Kerala Syllabus Class 7 Mathematics - Unit 4 Reciprocals - Questions and Answers | Teaching Manual 


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ഏഴാം ക്ലാസ്സ്‌  Mathematics - Unit 4 Reciprocals  എന്ന പാഠം ആസ്പദമാക്കി തയ്യാറാക്കിയ ചോദ്യോത്തരങ്ങള്‍. ഈ അധ്യായത്തിന്റെ Teachers Handbook, Teaching Manual എന്നിവ ഡൗൺലോഡ് ചെയ്യാനുള്ള ലിങ്ക് ചോദ്യോത്തരങ്ങളുടെ അവസാനം നൽകിയിട്ടുണ്ട്.

STD 7 - Maths: Unit 4 Reciprocals - Questions and Answers

♦ Textbook Activities - Page 64
(1)  Suma has 16 rupees with her. Safeer has 4 rupees.
(i)  What part of Suma's money does Safeer have?
(ii)  How many times Safeer's money does Suma have?
Answer:
i) 16 × ⁴⁄₁₆ = 4. That is ⁴⁄₁₆ part.
(simplifying ⁴⁄₁₆, we get ¼)
ii) 4 x ¹⁶⁄₄ = 16. That is ¹⁶⁄₄ times.
(simplifying ¹⁶⁄₄, we get 4 times)

(2) A large bag contains 9 kilograms of sugar. A small bag contains 6 kilograms.
(i) The weight of sugar in the heavier bag is how much times that in the lighter bag?
(ii) The weight of sugar in the lighter bag is what part of that in the heavier bag?
Answer: 
i) 6 x ⁹⁄₆ = 9, that is ⁹⁄₆ times.
(simplifying ⁹⁄₆ = ³⁄₂ = 1½ times)
ii) ⁶⁄₉ part. (simplifying, we get ⅔ part)

(3) The weight of an iron block is 6 kilograms. The weight of another block is 26 kilograms.
(i) The weight of the lighter block is what fraction of that of the heavier block?
(ii) The weight of the heavier block is how much times that of the lighter block?
Answer:
i) ⁶⁄₂₆ part (simplifying, we get ³⁄₁₃)
ii) ²⁶⁄₆ times (simplifying, we get ¹³⁄₃ = 4⅓)

(4) The length of a ribbon is 2⅔ times the length of a smaller ribbon. What part of the length of the large ribbon is the length of the small ribbon ?
Answer:
2⅔ times means ⁸⁄₃. 
So the part of the length of the smaller ribbon is ⅜ part of the length of the larg ribbon.
 Numarator and Denominator
When times are expressed as fractions, numarator will be greator than the denominator.
ie. that fraction will be greater than 1.
For example, ⁴⁄₃ times of 9 is 12.
If the part is exxpressed as fraction, it will be less than 1.
ie., numarator will be less than the denominator.
For example, ¾ th part of 12 is 9
Topsy-turvy
Simplification
If we divide numerator and denominator with the same number, we will get the simplified form of that fraction. Dividing with the highest common factor, will give the most simplified form.
Eg: ¹⁵⁄₃₀ = ½
If the denominator is less than numerator, then we can write the fraction as a counting number added with a fraction.
Eg: ¹³⁄₃
We cannot divide 13 with 3 exactly. But 12 can be divided by 3, giving 4, with a remainder. 1. This remainder when divided by 3 will be ⅓
So ¹³⁄₃ = ¹²⁺¹⁄₃ = 4 + ⅓ = 4⅓
♦ Textbook Activities - Page 67
(1) 27 students of a class got A plus in Maths. They form ¾ of the entire class. How many students are there in the class ?
Answer:
¾ part of the total students is 27
Then ⁴⁄₃ times of 27 will be the total students in the class.
ie, no students in the class is = 27 x ⁴⁄₃ = ¹⁰⁸⁄₃ = 36

(2) ⅔ of a bottle was filled with ½ litre of water. How many litres of water will the bottle hold ? 
Answer:
⅔ part of the volume of the bottle is ½ litre.
Then, ³⁄₂ times of ½ litre is the volume of the bottle.
∴ Volume = ½ x ³⁄₂ = ¾ litre

(3) ¾ of a vessel holds 1½ litres of water. What is the capacity of the vessel in litres if it is completely filled with water?
Answer:
¾ part of the total volume of the vessel = 1½ = ³⁄₂ litres.
Then ⁴⁄₃ times of ³⁄₂ is the volume of the vessel.
volume = 1½ x ⁴⁄₃ = ³⁄₂ x ⁴⁄₃ = ¹²⁄₆ = 2 litres

Method 2
¾ part of the vessel is 1½ litres
That is, ¼ part is ½ litre.
So the whole part is ½ + ½ + ½ + ½ = 4 x ½ = 2 litres
(4) Two of the three ribbons of the same length and half the third ribbon were placed end to end. It came to 1 metre. What is the length of a ribbon in centimetres ?
Answer:
2½ ribbons placed together has length = 1m
ie, 2½ times the length of a ribbon is 1m (2½ times = ⁵⁄₂)
So, the length of one ribbon will be ⅖ part of 1m
ie, length of the ribon = ⅖ x 1m = ⅖ x 100 cm
 = 2 x 20 cm = 40 cm
Reciprocal multiplication
A number when multiplied by is its reciprocal gives 1.
4 x ¼ = 1
⅓ x 3 = 1
⅖ x ⁵⁄₂ = 1
Fraction division
To devide a number with another number,  multiply with its reciprocal. That is, multiplication with the reciprocal is equal to division.
♦ Textbook Activities - Page 68
(1) A 16 metres long rod is cut into pieces of length ⅔ metre. How many such pieces will be there ?
Answer:
Here 16m rod should be cut into several pieces of length ⅔ m
∴ Number of pieces = 16 ÷ ⅔
= 16 x ³⁄₂ = ⁴⁸⁄₂ = 24 pieces.

(2) How many ¾ litre bottles are needed to fill 5¼ litres of water ? 
Answer:
We have to devide 5¼ litres into ¾ litre bottles.
ie, 5¼ ÷ ¾ = 5¼ x ⁴⁄₃
5¼ means ⁵ˣ⁴⁺¹⁄₄ = ²¹⁄₄
ie, ²¹⁄₄ x ⁴⁄₃ = ⁸⁴⁄₁₂ = 7 bottles

(3) 13½ kilograms of sugar is to be packed into bags with 2¼ kilograms sugar each. How many bags are needed ?
Answer:
We have devide 13½ kg into 2¼ kg bags.
ie, 13½ ÷ 2¼ = ²⁷⁄₂ ÷ ⁹⁄₄ = ²⁷⁄₂ x ⁴⁄₉ = ¹⁰⁸⁄₁₈ = 6 bags

(4) The area of a rectangle is 22½ square centimetres and one side is 3¾ centimetres long. What is the length of the other side ?
Answer:
Area = Side (1) x Side (2)
∴ Side (2) = area ÷ Side (1)
= 22½ ÷ 3¾
= ⁴⁵⁄₂ ÷ ¹⁵⁄₄ = ⁴⁵⁄₂ x ⁴⁄₁₅ = ¹⁸⁰⁄₃₀  = 6 cm

(5) How many pieces, each of length 2½ metres, can be cut off from a rope of length 11½ metres ? How many metres of rope will be left ?
Answer:
4 times of 2½ is 10. So 1½ m is left behind.
Division method
11½ ÷ 2½ = ²³⁄₂ + ⁵⁄₂ = ²³⁄₂ x ⅖ = ⁴⁶⁄₁₀ = 4⁶⁄₁₀ = 4 + ⁶⁄₁₀
This can be understood as:
11½ is equal to 4 parts of 2½ + ⁶⁄₁₀ part of 2½
⁶⁄₁₀ part of 2½ = 2½ x ⁶⁄₁₀
= ⁵⁄₂ x ⁶⁄₁₀ = ³⁰⁄₂₀ = ³⁄₂ = 1½ m
That is left over ⁶⁄₁₀ part is equal to 1½ m
It can also be explained like this.
11½ ÷ 2½ 
That is, how many 2½ s are there in 11½
There are four 2½ is 10
So the left over is 1½.
MORE QUESTIONS AND ANSWERS

1. Do the problems as a part and time.
a) The length of one pencil is 8 cm and the length of another pencil is 24 cm.
i) What part of the length of the larger pencil is the length of the smaller pencil?
ii) How many times the length of the smaller pencil is the length of the larger pencil?
Answer:
i) ⁸⁄₂₄ = ⅓ part
ii) ²⁴⁄₈ = 3 times

b) Sobha's weight is 36 kgs and Sanooja's weight is 54 kgs.
i) What part of Sanooja's weight is Sobha's weight?
ii) How many times of Sobha's weight is Sanooja's weight? 
Answer:
i) ³⁶⁄₅₄ = ⅔ part
ii) ⁵⁴⁄₃₆ = ³⁄₂ times

2. Answer the following using reciprocals.
a) The price of ½ kilogram rice is 100 rupees. What is the price of 1 kilogram of rice?
Answer:
The price of 2½ kg of rice is 2½ times that of 1 kg of rice, which is 100 rupees.
2½ times = ⁵⁄₂ times
That is price of 1kg of rice is ⅖ part of 100 rupees
ie, 100 x ⅖ = 40 rupees.

b) ²⁵⁄₁₀₀ part of number is 32, what is the number?
Answer:
²⁵⁄₁₀₀ part of a number = 32
∴ The number will be ¹⁰⁰⁄₂₅ times of 32.
ie, 32 x ¹⁰⁰⁄₂₅ = ³²ˣ¹⁰⁰⁄₂₅ = ³²⁰⁰⁄₂₅ = 128

3. Find by division
a) 20 litresof milk is poured equally into bottles of ⅓ litre. How many bottle are needed to fill the whole?
Answer:
20 ÷ ⅓ = 20 x 3 = 60 bottels

b) A watermelon weighing 7½ kg is cut into ¾ kg each. How many pieces will be there?
Answer:
7½ + ¾  = ¹⁵⁄₂ + ¾ = ¹⁵⁄₂ x ⁴⁄₃ = ⁶⁰⁄₆ = 10 pieces

c) If the length of a rectangle is 3½ m and its area is 5¼ sq. m, find its breath?
Answer:
Breadth = area ÷ length
= 5¼ + 3½ = ²¹⁄₄ + ⁷⁄₂ = ²¹⁄₄ x ²⁄₇ = ⁴²⁄₂₈ = ³ˣ¹⁴⁄₂ₓ₁₄ = ³⁄₂ = 1½ m

4) Calculate using the reciprocles and by the method of division.
a) How many bottles are needed to fill 22¼ litres of milk if the capacity of 1 bottle is ¼ litres?
Answer:
Division method
22¼ + ¼ = 22¼ x 4 = (22 + ¼) x 4
= (22 x 4) + (¼ x 4) = 88 + 1 = 89 bottles.
This is same as 22¼ + ¼ = 22¼ x 4 = ⁸⁹⁄₄ x 4 = 89

Reciprocles method
¼ part of the number of bottles is 22¼
Number of bottles = 4 times of 22¼
= 22¼ x 4 = 89 bottles.

b) ⅔ part of a vessel is filled with 5½ litres water. How much liters of water can the vessel hold?
Answer:
Division method
5½ + ⅔ = ¹¹⁄₂ + ⅔ = ¹¹⁄₂ x ³⁄₂ = ³³⁄₄ = 8¼ litres.

Reciprocles method
⅔ part of the vessel is 5½ litre
So, ³⁄₂ times of 5½ litres is the volume of the vessel.
= 5½ x ³⁄₂ = ¹¹⁄₂ x ³⁄₂ = ³³⁄₄ = 8¼ litres.