Class 7

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 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 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.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 7 Congruence of Triangles MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 7 Congruence of Triangles MCQ Questions

Multiple Choice Questions :

Question 1.
If ∠ABC ≅ ΔDEF, then which of the following statement is correct ?
(a) ∠A = ∠D
(b) ∠A = ∠E
(c) ∠B = ∠D
(d) ∠C = ∠E
Answer:
(a) ∠A = ∠D

Question 2.
If ΔABC ≅ ΔDEF, then which of the following statement is correct ?
(a) AB = EF
(b) BC = DE
(c) BC = EF
(d) AB = EF
Answer:
(c) BC = EF

Question 3.
Which of the following is congruent ?
(a) Shaving blades of the same company
(b) Sheets of the same letter pad.
(c) Biscuits of the same packet.
(d) All of above three are congruent
Answer:
(d) All of above three are congruent

Question 4.
Two line segments are congruents :
(a) Their shapes are same
(b) Direction is same
(c) Lengths are same
(d) All of the above.
Answer:
(c) Lengths are same

Question 5.
Out of two congruent angles measure of one angle is 70°, then the measure of other angle is :
(a) 70°
(b) 110°
(c) 90°
(d) 140°
Answer:
(a) 70°

Question 6.
When we write ∠A = ∠B then we really means :
(a) A = B
(b) m∠A = m∠B
(c) A and B are in the same direction
(d) A and B are of same shape
Answer:
(b) m∠A = m∠B

Question 7.
When we write ΔABC ≅ ΔDEF, we means:
(a) AB = DE
(b) BC = EF
(c) AC = DF
(d) All of the above
Answer:
(d) All of the above

Question 8.
If ΔABC ≅ ΔQPR, then which of the following statement is correct ?
(a) ∠A = ∠P
(b) ∠B = ∠R
(c) ∠B = ∠P
(d) ∠B = ∠Q
Answer:
(c) ∠B = ∠P

Fill in blanks :

Question 1.
When we write ∠A = ∠B, there we really means.
Answer:
m∠A = m∠B

Question 2.
Two lines segment are equal if their length are …………….
Answer:
equal

Question 3.
…………….. symbol is used to denote congruence between two figure.
Answer:
≅

Question 4.
The figures having same shape and size are called …………….. figures.
Answer:
congruent

Question 5.
…………….. stands for right angle hypotenuse side.
Answer:
RHS

Write True or False

Question 1.
Shaving blades of the same company are congruent. (True/False)
Answer:
True

Question 2.
Sheets of the same letter pad are congruent. (True/False)
Answer:
True

Question 3.
Two line segments are congruent if their shapes are same. (True/False)
Answer:
False

Question 4.
AAA is one of the laws of congruency of triangles. (True/False)
Answer:
False

Question 5.
Biscuits of the same packet are congruent. (True/False)
Answer:
True

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.4

1. Simplify each of following:

Question (a)
13 × 104
Solution:
13 × 104 = (10 + 3) (104)
= 10 × 104 + 3 × 104
= 10 × (100 + 4) + 3 × (100 + 4)
= 10 × 100 + 10 × 4 + 3 × 100 + 3 × 4
= 1000 + 40 + 300 + 12
= 1352

Question (b)
102 × 105
Solution:
102 × 105 = (100 + 2) × 105 = 100 × 105 + 2 × 105
= 100 × (100 + 5) + 2 × (100 + 5)
= 100 × 100 + 100 × 5 + 2 × 100 + 2 × 5
= 10000 + 500 + 200 + 10
= 10710

Question (c)
6 × 107
Solution:
6 × 107 = 6 × (100 + 7)
= 6 × 100 + 6 × 7
= 600 + 42
= 642

Question (d)
16 × 106
Solution:
16 × 106 = (10 + 6) × 106 = 10 × 106 + 6 × 106
= 10 × (100 + 6) + 6 × (100 + 6)
= 10 × 100 + 10 × 6 + 6 × 100 + 6 × 6
= 1000 + 60 + 600 + 36 = 1696

Question (e)
201 × 205
Solution:
201 × 205 = (200 + 1) × 205 = 200 × 205 + 1 × 205
= 200 × (200 + 5) + 1 × (200 + 5)
= 200 × 200 + 200 × 5 + 1 × 200 + 1 × 5
= 40000 + 1000 + 200 + 5 = 41205

Question (f)
22 × 102
Solution:
22 × 102 = (20 + 2) × 102 = 20 × 102 + 2 × 102
= 20 × (100 +-2) + 2 × (100 + 2)
= 20 × 100 + 20 × 2 + 2 × 100 + 2 x 2
= 2000 + 40 + 200 + 4 = 2244

Question (g)
6 × (4 + 3)
Solution:
6 × (4 + 3) = 6 × 4 + 6 × 3
= 24 + 18 = 42

Question (h)
(17 – 9) × 3
Answer:
(17 – 9) × 3
= 17 × 3 – 9 × 3
= 51 – 27 = 24

Question (i)
(20 + 4) ÷ 2
Solution:
= 20 ÷ 2 + 4 ÷ 2
= 10 + 2 = 12

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 7 Congruence of Triangles Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

1. In the following pair of triangles examine whether the triangles are congruent or not. Write the rule of congruence if triangles are congruent.

Question (i).

Answer:
ΔABC and ΔPQR
Side AB = Side PR ….(Given)
Side BC = Side PQ ….(Given)
Side AC = Side QR ….(Given)
So, By SSS congruence rule
ΔABC ≅ ΔPQR

Question (ii).

Answer:
In ΔABC and ΔEDF
∠B = ∠D ….(Each 90°)
Hypotenuse AC = Hypotenuse EF (given)
Side AB = Side DE ….(given)
So, by RHS congruence rule
ΔABC ≅ ΔEDF.

Question (iii).

Answer:
In ΔXYZ and ΔLMN
∠X = ∠L
Side XY = Side LN ….(given)
∠Y = ∠N …..(given)
So by ASA congruence rule
ΔXYZ ≅ ΔLNM

Question (iv).

Answer:
In ΔOPQ and ΔOSR
Side OQ = Side QR ….(given)
∠POQ = ∠SOR ….(Vertically opp. ∠s)
Side OP = Side OR ….(given)
So, by SAS congruence rule
ΔOPQ ≅ ΔORS

Question (v).

Answer:
In ΔOML and ΔMON
∠LOM = ∠OMN ….(given)
Side OM = Side MO ….(common)
∠OML = ∠MON ….(given)
So, by ASA congmence rule
ΔLOM ≅ ΔOMN

Question (vi).

Answer:
In ΔACD and ΔACB
Side AC = Side AC ….(common)
Side CD = Side AB ….(given)
Side AD = Side BC ….(given)
So, by SSS congruence rule
ΔACD ≅ ΔACB.

2. In fig. ΔAMP ≅ ΔAMQ, Give reason for the following steps.

Steps	Reasons
(i) PM = QM	………………………..
(ii) ∠PMA = ∠QMA	………………………..
(iii) AM = AM	………………………..
(iv) ΔAMP ≅ ΔAMQ	………………………..

Solution:

Steps	Reasons
(i) PM = QM	(i) given
(ii) ∠PMA = ∠QMA	(ii) given
(iii) AM = AM	(iii) common
(iv) ΔAMP ≅ ΔAMQ	(iv) By SAS congruence rule

3. In given figure AB = AC and BD = DC. Prove that :
(i) ΔABD as ΔACD
(ii) ∠B ≅ ∠C

Solution:
In ΔABD and ΔACD
Side AB = Side AC ….(given)
Side BD = Side DC ….(given)
Side AD = Side AD ….(common)
So, by SSS congruence rale
(i) ΔABD ≅ ΔACD
(ii) ∠B = ∠C ….(By the corresponding parts of congruent triangles are equal)

4. In the given (figure 7.20), AC = CE and BC = CD. Prove that ΔACB ≅ ΔECD.

Solution:
In ΔACB and ΔECD
Side AC = Side CE …..(given)
∠ACB = ∠ECD (vertically opposite angles)
Side BC = Side CD
So, by SAS congruence rule.

5. In the given figure

(i) Write three pairs of equal parts in ΔADC and ΔCBA
(ii) Is ΔADC ≅ ΔCBA ? Give reasons.
(iii) Is AD = CB ? Give reasons.
Solutions:
(i) In ΔADC and ΔCBA, three pairs of equal parts are
Side DC = Side AB (given)
∠ACD = ∠BAC (each 60°)
Side AC = Side CA (common).

(ii) Yes, from part (i) by using SAS congruence rules
We conclude that ΔADC ≅ ΔCBA with correspondence A → C, D → B, C → B.

(iii) Yes, from part (ii) ΔADC ≅ ΔCBA we know that, the corresponding parts of congruent triangle are equals.
Therefore AD = CB.

6. In the given figure PQ || RS and PQ = RS. Prove that
(i) ΔPOQ ≅ ΔSOR
(ii) ∠POQ = ∠SOR.

Solution:
In ΔPOQ and ΔSOR
∠OPQ = ∠OSR (Alternate interior angles)
Side PQ = Side RS (given)
∠OQP = ∠ORS (Alternate interior angles)

(i) By ASA congruence rule,
ΔPOQ ≅ ΔSOR

(ii) From part (i) ΔPOQ ≅ ΔSOR we know that, the corresponding parts of congruent triangles are equal
∴ ∠POQ = ∠SOR

7. In the given figure, M is midpoint of AD and ∠A = ∠D. Show that ΔAMB ≅ ΔDMC

Solution:
In ΔAMB and ΔDMC
∠A = ∠D ….(given)
Side AM = Side MD (∵ M is mid point of AD ∴)
∠AMB = ∠DMC (vertically opposite angles)
So, by ASA congruence
ΔAMB ≅ ΔDMC

8. In flie given figure SP ⊥ PQ, RQ ⊥ PQ and PR = QS.

(i) Write three parts of equal parts in ΔPQR and ΔSPQ
(ii) Prove that ΔPQR ≅ ΔQPS
Solution:
(i) In ΔPQR and ΔSPQ, three equal parts are
∠PQR = ∠SPQ (each 90°)
Hypotenuse PR = Hypotenuse SQ ….(given)
Side PQ = Side PQ (common)

(ii) From part (i) by using RHS congruence rule.
We conclude that ΔPQR = ΔQPS with correspondence
P ↔ Q, Q ↔ P, R ↔ S

9. In given figure AB ⊥ QR, AC ⊥ QP and QC = QB. Prove that
(i) ΔQAB ≅ ΔQAC
(ii) ∠AQB ≅ ∠AQC

Solution:
(i) In ΔQAB and ΔQAC
∠ABQ = ∠ACQ (Each 90°)
Hyp. AQ = Hyp AQ (common side)
Side QB = Side QC …(given)
So, by RHS congruence rule
ΔQAB ≅ ΔQAC

(ii) From part (i) ΔQAB = ΔQAC.
We know that the corresponding parts of congruent triangles are equal.
Therefore
∠AQB = ∠AQC.

10. Multiple Choice Questions :

Question (i).
Which of the following is not a congruence rule
(a) ASA
(b) SAS
(c) SSS
(d) AAA
Answer:
(d) AAA

Question (ii).
If ΔABC as ΔPQR, then the correct statement is
(a) ∠A = ∠Q
(b) ∠A = ∠R
(c) ∠A = ∠P
(d) AB = QR
Answer:
(c) ∠A = ∠P

Question (iii).
If ∠A = ∠D, ∠B = ∠E and AB = DE, then ΔABC ≅ ΔDEF, by the congruence rule :
(a) SSS
(b) ASA
(c) SAS
(d) RHS
Answer:
(b) ASA

11. ASA congruence criterion is same as SAS congruence criterion. (True/False)
Answer:
False

12. Two right angled triangles are always congruent. (True/False)
Answer:
False

13. ‘=’ symbol used for congruence of triangles. (True/False)
Answer:
False

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

1. Estimate each of the following using general rule:

Question (a)
837 + 987
Solution:
While rounding off to hundreds place
837 + 987 = 800 + 1000
= 1800

Question (b)
783 – 427
Solution:
While rounding off to hundreds place
783 – 427 = 800 – 400
= 400

Question (c)
1391 + 2783
Solution:
(i) While rounding off to thousands place
1391 + 2783 = 1000 + 3000
= 4000
(ii) While rounding off to hundreds place
1391 + 2783 = 1400 + 2800
= 4000

Question (d)
28292 – 21496.
Solution:
While rounding off to ten thousands place.
28292 – 21496 = 30000 – 20000
= 10000

2. Estimate the product using general rule:

Question (a)
898 × 785
Solution:
898 rounds off to hundreds place = 900
785 rounds off to hundreds place = 800
Estimated product = 900 × 800
= 720000

Question (b)
9 × 795
Solution:
9 rounding off to tens place = 10
795 rounding off to tens place = 800
Estimated product = 10 × 800
= 8000

Question (c)
(c) 87 × 317
Solution:
87 rounded off to hundreds place = 100
317 rounded off to hundreds place = 300
Estimated product = 90 × 300
= 27000

Question (d)
9250 × 29
Solution:
9250 rounds off to thousands place = 9000
29 rounds off to tens place = 30
Estimated product = 9000 × 30
= 270000

3. Estimate by rounding off to nearest hundred:

Question (a)
439 + 334 + 4317
Solution:
439 rounds off to nearest hundreds = 400
334 rounds off to nearest hundreds = 300
4317 rounds off to nearest hundreds = 4300
Estimated sum = 400 + 300 + 4300 = 5000

Question (b)
108734 – 47599.
Solution:
108734 rounds off to nearest hundreds = 108700
47599 rounds off to nearest hundreds = – 47600
Estimated difference = 61100

4. Estimate by rounding off to nearest tens:

Question (a)
439 + 334 + 4317
Solution:
439 + 334 + 4317
439 rounds off to nearest tens = 440
334 rounds off to nearest tens = + 330
4317 rounds off to nearest tens = + 4320
Estimated sum = 5090

Question (b)
108734 – 47599
Solution:
108734 rounds off to nearest tens = 108730
47599 rounds off to nearest tens = – 47600
Estimated difference = 61130