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These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers InText Questions

NCERT Intext Question Page No. 250

Question 1.
Write the following numbers in generalised form.;
(i) 25
(ii) 73
(iii) 129
(iv) 302
Answer:
(i) 25 = 20 + 5 = 2 × 10 + 5 × 1 = 10 × 2 + 5
(ii) 73 = 70 + 3 = 7 × 10 + 3 × 1 = 10 × 7 + 3
(iii) 129 = 100 + 20 + 9
= 1 × 100 + 2 × 10 + 9 × 1
= 100 × 1 + 10 × 2 + 9
(iv) 302 = 3 × 100 + 0 × 10 + 2 × 1
= 100 × 3 + 10 × 0 + 2

Question 2.
Write the following in the usual form.
(i) 10 × 5 + 6
(ii) 100 × 7 + 10 × 1 + 8
(iii) 100 × a + 10 × c + b
Answer:
(i) 10 × 5 + 6 = 50 + 6 = 56
(ii) 100 × 7 + 10 × 1 + 8
= 700 + 10 + 8 = 718
(iii) 100 × a + 10 × c + b
= 100a + 10c + b = a c b

NCERT Intext Question Page No. 257

Question 1.
If the division N + 5 leave a remainder of 3, what might be the one’s digit of N?
Answer:
The unit digit when divided by 5 must leave a remainder of 3.50, the one’s digit must be either 3 or 8.

Question 2.
If the division N + 5 leaves a remainder of 1, what might be the ones digit of N?
Answer:
The one’s digit when divided by 5 must have a remainder of 1. So the one’s digit must be either 1 or 6.

Question 3.
If the division N + 5 leaves a remainder of 4. What might be the one’s digit of N?
Answer:
The one’s digit, when divided by 5 must leave a remainder of 4. So the one’s digit must be either 4 or 9.

NCERT Intext Question Page No. 259

Question 1.
Check the divisiblity of the following number by 9.
(i) 108
(ii) 616
(iii) 294
(iv) 432
(v) 927
Answer:
(i) 108
Sum of the digits = 1 + 0 + 8 = 9 which is divisible by 9.
∴ 108 is divisible by 9.

(ii) 616
Sum of the digits = 6 + 1 + 6
= 13 which is not divisible by 9.
∴ 616 is not divisible by 9.

(iii) 294
Sum of the digits = 2 + 9 + 4 = 15
which is not divisible by 9.
∴ 294 is not divisible by 9.

(iv) 432
Sum of the digit = 4 + 3 + 2 = 9
which is divisible by 9.
∴ 432 is divisible by 9.

(v) 927
Sum of the digits = 9 + 2 + 7 = 18
which is divisible by 9.
∴ 927 is divisible by 9.

Question 9.
Check the divisibility of the following number by 3.
(i) 108
(ii) 616
(iii) 294
(iv) 432
(v) 927
Answer:
(i) 108
Sum of the digits =1 + 0 + 8 = 9
which is divisible by 3.
∴ 108 is divisible by 3.

(ii) 616
Sum of the digits = 6 + 1 + 6 = 13
which is not divisible by 3.
∴ 613 is also not divisible by 3.

(iii) 294
Sum of the digits = 2 + 9 + 4 = 15
which is divisible by 3.
∴ 294 is also divisible by 3.

(iv) 432
Sum of the digits = 4 + 3 + 2 = 9
which is divisible by 3.
∴ 432 is also divisible by 3.

(v) 927
Sum of the digits = 9 + 2 + 7 = 18 which is divisible by 3.
∴ Thus 927 is also divisible by 3.

NCERT Intext Question Page No. 251

Question 1.
Check what the result would have been if Sundaram had chosen the numbers shown below.
1. 27
2. 39
3. 64
4. 17
Answer:
1. Chosen number = 27
Number with reversed digits = 72
Sum of the two numbers = 27 + 72 = 99
Now, 99 = 11 [9] = 11 [2 + 7]
= 11 [Sum of the digits of the chosen number]

2. Chose number = 39
Number with reversed digits = 93
Sum of the two numbers = 39 + 93 = 132
Now, 132 ÷ 11 = 12
i. e„ 132 = 11 [12] = 11 [3 + 9]
= 11 [Sum of the digits of the chose number]

3. Chosen number = 17
Number with reversed digits = 71
Sum = 17 + 71 = 88
Now, 88 = 11 [8] = 11 [1 + 7]
= 11 [Sum of the digits of the chosen number]

Question 2.
Check what the result would have been if Sundaram had chosen the numbers shown.
1. 17
2. 21
3. 96
4. 37
Answer:
1. Chosen number =17
Number with reversed digits = 71
Difference = 71 – 17 = 54 = 9 × [6]
= 9 x [Difference of the digits of the chosen number (7 – 1 = 6)]

2. Chosen number = 21
Number with reversed digits = 12
Difference = 21 – 12 = 9 = 9 × [1]
= 9 x [Difference of the digits of the chosen number (12 -1 = 1)]

3. Chosen number = 96
Number with reversed digits = 69
Difference = 96 – 69 = 27 = 9 × [3]
=9 x [Difference of the digits of the chosen number (9 – 1 = 3)]

4. Chosen number = 37
Number with reversed digits = 73
Difference = 73 – 37 = 36 = 9 × [4]
=9 x [Difference of the digits of the chosen number (7-3 = 4)]

NCERT Intext Question Page No. 252

Question 1.
Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.
1. 132
2. 469
3. 737
4. 901
Answer:
1. 132 Chosen number = 132
Reversed number = 231
Difference = 231 – 132 = 99
We have 99 ÷ 99 = 1, remainder = 0

2. 469 Chosen number = 469
Reversed number = 964
Difference = 964 – 469 = 495
We have 99 ÷ 99 = 5, remainder = 0

3. Chosen number = 737
Reversed number = 737
We have Difference = 737 – 737 = 0
0 ÷ 99 =0, remainder = 0

4. Chosen number = 901
Reversed number = 109
Difference = 901 – 109 = 792
We have 792 ÷ 99 = 8, remainder = 0
Forming three-digit number with given three digits

NCERT Intext Question Page No. 253

Question 1.
Check what the result would have been if Sundaram had chosen the numbers shown below.
1. 417
2. 632
3. 117
4. 937
Answer:
1. Chosen number = 417
Two other numbers with the same digits are 741 and 174
Sum of the three numbers

We have 1332 ÷ 37 = 36, remainder = 0

2. Chosen number = 632
Two other numbers are 263 and 326
Sum of the three numbers

We have 1221 ÷ 37 = 33, remainder = 0

3. Chosen number =117
Other numbers are 711 and 171
Sum of the three numbers

We have 1221 ÷ 37 = 33, remainder = 0

4. Chosen number = 937
Other two numbers are 793 and 379
Sum of the three numbers

We have 2109 + 37 = 57, remainder = 0

These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.1

Question 1.
Find the values of the letters in each of the following and give reasons for the steps involved.


Answer:
1. Let us see the sum in units column. It is A + 5 and we get 2 from this
(A+ 5 = 7 + 5= 12). So, A has to be 7
For sum in ten’s column, we have
1 + 3 + 2 = B
∴ B = 6

Question 2.
Here, there are three letters A, B and C whose values are to be found out.
Answer:
Let us see the sums in units column.
It is A + 8 and we get 3 from that so A has to be 5 (A + 8 = 5 + 8 = 13)
Now, for the sum in tens column, we have
I + 4 + 9= 14
∴ B = 4, C = 1
∴ The given puzzle is solved,
i. e. A = 5, B = 4 and C = 1

Question 3.
Units digit of A x A is A
Answer:
∴ A = 1 or 5 or 6
When, A = 1
II ≠ 9A
∴ A ≠ 1
When, A = 5
15 x 5 ≠ 9A;
∴ A ≠ 5
When, A = 6
16 x 6 = 96
∴ A = 6

Question 4.
Here, there are two letters A and B whose values are to be found out.
Answer:
B + 7 gives A
and A + 3 gives 6
The possible values are 0 + 7 = 7
A = 7 but 7 + 3 ≠ 6 so it is not acceptable.
1 + 7 = 8
A = 8 but 8 + 3 ≠ 6 so not acceptable
2 + 7 = 9
A = 9 but 9 + 3 ≠ 6 so not acceptable
3 + 7= 10
A = 0 but 1+0 + 3 ≠ 6 so, not acceptable.
4 + 7= 11
A = 1 but 1 + 1 + 3 ≠ 6 so not acceptable.
5 + 7= 12
A = 2 Also 1 + 2 + 3 = 6
So B = 5 works and then we get A as 2
∴ the puzzle is solved as shown below

i. e. A = 2 and B = 5.

5. We need B x 3 = □ B
Since 5 x 3 = 15
∴ Possible value of B can be 5.
Also 0 x 3 = 0, i.e., B = 0 can be another possible value
∵ A x 3 = A + 0 = A
∴ Possible value of A = 5 or A = 0
∴ Since C ≠ 0
∴ Possible value of A = 5
30, B must be
Thus, B = 0.

6. B can be either 5 or 0.
∵ A x 5 = A
∴ B must be 0
Again A can either 5 or 0
∴ C ≠ 0
∴ A ≠ 0
∴ A must be equal to 5
Thus, we have
Therefore, A = 5, B = 0, C = 2

7. B can be 2, 4, 6 or 8
We need product 111 or 222, or 333 or 444 or 888 out of them 111 and 333 are rejected Possible products are 222, 444 or 888 To obtain
The possible value is B = 4

∴ A can be either 2 or 7

∴ A x 6 = 7 x 6 is the accepted value

Thus, A = 7 and B = 4

Thus, A = 7 and B = 9

9. ∵  8 – 1 = 7
∴ B = 7
7 + 4= 11 ∴ A = 4

Now, we have
Thus A = 4 and B = 7

10. 10 – 2 = 8
∴ A = 8
Also 9 – 8 = 1

∴ B = 1
Now, we have
Thus, A = 8 and B = 1

These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.2

Question 1.
It 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Answer:
21Y5 is a multiple of 9
∴ 2 + 1 + y + 5 = 8 + y must be divisible by 9
(8 + y) should be 9, 18, 27,… etc.
Since y is a digit.
8 + y = 9
y = 9 – 8 = 1
∴ The value of y = 1.

Question 2.
If 31z5 is a multiple of 9, where Z is a digit. What is the value of z? You will find that there are two answers for the last problem. Why is this so?
Answer.
31z5 = 3 + l + z + 5 = 9 + z
31z5 is divisible by 9
∴ 9 + z must be equal to 9, 18, 27,… etc.
Z is a single digit number.
9 + z is one of these numbers.
9 + z = 9 then z = 0
9 + z=18 thenz= 18-9 = 9
∴ The value of z = 0 or 9.

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
Answer:
Since 24x is a multiple of 3, so sum of digits i.e. 6 + x is a multiple of 3.
6 + x is one of these numbers: 0, 3,6, 9 12, 15, 18,… But since x is a digit,
∴ 6 + x = 6 or 9 or 12 or 15
∴ x = 0, or 3 or 6 or 9
The value of x = 0 or 3 or 6 or 9

Question 4.
31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer:
Sum of the digits =3+l+z+5=9+z 9 + z is a multiple of 3
So, 9 + z is one of the numbers: 0, 3, 6, 9, 12, 15,…
But since z is a digit, 9 + z = 9 or 12 or 15 or 18 or 21…
∴ The Value of z = 0 or 3 or 6 or 9 or 15 or 18.

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

Question 1.
Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a+ 3b) and (x + 5)
(v) (2pq + 3q²) and (3pq -2q²)
(vi) \(\left(\frac{3}{4} a^{2}+3 b^{2}\right)\) and \(\left(a^{2}-\frac{2}{3} b^{2}\right)\)
Answer:
(2x + 5) (4x – 3) = 2x (4x – 3) + 5 (4x – 3)
= 2x × 4x + 2x (-3) + 5 × 4x + 5 (- 3)
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15

(ii) (y – 8) (3y – 4) = y (3y – 4) – 8 (3y – 4)
= y × 3y + y (-4) – [8 × 3y + 8 (-4)]
= 3y² – 4y – (24y – 32)
= 3y² – 4y – 24y + 32
= 3y² – 28y + 32

(iii) (2.5l – 0.5m) (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) – 0.5m (2.5l + 0.5m)
= 2.5l × 2.5l + 2.5l × 0.5m – [0.5m × 2.5l + 0.5m × 0.5m]
= 6.25l² + 1.25lm – 1.25lm – 0.25m²
= 6.25l² – 0.25m²

(iv) (a + 3b) (x + 5) = a × (x + 5) + 3b × (x + 5)
= ax + 5a + 3bx + 15b
= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) × (3pq – 2q²)
= 2pq × (3pq – 2q²) + 3q² × (3pq – 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²q² + 5pq³ – 6q⁴


= 3a⁴ -2a²b² +12a²b² – 8b⁴
= 3a⁴ + 10a²b² – 8b⁴

Question 2.
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a² + b) (a + b²)
(iv) (p² – q²) (2p + q)
Answer:
(i) (5 – 2x) (3 + x) = 5 × (3 + x) – 2x (3 + x)
= 5 × 3 + 5 × x – (2x) × 3 – 2x (x)
= 15 + 5x – 6x – 2x² = 15 – x – 2x²

(ii) (x + 7y)(7x – y) = x × (7x – y) + 7y × (7x – y)
= x × 7x – x × y + 7y × 7x – 7y × y
= 7x² – xy + 49xy – 7y²
= 7x² + 48xy – 7y²

(iii) (a² + b)(a + b²) = a² × (a + b²) + b × (a + b²)
= a² × a + a² × b² + b × a + b × b²
= a^{3</sup + a2b2 + ab + b3}

(iv) (p² – q²) (2p + q)
= p² × (2p + q) – q² (2p + q)
= p² × 2p + p² × q – q² × 2p – q² × q
= 2p³ + p²q – 2pq² – q³

Question 3.
Simplify
(i) (x² – 5) (x + 5) + 25
(ii) (a² + 5) (b³ + 3) + 5
(iii) (t + s²) (t² – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y) (x² – xy + y²)
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 45x + 12y
(viii) (a + b + c) (a + b – c)
Answer:
(i) (x² – 5) (x + 5) + 25
= x² × (x + 5) – 5 (x + 5) + 25
= x²× x + x² × 5 – 5 × x – 5 × 5 + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x

(ii) (a² + 5) (b³ + 3) + 5
= a² × (b³ + 3) + 5 × (b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5³ + 20

(iii) (t + s²) (t² – s) = t × (t² – s) + s² × (t² – s)
= t × t² – t × s + s² × t² – S² × s
= t³ – ts + s²t² – s³

(iv) (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)
= a × (c – d) + b × (c – d) + a × (c + d) – b × (c + d ) + 2 × ac + 2 × bd
= a × c – a × d+ b × c – b × d + a × c + a × d – b × c – b × d + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac – ad + ad + bc – bc – bd – bd + 2bd
= 4ac – 2bd + 2bd = 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x × (2x + y) + y (2x + y) + x × (x – y) + 2y × (x – y)
= x × 2x + x × y + y × 2x + y × y + x × x + x × (-y) + 2y × x + 2y × (-y)
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y² = 2x² + x² + xy + 2xy + 2xy – xy + y² – 2y²
= 3x² + 5xy – xy – y²
= 3x² + 4xy – y²

(vi) (x + y) (x² – xy + y²)
= x × (x² – xy + y²) + y × (x² – xy + y²)
= x × x² + x × (-xy) + x × y² + y × x² + y × (-xy) + y × y²
= x³ – x²y + xy² + x²y – xy² + y³
= x³ – x²y + x²y + xy² – xy² + y³
= x³ + y³

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x × (1.5x + 4y + 3) – 4y × (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x × 1.5x + 1.5x × 4y + 1.5x × 3 – 4y × 1.5x – 4y × 4y – 4y × 3 – 4.5x + 12y
= 2.25xz + 6xy + 4.5x – 6xy – 16y2
– 12y – 4.5x + 12y
= 2.25x² + 6xy – 6xy + 4.5x – 4.5x – 16y² – 12y + 12y
= 2.25x² + 0 + 0 – 16y² + 0
= 2.25x² – 16y²

(viii) (a + b + c) (a + b – c)
= a × (a + b – c) + b × (a + b – c) + c × (a + b – c)
= a × a + a × b + a × -c + b × a + b × b – b × c + c × a + b × c – c × c
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + ab + ab – ac + ac – bc + bc + b² – c²
= a² + 2ab + o + o + b² – c²
= a² + b² – c² + 2ab

These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions

NCERT In-text Question Page No. 158

Question 1.
Find the percentage of children of different heights for the following data.
Answer:

Question 2.
A shop has the following number of shoe pairs of different sizes.
Size 2 : 20
Size 3 : 30
Size 4 : 28
Size 5 : 14
Size 6 : 8
Write this information in tabular form as done earlier and find the percentage of each shoe size available in the shop.
Answer:

NCERT In-text Question Page No. 159 & 160

Question 1.
A collection of 10 chips with different colours is given.
Answer:

Question 2.
Mala has a collection of bangles. She has 20 gold bangles and 10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as done in the above example?
Answer:
We have

Note: We can use percentages for comparing various quantities.

NCERT In-text Question Page No. 160

Question 1.
Look at the examples below and in each of them, discuss which is better for comparison.
In the atmosphere, 1 g of air contains:
(i) .78g Nitrogen – 78 % Nitrogen
.21 g Oxygen or 21 % Oxygen
.01g Nitrogen – 1 % Other Gas
(ii) A shirt has:
\(\frac{2}{5}\) Cotton 60% Cotton
or
\(\frac{2}{5}\) Polyester 40% Polyester
Answer:
The second one (i.e. percentage) is better for comparison in both the cases.

NCERT In-text Question Page No. 161

Question 1.
Convert the following to per cents:
(a) \(\frac{12}{16}\)
(b) 3.5
(c) \(\frac{49}{50}\)
(d) \(\frac{2}{2}\)
(e) 0.05
Answer:
(a) \(\frac{12}{16}\) = (\(\frac{12}{16}\) x 100) % = ( \(\frac{3}{4}\) x 100) %
= (3 x 25)% = 75%

(b) 3.5 = (3.5 x 100)% = 350%

(c) \(\frac{49}{50}\) ( \(\frac{49}{50}\) x 100)%
= (\(\frac{49}{50}\) x 100)%
= (49 x 2)% = 98%

(d) \(\frac{2}{2}\) = ( \(\frac{2}{2}\) x 100)%
= (1 x 100)%
= 100%
(e) 0.05 = (0.05 x 100) % = 5 %

Question 2.
(i) Out of 32 students, 8 are absent.
What per cent of the students are absent?
(ii) There are 25 radios, 16 of them are out of order. What per cent of radios are out of order?
(iii) A shop has 500 parts, out of which 5 are defective. What per cent are defective?
(iv) There are 120 voters, 90 of them voted yes. What per cent voted yes?
Answer:

= (3 x 25)% = 75%
75% voters voted ‘yes’

NCERT In-text Question Page No. 162

Question 1.
Fill in the blanks
(i) 35% + ……………. % = 100%.
(ii) 64% + 20% + ……………. % = 100%
(iii) 45% = 100% – ……………. %.
(iv) 70% = ……………. % – 30%
Answer:
(i) 100% – 35% = 65%
35% + 65% = 100%
(ii) 64%+ 20% =84%
100% – 84% = 16%
64% + 20% + 16% = 100%
(iii) 100%-45% =55%
45% = 100% – 55%
(iv) 70% + 30% = 100%
70% = 100% – 30%

Question 16.
If 65% of students in a class have a bicycle, what per cent of the student do not have bicycles?

Answer:
Part of students having bicycles = 65%
Remaining part of students
= 100% – 65% = 35%
Thus, 35% of students do not have bicycles.

Question 17.
We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what per cent are mangoes?
Answer:
Quantity of apples = 50%
Quantity of oranges = 30%
Quantity of mangoes in the basket
= 100% – (50% + 30%)
= 100% -80% = 20%
Thus, the percent of mangoes in the basket = 20%

NCERT In-text Question Page No. 163

Question 1.
What per cent of these figures are shaded?

Answer:
(i) Fraction which is shaded
\(=\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)=\left(\frac{1+1+1}{4}\right)=\frac{3}{4}\)
Percentage of shaded part
= \(\frac { 3 }{ 4 }\) x 100% = 75%
(ii) Fraction of diagram which is shaded
\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{2+1+1}{8}=\frac{4}{8}=\frac{1}{2}\)
Percentage of diagram which is shaded = \(\frac { 1 }{ 2 }\) x 100% = 50%

NCERT In-text Question Page No. 164

Question 1.
Find
(a) 50% of 164
(b) 75% of 12]
(c) 12 \(\frac { 1 }{ 2 }\) % of 64
Answer:
(a) 50% of 164 = \(\frac { 50 }{ 100 }\) x 164
= \(\frac { 1 }{ 2 }\) x 164 = 82

(b) 75% of 12 = \(\frac { 75 }{ 100 }\) x 12 = \(\frac { 3 }{ 4 }\) x 12
= 3 x 3 = 9

(c) 12\(\frac { 1 }{ 2 }\) % of 64 = \(\frac { 25 }{ 2 }\)% or 64
= \(\frac{2}{100} \times 64=\frac{25}{100} \times \frac{64}{2}\)
= \(\frac { 1 }{ 8 }\) x 64 = 8

Question 2.
8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain?
Answer:
Number of children who like getting wet in rain
= 8% of 25 = \(\frac { 8 }{ 100 }\) x 25 = 2

Question 3.
9 is 25% of what number?
Answer:
Let the required be P
25% of P = 9
\(\frac { 25 }{ 100 }\) x P = 9
25 P = 9 x 100
P = \(\frac { 9 x 100 }{ 25 }\) = 9 x 4 = 36
The required number is 36.

Question 4.
75% of what number is 15?
Answer:
Let the required number be p
75% of P = 15
= \(\frac { 75 }{ 100 }\) x P = 15
\(\frac { 3 }{ 4 }\) x P = 15
\(\frac { 3P }{ 4 }\) = 15
3P = 15 x 4
P = \(\frac{15 \times 4}{3}\) = 5 x 4 = 20
The required number is 20

NCERT In-text Question Page No. 166

Question 1.
Divide 15 sweets between Manu and Sonu so that they get 20% and 80% of them respectively.
Answer:
Share of Manu = 20% of 15 sweets
= \(\frac { 20 }{ 100 }\) x 15 sweets
= 3 sweets
Share of Sonu = 80% of 15 sweets
= \(\frac { 80 }{ 100 }\) x 15
= \(\frac { 4 }{ 5 }\) x 15
= 4 x 3 = 12 sweets
Manu share = 3 sweets and Sonu share =12 sweets.

Question 1.
If the angles of a triangle are in the ratio 2:3:4, find the value of each angle.
Answer:
We know that sum of three angles of a triangle =180°
Let the three angles be 2x, 3x, and 4x
2x +3x + 4x = 180°
9x = 180°
x = \(\frac { 180° }{ 9 }\) = 20°
The three angles are 2(20° ); 3(20° ) and 4(20°)
i.e. 40°, 60° and 80°

NCERT In-text Question Page No. 167

Question 1.
Find percentage of increase or decrease
(a) Price of shirt decreased from ₹ 80 to ₹ 60
(b) Marks in a test increased from 20 to 30
Answer:
(a) Initial price of the shirt = ₹ 80
Decrease price of the shirt = ₹ 60
Decrease in the price
= ₹80 – ₹60 = ₹20
Per cent of decrease in price 20
= \(\frac { 20 }{ 80 }\) x 100%
= ₹ 25%

(b) Initial marks = 20
Increased marks = 30
Increase in marks = 30 – 20 = 10
Percent of increase in marks
= \(\frac { 10 }{ 20 }\) x 100% = 50%

Question 2.
My mother says in her childhood petrol was ₹ 1 a litre. It is ₹ 52 per litre today. By what per centage has the price gone up?
Answer:
Initial Price = ₹ 1
Increase price = ₹ 52
Increase in price = ₹52 – ₹1 = ₹51
Percentage of increase in petrol
= \(\frac { 51 }{ 1 }\) x 100%
= 5100%

NCERT In-text Question Page No. 169

Question 1.
A shopkeeper bought a chair for ₹ 375 and sold it for ₹ 400. Find the gain percentage.
Answer:
C.P of a chair = ₹ 375
S.P of a chair = ₹ 400
Profit = S.P – C.P
= ₹ 400 – ₹ 375 = ₹ 25
Profit percentage = \(\frac { Profit }{ C.P }\) x 100%
= \(\frac { 25 }{ 375 }\) x 100%
= \(\frac { 1 }{ 15 }\) x 100% = \(\frac { 20 }{ 3 }\)%
= 6\(\frac { 2 }{ 3 }\)%

Question 2.
Cost of an item is₹50 It was sold with a profit of 12%. Find the selling price.
Answer:
We have
CP of the item = ₹50
Profit % = 12%
Profit = 12% of ₹50
= ₹\(\frac { 12 }{ 100 }\) x 50 = ₹6
SP = CP + Profit
= ₹50 + ₹6 = ₹56

Question 3.
An article was sold for ₹250 with a profit of 5%. What was its cost price?
Answer:
Let the cost price be ₹ x.
Profit % = 5%
∴ Profit = 5% of x = ₹ \(\frac { 5 }{ 100 }\) x x =₹ \(\frac { x }{ 20 }\)
∴ SP = CP + Profit
∴ ₹ 250 = ₹ x + \(\frac { x }{ 20 }\) [∵ ₹ = ₹ 250]
or ₹ 250 = ₹ \(\frac { 21 }{ 20 }\)x
or x = \(\frac{250 \times 20}{21}=\frac{5000}{21}=238 \frac{2}{21}\)
Thus cost price of the article = ₹ 238 \(\frac { 2 }{ 21 }\)

Question 4.
An item was sold for ₹ 540 at a loss of 5% what was its cost price?
Answer:
Let the cost rice be ₹ x
Loss% = 5%
Loss = 5% of ₹ x
= \(\frac { 5 }{ 100 }\) x x = \(\frac { x }{ 20 }\)
Selling price = Cost price – Loss
540 = x – \(\frac { x }{ 20 }\)
540 = \(\frac{20 x-x}{20}=\frac{19 x}{20}\)
19x = 540 x 20
x = \(\frac{540 \times 20}{19}\)
= \(\frac{10800}{19}=568 \frac{8}{19}\)
or ₹ 568.40 (approx.)
Cost price of the article
= ₹ 568\(\frac{8}{19}\) or ₹ 568.42

NCERT In-text Question Page No. 170

Question 1.
₹ 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one year.
Answer:
Principal (P) = ₹ 10,000
Rate (R) = 5% p.a.
Time (T) = 1 year

Question 2.
₹ 3,500 is given at 7% p.a. rate of interest. Find the interest which will be received at the end of two years.
Answer:
Principal (P) = ₹ 3,500; Rate (R) = 7% p.a. Time (T) = 2 years
∴ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{3,500 \times 7 \times 2}{100}\)
= ₹ 35 x 7 x 2 = ₹ 490.

Question 3.
₹ 6,050 is borrowed at 6.5% rate of interest p.a. Find the interest and the amount to be paid at the end of 3 years.
Answer:
Principal (P) = ₹ 60,50
Rate (R) = 6.5% p.a.
Time (T) = 3% years
∴ Simple interest

Since, Amount = Principal + Interest
= ₹ 6,050 + ₹ 1,17.75 = ₹ 7,229.75
Thus, the amount to be paid at the end of 3 years = 7,229.75.

Question 4.
₹ 7000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the amount to be paid at the end of the second year.
Answer:
Principal (P) = ₹ 7,000
Rate = 3.5% p.a.
Time (T) = 2 years
∴ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\begin{aligned}
&7,000 \times 3.5 \times 2\\
&0
\end{aligned}\) = ₹ 70 x \(\frac{35}{10}\) x 2
= ₹ 7 x 35 x 2 = ₹ 490
Since, Amount = Principal + Interest
Amount to be paid at th end of 2nd year = ₹ 7,000 + ₹ 490 = ₹ 7,490.

NCERT In-text Question Page No. 171

Question 1.
You have ₹ 2,400 in you account and the interest rate is 5%. After how many years would you earn ₹ 240 as interest.
Answer:
Here, principal (P) = 2,400
Rate (R) = 5% p.a.
Time (T) = ?
Simple interest = ₹ 2,40
∵ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
∴ 240 = \(\frac{2400 \times 5 \times \mathrm{T}}{100}\)
∴ T = \(\frac{240 \times 100}{2400 \times 5}\) = 2 years
Thus, an interest of ? 240 will be obtained after 2 years.

Question 2.
On a certain sum, the interest paid after 3 years is ₹ 450 at 5% rate of interest per annum. Find the sum.
Answer:
Let the sum be ‘P’
450 = \(\frac{3 \mathrm{P}}{20}\)
Interest (I) = ₹ 450
3P = 450 x 20
Rate (R) = 5% p.a
P = \(\)
Time (T) = 3 years = 150 x 20
I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ 3000
450 = \(\frac{\mathrm{P} \times 5 \times 3}{100}\)
The required sum is ₹ 3000.

These NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions

NCERT In-text Question Page No. 140 and 141
Question 1.
Lengths of the sides of the triangles are indicated. By applying the SSS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form:


Answer:
(i) InΔABC and ΔPQR such that
AB = 1.5 cm; PQ =1.5 cm
∴ AB = PQ
BC = 2.5 cm; QR = 2.5 cm
∴ BC = QR
AC = 2.2 cm and PR = 2.2 cm
∴ AC = PR
∴ The three sides of ΔABC are equal to the three sides of ΔPQR.
∴ ΔABC ≅ ΔPQR
( By SSS congruence criterion)

(ii) In ΔDEF and ΔLMN
DE = 3.2 cm, MN = 3.2
∴ DE = NM
EF = 3 cm; LM = 3 cm
∴ EF = LM
DF = 3.5 cm, LN = 3.5 cm
∴ DF = LN
∴ The three sides of ΔDEF are equal to three sides of ALMN.
ΔDEF ≅ ΔNML
(By SSS congruence criterion)

(iii) In ΔABC and ΔPQR,
∴ AC = 5 cm, PR = 5 cm
∴ AC = PR
BC = 4 cm, PQ = 4 cm
∴ BC = PQ
AB = 2 cm, QR = 2.5 cm
AB ≠ QR
AB ≠ QR
ΔABC and ΔPQR are not congruent,

(iv) We have ΔABD and ΔADC., such that
AB = 3.5 cm, AC = 3.5 cm
∴ AB = AC
BD = 2.5 cm, CD = 2.5 cm
AD is common ( AD = AD)
since three sides of ΔABD are equal to the three sides of ΔACD
∴ The two triangles are congruent.
ΔABD ≅ ΔACD
(SSS Congruence criterion)

Question 2.
In the given figure, AB = AC and D is the mid- point of \(\overline{\mathrm{BC}}\)
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC? Give reasons.
(iii) Is ∠B = ∠C ? Why?

Answer:
(i) Here, AB = AC and D is the midpoint of \(\overline{\mathrm{BC}}\)
∴ BD = DC
In ΔADB and ΔADC,
AB = AC, BD = DC
AD = AD (common)

(ii) The three sides of ΔADB are equal to the three sides of ΔADC
By SSS rule of congruence criterion
ΔADB ≅ ΔADC

(iii) Since ΔADB ≅ ΔADC,
∠B = ∠C (CPCT)

Question 3.
In the given figure, AC = BD and AD = BC. Which of the following statement is meaningfully written?
(i) ΔABC ≅ ΔABD
(ii) ΔABC ≅ ΔBAD

Answer:
Have AC = BD and AD = BC
In the given triangles ABC and ABD
AB = AB (common)
BD = AC (Given)
AD = BC (Given)
Three sides of ΔABD are equal to the three sides of ΔABC .So the two triangles are congruent,
∴ A ↔ B; B ↔ A; D ↔ C
(i) ΔABC ≅ ΔABD is not true or meaningless.
(ii) ΔABC ≅ ΔBAD is true or meaningful.

NCERT In-text Question Page No. 143 & 144
Question 1.
Which angle is included between the sides \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{EF}}\) is ΔDEF.
Answer:
In ADEF, the angle between \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{EF}}\), is ∠DEF.

Question 2.
By applying SAS congruence rule, you want to establish that ΔPQR ≅ ΔFED. It is given that PQ = FE and RP = DF. What additional information is needed to establish the congruence?
Answer:
We have ΔPQR = ΔFED
[Using SAS congruence rule]
∴ P ↔ F, Q ↔ E, R ↔ D
and PQ = FE and RP = DF.

Question 3.
The given measures of some parts of the triangles are indicated. By applying SAS congruence rule, state the pairs of congruence triangles, if any in each case. In case of congruent triangles, write them in symbolic form.


Answer:
(i) ΔABC and ΔDEF are congruent
AB = 2.5 cm; DE = 2.5 cm
∴ AB = DE
AC = 2.8 cm, DF = 2.8 cm
∴ AC = DF
∠A = 80°; ∠D= 70°
∴ ∠A ≠∠D
∴ ΔABC and ΔDEF are not congruent.

(ii) In ΔDEF and ΔPQR
EF = 3 cm; QR = 3 cm
∴ EF = QR
FD = 3.5 cm; PQ = 3.5 cm
∴ FD = PQ
∠F = 40°, ∠Q = 40°
∴ ∠F = ∠Q
∴ Two sides of ΔADE and their included angle are equal to the corresponding sides and their included angle of ΔPQR
∴ The two triangles are congruent.
ΔDEF ≅ ΔPQR
(SAS Congruence rule)

(iii) In ΔABC and ΔPQR.
AC = 2.5 cm, PR = 2.5 cm
∴ AC = PR
BC = 3 cm; PQ = 3 cm
∴ BC = PQ
∠C = 35° = ∠P = 35°
∴ ∠C = ∠P
∴ Two sides of ΔABC and their included angle are equal to the two corresponding sides and their included angle of ΔPQR.
∴ ΔABC ≅ ΔRQP

(iv) In the ΔPRS and ΔQPR
PQ = 3.5 cm, SR = 3.5 cm
∴ PQ = SR
PR = PR (PR is common)
∠QPR = 30°∠PRS = 30°
∴ Two sides of ΔQPR and their included angle are equal to the corresponding sides and their included angle of ΔPRS.
∴ ΔPQR ≅ ΔRPS
(SAS congruence rule)

Question 4.
In the given Figure, \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O.
(i) State the three pairs of equal parts in two triangles AOC and BOD.
(ii) Which of the following statements are true?
(a) ΔAOC ≅ ΔDOB
(b) ΔAOC ≅ ΔBOD

Answer:
Since \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O.
AO = BO and CO = DO
∠AOC = ∠BOD
(vertically opposite angles are equal)

(i) In ΔAOC and ΔBOD
we have, AO = BO and CO = DO
∠AOC = ∠BOD

(ii) From the above given relation
ΔAOC ≅ ΔBOD
(By SAS congruence rule)
(a) The statement ΔAOC ≅ ΔDOB is false.
(b) The statement ΔAOC ≅ ΔBOD is true.

NCERT In-text Question Page No. 145 & 146
Question 1.
What is the side included between the angles M and N of ΔMNP?
Answer:
In AMNP, the side included between M and N is \(\overline{\mathrm{MN}}\).

Question 2.
You want to establish ΔDEF ≅ ΔMNP, using the ASA congruence rule. You are given that ∠D = ∠M and ∠F = ∠P. What information is needed to establish the congruence? (Draw a rough figure and then try!)
Answer:
To establish ΔDEF ≅ ΔMNP, using ASA congruence rule, we need the side containing ∠D and ∠F to be equal to the side containing ∠M and ∠P.

i. e. We need \(\overline{\mathrm{DF}}=\overline{\mathrm{MP}}\).

Question 3.
In the given figure, PL⊥OB and PM⊥OA such that PL = PM. Prove that ΔPLO ≅ ΔPMO

Answer:
In ΔPLO and ΔPMO, we have
∠PLO = ∠PMO = 90°
(Given)
OP = OP (Hypotenuse – common side)
PL = PM (Given)
By using RHS Congruency, we get
ΔPLO ≅ ΔPMO

Question 4.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In case of congruence, write it in symbolic form.
ΔDEF
(i) ∠D = 60°, ∠F = 80°, ∠DF = 5 cm
(ii) ∠D = 60°, ∠F = 80°, DF = 6cm
(iii) ∠E = 80°, ∠F = 30°, EF = 5 cm

ΔPQR
(i) ∠Q = 60°, ∠R = 80°, QR = 5cm
(ii) ∠Q = 60°, ∠R = 80°, QP = 6cm
(iii) ∠P = 80°, PQ = 5 cm, ∠R = 30°
Answer:
(i) Since ∠D = ∠Q (each = 60°)
∠F = ∠R (each = 80°)
Included side DF = Included side QR [each = 5 cm]
∴ Using ASA congruence rule, we can say that two triangles are congruent.
Also, D ↔ Q, F ↔ R and E ↔ P
∴ ΔDEF ≅ ΔQPR

(ii) Here, QP is not the included side.
∴ The given triangles are not congruent.

(iii) Here, PQ is not the included side.
∴ The given triangles are not congruent.

Question 5.
In the given figure ray AZ bisects ∠DAB as well as ∠DCB.
(i) State the three pairs of equal parts in triangles BAC and DAC.
(ii) Is ΔBAC ≅ ΔDAC? Give reasons.
(iii) Is AB = AD? justify your answer.
(iv) Is CD = CB? Give reasons.

Answer:
(i) AC is the bisector of ∠DAB as well as of ∠DCB
∴ ∠DAC = ∠BAC and ∠DCA = ∠BCA
In ΔBCA and ΔDAC equal parts are
AC = AC (Common)
∠DAC = ∠BAC (AC is a bisector)
∠DCA = ∠BCA (AC is a bisector)

(ii) From the above relation, the two triangles are congruent (using ASA congruence rule)
A ↔ A C ↔ C and D ↔ B
∴ ΔADC ≅ ΔABC
or
ΔBAC ≅ ΔDAC

(iii) Since ΔBAC ≅ ΔDAC The corresponding parts are equal.
∴ AB = AD

(iv) Since ΔBAC ≅ ΔDAC
∴ CD = CB (Corresponding parts are equal)

NCERT In-text Question Page No. 148
Question 1.
In the given figure measures of some parts of triangles are given. By applying R.H.S congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.


Answer:
(i) In the right ΔPQR and right ΔDEF Hypotenuse PR = Hypotenuse DF (each 6 cm)
PQ = 3 cm, DE = 2.5 cm
PQ ≠ DE
∴ ΔPQR is not congruent to ΔDEF

(ii) Right ΔABC and right ΔABD we have two right triangles ABC and ABD such that
Hypotenuse AB = Hypotenuse BA (common)
AC = 2 cm, BD = 2 cm
∴ AC = BD ∠C = ∠D = 90°
Buy using RHS congruence rule, we can say that the two right triangles are congruent
∴ ΔABC ≅ ΔBAD

(iii) Right ΔABC and Right ΔACD, We have right ΔABC and right ΔACD such that Hypotenuse AC = Hypotenuse AC (common) side \(\overline{\mathrm{AB}}\) = side \(\overline{\mathrm{AD}}\) (each 3.6 cm)
∴ The two right triangles are congruent.
∴ ΔABC ≅ ΔADC

(iv) Right ΔPQS and Right ΔPRS
Hypotenuse PQ = Hypotenuse PR (each 3 cm)
side PS = side PS (common)
∴ using R.H.S congruence rule that two right triangles are congruent.
∴ i.e., ΔPQS ≅ ΔPRS

Question 2.
It is to be established by RHS congruence rule that ΔABC ≅ ΔRPQ. What additional information is needed, if it is given that ∠B = ∠P = 90° and AB = RP?
Answer:
We have ΔABC ≅ ΔRPQ
Since, ∠B = 90° ⇒ Side AC is hypotenuse and ∠P = 90° ≅ Side RQ is hypotenuse.
The required information needed is Hypotenuse
AC = Hypotenuse RQ.

Question 3.
In the given figure, BD and CE are altitudes of ΔABC such that BD = CE.
(i) State the three pairs of equal parts in ΔCBD and ΔBCE.
(ii) Is ΔCBD ≅ ΔBCE? why or why not?
(iii) Is ∠DCB = ∠EBC? why or why not?

Answer:
(i) Hypotenuse BC = Hypotenuse BC
(common)
side BD = side CE (Given)
∠BEC = ∠BDC= 90°
The above are the three equal parts in ΔCBD and ΔBCE

(ii) ∠D = ∠E and ∠B = ∠C
∴ ΔCBD ≅ ΔBCE (By RHS)

(iii) Since ΔCBD ≅ ΔBCE
∠DCB =∠EBC
( ∵ their corresponding parts are equal)

Question 4.
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC? why or why not?
(iii) Is ∠B = ∠C? why or why not?
(iv) Is BD = CD? why or why not?

Answer:
(i) The three pairs of equal parts in ΔADB and ΔADC are
AD = AD (common)
Hypotenuse AB = Hypotenuse AC
(Given)
∠ADB = ∠ADC (each 90°)

(ii) A ↔ A, B ↔ C, D ↔ D
∴ ΔADB ≅ ΔADC (RHS)

(iii) ΔADB ≅ ΔADC
As, corresponding parts are equal.
∴ ∠B = ∠C

(iv) ΔADB ≅ ΔADC
As corresponding parts are equal.
BD = CD