NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.
Construct the right-angled ∆ PQR where m ∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:
Steps of Construction:

1. Draw QR of length 8 cm.
   
2. At Q, draw QX ⊥ QR.
3. With R as centre, draw an arc of radius 10 cm.
4. Mark the meeting point of these two as P.

∆ PQR is now obtained.

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of Construction

1. Draw QR of length 4 cm.
2. At Q, draw QX ⊥ QR.
   
3. With R as centre, draw an arc of radius 6 cm.
4. Mark the meeting point of arc and QX as P.

∆ PQR is now obtained.

Question 3.
Construct an isosceles right-angled triangle ABC where m ∠ACB = 90° and AC = 6 cm.
Solution:
Steps of Construction

1. Draw AC of length 6 cm.
2. At C, draw CX ⊥ CA.
   
3. With C as centre, draw an arc of radius 6 cm to intersect CX at B.
4. Join AB. ∆ACB is now obtained.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2.

• Rational Numbers Class 7 Ex 9.1
• Rational Numbers Class 7 MCQ

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

Question 1.
Find the sum:

Solution:

Question 2.
Find :

Solution:

Question 3.
Find the product :

Solution:

Question 4.
Find the value of :

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3.

• Comparing Quantities Class 7 Ex 8.1
• Comparing Quantities Class 7 Ex 8.2
• Comparing Quantities Class 7 MCQ

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

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit percent or loss per cent in each case.

(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerater bought for ₹ 12,000 and sold at ₹ 13,500.
(c) Acupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.

Solution:

Question 2.
Convert each part of the ratio to percentage

(a) 3 : 1
(b) 2 : 3 : 5
(c) 1 : 4
(d) 1 : 2 : 5

Solution:

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of the price increase?
Solution:

Question 5.
1 buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
C.P. = ₹ 10,000
Profit = 20% of C.P.
= 20% of ₹ 10,000
= ₹ \(\frac { 20 }{ 100 } \) × 10,000 = ₹ 2,000
∴ S.P. = C.P. + Profit
= ₹ 10,000 + ₹ 2,000 = ₹ 12,000
Hence, I get ₹ 12,000 for it.

Question 6.
Juki sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she brought it?
Solution:

Question 7.
(i) Chalk contains calcium, carbon, and oxygen in the ratio of 10 : 3 : 12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Solution:

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:
C.P. = ₹ 275
Loss = 15% of C.P = \(\frac { 15 }{ 100 } \) × C.P
= \(\frac { 15 }{ 100 } \) × 275
= ₹ 41.25
S.P. = C.P. – Loss = 275 – 41.25 = ₹ 233.75
hence, she se11 it for ₹ 233.75.

Question 9.
Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹ 1,200 at 12% p.a.
(b) Principal = ₹ 7,500 aat 5% p.a.

Solution:

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:

Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2.

• Comparing Quantities Class 7 Ex 8.1
• Comparing Quantities Class 7 Ex 8.3
• Comparing Quantities Class 7 MCQ

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

EXERCISE 8.2

Question 1.
Convert the given fractional numbers to per cents.

(a) \(\frac { 1 }{ 8 } \)
(b) \(\frac { 5 }{ 4 } \)
(c) \(\frac { 3 }{ 40 } \)
(d) \(\frac { 2 }{ 7 } \)

Solution:

Question 2.
Convert the given decimal fractions to per cents.

(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35

Solution:

Question 3.
Estimate what part of the figures is coloured and hence find the percent which is coloured.

Solution:

Question 4.
Find:

(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg.

Solution:

Question 5.
Find the whole quantity if

(a) 5% of it is 600.
(b) 12% of it is ₹ 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 liters.

Solution:

Question 6.
Convert given percents to decimal fractions and also to fractions in the simplest forms.

(a) 25%
(b) 150%
(c) 20%
(d) 5%

Solution:

Question 7.
In a city, 30% are females, 40% are males, and the remaining are children. What percent are children?
Solution:
[100% (30% + 40%)] are children
⇒ [100% – 70% ] are children
⇒ 30% are children.

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Percentage of voters who voted = 60%
Percentage of voters who did not vote = (100 – 60)% = 40%
Total number of voters = 15000
Number of voters who did not vote = 40% of 15000
= ( \(\frac { 40 }{ 100 } \) × 15000 )
= 6000

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let her salary be ₹ P. Then, 10% of P = 400
⇒ \(\frac { 10 }{ 100 } \) × p = 400
⇒ \(\frac { P }{ 10 } \) = 400
⇒ P = 400 × 10
⇒ P = 4000
Hence, her salary is ₹ 4000.

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Solution:
Out of 100, 25% matches are won. Then, out of 20, the number of matches that the team won
= \(\frac { 25 }{ 100 } \) × 20
= \(\frac { 1 }{ 4 } \) × 20
= 5

We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2.

• Congruence of Triangles Class 7 Ex 7.1
• Congruence of Triangles Class 7 MCQ

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

Question 1.
Which congruence criterion do you use in the following?

Given:
So.
AC = DF
AB = DE
BC = EF
so ∆ ABC = ∆ DEF

Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆ PQR ≅ ∆ XYZ

Given : ∠MLN = ∠ FGH
∠NML = ∠GFH
ML = GF
So, ∆ LMN ≅ ∆ GFH

Given : EB = DB
AE = BC
∠A = ∠C = 90°
So, ∆ ABE ≅ ∆ CDB
Solution:

(a) SSS congruence criterion
(b) SAS congruence criterion
(c) ASA congruence criterion
(d) RHS congruence criterion.

Question 2.
You want to show that ∆ ART ≅ ∆ PEN,

(а) If you have to use SSS criterion, then you need to show
(i)AR = (ii) RT = (iii) AT =

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(i) RT = and (ii) PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ? (ii) ?
Solution:

Question 3.
You have to show that ∆ AMP = ∆ AMQ.
In the following proof, supply the missing reasons.


Solution:

Question 4.
In ∆ ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆ PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ∆ ABC = ∆ PQR? by AAA congruence criterion. Is he justified’? Why or why not?
Solution:
No! he is not justified because AAA is not a criterion for congruence of triangles.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆ RAT = ?

Solution:
∆ RAT ≅ ∆ WON

Question 6.
Complete the congruence statement:


Solution:
∆ BCA = ∆ BTA
∆ QRS = ∆ TPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that

1. the triangles are congruent
2. the triangles are not congruent. What can you say about their perimeters?

Solution:

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:

In ∆ ABC and ∆ DEF,
AB = DF (= 2 cm)
BC = ED (= 4 cm)
CA = EF (= 3 cm)
∠BAC = ∠EDF
∠ABC = ∠DEF
But ∆ ABC is not congruent to ∆ DEF.

Question 9.
If ∆ ABC and ∆ PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:
BC = RQ by ASA congruence rule.

Question 10.
Explain why ∆ ABC ≅ ∆ FED

Solution:
∠ABC = ∠FED (= 90°) BC = ED
∠ACB = ∠FDE
∵ The sum of the measures of the three angles of a triangle is 180°.
∆ ABC ≅ ∆ FED (By SAS congruence criterion)

We hope the NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2, drop a comment below and we will get back to you at the earliest.