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NMTC 2018 Question Papers With Solutions Junior Level For Class 9 and 10

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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. \sqrt{2}

B. \sqrt{3}

C. \sqrt{6}

D. \sqrt{18}

Solution:

nmtc 2018 question papers with solutions junior

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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 3:

One of the factors of 9x2 – 4z2 – 24xy + 16y2 + 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: 9x2 – 4z2 – 24xy + 16y2 + 20y – 15x + 10

The correct question is 9x2 – 4z2 – 24xy + 16y2 + 20y – 15x + 10z.

Thus,

9x2 – 4z2 – 24xy + 16y2 + 20y – 15x + 10z

(9x2 – 24xy + 16y2) – 4z+ 20y – 15x + 10z

= (3x – 4y) – (2z)+ 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 9x2 – 4z2 – 24xy + 16y2 + 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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 5:

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

Options:

A. 2

B. 4

C. 5

D. No Solution

Solution: BONUS

nmtc 2018 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 6:

If (x + 1)2 = x, the value of 11x3 + 8x2 + 8x – 2 is

Options:

A. 1

B. 3

C. 3

D. 4

Solution:

nmtc 2018 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 7:

There are two values of m for which the equation 4x2 + 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 question papers with solutions junior

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 question papers with solutions junior

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 question papers with solutions juniornmtc 2018 question papers with solutions junior

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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 11:

The four digit number 8ab9 is a perfect square. The value of a2 + b2 is

Options:

A. 52

B. 62

C. 54

D. 68

Solution:

nmtc 2018 question papers with solutions junior

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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 13:

a, b are real numbers. The least value of a2 + ab + b2 – a – 2b is

Options:

A. 1

B. 0

C. -1

D. 2

Solution:

Let y = f(a, b) = a2 + ab + b2 – 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 question papers with solutions junior

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 question papers with solutions junior

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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 18:

The number of natural numbers n for which n2 + 96 is perfect square is______

Solution:

nmtc 2018 question papers with solutions junior

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 question papers with solutions junior

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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 21:

If then the value of 2x3 – 6x is______

Solution:

nmtc 2018 question papers with solutions junior

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 question papers with solutions junior

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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 24:

A sphere is inscribed in a cube that has a surface area of 24 cm2. A second cube is then inscribed within the sphere. The surface area of the inner cube (in cm2) is_____

Solution:

nmtc 2018 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 25:

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

Solution:

nmtc 2018 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 26:

The value of x which satisfies the equation  is

Solution:

nmtc 2018 question papers with solutions junior

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 question papers with solutions junior

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 question papers with solutions junior

NMTC 2018 Paper For Junior Level Ques No 29:

When expanded the units place of (3127)173 is_____

Solution:

nmtc 2018 question papers with solutions junior

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:

nmtc 2018 question papers with solutions junior

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