﻿ TET Class 12 Mathematics Solved Question Paper 2014 | Previous Year Papers | Zigya

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# TET Class 12 Mathematics Solved Question Paper 2014

#### Multiple Choice Questions

1.

If x is an integer, then (x+1)4-(x-1)4 is always divisible by

• 6

• 8

• 9

• 12

B.

8

Consider (x+1)4-(x-1)4

$⇒$[(x+1)2]2 - [(x-1)2]2

$⇒$[x2+1+2x]2- [x2+1-2x]2

$⇒$(x2+1+2x+x2+1-2x) (x2+1+2x-x2-1+2x)

$⇒$(2x2+2)(4x)

$⇒$2(x2+1)(4x)

$⇒$8x(x2+1)

Hence if x is an integer , then (x+1)4-(x-1)will be always divisible by 8 .

2.

What should be subtracted from $\frac{-5}{7}$to get -1?

• $\frac{-2}{7}$

• $\frac{4}{7}$

• $\frac{2}{7}$

• $\frac{-4}{7}$

C.

$\frac{2}{7}$

Let x should be subtracted from $\frac{-5}{7}$ to get -1

Then ,

$⇒$-x = -1+$\frac{5}{7}$

$⇒$-x = $\frac{-7+5}{7}$

$⇒$-x = $\frac{-2}{7}$

$\therefore$ x = $\frac{2}{7}$

3.

The hundreds digit of a three-digit number is 7 more than the units digit. The digits of the number are reversed and the resulting number is subtracted from the original three-digit number. The units digit of the final number so obtained is

• 0

• 1

• 2

• 3

D.

3

Let the three-digit number be 100x +10y+ z

As per question , x = 7 + z .........(i)

On Reversing the number we get : 100z + 10 y + x....(ii)

Now subtracting  (ii) From (i)

100x+10y+z-100z-10y-x

=99x-99z

=99(x-z)

=99 x 7 [ from eqn (i) ]

=693

Hence the unit digit will be 3

4.

What is the probability that a randomly selected factor from positive factors of 72 is less than 11?

• $\frac{5}{12}$

• $\frac{7}{11}$

• $\frac{7}{12}$

• $\frac{7}{10}$

C.

$\frac{7}{12}$

Factors of 72 are

1 x72

2 x36

3 x 24

4 x18

6 x12

8 x9

$\therefore$ Total number of the possible outcomes =12

Number of favorable outcomes less than 11=7

$\therefore$ Probability = $\frac{7}{12}$

5.

The value of $\sqrt{500}×\sqrt{16}$ is

• 16

• 20

• 25

• 18

B.

20

Given $\sqrt{500}×\sqrt{16}$

$\sqrt{8000}$

$\sqrt{{\left(2×10\right)}^{3}}$

= 2 x 10

=  20

6.

Numbers $\frac{-11}{20}$,$\frac{7}{-15}$,$\frac{17}{30}$and$\frac{-3}{10}$are written in descending order as

• $\frac{17}{-30}$>$\frac{-11}{20}$>$\frac{-3}{10}$>$\frac{7}{-15}$

• $\frac{-3}{10}$>$\frac{7}{-15}$>$\frac{-11}{20}$>$\frac{17}{-30}$

• $\frac{-3}{10}$>$\frac{-11}{20}$>$\frac{7}{-15}$>$\frac{17}{-30}$

• $\frac{-11}{20}$>$\frac{17}{-30}$>$\frac{-3}{10}$>$\frac{7}{-15}$

B.

$\frac{-3}{10}$>$\frac{7}{-15}$>$\frac{-11}{20}$>$\frac{17}{-30}$

We have, $\frac{-11}{20}$,$\frac{7}{-15}$,$\frac{17}{-30}$,$\frac{-3}{10}$

LCM of the denominators , 20, 15, 30 and 10 = 60

$⇒$$\frac{-11}{20}=\frac{-33}{60}$$\frac{7}{-15}=\frac{-28}{60}$,$\frac{17}{-30}=\frac{-34}{60}$ and$\frac{-3}{10}=\frac{-18}{60}$

Therefore the correct descending order on comparing the numerator will be:

$\frac{-18}{60}>\frac{-28}{60}>\frac{-34}{60}>\frac{-18}{60}$

i.e. $\frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}$

7.

LCM of two prime numbers x and y, ( x > y), is 161. The value of 3y-x is

• 2

• -2

• -5

• 62

B.

-2

Let the two prime numbers be x and y

LCM of the two numbers: 161

$⇒$xy = 161 [ The LCM of the two prime number will be their product since they will not have any common factor ]

Now 161 = 23 x 7

$\therefore$ 3y - x = 3x7 - 23

= 21-23

= -2

8.

If a= $\sqrt{{\left(2013\right)}^{2}+2013+2014}$, then the value of a is

• 1002

• 1007

• 2013

• 2014

D.

2014

Given : a = $\sqrt{{\left(2013\right)}^{2}+2013+2014}$

$\sqrt{{\left(2013\right)}^{2}+2013+2013+1}$

$\sqrt{{\left(2013\right)}^{2}+2×2013×1+{\left(1\right)}^{2}}$

=$\sqrt{{\left(2013+1\right)}^{2}}$

=2013+1

=2014

9.

If a, b and c are different integers such that a < b < c < 0, then which of the following statements is true?

• a + c < b

• ab < c

• a + b > c

• ac > ab

A.

a + c < b

Given that, a < b < c < 0

Let a=-3, b=-2, C = -1

then -3 < -2 < -1 < 0

From option (1), a + c < b

-3 + (-1) < -2

-4  < -2

which is True

10.

In standard form, the number 829030000 is expressed as k x 10n. The value of k + n is

• 90.903

• 16.2903

• 15.2903

• 91.903

B.

16.2903

Given  829030000

In standard form k x 10n,  829030000 = 8.2903 x108

On comparing we get k=8.2903, and n=8

$\therefore$ k+n = 8.2903+8 = 16.2903