Complusory Subjects - Page 3 of 11

Q: 6(b) CSS GSA 2023 Solved – Find the missing number in the series below:

(i) 1, 8, 4, 27, 9, ?

(ii) 3, 6, 8, 16, 18, ?

(iii) 2, 8, 512, ?

(iv) 81, 9, 64, 8, ?, 12

(v) 6, 11, 21, 36, 56, ?

(i) 1, 8, 4, 27, 9, ?

Square and Cube Series

1 square of 1

8 cube of 2

4 square of 2

27 cube of 3

9 square of 3

64 cube of 4

1	8	4	27	9	64	16
Square of 1	Cube of 2	Square of 2	Cube of 3	Square of 3	Cube of 4	Square of 4
12	23	22	33	32	43	42

(ii) 3, 6, 8, 16, 18, ?

Multiplication and addition series

3 x 2 = 6

6 + 2 = 8

8 x 2 = 16

16 + 2 = 18

18 x 2 =36

3	6	8	16	18	36
Multiply by 2	Addition of 2	Multiply by 2	Addition of 2	Multiply by 2	Addition of 2
3 x 2	6 + 2	8 x 2	16 + 2	18 x 2	36 + 2

(iii) 2, 8, 512, ?

Cube Series:

2 x 2 x 2 = 8

8 x 8 x 8 = 512

512 x 512 x 512 = 134217728

2	8	512
Cube of 2	Cube of 8	Cube of 512
23	83	5123

(iv) 81, 9, 64, 8, ?, 12

Square Root Series:

√81=9

√64=8

√144=12

144

Square Root of 9

√81

√64

√144

(v) 6, 11, 21, 36, 56, ?

Addition of Multiples of 5

6 + 5 = 11

11 + 10 = 21

21 + 15 = 36

36 + 20 = 56

56 + 25 = 81

6	11	21	36	56	81
Addition of 5	Addition of 10	Addition of 15	Addition of 20	Addition of 25	Addition of 30
6 + 5	11 + 10	21 + 15	36 + 20	56 + 25	81 + 30

Q: 6(b) CSS GSA 2023 Solved – Find the missing number in the series below: Read More »

Q: 8(a) CSS GSA 2023 Solved – Ali buys an oven for Rs.36800 and sells it at a gain of 8.5%. For how much does he sell it?

Formulas To be Used In Order To Solve This Problem

Gain % = (Gain X 100)/(C.P)

Gain = S.P – C.P

Given:

C.P = Rs. 36800

Gain% = 8.5%

To Find:

S.P = ?

Solution:

Gain % = (Gain X 100)/(C.P)

8.5 = (Gain X 100)/(36800 )

(8.5 X 36800)/100 = Gain

Gain = 3128

So,

Gain=S.P-C.P

Gain + C.P = S.P

3128+36800= S.P

S.P = 39928

Ali Sold The Oven For Rs. 39928 at a gain of 8.5%.

Q: 8(a) CSS GSA 2023 Solved – Ali buys an oven for Rs.36800 and sells it at a gain of 8.5%. For how much does he sell it? Read More »

Q:8(b) CSS GSA 2023 Solved – A card is drawn at random from a box containing 12 cards numbered 1,2,3,4,5,…,12. Find the probability of drawing.

(i) ‘8’

(ii) an even number

(iii) a perfect square

(iv) a negative number

(v) a number less than 13

Since, Sample Space or

S=1,2,3,4,5,6,7,8,9,10,11,12

Then,

(i) ‘8’

Solution:

S=1,2,3,4,5,6,7,8,9,10,11,12

Let A be the event selecting ‘8’, then n(A)=8

P(A)=(n(A))/(n(S))

P(A)=1/12

As 8 numbered card is only one in the box, so the Probability of drawing a

card ‘8’ from box will be 1/12

(ii) an even number

Solution:

S=1,2,3,4,5,6,7,8,9,10,11,12

Let A be the event selecting ‘even number’, then n(A)=2,4,6,8,10,12

P(A)=(n(A))/(n(S))

P(A)=6/12

P(A)=1/2

As Even number cards is six in the box, so the Probability of drawing a card an even number from box will be 1/2

(iii) a perfect square

Solution:

S=1,2,3,4,5,6,7,8,9,10,11,12

Let A be the event selecting ‘a perfect square’, then n(A)=1, 4, 9

P(A)=(n(A))/(n(S))

P(A)=3/12

P(A)=1/4

As a perfect square number cards is 3 in the box, so the Probability of drawing a card a

perfect square from box will be 1/4

(iv) a negative number

Solution:

S=1,2,3,4,5,6,7,8,9,10,11,12

Let A be the event selecting ‘a negative number’, then n(A)=0

P(A)=(n(A))/(n(S))

P(A)=0/12

P(A)=0

As there is no any negative number card in the box, so the Probability of drawing a card a negative number from box will be 0

(v) a number less than 13

Solution:

S=1,2,3,4,5,6,7,8,9,10,11,12

Let A be the event selecting ‘a number less than 13’, then n(A)=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

P(A)=(n(A))/(n(S))

P(A)= 12/12

P(A)= 1

As all numbers in the box are less than 13, so the Probability of drawing a card a

Number less than 13 from box will be 1.

Q:8(b) CSS GSA 2023 Solved – A card is drawn at random from a box containing 12 cards numbered 1,2,3,4,5,…,12. Find the probability of drawing. Read More »