## Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions (Decimals)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions(Decimals)

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

POINTS TO REMEMBER

1. Decimal fraction (or a decimal number)
A decimal fraction is a fraction whose denominator is 10 or a higher power of 10.
In order to express a given decimal fraction in shorter form, the denominator is not written but its absence is shown by a dot which is called a decimal point inserted in a proper place.
Note :
(i) When there is no number is the left of the decimal point, generally, a zero is written.
(ii) Generally, a decimal fraction has two parts, the first part which is on the right ox the decimal point is called decimal part and the part on the left side of the decimal point, is called integeral part.
(iii) The decimal part is always less than 1.
The integral part is read according to its value and the decimal part is read by naming each digit, in order, separately
3. Converting a decimal number into a vulgar fraction :
Remove the decimal point from the decimal number and write in its denominator with as many zeros as the number of digits are in the decimal parts to the right of 1.
In the decimal number, the name of each place is given as under is the place value chart:

 Thousands Hundreds Tens Units • Tenths Hundredths Thousandth Decimal point and so on
4. Converting a given fraction in to a decimal fraction :
(a) When the denominator in given fraction is 10. 100, 1000 etc., then count from extreme right to left, mark decimal point after as many digits of the numerator as there are zeroes in the denominator.
(b) When the denominator of the given number is other then 10 or higher power of 10, then divide in an ordinary way and mark the decimal point in the quotient just after the division of unit digit is completed. After this, any number of zeroes can be borrowed to.complete the division.
Note : The number of figures that follow the decimal part is called the number of decimal places.
5. Addition and Subtraction of decimal numbers.
(a) Addition : Write the given decimal numbers in such a way, that the decimal points of all the numbers fall in the same vertical line. Digits with the same place value are placed one below the other units are written below units, tens below tens and so on.
Addition is started from the right side, as done in the usual addition (empty places may be filled up by zeroes). In the result (total), the decimal point is placed under decimal points of the numbers added.
(b) Subtraction : In subtraction also, the numbers are written in such a way that their decimals are in ihe same vertical line. Digits with the same place value are placed one below the other (empty places may be filled by zeroes).
Subtraction is started from the right side, as in the case of normal subtraction. In the result, decimal point is placed just under the other decimal points.
6. Multiplication of decimal numbers :
(1) Multiplication by 10,100, 1000 etc. shift the decimal point, in the multiplicand, to the right by as many digits as there are zeroes in the multiplier.
(2) Multiplication by a whole number : Multiply in an ordinary way, without considering the decimal point. In the product, the decimal point should be fixed by counting as many digits from the right as there are decimal places in the multiplicand.
Thus, (i) 0.3 x 6 = 1.8 (ii) 0.26 x 18 = 4.68 and so on.
(3) Multiplication of a decimal number by another decimal number :
Multiply the two numbers in a normal way, ignoring their decimals. In the product, decimal point is fixed counting from right, the digits equal to the sum of decimal places in the multiplicand and the multiplier.
Thus, 32.5 x ().()7 = 2.275
Since, the multiplicand (32.5) has one decimal place and multiplier (0.07) has two decimal places, their product will have 1+2 = 3 decimal places.
7. Division of decimal numbers :
(1) Division by 10, 100, 1000 etc : Shift the decimal points to the left as many digits as there are
zeroes in the divisor. .
(2) Division by a whole number : Divide in the normal manner, ignoring the decimal, and mark tire decimal point; in the quotient, while just crossing over the decimal point in the dividend.
8. Recurring Decimals :
On performing a division, sometimes we find that the same remainder is left, no matter how long we continue the division process. For this reason, the same digit appeares again and again in the quotient. This fact is shown by puting a dot as a bar over the repeating digits.
9. Rounding off of decimal numbers :
(i) If the answer required is correct to two decimal places, we retain digits upto three decimal places.
(ii) If the digit in the third decimal place is five or more than five, then the digit in the second decimal place is increased by one and, if the digit in the third decimal place is less than five, then the digit in the second decimal place is not altered.
(iii)The third digit which was retained is now omitted.
Thus, for getting 8.4813 correct to two decimal places.
Write the given number upto three decimal places i.e. 8.481.
Since, the digit in the third decimal place is 1 which is less than 5.
∴The digit in the second decimal place is not altered.
And, so 8.4813 = 8.48, correct to two decimal places.
In the same way, to get 3.946824 correct to nearest thousandths i.e., correct to three decimal places, first write it as 3.9468.
Then according to the rule, the digit in the third place changes from 6 to 7.
3.9468 = 3.947, correct to three decimal places.

### Decimal Fractions Exercise 4A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Convert the following into fractions in their lowest terms :
(i) 3.75
(ii) 0.5
(iii) 2.04
(iv) 0.65
(v) 2.405
(vi) 0.085
(vii) 8.025

Question 2.
Convert into decimal fractions

Question 3.
Write the number of decimal places in :
(i) 0.4762
(ii) 7.00349
(iii) 8235.403
(iv) 35.4
(v) 2.608
(vi) 0.000879

(i) In 0.4762, there are four places.
(ii) In 7.00349, there are five places.
(iii) In 8235.403, there are three places.
(iv) In 35.4, there is one place.
(v) In 2.608, there are three places.
(vi) In 0.000879, there are six places.

Question 4.
Write the following decimals as word statements :
(i) 0.4,0.9,0.1
(ii) 1.9, 4.4, 7.5
(iii) 0.02, 0.56, 13.06
(iv) 0.005,0.207, 111.519
(v) 0.8, 0.08, 0.008, 0.0008
(vi)256.1, 10.22, 0.634

(i) 0.4 = zero point four, 0.9 = zero point nine, 0.1 = zero point one.
(ii) 1 .9 = one point nine, 4.4 = four point four, 7.5 = seven point five.
(iii) 0.02 = zero point zero two, 0.56 = zero point five six, 13.06 = thirteen point zero six.
(iv) 0.005 = zero point zero zero five, 0.207 = zero point two zero seven, 111.519 = one hundred eleven point five one nine.
(v) 0.8 = zero point eight, 0.08 = zero point zero eight, 0.008 = zero point zero zero eight, 0.0008 = zero point zero zero zero eight
(vi) 256.1 = Two hundred fifty six point one, 10.22 = Ten point two two, 0.634 = zero point six three four.

Question 5.
Convert the given fractions into like fractions:
(i) 0.5,3.62,43.987 and 232.0037
(ii) 215.78, 33.0006, 530.3 and 0.03569

(i) 0.5, 3.62, 43.987 and 232.0037
In these decimals, the greatest places of decimal is 4
∴0.5 = 0.5000
3.62 = 3.6200
43.987 = 43.9870
232.0037 = 232.0037

(ii) 215.78, 33.0006, 530.3 and 0.03569
In these decimals, the greatest places of decimal is 5
∴215.78 = 215.78000
33.0006 = 33.00060
530.3 = 530.30000
0.03569 = 0.03569

### Decimal Fractions Exercise 4B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
(i) 0.5 and 0.37 (ii) 3.8 and 8.7
(iii) 0.02, 0.008 and 0.309
(iv) 0. 4136, 0. 3195 and 0.52
(v) 9.25, 3.4 and 6.666
(vi) 3.007, 0.587 and 18.341
(vii) 0.2, 0.02 and 2.0002
(viii) 6. 08, 60.8, 0.608 and 0.0608
(ix) 29.03, 0.0003, 0.3 and 7.2
(x) 3.4, 2.025, 9.36 and 3.6221

Question 2.
Subtract the first! number from the second :
(i) 5.4, 9.8
(ii) 0.16, 4.3
(iii) 0.82, 8.6
(v) 2.237, 9.425
(vi) 41 .03, 59.46
(vii) 3.92. 26.86
(viii) 4.73, 8.5
(ix) 12.63, 36.2
(x) 0.845, 3.71

Question 3.
Simplify :
(i) 28.796 -13.42 – 2.555
(ii) 93.354 – 62.82 – 13.045
(iii) 36 – 18.59 – 3.2
(iv) 86 + 16.95 – 3.0042
(v) 32.8 – 13 – 10.725 +3.517
(vi) 4000 – 30.51 – 753.101 – 69.43
(vii) 0.1835 + 163.2005 – 25.9 – 100
(viii) 38.00 – 30 + 200.200 – 0.230
(ix) 555.555 + 55.555 – 5.55 – 0.555

Question 4.
Find the difference between 6.85 and 0.685.

Question 5.
Take out the sum of 19.38 and 56.025 then subtract it from 200. 111.

Question 6.
Add 13.95 and 1.003 ; and from the result, subtract the sum of 2.794 and 6.2.

Question 7.
What should be added to 39.587 to give 80.375 ?

Question 8.
What should be subtracted from 100 to give 19.29?

Question 9.
What is the excess of 584.29 over 213.95 ?

Question 10.
Evaluate:
(i) (5.4 – 0.8) + (2.97 -1.462)
(ii) (6.25 + 0.36) -(17.2 – 8.97)
(iii) 9.004 + (3 -2.462)
(iv) 879.4 – (87.94 – 8 .794)

Question 11.
What is the excess of 75 over 48.29?

Question 12.
If A = 237.98 and B = 83.47.
Find :
(i) A – B
(ii) B – A.

Question 13.
The cost of one kg of sugar increases from ?28.47 to T32.65. Find the increase in cost.

### Decimal Fractions Exercise 4C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Multiply:
(i) 0.87 by 10
(ii) 2.948 by 100
(iii) 6.4 by 1000
(iv) 5.8 by 4
(v) 16.32 by 28
(vi) 5. 037 by 8
(vi) 4.6 by 2.1
(viii) 0.568 by 6.4

Question 2.
Multiply each number by 10, 100, 1000 :
(i) 0.5
(ii) 0.112
(iii) 4.8
(iv) 0.0359
(v) 16.27
(vi) 234.8

(i) 0.5 x 10 = 5,0.5 x 100 = 50,
0.5 x 1000 = 500
(ii) 0.112 x 10= 1.12,0.112 x 100
= 11.2, 0.112 x 1000= 112
(iii) 4.8 x 10 = 48, 4.8 x 100 = 480,
4.8 x 1000 = 4800
(iv) 0.0359 x 10 = 0.359,0.0359 x 100 = 3.59, 0.0359 x 1000 = 35-9
(v) 16.27 x 10 = 162.7, 16.27 x 100 = 1627, 16.27 x 1000= 16270
(vi) 234.8 x 10 = 2348, 234.8 x 100 = 23480, 234.8 x 1000 = 234800

Question 3.
Evaluate:
(i) 5.897 x 2.3
(ii) 0.894 x 87
(iii) 0.01 x 0.001
(iv) 0.84 x 2.2 x 4
(v) 4.75 x 0.08 x 3
(vi) 2.4 x 3.5 x 4.8
(vii) 0.8 x 1.2 x 0.25
(viii) 0.3 x 0.03 x 0.003
(ix) 12.003 x (0.2)5

Question 4.
Divide :
(i) 54.9 by 10
(ii) 7.8 by 100
(iii) 324.76 by 1000
(iv) 12.8 by 4
(v) 27.918 by 9
(vi) 4.672 by 8
(vii) 4.32 by 1.2
(viii) 7.644 by 1.4
(ix) 4.8432 by 0.08

Question 5.
Divide each of the given numbers by 10, 100, 1000 and 10000
(i) 2.1
(ii) 8.64
(iii) 5-01
(iv) 0.0906
(v) 0.125
(vi) 111.11
(vii) 0.848 x 3
(viii)4.906 x (0.2) ²
(ix) (1.2)² x(0.9)²

(i) 2.1 ÷ 10 = 0.21, 2.1 ÷ 100 = 0.021,
2.1 ÷ 1000 = 0.0021
and 2.1 ÷10000 = 0.00021

(ii) 8.64 ÷ 10 = 0-864, 8.64 ÷ 100 = 0-0864,
8.64 ÷ 1000 = 0-00864
and 8.64 ÷ 10000 = 0.000864

(iii)5.01 ÷ 10 = 0.501,
5.01 ÷ 100 = 0.0501,
5.01 ÷1000 = 0.00501,
5.01 ÷ 10000 = 0.000501

(iv) 0.0906 ÷ 10 = 0.00906,
0.0906 ÷ 100 = 0.000906,
0.0906 ÷ 1000 = 0.0090906,
0.0906 ÷ 10000 = 0.00000906

(v) 0.125
Now 0.125 + 10 = 0.0125,
0.125 ÷ 100 = 0-00125,
0.125 ÷1000 = 0.000125,
0.125 ÷ 10000 = 0.0000125

(vi) 111.11÷ 10= 11.111,
111.11÷ 100 = 1.1111,
111.11 ÷ 1000 = 0.11111,
111.11 ÷ 10000 = 0.011111

(vii) 0.848 x 3 = 2.544 ,
Now 2.544 ÷ 10 = 0.2544,
2.544 ÷ 100 = 0-02544,
2-544 ÷ 1000 = 0-002544,
2-544 ÷ 10000 = 0-0002544

(viii) 4.906 x (0.2)² = 4.906 x 0.2 x 0.2
= 4.906 x 0.04 = 0.19624
Now 0.19624 + 10 = 0.019624,
0.19624 + 100 = 0.0019624,
0.19624 + 1000 = 0.00019624,
0.19624 + 10000 = 0.000019624

(ix) (1.2)² x (0.9)² = 1.2 x 1.2 x 0.9 x 0.9 = 1.44 x 0.81 = 1.1664
Now 1 .1664 + 10 = 0.11664,
1.1664 ÷ 100 = 0.011664,
1.1664 ÷ 1000 = 0.0011664,
1.1664 ÷ 10000 = 0.00011664

Question 6.
Evaluate :
(i) 9.75 + 5
(ii) 4.4064 + 4
(iii) 27.69 + 30
(iv) 19.25 + 25
(v) 20.64+ 16
(vi) 3.204 + 9
(vii) 0.125 + 25
(viii) 0.14616 + 72
(ix) 0.6227+ 1300
(x) 257.894+ 0-169
(xi) 6.3 + (0.3)²

Question 7.
Evaluate:
(i) 4.3 x 0.52 x 0.3
(ii) 3.2 x 2.5 x 0.7
(iii) 0.8 x 1.5 x 0.6
(iv) 0.3 x 0.3 x 0.3
(v) 1.2 x 1.2 x 0.4
(vi) 0.4 x 0.04 x 0.004
(vii) 0.5 x 0.6 x 0.7
(Viii) 0.5 x 0.06 x 0.007

Question 8.
Evaluate:
(i) (0.9)²
(ii) (0.6)² x 0.5
(iii) 0.3 x (0.5)²
(iv) (0.4)³
(v) (0.2)3 x 5
(vi) (0.2)3 x 0.05

(i) (0.9)2
⇒ 0.9 x 0.9 = 0.81
(Sum of decimal places 1 + 1=2)

(ii) (0.6)2 x 0.5
⇒ 0.6 x 0.6 x 0.5
⇒ 0.36 x 0.5 = 0.180 or 0.18
(Sum of decimal places = 1 + 1 + 1 = 3)

(iii) 0.3 x (0.5)2
⇒ 0.3 x 0.5 x 0.5
⇒ 0.3 x 0.25 = 0.075
(Sum of decimal places 1 + 1 + 1 = 3)
(iv) (0.4)3
⇒ 0.4 x 0.4 x 0.4
⇒ 0.16 x 0.4 = 0.064
(Sum of decimal places 1 + 1 + 1 = 3)

(v) (0.2)3 x 5
⇒ 0.2 x 0.2 x 0.2 x 5
⇒ 0.08 x 5 = 0.40 or 0.4
(Sum of decimal places 1 + 1 + 1 = 3)

(vi) (0.2)3 x 0.05
⇒ 0.2 x 0.2 x 0.2 x 0.05
⇒ 0.008 x 0.05 = 0.00040
(Sum of decimal places = 5)

Question 9.
Find the cost of 36.75 kg wheat at the rate of ₹12.80 per kg.

Question 10.
The cost of a pen is ₹56.15. Find the cost of 16 such pens.

Question 11.
Evaluate:
(i) 0.0072 ÷ 0.06
(ii) 0.621 ÷ 0.3
(iii) 0.0532 ÷ 0.005
(iv) 0.01162 ÷ 0.14
(v) (7.5 x 40.4) ÷ 25
(vi) 2.1 ÷ (0.1 x 0.1)

Question 12.
Fifteen indentical articles weigh 31.50 kg. Find the weight of each article.
Weight of 15 articles = 31.50 kg
∴ Weight of one article
= 31.50- 15 = 2.1 kg

Question 13.
The product of two numbers is 211.2. If one of these two numbers is 16.5, find the other number.

Question 14.
One dozen identical articles cost ₹45.96. Find the cost of each article.
∴ Weight of one dozen articles = ₹45.96
One dozen = 12
∴ Cost of one article = 45.96 + 12 = ₹3.83

### Decimal Fractions Exercise 4D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find whether the given division forms a terminating decimal or a non-terminating decimal:
(i) 3 ÷ 8
(ii)8 ÷ 3
(iii) 6÷ 5
(iv) 5 ÷ 6
(v) 12.5 ÷ 4
(vi) 23 ÷ 0.7
(vii) 42 ÷ 9
(viii) 0.56÷ 0.11

Question 2.
Express as recurring decimals :

Question 3.
Convert into vulgar fraction :
(i) 0.$$\bar { 3 }$$
(ii) 0.$$\bar { 8 }$$
(iii) 4.$$\bar { 4 }$$
(iv) 23.$$\bar { 7 }$$

Question 4.
Convert into vulgar fraction :
(i) 0.$$\bar { 35 }$$
(ii) 2.$$\bar { 23 }$$
(iii) 1.$$\bar { 28 }$$
(iv) 5.$$\bar { 234 }$$

Question 5.
Convert into vulgar fraction :
(i) 0 0.3$$\bar { 7 }$$
(ii) 0.2$$\bar { 45 }$$
(iii) 0.68$$\bar { 5 }$$
(iv) 0.4$$\bar { 42 }$$

### Decimal Fractions Exercise 4E – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Round off:
(i) 0 .07, 0.112, 3.59, 9.489 to the nearest tenths.
(ii) 0.627, 100.479, 0 065 and 0.024 to the nearest hundredths.
(iii) 4.83,0.86,451 .943 and 9.08 to the nearest whole number.

(i) 0.07 = 0.1,
0.112 = 0.1
3 . 59 = 3.6, 9.489 = 9.5
(ii) 0.627 = 0.63,
100.479 = 100.48
0.065 = 0.07,
0.024 = 0.02
(iii) 4.83 = 5,
0.86= 1,
451.943 = 452
9.08 = 9

Question 2.
(i) 18 .35 x 1.2
(ii) 62.89 x 0.02

Question 3.
Write the number of significant figures (digits) in:
(i) 35.06
(ii) 0.35
(iii) 7.0068
(iv) 19 .0
(v) 0.0062
(vi) 0 4.2 x 0.6
(vii) 0.08 x 25
(viii) 3.6 ÷ 0.12 .

(i) 35.06 : In this significant figures i.e. digits are 4
(ii) In 0.35, significant figures are 2
(iii) In 7.0068, significant figures are 5
(iv) In 19.0, significant figures are 3
(v) In 0.0062, significant figures are 2
(vi) In 4.2 x 0.6 = 2.52, significant figure are 3
(vii) In 008 x 25 = 2.00 = 2 significant figure is 1
(viii) In 3.6 ÷0 .12 or 360 ÷ 12 = 30, significant figure are 2.

Question 4.
Write :
(i) 35.869,0 008426,4.952 and 382.7, correct lo three significant figures.
(ii) 60.974. 2.8753, 0.001789 and 400.04, correct to four significant figures.
(iii) 14.29462, 19.2, 46356.82 and 69, correct to five significant figures.

(i) Correct to three significant figures are
35.869 → 35.9
0.008426 →0.00843
4. 952→ 4.95
382.7 →383
(ii) Correct to four significant figures
60.974 →60.97
2. 8753 → 2.875
0.001789 → 0.001789
400.04 → 400.0
(iii) Correct to five significant figures
14.29462→ 14.295
19.2 → 19.200
46356.82 →46357
69 → 69.000

### Decimal Fractions Exercise 4F – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
The weight of an object is 3 .06 kg. Find the total weight of 48 similar objects.

Question 2.
Find die cost of 17.5 m cloth at the rate of Rs. 112.50 per metre.

Question 3.
One kilogramme of oil costs Rs. 73.40. Find the cost of 9.75 kilogramme of the oil.

Question 4.
Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.
Weight of 8 objects = 51-2 kg
∴ Weight of 1 object = 51 -2 + 8 kg = 6-4 kg Ans.

Question 5.
18.5 m of cloth costs Rs. 666. Find the cost of 3.8 m cloth.

Cost of 18.5 m cloth = Rs. 666
Cost of 1 m cloth = Rs. 666 ÷18.5 and cost of 3.8 m cloth
= Rs. (666 ÷18.5) x 3-8 = Rs. (6660 ÷ 185) x 3.8 = Rs. 36 x 3.8 = Rs. 136.80

Question 6.
Find die value of:
(i) 0.5 of Rs. 7.60 + 1.62 of Rs. 30
(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg
(iii) 6.25 of 8.4 – 4.7 of 3.24
(iv) 0.98 of 235 – 0 .09 of 3.2

Question 7.
Evaluate:
(i) 5.6 – 1 .5 of 3.4
(ii) 4.8 ÷ 0.04 of 5
(iii) 0.72 of 80 + 0.2
(iv) 0.72 ÷ 80 of 0.2
(v) 6.45 + (3.9 – 1.75)
(vi) 0.12 of(0.104 – 0.02)+ 0.36 x 0.5

## Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

EXERCISE 2 (A)

Question 1.
Write down a rational number whose numerator is the largest number of two digits and denominator is the smallest number of four digits.

Solution:
Largest two digit = 99
Smallest, number of four digit = 1000 Now numerator = 99 and denominator = 1000
∴ Rational number = $$\frac { 99 }{ 1000 }$$

Question 2.
Write the numerator of each of the following rational numbers:

Solution:

Question 3.
Write the denominator of each of the following rational numbers:

Solution:

Question 4.
Write down a rational number numerator (-5) x (-4) and denominator (28 – 27) x (8 – 5).

Solution:

Question 5.

Solution:

Question 6.
Separate positive and negative rational numbers from the following :

Solution:

Question 7.
Find three rational numbers equivalent to

Solution:

Question 8.
Which of the following are not rational numbers :

Solution:

Question 9.
Express each of the following integers as a rational number with denominator 7 :
(i) 5
(ii) -8
(iii) 0
(iv) -16
(v) 7

Solution:

Question 10.
Express $$\frac { 3 }{ 5 }$$ as a rational number with denominator:

Solution:

Question 11.
Express $$\frac { 4 }{ 7 }$$ as a rational number with numerator :

Solution:

Question 12.
Find x, such that:

Solution:

Question 13.
Express each of the following rational numbers to the lowest terms :

Solution:

Question 14.
Express each of the following rational numbers in the standard form.

Solution:

EXERCISE 2 (B)

Question 1.
Mark the following pairs of rational numbers on the separate number lines :

Solution:

Question 2.
Compare:

Solution:

Question 3.
Compare:

Solution:

Question 4.
Arrange the given rational numbers in ascending order :

Solution:

Question 5.
Arrange the given rational numbers in descending order:

Solution:

Question 6.
Fill in the blanks :

Solution:

EXERCISE 2 (C)

Question 1.

Solution:

Question 2.

Solution:

Question 3.
Evaluate:

Solution:

Question 4.
Evaluate:

Solution:

Question 5.
Subtract :

Solution:

Question 6.
Subtract :

Solution:

Question 7.
The sum of two rational numbers is $$\frac { 11 }{ 24 }$$. If one of them is $$\frac { 3 }{ 8 }$$, find the other.

Solution:

Question 8.
The sum of two rational numbers is $$\frac { -7 }{ 11 }$$. If one of them is $$\frac { 13 }{ 24 }$$, find the other.

Solution:

Question 9.
The sum of two rational numbers is -4. If one of them is $$-\frac { 13 }{ 12 }$$ , find the other.

Solution:

Question 10.
What should be added to $$-\frac { 3 }{ 6 }$$ to get $$-\frac { 11 }{ 24 }$$ ?

Solution:

Question 11.
What should be added to $$\frac { -3 }{ 5 }$$ to get 2?

Solution:

Question 12.
What should be subtracted from $$\frac { -4 }{ 5 }$$ to get 1?

Solution:

Question 13.
The sum of two numbers is $$-\frac { 6 }{ 5 }$$. If one of them is -2, find the other.

Solution:

Question 14.
What should be added to $$\frac { -7 }{ 12 }$$ to get $$\frac { 3 }{ 8 }$$?

Solution:

Question 15.
What should be subtracted from $$\frac { 5 }{ 9 }$$to get $$\frac { 9 }{ 5 }$$ ?

Solution:

EXERCISE 2 (D)

Question 1.
Evaluate:

Solution:

Question 2.
Multiply:

Solution:

Question 3.
Evaluate:

Solution:

Question 4.
Find the cost of 3 $$\frac { 1 }{ 2 }$$ m cloth, if one metre cloth costs ₹325 $$\frac { 1 }{ 2 }$$.

Solution:

Question 5.
A bus is moving with a speed of 65 $$\frac { 1 }{ 2 }$$ km per hour. How much distance will it cover in 1 $$\frac { 1 }{ 3 }$$ hours.

Solution:

Question 6.
Divide:

Solution:

Question 7.
Evaluate:

Solution:

Question 8.
The product of two numbers is 14. If one of the numbers is $$\frac { -8 }{ 7 }$$, find the other.

Solution:

Question 9.
The cost of 11 pens is ₹3 $$\frac { 2 }{ 3 }$$. Find the cost of one pen.

Solution:

Question 10.
If 6 identical articles can be bought for ₹2 $$\frac { 6 }{ 17 }$$. Find the cost of each article.

Solution:

Question 11.
By what number should $$\frac { -3 }{ 8 }$$ be multiplied so that the product is $$\frac { -9 }{ 16 }$$ ?

Solution:

Question 12.
By what number should $$\frac { -5 }{ 7 }$$ be divided so -15 that the result is$$\frac { -15 }{ 28 }$$ ?

Solution:

Question 13.
Evaluate :

Solution:

Question 14.
Seven equal piece are made out of a rope 5 of 21 $$\frac { 5 }{ 7 }$$ m. Find the length of each piece.

Solution:

EXERCISE 2 (E)

Question 1.
Evaluate:

Solution:

Question 2.
The sum of two rational numbers is $$\frac { -3 }{ 8 }$$. If one of them is $$\frac { 3 }{ 16 }$$, find the other,

Solution:

Question 3.
The sum of two rational numbers is -5. If one of them is $$\frac { -52 }{ 25 }$$ , find the other.

Solution:

Question 4.
What rational number should be added to $$-\frac { 3 }{ 16 }$$ to get $$\frac { 11 }{ 24 }$$

Solution:

Question 5.
What rational number should be added to $$-\frac { 3 }{ 5 }$$ to get 2?

Solution:

Question 6.
What rational number should be subtracted from $$-\frac { 5 }{ 12 }$$to get $$\frac { 5 }{ 24 }$$

Solution:

Question 7.
What rational number should be subtracted from $$\frac { 5 }{ 8 }$$ to get $$\frac { 8 }{ 5 }$$ ?

Solution:

Question 8.
Evaluate:

Solution:

Question 9.
The product of two rational numbers is 24. If one of them is $$-\frac { 36}{ 11 }$$, find the other.

Solution:

Question 10.
By what rational number should we multiply $$\frac { 20 }{ -9 }$$ ,so that the product may be $$\frac { -5 }{ 9 }$$ ?

Solution:

## Selina Concise Mathematics class 7 ICSE Solutions – Ratio and Proportion (Including Sharing in a Ratio)

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

POINTS TO REMEMBER

1. Ratio
A ratio is a method to compare two quantities of the same kind with same unit; by dividing the first quantity by the second. The symbol (:) is used for ratio between two quantities e.g. a : b.
Note:
(i) A ratio is a pure number and has no unit.
(ii) A ratio must always be expressed in its lowest terms in simplest form.
(iii) If each term of a ratio is multiplied or divided by the same number or quantity, the ratio remains the same.
2. Proportion :
Proportion is equality of two ratios : e.g. a : b = c : d
i.e. Ratio between first and second is equal to ratio between third and fourth term.
(ii) a and d are called extreme terms and b and c are called mean terms
and a x d = b x c
(iii) Fourth term is called fourth proportional.
3. Continued Proportion
Three quantities are called in continued proportion if the ratio between first and second is equal to the ratio between second and third i. e.
a, b, c are in continued proportion if a : b = b : c
b the middle term is called the mean proportional between a and c and c, the third term is called the third proportional to a and b.

EXERCISE 6 (A)

Question 1.
Express each of the given ratio in its simplest form :

Question 2.
Divide 64 cm long string into two parts in the ratio 5 : 3.

Sum of ratios = 5 + 3 = 8
∴ first part = $$\frac { 5 }{ 8 }$$ of 64 cm = 40 cm
Second part = $$\frac { 3 }{ 8 }$$ of 64 cm = 24 cm

Question 3.
Rs. 720 is divided between x and y in the ratio 4:5. How many rupees will each get?

Sol. Total amount = Rs. 720 Ratio between x, y = 4 : 5
Sum of ratios = 4 + 5 = 9
x’s share = $$\frac { 4 }{ 9 }$$ of Rs. 720 = Rs. 320
y’s share =$$\frac { 5 }{ 9 }$$ of Rs. 720 = Rs. 400

Question 4.
The angles of a triangle are in the ratio 3 :2 : 7. Find each angle.

Ratio in angles of a triangle = 3:2:7
Sum of ratios = 3 + 2 + 7=12
Sum of angles of a triangle = 180°
∴ First angle = $$\frac { 3 }{ 12 }$$x 180°= 45°
Second angle = $$\frac { 2 }{ 12 }$$ x 180°= 30°
Third angle = $$\frac { 7 }{ 12 }$$ x 180°= 105°

Question 5.
A rectangular field is 100 m by 80 m. Find the ratio of

Length of field (l) = 100 m
∴Perimeter = 2 (l + b) = 2 (100 + 80) m = 2 x 180 = 360 m
(i) Ratio between length and breadth
= 100 : 80 = 5 : 4
(Dividing by 20, the HCF of 100 and 80)

(ii) Ratio between breadth and its perimeter
= 80 : 360 = 2 : 9
(Dividing by 40, the HCF of 80 and 360)

Question 6.
The sum of three numbers, whose ratios are 3 $$\frac { 1 }{ 3 }$$ : 4 $$\frac { 1 }{ 5 }$$ : 6 $$\frac { 1 }{ 8 }$$ is 4917.Find the numbers.

Question 7.
The ratio between two quantities is 3 : the first is Rs. 810, find the second.

Ratio between two quantities = 3 : 4
Sum of ratio = 3+4 = 7
∴ Second quantity = Rs. $$\frac { 810 x 4 }{ 3 }$$
= Rs. 270 x 4 = Rs. 1080

Question 8.
Two numbers are in the ratio 5 : 7. Their difference is 10. Find the numbers.

Ratio between two numbers = 5:7
Difference = 7-5 = 2
If difference is 2, then first number = 5
and if difference is 10, then first number
= $$\frac { 5 }{ 2 }$$ x 10=25
and second number = $$\frac { 7 }{ 2 }$$ x 10 = 35

Question 9.
Two numbers are in the ratio 10 : 11. Their sum is 168. Find the numbers.

Ratio between two numbers = 10 : 11
Sum of ratios = 10 + 11=21
Total sum = 168
∴first number = $$\frac { 168 }{ 21 }$$x 10 =80
Second number = $$\frac { 168 }{ 21 }$$x 11 =88 Ans.

Question 10.
A line is divided in two parts in the ratio 2.5 : 1.3. If the smaller one is 35T cm, find the length of the line.

Ratio between two parts of a line
= 2-5 : 1-3 =25 : 13
Sum of ratios = 25 + 13 = 38
Length of smaller part = 35.1 cm 38
Now length of line = $$\frac { 38 }{ 13 }$$ x 35.1 cm
= 38 x 2.7 cm = 102.6 cm

Question 11.
In a class, the ratio of boys to the girls is 7:8. What part of the whole class are girls.

Ratio between boys and girls = 7:8
Sum of ratios = 7 + 8 = 15
∴ Girls are $$\frac { 8 }{ 15 }$$ of the whole class.

Question 12.
The population of a town is ’ 50,000, out of which males are $$\frac { 1 }{ 3 }$$ of the whole population. Find the number of females. Also, find the ratio of the number of females to the whole population.

Total population = 180,000
Population of males = $$\frac { 1 }{ 3 }$$ of 180,000 = 60,000
∴ Population of females = 180,000 – 60,000 = 120,000
Ratio of females to whole population
= 120,000 : 180,000 = 2:3

Question 13.
Ten gram of an alloy of metals A and B contains 7.5 gm of metal A and the rest is metal B. Find the ratio between :
(i) the weights of metals A and B in the alloy.
(ii) the weight of metal B and the weight of the alloy.

Total weight of A and B metals = 10 gm A’s weight = 7.5 gm B’s weight = 10 – 7.5 = 2.5 gm

(i) Ratio between A and B = 7.5 : 2.5
= $$\frac { 75 }{ 10 }$$ : $$\frac { 25 }{ 10 }$$ =3:1

(ii) Ratio between B and total alloy
= 2.5 : 10 = $$\frac { 25 }{ 10 }$$ : 10
⇒ 25 : 100 = 1 : 4

Question 14.
The ages of two boys A and B are 6 years 8 months and 7 years 4 months respectively. Divide Rs. 3,150 in the ratio of their ages.

A’s age = 6 years 8 months
= 6 x 12 + 8 = 72 + 8 = 80 months
B’s age = 7 years 4 months = 7 x 12 + 4 = 84 + 4 = 88 months
∴ Ratio between them = 80 : 88 = 10 : 11
Amount = Rs. 3150
Sum of ratios = 10 + 11 =21
∴A’s share = $$\frac { 3150 x 10 }{ 21 }$$ = 1500 = Rs. 1500

B’s share = $$\frac { 3150 x 11 }{ 21 }$$ = 1650 = Rs. 1650

Question 15.
Three persons start a business and spend Rs. 25,000; Rs. 15,000 atid Rs. 40,000 respectively. Find the share of each out of a profit of Rs. 14,400 in a year.

A’s investment = Rs. 25000
B’s investment = Rs. 15000
C’s investment = Rs. 40000
∴ Ratio between their investment
= 25000 : 15000 : 40000
=5 : 3 : 8
Sum of ratios = 5 + 3 + 8=16 Total profit = ₹ 14400
∴ A’s share = $$\frac { 14400 }{ 16 }$$ x 5 = ₹ 4500
B’s share = $$\frac { 14400 }{ 16 }$$ x 3 = ₹ 2700
C’s share = $$\frac { 14400 }{ 16 }$$ x 8 = ₹ 7200

Question 16.
A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifth of what the second gets. Find the share of each.

Question 17.
Two poles of different heights are standing vertically on a horizontal field. At a particular time, the ratio between the lengths of their shadows is 2 :3. If the height of the smaller pole is 7.5 m, find the height of the other pole.

Question 18.
Two numbers are in the ratio 4 : 7. If their L.C.M. is 168, find the numbers.

Given, Ratio in two numbers = 4:7
and their L.C.M. = 168
Let first number = 4x
and second number = 7x
Now, L.C.M. of 4x and 7x
= 4 x 7 x x = 28x
∴ 28x = 168
x = $$\frac { 168 }{ 28 }$$
x = 6
∴ Required numbers = 4x and 7x = 4 x 6 = 24 and 7 x 6 = 42

Question 19.
is divided between A and B in such a way that A gets half of B. Find :
(i) the ratio between the shares of A and B.
(ii) the share of A and the share of B.

Question 20.
The ratio between two numbers is 5 : 9. Find the numbers, if their H.C.F. is 16.

Let the first number be 5x and second number be 9x
H.C.F. of 5x and 9x = Largest number common to 5x and 9x = x
Given H.C.F. = 16 ⇒ x = 16
∴Required numbers = 5x and 9x = 5×16 and 9×16 = 80 and 144

Question 21.
A bag contains ₹ 1,600 in the form of ₹10 and ₹20 notes. If the ratio between the numbers of ₹10 and ₹20 notes is 2 : 3; find the total number of notes in all.

Question 22.
The ratio between the prices of a scooter and a refrigerator is 4 : 1. If the scooter costs ₹45,000 more than the refrigerator, find the price of the refrigerator.

Ratio between the prices of scooter and a refrigerator = 4:1
Cost price of scooter = ₹45,000
Let the cost of scooter = 4x
Cost of refrigerator = 1x
According to condition,
Cost of scooter > Cost of refrigerator
⇒ 4x- 1x = 45000
⇒ 3x = 45000
x = $$\frac { 45000 }{ 3}$$
⇒ x = ₹15000
.’. Price of refrigerator = ₹15000

EXERCISE 6 (B)

Question 1.
Check whether the following quantities form a proportion or not ?

Question 2.
Find the fourth proportional of

Question 3.
Find the third proportional of

Question 4.
Find the mean proportional between

Question 5.

Question 6.
If x: y – 5 :4 and 2 : x = 3 :8, find the value of y.

Question 7.
Find the value of x, when 2.5 : 4 = x : 7.5.

Question 8.
Show that 2, 12 and 72 are in continued proportion.

## Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

### Probability Exercise 22A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
A coin is tossed once. Find the probability of
Solution:
Total number of possible outcomes are Head (H) and Tail (T) i.e. 2
(i) P (Getting a head) =$$\frac { 1 }{ 2 }$$
(ii) P (Not getting a head) = $$\frac { 1 }{ 2 }$$

Question 2.
A coin is tossed 80 times and the head is obtained 38 times. Now, if a coin tossed once, what will the probability of getting:
(i) a tail
Solution:

Question 3.
A dice is thrown 20 times and the outcomes are noted as shown below :

Now a dice is thrown at random, find the probability of getting :
Solution:

Question 4.
A survey of 50 boys showed that 21 like tea while 29 dislike it. Out of these boys, one boy is chosen at random. What is the probability that the chosen boy
(i) likes tea
(ii) dislikes tea
Solution:

Question 5.
In a cricket match, a batsman hits a boundary 12 times out of 80 balls he plays, further, if he plays one ball more, what will be the probability that:
(i) he hits a boundary
(ii) he does not hit a boundary
Solution:

Question 6.
There are 8 marbles in a bag with numbers from 1 to 8 marked on each of them. What is the probability of drawing a marble with number
(i) 3
(ii) 7
Solution:
Total number of marbles = 8
(i) Probability (of getting a marble with number 3) = $$\frac { 1 }{ 8 }$$
(ii) Probability (of getting a marble with number 7) = $$\frac { 1 }{ 8 }$$

Question 7.
Two coins are tossed simultaneously 100 times and the outcomes are as given below:

If the same pair of coins is tossed again at random, find the probability of getting :
Solution:

Question 8.
A bag contains 4 white and 6 black balls,- all of the same shape and same size. A ball is drawn from the bag without looking into the bag. Find the probability that the ball drawn is :
(i) a black ball
(ii) a white ball
(iii) not a black ball
Solution:

Question 9.
In a single throw of a dice, find the probability of getting a number:
(i) 4
(ii) 6
(iii) greater than 4
Solution:

Question 10.
Hundred identical cards are numbered from 1 to 100. The cards are well shuffled and then a card is drawn. Find the probability that the number on the card drawn is :
(i) 50
(ii) 80
(iii) 40
Solution:

### Probability Exercise 22B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Suppose S is the event that will happen tommorow and P(S) = 0.03.
(i) State in words, the complementary event S’.
(ii) Find P(S’)
Solution:
Given, P(S) = 0.03
(i) The event will not happen tommorow.
(ii) P(S’) = 1 – P(S)
P(S’) = 1 – 0.03 [∵ P(S) + P(S’) = 1]
P(S’) = 0.97

Question 2.
Five Students A, B, C, D and E are competing in a long distance race. Each student’s probability of winning the race is given below:
A → 20 %, B → 22 %, C → 7 %, D → 15% and E → 36 %
(i) Who is most likely to win the race ?
(ii) Who is least likely to win the race ?
(iii) Find the sum of probabilities given.
(iv) Find the probability that either A or D will win the race.
(v) Let S be the event that B will win the race.
(a) Find P(S)
(b) State, in words, the complementary event S’.
(c) Find P(S’)
Solution:

Question 3.
A Ticket is randomly selected from a basket containing3 green, 4 yellow and 5 blue tickets. Determine the probability of getting:
(i) a green ticket
(ii) a green or yellow ticket.
(iii) an orange ticket.
Solution:

Question 4.
Ten cards with numbers 1 to 10 written on them are placed in a bag. A card is chosen from the bag at random. Determine the probability of choosing:
(i) 7
(ii) 9 or 10
(iii) a number greater than 4
(iv) a number less than 6
Solution:

Question 5.
A carton contains eight brown and four white eggs. Find the probability that an egg selected at random is :
(i) brown
(ii) white
Solution:

Question 6.
A box contains 3 yellow, 4 green and 8 blue tickets. A ticket is chosen at random. Find the probability that the ticket is :
(i) yellow
(ii) green
(iii) blue
(iv) red
(v) not yellow
Solution:

Question 7.
The following table shows number of males and number of females of a small locality in different age groups.

If one of the persons, from this locality, is picked at random, what is the probability that
(a) the person picked is a male ?
(b) the person picked is a female ?
(c) the person picked is a female aged 21-50 ?
(d) the person is a male with age upto 50 years?
Solution:

## Selina Concise Mathematics Class 7 ICSE Solutions Chapter 19 Congruency: Congruent Triangles

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 19 Congruency: Congruent Triangles

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

POINTS TO REMEMBER

1. Meaning of Congruency : If two geometrical figures coincide exactly, by placing one over the other, the figures are said to be congruent to each other.
1. Two lines AB and CD are said to be congruent if, on placing AB on CD, or CD on AB ; the two AB and CD exactly coincide.
It is possible only when AB and CD are equal in length.Two figures ABCD and PQRS are said to be

congruent if, on placing ABCD on PQRS or PQRS on ABCD the two figures exactly coincide i.e. A and P coincide. B and Q coincide, C and R coincide and D and S coincide.
It is possible only when :
AB = PQ, BC = QR, CD = RS and AD = PS
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R and ∠D = ∠S.

2. Congruency in Triangles : Let triangle ABC is placed over triangle DEF ; such that, vertex A falls on vertex D and side AB falls on side DE ; then if the two triangles coincide with each other in such a way that B falls on E ; C falls on F ; side BC coincides with side EF and side AC coincides with side DF, then the two triangles are congruent to each other.
The symbol used for congruency is “ ≡ ” or “ ≅ ”
∴∆ ABC is congruent to ∆ DEF is written as :
∆ ABC = ∆ DEF or ∆ABC = ∆DEF.

3. Corresponding Sides and Corresponding Angles : In case of congruent triangles ABC and DEF, drawn above ; when A ABC is placed over A DEF to cover it exactly; then the sides of the two triangles, which coincide with each other, are called corresponding sides. ‘ Thus, the side AB and DE are corresponding sides; sides BC and EF are corresponding sides and sides AC and DF are also corresponding sides.
In the same way. the angles of the two triangles which coincide with each other, are called corresponding angles. Thus, three pairs of corresponding angles are ∠A and ∠D ; ∠B and ∠E and also ∠C and ∠F.
Note : The corresponding parts of congruent triangles are always equal (congruent).
∴(i) AB = DE, BC = EF and AC = DF. i.e. corresponding sides are equal.
Also (ii) ∠A = ∠D, ZB = ∠E and ∠C = ∠F
i.e. corresponding angles are equal.
4. Conditions of Congruency:
1. If three sides of one triangle are equal to three sides of the other triangle, each to each, then the two triangles are congruent.
The test is known as : side, side, side and is abbreviated as S.S.S.
In triangle ABC and PQR, given alongside:
AB = PQ ; BC = QR and AC = PR
And, so A ABC is congruent to ∆ PQR e. ∆ ABC = ∆ PQR by S.S.S.
Because in congruent triangles, corresponding sides and corresponding angles are equal.
∴ ∠A = ∠P : ∠B = ∠Q and ∠C =∠R

2. If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, each to each, then the triangles are congruent.
This test is known as : side, angle, side and is abbreviated as S.A.S. In the given triangles, AB = XZ ; BC = XY and ∠ABC = ∠ZXY
∴ ∆ ABC ≅ ∆ ZXY
Note : Triangles will be congruent by S.A.S., only when the angles included by the corresponding equal sides are equal.
The pairs of corresponding sides of these two congruent triangles are : AB and ZX ; BC and XY ; AC and ZY
The pairs of corresponding angles are :
∠B and ∠X ; ∠A and ∠Z : ∠C and ∠Y.

3. If two angles and the included side of one triangle are equal to the two angles and the included side of the other triangle ; then the triangles are congruent.
This test is known as : angle, side, angle and is abbreviated as A.S.A.
In the given figure :
BC = QR;
∠B = ∠Q and ∠C = ∠R
∴ ∆ABC = ∆ PQR.. (by A.S.A.)

4. If the hypotenuse and one side of a right angled triangle are equal to the hypotenuse and one side of another right angled triangle, then the two triangles are congruent.
This test is known as : right angle, hypotenuse, side and is abbreviated as R.H.S.
In the given figure :
∠B = ∠E = 90° ; AB = FE
and hypotenuse AC = hypotenuse FD
∴ ∆ ABC = ∆ FED (byRH.S.)
The corresponding angles in this case are :
∠A and ∠F ; ∠B and ∠E ; ∠C and ∠D and the corresponding sides are :
AB and EF ; AC and FD ; BC and ED.
Since the triangles are congruent, therefore all its corresponding sides are equal and corresponding angles are also equal.
∴ BC = ED ; ∠A = ∠F and ∠C = ∠D
Note : If three angles of a triangle are equal to the three angles of the other triangle, then the triangle are not necessarily congruent.
For congruency at least one pair of corresponding sides must be equal.
∴A.A.A. is not a test of congruency.

### Congruency: Congruent Triangles Exercise 19 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
State, whether the pairs of triangles given in the following figures are congruent or not:

Solution:
(i) In these triangles, corresponding sides are not equal. Hence these are not congruent triangles.
(ii) In the first A, third angle
= 180°-(40°+ 30°)
= 180° – 70°
= 110°
Now in these two triangles the sides and included angle of the one are equal to the corresponding sides and included angle.
Hence these are congruent triangles
(S.A.S. axiom)
(iii) In these triangles, corresponding two sides are equal but included angles are not-equal. Hence these are not congruent triangles.
(iv) In these triangles, corresponding three sides are equal.
Hence these are congruent triangles.
(S.S.S. Axiom)
(v) In these right triangles, one side and diagonal of the one, are equal to the corresponding side and diagonal are equal. Hence these are congruent triangles. –
(R.H.S. Axiom)
(vi) In these triangles two sides and one angle of the one are equal to the corresponding sides and one angle of the other are equal.
Hence these are congruent triangles.
(S.S. A. Axiom).
(vii) In A ABC. AB = 2 cm, BC = 3.5 cm and ∠C = 80° and in ∆ DEF,
DE = 2 cm, DF = 3.5 cm and ∠D = 80°

From the figure  we see that two corresponding sides  are equal but their included angles are not equal.
Hence, these are not congruent triangles

Question 2.
In the given figure, prove that:
∆ABD ≅ ∆ ACD

Solution:

Question 3.
Prove that:
(ii) ∠B = ∠D
(iii) AC bisects angle DCB

Solution:

Question 4.
Prove that:
(i) ∆ABD  ≡ ∆ACD
(ii) ∠B = ∠C

Solution:

Question 5.
In the given figure, prove that:
(i) ∆ACB ≅ ∆ECD
(ii) AB = ED

Solution:

Question 6.
Prove that:
(i) ∆ ABC ≅ ∆ ADC
(ii) ∠B = ∠D

Solution:

Question 7.
In the given figure, prove that: BD = BC.

Solution:

Question 8.
In the given figure ;
∠1 = ∠2 and AB = AC. Prove that:
(i) ∠B = ∠C
(ii) BD = DC
(iii) AD is perpendicular to BC.

Solution:

Question 9.
In the given figure prove tlyat:
(i) PQ = RS
(ii) PS = QR

Solution:

Question 10.
(i) ∆ XYZ ≅ ∆ XPZ
(ii) YZ = PZ
(iii) ∠YXZ = ∠PXZ

Solution:

Question 11.
In the given figure, prove that:
(i) ∆ABC ≅ ∆ DCB
(ii) AC=DB

Solution:

Question 12.
In the given figure, prove that:
(i) ∆ AOD ≅ ∆ BOC

Solution:

Question 13.
ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:
(ii)BD = CE

Solution:

Question 14.
Use the informations given in the following figure to prove triangles ABD and CBD are congruent.
Also, find the values of x and y.

Solution:

Question 15.
The given figure shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that:
(i) ∆ABD ≅ ∆ACD
(ii) AB=AC
(iii) ∠B = ∠C

Solution: