18 + Interview Questions in GENERAL QA IN NMTC Page 1 InterviewSolution

1.	The number of ways in which 26 identical chocolates be distributed between Amy, Bob, Cathy and Daniel so that each receives at least one chocolate and Amy receives more chocolates than Bob is _______.
Answer» Total number of natural solution of equation x₁ + x₂ + x₃ + x₄ = 26 is ²⁵C₃ = 2300. total number of natural solution of equation 2x₁ + x₃ + x₄ = 26 = 1 + 3 + + 23 = 144 Total number of natural solution of equation x₁+ x₂ + x₃ + x₄ = 26 when x₁ > x₂ is 2300 - 144/2 = 1078

1.

The number of ways in which 26 identical chocolates be distributed between Amy, Bob, Cathy and Daniel so that each receives at least one chocolate and Amy receives more chocolates than Bob is _______.

Answer»

Total number of natural solution of equation x₁ + x₂ + x₃ + x₄ = 26 is ²⁵C₃ = 2300. total number of natural solution of equation 2x₁ + x₃ + x₄ = 26 = 1 + 3 + + 23 = 144

Total number of natural solution of equation x₁+ x₂ + x₃ + x₄ = 26 when x₁ > x₂ is 2300 - 144/2 = 1078

Discussion

2.	The numbers 2, 3, 12, 14, 15, 20, 21 may be divided into two sets so that the product of the numbers in each set is the same. What is this product?(A) 420 (B) 1260 (C) 2520 (D) 6720
Answer» Correct option: (C) 2520 Explanation: P² = 2 × 3 × 12 × 14 × 15 × 20 × 21 => P = 2520

2.

The numbers 2, 3, 12, 14, 15, 20, 21 may be divided into two sets so that the product of the numbers in each set is the same. What is this product?(A) 420 (B) 1260 (C) 2520 (D) 6720

Answer»

Correct option: (C) 2520

Explanation:

P² = 2 × 3 × 12 × 14 × 15 × 20 × 21

=> P = 2520

Discussion

3.	We have four sets S1, S2, S3, S4 each containing a number of parallel lines. The set Si contains i + 1 parallel lines i = 1,2,3,4. A line in Si is not parallel to lines in Sj when i ≠ j. In how many points do these lines intersect ?(A) 54 (B) 63 (C) 71(D) 95
Answer» Correct option: (C) 71 Explanation: Number of point of intersection = ²C₁ (³C₁ + ⁴C₁ + ⁵C₁) + ³C₁ (⁴C₁ + ⁵C₁) + ⁴C₁ × ⁵C₁ = 2(12) + 3(9) + 20 = 24 + 27 + 20 = 71

3.

We have four sets S1, S2, S3, S4 each containing a number of parallel lines. The set Si contains i + 1 parallel lines i = 1,2,3,4. A line in Si is not parallel to lines in Sj when i ≠ j. In how many points do these lines intersect ?(A) 54 (B) 63 (C) 71(D) 95

Answer»

Correct option: (C) 71

Explanation:

Number of point of intersection = ²C₁ (³C₁ + ⁴C₁ + ⁵C₁) + ³C₁ (⁴C₁ + ⁵C₁) + ⁴C₁ × ⁵C₁

= 2(12) + 3(9) + 20

= 24 + 27 + 20

= 71

Discussion

4.	The four digit number 8ab9 is a perfect square. The value of a2 + b2 is(A) 52 (B) 62 (C) 54 (D) 68
Answer» Correct option: (A) 52 Explanation: 93² = 8649 ∴ a = 6, b = 4 ∴ a² + b² = 6² + 4² = 52.

4.

The four digit number 8ab9 is a perfect square. The value of a2 + b2 is(A) 52 (B) 62 (C) 54 (D) 68

Answer»

Correct option: (A) 52

Explanation:

93² = 8649

∴ a = 6, b = 4

∴ a² + b² = 6² + 4² = 52.

Discussion

5.	A cube has edge length x (an integer). three faces meeting at a corner are painted blue. The cube is then cut into smaller cubes of unit length. If exactly 343 of these cubes have no faces painted blue, then the value of x is ________ .
Answer» (x 1)³ = 343 ⇒ x 1 = 7 ⇒ x = 8

5.

A cube has edge length x (an integer). three faces meeting at a corner are painted blue. The cube is then cut into smaller cubes of unit length. If exactly 343 of these cubes have no faces painted blue, then the value of x is ________ .

Answer»

(x 1)³ = 343

⇒ x 1 = 7

⇒ x = 8

Discussion

6.	There are 4 coins in a row and all are showing heads to start with. The coins can be flipped with the following rules :(a) The fourth coin (from the left) can be flipped any time(b) An intermediate coin can be changed to tail only if its immediate neighbor on the right is heads and all other coins (if any) to its right are tails.(c) Only one coin can be flipped in one step.The minimum number of steps required to bring all coins to show tails is _______.
Answer» (8) HHHH, HHTH, HHTT, THTT, THHT, TTHT, TTHH, TTTH, TTTT

6.

There are 4 coins in a row and all are showing heads to start with. The coins can be flipped with the following rules :(a) The fourth coin (from the left) can be flipped any time(b) An intermediate coin can be changed to tail only if its immediate neighbor on the right is heads and all other coins (if any) to its right are tails.(c) Only one coin can be flipped in one step.The minimum number of steps required to bring all coins to show tails is _______.

Answer»

(8)

HHHH, HHTH, HHTT, THTT, THHT,

TTHT, TTHH, TTTH, TTTT

Discussion

7.	In an elementary school 26% of the students are girls. If there are 240 less girls than boys, then the strength of the school is _________
Answer» Girls = 26% Boys = 100 – 26 = 74% Given 74% – 26% = 240 48% = 240 1% = 240/48 students 100% = 240/48 x 100 = 500 students

7.

In an elementary school 26% of the students are girls. If there are 240 less girls than boys, then the strength of the school is _________

Answer»

Girls = 26%

Boys = 100 – 26 = 74%

Given 74% – 26% = 240

48% = 240

1% = 240/48 students

100% = 240/48 x 100

= 500 students

Discussion

8.	Vishva plays football every 4th day. He played on a Tuesday . He plays football on a Tuesday again in _________ days.
Answer» Number of days in week = 7 days Vishva plays football in = 4 days He will play football on a Tuesday again in 7 × 4 = 28 days

8.

Vishva plays football every 4th day. He played on a Tuesday . He plays football on a Tuesday again in _________ days.

Answer»

Number of days in week = 7 days

Vishva plays football in = 4 days

He will play football on a Tuesday again in 7 × 4 = 28 days

Discussion

9.	Samrud had to multiply a number by 35. By mistake he multiplied by 53 and got a result 720 more. The new product is _________.
Answer» Let the number be x incorrect product = x × 53 correct product = x × 35 x × 53 – x × 35 = 720 18x = 720 x = 40 New product x × 53 => 40 × 53 = 2120

9.

Samrud had to multiply a number by 35. By mistake he multiplied by 53 and got a result 720 more. The new product is _________.

Answer»

Let the number be x

incorrect product = x × 53

correct product = x × 35

x × 53 – x × 35 = 720

18x = 720

x = 40

New product x × 53

=> 40 × 53 = 2120

Discussion

10.	ABC is a triangle in which the angles are in the ratio 3 : 4 : 5 . PQR is a triangle in which the angles are in the ratio 5 : 6 : 7. The difference between the least angle of ABC and the least angle of PQR is aº. Then a = _________
Answer» 3x + 4x + 5x = 180° 12x = 180 x = 15 Least angle of triangle ABC is = 3 × 15 = 45° 5x + 6x + 7x = 180° 18x = 180° x = 10 Least angle of triangle PQR = 5 × 10 = 50 Difference a° = 50 – 45 a° = 5

10.

ABC is a triangle in which the angles are in the ratio 3 : 4 : 5 . PQR is a triangle in which the angles are in the ratio 5 : 6 : 7. The difference between the least angle of ABC and the least angle of PQR is aº. Then a = _________

Answer»

3x + 4x + 5x = 180°

12x = 180

x = 15

Least angle of triangle ABC is = 3 × 15 = 45°

5x + 6x + 7x = 180°

18x = 180°

x = 10

Least angle of triangle PQR = 5 × 10 = 50

Difference a° = 50 – 45

a° = 5

Discussion

11.	When a two digit number divides 265, the remainder is 5. The number of such two digit numbers is ________
Answer» 265 – 5 = 260 260 = 2 × 2 × 5 × 13 two digits such numbers will be => 1 × 13 = 13 => 2 × 13 = 26 => 2 × 5 = 10 => 4 × 5 = 20 => 4 × 13 = 52 => 5 × 13 = 65 6

11.

When a two digit number divides 265, the remainder is 5. The number of such two digit numbers is ________

Answer»

265 – 5 = 260

260 = 2 × 2 × 5 × 13

two digits such numbers will be

=> 1 × 13 = 13

=> 2 × 13 = 26

=> 2 × 5 = 10

=> 4 × 5 = 20

=> 4 × 13 = 52

=> 5 × 13 = 65

6

Discussion

12.	There are 5 cards. Five positive integers (may be different or equal) are written on these cards, one on each card. Abhiram finds the sum of the numbers on every pair of cards. He obtains only three different totals 57, 70, 83. Find the largest integer written on a card.
Answer» As the three sum are obtained 57,70,83 We have combination of three numbers same and two numbers different a, a, a, b, c a + b = 57 b + c = 70 c + a = 83 2a = 70 a = 35 b = 22 c = 48 Largest integer is 48

12.

There are 5 cards. Five positive integers (may be different or equal) are written on these cards, one on each card. Abhiram finds the sum of the numbers on every pair of cards. He obtains only three different totals 57, 70, 83. Find the largest integer written on a card.

Answer»

As the three sum are obtained 57,70,83

We have combination of three numbers same and two numbers different

a, a, a, b, c

a + b = 57

b + c = 70

c + a = 83

2a = 70

a = 35

b = 22

c = 48

Largest integer is 48

Discussion

13.	a, b, c are three natural numbers such that a x b x c = 27846. If a/6 = b + 4 = c - 4, find a + b + c.
Answer» a ×b × c = 27846 ...... (1) a/b = b + 4 = c - 4 a = 6b + 24 b = b c = b + 8 Put value of a, b, c in (1) (6b + 24) (b) (b + 8) = 27846 6(b + 4) (b) (b + 8) = 27846 (b + 4) (b) (b + 8) = 4641 = 13 × 17 × 3 × 7 = 13 × 17 × 21 ∴ b = 13 a = 6 × 13 + 24 = 78 + 24 = 102 c = b + 8 = 13 + 8 = 21 ∴ a + b + c = 102 + 13 + 21 = 136

13.

a, b, c are three natural numbers such that a x b x c = 27846. If a/6 = b + 4 = c - 4, find a + b + c.

Answer»

a ×b × c = 27846 ...... (1)

a/b = b + 4 = c - 4

a = 6b + 24

b = b

c = b + 8

Put value of a, b, c in (1)

(6b + 24) (b) (b + 8) = 27846

6(b + 4) (b) (b + 8) = 27846

(b + 4) (b) (b + 8) = 4641 = 13 × 17 × 3 × 7 = 13 × 17 × 21

∴ b = 13

a = 6 × 13 + 24

= 78 + 24

= 102

c = b + 8 = 13 + 8 = 21

∴ a + b + c = 102 + 13 + 21 = 136

Discussion

14.	The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the sum of the ages of their children. Six years hence the sum of their ages will be three times the sum of the ages of their children. How many children do they have?
Answer» Let present age of man = M Let present age of wife = W Let the no. of children = x Let the sum of the ages of children = C ATQ M + W = 6 C ... (1) M - 2 + W - 2 = 10 (C - 2x ) ... (2) M + 6 + W + 6 = 3 (C + 6x) ... (3) From (2) M + W - 4 = 10 C - 20 x By using (1) 6 C - 4 = 10 C - 20 x - 4C + 20x = 4 -C + 5x = 1 .... (5) From (3) M + W + 12 = 3C + 18x By using (1) 6C + 12 = 3C + 18x 3C - 18x = - 12 C - 6x = - 4 .... (6) From (5) & (6) - C + 5x = 1 C - 6x = - 4 ∴ - x = - 3 x = 3

14.

The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the sum of the ages of their children. Six years hence the sum of their ages will be three times the sum of the ages of their children. How many children do they have?

Answer»

Let present age of man = M

Let present age of wife = W

Let the no. of children = x

Let the sum of the ages of children = C

ATQ

M + W = 6 C ... (1)

M - 2 + W - 2 = 10 (C - 2x ) ... (2)

M + 6 + W + 6 = 3 (C + 6x) ... (3)

From (2) M + W - 4 = 10 C - 20 x

By using (1) 6 C - 4 = 10 C - 20 x

- 4C + 20x = 4

-C + 5x = 1 .... (5)

From (3) M + W + 12 = 3C + 18x

By using (1) 6C + 12 = 3C + 18x

3C - 18x = - 12

C - 6x = - 4 .... (6)

From (5) & (6)

- C + 5x = 1

C - 6x = - 4

∴ - x = - 3

x = 3

Discussion

15.	If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b c prove that a2 + b2 + c2 + 2abc = 1
Answer» If x²(y + z) = a², y²(z + x) = b², z²(x + y) = c², xyz = a b c x² (y + z) y² (z + x) z² (x + y) = a²b²c² x²y²z² (x + y) (y + z) (z + x) = a²b²c² xyz = abc (x + y) (y + z) (z + x) = 1 a²+ b²+ c²+ 2abc => x²(y + z) + y² (z + x) + z²(x + y) + 2xyz Put y = - z 0 + z²(z + x) + z²(x - z) + 2x (-z)z z³ + z²x + z²x - z³ - 2z²x = 0 ∴ (y + z) is factors Similarly (x + y) & (x + z) are factors x²(y + z) + y²(z + x) + z²(x + y) + 2xyz = k (x + y) (y + z) (x + z) For k put x = 0 y = 1 z = 1 0 + 1 + 1 + 0 = k (1) (2) (1) 2 = 2 k k = 1 x²(y + z) + y²(z + x) + z²(x + y) + 2xyz = (x + y) (y + z) (z + x) and (x + y) (y + z) (z + x) = 1 ∴ x²(y + z) + y²(z + x) + z² (x + y) + 2xyz = 1

15.

If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b c prove that a2 + b2 + c2 + 2abc = 1

Answer»

If x²(y + z) = a², y²(z + x) = b², z²(x + y) = c², xyz = a b c

x² (y + z) y² (z + x) z² (x + y) = a²b²c²

x²y²z² (x + y) (y + z) (z + x) = a²b²c²

xyz = abc

(x + y) (y + z) (z + x) = 1

a²+ b²+ c²+ 2abc

=> x²(y + z) + y² (z + x) + z²(x + y) + 2xyz

Put y = - z

0 + z²(z + x) + z²(x - z) + 2x (-z)z

z³ + z²x + z²x - z³ - 2z²x = 0

∴ (y + z) is factors

Similarly (x + y) & (x + z) are factors

x²(y + z) + y²(z + x) + z²(x + y) + 2xyz

= k (x + y) (y + z) (x + z)

For k put x = 0 y = 1 z = 1

0 + 1 + 1 + 0 = k (1) (2) (1)

2 = 2 k

k = 1

x²(y + z) + y²(z + x) + z²(x + y) + 2xyz

= (x + y) (y + z) (z + x)

and (x + y) (y + z) (z + x) = 1

∴ x²(y + z) + y²(z + x) + z² (x + y) + 2xyz = 1

Discussion

16.	Prove that x4 + 3x3 + 6x2 + 9x + 12 cannot be expressed as a product of two polynomials of degree 2 with integer coefficients.
Answer» Let x⁴+ 3x³ + 6x² + 9x + 12 = (x² + Ax + B) (x² + Cx + D) = x⁴ + Cx³ + Dx² + Ax³ + ACx² + ADx + Bx² + BCx + BD = x⁴ + (A + C)x³ + (D + AC + B) x² + (AD + BC)x + BD Now by comparing coefficient A + C = 3 B + D + AC = 6 AD + BC = 9 BD = 12 Case - I : B = 1, D = 12 ∴ A + C = 3 12A + C = 9 have no integer solution. Case - II : B = - 1, D = - 12 C + 12 A = - 9 C + A = 3 have no integer solution. Case - III : B = 2, D = 6 2C + 6A = 9 C + A = 3 have no integer solution. Case - IV : B = - 2, D = - 6 2C + 6A = - 9 A + C = 3 have no integer solution. So, x⁴ + 3x³ + 6x²+ 9x + 12 cannot be expressed as a product of two polynomial of degree 2 with integer coefficient.

16.

Prove that x4 + 3x3 + 6x2 + 9x + 12 cannot be expressed as a product of two polynomials of degree 2 with integer coefficients.

Answer»

Let x⁴+ 3x³ + 6x² + 9x + 12

= (x² + Ax + B) (x² + Cx + D)

= x⁴ + Cx³ + Dx² + Ax³ + ACx² + ADx + Bx² + BCx + BD

= x⁴ + (A + C)x³ + (D + AC + B) x² + (AD + BC)x + BD

Now by comparing coefficient

A + C = 3

B + D + AC = 6

AD + BC = 9

BD = 12

Case - I : B = 1, D = 12

∴ A + C = 3

12A + C = 9 have no integer solution.

Case - II : B = - 1, D = - 12

C + 12 A = - 9

C + A = 3 have no integer solution.

Case - III : B = 2, D = 6

2C + 6A = 9

C + A = 3 have no integer solution.

Case - IV : B = - 2, D = - 6

2C + 6A = - 9

A + C = 3 have no integer solution.

So, x⁴ + 3x³ + 6x²+ 9x + 12 cannot be expressed as a product of two polynomial of degree 2 with integer coefficient.

Discussion

17.	Let a, b, c be real numbers, not all of them are equal. Prove that if a + b + c = 0, then a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.Prove the converse, if a2 + ab + b2 = b2 + bc + c2 = c2 = ca + a2, then a + b + c = 0.
Answer» I. a² + ab + b² = (– b – c)² + (– b – c) b + b² {as a + b + c = 0, a = – b – c} = (b² + c² + 2bc) – b² – bc + b² = b² + c² + bc Similarly b² + c² + bc = c² + ca + a² ∴ a² + ab + b² = b² + c² + bc = c² + a² + ac. II. a² + ab + b² = b² + bc + c² a²+ ab – bc – c² = 0 a² – c² + b (a – c) = 0 (a – c) (a + c + b) = 0 a – c = 0 or a + b + c = 0 as a – c = 0 is not possible as a, b, c are not equal. ∴ a + b + c = 0.

17.

Let a, b, c be real numbers, not all of them are equal. Prove that if a + b + c = 0, then a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.Prove the converse, if a2 + ab + b2 = b2 + bc + c2 = c2 = ca + a2, then a + b + c = 0.

Answer»

I. a² + ab + b² = (– b – c)² + (– b – c) b + b² {as a + b + c = 0, a = – b – c}

= (b² + c² + 2bc) – b² – bc + b²

= b² + c² + bc

Similarly b² + c² + bc = c² + ca + a²

∴ a² + ab + b² = b² + c² + bc = c² + a² + ac.

II. a² + ab + b² = b² + bc + c²

a²+ ab – bc – c² = 0

a² – c² + b (a – c) = 0

(a – c) (a + c + b) = 0

a – c = 0 or a + b + c = 0

as a – c = 0 is not possible as a, b, c are not equal.

∴ a + b + c = 0.

Discussion

18.	The difference between the biggest and the smallest three digit number each of which has different digits is(A) 864 (B) 875 (C) 885 (D) 895
Answer» Correct option: (C) 885 Explanation: Biggest three digit number with distinct digit = 987 Smallest three digit number with distinct digit = 102 Difference = 987 – 102 = 885. Option C is correct The biggest three digit number having three different digit is 987 The smallest three digit number having three different digit is 102 So the difference = 987-102=885

18.

The difference between the biggest and the smallest three digit number each of which has different digits is(A) 864 (B) 875 (C) 885 (D) 895

Answer»

Correct option: (C) 885

Explanation:

Biggest three digit number with distinct digit = 987

Smallest three digit number with distinct digit = 102

Difference = 987 – 102 = 885.

Option C is correct

The biggest three digit number having three different digit is 987

The smallest three digit number having three different digit is 102

So the difference = 987-102=885

Discussion

Explore topic-wise InterviewSolutions in .

The number of ways in which 26 identical chocolates be distributed between Amy, Bob, Cathy and Daniel so that each receives at least one chocolate and Amy receives more chocolates than Bob is _______.

The numbers 2, 3, 12, 14, 15, 20, 21 may be divided into two sets so that the product of the numbers in each set is the same. What is this product?(A) 420 (B) 1260 (C) 2520 (D) 6720

We have four sets S1, S2, S3, S4 each containing a number of parallel lines. The set Si contains i + 1 parallel lines i = 1,2,3,4. A line in Si is not parallel to lines in Sj when i ≠ j. In how many points do these lines intersect ?(A) 54 (B) 63 (C) 71(D) 95

The four digit number 8ab9 is a perfect square. The value of a2 + b2 is(A) 52 (B) 62 (C) 54 (D) 68

A cube has edge length x (an integer). three faces meeting at a corner are painted blue. The cube is then cut into smaller cubes of unit length. If exactly 343 of these cubes have no faces painted blue, then the value of x is ________ .

In an elementary school 26% of the students are girls. If there are 240 less girls than boys, then the strength of the school is _________

Vishva plays football every 4th day. He played on a Tuesday . He plays football on a Tuesday again in _________ days.

Samrud had to multiply a number by 35. By mistake he multiplied by 53 and got a result 720 more. The new product is _________.

ABC is a triangle in which the angles are in the ratio 3 : 4 : 5 . PQR is a triangle in which the angles are in the ratio 5 : 6 : 7. The difference between the least angle of ABC and the least angle of PQR is aº. Then a = _________

When a two digit number divides 265, the remainder is 5. The number of such two digit numbers is ________

There are 5 cards. Five positive integers (may be different or equal) are written on these cards, one on each card. Abhiram finds the sum of the numbers on every pair of cards. He obtains only three different totals 57, 70, 83. Find the largest integer written on a card.

a, b, c are three natural numbers such that a x b x c = 27846. If a/6 = b + 4 = c - 4, find a + b + c.

If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b c prove that a2 + b2 + c2 + 2abc = 1

Prove that x4 + 3x3 + 6x2 + 9x + 12 cannot be expressed as a product of two polynomials of degree 2 with integer coefficients.

Let a, b, c be real numbers, not all of them are equal. Prove that if a + b + c = 0, then a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.Prove the converse, if a2 + ab + b2 = b2 + bc + c2 = c2 = ca + a2, then a + b + c = 0.

The difference between the biggest and the smallest three digit number each of which has different digits is(A) 864 (B) 875 (C) 885 (D) 895