Woksheet Solutions: Logical Reasoning | Mental Mathematics for Class 8 PDF Download

Q1:
(a) 6
(b) 3
(c) 12
(d) 33333
Ans: Step 1

Step 2
Using algebraic identities (a + b)² = a² + b² + 2ab (a − b)² = a² + b² − 2ab

Step 3
Therefore, the value of


Q2: P(x) is a polynomial such that  If P(9) = 344, Find the value of P(21).
(a) 344
(b) 348
(c)
(d)
Ans: (a)
Step 1:
It is given that,

It means that value of polynomial P for any x is same as its value for .
Step 2:
It also means that if we keep adding 3/2 to the the x value will remain unchanged.

 … [Where n is any natural number]
Step 3:
Now we can write P(21) in terms of P(9)as following,
P(21) = P(9+12)

⟹ P(21) = P(9) [Using relation (1)]

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

Q3: 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
Ans: (a)

• Step 1: According to the question, Tina introduces Rajesh as the son of the brother of her mother.
• Step 2: Or, we can say that the brother of Tina's mother is the maternal uncle of Tina.
• Step 3: Or, we can say that Rajesh is the son of Tina's maternal uncle.
• Step 4: Or, we can say that Rajesh is the cousin of Tina.
• Step 5: Hence, option a is the correct answer

Q4: 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+).
Ans: 6
Step 1: 
It is given that x+ = 1 + 2 + 3 + ... + x

Step 2: 
According to the question, (21+) - (20+) = (k+)
⇒ (1+2+3+ ... +21) - (1+2+3+ ... +20) = (1+2+3+ ... +k)
⇒ (1+2+3+ ... +20) + 21 - (1+2+3+ ... +20) = (1+2+3+ ... +k)
⇒ 21 = (1+2+3+...+k)

Step 3: 
Let us find the value of k for which sum of first k natural numbers is 21.
Let us start with k = 2
For k = 2, 1 + 2 = 3
For k = 3, 1 + 2 + 3 = 6
For k = 4, 1 + 2 + 3 + 4 = 10
For k = 5, 1 + 2 + 3 + 4 + 5 = 15
For k = 6, 1 + 2 + 3 + 4 + 5 + 6 = 21

Step 4:  
Hence, the value of k is 6.

Q5: 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?
Ans: 44
Step 1:
Let us look at the clock at 9 am:
We can see that the hour hand and the minute hand are making a right angle.

Step 2: 
In a clock, while the minute hand moves, the hour hand also moves, although a lot slowly. It is easy to see that in a 12 hour period, the minute hand make 12 revolutions while, the hour hand makes one.
We can visualize the above statement this way:
If we hold the clock in our hands and always keep on rotating it slowly, such that the hour hand always stay on the same position, the minute hand will make 12 - 1 = 11 revolutions. In other words, the minute hand makes 11 revolutions around the hour hand in a 12 hour period.

Step 3:
For each revolution around the hour hand, the minute hand makes a right angle twice with it. The total number of times we see the two hands making a right angle is 11 × 2 = 22.

Step 4:
In 12 hours, the number of times the two hands make a right angle = 22
In 24 hours, the number of times the two hands make a right angle = 22 × 2 = 44

Q6: How many more squares need to be shaded to cover 75% of the total area?
(a) 6
(b) 42
(c) 24
(d) 48
Ans: (b)
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

= 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

Q7: 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
Ans: (a)
Step 1: 
1, 2, 4, 3, 9, 4, 16, __25, 6, 36, 7
If we look at the given series carefully, we notice that the given series is like:
1, n, n², (n+1), (n+1)², (n+2), (n+2)² ........
(Where, n = 2, 3, 4.....). So, we can say that the missing number in the series 1, 2, 4, 3, 9, 4, 16, __25, 6, 36, 7 is 5.

Step 2:
Hence, option a is the correct answer.

Q8: 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
Ans: (c)
Step 1:
It is given that x+ = 1 + 2 + 3 + ... + x

Step 2:
According to the question, (15+) - (14+) = (k+)
⇒ (1+2+3+ ... +15) - (1+2+3+ ... +14) = (1+2+3+ ... +k)
⇒ (1+2+3+ ... +14) + 15 - (1+2+3+ ... +14) = (1+2+3+ ... +k)
⇒ 15 = (1+2+3+...+k)

Step 3:
Let us find the value of k for which sum of first k natural numbers is 15.
Let us start with k = 2
For k = 2, 1 + 2 = 3
For k = 3, 1 + 2 + 3 = 6
For k = 4, 1 + 2 + 3 + 4 = 10
For k = 5, 1 + 2 + 3 + 4 + 5 = 15
Step 4:
Therefore, the value of k is 5. Hence, option c is the correct answer.

Q9: If N = 11111², find the 5^th digit in the expansion of N from the right.
Ans: 5

• Step 1: The square of 11 is equal to 121.
• Step 2: The square of 111 is equal to 12321.
• Step 3: We can see that the squares of such numbers follow a pattern. Based on the pattern the square of 11111 will be 123454321.
• Step 4: We can see that the 5th digit in the square of 11111 from the right is 5

Q10: 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?
Ans: 625
Step 1:
If we look at the pattern, we notice that Ashish likes only those numbers which have complete squares.
144 = 12²
1600 = 40²
5625 = 75²

Step 2:
Now, observe the given options and see if there is any number which has complete square. Among the given options, we notice that 625 = 25².
Hence, 625 has complete square of 25.

Step 3:
Therefore, Ashish will like 625.