Raju

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 12 Worksheet 2 Solutions of Percentage offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 12 Worksheet 2 Solutions

Question 1.
Express the following as a fraction in the lowest term.
(a) 3%
Solution:
3% = \(\frac{3}{100}\)

(b) 35%
Solution:
35% = \(\frac{35}{100}\) (Divide 35 and 100 by HCF 5)
= \(\frac{7}{20}\)

(c) 138%
Solution:
138% = \(\frac{138}{100}\) (Divide 138 and 100 by HCF 2)
= \(\frac{69}{50}\)
= 1\(\frac{19}{50}\)

(d) 675%
Solution:
675% = \(\frac{675}{100}\) (Divide 675 and 100 by HCF 25)
= \(\frac{27}{4}\)
= 6\(\frac{3}{4}\)

(e) 80%
Solution:
80% = \(\frac{80}{100}\) (Divide 80 and 100 by HCF 20)
= \(\frac{4}{5}\)

(f) 150%
Solution:
150% = \(\frac{150}{100}\) (Divide 150 and 100 by HCF 50)
= \(\frac{3}{2}\)
= 1\(\frac{1}{2}\)

Question 2.
Express the following percentages into decimals.
(a) 0.2%
Solution:
The Decimal shifted by two places to the left when we converted percentages into decimals.
0.2% = \(\frac{0.2}{100}\) = .002

(b) 5.6%
Solution:
The Decimal shifted by two places to the left when we converted percentages into decimals.
5.6% = \(\frac{5.6}{100}\) = .056

(c) 2.75%
Solution:
The Decimal shifted by two places to the left when we converted percentages into decimals.
2.75% = \(\frac{2.75}{100}\) = .0275

(d) 3.8%
Solution:
The Decimal shifted by two places to the left when we converted percentages into decimals.
3.8% = \(\frac{3.8}{100}\) = .038

(e) 1.9%
Solution:
The Decimal shifted by two places to the left when we converted percentages into decimals.
1.9% = \(\frac{1.9}{100}\) = .019

(f) 9.5%
Solution:
The Decimal shifted by two places to the left when we converted percentages into decimals.
9.5% = \(\frac{9.5}{100}\) = .095

Question 3.
Complete the table given below. The first is done for you.

Solution:

Question 4.
Convert the following into fractions in the lowest term.
(a) \(\frac{3}{4}\)%
Solution:

(b) 16\(\frac{2}{3}\)%
Solution:

(c) 33\(\frac{1}{3}\)%
Solution:

\(\frac{1}{3}\)

(d) 7\(\frac{1}{7}\)%
Solution:

(e) 11\(\frac{1}{4}\)%
Solution:

(f) 12\(\frac{1}{2}\)%
Solution:

DAV Class 5 Maths Chapter 12 Worksheet 2 Notes

Convert percentages into fractions in the lowest term.

Example 1.
Convert 25% into a fraction.
Solution:
25% = \(\frac{25}{100}\) (Divide 25 & 100 100 by HCF 25)
= \(\frac{1}{4}\)

Convert percentages into decimal numbers.

Example 2.
Convert 35% into a decimal number.
Solution:
35% = \(\frac{35}{100}\)
= 0.35 (Decimal shifted by two places to the left)

Example 3.
Convert 343% into a decimal number.
Solution:
343% = \(\frac{343}{100}\)
= 3.43 (Decimal shifted by two places to the left)

The DAV Maths Book Class 6 Solutions and DAV Class 6 Maths Chapter 9 Worksheet 1 Solutions of Line Segments offer comprehensive answers to textbook questions.

DAV Class 6 Maths Ch 9 WS 1 Solutions

Question 1.
Compare the following pairs of line segments using divider:

(a) AB ____ CD
Answer:
AB > CD

(b) PQ ____ RS
Answer:
PQ = RS

(c) DE ____ FG
Answer:
DE < FG

Question 2.
Measure the two line segments using the:
(a) Ruler

(b) Divider

Answer:
Answer is not required. It is an activity to be done by the students themselves.

Question 3.
Construct the line segments of the following lengths using compasses:
(a) 3.1 cm
(b) 6.2 cm
(c) 7.4 cm
(d) 9.3 cm
Answer:
(a) 3.1 cm
Step 1: Draw a line l and take a point A on it.

Step 2: Open the compass to a length of 3.1 cm and draw an arc B on the line.

Taking A as centre make a line segment AB.

(b) 6.2 cm
Step 1: Draw a line r and take a point P on it.

Step 2: Open the compass upto 6.2 cm and with centre P, draw an arc Q making a line segment PQuestion

Here, PQ is the required line segment of length 6.2 cm.

(c) 7.4 cm
Step 1: Draw a line m and take any point S on it.

Step 2: Open the compass upto 7.4 cm and draw an arc T, taking S as centre making a line segment \(\overline{\mathrm{ST}}\).

Here, \(\overline{\mathrm{ST}}\) is the required line segment of length 7.4 cm.

(d) 9.3 cm
Step 1: Draw a line q and take any point A on it.

Step 2: Open the compass upto 9.3 cm and draw an arc B, taking A as centre making a line segment \(\overline{\mathrm{AB}}\).

Here, \(\overline{\mathrm{AB}}\) is the required line segment of length 9.3 cm.

Question 4.
Construct the line segments equal in lengths to the segments given below:

Answer:
Step 1: Draw a line r and take any point P on it.

Step 2: Open the compass and put in two pin point on the two ends of the given line segments A and B.
Step 3: Without disturbing the arms of the compass put its one pin on P and mark another arc at Q with its other arm pin making \(\overline{\mathrm{PQ}}\) line segment.

Here, PQ is the required line segment.

Answer:
Step 1: Draw a line s and take a point A on it.

Step 2: Open the compass and put its two pin points on the two ends of the given line segment, P and Question
Step 3: Without disturbing the arms of the compass put its one pin on A and mark another arc at B with its other pin making AB line segment.

Question 5.
Construct a line segment AB of length 7.5 cm. From this line segment, cut off a line segment AC of length 3.2 cm. Measure the length of the remaining line segment CB.
Answer:

Step 1: Draw line and take any point A on it.
Step 2: Open the compass upto 7.5 cm and put one pin point on A and mark an arc which cuts the line at B. So AB = 7.5 cm.
Step 3: Now again put the pin point of the compass on A and cut an arc equal to 3.2 cm and mark it as C.
Step 4:

Measure the length of CB with the help of ruler or compass which is formed to be 4.3 cm.
Hence the length of the remaining segment = 4.3 cm.

Question 6.
If AB = 3.8 cm and CD = 2.5 cm, construct a line segment whose length is the sum of the lengths of these line segments and measure it.
Answer:

Step 1: Draw a line and take any point P on it.
Step 2: Open the compass and measure the length AB.
Step 3: Put the pin-point of the compass at P and mark an arc at Q so that PQ = 3.8 cm.
Step 4: Put the pin-point of the compass at Q and mark an arc at R on the line at a distance of 2.5 cm.

Step 5: \(\overline{\mathrm{PR}}=\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}\) = 3.8 cm + 2.5 cm = 6.3 cm Hence the required length is 6.3 cm.

Question 7.
If PQ = 5.4 cm and RS = 2.7 cm, construct a line segment whose length is the difference of the length of these line segments.
Answer:

Step 1 : Draw a line l and take a point A on it.

Step 2: Using compass draw a line segment AB on line l equal to the length of given line PQ starting from point A.

Step 3: Now starting from point B (to the left) construct a line segment BC equal to the length of the given line segments RS with the help of compass.
Now AC is the required line segment.
\(\overline{\mathrm{AC}}=\overline{\mathrm{AB}}-\overline{\mathrm{CB}}\) = 5.4 cm – 2.7 cm = 2.7 cm

Question 8.
If AB = 4.5 cm, CD = 2.3 cm, then construct the following line segments:
(a) 2 CD
Answer:
2 CD
Step 1: Draw any line l and take any point R on it.
Step 2: Using compass starting from P construct a line segment PE equal to the length of CD i.e. 2.3 cm.

Step 3: Starting from E draw a line segment EQ equal to the length of CD i.e.

Hence PQ = PE + EQ
= 2.3 cm + 2.3 cm = 4.6 cm

(b) 3 AB
Answer:
3 AB
Step 1: Draw any line l and take any point H on it.

Step 2: Using compass and starting from H, construct a line segment HT equal to the length of AB i.e. 4.5 cm.

Step 3: Now starting from T (to the right) draw a line segment TR equal to the length of AB i.e. 4.5 cm

Step 4: Again starting from R(to the right) draw a line segment RS equal to the length of AB i.e. 4.5 cm.
Hence the required line segement is HS
where HS = HT + TR + RS
= 4.5 cm + 4.5 cm + 4.5 cm = 13.5 cm

(c) AB – CD
Answer:
AB – CD
Step 1: Draw any line l and take a point P on it.

Step 2: Using compass draw a line segment PQ on line l equal to the length of AB i.e. 4.5.

Step 3: Now with the help of compass starting from Q (to the left), construct a line segment QS on l equal to the length CD i.e. 2.3 cm.

Here PS is the line segment equal to the difference of AB – CD = 4.5 cm – 2.3 cm = 2.2 cm.

(d) AB + 2CD
Answer:
AB + 2 CD
Step 1 : Draw any line l and take a point P on it.

Step 2: Using compass, starting from P, draw a line segment PQ on l equal to the length of AB i.e. 4.5 cm.

Step 3: Again using compass, starting from Q (to the right) construct a line segment QS on l equal to double the length of CD i.e. 2 x 2.3 = 4.6 cm.
Here PS is the required line segment
where PS = AB + 2 CD
= 4.5 + 2 × 2.3 = 4.5 + 4.6 = 9.1 cm

DAV Class 6 Maths Chapter 9 Value Based Questions

Question 1.
A group of four children were given a project where they were asked to bring some eco-friendly fibres. Neha forgot to bring the fibre and was upset. Two of her friends Sneha and Swati shared their fibres with Neha. Sneha gave 16.5 cm of fibre and Swati gave 8.5 cm of fibre to her.
(a) How much more fibre did Sneha give to Neha than Swati? Construct the line segment representing this length.
(b) What value is exhibited by the children?
Answer:
(a) Sneha gave 16.5 cm of fibre and Swati gave 8.5 cm of fibre

Difference between lengths of fibres of Sneha and Swati is
= 16.5 cm – 8.5 cm = 8 cm
So, Sneha gave 8.0 cm more fibre to Neha than Swati.

(b) The value of cooperation is exhibited by the children.

DAV Class 6 Maths Chapter 9 Worksheet 1 Notes

Line segment is a portion of a line having two end points. It has definite length. It is represented by \(\overline{\mathrm{AB}}\).

Example: AB is a segment. A and B are its two ends.

Comparison of line segments:
The two line segments can be compared by measuring their lengths with the help of divider.

Here, we can observe that \(\overline{\mathrm{CD}}\) is longer than \(\overline{\mathrm{AB}}\) or \(\overline{\mathrm{AB}}\) is shorter than \(\overline{\mathrm{CD}}\).

Length of the line segment can be measured with the help of ruler or scale.

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 12 Worksheet 1 Solutions of Percentage offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 12 Worksheet 1 Solutions

Question 1.
Fill in the boxes.
(a)

Solution:
\(\frac{55}{100}\) = 55

(b)

Solution:
\(\frac{4}{100}\) = 4

(c)

Solution:
\(\frac{93}{100}\) = 93

(d)

Solution:
\(\frac{125}{100}\) = 125

(e)

Solution:
\(\frac{41}{100}\) = 41%

(f)

Solution:
\(\frac{11}{100}\) = 11%

(g)

Solution:
\(\frac{11}{100}\) = 11%

(h)

Solution:
\(\frac{132}{100}\) = 132%

Question 2.
Solve the following questions.
(a) In a box, there are 100 chalks. Out of these 65 chalks are blue and the rest are pink in colour. Complete the following statements.

Solution:
Fraction of blue chalks = \(\frac{65}{100}\) = 65%
Number of pink chalks = 35
Fraction of pink chalks = \(\frac{35}{100}\) = 35%

(b) Anuj had 100 stamps. Out of these, 32 are foreign stamps and the rest are Indian stamps. Find the percentage of Indian stamps that Anuj had.
Solution:
% of Indian stamps = ?
Foreign stamps = 32
No, of Indian stamps = 100 – 32 = 68
% of Indian stamps = 68%

(c) Here is a big square divided into 100 small squares. Five squares are shaded black to show 5%.

Now shade the squares to represent the following percent.

Solution:

DAV Class 5 Maths Chapter 12 Worksheet 1 Notes

Fractions whose denominator is equal to hundred is called percentage.
e.g., \(\frac{52}{100}\), \(\frac{33}{100}\)

Cent means 100. ‘Percent’ means ‘for every 100’.

By two methods a fraction can be converted into a percentage:

• By converting a fraction into an equivalent fraction with a denominator of 100.
  e.g., \(\frac{7}{25}=\frac{7 \times 4}{25 \times 4}=\frac{28}{100}\) = 28%
• By multiplying the fraction by 100.
  e.g., \(\frac{7}{25}\) = \(\frac{7}{25}\) × 100 = \(\frac{700}{25}\) = 28%

By two methods a decimal can be converted into a percentage:

• By converting decimals into equivalent fractions with denominator 100.
  e.g., 0.4 = \(\frac{4 \times 10}{10 \times 10}=\frac{40}{100}\) = 40%
• By multiplying decimal by 100.
  e.g., 0.4 = (0.4 × 100)% = 40%

By two methods a whole number can be converted into a percentage:

• By converting whole numbers into equivalent fractions with denominator 100.
  e.g., 8 = \(\frac{8 \times 100}{100}\) = 800%
• By multiplying the whole number by 100.
  e.g., 8 = (8 × 100)% = 800%

The DAV Maths Book Class 6 Solutions and DAV Class 6 Maths Chapter 10 Brain Teasers Solutions of Angles offer comprehensive answers to textbook questions.

DAV Class 6 Maths Ch 10 Brain Teasers Solutions

Question 1.
A. Tick (✓) the correct answer.
What kind of angle do you get when you open any two adjacent fingers of your hand?
(i) right angle
(ii) obtuse angle
(iii) acute angle
(iv) straight angle
Answer:
(ii) acute angle
When we open any two adjacent fingers of our hand, we get an acute angle between them.

(b) An angle measuring 240° is:
(i) obtuse angle
(ii) acute angle
(iii) supplementary angle
(iv) reflex angle
Answer:
(iv) reflex angle
The angle which is more than 180° and less than 360° is called reflex angle.

(c) An supplement of an angle of 75° is
(i) 95°
(ii) 105°
(iii) 100°
(iv) 115°
Answer:
(iii) 105°
When the sum of two angles is equal to 180°, then they are called supplementary angles to each other.

(d) The Complement of an angle of 55° is
(i) 45°
(ii) 125°
(iii) 35°
(iv) 135°
Answer:
(iii) 35°
When the sum of two angles is equal to 90°, then they are called complementary angles to each other.

(e) How many right angles make one complete angle?
(i) 2
(ii) 3
(iii) 1
(iv) 4
Answer:
(iv) 4
One complete angle has 360° i.e. 4 x 90°. This means that complete angle is made up of 4 right angles.

B. Answer the following questions.
(a) How many degrees are there in the angle made by the hour hand and the minute hand of a clock, when it is 9 o’clock?
Answer:
90°.
When a clock strikes 9, its hour hand and minute hand stand vertical to each other. At 9 o’clock, hour hand is at 9 indicator while the minute hand is at 12 indicator and they form a right angle.

(b) Name of points in the exterior of ∠BCA but, in the interior of ∠ACD.

Answer:
Point S and point P.
Points S, P and R are in the exterior of ∠BCA but only points S and P are in the interior of ∠ACD. Point R is in the exterior of ∠ACD.

(c) State the kind of angle between the directions South-East and North.
Answer:
Obsute angle.

The kind of angle between South-East and North is 135° which is greater than 90° and smaller than 180°. So, it forms an obtuse angle.

(d) If the measure of one angle of a linear pair is of 85°, find the other angle. Find x is the given figure.
Answer:
95°
If the two adjacent angles combine together to form an angle of 180°, then the adjacent angles is called a linear pairs.

(e) Find x is the given figure.

Answer:
320°
Here, x + 40° = 360°
⇒ x = 360° – 40° = 320°

Question 2.
How many degrees are there in
(a) \(\frac{2}{5}\) right angle?
Answer:
\(\frac{2}{5}\) right angle

(b) \(\frac{5}{6}\) straight angle?
Answer:
\(\frac{5}{6}\) straight angle

(c) \(\frac{2}{5}\) complete angle?
Answer:
\(\frac{2}{5}\) complete angle

Question 3.
Name the given angle in six different ways.

Answer:
∠a, ∠YXZ, ∠YXN, ∠MXN, ∠MXZ and ∠ZXY.

Question 4.
Complete the following using given figure:

(a) Vertex of ∠1 is .
(b) Another name for ∠2 is .
(c) Another name for ∠1 + ∠2 is .
(d) Common arm of ∠1 and ∠2 is .
(e) ∠1 and ∠2 are angles.
Answer:
(a) A
(b) ∠CAD
(c) ∠BAD
(d) AC
(e) adjacent

Question 5.
While doing drill, Rohan was asked to about turn. After two minutes, he was again asked to about turn. How many degrees did Rohan turn in the activity?
Answer:
In first turn, he moved through 180° and in second turn he again moved through 180°.
∴ Total angle moved in this activity = 180° + 180° = 360°.

Question 6.
Answer in ‘Yes’ or ‘No’.
(a) Can 2 acute angles be supplementary?
Answer:
No

(b) Can 2 obtuse angles form a linear pair?
Answer:
No

(c) Can linear pair be supplementary?
Answer:
Yes

(d) Can 2 acute angles be complementry?
Answer:
Yes

(e) Can vertically opposite angles be adjacent?
Answer:
No

Question 7.
Look at the given figure and answer the following:

(a) Sum of ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7, ∠8.
(b) ∠3 = ____________
(Vertical angles).
(c) ∠5 = ____________
(Vertical angles).
(d) ∠4 and are adjacent angles.
(e) ∠1 + ∠2 + ∠3 + ∠5 + ∠6 + ∠7 + ∠8 = 320°.
What is the measure of ∠4?
Answer:
(a) ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
(b) ∠3 = ∠7
(c) ∠5 = ∠1
(d) ∠4 and ∠5; ∠4 and ∠3
(e) ∠4 = 360° – 320° = 40°

Question 8.
Fill in the blanks:
(а) Sum of the angles forming a linear pair is always equal to ____________ degrees.
Answer:
180°

(b) A reflex angle lies between ____________ and ____________.
Answer:
180°; 360°

(c) An angle whose measure is equal to 360° is called a angle.
Answer:
Complete

(d) The angle turned by the minute hand of a clock from the digit 2 to 8 is a angle and measures ____________ degrees.
Answer:
straight, 180°

(e) The complement of 0.5° is ____________ degrees.
Answer:
89.5°

DAV Class 6 Maths Chapter 10 HOTS

Question 1.
Reena was standing facing North-West direction. She turns by 90° in anti-clockwise direction. Again she turns by 180° in anti-clockwise direction. Now she turns 45° in clockwise direction. Which direction is Reena facing now?
Answer:

Reena was standing facing North-West. She turns by 90° in anti-clockwise direction and she is facing South-West now. After that she turns by 180° in anti-clockwise direction and she is facing North-East direction. On turnning 45° in clockwise direction, she is facing East direction now.

Question 2.
Find the complement of the supplement of twice of an angle of 60°.
Answer:
Twice of angle
60° = 2 x 60° = 120°
Thus, supplementary angle of 120° = 180° – 120° = 60°
Now, the complementary of 60° = 90° – 60° = 30°.

DAV Class 6 Maths Chapter 10 Enrichment Questions

Question 1.
Observe the following letters of English alphabet.

We find a linear pair of angles in letter E and vertically opposite angles in letter X.
Now, make a list of letters in English alphabet which have linear pair and vertically opposite angles. Colour linear pair with green and vertically opposite angles with orange.
Answer:
List of English Alphabet:

All these English Alphabet form linear pairs. Only X forms vertically opposite angles.

Additional Questions

Question 1.
Observe the given figure and answer the following:

(a) The points in the interior of the angle.
(b) The points in the exterior of the angle.
(c) Points in the angular region.
Answer:
(a) The points in the interior of angle is P, Q and S.
(b) The points in the exterior of the angle are T, U, V and R.
(c) Angular points are Y, X, O, M, N, S, Q and P.

Question 2.
Name the following angles:

Answer:
∠O

Answer:
∠ABC

Answer:
∠PQR

Answer:
∠b

Answer:
∠MNO

Answer:
∠STU

Question 3.
Draw the following angles using pencil and scale only:
(a) Right angle
Answer:

(b) Straight angle
Answer:

(c) Obtuse angle
Answer:

(d) Reflex angle
Answer:

(e) Acute angle
Answer:

Question 4.
Fill in the blanks:
(a) Acute angle is always less than ____________.
Answer:
90°

(b ) Right angle is always equal to ____________.
Answer:
90°

(c) Measure of straight angle is ____________.
Answer:
180°

(d) Reflex angle is always greater than ____________.
Answer:
180°

(e) Obtuse angle is always less than ____________.
Answer:
180°

Question 5.
Write the kind of angles marked in the given figure:
Answer:

∠1 is acute angle
∠2 is right angle
∠3 is obtuse angle
∠4 is reflex angle
∠5 is acute angle
∠6 is obtuse angle
∠7 is acute angle
∠8 is acute angle

Question 6.
Identify the complementary and supplementary angles from the following:
(a) 33°, 57°
(b) 120°, 60°
(c) 22\(\frac{1}{2}\), 67\(\frac{1}{2}\)
(d) 0°, 90°
(e) 25°, 65°
(f) 89°, 1°
Answer:
(a) 33° + 57° = 90°
they are complementary angles

(b) 120° + 60° = 180°
they are supplementary angles

(c) 22\(\frac{1}{2}\) + 67\(\frac{1}{2}\) = 90°
they are complementary angles

(d) 0° + 90° = 90°
they are complementary angles

(e) 25° + 65° = 90°
they are complementary angles

(f) 89° + 1° = 90°
they are complementary angles

Question 7.
Write the supplement of the given angles:
(a) 110°
Answer:
180° – 110° = 70°

(b) 120°
Answer:
180° – 120° = 60°

(c) 105°
Answer:
180° – 105° = 75°

(d) 67\(\frac{1}{2}\)°
Answer:
180° – 67\(\frac{1}{2}\)° = 112\(\frac{1}{2}\)°

(e) 38.5°
Answer:
180° – 38.5° = 141.5°

(f) 25°
Answer:
180° – 25° = 155°

(g) 65°
Answer:
180° – 65° = 115°

(h) 0°
Answer:
180° – 0° = 180°

(i) 1°
Answer:
180° – 1° = 179°

(j) 63°
Answer:
180° – 63° = 117°

Question 8.
Write true or false for the given statements:
(a) Reflex angle is always more than 90°.
Answer:
False

(b) The sum of two angles of linear pair is 180°.
Answer:
True

(c) The sum of supplementary angles is 180°.
Answer:
True

(d) Vertically opposite angles are always equal.
Answer:
True

(e) Two adjacent angles have a common arm.
Answer:
True

(f) The measure of a complete angle is 360°.
Answer:

(g) The complement of 55° is 135°.
Answer:
False

(h) Angles of linear pair are not straight angles.
Answer:
False

Question 9.
Observe the given figure and find the value of x.

Answer:
(3x + 5)° + (2x + 15)° = 180° (Linear pair)
⇒ 3x + 5° + 2x + 15° = 180°
⇒ 5x + 20° = 180°
⇒ 5x + 20° – 20° = 180° – 20°
⇒ 5x = 160°
⇒ 5x ÷ 5 = 160° ÷ 5
⇒ x = 32°

Question 10.
Find the value of x in the following figures:

Answer:
x + 50° = 180° (Linear pair)
∴ x = 180° – 50° = 130°

Answer:
x = 110°(Vertically opposite angles)

Answer:
x + 35° = 90° (Complementary angles)
∴ x = 90° – 35° = 55°

Answer:
x + 180° = 360° (Complete angles)
∴ x = 360° – 180° = 180°

The DAV Maths Book Class 6 Solutions and DAV Class 6 Maths Chapter 10 Worksheet 2 Solutions of Angles offer comprehensive answers to textbook questions.

DAV Class 6 Maths Ch 10 WS 2 Solutions

Question 1.
In which of the following are ∠1 and ∠2 adjacent angles?

Answer:
∠1 and ∠2 are not adjacent angles as they have no common arm.

Answer:
∠1 and ∠2 are adjacent angles as they have common arm and common vertex.

Answer:
∠1 and ∠2 are not adjacent angles as the non-common arms are not on opposite sides.

Answer:
∠1 and ∠2 are adjacent angles as they have common arm and common vertex.

Answer:
∠1 and ∠2 are not adjacent angles as the non-common are not on the opposite sides.

Answer:
∠a and ∠b are not adjacent angles as they have no common arm and no common vertex.

Question 2.
Write down three pairs of adjacent angles in the following figure.

Answer:
∠CAB and ∠DAC
∠EAD and ∠DAC
∠EAC and ∠CAB are the pairs of adjacent angles.

Question 3.
In the following figure, are ∠a and ∠b adjacent? Why or Why not?

Answer:
∠a and ∠b are not adjacent angles as they have no common arm and no common vertex.

Question 4.
Identify the following pairs as complementary and supplementary angles:
(a) 130°, 50°
Answer:
130° + 50° = 180°, they are supplementary angles.

(b) 55°, 35
Answer:
55° + 35° = 90°, they are complementary angles.

(c) 95°, 85°
Answer:
95° + 85° = 180°, they are supplementary angles.

(d) 60.5°, 29.5°
Answer:
60.5° + 29.5° = 90°, they are complementary angles.

(e) 74°, 106
Answer:
74° +106° = 180°, they are supplementary angles.

(f) 135\(\frac{1^{\circ}}{2}\) + 44 \(\frac{1^{\circ}}{2}\)
Answer:
135\(\frac{1^{\circ}}{2}\) + 44 \(\frac{1^{\circ}}{2}\) = 180°, they are supplementary angles.

(g) 33°, 57
Answer:
33° + 57° = 90°, they are complementary angles.

(h) 0°, 90°
Answer:
0° + 90° = 90°, they are complementary angles.

Question 5.
Write the complement of the following
(a) 20°
Answer:
90° – 20° = 70°, so 70° is the complement of 20°.

(b) 55°
Answer:
90° – 55° = 35°, so 35° is the complement of 55°.

(c) 71°
Answer:
90° – 71° = 19°, so 19° is the complement of 71°.

(d) 6°
Answer:
90° – 6° = 84°, so, 84° is the complement of 6°.

(e) 59\(\frac{1^{\circ}}{2}\)
Answer:
90° – 59\(\frac{1^{\circ}}{2}\) = 30\(\frac{1^{\circ}}{2}\), so 30\(\frac{1^{\circ}}{2}\) is the complement of 59\(\frac{1^{\circ}}{2}\)

(f) 60.5°
Answer:
90° – 60.5° = 29.5° so 29.5° is the complement of 60.5°.

Question 6.
Write the supplement of the following angles.
(a) 110°
Answer:
180° – 110° = 70°, so, 70° is the supplement of 110°.

(b) 82°
Answer:
180° – 82° = 98°, so 98° is the supplement of 82°.

(c) 31 .5°
Answer:
180° – 31.5° = 148.5°, so 148.S »° is the supplement of 31.5 °

(d) 180°
Answer:
180° – 180° = 0°, so 0° is the supplement of 180°.

(e) 96.5°
Answer:
180° – 96.5° = 83.5°, i so 83.5 ° is the supplement of 96.5°

(f) 100°
Answer:
180° – 100° = 80°, so 80° is the supplement of 100°.

Question 7.
Draw the following using scale and pencil only. Also name them.
(a) One pair of adjacent angles
Answer:
∠1 and ∠2 are adjacent angles

(b) One linear pair
Answer:
∠1 and ∠2 are the linear pairs

(c) One pair of complementary adjacent angles
Answer:
∠1 and ∠2 are complementary angles

(d) One pair of supplementary non- adjacent angles
Answer:
∠x and ∠y are supplementary angles

(e) Vertically opposite angles
Answer:
∠x, ∠y and ∠p and ∠q are pairs of vertically opposite angles.

Question 8.
How many linear pairs are there in the figure? Name any four pairs.

Answer:
There are 8 linear pairs. ∠1 and ∠2, ∠2 and ∠3, ∠3 and ∠4, ∠5 and ∠6 are the four linear pairs.

Question 9.
One of the angles of a linear pair is obtuse. What kind of angle is the other?
Answer:

As the sum of the angles of a linear pair is 180° and if one angle is obtuse then other angle will be acute.

Question 10.
Write the pairs of vertically opposite angles in the following figures:

Answer:
∠a and ∠c, ∠b and ∠d are the pairs of vertically opposite angles.

Answer:
∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7, ∠4 and ∠8 are the vertically opposite angles.

Question 11.
Which of the following cannot be a pair of complementary angles?
(a) 2 acute angles
(b) 2 obtuse angles
(c) 2 right angles
Answer:
(a) 2 acute angles can form a pair of complementary angles.
(b) 2 obtuse angles cannot form a pair of complementary angles.
(c) 2 right angles cannot form a pair of complementary angles.

Question 12.
Look carefully at the figure and write down:

(a) One pair of complementary angles
(b) Two pairs of supplementary angles.
Answer:
(a) ∠AOB and ∠COB is the pair of complementary angles.
(b) ∠AOB and ∠BOD, ∠COD and ∠AOC are the pairs of supplementary angles.

Question 13.
Answer the questions from the given figure:

(a) Are ∠1 and ∠2 vertically opposite angles?
(b) Are ∠3 and ∠6 vertically opposite angles?
(c) Are ∠1 and ∠3 linear pairs?
(d) Are ∠6 and ∠4 adjacent angles?
Answer:
(a) Yes, ∠1 and ∠2 opposite angles.
(b) No, ∠3 and ∠6 are not vertically opposite angles.
(c) ∠1 and ∠3 are not linear pairs.
(d) Yes, ∠6 and ∠4 are adjacent angles.

Question 14.
Write ‘True’ or ‘False’ for the following statements:
(a) Two obtuse angles can be supplementary.
Answer:
False

(b) Supplement of a right angle is a right angle.
Answer:
True

(c) Adjacent supplementary angles form a linear pair.
Answer:
True

(d) Complementary angles are always adjacent.
Answer:
False

(e) Vertically opposite angles have a common vertex and a common arm.
Answer:
False

(f) There are 2 pairs of vertically opposite angles in the plus sign.
Answer:
True

(g) Adjacent angles can be complementary.
Answer:
True

Question 15.
Find the measure of ∠x in the following figures. Give reason also.

Answer:
x + 40° = 180° (Linear pairs)
180° – 40° = 140°

Answer:
x + 47° = 90° (Complementary angles)
x = 90° – 47° = 43°

Answer:
x = 110° (Vertically opposite angles)

Answer:
x + 60° = 180° (Supplementary angles)
x = 180° – 60° = 120°

DAV Class 6 Maths Chapter 10 Value Based Questions

Question 1.
We all know that Yoga is good for our body, mind and soul. June 21 has been declared as the ‘International Day of Yoga’.
(а) In the above yoga posture, write the type of ∠1, ∠2 and ∠3.
(b) Discuss the benefits of yoga.
Answer:
(a) ∠1 is an obtuse angle.
∠2 is a right angle.
∠3 is also an obtuse angle.

(b) Yoga energizes our mind, body and soul. In this way, it keeps us fit and healthy.

DAV Class 6 Maths Chapter 10 Worksheet 2 Notes

1. Adjacent angles: Two angles are said to be adjacent if:

• They have common vertex
• They have common arm.
• The two non-common arms are on the opposite sides of the common arms ∠QOS and ∠ROS are adjacent angles.

2. Linear pairs: Two adjacent angles whose sum is equal to 180° are called linear pairs.

∠POR + ∠QOR = 180°
∴ ∠POR and ∠QOR from a linear pair

3. Vertically opposite angles: Two angles obtained by intersections the two lines having no common arms but a common vertex are called vertically opposite angles.

∠AOD and ∠COB are vertically opposite angles.
∠AOC and ∠BOD are vertically opposite angles.
Vertically opposite angles are equal to each other.

4. Complementary angles: The two angles whose sum is 90° is called complementary angles.

∠AOB + ∠COB = 90°
∴ ∠AOB and ∠COB are complementary angles.

5. Supplementary angles: The two angles whose sum is 180° are called supplementary angles.

∠AOB + ∠PQR = 70° + 110° = 180°
∴ ∠AOB and ∠PQR are supplementary angles.

The DAV Maths Book Class 6 Solutions and DAV Class 6 Maths Chapter 10 Worksheet 1 Solutions of Angles offer comprehensive answers to textbook questions.

DAV Class 6 Maths Ch 10 WS 1 Solutions

Question 1.
Name the following angles:

Answer:
(a) ∠YXZ
(b) ∠5
(c) ∠a
(d) ∠P
(e) ∠RPQ
(f) ∠3

Question 2.
Draw the following angles as per their types (use scale and pencil only):
(a) ∠LMN (Reflex angle)
Answer:

(b) ∠y (Obtuse angle)
Answer:

(c) ∠XYZ (complete angle)
Answer:

(d) ∠3 (Reflex angle)
Answer:

Question 3.
Arrange the measure of the following angles in ascending order. Obtuse angle, Straight angle, Zero angle, Right angle, Reflex angle, Complete angle, Acute angle.
Answer:
Zero angle, Acute angle, Right angle, Obtuse angle, Straight angle, Reflex angle, Complete angle.

Question 4.
Look at the figure and answer the questions:

(a) Name the points in the interior of ∠LMN
(b) Name the points which lie in the angular region of ∠LMN
(c) Name the points which lie in the exterior of ∠LMN.
Answer:
(a) Points B, Y and C are in the interior of ∠LMN.
(b) Points L, M, N, B, Y and C are in the angular region.
(c) Points A, X and P lie in the exterior of ∠LMN.

Question 5.
Write another name for the following angles in the figure:

(a) ∠1
(b) ∠AFG
(c) ∠2
(d) ∠5
(e) ∠BGD
Answer:
(a) ∠1 = ∠ABG
(b) ∠AFG = ∠4
(c) ∠2 = ∠GED
(d) ∠5 = ∠EGD
(e) ∠BGD = ∠3

Question 6.
Look at the figure and answer the following questions:

(a) Is the point P in the exterior of ∠SOR?
(b) Is the point R in the interior of ∠SOR?
(c) Is the point Q in the interior of ∠ROP?
(d) Is the point S in the exterior of ∠POQ?
(e) Name the point which is in the interior of ∠POR but in the exterior of ∠SOR.
Answer:
(a) Yes, it is in the exterior of ∠SOR.
(b) No, R is not in the interior of ∠SOR.
(c) Yes, Q is in the interior of ∠ROP.
(d) Yes, S is in the exterior of ∠POQ.
(e) Q is in the interior of ∠POR points P and Q are in the exterior of ∠SOR.

Question 7.
Write the kind of angle formed between the following directions:
(a) North and South
Answer:

∠NOS = 180°
So, it is a straight angle.

(b) North and North-West
Answer:

∠NOP = 45°
∠NOP is an acute angle

(c) South and East
Answer:

∠SOE = 90°
So, it is right angle.

(d) East and West
Answer:

∠EOW = 180°
So, it is a straight angle.

Question 8.
Fill in the following blanks:
(a) A reflex angle measures more than _________ and less than _________.
Answer:
180°, 360°

(b) One complete angle = _________ right angles.
Answer:
4

(c) The minute hand turns through _________ degrees from the digit 2 to 11 on a clock.
Answer:
270

(d) An angle of measure 345.5° is a _________ angle.
Answer:
Reflex

DAV Class 6 Maths Chapter 10 Worksheet 1 Notes

Look at the figure, B is the vertex of ∠ABC, BA and BC are the arms of ∠ABC. Unit of an angle is ‘degree’. Angle is measured by ‘protractor’.

Types of Angles
1. Acute angle: The angle whose measure is less than 90° is called an acute angle.
2. Obtuse angle: The angle whose measure is more than 90° but less than 180° is called obtuse angle.
3. Right angle: The angle whose measure is equal to 90° is called right angle.

4. Straight angle: The angle whose measure is equal to 180° is called straight angle.
5. Complete angle: The angle whose measure is 360° is called complete angle.
6. Reflex angle: The angle whose measure is more than 180° but less than 360° is called reflex angle.

7. Zero angle: The angle whose measure is equal to O’ is called zero angle.

Interior and exterior of angles:
The region falling between the arms of the angle is called the interior of the angle and the region which does not fall between the arms of the angle is called exterior of the angle.

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 11 Brain Teasers Solutions of Profit and Loss offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 11 Brain Teasers Solutions

Question 1.
Tick (✓) the correct answer.
(a) If selling price = ₹ 6,085, profit = ₹ 1,289 then cost price is-
(i) ₹ 7,374
(ii) ₹ 7,264
(iii) ₹ 4,796
(iv) ₹ 4,804
Solution:
S.P. = ₹ 6,085
Profit = ₹ 1,289
C.P. = S.P. – Profit
= ₹ 6,085 – ₹ 1,289
= ₹ 4,796
Option (iii)

(b) If there is a loss in a transaction, which of the following is true?
(i) S.P. > C.P.
(ii) S.P. < C.P.
(iii) S.P. = C.P.
(iv) None of these
Solution:
S.P. < C.P.
i.e. Option (ii)

(c) If cost price = ₹ 4,273, profit = ₹ 729 then, selling price is-
(i) ₹ 4,996
(ii) ₹ 5,002
(iii) ₹ 4,902
(iv) ₹ 5,992
Solution:
C.P. = ₹ 4,273
Profit = ₹ 729
S.P. = ₹ 4,273 + ₹ 729 = ₹ 5,002
Option (ii)

(d) Which of the following is true?
(i) S.P. + L = C.P.
(ii) S.P. -L = C.P.
(iii) C.P. – G = S.P.
(iv) S.P. + G = C.P.
Solution:
S.P. + L = C.P.
i.e. Option (i)

(e) If C.P. = ₹ 2,500, Transport Charges = ₹ 75, Gain = ₹ 225 the S.P. is-
(i) ₹ 2,350
(ii) ₹ 2,800
(iii) ₹ 2,725
(iv) ₹ 2,275
Solution:
C.P. = ₹ 2,500
Transport Charges = + ₹ 75
Actual C.P. = ₹ 2,575
Gain = ₹ 225
so S.P. > C.P.
S.P. = ₹ 2575 + ₹ 225 = ₹ 2,800
Option (ii)

Question 2.
Fill in the blanks.

Solution:
(a) C.P. = ₹ 750
Transport Charges = ₹ 30
Actual C.P. = ₹ 750 + ₹ 30 = ₹ 780
S.P. = ₹ 900
Since S.P. > C.P., Profit will occur
Profit = SP. – C.P.
= ₹ 900 – ₹ 780
= ₹ 120

(b) C.P. = ₹ 10,200
Transport Charges = ₹ 75
Actual C.P. = ₹ 10,200 + ₹ 75 = ₹ 10,275
S.P. = ₹ 10,075
Since C.P. > S.P., Loss will occur
Loss = C.P. – S.P.
= ₹ 10,275 – ₹ 10,075
= ₹ 200

Question 3.
Fill in the blanks.

Solution:
(a) Cost Price = ₹ 410
Selling price = ?
Profit = ₹ 75
Selling price = C.P. + Profit
= ₹ 410 + ₹ 75
= ₹ 485

(b) Cost Price = ₹ 3,400
Loss = ₹ 250
Selling Price = C.P. – Loss
= ₹ 3,400 – ₹ 250
= ₹ 3,150

(c) Cost Price = ₹ 70,800
Selling Price = ₹ 75,000
Since S.P. > C.P., Profit will occur
Profit = ₹ 75,000 – ₹ 70,800 = ₹ 4,200

(d) Selling Price = ₹ 890
Profit = ₹ 110
Cost Price = S.P. – Profit
= ₹ 890 – ₹ 110
= ₹ 780

(e) Selling Price = ₹ 4,850
Loss = ₹ 1,050
C.P. = S.P. + Loss
= ₹ 4,850 + ₹ 1,050
= ₹ 5,900

Question 4.
Sushil bought six dozen bananas at ₹ 35 per dozen. He sold all the bananas for ₹ 300. Did he have a gain or loss and how much?
Solution:
C.P. of 1 dozen bananas = ₹ 35
C.P. of 6 dozen bananas = ₹ 210
S.P. of 6 dozen bananas = ₹ 300
Since S.P. > C.P. Sushil makes a profit
Profit = ₹ 300 – ₹ 210 = ₹ 90
A profit of ₹ 90 will occur.

Question 5.
A man purchased seven stools and eight tables. The cost of one stool was ₹ 150 and the cost of one table was ₹ 280. If he sold all the stools and tables for ₹ 3100, find his gain or loss.
Solution:
Cost of 1 stool = ₹ 150
Cost of 7 stools = 150 × 7 = ₹ 1050
Cost of 1 table = ₹ 280
Cost of 8 tables = 8 × ₹ 280 = ₹ 2240
Total cost price of 7 stool & 8 tables = ₹ 1050 + ₹ 2240 = ₹ 3,290
S.P. of all the stools and tables = ₹ 3,100
Since S.P. < C.P. So loss will occur to man.
Loss = C.P. – S.P.
= ₹ 3290 – ₹ 3,100
= ₹ 190

Question 6.
An engineer spent ₹ 24,500 to assemble a computer. He sold it at a profit of ₹ 8,250. At what price did he sell it?
Solution:
C.P. of computer = ₹ 24,500
Profit = + ₹ 8,250
S.P. = ?
S.P. = C.P. + Profit
= ₹ 24,500 + ₹ 8,250
= ₹ 32,750

Question 7.
A man sold a bag for ₹ 75, at a loss of ₹ 12. Find the cost price of the bag.
Solution:
S.P. of bag = ₹ 75
Loss = ₹ 12
C.P. should be greater than S.P.
Cost Price = S.P. + Loss
= ₹ 75 + ₹ 12
= ₹ 87

Question 8.
Three dozen oranges were sold for ₹ 120. If there was a loss of ₹ 9, find the cost price of one dozen oranges.
Solution:
S.P. of 3 dozen oranges = ₹ 120
S.P. of 1 dozen oranges = ₹ 120 ÷ 3 = ₹ 40
Loss = 9
So C.P. > S.P.
C.P. = S.P. + Loss
= ₹ 40 + ₹ 9
= ₹ 49

Question 9.
Fill in the blanks.
(a) If the cost price of an article is greater than the selling price, we have a _________
Solution:
Loss

(b) Selling price = Cost price – _________
Solution:
Loss

(c) Cost price = Selling price – _________
Solution:
Profit

(d) Cost price + _________ = Selling price.
Solution:
Profit

(e) Selling price + _________ = Cost Price.
Solution:
Loss

Additional Questions

Question 1.
Solve the following questions.
(a) C.P. = ₹ 340, S.P. = ₹ 550, Profit = ?, Loss = ?
Solution:
C.P. = ₹ 340
S.P. = ₹ 550
Since S.P. > C.P. we have a profit
Profit = S.P. – C.P.
= ₹ 550 – ₹ 340
= ₹ 210

(b) C.P. = ₹ 50, S.P. = ₹ 49, Profit = ?, Loss = ?
Solution:
C.P. = ₹ 50
S.P. = ₹ 49
Since S.P. < C.P. we have a loss
Loss = C.P. – S.P.
= ₹ 50 – ₹ 49
= ₹ 1

(c) C.P. = ₹ 176, S.P. = ₹ 150, Profit = ?, Loss = ?
Solution:
C.P. = ₹ 176
S.P. = ₹ 150
Since C.P. > S.P. loss has occurred
Loss = C.P. – S.P.
= ₹ 176 – ₹ 150
= ₹ 26

(d) C.P. = ₹ 2,340, ₹ S.P. = ₹ 2,130, Profit = ?, Loss = ?
Solution:
C.P. = ₹ 2,340
S.P. = ₹ 2,130
Since C.P. > S.P. we have a loss
Loss = C.P. – S.P.
= ₹ 2,340 – ₹ 2,130
= ₹ 210

(e) C.P. = ₹ 46.50, S.P. = ₹ 41.75, Profit = ?, Loss = ?
Solution:
C.P. = ₹ 46.50
S.P. = ₹ 41.75
Since C.P. > S.P. we have a loss
Loss = C.P. – S.P.
= ₹ 46.50 – ₹ 41.75
= ₹ 4.75

Question 2.
Find the selling price of the following.
(a) A shopkeeper purchased a toy bus for ₹ 275 and sold it at a gain of ₹ 25. Find the selling price of the bus.
Solution:
C.P. of toy bus = ₹ 275
Gain = ₹ 25
Since gain occurs S.P. > C.P.
S.P. of bus = ₹ 275 + ₹ 25 = ₹ 300

(b) C.P. = ₹ 1120, Gain = ₹ 50, S.P. = ?
Solution:
C.P. = ₹ 1120
Gain = ₹ 50
Since gain occurs S.P. > C.P.
S.P. = ₹ 1120 + ₹ 50 = ₹ 1170

Question 3.
Deepak sold a tape recorder for ₹ 1,700. He made a profit of ₹ 170. Find the cost price of the tape recorder.
Solution:
Deepak has made a profit of ₹ 170.
So the cost price will be less than S.P.
Selling Price = ₹ 1700
Profit = – ₹ 170
Cost Price = ₹ 1700 – ₹ 170 = ₹ 1530
Thus, the C.P. of the tape recorder is ₹ 1530

Question 4.
Sahil sold a toothpaste for ₹ 95 at a loss of ₹ 12. Find the cost price of the toothpaste.
Solution:
Sahil has made a loss of ₹ 12.
So the cost price should be more than the selling price.
Selling Price = ₹ 95
Loss = ₹ 12
Cost Price = ₹ 95 + ₹ 12 = ₹ 107
Thus C.P. of toothpaste is ₹ 107

Question 5.
Find S.P. if C.P. = ₹ 3,500, transport charges = ₹ 75, Gain = ₹ 220.
Solution:
C.P. = ₹ 3,500
Transport Charges = ₹ 75
Actual C.P. = ₹ 3575
As gain occurs so S.P. > C.P.
S.P. = C.P. + Gain
= ₹ 3575 + ₹ 220
= ₹ 3795

Question 6.
Fill in the blanks.
(a) If the C.P. of the article is less than the selling price we have a _________
Solution:
Gain

(b) Cost Price = Selling Price + _________
Solution:
Loss

(c) Selling price – Gain = _________
Solution:
Cost Price

(d) Cost Price – _________ = Selling Price.
Solution:
Gain

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 11 Worksheet 4 Solutions of Profit and Loss offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 11 Worksheet 4 Solutions

Question 1.
Complete the table by filling the column of cost price. The first one is done for you.

Solution:
(b) Selling Price = ₹ 95
Profit = – ₹ 12
Cost Price = ₹ 83
∴ The cost price will be less

(c) Selling Price = ₹ 123
Profit = – ₹ 18
Cost Price = ₹ 105
∴ C.P. < S.P.

(d) Selling Price = ₹ 675
Loss = + ₹ 155
Cost Price = ₹ 830
As Loss Occurs, C.P. will be more

(e) Selling Price = ₹ 5,000
Profit = – ₹ 840
Cost Price = ₹ 4160
As Profit occurs, C.P. will be less

Question 2.
Solve the following questions.
(a) Salim is a carpenter. He sold a chair for ₹ 900 making a profit of ₹ 250. Find the cost price of the chair.
Solution:
Selling Price = ₹ 900
Profit = – ₹ 250
C.P. = ₹ 650
As profit occurs so cost price will be less
So the C.P. of the chair is ₹ 650

(b) A merchant sold a fan for ₹ 1,250 at a loss of ₹ 200. Find the cost price of the fan.
Solution:
Selling Price = ₹ 1,250
Loss = + ₹ 200
C.P. = ₹ 1,450
As losses occur so cost price will be more
So the C.P. of the fan is ₹ 1,450

(c) Mr Goel sold a scooter for ₹ 6,200. If he had made a profit of ₹ 1,050 on the scooter, find the cost price of the scooter.
Solution:
Selling Price = ₹ 6,200
Profit = – ₹ 1,050
C.P. = ₹ 5,150
As profit occurs so cost price should be less
So C.P. of scooter is ₹ 5,150

DAV Class 5 Maths Chapter 11 Value Based Questions

It was Diwali time. Nisha’s father bought show-pieces worth ₹ 54,500 for selling at his shop during the festival. He sold all showpieces for ₹ 59,540. He decided to share his profit. As winter was approaching, Nisha suggested her father buy some blankets for the poor people. On Diwali night, Nisha with her father went to the ‘Night Shelter’ and distributed blankets to the needy. Nisha was happy. Her father also felt satisfied sharing his profit in this way.

Question 1.
How much profit was made by Nisha’s father?
Solution:
C.P. of showpieces = ₹ 54,500
S.P. of showpieces = ₹ 59,540
As S.P. > C.P. So S.P. – C.P. = Profit
∴ Profit = ₹ 59,540 – ₹ 54,500 = ₹ 5,040
A profit of ₹ 5,040 was made by Nisha’s father.

Question 2.
Suggest different things that you can donate to poor people.
Solution:
We can donate

• Blankets
• Sweets
• Clothes (winter clothes)
• Toys

Question 3.
With whom do you share your Diwali gifts?
Solution:
Do it yourself.

Word Problems

Example 1.
Mona sold a tape recorder for ₹ 1,700. She made a profit of ₹ 120. Find the cost price of the tape recorder.
Solution:
Mona has made a profit of ₹ 120.
So, the cost price will be less than the selling price.
Selling price = ₹ 1700
Profit = – ₹ 120
Cost Price = ₹ 1580
Thus, the cost price of the tape recorder is ₹ 1580.

Example 2.
Anil sold a perfume for ₹ 90 at a loss of ₹ 10. Find the cost price of the perfume.
Solution:
Anil has made a loss of ₹ 10 so the cost price of the perfume should be more than the selling price.
Selling price = ₹ 90
Loss = + ₹ 10
Cost price = ₹ 100
Thus, the cost price of the perfume is ₹ 100.

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 11 Worksheet 3 Solutions of Profit and Loss offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 11 Worksheet 3 Solutions

Question 1.
Complete the table by filling in the column of the selling price. The first one is done for you.

Solution:
(b) Loss = ₹ 11
S.P. = C.P. – Loss
= ₹ 85 – ₹ 11
= ₹ 74

(c) C.P. = ₹ 480
Profit = ₹ 75
S.P. = C.P. + Profit
= ₹ 480 + ₹ 75
= ₹ 555

(d) C.P. = ₹ 1,200
Loss = ₹ 195
S.P. = C.P. – Loss
= ₹ 1,200 – ₹ 195
= ₹ 1,005

(e) C.P. = ₹ 4,250
Profit = ₹ 300
S.P. = C.P. + Profit
= ₹ 4,250 + ₹ 300
= ₹ 4,550

Question 2.
Solve the following questions.
(a) A shopkeeper purchased a saree for ₹ 375 and sold it at a gain of ₹ 90. Find the selling price of the saree.
Solution:
C.P. of saree = ₹ 375
Gain = ₹ 90
S.P. of saree = C.P. + Gain
= ₹ 375 + ₹ 90
= ₹ 465

(b) A watchmaker bought an old watch for ₹ 420 and spent ₹ 85 on repairs. If he sold the watch at a gain of ₹ 60, find the selling price of the watch.
Solution:
C.P. of old watch = ₹ 420
Amount spent on repairing = ₹ 85
Actual cost price of the watch = ₹ 505
Gain = ₹ 60
S.P. of watch = C.P. + Gain
= ₹ 505 + ₹ 60
= ₹ 565

(c) Rahul purchased a storybook for ₹ 225. After reading it, he sold it to his friend at a loss of ₹ 120. Find the amount paid by his friend for the book.
Solution:
C.P. of storybook = ₹ 225
Loss = ₹ 120
S.P. of storybook = C.P. – Loss
= ₹ 225 – ₹ 120
= ₹ 105

Word Problems

Example 1.
Sahil buys oranges from the wholesale market at ₹ 40 per kilogram. By selling it he gains ₹ 3 per kilogram. What is the selling price of oranges?
Solution:
Cost price = ₹ 40
Profit = + ₹ 3
Selling price = ₹ 43
Thus the selling price of the oranges is ₹ 43.

Example 2.
Ashi buys a pen for ₹ 50. She had to sell it at a loss of ₹ 6. Find the selling price.
Solution:
Ashi has made a loss of ₹ 6. So, the selling price should be less than the cost price.
Solution:
Cost price = ₹ 50
Loss = – ₹ 6
Selling price = ₹ 44
Thus the selling price of a pen is ₹ 44.

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 11 Worksheet 2 Solutions of Profit and Loss offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 11 Worksheet 2 Solutions

Question 1.
Solve the following questions.
(a) A vegetable seller bought potatoes for ₹ 200. He spent ₹ 20 on transport and then sold them for ₹ 290. Find the profit or loss.
Solution:
C.P. of potato = ₹ 200
Transport Charges = ₹ 20
Actual cost price of the potatoes = ₹ 220
Selling price of the potatoes = ₹ 290
Since the selling price is more than C.P. vegetable seller makes a profit.
Profit = S.P. – C.P.
= ₹ 290 – ₹ 220
= ₹ 70
Thus vegetable seller makes a profit of ₹ 70

(b) Mr Gupta bought a car for ₹ 40,000. He spent ₹ 8,000 for painting it. If he sold the car for ₹ 53,000 find the gain or loss.
Solution:
C.P. of car = ₹ 40,000
Money spent on painting = ₹ 8,000
Actual C.P. of car = ₹ 48,000
S.P. of the car = ₹ 53,000
Since, S.P. > C.P., he makes a profit
Gain = S.P. – C.P.
= ₹ 53,000 – ₹ 48,000
= ₹ 5,000
Thus, Mr Gupta gains a profit ₹ 5,000.

(c) Ramlal purchased a plot for ₹ 1,35,000. He constructed a boundary wall around it which cost him ₹ 25,500. If he sold the plot for ₹ 1,90,500. Find the gain or loss.
Solution:
C.P. of plot = ₹ 1,35,000
Cost of boundary wall = ₹ 25,500
Actual cost of plot = ₹ 1,60,500
S. P. of plot = ₹ 1,90,500
Since, S.P. > C.P., Ramlal make a profit
Gain = S.P. – C.P.
= ₹ 1,90,500 – ₹ 1,60,500
= ₹ 30,000
So, Ramlal gained a profit of ₹ 30,000.

(d) A merchant bought a sofa set for ₹ 18,000. He sold it to his friend for ₹ 15,750. If the merchant had spent ₹ 2,500 on transport, find his gain or loss.
Solution:
C.P. of sofa set = ₹ 18,000
S.P. of sofa set = ₹ 15,750
Transport charges = ₹ 2,500
Actual C.P. of sofa set = ₹ 18,000 + ₹ 2,500 = ₹ 20,500
Since, S.P. < C.P., so, the merchant has a loss
Loss = C.P. – S.P.
= ₹ 20,500 – ₹ 15,750
= ₹ 4,750
So, the merchant had a loss of ₹ 4,750.

(e) A property dealer purchased a house for ₹ 1,50,000. He spent ₹ 15,000 on repairing the house. After six months, he sold the house for ₹ 2,15,000. Find the gain or loss of the property dealer.
Solution:
C.P. of property = ₹ 1,50,000
Amount spent on repairing = ₹ 15,000
Actual cost of property = ₹ 1,65,000
S.P. of property = ₹ 2,15,000
Since, S.P. > C.P., property dealer makes a profit
Gain = S.P. – C.P.
= ₹ 2,15,000 – ₹ 1,65,000
= ₹ 50,000
So, the property dealer gained a profit of ₹ 50,000.

Word Problem

Example.
Rohit buys a chair for ₹ 400. He pays ₹ 50 as transport charges and sells it to Mr. Varan for ₹ 520. Find the profit or loss.
Solution:
C.P. of chair = ₹ 400
Transport charges = ₹ 50
Actual cost price of the chair = ₹ 450
Selling price of the chair = ₹ 520
Since S.P. is more than C.P. Rohit makes a profit.
Profit = S.P. – C.P.
= ₹ 520 – ₹ 450
= ₹ 70
Thus Rohit made a profit of ₹ 70.