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**Part A: Instruction:**

**Only One of the choices A, B, C, D is correct. For each correct response, you get 1 mark. For each incorrect response, you lose 1/2 mark.**

**NMTC 2018 Paper For ****Junior**** Level Ques No 1:**

The value of

is

**Options:**

A.

B.

C.

D.

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 2:**

A train moving with a constant speed crosses a stationary pole in 4 seconds and a platform 75 m long in 9 seconds. The length of the train is (in meter)

**Options:**

A. 56

B. 58

C. 60

D. 62

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 3:**

One of the factors of 9x^{2} – 4z^{2} – 24xy + 16y^{2} + 20y – 15x + 10 is

**Options:**

A. 3x – 4y – 2z

B. 3x + 4y – 2z

C. 3x + 4y +2z

D. 3x – 4y + 2z

**Solution: BONUS**

The given question is wrong: 9x^{2} – 4z^{2} – 24xy + 16y^{2} + 20y – 15x + 10

The correct question is 9x^{2} – 4z^{2} – 24xy + 16y^{2} + 20y – 15x + 10z.

Thus,

9x^{2} – 4z^{2} – 24xy + 16y^{2} + 20y – 15x + 10z

= (9x^{2} – 24xy + 16y^{2}) – 4z^{2 }+ 20y – 15x + 10z

= (3x – 4y)^{2 } – (2z)^{2 }+ 5(4y – 3x + 2z)

= (3x – 4y + 2z)(3x – 4y – 2z)^{ }– 5(3x – 4y – 2z)

= (3x – 4y – 2z)(3x – 4y + 2z – 5)

Therefore, one of the factors of 9x^{2} – 4z^{2} – 24xy + 16y^{2} + 20y – 15x + 10 is (3x – 4y – 2z).

**NMTC 2018 Paper For ****Junior**** Level Ques No 4:**

The natural number which is subtracted from each of the four numbers 17, 31, 25, 47 to give four numbers in proportion is

**Options:**

A. 1

B. 2

C. 3

D. 4

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 5:**

The solution to the equation 5(3^{x}) + 3(5^{x}) = 510 is

**Options:**

A. 2

B. 4

C. 5

D. No Solution

**Solution: BONUS**

**NMTC 2018 Paper For ****Junior**** Level Ques No 6:**

If (x + 1)^{2} = x, the value of 11x^{3} + 8x^{2} + 8x – 2 is

**Options:**

A. 1

B. 3

C. 3

D. 4

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 7:**

There are two values of m for which the equation 4x^{2} + mx + 8x + 9 = 0 has only one solution for x. The sum of these two values of m is

**Options:**

A. 1

B. 3

C. 3

D. 4

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 8:**

The number of zeros in the product of the first 100 natural numbers is

**Options:**

A. 12

B. 15

C. 18

D. 24

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 9:**

The length of each side of a triangle is increased by 20% then the percentage increase of area is

**Options:**

A. 60%

B. 120%

C. 80%

D. 44%

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 10:**

The number of pairs of relatively prime positive integers (a, b) such that a/b + 15b/4a is an integer is

**Options:**

A. 1

B. 3

C. 3

D. 4

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 11:**

The four digit number 8ab9 is a perfect square. The value of a^{2} + b^{2} is

**Options:**

A. 52

B. 62

C. 54

D. 68

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 12:**

a, b are positive real numbers such that 1/a + 9/b = 1. The smallest value of a + b is

**Options:**

A. 15

B. 16

C. 17

D. 18

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 13:**

a, b are real numbers. The least value of a^{2} + ab + b^{2} – a – 2b is

**Options:**

A. 1

B. 0

C. -1

D. 2

**Solution:**

Let y = f(a, b) = a^{2} + ab + b^{2} – a – 2b … (1)

Differentiate y with respect to a and b separately.

On differentiating (1) w.r.t. a, we get dy/da = 2a + b – 1. For least, dy/da = 0.

Thus, 2a + b – 1 = 0 … (2)

On differentiating (1) w.r.t. b, we get dy/db = a + 2b – 2. For least, dy/db = 0.

Thus, a + 2b – 2 = 0 … (3)

On solving equations (2) and (3), we get a = 0 and b = 1.

Put a = 0 and b = 1 in equation (1). we get least value of given expression as

Minimum value of y = (0)^{2} + (0)(1) + (1)^{2} – (0) – 2(1) = 1 – 2 = -1

**NMTC 2018 Paper For ****Junior**** Level Ques No 14:**

I is the incenter of a triangle ABC is which A = 80 Degree. BIC =

**Options:**

A. 120 Degree

B. 110 Degree

C. 125 Degree

D. 130 Degree

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 15:**

In the adjoining figure, ABCD is a square and DFEB is a rhombus. The angle CDF = ?

**Options:**

A. 15 Degree

B. 18 Degree

C. 20 Degree

D. 30 Degree

**Solution:**

**Part B: Instruction:**

**Write the correct answer in the space provided in the responsive sheet. For each correct response, you get 1 mark. For each incorrect response, you lose 1/4 mark.**

**NMTC 2018 Paper For ****Junior**** Level Ques No 16:**

ABCD is a square. E, F are points on BC, CD respectively and EAF = 45 Degree. The value of BF/(BE + DF) is______

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 17:**

The average of 5 consecutive natural numbers is 10. The sum of the second and fourth of these numbers is_____

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 18:**

The number of natural numbers n for which n^{2} + 96 is perfect square is______

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 19:**

n is an integer and

is also an integer. The sum of all such n is__

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 20:**

a/b is a fraction where a, b have no common factors other than 1. b exceeds a by 3. If the numerator is increased by 7, the fraction is increased by unity. The value of a + b_____

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 21:**

If then the value of 2x^{3} – 6x is______

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 22:**

The angles of a heptagon are 160 Degree, 135 Degree, 185 Degree, 140 Degree, 125 Degree, x Degree, x Degree. The value of x is _____

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 23:**

ABC is a triangle and AD is its altitude. If BD = 5DC, then the value of

is

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 24:**

A sphere is inscribed in a cube that has a surface area of 24 cm^{2}. A second cube is then inscribed within the sphere. The surface area of the inner cube (in cm^{2}) is_____

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 25:**

A positive integer n is a multiple of 7. If lies between 15 and 16, the number of possible values(s) of n is_____

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 26:**

The value of x which satisfies the equation is

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 27:**

M man do a work in m days. If there had been N men more, the work would have been finished n days earlier, then the value of m/n – M/N is_____

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 28:**

The sum of the digits of a two number is 15. If the digit of the given number are reserved, the number is increased by the square of 3. The original number is______

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 29:**

When expanded the units place of (3127)^{173} is_____

**Solution:**

**NMTC 2018 Paper For ****Junior**** Level Ques No 30:**

If a : (b + c) = 1 : 3 and c : (a + b) = 5 : 7, then b : (c + a) is_______

**Solution:**

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