Class 6

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 4 Geometry Additional Questions

Question 1.
List any three letters having same mirror image as the object or letters?
Solution:
A, O and T

Question 2.
In the word SYMMETRY which letters do not have identical mirror images?
Solution:
S, E and R

Question 3.
Is Indian National flag symmetrical ?
Solution:
Yes, Horizontal

Question 4.
In the word ‘FIGURE’ mention the symmetrical letters.
Solution:
I, U and E

Question 5.
Are the number of symmetrical lines equal in rectangle and square?
Solution:
No, not equal.

Question 6.
Find the number of lines of symmetry in

i) The number of lines of symmetry
ii) The number of lines of symmetry
Solution:
i) The number of lines of symmetry for this figure are 2.

ii) The number of lines of symmetry for this figure are 2.

Question 7.
Give the reflection of the letter M as shown in the figure.

Solution:

Question 8.
Find the number of lines of symmetry of the following.

Solution:

Question 9.
Are human faces symmetrical ?
Solution:
Yes

Question 10.
List some symmetrical objects from your home?
Solution:
Wall clock, dining table top, bucket, photo frame etc.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Additional Questions

Question 1.
Convert into indicated units: 200 mg into g.
Solution:
200 mg = \(200.0 \times \frac{1}{1000}\) = 0.2 g

Question 2.
(a) What are the universally accepted basic metric units?
(b) A cow gives 10 litres of milk in the morning and 8 litres in the evening. Find the total milk it gives for a week in ml?
Solution:
(a) Universally accepted basic metric units are:
Length in metre, Weight in gram and Capacity (Volume) in litre.
(b) Litres of milk the cow gives in the morning = 10 l.
Litres of milk the cow gives in the evening = 8 l.
Total milk per day = 10 + 8 = 18 l
Total milk for a week = 18 × 7 l = 126 l = 126 × 1000 ml = 126000 ml.
Total milk the cow gives for a week = 1,26,000 ml.

Question 3.
A box contains 20 kg 280 g of sugar. After transferring into a smaller box of 2 kg 720 g 7 times find the remaining sugar left in the big box in cg.
Solution:
Total sugar in the box = 20 kg 280 g.
Measurement of smaller box = 2 kg 720 g.
7 times the smaller box = 2 kg 720 g × 7 = 19 kg 040 g.
Remaining sugar = 20 kg 280 g – 19 kg 40 g = 1 kg 240 g = 1240 g. = 1240 × 100 cg = 124000 eg.
Sugar left in the box = 1,24,000 eg.

Question 4.
Sneha needs 1 m 20 cm of cloth for her blouse. If she bought 20 such cloths find the total cloth she bought in cm?
Solution:
For one blouse the cloth needed = 1 m 20 cm
Cloth needed for 20 blouse = 1 m 20 cm × 20
= 20 m 400 cm
= 20 × 100 + 400 cm
= 2000 + 400 cm
= 2400 cm.
Total cloth bought = 2400 cm.

Question 5.
Can you add 2 m 20 cm and 4 kg 80 g? Why?
Solution:
No, we can’t. To add or subtract, the measures should be in the same unit.

Question 6.
If you serve a cup of coffee measuring 150 ml each for 200 people find the volume of coffee needed in litres?
Solution:
For one cup of coffee it needs 150 ml.
For 200 people it needs 150 × 200 ml = 30000 ml = 30000 × \(\frac{1}{1000}\) l = 30 l
Coffee needed = 30 l.

Question 7.
A lorry full of fertiliser is bought to the garden having 120 coconut trees. If the lorry contain 1920 kg, then find the measure of fertiliser each coconut tree can have?
Solution:
Total measure of fertiliser = 1920 kg.
Number of coconut tress = 120
For one tree = \(\frac{1920}{120}\) kg = 16 kg.
One tree can have 16 kg of fertiliser

Question 8.
Convert 7 kg 250 g in to mg, g and cg?
Solution:
7 kg 250 g.
(i) 7 kg 250 g = 7 × 1000 + 250 g. = 7000 + 250 = 7250 g.
(ii) 7 kg 250 g = 7250 g. = 7250 × 1000 mg = 72,50,000 mg.
(iii) 7 kg 250 g = 7250 g. = 7250 × 100 cg = 7,25,000 cg.

Question 9.
Convert 20 m 20 cm into cm, m and km.
Solution:
20 m 20 cm.
(i) 20 m 20 cm = 20 × 100 + 20 cm = 2000 + 20 cm = 2020 cm
(ii) 20 m 20 cm = 2020 cm = 2020 × \(\frac{1}{100}\) m = 20.20 m
(iii) 20 m 20 cm = 20.20 m = 20.20 × \(\frac{1}{1000}\) km = 0.02020 km

Question 10.
Convert 85 l 720 ml into ml, l and kl?
Solution:
85 l 720 ml
(i) 85 l 720 ml = 85 × 1000 + 720 ml = 85000 + 720 ml = 85,720
(ii) 85 l 720 ml = 85 l + 720 × \(\frac{1}{1000}\) l = 85 l + 0.720 l = 85.72 l.
(iii) 85 l 720 ml = 85.72 l = 85.72 × \(\frac{1}{1000}\) kl = 0.08572 kl.

Question 11.
Which is greater 0.007 g, 0.7 kg or 70 mg?
Solution:
0.007 g = 0.007 × 1000 mg = 7 mg
0.7 kg = 0.7 × 10,00,000 mg = 7,00,000 mg
70 mg = 70 mg
∴ 0.7 kg is greater.

Question 12.
A bus leaves for Kanchipuram from Chennai at 4.30 p.m. It takes 1 hr 25 min. to reach there. At what time will it reach at Kanchipuram?
Solution:
Starting time from Chennai = 4 : 30 pm.
Duration of travel = 01 hr : 25 min
Arrival at Kanchipuram = 5 : 55 pm
The bus will reach Kanchipuram at 5 : 55 pm.

Question 13.
The duration of a film show is 3 hrs 15 min. It starts at 6 : 30 p.m. When will it end?
Solution:
The film starts at = 6 : 30 pm.
Duration of the film show = 3 hr : 15 min
The show end at = 9 : 45 p.m.

Question 14.
A train arrive Chennai Central at 11 : 55 am. It reached Chennai 1 hr 25 min late. What is the scheduled arrival time of the train at Chennai?
Solution:
The train arrived Chennai at 11 : 55 am = 11 : 55 hours
The train is late by = 1 hour 25 min
Scheduled Time = 10 : 30 hours
The scheduled arrival time = 10.30 a.m

Question 15.
Raju visited a fashion show. He stayed there for 2 hr 30 min and came back to home. If he reached in the fashion show at 8 : 45 pm. When did he leave for his home?
Solution:
Raju reached the fashion show at 8 : 45 p.m
He stayed there for 2 hr : 30 min
He will leave the show at 10 : 75 min
= 10 hr 75 min
= 10 hrs (60 + 15) min
= 10 hrs + 1 hr 15 min
= 11 hrs 15 min
Raju will leave the show at 11 : 15 p.m

Question 16.
Mammalia travelled 4 hrs 45 min by bus and 4 hrs 45 min by train. Calculate the time she spent in travelling?
Solution:
Time travelled by bus = 4 hrs 45 min
Time travelled by train = 4 hrs 45 min
Total time she travelled = 8 hrs 90 min
= 8 hrs (60 + 30) min 8 hrs + 1 hr 30 min
Total time she travelled = 9 hr 30 min

Question 17.
David left home at 4 : 30 p.m to meet his friend. He came back after 3 hr 25 min At what time did he came back?
Solution:
Time at which David left home = 4 : 30 p.m.
Time he spent outside = 3 hr : 25 min
Time at which he came back = 7 : 55 p.m

Question 18.
Malar reached her school at 7 : 30 a.m. and left for home at 12 : 45 p.m. How long did she stay in school?
Solution:
Malar left for home at 12 : 45 p.m. = 12 : 45 hrs
She reached school at = 7 : 30 hrs
She stayed at school for = 5 : 15 hrs

Question 19.
Mathi’s date of birth was 25.07.2000. What is her age on 29.08.2017?
Solution:
Convert the given date in YY MM DD format
Age = 2017.08.29
Date of birth = 2000 .07 .25
Age = 17. 01. 04
She is 17 years one month 04 days old.

Question 20.
Find whether 2000,2018 and 2012 are leap years or not.
Solution:
2000 is a leap year (divisible by 400)
2018 is a not leap year (not divisible by 4)
2012 is a leap year (divisible by 4)

Question 21.
Find the railway time for 12 : 05 a.m, 4 : 20 pm
Solution:
12 : 05 am is 00 : 05 hrs
4 : 20 p.m. is (12 + 4) : 20 hrs = 16 : 20 hrs

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Intext Questions

Exercise 5.1

Try These (Text book Page No. 74)

Question 1.
Find the 10th term of the Fibonacci sequence
Solution:
We have the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,….
The 10th term is 55.

Question 2.
If the 11^th term of the Fibonacci sequence is 89 and 13^th term is 233 then, what is the 12^th term?
Solution:
Given 11^th term of the Fibonacci sequence is 89 and 13^th term of is 233
We know that a number in Fibonacci sequence is got by adding the previous two consecutive numbers.
11^th term + 12^th term = 13^th term
89 +12^th term =233
∴ 12^th term = 233 – 89 = 144
∴ 12^th term in Fibonacci sequence = 144.

Try this (Text book Page No.75)

Question 1.
Are two consecutive Fibonacci numbers relatively prime?
Solution:
GCD of two consecutive numbers in Fibonacci sequence is 1.
∴ They are relatively prime.

Try These (Text book Page No. 78)

Question 1.
The teacher should give oral instructions to the students to draw the geometrical figure already drawn by him / her.
i) Draw a square. In the middle of the square, draw a circle in such a way that the circle does not touch any side of the square. Divide the circle into four equal parts. Shade the bottom right part of the circle. Ask the students to show that figure drawn by them.
ii) Draw a triangle on a piece of paper. Make it crazy looking by adding some features. Give instructions to your friend to draw it exactly the same.

Solution:
i) The figure will be as drawn below.
ii) The instructions may be:

1. Draw a square of side 3 cm.
2. Draw an equilateral triangle of side 2 cm inside the square without touching the sides of the square.
3. Draw a small circle at the middle of the triangle.
4. Draw two eyes above the circle.
5. Draw a mouth below the circle.
6. Draw two legs at the base of the triangle.
7. Draw hands to the other to sides of the triangle stretching inside the square. Draw two ears above the hands.

Question 2.
Suppose your friend wants to come to your house from his/her house. Give clear instructions in order, to reach your house.
Solution:
Activity to be done by the students themselves

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Intext Questions

Question 1.
Check whether the Tree diagrams are equal or not.

Solution:
(i) Their algebraic expressions are a × (b – c) and (a × b) – (a × c)
∴ distributive property of multiplication over subtraction
∴ They are equal
(ii) Their algebraic expressions are a × (b – c) and (a × b) – c
Both are not equal [By BODMAS rule]

Question 2.
Check whether the following algebraic expressions are equal or not by using Tree diagrams.
(i) (x + y) + z and x + (y + z)
(ii) (p × q) × r and p × (q × r)
(iii) a – (b – c) and (a – b) – c
Solution:
(i) (x + y) + z and x + (y + z)
The tree diagrams are

Subtraction is not associative and the expressions are not equal.
(ii) (p × q) × r and p × (q × r)
The tree diagram is

Multiplication is associative
∴ Both expressions are equal.

(iii) a – (b – c) and (a – b) – c
Their tree diagrams are

Subtraction is not associative
∴ Both expressions are not equal.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2

Question 1.
Draw a line segment AB = 7 cm and Mark a point P on it. Draw a line perpendicular to the given line segment at P.
Solution:
Construction:
(i) Drawn a segment \(\overline{\mathrm{AB}}\) such that \(\overline{\mathrm{AB}}\) = 7 cm and took a point P anywhere on the line.
(ii) Placing the set square on the line in such a way that the vertex of its right angle coincides with P and one arm of the right angle coincides with the line AB.
(iii) Drawn a line PQ through P along the other arm of the right angle of the set square,
(iv) The line PQ is perpendicular to the line AB at P. ie PQ ⊥ AB and ∠APQ = ∠BPQ = 90°

Question 2.
Draw a line segment LM = 6.5 cm and take a point P not lying on it. Using set square construct a line perpendicular to LM through P.
Solution:
Construction:
(i) Drawn a line segment \(\overline{\mathbf{L M}}\) such that \(\overline{\mathbf{L M}}\) = 6.5 cm and marked a point P anywhere above \(\overline{\mathbf{L M}}\)
(ii) Placing one of the arms of the right angle of the set square along the line segment LM.
(iii) Sliding the set square along the line segment in such a way that the other arm of its right angle touches the point P. Draw a line along this side, passing through point P meeting \(\overline{\mathbf{L M}}\) at Q.
(iv) The line PQ is perpendicular to the line segment LM. ie LM ⊥ PQ.

Question 3.
Find the distance between the given lines using a set square at two different points on each of the pairs of lines and check whether they are parallel.

Solution:
Making the points P, Q, R, S and A, B, C, D on the given lines
PQ = RS = 0.9 cm
AB = CD = 1 cm
Distance between the two lines are equal.
They are parallel lines.

Question 4.
Draw a line segment measuring 7,8cm. Mark a point B above it at a distance of 5 cm. Through B draw a line parallel to the given line segment.
Solution:
Construction:
(i) Using a scale drawn a line segment \(\overline{\mathrm{PQ}}\) = 7.8 cm. Marked a point A on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with A and one of the edges of right angle lies along the line segment PQ. Mark a point B. Such that AB = 5 cm above the line PQ.
(iii) Placed the scale and the set square in such a way the set square is below PQ and one edge that form right angle with PQ. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and sliding the set square along the edge of the scale until the edge touches the point B. Drawn the line BC through B.
(v) Now the line BC is parallel to PQ i.e, BC || PQ.

Question 5.
Draw a line. Mark a point R below it at a distance of 5.4 cm. Through R draw a line parallel to the given line.
Solution:
(i) Using a scale drawn a line AB and marked a point Q on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with Q and one of the edges of the right angle lies along AB. Marked the Point R such that QR = 5.4 cm.
(iii) Placing the set square above AB in such a way that one of the edges that form a right angle with AB. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and slide the set square along the edge of the scale until the edge touches the point R. Drawn the line RS through R. The line RS is parallel to AB i.e RS || AB.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Additional Questions

Question 1.
Draw tree diagrams for the following questions:
(i) The number of books sold in a book fair is as follows. First day 1,82,192, Second day 1,28,194, Third day 80,520 fourth day 92,004 and the fifth day 50,020. Find the total number of the book sold.
(ii) A water purification project cost 1,82,71,000. The machinery was bought for ₹ 69,12,000. What is the amount needed to complete the project?
(iii) The number of flowers needed to arrange in a flower pot is 62. Find the number of flowers needed to arrange in 55 pots?
(iv) If the total scholarship money sanctioned for 50 students are ₹ 62,000. Find the amount that each student can get?
Solution:

Question 2.
Convert into tree diagrams
(i) (10 × 5) + (2 × 16)
(ii) (5 × 3) – (8 × 6) + 9
(iii) [9 + (3 × 2)] – [(6 × 4) + 5]
(iv) [(4 – 1) × 16] + [(16 + 9) × 3]
(v) {[(10 × 6) + 5] × [ 4 + (3 – 2)]} ÷ [4 × (2 + 9)]
(vi) 4 + [8 × 6 + {(4 × 3) – (10 ÷ 4)}]
Solution:

Question 3.
Convert the following tree diagrams into numerical expressions.

Solution:
(a) [4 × (6 + 2)] – [(4 – 2) ÷ 2]
(b) (12 × 6) + (6 ÷ 3)
(c) [4 × (10 – 2)] + [(4 + 9) × 3]

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 1 Fractions Ex 1.2

Miscellaneous Practice Problems

Question 1.
Sankari purchased 2\(\frac{1}{2}\) m cloth to stitch a long skirt and 1\(\frac{3}{4}\) m cloth to stitch blouse. If the cost is ₹ 120 per metre then find the cost of cloth purchased by her
Solution:
Cloth to stitch a long skirt = 2\(\frac{1}{2}\) m
Cloth to stitch a blouse = 1\(\frac{3}{4}\) m
Total length of the cloth = \(2 \frac{1}{2}+1 \frac{3}{4} m=2+\frac{1}{2}+1+\frac{3}{4}\) m

Cost of cloth purchased by Sankari = ₹ 510

Question 2.
From his office, a person wants to reach his house on foot which is at a distance of 5\(\frac{3}{4}\) km. If he had walked 2\(\frac{1}{2}\) km, how much distance still he has to walk to reach his house?
Solution:

Distance still he has to be walked = 3\(\frac{1}{4}\) km

Question 3.
Which is smaller? The difference between \(\frac{1}{2}\) and 3\(\frac{2}{3}\) or the sum of 1\(\frac{1}{2}\) and 2\(\frac{1}{4}\)
Solution:

Question 4.
Mangai bought 6\(\frac{3}{4}\) kg of apples. If Kalai bought 1\(\frac{1}{2}\) times as Mangai bought, then how many kilograms of apples did Kalai buy?
Solution:
Weight of apples Mangai bought = 6\(\frac{3}{4}\) kg

Weight of apples Kalai bought = 10\(\frac{1}{8}\) kg

Question 5.
The length of the staircase is 5 \(\frac{1}{2}\) m. If one step is set at \(\frac{1}{4}\) m, then how many steps will be there in the staircase?
Solution:
Length of the staircase = 5\(\frac{1}{2}\) m
Distance between each step = \(\frac{1}{4}\) m
∴ Number of steps in the staircase = \(5 \frac{1}{2} \div \frac{1}{4}=\frac{11}{2} \div \frac{1}{4}=\frac{11}{2} \times \frac{4}{1}\) = 22
There will be 22 steps in the staircase

Challenge Problems

Question 6.
By using the following clues, find who am I?
(i) Each of my numerator and denominator is a single-digit number.
(ii) The sum of my numerator and denominator is a multiple of 3.
(iii) The product of my numerator and denominator is a multiple of 4
Solution:
The numerator may be any one of!, 2, 3,4, 5, 6, 7, 8, 9 and the denominator may be any one of 1, 2,3,4, 5,6, 7,8,9. Sum of numerator and denominator is a multiple of 3.
∴ Possible proper fractions are \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{2}{4}, \frac{2}{7}, \frac{3}{6}, \frac{3}{9}, \frac{4}{5}, \frac{4}{8}, \frac{5}{7}, \frac{6}{9}\)
Also given the product of numerator and denominator is a multiple of 4.
∴ Possible fractions are \(\frac{1}{8}, \frac{2}{4}, \frac{4}{5}, \frac{4}{8}\)

Question 7.
Add the difference between 1\(\frac{1}{3}\) and 3\(\frac{1}{6}\) and the difference between 4\(\frac{1}{6}\) and 2\(\frac{1}{3}\)
Solution:

Question 8.
What fraction is to be subtracted from 9\(\frac{3}{7}\) to get 3\(\frac{1}{5}\) ?
Solution:

Question 9.
The sum of two fractions is 5\(\frac{3}{9}\). If one of the fractions is 2\(\frac{3}{4}\), find the other fraction.
Solution:

The other number is 2\(\frac{7}{12}\)

Question 10.
By what number should 3\(\frac{1}{16}\) be multiplied to get 9\(\frac{3}{16}\) ?
Solution:

The number to be multiplied is 3.

Question 11.
Complete the fifth row in the Leibnitz triangle which is based on subtraction.

Solution:

Question 12.
A painted \(\frac{3}{8}\) of the wall of which one third is painted in yellow colour. What fraction is the yellow colour of the entire wall?

Solution:

\(\frac{1}{8}\) of the wall is painted yellow.

Question 13.
A rabbit has to cover 26\(\frac{1}{4}\) m to fetch its food. If it covers 1\(\frac{3}{4}\) m in one jump, then how many jumps will it take to fetch its food?

Solution:
Total distance to be covered by the rabbit = 26\(\frac{1}{4}\)m
Distance covered in one jump = 1\(\frac{3}{4}\) m

∴ The rabbit jumps 15 times to fetch its food.

Question 14.
Look at the picture and answer the following questions :
(i) What is the distance from School to the library via bus stop?
(ii) What is the distance between school and library via Hospital?
(iii) Which is the shortest distance between (i) and (ii)?
(iv) The distance between School and Hospital is ____ times the distance between the school and Bus stop.

Solution:


The distance between school and Hospital is 6 times the distance between school and bus stop.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Intext Questions

Try These (Textbook Page No.49)

Question 1.
Arrange in ascending order
(i) C.P, M.P, Discount
(ii) M.P., S.P., Discount
Solution:
(i) Discount < C.P < M.P.
(ii) Discount < S.P < M.P

Question 2.
Which is greater S.P or M.P?
Solution:
M.P is greater than S.P

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Additional Questions

Question 1.
______ is the difference between S.P and C.P., when S.P > C.P.
Solution:
Profit

Question 2.

(a) What is the total bill amount?
(b) Which item is the costliest one?
(c) Write the name of the shop?
Solution:
(a) ₹ 4,050
(b) Tables
(c) ABC Furnitures, Madurai

Question 3.
A vegetable vendor bought a dozen drum sticks for ₹ 42. Two of them become dry. If he has to get a profit of ₹ 8. Find the S.P of each drumstick.
Solution:
12 drumsticks cost price = ₹ 42
Profit = ₹ 8
Profit = S.P – C.P
⇒ 8 = S.P – 42
⇒ S.P = 8 + 42 = ₹ 50
S.P of 10 drumsticks = ₹ 50
S.P of each drumstick = \(\frac { 50 }{ 10 }\) = ₹ 5

Question 4.
Rice is being sold at ₹ 1800 per bag of 25kg at profit of ₹ 250. Find the Cost price of the rice bag.
Solution:
S.P = ₹ 1800
Profit = ₹ 250
Profit = S.P – C.P
⇒ 250 = 1800 – C.P
⇒ C.P = 1800 – 250 = ₹ 1550

Question 5.
Mangai buys a book for ₹ 300. She wants to sell it at a profit of ₹ 50 after making a discount of ₹ 30. What is the M.P of a book?
Solution:
Profit = S.P – C.P
⇒ 50 = S.P – 300
⇒ S.P = 50 + 300 = 350
S.P = M.P – Discount
⇒ 350 = M.P – 30
⇒ M.P = 350 + 30 = ₹ 380

Question 6.
Manila buys a dress for ₹ 750. She wants to sell it at a profit of ₹ 75 on sales and she marks ₹ 900. What is the discount that she will give to her customers?
Solution:
Profit = S.P – C.P
⇒ 75 = S.P – 750
⇒ S.P = 75 + 750 = ₹ 825
S.P = M.P – Discount
⇒ 825 = 900 – Discount
⇒ Discount = 900 – 825 = ₹ 75

Question 7.
Complete the following table:

Solution:
(i) Profit = S.P – C.P = 60 – 50 where CP < SP = ₹ 10
(ii) S.P = C.P – loss = 80 – 15 = ₹ 65

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Additional Questions

Question 1.
A pair of prime numbers whose difference is 2, is called _____
Solution:
Twin primes

Question 2.
The first 4 digit number divisible by 3 ______
Solution:
1002

Question 3.
Prime triplet ______
Solution:
(3, 5, 7)

Question 4.
100 years = ______
Solution:
100 years = 1 Century

Question 5.
Loss = ____
Solution:
Loss = CP – SP

Question 6.
526 ml _____
Solution:
526 ml = 0.526 L

Question 7.
No sides are equal _____
Solution:
Scalene Triangle

Question 8.
What is the total number of primes up to 100?
Solution:
25

Question 9.
Check whether (37, 39) is a twin prime?
Solution:
No, because 39 is not a prime number.

Question 10.
Check the divisibility by 11 of 684398?
Solution:
In 684398
Sum of digits in odd places = 8 + 3 + 8 = 19
Sum of digits in even places = 6 + 4 + 9 = 19.
Difference = 19 – 19 = 0.
684398 is divisible by 11.

Question 11.
Is 53249624 is divisible by 8? How?
Solution:
In 53249624, consider the last three digits 624, which is divisible by 8.
53249624 is divisible by 8.

Question 12.
Factorise 1056
Solution:

1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11

Question 13.
Express 42 and 100 as the sum of two consecutive primes?
Solution:
42 = 19 + 23
100 = 47 + 53

Question 14.
A heap of stones can be made up into groups of 21. When made up into groups of 16, 20, 25 and 45 there are 3 stones left in each case. How many stones at least can there be in the heap?
Solution:
LCM of 16, 20, 25, 45 = 2 × 5 × 2 × 5 × 2 × 2 × 3 × 3 = 3600

The heap contain 3600 + 3 = 3603 stones at least.
3603 stones at least can there be in the heap.

Question 15.
Find the largest number of four digits exactly divisible by 12, 15, 18 and 27
Solution:
The largest number of four digits = 9999
lcm of 12, 15, 18, 27 is 540
Dividing 9999 by 540
We get 279 as the remainder


LCM = 2 × 3 × 3 × 2 × 5 × 3 = 540
Required number = 9999 – 279 = 9720

Question 16.
Find the least number which when divided by 6, 7, 8, 9 and 12 leaves the same remainder 1 in each case.
Solution:
Required number = [LCM (6, 7, 8, 9, 12)] + 1
LCM (6, 7, 8, 9, 12) = 3 × 2 × 2 × 2 × 3 × 7 = 504
Required number = 504 + 1 = 505

Question 17.
Product of two co-prime numbers is 117. Then what will be their LCM?
Solution:
We know that LCM × HCF = Product of two numbers
Also, we know that HCF of two co-primes = 1
LCM × 1 = 117
LCM = 117

Question 18.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Solution:
LCM of 2, 4, 6, 8, 10 and 12 is 120
So the bell will toll together after every 120 seconds i.e 2 minutes.
In 30 minutes, they will toll together \(\frac{30}{2}\) + 1 = 16 times.
LCM = 2 × 2 × 3 × 2 × 5 = 120