 10, 10, 10, 10; 10, 10, 10, 20; 10, 10, 20, 10; 10, 20, 10, 10; 20, 10, 10, 10; 10, 20, 20; 20, 10, 20; 20, 20, 10.

8) A tree of height 36m is on one edge of a road broke at a certain height. It fell in such a way that the top of the tree touches the other edge of the road. If the breadth of the road is 12m, then what is the height at which the tree broke?

16

24

12

18

The height at which the tree broke can be determined by using the concept of similar triangles. Since the top of the tree touches the other edge of the road, we can form a right triangle with the height of the tree as the vertical side and the breadth of the road as the horizontal side. The height of the tree broken can be considered as the vertical side of a similar triangle formed by the remaining part of the tree. By setting up a proportion between the two triangles, we can solve for the height at which the tree broke. In this case, the proportion would be 36/12 = x/16, where x represents the height at which the tree broke. Solving this proportion gives us x = 16.

Explanation

The height at which the tree broke can be determined by using the concept of similar triangles. Since the top of the tree touches the other edge of the road, we can form a right triangle with the height of the tree as the vertical side and the breadth of the road as the horizontal side. The height of the tree broken can be considered as the vertical side of a similar triangle formed by the remaining part of the tree. By setting up a proportion between the two triangles, we can solve for the height at which the tree broke. In this case, the proportion would be 36/12 = x/16, where x represents the height at which the tree broke. Solving this proportion gives us x = 16.

9) What word will be in 32^nd place when the letters of the word, "WAITING," are rearranged without repetition in alphabetical order?

AGNTIIW

GANTIWI

AGTNIWI

AGNTIWI

The given word is "WAITING". When the letters of this word are rearranged without repetition in alphabetical order, the 32nd word will be "AGNTIWI".

Explanation

The given word is "WAITING". When the letters of this word are rearranged without repetition in alphabetical order, the 32nd word will be "AGNTIWI".

10) A sealed envelope contains a card with a single digit written on it. Three of the following statements are true and one is false.I. The digit is 1.II. The digit is not 2.III. The digit is not 9.IV. The digit is 8.Which one of the following must necessarily be correct?

II is false

III is true

IV is false

The digit is even.

I is true

Based on the given statements, we can determine that three of the statements are true and one is false. The false statement is IV, which states that the digit is 8. Therefore, the digit cannot be 8. The only statement that is left is III, which states that the digit is not 9. Since we know that III is true and the other statements are false, it necessarily means that III is correct. Hence, III is true.

Explanation

Based on the given statements, we can determine that three of the statements are true and one is false. The false statement is IV, which states that the digit is 8. Therefore, the digit cannot be 8. The only statement that is left is III, which states that the digit is not 9. Since we know that III is true and the other statements are false, it necessarily means that III is correct. Hence, III is true.

11) A can complete a piece of work in 8 hours, B can complete in 10 hours and C in 12 hours. If A,B, C start the work together but A laves after 2 hours. Find the time taken by B and C to complete the remaining work( in hours)

2 (1/11)

4 (1/11)

2 (6/11)

2

A can complete 1/8 of the work in 1 hour, B can complete 1/10 of the work in 1 hour, and C can complete 1/12 of the work in 1 hour. Together, they can complete (1/8 + 1/10 + 1/12) = 11/60 of the work in 1 hour. Since A leaves after 2 hours, the remaining work is (1 - 2/8) = 6/8. B and C together can complete (11/60) of the work in 1 hour, so they will take (6/8) / (11/60) = 2 (1/11) hours to complete the remaining work.

Explanation

A can complete 1/8 of the work in 1 hour, B can complete 1/10 of the work in 1 hour, and C can complete 1/12 of the work in 1 hour. Together, they can complete (1/8 + 1/10 + 1/12) = 11/60 of the work in 1 hour. Since A leaves after 2 hours, the remaining work is (1 - 2/8) = 6/8. B and C together can complete (11/60) of the work in 1 hour, so they will take (6/8) / (11/60) = 2 (1/11) hours to complete the remaining work.

12) Three generous friends, each with some money, redistribute the money as follows: Sandra gives enough money to David and Mary to double the amount of money each has. David then gives enough to Sandra and Mary to double their amounts. Finally, Mary gives enough to Sandra and David to double their amounts. If Mary had 11 rupees at the beginning and 17 rupees at the end, what is the total amount that all three friends have?

105

60

88

71

not-available-via-ai

Explanation

not-available-via-ai

13) What is the value of 77!*(77!-2*54!)^3/(77!+54!)^3+54!*(2*77!-54!)^3/(77!+54!)^3

77! - 54!

77! + 54!

77!^2 - 54!^2

77!^2 + 54!^2

The given expression can be simplified by factoring out a common term of (77! - 54!)^3 / (77! + 54!)^3 from both terms. This leaves us with (77! - 54!) / (77! + 54!). Since the numerator and denominator have a common factor of (77! - 54!), they cancel out, leaving us with 1 as the final answer.

Explanation

The given expression can be simplified by factoring out a common term of (77! - 54!)^3 / (77! + 54!)^3 from both terms. This leaves us with (77! - 54!) / (77! + 54!). Since the numerator and denominator have a common factor of (77! - 54!), they cancel out, leaving us with 1 as the final answer.

14) On the fabled Island of Knights and Knaves, we meet three people, A, B, and C, one of whom is a knight, one a knave, and one a spy. The knight always tells the truth, the knave always lies, and the spy can either lie or tell the truth.A says: "C is a knave."B says: "A is a knight."C says: "I am the spy."Who is the knight, who the knave, and who the spy?

A

B

C

Cannot be determined

Based on the statements given, we can determine that A cannot be the knight because if A were the knight, they would be telling the truth about C being a knave, which contradicts what C says. We can also determine that C cannot be the knight because if C were the knight, they would be telling the truth about being the spy, which contradicts what the spy can do (lie or tell the truth). Therefore, B must be the knight because if B were the knight, they would be telling the truth about A being a knight.

Explanation

Based on the statements given, we can determine that A cannot be the knight because if A were the knight, they would be telling the truth about C being a knave, which contradicts what C says. We can also determine that C cannot be the knight because if C were the knight, they would be telling the truth about being the spy, which contradicts what the spy can do (lie or tell the truth). Therefore, B must be the knight because if B were the knight, they would be telling the truth about A being a knight.

15) How many different integers can be expressed as the sum of three distinct numbers from the set {3, 8, 13, 18, 23, 28, 33, 38, 43, 48}?

8

56

120

22

The set {3, 8, 13, 18, 23, 28, 33, 38, 43, 48} contains 10 numbers. To find the number of different integers that can be expressed as the sum of three distinct numbers from this set, we can use the combination formula. The number of combinations of 3 distinct numbers that can be selected from a set of 10 is given by 10C3 = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120. Therefore, there are 120 different integers that can be expressed as the sum of three distinct numbers from the given set.

Explanation

The set {3, 8, 13, 18, 23, 28, 33, 38, 43, 48} contains 10 numbers. To find the number of different integers that can be expressed as the sum of three distinct numbers from this set, we can use the combination formula. The number of combinations of 3 distinct numbers that can be selected from a set of 10 is given by 10C3 = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120. Therefore, there are 120 different integers that can be expressed as the sum of three distinct numbers from the given set.