Logical Reasoning Class 8 - Edugain Math

ID : in-8-Logical-R easoning [1]

Class 8 Logical Reasoning

For more such worksheets visit

Choose correct answer(s) from the given choices

( (1) 99999 +

1

) (99999 +

1

) ( - 99999 -

1

) (99999 -

1

)= ?

33333

33333

33333

33333

a. 6

b. 3

c. 12

d. 33333

(2) How many more squares need to be shaded to cover 75% of the total area?

a. 6 c. 24

b. 42 d. 48

3

(3) P (x) is a polynomial such that P (x + ) = P (x). If P (9) = 344, find the value of P (21).

2

a. 344

b. 348

3

c. 344

2

1

d. 344

2

(4) What is the missing number in the series: 1, 2, 4, 3, 9, 4, 16, __25, 6, 36, 7 ?

a. 5

b. 4

c. 25

d. 6

(5) Tina introduces Rajesh as the son of the brother of her mother. How is Tina related to Rajesh?

a. Cousin

b. Nephew

c. Aunt

d. Daughter

(6) If we use (x+) to indicate the following sum:

1 + 2 + 3 + ... + x

then find the value of k in the following equation:

(15+) - (14+) = (k+).

a. 6

b. 7

c. 5

d. 3

C opyright 2021 w w w .

Fill in the blanks

(7) If we use (x+) to indicate the following sum: 1 + 2 + 3 + ... + x

then the value of k in the following equation is

.

(21+) - (20+) = (k+).

ID : in-8-Logical-R easoning [2]

Answer the questions

(8) If N = 111112, find the 5th digit in the expansion of N from the right.

(9) During a day, the hour hand and the minute hand of a clock form a right angle, at multiple times. For example, the two hands form a right angle at 9 am. How many times during a day (24 hours) will the two hands form a right angle?

(10) Ashish likes 144 but not 134, 1600 but not 1700, and 5625 but not 5610. Which of the following four numbers would he like: 629, 618, 625 or 622?

? 2021 Edugain (). All Rights Reserved

Many more such worksheets can be generated at

C opyright 2021 w w w .

Solutions

(1) c. 12

ID : in-8-Logical-R easoning [3]

Step 1

(99999 +

1

) (99999 +

1

) ( - 99999 -

1

) (99999 -

1

)

33333

33333

33333

33333

( = 99999 +

2

1

) ( - 99999 -

2

1

)

33333

33333

Step 2

Using

algebraic

identities

(a

+

2

b)

=

2

a

+

2

b

+

2ab

and

(a

-

2

b)

=

2

a

+

2

b

-

2ab

(99999 +

2

1

) ( - 99999 -

2

1

)

33333

33333

2

2

1

1

1

1

( 2

=99999 +

) + 2 ? 99999 ?

[ ( 2

- 99999 +

) - 2 ? 99999 ?

]

33333

33333

33333

33333

2

1

( 2

=99999 +

) + 2 ? 99999 ?

33333

1 33333

2

1

( 2

- 99999 -

) + 2 ? 99999 ?

33333

1 33333

=2 ? 3 + 2 ? 3 =12

Step 3

Therefore, the value of

(99999 +

) ( 1 99999 +

) ( 1 - 99999 -

) ( 1 99999 -

1)

33333

33333

33333

33333

is 12.

C opyright 2021 w w w .

(2) b. 42

ID : in-8-Logical-R easoning [4]

Step 1 The total number of squares = number of rows ? number of columns = 8 ? 8 = 64

Step 2 Let us now find the required number of shaded squares, which is 75% of the total number of squares, or 75% of 64

75 = 64 ?

100 = 48

Step 3 We can see in the picture that 6 squares are already shaded.

Step 4 Therefore, the number of more squares which need to be shaded = The required number of shaded squares - The number of squares already shaded = 48 - 6 = 42

Step 5 Hence, 42 more squares need to be shaded to cover 75% of the total area.

C opyright 2021 w w w .

(3) a. 344

ID : in-8-Logical-R easoning [5]

Step 1

It is given that,

3

( ) P x +

= P (x)

2

3

It means that value of polynomial P for any x is same as its value for (x + ) .

2

Step 2

3

It also means that if we keep adding to the x, the value will remain unchanged.

2

3

( ) i. e. P (x) = P x +

2

3

3

3

[( ) ] ( ) P (x) = P x +

+

= P x+2?

2

2

2

3

3

3

[( ) ] ( ) P (x) = P x + 2 ?

+

= P x+3?

2

2

2

3

3

3

[( ) ] ( ) P (x) = P x + 3 ?

+

= P x+4?

2

2

2

and so on..

3

3

3

[( ) ] ( ) P (x) = P x + 3 ?

+

= P x+n?

... [ Where n is any natural number]

2

2

2

Step 3

Now we can write P (21) in terms of P (9) as following,

P (21) = P (9 + 12)

3 P (21) = P (9 + 8 ? )

2 P (21) = P (9) [Using relation (1)]

Step 4 Since P (21) = P (9), value of P (21) = 344

C opyright 2021 w w w .

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download