Class 6

Samacheer Kalvi 6th Books Solutions Guide

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Samacheer Kalvi 6th Maths Book Answers Solutions Guide

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Students can also read Samacheer Kalvi 6th Science Book Solutions.

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Samacheer Kalvi 6th English Book Answers Solutions Guide

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Samacheer Kalvi 6th Social Science Book Answers Solutions Guide

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Samacheer Kalvi 6th Tamil Book Answers Solutions Guide

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 1 Fractions Ex 1.1

Question 1.
Fill in the blanks.
(i) 7$$\frac{3}{4}$$ + 6$$\frac{1}{2}$$ = _______
(ii) The sum of whole number and a proper fraction is called ______
(iii) 5$$\frac{1}{3}$$ – 3$$\frac{1}{2}$$ = ______
(iv) 8 ÷ $$\frac{1}{2}$$ = ______
(v) The number which has its own reciprocal is _______.
Solution:
(i) 14$$\frac{1}{4}$$
(ii) Mixed Fraction
(iii) 1$$\frac{5}{6}$$
(iv) 16
(v) 1

Question 2.
Say True or False.
(i) 3$$\frac{1}{2}$$ can be written as 3 + $$\frac{1}{2}$$.
(ii) The sum of any two proper fractions is always an improper fraction.
Hint: $$\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$$
(iii) The mixed fraction of $$\frac{13}{4}$$ is 3$$\frac{1}{4}$$
(iv) The reciprocal of an improper fraction is always a proper fraction.
(v) 3$$\frac{1}{4}$$ × 3$$\frac{1}{4}$$ = 9$$\frac{1}{16}$$
Solution:
(i) True
(ii) False
(iii) True
(iv) True
(v) False

The mixed number 2 2 3 as an improper fraction.

Question 3.
Solution:
(i) Find the sum of $$\frac{1}{7}$$ and $$\frac{3}{9}$$

(ii) What is the total of 3$$\frac{1}{3}$$ and 4$$\frac{1}{6}$$.

(iii) Simplify : 1$$\frac{3}{5}$$ + 5$$\frac{4}{7}$$

(iv) Find the difference between $$\frac{8}{9}$$ and $$\frac{2}{7}$$

(v) Subtract 1$$\frac{3}{5}$$ and 2$$\frac{1}{3}$$

(vi) Simplify: 7$$\frac{2}{7}$$ – 3$$\frac{4}{21}$$

fraction and whole number calculator

Question 4.
Convert mixed fraction into improper fractions and vice versa:

Solution:

Question 5.
Multiply the following :

Solution:

Question 6.
Divide the following:

Solution:

Question 7.
Gowri purchased 3$$\frac{1}{2}$$ kg of tomatoes, $$\frac{3}{4}$$ kg of brinjal and 1$$\frac{1}{4}$$ kg of onion, what is the total weight of the vegetables she bought?
Solution:
Weight of tomatoes Gowri purchased = 3$$\frac{1}{2}$$ kg
Weight of Brinjal purchased = $$\frac{3}{4}$$ kg

Total weight of vegetables that Gowri purchased = 5$$\frac{1}{2}$$ kg

Question 8.
An oil tin contains 3$$\frac{3}{4}$$ litres of oil of which 2$$\frac{1}{2}$$ litres of oil is used. How much oil is left over?
Solution:

Quantity of oil leftover = 1$$\frac{1}{4}$$ litres.

Question 9.
Nilavan can walk 4$$\frac{1}{2}$$km in an hour. How much distance will he cover in 3$$\frac{1}{2}$$ hours?
Solution:
Distance walked by Nilavan in one hour = 4$$\frac{1}{2}$$ km.

Nilavan walks 15$$\frac{3}{4}$$ km in 3$$\frac{1}{2}$$ hours

Question 10.
Ravi bought a curtain of length 15$$\frac{3}{4}$$ m. If he cut the curtain into small pieces each of length 2$$\frac{1}{4}$$ m, then how many small curtains will he get?
Solution:

Objective Type Questions

Question 11.
Which of the following statement is incorrect?

Solution:
(d)$$\frac{10}{11}<\frac{9}{10}$$
Hint:

Question 12.
The difference between $$\frac{3}{7}$$ and $$\frac{2}{7}$$ is

Solution:
(a) $$\frac{13}{63}$$
Hint:

Question 13.
The reciprocal of $$\frac{53}{17}$$ is

Solution:
(c) $$\frac{17}{53}$$
Hint:
$$\frac{\frac{1}{53}}{\frac{53}{17}}=\frac{17}{53}$$

Question 14.
If $$\frac{6}{7}$$ = $$\frac{A}{49}$$, then the value of A is
(a) 42
(b) 36
(c) 25
(d) 48
Solution:
(a) 42

Question 15.
Pugazh has been given four choices for his pocket money by his father. Which of the choices should he take in order to get the maximum money?

Solution:
(c) $$\frac{4}{5}$$ of ₹150

Samacheer Kalvi 6th Science Book Answers Solutions Guide

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6th Standard Science Book Answers பருவம் – 3

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.2

The LCM of 12 and 18 is 36.

Question 1.
Fill in the blanks.
(i) The HCF of 45 and 75 is ______
(ii) The HCF of two successive even numbers is ______
(iii) If the LCM of 3 and 9 is 9, then their HCF is ______
(iv) The LCM of 26, 39 and 52 is ______
(v) The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is ______
Solution:
(i) 15
(ii) 156
(iii) 2
(iv) 3
(v) 3

Question 2.
Say True or False
(i) The numbers 57 and 69 are co-primes.
(ii) The HCF of 17 and 18 is 1.
(iii) The LCM of two successive numbers is the product of the numbers.
(iv) The LCM of two co-primes is the sum of the numbers.
(v) The HCF of two numbers is always a factor of their LCM.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True

Question 3.
Find the HCF of each set of numbers using the prime factorisation method.
(i) 18, 24
(ii) 51, 85
(iii) 61, 76
(iv) 84, 120
(v) 27, 45, 81
(vi) 45, 55, 95
Solution:
(i) 18, 24.
Prime factorisation of 18 = 2 × 3 × 3
Prime factorisation of 24 = 2 × 2 × 2 × 3
Common factors of 18 and 24 = 2 × 3 = 6
HCF (18, 24) = 6

(ii) 51, 85
Prime factorisation of 51 = 3 × 17
Prime factorisation of 85 = 5 × 17
Common factors of 51 and 85 = 17
HCF (51, 85) = 17

(iii) 61, 76
Prime factorisation of 61 = 1 × 61
Prime factorisation of 76 = 2 × 2 × 19 × 1
Common factors of 61 and 76 = 1
HCF (61, 76) = 1

(iv) 84, 120
Prime factorisation of 84 = 2 × 2 × 3 × 7
Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5
Common factors of 84 and 120 = 2 × 2 × 3
HCF (84, 120) = 12

(v) 27, 45, 81
Prime factorisation of 27 = 3 × 3 × 3
Prime factorisation of 45 = 3 × 3 × 5
Prime factorisation of 81 = 3 × 3 × 3 × 3
Common factors of 27, 45, 81 = 3 × 3 = 9
HCF (27, 45, 81) = 9

(vi) 45, 55, 95
Prime factorisation of 45 = 3 × 3 × 5
Prime factorisation of 55 = 5 × 11
Prime factorisation of 95 = 5 × 19
Common factors of 45, 55, 95 = 5
HCF (45, 55, 95) = 5

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Question 4.
Find the LCM of each set of numbers using the prime factorisation method.
(i) 6, 9
(ii) 8, 12
(iii) 10, 15
(iv) 14, 42
(v) 30, 40, 60
(vi) 15, 25, 75
Solution:
(i) 6, 9
Prime factorisation of 6 = 2 × 3
Prime factorisation of 9 = 3 × 3
Product of common factors = 3
Product of other factors = 2 × 3 = 6
LCM (6, 9) = 3 × 6 = 18

(ii) 8, 12
8 = 2 × 4 = 2 × 2 × 2
12 = 2 × 6 = 2 × 2 × 3
Product of common factors = 2 × 2 = 4
Product of other factors = 2 × 3 = 6
LCM = Product of common factors × Product of other factors = 4 × 6 = 24
LCM (8, 12) = 24.

(iii) 10, 15
10 = 2 × 5
15 = 3 × 5
Product of common factors = 5
Product of other factors = 2 × 3 = 6
LCM (10, 15) = Product of common factors × Product of other factors = 5 × 6 = 30

(iv) 14, 42
14 = 2 × 7
42 = 2 × 21 = 2 × 3 × 7
Product of common factors = 2 × 7
Product of other factors = 3
LCM (14, 42) = Product of common factors × Product of other factors = 2 × 7 × 3 = 42
LCM (14, 42) = 42

(v) 30, 40, 60
30 = 3 × 2 × 5
40 = 2 × 2 × 2 × 5
60 = 2 × 3 × 2 × 5
Product of highest powers of the common factors = 3 × 23 × 5 = 120
LCM (30, 40, 60) = 120

(vi) 15, 25, 75
15 = 5 × 3
25 = 5 × 5
75 = 5 × 5 × 3
Product of the highest powers of the common factors = 3 × 52 = 3 × 25 = 75
LCM (15, 25, 75) = 75

Question 5.
Find the HCF and the LCM of the numbers 154, 198 and 286.
Solution:
Prime factorisation of 154 = 2 × 7 × 11
Prime Factorisation of 198 = 2 × 3 × 3 × 11
Prime factorisation of 286 = 2 × 11 × 13

To find HCF
Product of common factors of 154, 198 and 286 = 2 × 11 = 22
HCF (154, 198, 286) = 22

To find LCM
Product of common factors of atleast two numbers = 2 × 11 = 22
Product of other factors = 7 × 3 × 3 × 13 = 819
LCM (154, 198, 286) = Product of common factors × Product of other factors = 22 × 819 = 18,018
LCM (154, 198,286) = 18,018

Question 6.
What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?
Solution:
This Problem is HCF related problem
Prime factorisation of 80 = 2 × 2 × 2 × 2 × 5
Prime factorisation of 100 = 2 × 2 × 5 × 5
Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5
Product of common factors 80, 100 and 120 = 2 × 2 × 5 = 20
HCF (80, 100, 120) = 20
The volume of the vessel = 20 litres

Question 7.
The traffic lights at three different road junctions change after every 40 seconds, 60 seconds and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?
Solution:
This is an LCM related problem
Finding the LCM of 40, 60 and 72
60 seconds = 1 min
360 min = 6 min

LCM (40, 60, 72) = 2 × 2 × 3 × 5 × 2 × 3 = 360
After 360 seconds they will change again i.e after six minutes they will change again
i.e at 8.06 am they will change again simultaneously.

Question 8.
The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible?
Solution:
Let the numbers be 14x and 14y
14x × 14y = 14 × 210
⇒ $$x y=\frac{210}{14}=\frac{30}{2}=15$$
x = 1, y = 15; x = 3, y = 5
(14, 210), (42, 70) Two pairs

Question 9.
The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers is 36, then find the other number.
Solution:
HCF = 12
Product of two numbers = LCM × HCF
36 × other number = 72 × 12
Other number = $$\frac{72×12}{36}$$
Other number = 24

Objective Type Questions

Question 10.
Which of the following pairs is co-prime?
(a) 51, 63
(b) 52, 91
(c) 71, 81
(d) 81, 99
Solution:
(c) 71, 81

Question 11.
The greatest 4 digit number which is exactly divisible by 8, 9, and 12 is
(a) 9999
(b) 9996
(c) 9696
(d) 9936
Solution:
(d) 9936

Question 12.
The HCF of two numbers is 2 and their LCM is 154. If the difference between numbers is 8, then the sum is
(a) 26
(b) 36
(c) 46
(d) 56
Solution:
(b) 36

Question 13.
Which of the following cannot be the HCF of two numbers whose LCM is 120?
(a) 60
(b) 40
(c) 80
(d) 30
Solution:
(c) 80

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.1

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Therefore, 24 has 8 factors.

Question 1.
Fill in the blanks:
(i) The number of prime numbers between 11 and 60 is ……..
(ii) The numbers 29 and ……….. are twin primes.
(iii) 3753 is divisible by 9 and hence divisible by ……….
(iv) The number of distinct prime factors of the smallest 4 digit number is ………
(v) The sum of distinct prime factors of 30 is ………
Solution:
(i) 12
(ii) 31
(iii) 3
(iv) 2
(v) 10

Question 2.
Say True or False
(i) The sum of any number of odd numbers is always even.
(ii) Every natural number is either prime or composite.
(iii) If a number is divisible by 6, then it must be divisible by 3.
(iv) 16254 is divisible by each of 2, 3, 6, and 9.
(v) The number of distinct prime factors of 105 is 3.
Solution:
(i) False
(ii) False
(iii) True
(iv) True
(v) True

The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, and 144.

Question 3.
Write the smallest and the biggest two-digit prime number.
Solution:
Smallest is 11
Biggest is 97

Question 4.
Write the smallest and the biggest three-digit composite number.
Solution:
Smallest three-digit number = 100, divisible by 2, 4, 5, 10, 20, 25, 50 and so it is the composite biggest three-digit number is 999, composite.
Smallest is 100
Biggest is 999

Question 5.
The sum of any three odd natural numbers is odd. Justify this statement with an example.
Solution:
True, as we know, that,“the sum of any three odd numbers is always an odd number”.
Example: 3 + 7 + 9 = 19 is odd.

Question 6.
The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pairs exist up to 100.
Solution:
The required pairs are
(i) 17 and 71
(ii) 37 and 73
(iii) 79 and 97

Question 7.
Your friend says that every odd number is prime. Give an example to prove him/her wrong.
Solution:
15 is an odd number not prime.

Question 8.
Each of the composite numbers has at least three factors. Justify this statement with an example.
Solution:
True. The composite number 4 has 3 factors namely 1, 2, and 4.

Question 9.
Find the dates of any month of a calendar which are divisible by both 2 and 3.
Solution:
6, 12, 18, 24, 30

Find the factors of 18. 18. 1 x 18. 2 x 9. 3 x 6. 4 x. 5 x. 6 repeats.

Question 10.
I am a two-digit prime number and the sum of my digits is 10.1 am also one of the factors of 57. Who am I?
Solution:
19

Question 11.
Find the prime factorisation of each number by factor tree method and division method.
(a) 60
(b) 128
(c) 144
(d) 198
(e) 420
(f) 999
Solution:
(a) 60

∴ 60 = 2 × 2 × 3 × 5
Also

∴ 60 = 2 × 2 × 3 × 5
(b) 128

∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Also

∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
(c) 144

∴ 144 = 2 × 2 × 2 × 2 × 3 × 3
Also

∴ 144 = 2 × 2 × 2 × 2 × 3 × 3
(d) 198

∴ 198 = 2 × 3 × 3 × 11
Also

∴ 198 = 2 × 3 × 3 × 11
(e) 420

∴ 420 = 2 × 2 × 3 × 5 × 7
Also

∴ 420 = 2 × 2 × 3 × 5 × 7
(f) 999

∴ 999 = 3 × 3 × 3 × 37
Also

∴ 999 = 3 × 3 × 3 × 37

Question 12.
If there are 143 math books to be arranged in equal numbers in all the stacks, then find the number of books in each stack and also the number of stacks.
Solution:
Total number of books = 143
Factorizing 143 = 11 × 13

The number of stacks and the number of books in each stack may be (11, 13) or (13, 11).

Objective Type Questions

Question 13.
The difference between two successive odd number is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(b) 2

Question 14.
The only even prime number is
(a) 4
(b) 6
(c) 2
(d)
Solution:
(c) 2

Question 15.
Which of the following numbers is not prime?
(a) 53
(b) 92
(c) 97
(d) 71
Solution:
(b) 92

Question 16.
The sum of the factors of 27 is
(a) 28
(b) 37
(c) 40
(d) 31
Solution:
(c) 40

Question 17.
The factors of a number are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. What is the number?
(a) 80
(b) 100
(c) 128
(d) 160
Solution:
(a) 80

Question 18.
The prime factorisation of 60 is 2 × 2 × 3 × 5. Any other number which has the same prime factorisation as 60 is
(a) 30
(b) 120
(c) 90
(d) impossible
Solution:
(d) impossible

Question 19.
If the number 6354*97 is divisible by 9, then the value * is
(a) 2
(b) 4
(c) 6
(d) 7
Solution:
(a) 2

Question 20.
The number 87846 is divisible by
(a) 2 only
(b) 3 only
(c) 11 only
(d) all of these
Solution:
(d) all of these

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 1.
Say the time in two ways:

Solution:
(i) 10 : 15 hours, quarter past 10
(ii) 6 : 45 hours, quarter to 7
(iii) 4 : 10 hours, 10 minutes past 4
(iv) 3 : 30 hours, half past 3
(v) 9 : 40 hours, 20 minutes to 10

Question 2.
Match the following:
(i) 9.55 – (a) 20 minutes past 2
(ii) 11.50 – (b) quarter past 4
(iii) 4.15 – (c) quarter to 8
(iv) 7.45 – (d) 5 minutes to 10
(v) 2.20 – (e) 10 minutes to 1?
Solution:
(i) 9.55 – (d) 5 minutes to 10
(ii) 11.50 – (e) 10 minutes to 12
(iii) 4.15 – (b) quarter past 4
(iv) 7.45 – (c) quarter to 8
(v) 2.20 – (a) 20 minutes past 2

Question 3.
Convert the following :
(i) 20 minutes into seconds
(ii) 5 hours 35 minutes 40 seconds into seconds
(iii) 3$$\frac { 1 }{ 2 }$$ hours into minutes
(iv) 580 minutes into hours
(v) 25200 seconds into hours
Solution:
(i) 20 minutes into seconds
20 minutes = 20 × 60 seconds = 1200 seconds
(ii) 5 hours 35 minutes 40 seconds into seconds

∴ 5 hours 35 minutes 40 seconds = 20,140 seconds
(iii) 3$$\frac { 1 }{ 2 }$$ hours into minutes
3$$\frac { 1 }{ 2 }$$ hours = 3 hours 30 minutes
= 3 × 60 minutes + 30 minutes
= 180 minutes + 30 minutes
= 210 minutes
∴ 3$$\frac { 1 }{ 2 }$$ hours = 210 minutes
(iv) 5580 minutes into hours
580 minutes = $$\frac{580}{60}$$ hours = 9 hours 40 minutes
∴ 580 minutes = 9 hours 40 minutes
(v) 25200 seconds into hours
25200 seconds = $$\frac{25200}{60}$$ minutes = 420 minutes = $$\frac{420}{60}$$ hours = 7 hours
∴ 25200 seconds = 7 hours

Question 4.
The duration of electricity consumed by the farmer for his pump set on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity.
Solution:
Electricity consumption on Monday = 7 hours + 20 minutes + 35 seconds
Electricity consumption on Tuesday = 3 hours + 44 minutes + 50 seconds
Total consumption = 10 hours + 64 minutes + 85 seconds
= 10 hours + (60 + 4) minutes + (60 + 25) seconds
= 1o hours + 1 hrs 4 minutes + 1 minutes 25 seconds
= 11 hours 5 minutes 25 seconds
∴ Total consumption of electricity = 11 hours 5 minutes 25 seconds

Question 5.
Subtract 10 hrs 20 min 35 sec from 12 hrs 18 min 40 sec.
Solution:

1 hour 58 minutes 05 seconds

Question 6.
Change the following into 12 hour format:
(i) 02 : 00 hours
(ii) 08 : 45 hours
(iii) 21 : 10 hours
(iv) 11 : 20 hours
(v) 00 : 00 hours
Solution:

Question 7.
Change the following into 24-hour format.
(i) 3.15 a.m.
(ii) 12.35 p.m.
(iii) 12.00 noon
(iv) 12.00 midnight.
Solution:

Question 8.
Calculate the duration of time.
(i) from 5.30 a.m to 12.40 p.m
(ii) from 1.30 p.m to 10.25 p.m
(iii) from 20:00 hours to 4:00 hours
(iv) from 17:00 hours to 5 :15 hours
Solution:
(i) From 5.30 am to 12.40 pm
= (5.30 am to 12 pm) + 12 pm to 12.40 pm
= 6 hrs 30 min + 40 min
= 6 hrs 70 min
= 7 hrs 10 min
(ii) from 1.30 pm to 10.25 p.m.

(iii) from 20 : 00 hrs to 4 : 00 hrs.

(iv) from 17 : 00 hours to 5 : 15 hours.

Question 9.
The departure and arrival timing of the Vaigai Superfast Express (No. 12635) from Chennai Egmore to Madurai Junction are given. Read the details and answer the following.

(i) At what time does the Vaigai Express start from Chennai and arrive at Madurai?
(ii) How many halts are between there Chennai and Madurai?
(iii) How long does the train halt at the Villupuram junction?
(iv) At what time does the train come to Sholavandan?
(v) Find the journey time from Chennai Egmore to Madurai?
Solution:
(i) It starts from Chennai at 13 : 40 hrs and arrive at Madurai at 21 : 20 hrs.
(ii) There are 8 halts.
(iii) Departure from Villupuram = 15 hours 55 minutes
Arrival at Villupuram = 15 hours 50 minutes
The train halt at Villupuram for = 05 minutes
(iv) At 20 : 34 hours the train come to Sholavandan
(v) Arrival time at Madurai = 20 hours 80 (20 + 60) minutes
Departure time from Chennai Egmore = 13 hours 40 minutes
Journey Time = 07 hours 40 minutes

Question 10.
Manickam joined a chess class on 20.02.2017 and due to exam, he left practice after 20 days. Again he continued practice from 10.07.2017 to 31.03.2018. Calculate how many days did he practice?
Solution:
Before Exams, he practiced for 20 days.
Days From 10.07.2017 to 31.03.2018
July – 22 Days (From 10.7.2017)
August – 31 Days
September – 30 Days
October – 31 Days
November – 30 Days
December – 31 Days
January 2018 – 31 Days
February – 28 Days
March – 31 Days (upto 31.3.2018)
Total – 265 Days
Total number of days practiced = Number of days practiced before exam + Number of days practiced after exam
= 20 + 265
= 285 days.
∴ He practiced for 285 days.

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Question 11.
A clock gains 3 minutes every hour. If the clock is set correctly at 5 a.m, find the time shown by the clock at 7 p.m?
Solution:
5 a.m. = 5.00 hours
7 p.m = 19.00 hours
Difference = 19.00 – 5.00 = 14.00 hours
Time gain = 14 × 3 = 42 minutes
Time shown by the clock = 7.42 p.m

Question 12.
Find the number of days between republic day and kalvi valarchi day in 2020.
Solution:
2020 is a leap year republic day – 26.01.2020
Kalvi valarchi day – 15.07.2020
Jan – 5
Feb – 29
Mar – 31
April – 30
May – 31
June – 30
July – 14
Total – 170 days

Question 13.
If the 11th of January 2018 is Thursday, what is the day on 20th July of the same year?
Solution:
January – 21 Days (31 – 10)
February – 28 Days
March – 31 Days
April – 30 Days
May – 31 Days
June – 30 Days
July – 19 Days
Total – 190 Days
190 days ÷ 7
190 days = 27 weeks + 1 day
The required day is the first day after Thursday.
∴ 20th of July is Friday.

Question 14.
(i) Convert 480 days into years
(ii) Convert 38 months into years
Solution:
(i) Convert 480 days into years.
We know that 1 year = 365 days. [∵ 480 – 365 = 115 days ]
480 days = 1 year 115 days.
115 days = $$\frac{115}{30}$$ months = 3 months 25 days
∴ 480 days = 1 year 3 months 25 days.
(ii) Convert 38 months into years
1 year has 12 months.
38 months = $$\frac{38}{12}$$ years = 3 years 2 months
∴ 38 months = 3 years 2 months.

Question 15.
Calculate your age as on 01.06.2018
Solution:

Age is 12 years 2 months

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Objective Type Questions

Question 16.
2 days = …….. hours
(i) 38
(ii) 48
(iii) 28
(iv) 40
Solution:
(ii) 48

Question 17.
3 weeks = _____ days.
(a) 21
(b) 7
(c) 14
(d) 28
Solution:
(a) 21

Question 18.
Number of ordinary years between two consecutive leap years is
(i) 4 years
(ii) 2 years
(iii) 1 year
(iv) 3 years
Solution:
(iv) 3 years

Question 19.
What time will it 5 hours after 22 : 35 hours?
(a) 2 : 30 hrs
(b) 3 : 35 hrs
(c) 4 : 35 hrs
(d) 5 : 35 hrs
Solution:
(b) 3 : 35 hrs

Question 20.
2 $$\frac{1}{2}$$ years is equal to months.
(i) 25
(ii) 30
(iii) 24
(iv) 5
Solution:
(ii) 30