NCERT Solutions for Class 7 Maths PDF

NCERT Solutions Class 7 Maths PDF format. Download Exemplar problems book, NCERT solutions of Science and answers. Complete description of each question is given in the solutions. Practice maths with Vedic Maths to improve your calculations and make it faster. Download 7 Maths Offline App in English Medium and 7 गणित Offline App in हिंदी मीडियम. These Apps work without internet, once downloaded.



NCERT Solutions Class 7 Maths PDF

Download 7 Maths Offline App in English Medium and 7 गणित Offline App in हिंदी मीडियम.

Table of Contents

NCERT Solutions for Class 7 Maths




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Hindi Medium NCERT Solutions for class 7 Maths is now available for the current academic year 2019-20 onward. यदि विद्यार्थियों या अभिभावकों की तरफ से कोई सलाह वेबसाइट को सुधारने के लिए हो तो अवश्य दे। आपका योगदान अन्य विद्यार्थिओं के लिए मददगार होगा। इस वर्ष 2019-20 में लगभग सभी विषयों के हिंदी और अंग्रेजी माध्यम के हल उपलब्ध होंगे। Your suggestion is always valuable for us in improving website as well as contents. We are also working for UP Board and Bihar Board also. The contents will be online April onward. NCERT Books for all Subjects are available in PDF as well as ZIP format. NCERT Solutions for Class 7 Science in Hindi will be available for the session 2019-20.  Download 7 Maths Offline App in English Medium and 7 गणित Offline App in हिंदी मीडियम.

Table of Contents

A certain freezing process requires that room temperature be lowered from 40 oC at the rate of 5 oC every hour. What will be the room temperature 10 hours after the process begins?

Given: Present room temperature = 40 oC
Decreasing the temperature every hour = 5 oC
Room temperature after 10 hours
= 40 oC + 10 x (–5 oC )
= 40 oC – 50 oC
= – 10 oC
Thus, the room temperature after 10 hours is – 10 oC after the process begins.

विद्या और प्रताप पिकनिक पर गए । उनकी माँ ने उन्हें 5 लीटर पानी वाली एक बोतल दी। विद्या ने कुल पानी का 2/5 उपयोग किया। शेष पानी प्रताप ने पिया। विद्या ने कितना पानी पिया?

दिया है:
बोतल में पानी की कुल मात्रा = 5 लीटर
विद्या द्वारा पानी का उपयोग = 5 लीटर का 2/5 = 2/5 x 5 = 2 लीटर
अतः, विद्या ने बोतल से 2 लीटर पानी पिया।

Shyam bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?

Total weight of fruits bought by Shyam
= 5 kg 300 g + 3 kg 250 g
= 8 kg 550 g
Total weight of fruits bought by Sarala
= 4 kg 800 g + 4 kg 150 g
= 8 kg 950 g

On comparing the quantity of fruits,
8 kg 550 g < 8 kg 950 g Therefore, Sarala bought more fruits.

Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14.

Arranging the given data in ascending order,
12, 12, 13, 13, 14, 14, 14, 16, 19
Mode is the observation occurred the highest number of times = 14
Median is the middle observation = 14

इरफ़ान कहता है कि उसके पास परमीत के पास जितने कंचे हैं उनके पाँच गुने से 7 अधिक कंचे हैं। इरफ़ान के पास 37 कंचे हैं। परमीत के पास कितने कंचे हैं?

माना, परमीत के पास कंचों की कुल संख्या = m
प्रश्न के अनुसार,
5 m + 7 = 37
So, 5m = 37 – 7 = 30
and m = 30/5 = 6
अतः, परमीत के पास कुल 6 कंचे हैं।

Can two angles be supplementary if both of them are acute.

No, because sum of two acute angles is less than 180.

ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Given: AB = 25 cm, AC = 7 cm
Let BC be x cm.
In right angled triangle ACB,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem] (AB)^2 = (AC)^2 + (BC)^2
25^2 = 7^2 + x^2
625 = 49 + x^2
x^2 = 625 – 49 = 576
x = 24 cm
Thus, the length of BC is 24 cm.

Give any two real time examples for congruent shapes.

Two footballs with same size and different colour.

I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?

The cost price of T.V. = ₹ 10,000
Profit percent = 20%
Now, Profit = Profit% of C.P.
= 20/100 x 1000 = ₹ 2,000
Selling price = C.P. + Profit = ₹10,000 + ₹2,000 = ₹ 12,000
Hence, he gets ₹12,000 on selling his T.V.

What is the standard form of a rational number?

A rational number is said to be in the standard form, if its denominator is a positive integer and the numerator and
denominator have no common factor other than 1.

Examine whether you can construct DEF such that EF = 7.2 cm, E = 110 and F = 80. Justify your answer.

Given: In DEF, E = and F =
Using angle sum property of triangle, ∠D + ∠E + ∠F= 180°
⟹∠D+〖110° +〖80〗° =〖180〗°
⟹∠D + 190° = 180°
⟹∠D = 180° – 190° = -10°
Which is not possible.

The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also, find the area of the rectangle.

Perimeter of rectangle = 130 cm
2 (length + breadth) = 130 cm
2 (length + 30) = 130
length + 30 = 130/2
length + 30 = 65
length = 65 – 30 = 35 cm
Now area of rectangle
= length x breadth
= 35 x 30
= 1050 cm^2
Thus, the area of rectangle is 1050 cm^2.

What are monomial, binomial or trinomial in Algebraic Expressions?

• Expression with one term is called a ‘Monomial’.
• Expression with two unlike terms is called a ‘Binomial’.
• Expression with three unlike terms is called a ‘Trinomial’.

क्या हमें कोई ऐसी क्रम 1 से अधिक की घूर्णन सममिति प्राप्त हो सकती है, जिसके घूर्णन के कोण 45° हों?

यदि घूर्णन का कोण 45° हो तो, घूर्णन सममिति संभव है। क्योंकि 360° का कोण 45° से विभाज्य है।

जाँच कीजिए कि क्या ये कथन सत्य हैं। एक घन एक आयत के आकार की छाया दे सकता है।

सत्य

5 thoughts on “NCERT Solutions for Class 7 Maths PDF”

  1. एक किसान के 4200 किलोमीटर भु भाग मूँग है वह तारबदी करवाना चाहता है खेत की चोङाई 30 मिटर है तार कितना लबा होगा

    1. खेत की ल. = 4200/30 = 140 किमी
      खेत का परिमाप = 2(ल. + चौ.‌) = 2 (140+30) = 340 किमी
      खेत की तारबन्दी के लिये आवश्यक तार की ल. = खेत का परिमाप = 340 किमी.

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