Class 7

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

1. Construct ΔABC in which AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
Solution:
Given : Three sides of a triangle as AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
To construct : A triangle with these three sides.
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment BC = 5 cm.

Step 3 : From B; point A is at a distance 3.5 cm. So, with B as centre, draw an arc of radius
3.5 cm (Now point A will be some where on this arc. Our job is to find where exactly A is.)

Step 4. From C, point A is at a distance of 7 cm. So, with C as centre; draw an arc of radius 7 cm. (A will be some where on this arc. We have to fix it).

Step 5. A has to be on both the arcs drawn. So it is the point of intersection of arcs.
Mark the point of intersection of arcs as A. Join AB and AC.
Thus we obtain ΔABC.

2. Construct a triangle ABC in which AB = BC = 6.5 cm and CA = 4 cm. Also name the kind of triangle drawn.
Solution:
Given : Three sides of triangle as AB = BC = 6.5 cm. and CA = 4 cm.
To construct : A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment AC = 4 cm.

Step 3. From A; point B is at a distance of 6.5 cm. So, with A as centre, draw an arc of radius 6.5 cm. (Now point B will be somewhere on this arc. Our job is to find where exactly A is.)

Step 4. From C; point B is at a distance of 6.5 cm. So; with C as centre; draw an arc of radius 6.5 cm. (B will be some where on this arc. We have to fix it.)

Step 5. B has to be on both the arcs drawn. So it is the point of intersection of arcs. Mark the point of intersection of arcs as B. Join AB and BC.

We observe that AB = BC = 6.5 cm.
Since two sides are of equal length. Thus we obtain an isosceles ΔABC.

3. Construct a triangle XYZ such that length of each side is 5 cm. Also name the kind of triangle drawn.
Solution:
Given : A triangle XYZ in which XY = YZ = ZX = 5 cm..
To Construct. A triangle XYZ with each side 5 cm.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle XYZ and indicate the lengths of the three sides.

Step 2. Draw a line segment YZ = 5 cm.

Step 3. From Y; point X is at a distance of 5 cm. So, with Y as centre, draw an arc of radius 5 cm. (Now point X will be somewhere on this arc. Our job is to find where exactly X is.)

Step 4. From Z, point X is at a distance of 5 cm. So, with Z as centre, draw an arc of radius 5 cm. (X will be somewhere oh this arc. We have to fix it.)

Step 5. Point X has to be on both the arcs drawn. So, it is the point of intersection of arcs.
Mark the point of intersection of arcs as X. Join XY and XZ.

Thus we obtain an equilateral ΔXYZ each of whose side is 5 cm.

4. Construct a triangle PQR such that PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm. Measure ∠PQR and also name the kind of triangle drawn.
Solution:
Given. Three sides of triangle as PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm.
To construct. A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle PQR and indicate the lengths of the three sides.

Step 2. Draw a line segment QR of length 6 cm.

Step 3. From Q; point P is at a distance of 2.5 cm. So, with Q as centre, draw an arc of radius 2.5 cm. (Now point P will be some where on this arc. Our job is to find where exactly P is.)

Step 4. From R; point P is at a distance of 6.5 cm. So; with R as centre; draw an arc of radius 6.5 cm. (P will be somewhere on this arc. We have to fix it.)

Step 5. Point P has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark a point of intersection of arcs as P. Join PQ and PR.

Thus we obtain ΔPQR; on measuring ∠PQR, we observe that ∠PQR = 90° so it is a right angled triangle.

5. Construct a triangle ABC, in which AB = 6 cm, BC = 2 cm, CA = 3 cm. (If possible). If not possible give the reason.
Solution:
Since AB = 6 cm, BC = 2 cm, CA = 3 cm
Here BC + CA = 2 cm + 3 cm
= 5 cm < 6 cm < AB
Which is not possible because the sum of two sides of a triangle is always greater than third side of the triangle.

6. Question (i).
Which of the following can be used to construct a triangle ?
(a) The lengths of the three sides
(b) The perimeter of the triangle
(c) The measures of three angles
(d) The name of three vertices
Answer:
(a) The lengths of the three sides

Question (ii).
A triangle can be constructed by taking its sides as :
(a) 1.8 cm, 2.6 cm, 4.4 cm
(b) 3 cm, 4 cm, 8 cm
(c) 4 cm, 7 cm, 2 cm
(d) 5 cm, 4 cm, 4 cm.
Answer:
(d) 5 cm, 4 cm, 4 cm.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1

1. Draw a line, l, take a point p outside it. Through p, draw a line parallel to l using ruler and compass only.
Solution:
Steps of Construction :
Step 1. Draw a line l of any suitable length Mid a point ‘p’ outside l [see Fig. (i)].

Step 2. Take a point ‘q’ on l and join q to p [see Fig. (ii)].

Step 3. With q as centre and a convenient radius, draw an arc cutting l at E and pq at F [see Fig. (iii)].

Step 4. Now with p as a centre and the same radius as in step 3, draw an arc GH cutting pq at I [see Fig. (iv)].

Step 5. Place the pointed tip of the compasses at E and adjust the opening so that the pencil tip is at F [see Fig. (v)].

Step 6. With the same opening as in step 5 and with I as centre, draw an arc cutting the arc GH at J. [see Fig. (vi)].

Step 7. Now, join pand J to draw a line ‘m’ [see Fig. (vii)],

Note that : ∠Jpq and ∠pqE are alternate interior angles and ∠pqE = ∠qpJ
∴ m || l

2. Draw a line parallel to a line l at a distance of 3.5 cm from it.
Solution :
Steps of construction :
Step 1. Take a line ‘l’ and any point say O on it [see Fig. (i)].

Step 2. At O draw ∠AOB = 90°. [see Fig. (ii)].

Step 3. Place the pointed tip of the compasses at ‘0’ (zero) mark on ruler and adjust the opening so that the pencil tip is at 3.5 cm [see Fig. (iii)]

Step 4. With the same opening as in step 3 and with O as centre draw an arc cutting ray OB at X. [see Fig. (iv)].

Step 5. At X draw a line ‘m’ perpendicular to OB. In other words, draw ∠CXO = 90° [see Fig. (v)].

In this way, line m is parallel to l.
Note that. ∠AOX and ∠CXO are alternate angles and ∠AOX = ∠CXO (each = 90°).
∴ m || l.
Note. We may use any of three properties regarding the transversal OX and parallel lines l and m.

3. Let l be a line and P be a point not on l. Through P, draw a line ‘m’ parallel to l. Now, join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meets l at S. What shape do the two sets of parallel lines enclose ?
Solution:
Steps of Construction :
Step 1. Take a line ‘l’ and a point ‘P’ outside l. [see Fig . (i)]

Step 2. Take any point A on l and join P to A [see Fig. (ii)]

Step 3. With A as centre and convenient radius draw an arc cutting l at B and AP at C. [see Fig. (iii)]

Step 4. Now with P as centre and the same radius as in step 3, draw an arc DE cutting PA at F [see Fig. (iv)].

Step 5. Place the pointed tip of the compasses at B and adjust the opening so that the pencil tip is at C [see Fig. (v)].

Step 6. With the same opening as in step 5 and with F as centre, draw an arc cutting the arc DE at G [see Fig. (vi)].

Step 7. Now join PG to draw line ‘m’ [see Fig. (vii)].

Note that. ∠PAB and ∠APG are alternate interior angles and ∠PAB = ∠APG
∴ m || l
Step 8. Take any point Q on l. Join PQ [see Fig. (viii)].

Step 9. Take any other point R on m [see Fig. (ix)].

Step 10. With P as centre and convenient radius, draw an arc cutting line m at H and PQ at I [see Fig. (x)].

Step 11. Now with R as centre and the same radius as in step 10, draw an arc JK [see Fig. (xi)].

Step 12. Place the pointed tip of compasses at H and adjust the opening so that the pencil tip is at I.

Step 13. With the same opening as in step 12 and with R as centre, draw an arc cutting the arc JK at L [see Fig. (xii)].

Step 14. Now join RL to draw a line parallel to PQ. Let this meet l at S. [see Fig. (xiv)]

Note that. ∠RPQ and ∠LRP are alternate interior angles
and ∠RPQ = ∠LRP
∴ RS || PQ.
Now we have
PR || QS
[∵ m || l and PR is part of m and QS is part of line l]
and PQ || RS
∴ PQSR is a parallelogram.

4.

Question (i).
How many parallel lines can be drawn, passing through a point not lying on the given line ?
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(c) 1

Question (ii).
Which of the following is used to draw a line parallel to a given line ?
(a) A protractor
(b) A ruler
(c) A compasses
(d) A ruler and compasses.
Answer:
(d) A ruler and compasses.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 9 Rational Numbers MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 9 Rational Numbers MCQ Questions

Multiple Choice Questions :

Question 1.
Look at the above number line and tell which of the following values is the greatest.
(a) a + b
(b) b – a
(c) a × b
(d) a ÷ b
Answer:
(a) a + b

Question 2.
The product of a rational number and its reciprocal is always :
(a) -1
(b) 0
(c) 1
(d) equal to itself.
Answer:
(c) 1

Question 3.
Which of the following is not an natural number ?
(a) 0
(b) 2
(c) 10
(d) 105.
Answer:
(a) 0

Question 4.
The ascending order of \(\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}\) is :
(a) \(\frac{-1}{5}<\frac{3}{5}<\frac{2}{5}\)
(b) \(\frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}\)
(c) \(\frac{-3}{5}<\frac{-1}{5}<\frac{-2}{5}\)
(d) \(\frac{-2}{5}<\frac{-3}{5}<\frac{-1}{5}\)
Answer:
(b) \(\frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}\)

Question 5.
Write \(\frac {1}{4}\) as a rational number with 4 denominator 20.
(a) \(\frac{2}{20}\)
(b) \(\frac{3}{20}\)
(c) \(\frac{5}{20}\)
(d) \(\frac{4}{20}\)
Answer:
(c) \(\frac{5}{20}\)

Question 6.
The value of \(\frac{-13}{7}+\frac{6}{7}\) :
(a) \(\frac{19}{7}\)
(b) \(\frac{-7}{7}\)
(c) \(\frac{7}{7}\)
(d) None of these
Answer:
(b) \(\frac{-7}{7}\)

Question 7.
The additive inverse of \(\frac{-9}{11}\) is :
(a) \(\frac{9}{11}\)
(b) \(\frac{-9}{11}\)
(c) \(\frac{11}{9}\)
(d) \(\frac{-11}{9}\)
Answer:
(a) \(\frac{9}{11}\)

Question 8.
The additive inverse of \(\frac{5}{7}\) is :
(a) \(\frac{7}{5}\)
(b) \(\frac{-5}{7}\)
(c) \(\frac{-7}{5}\)
(d) \(\frac{5}{7}\)
Answer:
(b) \(\frac{-5}{7}\)

Fill in the blanks :

Question 1.
The smallest natural number is ………………..
Answer:
1

Question 2.
The numbers which are used for counting are called ………………..
Answer:
natural number

Question 3.
All natural number along with zero (0) are called ………………..
Answer:
whole number

Question 4.
Reciprocal of \(\frac {5}{3}\) is ………………..
Answer:
\(\frac {3}{5}\)

Question 5.
Additive inverse of –\(\frac {3}{4}\) is ………………..
Answer:
\(\frac {3}{4}\)

Write True or False :

Question 1.
The smallest natural number is zero (0). (True/False)
Answer:
False

Question 2.
0 is an integer which is neither negative nor positive. (True/False)
Answer:
True

Question 3.
The smallest whole number is 1. (True/False)
Answer:
False

Question 4.
–\(\frac {2}{5}\) is a fraction. (True/False)
Answer:
False

Question 5.
\(\frac {1}{0}\) not a rational number. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 9 Rational Numbers Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

1. Find the sum

Question (i).
\(\frac{6}{9}+\frac{2}{9}\)
Solution:
\(\frac{6}{9}+\frac{2}{9}\) = \(\frac{6+2}{9}\)
= \(\frac {8}{9}\)

Question (ii).
\(\frac{-15}{7}+\frac{9}{7}\)
Solution:
\(\frac{-15}{7}+\frac{9}{7}\) = \(\frac{-15+9}{7}\)
= \(\frac{-6}{7}\)

Question (iii).
\(\frac{17}{11}+\left(\frac{-9}{11}\right)\)
Solution:
\(\frac{17}{11}+\left(\frac{-9}{11}\right)\) = \(\frac{17-9}{11}\)
= \(\frac{8}{11}\)

Question (iv).
\(\frac{-5}{6}+\frac{3}{18}\)
Solution:
\(\frac{-5}{6}+\frac{3}{18}\)
Now, \(\frac{-5}{6}=\frac{-5}{6} \times \frac{3}{3}=\frac{-15}{18}\)

L.C.M. of 6 and 18
= 2 × 3 × 3 = 18
Thus, \(\frac{-5}{6}+\frac{3}{18}=\frac{-15}{18}+\frac{3}{18}\)
= \(\frac{-15+3}{18}\)
= \(\frac {-12}{18}\)
= \(\frac {-2}{3}\)

Question (v).
\(\frac{-7}{19}+\frac{-3}{38}\)
Solution:
\(\frac{-7}{19}+\frac{-3}{38}\)
Now, \(\frac{-7}{19}=\frac{-7}{19} \times \frac{2}{2}\)
= \(\frac {-14}{38}\)
\(\begin{array}{l|l}
2 & 19,38 \\
\hline 19 & 19,19 \\
\hline & 1,1 \\
\hline
\end{array}\)
L.C.M. = 2 × 19
= 38
Thus, \(\frac{-7}{19}+\frac{-3}{38}=\frac{-14}{38}+\frac{-3}{38}\)
= \(\frac{-14-3}{38}\)
= \(\frac{-17}{38}\)

Question (vi).
\(-3 \frac{4}{7}+2 \frac{3}{7}\)
Solution:
\(-3 \frac{4}{7}+2 \frac{3}{7}\)
= \(-\frac{25}{7}+\frac{17}{7}\)
= \(\frac{-25+17}{7}\)
= \(\frac{-8}{7}\)

Question (vii).
\(\frac{-5}{14}+\frac{8}{21}\)
Solution:
\(\frac{-5}{14}+\frac{8}{21}\)
Now, \(\frac{-5}{14}=\frac{-5}{14} \times \frac{3}{3}\)
= \(\frac{-15}{42}\)
\(\begin{array}{l|l}
2 & 14,21 \\
\hline 3 & 7,21 \\
\hline 7 & 7,7 \\
\hline & 1,1
\end{array}\)
L.C.M of 14, 21 = 2 × 3 × 7
= 42
\(\frac{8}{21}=\frac{8}{21} \times \frac{2}{2}\)
= \(\frac{16}{42}\)
Thus, \(\frac{-5}{14}+\frac{8}{21}\)
= \(\frac{-15}{42}+\frac{16}{42}\)
= \(\frac{-15+16}{42}\)
= \(\frac{1}{42}\)

Question (viii).
\(-4 \frac{1}{15}+3 \frac{2}{20}\)
Solution:

2. Find

Question (i).
\(\frac{7}{12}-\frac{11}{36}\)
Solution:
\(\frac{7}{12}-\frac{11}{36}\) = \(\frac{7}{12}\) + (Additive inverse of \(\frac{11}{36}\))
= \(\frac{7}{12}+\left(\frac{-11}{36}\right)\)
= \(\frac{21+(-11)}{36}\)
\(\begin{array}{l|l}
2 & 12,36 \\
\hline 2 & 6,18 \\
\hline 3 & 3,9 \\
\hline 3 & 1,3 \\
\hline & 1,1
\end{array}\)
L.C.M of 12 and 36
= 2 × 2 × 3 × 3
= 36
= \(\frac{10}{36}=\frac{5}{18}\)

Question (ii).
\(\frac{-5}{9}-\frac{3}{5}\)
Solution:
\(\frac{-5}{9}-\frac{3}{5}\) = \(\frac {-5}{9}\) + (additive inverse of \(\frac {3}{5}\))
= \(\frac{-5}{9}+\left(\frac{-3}{5}\right)\)
= \(\frac{-25+(-27)}{45}\)
L.C.M of 9 and 5 is 45 = \(\frac {-52}{45}\)

Question (iii).
\(\frac{-7}{13}-\left(\frac{-5}{91}\right)\)
Solution:
\(\frac{-7}{13}-\left(\frac{-5}{91}\right)\) = \(\frac {-7}{13}\) + (additive inverse of \(\frac {-5}{91}\))
= \(\frac{-7}{13}+\left(\frac{5}{91}\right)\)
= \(\frac{-49+(5)}{91}\)
L.C.M of 13 and 91 = 7 × 13 = 91
\(\begin{array}{l|l}
7 & 13,91 \\
\hline 13 & 13,13 \\
\hline & 1,1
\end{array}\)
= \(\frac {-44}{91}\)

Question (iv).
\(\frac{6}{11}-\frac{-3}{4}\)
Solution:
\(\frac{6}{11}-\frac{-3}{4}\) = \(\frac {6}{11}\) + (additive inverse of \(\frac {-3}{4}\))
= \(\frac{6}{11}+\left(\frac{3}{4}\right)\)
= \(\frac{24+33}{44}\)
L.C.M of 11 and 4 is 44 = \(\frac {57}{44}\)

Question (v).
\(3 \frac{4}{9}-\frac{28}{63}\)
Solution:

3. Find the product of :

Question (i).
\(\frac{5}{9} \times \frac{-3}{8}\)
Solution:
\(\frac{5}{9} \times \frac{-3}{8}\)
= \(\frac{5 \times-3}{9 \times 8}\)
= \(\frac {-5}{24}\)

Question (ii).
\(\frac{-3}{7} \times \frac{7}{-3}\)
Solution:
\(\frac{-3}{7} \times \frac{7}{-3}\)
= \(\frac{-3 \times 7}{7 \times-3}\)
= 1

Question (iii).
\(\frac{3}{13} \times \frac{5}{8}\)
Solution:
\(\frac{3}{13} \times \frac{5}{8}\)
= \(\frac{3 \times 5}{13 \times 8}\)
= \(\frac {15}{104}\)

Question (iv).
\(\frac {3}{10}\) × (-18)
Solution:
\(\frac {3}{10}\) × (-18)
= \(\frac{3 \times-18}{10}\)
= \(\frac{-27}{5}\)

4. Find the value of:

Question (i).
-9 ÷ \(\frac {3}{5}\)
Solution:
-9 ÷ \(\frac {3}{5}\)
= -9 × (Reciprocal of \(\frac {3}{5}\))
= -9 × \(\frac {5}{3}\)
= -15

Question (ii).
\(\frac {-4}{7}\) ÷ 4
Solution:
\(\frac {-4}{7}\) ÷ 4
= \(\frac {-4}{7}\) × (Reciprocal of 4)
= \(\frac{-4}{7} \times \frac{1}{4}\)
= \(\frac {-1}{7}\)

Question (iii).
\(\frac{7}{18} \div \frac{5}{6}\)
Solution:
\(\frac{7}{18} \div \frac{5}{6}\)
= \(\frac {7}{18}\) × (Reciprocal of \(\frac {5}{6}\))
= \(\frac{7}{18} \times \frac{6}{5}\)
= \(\frac {7}{15}\)

Question (iv).
\(\frac{-8}{35} \div\left(\frac{-2}{7}\right)\)
Solution:
\(\frac{-8}{35} \div\left(\frac{-2}{7}\right)\)
= \(\frac {-8}{35}\) × (Reciprocal of \(\frac {-2}{7}\))
= \(\frac{-8}{35} \times \frac{7}{-2}\)
= \(\frac {4}{5}\)

Question (v).
\(\frac {-9}{15}\) ÷ -18
Solution:
\(\frac {-9}{15}\) ÷ -18
= \(\frac {-9}{15}\) × (Reciprocal of -18)
= \(\frac{-9}{15} \times \frac{1}{-18}\)
= \(\frac {1}{30}\)

5. What ratonal number should be added to \(\frac {-5}{12}\) to get \(\frac {-7}{8}\)?
Solution:
Let the required number to be added be x.
then, \(\frac {-5}{12}\) + x = \(\frac {-7}{8}\)

L.C.M of 8, 12 = 2 × 2 × 2 × 3
= 24
= \(\frac{-7 \times 3+5 \times 2}{24}\)
= \(\frac{-21+10}{24}\)
= \(\frac {-11}{24}\)

6. What number should be subtracted from \(\frac {-2}{3}\) to get \(\frac {-5}{6}\) ?
Solution:
Let the required number to be subtracted be x, then
\(\frac{-2}{3}-x=\frac{-5}{6}\)
⇒ \(\frac{-2}{3}-\left(\frac{-5}{6}\right)\) = x
x = \(\frac{-2}{3}+\frac{5}{6}\)
= \(\frac{-4+5}{6}\)
= \(\frac {1}{6}\)

7. The product of two rational numbers is \(\frac {-11}{2}\). If one of them is \(\frac {33}{8}\), find the other number.
Solution:
Let the required number be x, then

8. Multiple Choice Questions

Question (i).
The sum of \(\frac{5}{4}+\left(\frac{25}{-4}\right)\) =
(a) -5
(b) 5
(c) 4
(d) -4
Answer:
(a) -5

Question (ii).
\(\frac{17}{11}-\frac{6}{11}\) =
(a) 1
(b) -1
(c) 6
(d) 3
Answer:
(a) 1

Question (iii).
\(\frac{2}{-5} \times \frac{-5}{2}\) =
(a) 1
(b) -1
(c) 2
(d) -5
Answer:
(a) 1

Question (iv).
\(\frac{7}{12} \div\left(\frac{-7}{12}\right)\) =
(a) 1
(b) -1
(c) 7
(d) -7
Answer:
(b) -1

Question (v).
Which of the following is value of (-4) × [(-5) + (-3)]
(a) -32
(b) 120
(c) 32
(d) -23
Answer:
(c) 32

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 9 Rational Numbers Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.1

1. Write two equivalent rational numbers of the following :

Question (i).
\(\frac {4}{5}\)
Solution:
\(\frac {4}{5}\) = \(\frac {4}{5}\) × \(\frac {2}{2}\)
= \(\frac {8}{10}\)
\(\frac {4}{5}\) = \(\frac {4}{5}\) × \(\frac {3}{3}\)
= \(\frac {12}{15}\)
∴ Equivalent rational numbers of \(\frac {4}{5}\) are \(\frac {8}{10}\) and \(\frac {12}{15}\)

Question (ii).
\(\frac {-5}{9}\)
Solution:
\(\frac {-5}{9}\) = \(\frac {-5}{9}\) × \(\frac {2}{2}\)
= \(\frac {-10}{18}\)
\(\frac {-5}{9}\) = \(\frac {-5}{9}\) × \(\frac {3}{3}\)
= \(\frac {-15}{27}\)
∴ Equivalent rational numbers of \(\frac {-5}{9}\) are \(\frac {-10}{18}\) and \(\frac {-15}{27}\)

Question (iii).
\(\frac {3}{-11}\)
Solution:
\(\frac {3}{-11}\) = \(\frac {3}{-11}\) × \(\frac {2}{2}\)
= \(\frac {6}{-22}\)
\(\frac {3}{-11}\) = \(\frac {3}{-11}\) × \(\frac {3}{3}\)
= \(\frac {9}{-33}\)
∴ Equivalent rational numbers of \(\frac {3}{-11}\) are \(\frac {6}{-22}\) and \(\frac {9}{-33}\)

2. Find the standard form of the following rational numbers :

Question (i).
\(\frac {35}{49}\)
Solution:
\(\frac {35}{49}\)
∵ H.C.F. of 35 and 49 is 7

So dividing both the numerator and denominator by 7 we get.
\(\frac {35}{49}\) = \(\frac{35 \div 7}{49 \div 7}\) = \(\frac {5}{7}\)
∴ Standard form of \(\frac {35}{49}\) is \(\frac {5}{7}\)

Question (ii).
\(\frac {-42}{56}\)
Solution:
\(\frac {-42}{56}\)
∵ H.C.F. of -42 and 56 is 14

So dividing both the numerator and denominator by 14 we get.
\(\frac {-42}{56}\) = \(\frac{-42 \div 14}{56 \div 14}\) = \(\frac{-3}{4}\)
∴ Standard form of \(\frac {-42}{56}\) is \(\frac{-3}{4}\)

Question (iii).
\(\frac {19}{-57}\)
Solution:
\(\frac {19}{-57}\)
∵ H.C.F. of 59 and 57 is 19

So dividing both the numerator and denominator by 19 we get.
\(\frac {19}{-57}\) = \(\frac{-19 \div 19}{-57 \div 19}\) = \(\frac{1}{-3}\)
∴ Standard form of \(\frac {19}{-57}\) is \(\frac{1}{-3}\)

Question (iv).
\(\frac{-12}{-36}\)
Solution:
\(\frac{-12}{-36}\)
∵ H.C.F. of 12 and 36 is 12.

So dividing both the numerator and denominator by 12 we get.
\(\frac{-12}{-36}\) = \(\frac{-12 \div 12}{-36 \div 12}\) = \(\frac{1}{3}\)
Standard form of \(\frac{-12}{-36}\) is \(\frac{1}{3}\)

3. Which of the following pairs represent same rational number ?

Question (i).
\(\frac{-15}{25}\) and \(\frac{18}{-30}\)
Solution:
\(\frac{-15}{25}\) = \(\frac{-15 \div 5}{25 \div 5}\)
= \(\frac{-3}{5}\)
\(\frac{18}{-30}\) = \(\frac{18 \div-6}{-30 \div-6}\)
= \(\frac{-3}{5}\)
∴ \(\frac{-15}{25}\) and \(\frac{18}{-30}\) represents the same number.

Question (ii).
\(\frac{2}{3}\) and \(\frac{-4}{6}\)
Solution:
\(\frac{2}{3}\) = \(\frac{2}{3}\) × \(\frac{1}{1}\)
= \(\frac{2}{3}\)
\(\frac{-4}{6}\) = \(\frac{-4 \div 2}{6 \div 2}\)
= \(\frac{-2}{3}\)
∴ \(\frac{-2}{3}\) and \(\frac{-4}{6}\) doesnot represents the same rational numbers.

Question (iii).
\(\frac{-3}{4}\) and \(\frac{-12}{16}\)
Solution:
\(\frac{-3}{4}\) = \(\frac{-3}{4}\) × \(\frac{4}{4}\)
= \(\frac{-12}{16}\)
\(\frac{-12}{16}\) = \(\frac{-12}{16}\)
∴ \(\frac{-3}{4}\) and \(\frac{-12}{16}\) represents the same rational number.

Question (iv).
\(\frac{-3}{-7}\) and \(\frac{3}{7}\)
Solution:
\(\frac{-3}{4}\) = \(\frac{-3 \div-1}{-7 \div-1}\)
= \(\frac{-3}{4}\)
∴ \(\frac{-3}{-7}\) and \(\frac{3}{7}\) represents the same rational number.

4. Which is greater in each of the following ?

Question (i).
\(\frac{3}{7}\), \(\frac{4}{5}\)
Solution:
Given rational nrnnbere are \(\frac{3}{7}\) and \(\frac{4}{5}\)
L.C.M. of 7 and 5 is 35
∴ \(\frac{3}{7}\) = \(\frac{3 \times 5}{7 \times 5}\)
= \(\frac{15}{35}\)
and \(\frac{4}{5}\) = \(\frac{4 \times 7}{5 \times 7}\)
= \(\frac{28}{35}\)
∵ Numerator of second is greater than first i.e. 28 > 15
So \(\frac{4}{5}\) > \(\frac{3}{7}\)

Question (ii).
\(\frac{-4}{12}\), \(\frac{-8}{12}\)
Solution:
Given rational numbere are \(\frac{-4}{12}\) and \(\frac{-8}{12}\)
∵ Numerator of first is greater than second i.e. -4 > – 8
∴ \(\frac{-4}{12}\) > \(\frac{-8}{12}\)

Question (iii).
\(\frac{-3}{9}\), \(\frac{4}{-18}\)
Solution:
Given rational numbers are \(\frac{-3}{9}\), \(\frac{4}{-18}\)
\(\frac{-3}{9}\) = \(\frac{-3 \times 2}{9 \times 2}\)
= \(\frac{-6}{18}\)
\(\frac{4}{-18}\) = \(\frac{4 \times-1}{-18 \times-1}\)
\(\frac{-4}{18}\)
Since -4 > – 6.
\(\frac{4}{-18}\) > \(\frac{-3}{9}\)

Question (iv).
-2\(\frac{3}{5}\), -3\(\frac{5}{8}\)
Solution:
-2\(\frac{3}{5}\) = \(\frac{-13}{5} \times \frac{8}{8}\)
= \(\frac{-104}{40}\)
-3\(\frac{5}{8}\) = \(\frac{-29}{8} \times \frac{5}{5}\)
= \(\frac{-135}{40}\)
∵ -104 > -135
∴ \(\frac{-13}{5}\) > \(\frac{-29}{8}\)
Thus, -2\(\frac{3}{5}\) > -3\(\frac{5}{8}\)

5. Write the following rational numbers in ascending order.

Question (i).
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Solution:
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Here -5 < -3 < -1
i.e. \(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Therefore, the ascending order is:
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)

Question (ii).
\(\frac{-1}{5}, \frac{-2}{15}, \frac{-4}{5}\)
Solution:
\(\frac{-1}{5}, \frac{-2}{15}, \frac{-4}{5}\)
L.C.M of 5, 15, 5 is 15

Question (iii).
\(\frac{-3}{8}, \frac{-2}{4}, \frac{-3}{2}\)
Solution:
\(\frac{-3}{8}, \frac{-2}{4}, \frac{-3}{2}\)
L.C.M of 8, 4, 2 is 8
∴ \(\frac{-3}{8}=\frac{-3}{8} \times \frac{1}{1}=\frac{-3}{8}\)
\(\frac{-2}{4}=\frac{-2 \times 2}{4 \times 2}=\frac{-4}{8}\)
\(\frac{-3}{2}=\frac{-3 \times 4}{2 \times 4}=\frac{-12}{8}\)
∴ -12 < -4 < -3
or \(\frac {-12}{8}\) < \(\frac {-4}{8}\) < \(\frac {-3}{8}\)
Hence assending order is \(\frac{-3}{2}, \frac{-2}{4}, \frac{-3}{8}\)

6. Write five rational numbers between following rational numbers.

Question (i).
-2 and -1
Solution:
Given rational numbers are -2 and -1
Let us write -2 and -1 as rational numbers with 5 + 1 = 6 as denominator.
We have -2 = -2 × \(\frac {6}{6}\)
= \(\frac {-6}{6}\)
\(\frac {-12}{6}\) < \(\frac {-11}{6}\) < \(\frac {-10}{6}\) < \(\frac {-9}{6}\) < \(\frac {-8}{6}\) < \(\frac {-7}{6}\) < \(\frac {-6}{6}\)
Hence five rational numbers between -2 and -1 are :
\(\frac {-11}{6}\),\(\frac {-10}{6}\),\(\frac {-9}{6}\),\(\frac {-8}{6}\),\(\frac {-7}{6}\)
i.e. \(\frac {-11}{6}\),\(\frac {-5}{3}\),\(\frac {-3}{2}\),\(\frac {-4}{3}\),\(\frac {-7}{6}\)

Question (ii).
\(\frac {-4}{5}\) and \(\frac {-2}{3}\)
Solution:
Given rational numbers are \(\frac {-4}{5}\) and \(\frac {-2}{3}\)
First we find equivalent rational numbers having same denominator
Thus \(\frac {-4}{5}\) = \(\frac{-4 \times 9}{5 \times 9}\)
= \(\frac {-36}{45}\)
and \(\frac {-2}{3}\) = \(\frac{-2 \times 15}{3 \times 15}\)
= \(\frac {-30}{45}\)
Now, we choose any five integers -35, -34, -33, -32, -31 between the numerators -36 and -30
Then the five rational numbers between \(\frac {-36}{45}\) and \(\frac {-30}{45}\) are:
\(\frac{-35}{45}, \frac{-34}{45}, \frac{-33}{45}, \frac{-32}{45}, \frac{-31}{45}\)
Hence, five rational numbers between \(\frac {-4}{5}\) and \(\frac {-2}{3}\) are
\(\frac{-35}{45}, \frac{-34}{45}, \frac{-33}{45}, \frac{-32}{45}, \frac{-31}{45}\)
i.e. \(\frac{-7}{9}, \frac{-34}{45}, \frac{-11}{15}, \frac{-32}{45}, \frac{-31}{45}\)

Question (iii).
\(\frac {1}{3}\) and \(\frac {5}{7}\)
Solution:
Given rational numbers are \(\frac {1}{3}\) and \(\frac {5}{7}\)
First we find equivalent rational numbers having same denominator

\(<\frac{4}{7}<\frac{13}{21}<\frac{2}{3}<\frac{5}{7}\)
Hence, five rational numbers between \(\frac {1}{3}\) and \(\frac {5}{7}\) are
\(\frac{8}{21}, \frac{3}{7}, \frac{10}{21}, \frac{4}{7}, \frac{13}{21}\).

7. Write four more rational numbers in each of the following.

Question (i).
\(\frac{-1}{5}, \frac{-2}{10}, \frac{-3}{15}, \frac{-4}{20}, \ldots\)
Solution:
The given rational numbers are :
\(\frac{-1}{5}, \frac{-2}{10}, \frac{-3}{15}, \frac{-4}{20}, \ldots\)
\(\frac {-1}{5}\) is the rational number in its lowest form
Now, we can write
\(\frac{-2}{10}=\frac{-1}{-5} \times \frac{2}{2}\),
\(\frac{-3}{15}=\frac{-1}{5} \times \frac{3}{3}\) and \(\frac{-1}{5}=\frac{-1}{5} \times \frac{4}{4}\)
Thus, we observe a pattern in these numbers.
The next four rational numbers would be
\(\frac{-1}{5} \times \frac{5}{5}=\frac{-5}{25}\),
\(\frac{-1}{5} \times \frac{6}{6}=\frac{-6}{30}\),
\(\frac{-1}{5} \times \frac{7}{7}=\frac{-7}{35}\)
\(\frac{-1}{5} \times \frac{8}{8}=\frac{-8}{40}\)
Hence required four more rational numbers are :
\(\frac{-5}{25}, \frac{-6}{30}, \frac{-7}{35}, \frac{-8}{40}\)

Question (ii).
\(\frac{-1}{7}, \frac{2}{-14}, \frac{3}{-21}, \frac{4}{-28}, \ldots\)
Solution:
The given rational numbers are
\(\frac{-1}{7}, \frac{2}{-14}, \frac{3}{-21}, \frac{4}{-28}, \ldots\)
\(\frac {-1}{7}\) is the rational number in its lowest form
Now, we can write
\(\frac{2}{-14}=\frac{-1}{7} \times \frac{-2}{-2}=\frac{2}{-14}, \frac{3}{-21}\)
= \(\frac{-1}{7} \times \frac{-3}{-3}\) and \(\frac{4}{-28}=\frac{-1}{7} \times \frac{-4}{-4}\)
Thus, we observe a pattern in these numbers.
The next four rational numbers would be :
\(-\frac{1}{7} \times \frac{-5}{-5}=\frac{5}{-35}\), \(\frac{-1}{7} \times \frac{-6}{-6}=\frac{6}{-42}\),
\(\frac{-1}{7} \times \frac{-7}{-7}=\frac{7}{-49}\), \(\frac{-1}{7} \times \frac{-8}{-8}=\frac{8}{-56}\)
Hence required four more rational numbers are :
\(\frac{5}{-35}, \frac{6}{-42}, \frac{7}{-49}, \frac{8}{-56}\)

8. Draw a number line and represent the following rational number on it.

Question (i).
\(\frac {2}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0). We choose a point A to the right of 0 to represent 1.
Divide the segment OA into four equal parts. Second part from O to the right represents the rational number \(\frac {2}{4}\) as shown in the figure.

Question (ii).
\(\frac {-3}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0). We choose a point A to the right of 0 to represent -1.
Divide the segment OA into four equal parts. Third part from O to the left represents the rational number \(\frac {-3}{4}\) as shown in the figure.

Question (iii).
\(\frac {5}{8}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0).
We choose a point A to the right of 0 to represent 1.
Divide the segment OA into eight equal parts. Fifth part from O to the right represents the rational number as shown in the figure.

Question (iv).
\(\frac {-6}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0).
We choose a point A to the left of O to represent -2.
Divide the segment OA into eight equal parts. Sixth part from O to the left represents the rational number \(\frac {-6}{4}\) as shown in the figure.

9. Multiple Choice Questions :

Question (i).
\(\frac{3}{4}=\frac{?}{12}\) then ? =
(a) 3
(b) 6
(c) 9
(d) 12.
Answer:
(c) 9

Question (ii).
\(\frac{-4}{7}=\frac{?}{14}\) then ? =
(a) -4
(b) -8
(c) 4
(d) 8
Answer:
(b) -8

Question (iii).
The standard form of rational number \(\frac {-21}{28}\) is
(a) \(\frac {-3}{4}\)
(b) \(\frac {3}{4}\)
(c) \(\frac {3}{7}\)
(d) \(\frac {-3}{7}\)
Answer:
(a) \(\frac {-3}{4}\)

Question (iv).
Which of the following rational number is not equal to \(\frac {7}{-4}\) ?
(a) \(\frac {14}{-8}\)
(b) \(\frac {21}{-12}\)
(c) \(\frac {28}{-16}\)
(d) \(\frac {7}{-8}\)
Answer:
(d) \(\frac {7}{-8}\)

Question (v).
Which of the following is correct ?
(a) 0 > \(\frac {-4}{9}\)
(b) 0 < \(\frac {-4}{9}\)
(c) 0 = \(\frac {4}{9}\)
(d) None
Answer:
(a) 0 > \(\frac {-4}{9}\)

Question (vi).
Which of the following is correct ?
(a) \(\frac{-4}{5}<\frac{-3}{10}\)
(b) \(\frac{-4}{5}>\frac{3}{-10}\)
(c) \(\frac{-4}{5}=\frac{3}{-10}\)
(d) None
Answer:
(a) \(\frac{-4}{5}<\frac{-3}{10}\)

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 8 Comparing Quantities MCQ Questions

Multiple Choice Questions

Question 1.
Find the ratio of ₹ 10 to 10 paise.
(a) 1 : 1
(b) 100 : 1
(c) 1000 : 1
(d) 1000 : 10
Answer:
(b) 100 : 1

Question 2.
The ratio of ₹ 5 to 50 Paise is :
(a) 5 : 50
(b) 1 : 10
(c) 10 : 1
(d) 50 : 5.
Answer:
(c) 10 : 1

Question 3.
The ratio of 15 kg to 210 g is :
(a) 15 : 210
(b) 15 : 21
(c) 500 : 7
(d) 7 : 500.
Answer:
(c) 500 : 7

Question 4.
The Percentage of \(\frac {12}{16}\) is :
(a) 25%
(b) 12%
(c) 75%
(d) 16%
Answer:
(c) 75%

Question 5.
Convert \(\frac {5}{4}\) into percent.
(a) 100%
(b) 125%
(c) 75%
(d) 16%
Answer:
(b) 125%

Question 6.
Convert 12.35 into percent.
(a) 12.35%
(b) 123.5%
(c) 1235%
(d) 1.235%
Answer:
(c) 1235%

Question 7.
What percent part of figure is shaded ?

(a) 30%
(b) 50%
(c) 60%
(d) 20%
Answer:
(c) 60%

Question 8.
15% of 250 is :
(a) 250
(b) 375
(c) 37.5
(d) 3750
Answer:
(c) 37.5

Fill in Blanks :

Question 1.
25% of 120 litres is ………….. litres.
Answer:
30

Question 2.
The ratio of 4 km to 300 m is …………..
Answer:
40

Question 3.
The price at which an article is purchased is called …………..
Answer:
Cost price

Question 4.
If the selling price of an article is less than to cost prices then there is …………..
Answer:
loss

Question 5.
The symbol ………….. stands for percent
Answer:
%

Write True or False

Question 1.
The Ratio 1 : 5 and 2 : 15 are equivalent. (True/False)
Answer:
False

Question 2.
A ratio remains unchanged, if both of its terms are multiplied or divided by the same number. (True/False)
Answer:
True

Question 3.
If the selling price of an article is more than its cost price then there is a profit. (True/False)
Answer:
True

Question 4.
If cost price and selling price both are equal then three is profit. (True/False)
Answer:
False

Question 5.
Profit loss percentage is calculated on the cost price. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

1. Find what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.

Question (i).
Gardening shears bought for ₹ 250 and sold for ₹ 325
Solution:
C.P. of gardening shears = ₹ 250
S.P. of gardening shears = ₹ 325
Profit = S.P. – C.P.
= ₹ 325 – ₹ 250
= ₹ 75
Profit percentage = \(\left[\frac{\text { Profit }}{\text { Cost price }} \times 100\right] \%\)
= \(\left[\frac{75}{250} \times 100\right] \%\)
= 30%

Question (ii).
A refrigerater bought for ₹ 12,000 and sold at ₹ 13,500
Solution:
C.P. of refrigerator = ₹ 12,000
S.P. of refregerator = ₹ 13,500
Profit = S.P. – C.P.
= ₹ 13,500 – ₹ 12,000
= ₹ 1500
Profit percentage = \(\left(\frac{\text { Profit }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{1500}{12000} \times 100\right) \%\)
= 12.5%

Question (iii).
A cupboard bought for ₹ 2,500 and old at ₹ 3,000.
Solution:
C.P. of card board = ₹ 2,500
S.P. of card board = ₹ 3,000
Profit = S.P. – C.P.
= ₹ 3000 – ₹ 2500
= ₹ 500
Profit percentage = \(\left(\frac{\text { Profit }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{500}{2500} \times 100\right) \%\)
= 20%

Question (iv).
A shirt bought for ₹ 250 and sold at ₹ 150
Solution:
C.P. of shirt = ₹ 250
S.P. of shirt a ₹ 150
Since S.P. is less than C.P.
So, there will be a loss
Loss = C.P. – S.P.
= ₹ 250 – ₹ 150
= ₹ 100
Loss percentage = \(\left(\frac{\text { Loss }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{100}{250} \times 100\right) \%\)
= 40%

2. A shopkeeper buys an article for ₹ 735 and sold it for ₹ 850. Find his profit or loss.
Solution:
C.P. of an article = ₹ 735
S.P. of an article = ₹ 850
Profit = ₹ 850 – ₹ 735
= ₹ 115

3. Kirti bought a saree for ₹ 2500 and sold it for ₹ 2300. Find her loss and loss percent.
Solution:
C.P. of saree = ₹ 2500
S.P. of saree = ₹ 2300
Loss = C.P. – S.P.
= ₹ 2500 – ₹ 2300
= ₹ 200
Loss percentage = \(\left(\frac{\text { Loss }}{\text { C.P. }} \times 100\right) \%\)
= \(\frac {200}{2500}\) × 100
= 8%

4. An article was sold for ₹ 252 with a profit of 5%. What was its cost price ?
Solution:
S.P. of an article = ₹ 252
Profit = 5%
Let C.P. of article = ₹ 100
Profit = 5% of ₹ 100
= ₹ 5
S.P. of article = ₹ 100 + ₹ 5
= ₹ 105
If S.P. is ₹ 105, then C.P. = ₹ 100
If S.P. is ₹ 1 then C.P. = ₹ \(\frac {100}{105}\)
If S.P. is ₹ 252, then C.P. = ₹ \(\frac {100}{105}\) × 252
= ₹ 240

5. Amrit buys a book for ₹ 275 and sells it at a loss of 15%. For how much does she sell it ?
Solution:
C.P. of book = ₹ 275
Loss = 15%
∴ Loss on ₹ 275 = ₹ \(\frac {15}{100}\) × 275
= ₹ 41.25
Thus, S.P. of book = ₹ 275 – ₹ 41.25
= ₹ 233.75

6. Juhi sells a washing machine for ₹ 13500. She losses 20% in the bargain. What was the price at which she bought it ?
Solution:
S.P. of washing machine = ₹ 13500
Let C.P. = ₹ 100
Loss = 20%
S.P. = ₹ 100 – ₹ 20
= ₹ 80
If S.P. of washing machine is ₹ 80 then its cost price = ₹ 100
If S.P. of washing machine is ₹ 1 then its cost price = ₹ \(\frac {100}{80}\)
If S.P. of washing machine is ₹ 13500
then its cost price = ₹ \(\frac {100}{80}\) × 13500
= ₹ 16875

7. Anita takes a loan of ₹ 5000 at 15% per year as rate of interest. Find the interest she has to pay at the end of one year.
Solution:
Here, Principal (P) = ₹ 5000
Rate (R) = 15% per year
Time (T) = 1 year
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{5000 \times 15 \times 1}{100}\)
= ₹ 750

8. Find the amount to be paid at the end of 3 years in each case :

Question (i).
Principal = ₹ 1200 at 12% p.a.
Solution:
P = ₹ 1200, R = 12% p.a.
T = 3 years
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{1200 \times 12 \times 3}{100}\)
= ₹432
Amount = Principal + Simple Interest
= ₹ 1200 + ₹ 432
= ₹ 1632

Question (ii).
Principal = ₹ 7500 at 5% p.a.
Solution:
P = ₹ 7500, R = 5% p.a. T = 3 years
SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹\(\frac{7500 \times 5 \times 3}{100}\)
= ₹ 1125
Amount = P + S.I. = ₹ 7500 + ₹ 1125
= ₹ 8625

9. Find the time when simple interest on ₹ 2500 at 6% p.a. is ₹ 450
Solution:
P = ₹ 2500, R = 6% p.a. Time = T = ?,
S.I. = ₹ 450
T = \(\frac{\mathrm{SI} \times 100}{\mathrm{P} \times \mathrm{R}}\)
T = \(\frac{450 \times 100}{2500 \times 6}\)
= 3 years

10. Find the rate of interest when simple interest on ₹ 1560 in 3 years is ₹ 585.
Solution:
Principal (P) = ₹ 1560
Time (T) = 3 years
S.I. = ₹ 585
R = \(\frac{\text { SI } \times 100}{\mathrm{P} \times \mathrm{T}}\)
= \(\frac{585 \times 100}{1560 \times 3}\)
= \(\frac {125}{10}\)
= 12.5
Thus Rate of interest is 12.5% p.a.

11. If Nakul gives an interest of ₹ 45 for, one year at 9% rate p.a. what is the sum he borrowed ?
Solution:
Here S.I.= ₹ 45, R = 9% p.a.
T = 1 year, P = ?
P = \(\frac{\text { S.I. } \times 100}{\mathrm{R} \times \mathrm{T}}\)
= \(\frac{45 \times 100}{9 \times 1}\)
= ₹ 500

12. If ₹ 14,000 is invested at 4% per annum simple interest, how long will it take for the amount to reach ₹ 16240 ?
Solution:
P = ₹ 14,000
R = 4% p.a.
T = ?
A = ₹ 16240
S.I. = A – P
= ₹ 16240 – ₹ 14,000
= ₹ 2240
T = \(\frac{\text { SI } \times 100}{\mathrm{R} \times \mathrm{P}}\)
= \(\frac{2240 \times 100}{14000 \times 4}\)
= 4 years

13. Multiple Choice Questions :

Question (i).
If a man buys an article for ₹ 80 and sells it for ₹ 100, then gain percentage is
(a) 20%
(b) 25%
(c) 40%
(d) 125%
Answer:
(b) 25%

Question (ii).
If a man buys an article for ₹ 120 and sells it for ₹ 100, then his loss percentage is
(a) 10%
(b) 20%
(c) 25%
(d) 16\(\frac {2}{3}\)%
Answer:
(d) 16\(\frac {2}{3}\)%

Question (iii).
The salary of a man is ₹ 24000 per month. If he gets an increase of 25% in the salary, then the new salary per month is
(a) ₹ 2,500
(b) ₹ 28,000
(c) ₹ 30,000
(d) ₹ 36,000
Answer:
(c) ₹ 30,000

Question (iv).
On selling an article for ₹ 100, Renu gains ₹ 20 Her gain percentage is
(a) 25%
(b) 20%
(c) 15%
(d) 40%
Answer:
(a) 25%

Question (v).
The simple interest on ₹ 6000 at 8% p.a. for one year is
(a) ₹ 600
(b) ₹ 480
(c) ₹ 400
(d) ₹ 240
Answer:
(b) ₹ 480

Question (vi).
If Rohini borrows ₹ 4800 at 5% p.a. simple interest, then the amount she has to return at the end of 2 years is.
(a) ₹ 480
(b) ₹ 5040
(c) ₹ 5280
(d) ₹ 5600
Answer:
(c) ₹ 5280

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

1. Convert the following fractions into percents

(i). \(\frac {1}{8}\)
Solution:
\(\frac {1}{8}\) = \(\frac {1}{8}\) × 100
= \(\frac {25}{2}\)
= 12.5
Thus, \(\frac {1}{8}\) = 12.5%

(ii). \(\frac {49}{50}\)
Solution:
\(\frac {49}{50}\) = \(\frac {49}{50}\) × 100
= 98
Thus, \(\frac {49}{50}\) = 98%

(iii). \(\frac {5}{4}\)
Solution:
\(\frac {5}{4}\) = \(\frac {5}{4}\) × 100
= 125
Thus, \(\frac {5}{4}\) = 125%

(iv). 1\(\frac {3}{8}\)
Solution:
1\(\frac {3}{8}\) = \(\frac {11}{8}\) × 100
= \(\frac {275}{2}\)
= 137\(\frac {1}{2}\)
Thus, 1\(\frac {3}{8}\) = 137\(\frac {1}{2}\)%

2. Convert the following percents into fractions in simplest form

(i). 25%
Solution:
25% = \(\frac {25}{100}\)
= \(\frac {1}{4}\)

(ii). 150%
Solution:
150% = \(\frac {150}{100}\)
= \(\frac {3}{2}\)

(iii). 7\(\frac {1}{2}\)
Solution:
7\(\frac {1}{2}\) = \(\frac {15}{2}\) × \(\frac {1}{100}\)
= \(\frac {3}{40}\)

3. (i) Anita secured 324 marks out of 400 marks. Find the percentage of marks secured by Anita.
(ii) Out of 32 students, 8 are absent from the class. What is the percentage of students who are absent ?
(iii) There are 120 voters, 90 out of them voted. What percent did not vote ?
Solution:
(i) Anita scored 324 marks out of 400 marks.
∴ Percentage of marks secured by Anita = \(\left(\frac{324}{400} \times 100\right) \%\) = 50%

(ii) Out of 32 students 8 students are absent
∴ Percentage of students who are absent = \(\frac {8}{32}\) × 100% = 25%

(iii) Total voters = 120
Voters who voted = 90
The voters who did not vote = 120 – 90
= 30
Percentage voters who did not vote = \(\frac {30}{120}\) × 100%
= 25%

4. Estimate the part of figure which is shaded and hence find the percentage of the part which is shaded.

Solution:
(i) Shaded part = \(\frac {2}{4}\) = \(\frac {1}{2}\)
Percentage of shaded part = \(\left(\frac{1}{2} \times 100\right) \%\)
= 50%

(ii) Shaded part = \(\frac {2}{6}\) = \(\frac {1}{3}\)
Percentage of shaded part = \(\left(\frac{1}{3} \times 100\right) \%\)
= 33\(\frac {1}{3}\)%

(iii) Shaded part = \(\frac {5}{8}\)
Percentage of shaded part = \(\left(\frac{5}{8} \times 100\right) \%\)
= \(\frac {125}{2}\)%
= 62.5%

5. Convert the following percentages into ratios in simplest form :

(i). 14%
Solution:
14% = 14 × \(\frac {1}{100}\)
= \(\frac {7}{50}\)
= 7 : 50

(ii). 1\(\frac {3}{4}\)%
Solution:
1\(\frac {3}{4}\)% = \(\frac {7}{4}\) × \(\frac {1}{100}\)
= \(\frac {7}{400}\)
= 7 : 400

(iii). 33\(\frac {1}{3}\)%
Solution:
33\(\frac {1}{3}\)% = \(\frac {100}{3}\) × \(\frac {1}{100}\)
= \(\frac {1}{3}\)
= 1 : 3

6. Express the following ratios as percentages :

(i). 5 : 4
Solution:
5:4 = \(\frac {5}{4}\) × 100
= 125%

(ii). 1 : 1
Solution:
1 : 1 = \(\frac {1}{1}\) × 100
= 100%

(iii). 2 : 3
Solution:
2 : 3 = \(\frac {2}{3}\) × 100
= \(\frac {200}{3}\)%
= 66\(\frac {2}{3}\)%

(iv). 9 : 16
Solution:
9 : 16 = \(\frac {9}{16}\) × 100
= \(\frac {225}{4}\)%
= 56\(\frac {1}{4}\)%

7. Chalk contains calcium, carbon and sand in the ratio 12 : 3 : 10. Find the percentage of carbon in the chalk.
Solution:
Calcium : Carbon : sand
= 12 : 3 : 10
Total of ratios = 12 + 3 + 10
= 25
Percentage of carbon is chalk
= \(\frac {3}{25}\) × 100
= 12%

8. Convert each part of the following ratios into percentage :

(i). 3 : 1
Solution:
Total of ratios = 3 + 1 = 4
Percentage of first part = \(\frac {3}{4}\) × 100
= 75%
Percentage of second part = \(\frac {1}{4}\) × 100
= 25%

(ii). 1 : 4
Solution:
Total of ratios = 1 + 4 = 5
Percentage of first part = \(\frac {1}{5}\) × 100
= 20%
Percentage of second part = \(\frac {4}{5}\) × 100
= 80%

(iii). 4 : 5 : 6.
Solution:
Total of ratios = 4 + 5 + 6 = 15
Percentage of first part = \(\frac {4}{15}\) × 100
= \(\frac {8}{3}\)%
= 26\(\frac {2}{3}\)%
Percentage of second part = \(\frac {5}{15}\) × 100
= \(\frac {100}{3}\)%
= 33\(\frac {1}{3}\)%
Percentage of third part = \(\frac {6}{15}\) × 100
= 40%

9. Convert the following percentages to decimals :

(i). 28%
Solution:
28% = \(\frac {28}{100}\)
0.28

(ii). 3%
Solution:
3% = \(\frac {3}{100}\)
= 0.03

(iii). 37\(\frac {1}{2}\)%
Solution:
37\(\frac {1}{2}\)%
= \(\frac {75}{2}\) × \(\frac {1}{100}\)=
= \(\frac {3.75}{100}\)
= 0.375

10. Convert the following decimals to percentage :

(i). 0.65
Solution:
0.65 = (0.65 × 100)%
= \(\left(\frac{65}{100} \times 100\right) \%\)
= 65%

(ii). 0.9
Solution:
0.9 = (0.9 × 100)%
= \(\left(\frac{9}{10} \times 100\right) \%\)
= 90%

(iii). 2.1
Solution:
2.1 = (2.1 × 100)%
= \(\left(\frac{21}{10} \times 100\right) \%\)
= 210%

11. (i) If 65% of students in a class have a bicycle, then what percent of the students do not have a bicycle ?
(ii) We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what percent are mangoes ?
Solution:
(i) Percentage of student having a bicycle = 65%
Percentage of students do not have a bicycle = (100 – 65)%
= 35%

(ii) Percentage of apples = 50%
Percentage of oranges = 30%
Percentage Of mangoes = (100 – (50% + 30%)
= (100 – 80)%
= 20%

12. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Original population = 25000
Decreased population = 24,500
Decrease in population = (25000 – 24500)
= 500
Percentage decrease = \(\frac{\text { Decrease in population }}{\text { Original population }} \times 100 \%\)
= \(\frac {500}{25000}\) × 100%
= 2%

13. Arun bought a plot for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase ?
Solution:
Original price of plot = ₹ 3,50,000
The increased price of plot = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000.
Percentage of price increase
= \(\left(\frac{\text { Increase in price }}{\text { Original price }} \times 100\right)\)
= \(\frac {20,000}{350000}\) × 100%
= \(\frac {40}{7}\)%
= 5\(\frac {5}{7}\)

14. Find :

(i). 15% of 250
Solution:
15% of 250 = \(\frac {15}{100}\) × 250
= \(\frac {375}{10}\)
= 37.5

(ii). 25% of 120 litres
Solution:
25% of 120 litres = \(\frac {25}{100}\) × 20 litres
= 30 liters.

(iii). 4% of 12.5
Solution:
4% of 12.5 = \(\frac {4}{100}\) × \(\frac {125}{10}\)
= \(\frac {5}{10}\)
= 0.5

(iv). 12% of ₹ 250.
Solution:
12% of ₹ 250 = ₹\(\frac {12}{100}\) × 250
= ₹300

15. Multiple Choice Questions :

Question (i).
The ratio 2 : 3 expressed as percentage is
(a) 40%
(b) 60%
(c) 66\(\frac {2}{3}\)%
(d) 33\(\frac {1}{3}\)%
Answer:
(c) 66\(\frac {2}{3}\)%

Question (ii).
If 30% of x is 72, then x is equal to
(a) 120
(b) 240
(c) 360
(d) 480
Answer:
(b) 240

Question (iii).
0.025 when expressed as a percent is
(a) 250%
(b) 25%
(c) 4%
(d) 2.5%
Answer:
(d) 2.5%

Question (iv).
In a class, 45% of students are girls. If there are 22 boys in the class, then the total number of students in the class is
(a) 30
(b) 36
(c) 40
(d) 44
Answer:
(c) 40

Question (v).
What percent of \(\frac {1}{7}\) is \(\frac {2}{35}\) ?
(a) 20%
(b) 25%
(c) 30%
(d) 40%
Answer:
(d) 40%

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

1. Find the ratio of :

Question (i).
₹ 5 to 50 paise
Solution:
To find ratio of ₹ 5 to 50 paise.
Firstly convert both the quantities into same units.
₹ 1 = 100 paise
₹ 5 = 5 × 100 paise = 500 paise.
So, ratio of 500 paise to 50 paise
= \(\frac{500}{50}=\frac{10}{1}\)
= 10 : 1
Hence, required ratio is 10 : 1

Question (ii).
15 kg to 210 g
Solution:
To find ratio of 15 kg to 210 g
Firstly, convert both the quantities into same unit
1 kg = 1000 g
15 kg = 15 × 1000 g = 15000 g.
So, ratio of 15000 to 210 g
= \(\frac{15000}{210}=\frac{500}{7}\)
= 500 : 7
Hence, required ratio is 500 : 7

Question (iii).
4 m to 400 cm
Solution:
To find ratio of 4 m to 400 cm
Firstly convert both the quantities into same unit.
1 m = 100 cm
4 m = 4 × 100 cm = 400 cm
So, ratio of 400 cm to 400 cm
= \(\frac{400}{400}=\frac{1}{1}\)
= 1 : 1
Hence, required ratio is 1 : 1

Question (iv).
30 days to 36 hours
Solution:
To find ratio of 30 days to 36 hours
1 day = 24 hours
30 days = 30 × 24 hours
= 720 hours
So, ratio of 720 hours to 36 hours
= \(\frac{720}{36}=\frac{20}{1}\)
= 20 : 1
Hence required ratio is 20 : 1

2. Are the ratios 1: 2 and 2 :3 equivalent ?
Solution:
To check this we need to know whether
1 : 2 and 2 : 3 are equal
First convert the ratios into fractions
1 : 2 is written as \(\frac {1}{2}\)
2 : 3 is written as \(\frac {2}{3}\)
Now to change these fraction into like fraction.
Make denominator of both the fraction same.
\(\frac{1}{2}=\frac{1}{2} \times \frac{3}{3}=\frac{3}{6}\) and \(\frac{2}{3}=\frac{2}{3} \times \frac{2}{2}=\frac{4}{6}\)
4 > 3
\(\frac{4}{6}>\frac{3}{6}\)
Hence 1 : 2 and 2 : 1 are not equivalent

3. If the cost of 6 toys is ₹ 240, find the cost of 21 toys.
Solution:
The more the number of toys one purchases, the more is the amount to be paid.
Therefore, there is a direct proportion between the number of toys and the amount to be paid.
Let x be number of toys to be purchased
∴ 6 : 240 : : 21 : x
\(\frac{6}{240}=\frac{21}{x}\)
x = \(\frac{21 \times 240}{6}\) = ₹ 840
Thus cost of 21 toys is ₹ 840.

4. The car that I own can go 150 km with 25 litres of petrol. How far can it go with 30 litres of petrol ?
Solution:
More km ↔ More petrol
So, the consumption of petrol and the distance travelled by a car is a case of direct proportion.
Let the distance travelled be x km.
∴ 150 : 25 : : x : 30
\(\frac{150}{25}=\frac{x}{30}\)
x = \(\frac{150 \times 30}{25}\)
x = 180
Thus, it can go 180 km is 30 litres of petrol.

5. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students ?
Solution:
The more the number of students, the more is the number of computer needed.
More students ↔ More will be the computed needed.
There is a direct proportion between the number of students and the number of computers needed. Let computer needed will be x.
6 : 3 : : 24 : x
\(\frac{6}{3}=\frac{24}{x}\)
6 × x = 24 × 3
x = \(\frac{24 \times 3}{6}\) = 12
Thus, 12 computers will be needed.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.5

1. Which of the following are meaningless:

Question (a)
IC
Solution:
Since I can be subtracted only from V and X not from C
So IC is meaningless.

Question (b)
VD
Solution:
VD, since V cannot be subtracted
So VD is meaningless.

Question (c)
XCVII
Solution:
XCVII = XC + VII = 90 + 7 = 97
So XCVII is meaningful.

Question (d)
IVC
Solution:
Since IV cannot be subtracted from C.
So IVC is meaningless.

Question (e)
XM.
Solution:
Since X can be subtracted only from L and C not from M.
So XM is meaningless.

2. Write the following in Hindu Arabic Numerals:

Question (a)
XXV
Solution:
XXV = X + X + V
= 10 + 10 + 5 = 25

Question (b)
XLV
Solution:
XLV = XL + V = 40 + 5 = 45

Question (c)
LXXIX
Solution:
LXXIX = L + X + X + IX
= 50 + 10 + 10 + 9 = 79

Question (d)
XCIX
Solution:
XCIX = XC + IX = 90 + 9 = 99

Question (e)
CLXIX
Solution:
CLXIX = CL + X + IX = 40 + 10 + 9 = 59

Question (f)
DCLXII
Solution:
DCLXII = D + C + L + X + II
= 500 + 100 + 50 + 10 + 2 = 662

Question (g)
DLXIX
Solution:
DLXIX = D + L + X + IX
= 500 + 50 + 10 + 9
= 569

Question (h)
DCCLXVI
Solution:
DCCLXVI = D + CC + L + X + VI
= 500 + 200 + 50 + 10 + 6 = 766

Question (i)
CDXXXVIII
Solution:
CDXXXVIII = CD + XXX + VIII
= 400 + 30 + 8 = 438

Question (j)
MCCXLVI = M + CC + XL + VI
= 1000 + 200 + 40 + 6
= 1246

3. Express each of the following as Roman numerals:

Question (i)
(a) 29
(b) 63
(c) 94
(d) 99
(e) 156
(f) 293
(g) 472
(h) 638
(i) 1458
(j) 948
(k) 199
(l) 499
(m) 699
(n) 299
(o) 999
(p) 1000
Solution:
(a) 29 = 20 + 9 = XXIX
(b) 63 = 60 + 3 = LXIII
(c) 94 = 90 + 4 = XCIV
(d) 99 = 90 + 9 = XCIX
(e) 156 = 100 + 50 + 6 = CLVI
(f) 293 = 200 + 90 + 3 = CCXCffl
(g) 472 = 400 + 70 + 2 = CDLXXII
(h) 638 = 600 + 30 + 8 = DCXXXVIII
(i) 1458 = 1000 + 400 + 50 + 8 = MCDLVIII
(j) 948 = 900 + 40 + 8 = CMXLVIH
(k) 199 = 100 + 90 + 9 = CXCIX
(l) 499 = 400 + 90 + 9 = CDXCIX
(m) 699 = 600 + 90 + 9 = DCXCIX
(n) 299 = 200 + 90 + 9 = CCXCIX
(o) 999 = 900 + 90 + 9 = CMXCIX
(p) 1000 = M.