Posted by Anjali Kaur on Oct 04, 2021

# Sample Paper Applied Maths Class XII

In this post, you will find 15 questions based on the class XII Applied Maths curriculum. It includes topics such as number and quantification, algebra, probability distribution, index numbers.

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## (a) -1 (b) 0 (c) 7 (d) 9

Solution. For such questions, use the following steps and find the remainder.

(5 x 11) mod 8 = 55 mod 8

Now divide 55 by 8

The remainder is 7. So, option c.

## (a) 7/81 (b) 5/81 (c) 2/81 (d) 1/81

Solution.

P(X) = Summation of all probabilities = 1

a+3a+5a+7a+9a+11a+13a+15a+17a = 1

81a=1

a= 1/81

So, option is (d)

## (a) 2 km/h (b) 3 km/h (c) 4 km/h (d) 6 km/h

Solution. Speed of boat in still water = xkm/h

Speed of the stream = y km/h

x = 8 (given)

Upstream is v = (x-y)

Downstream is u = x+y

x+y = 3(x-y) [ATQ]

8 + y = 3(8- y)

8 + y = 24 – 3y

4y = 16

y = 4

So, correct option is (c) 4km/h

## 4.

Solution.

x + 4 = 0(mod3)

So, the number should be fully divisible by 3

35 + 4 is 39 which is fully divisible by 3.

Hence, option d

## 5.

Solution: In the case of a square matrix, then the determinant value of A and its adj(A) is the same.

Option (b) -2.

Solution. (a) 0

## (a) 8 (b) -8 (c) -3 (d)16

Solution:

|AB| = 12

|A||B| = 12 (since its a square matrix)

|A| (-4) = 12

|A| = -3

Hence, option (c)

## 8.

Solution. (a) no solution

## (a) Base year prices (b) Current year prices(c)Base year quantities (d) Current year quantities

Solution.

So, its option (c)

## 10.

Solution.

Using the Paasche formula.

P01 = (506 x 100)/ 451

= 112.19

Hence, option (a)

Solution.

Hence, option c

## 12.

Solution.

Let’s define each event.

P(E) = { 2, 3, 5, 7}

P(F) = {1,2,3)

P(E U F) = P(E) + P(F) – P (E intersection F)

P (E intersection F) = {2,3}

P (E U F) = {2,3,5,7} + {1,2,3} – {2,3}

P (E U F) = {2} + {3} + {5} + {7} + {1}

All these probabilities from the table

P (EUF) = .77

Hence, option (b)

## (a) 1:2:3 (b) 3:2:1 (c) 3:1:1 (d)2:1:1

Solution.

Let the total capital be x

Let’s C contribution be y

Then, B’s contribution = 1/3 of x

A = y + x/3

A + B + C = y + x/3 + x/3 + y = x

2y + 2x/3 = x

2y = x – 2x/3

2y = (3x – 2x) / 3

2y = x/3

y = x /6

x = 6y

Now, each partner contribution

A = y + x/3 = y + 6y/3 = y + 2y = 3y

B = x/3 = 6y/3 = 2y

C = y

Hence, the ratio is 3:2:1

Option (b)

## 14.

Solution. (d) an equivalence relation

## (a) 4:5:6 (b) 9:7:6 (c) 8:7:6 (d) 12:21:32

Solution.

Investment ratio = Profit / Time

Investment Ratio = 6/2 : 7/3 : 8/4

Take LCM which will be 12

then ratio will become 36 : 28: 24

Reduce it, by dividing it be common factor 4

9 : 7 : 6

Hence, option (b)

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