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SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion InText Questions

[Page No. 1]

There are several situations in our daily life where we use ratio and proportion. Let us look at the following pictures and answers to the given questions:
Question 1.
Can we say the ratio of speeds of Cheetah to Man?

Answer:
Speed of Cheetah = 120 kmph
Speed of a man = 20 kmph
∴ Ratio of speed of Cheetah to a man
= 120 : 20
= 6 : 1

Question 2.
What will be the ratio of heights of HematoAmir?

Answer:
Height of Hema = 150 cm
Height of Amir 75 cm
∴ Ratio of heights 150 : 75
= 2 : 1

Question 3.

Answer:
Cost of 2 Hands of banana = ₹ 80
∴ Cost of 3 Hands of banana
= \(\frac{3}{2}\) × 80 = ₹ 120

Question 4.
Do you know which aspect ratio makes constructions more attractive and beautiful?
Answer:
Golden ratio is more beautiful. It is the ratio of a line segment cut in to two pieces of different lengths such that the ratio of whole segment to that of the longer segment is equal to that of longer segment to the shorter segment.

Check your Progress [Page No. 4]

Question 1.
Write the compound ratio of the following given ratios.
(i) 3 : 5 and 4 : 3
Answer:
Given ratios are 3 : 5 and 4 : 3.
If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes.
that is a × c : b : d
3 × 4 : 5 × 3
12 : 15
∴ Compound ratio = 4 : 5

(ii) 8 : 3 and 6 : 5
Answer:
Given ratios are 8 : 3 and 6 : 5.
If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes, that is
a × c : b : d
8 × 6 : 3 × 5
∴ Compound ratio = 16 : 5

(iii) 2: 1 and 8: 7
Answer:
Given Ratios are 2 : 1 and 8 : 7.
If a : b and c: d are ratios, then their compound ratio is product of antecedents : product of extremes that is a × c : b : d
2 × 8 : 1 × 7
16 : 7
∴ Compound ratio= 16 : 7

Question 2.
Fill in the boxes with correct answers.

Answer:

[Page No. 6]

Discuss and answer the following questions related to real-life situations:
Question 1.
If the cost of 3 ball pens is ₹ 15, then what is the cost of 6 such pens?
Answer:
3 : 15 : : 6 : cost
Cost x 3 = 15 × 6
∴ Cost = \(\frac{15 \times 6}{3}\) = ₹ 30

Question 2.
In a school, for implementing ‘Mid day Meal’ scheme for class VII of 40 students they need 6 kgs of rice for one day.
(i) How much rice is needed for 6 days?
Answer:
Rice required for 40 students for one day = 6 kgs
Rice required for 6 days = 6 × 6 = 36 kg

(ii) How much rice is needed for 10 days?
Answer:
Rice required for 10 days = 10 × 6 = 60 kg

(iii) How much rice is needed for 5 days?
Answer:
Rice required for 5 days = 5 × 6 = 30 kg

Let’s Explore [Page No. 6]

Question 1.
Given below the number of workers working for construction of a house and their wages in total.

Answer:

Let’s Think [Page No. 7]

Question 1.
If the cost of 4 note books is 80. What would be the cost of7 note books? The above problem can be solved by using unitary method. Can you solve? How? Think.
Answer:
Given the cost of 4 note books = ₹ 80
Cost of 1 note book = 80 ÷ 4 = ₹ 20
Cost of 7 note books = ₹ 20 × 7
= ₹ 140

Check Progress [Page No 7]

Question 1.
Fill in the blanks, if the given quantities are in direct proportion.

Answer:

Let’s’ Do Activity [Page No. 8]

Take a clock and fix Its minute hand at a particular number on clock (if it is 12 It will be easy). Then find and note the angles made by minute hand in every 15 minutes interval of time:


Check whether they are in direct proportion or not?
Answer:

15 : 30 and 90 : 180
Here 1 :2 = 1 :2
If ratios are equal, then they are in direct proportion.
So, time passed and angle are in direàt proportion.

Let’s Think [Page No. 8]

Question 1.
What is the angle made by minutes hand in a minute?
Answer:
We know angle made by minutes hand in 15 m = 90°
Let angle made by minutes hand in 1 m = x°
Then 15 : 1 = 90°: x°
If the ratios are equal,
The product of means = The product of extremes
⇒ 15 × x = 1 × 90°
⇒ \(\frac{15 x}{15}=\frac{90^{\circ}}{15}\)
⇒ x = 6°
∴ Angle made by minutes hand in 1 minute = 6°

Question 2.
What is the angle made by hours hand in one minute?
Answer:
We know angle made by hours hand in 1 hour (60 minutes) = 30°
Let angle made by hours hand 1 minute = x°
Then 60 : 1 = 30°: x°
If the ratios are equal,
The product of means = The product of the extremes
⇒ 60 × x = 1 × 30
⇒ \(\frac{60 x}{60}=\frac{30}{60}=\frac{1}{2}\)
⇒ x = \(\frac{1}{2}\) (or) 0.5°
∴ Angle made, by hours hand in 1 minute = \(\frac{1}{2}\)° (or) 0.5°

Check your Progress [Page No. 9]

Question 1.
Fill in the blanks if the given quantities are in inverse proportion.

Answer:

Check your Progress [Page No. 13]

Question 1.
Analyse how the three quantities given below are related and find ‘x’.

Answer:
From the above,

• Ration and the number of days are in inversely proportional.
• Ration and number of men are in directly proportional.
• Number of men is depends on both Ration and number of days.

So we have to take compound ratio of 108 : 70 and 25 : 15
i. e., 108 × 25 : 70 × 15
∴ 18 : x = 108 × 25 : 70 × 15
If the ratios are equal.
The Product of means = The Product of extremes
⇒ 108 × 25 × x = 18 × 70 × 15

∴ x = 7

Let’s Think [Page No. 18]

Question 1.
A man buys chocolates 10 for 10 rupees and sells them at 10 for ₹ 12. Does this result profit or loss ? What percent?
Answer:
Given cost price of 10 chocolates = ₹ 10
Selling price of 10 chocolates = ₹ 12
SP > CP, then man gets profit.
Profit = S.P – C.P
= 12 – 10 = ₹ 2
We know profit percent

Profit percentage = 20%

Question 2.
If a shopkeeper bought sofa sets and increases their prices by 50% and sell at 50% less, is it a loss or gain?
Answer:
Let the cost price of sofa set = ₹ 100

= ₹ 5o
Total printed price = ₹ 100 + ₹ 50
= ₹ 150
Loss percent = 50%

Loss = ₹ 75
Selling price Printed Price – Loss
= 150 – 75 = ₹ 75
C.P= ₹ 100; S.P = ₹ 75
and C.P > S.P
So, the shopkeeper will get loss of 25%.

Check Your Progress [Page No. 20]

Answer:

Check Your Progress [Page No. 24]

Answer:

Let’s Think [Page No. 24]

Question 1.
At what rate per annum will the principle doubles in 10 years?
Answer:
Method 1:
Let the principle = 100
Interest rate = x% ; Time = 10 years

Amount = Principle + Interest
= Double the principle
= 100 + 10x = 200
⇒ 100 + 10x – 100 = 200 – 100
⇒ 10x = 100
⇒ \(\frac{10 x}{10}=\frac{100}{10}\)
⇒ x = 10%
∴ Rate of interest = 10%

Method 2:
Let the principle =?
Rate of Interest = R%
T = 10 years
Given that A = 2P = P + P [∵ A = P + I]
∴ I = P
∴ \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}\) = P
⇒ \(\frac{\mathrm{P} \times 10 \times \mathrm{R}}{100}\) = P
R = 10%
∴ The principle become double in 10 years at 10% rate of interest.

Question 2.
At what rate per annum will the principle be 4 times in 15 years?
Answer:
Method 1:
Let the principle = ₹ 100
Interest rate = x %
Tirne = 15 years
Interest = \(\frac{\mathrm{PTR}}{100}\)

Amount = Principle + Interest
= 4 times the principle
= 100 + 15x = 400
⇒ 100 + 15x – 100 = 400 – 100
⇒ 15x = 300
⇒ \(\frac{15 x}{15}=\frac{300}{15}\) = 20%
∴ x = 20%

Method 2:
Let the principle = P
Time = 15 years
A = 4timesP = 4P
⇒ A = P + 3P
But A = P + I
So I = 3P

We know that I = \(\frac{\text { PTR }}{100}\)
⇒ 3P = \(\frac{P \times 15 \times R}{100}\)
⇒ R = \(\frac{100 \times 3}{15}\) = 20%
∴ The principle become 4 times in 15 years at 20% rate of interest.

Examples:

Question 1.
Find the compound ratio of 8 : 7 and 9 : 13.
Answer:
Given ratio = 8 : 7 :: 9 : 13
Compound ratio = 8 × 9 : 7 × 13 = 72 : 91
(∵ Compound ratio product of antecedent product consequents).

Question 2.
Two friends Prabhu and Suresh started a business with ₹ 1,00,000 each. After 3 months Suresh left the business. At the end of the year, there was a profit of ₹ 20,000. Calculate the profits shared by Prabhu and Suresh?
Answer:
Here Prabhu and Suresh started a business with ₹ 1,00,000.
Prabhu continued till the end of the year.

Suresh continued only for 3 months.
The ratio of the contributions
= 100000: 100000 = 1 : 1

Ratio of their period of business
= 12 : 3 = 4 : 1
So, their profits sho’uld be divided on the basis of compound ratio
Compound ratio = 1 × 4 : 1 × 1
Profit = ₹ 20,000
Total parts= 4 + 1 = 5
Suresh’s profit = 20000 × \(\frac{1}{5}\)
= ₹ 4000
Prabhus profit = 20000 – 4000
= ₹ 16000

Question 3.
Rani started a beauty parlour with an amount ₹ 75,000. After 4 months, Vani also joined with Rani with an amount ₹ 50,000. After one year, they got a profit of ₹ 52,000. Calculate the profits shared by Rani and Vani.
Answer:
Rani’s investment = ₹ 75,000
Rani’s period in business = 1 year = 12 months
Vani’s investment = ₹ 56,000
Vani’s period in business = 8 months
The ratio of investments of Rani to Vani = 75,000 : 50,000 = 3:2
The ratio of periods of business of Rani to Vani =12:8 = 3:2

The profit should be distributed on the, basis of compound ratio.
Compound ratio = 3 ×. 3 : 2 × 2
= 9 : 4
Profit = ₹ 52,000

Total parts = 9 + 4 = 13
Rani’s profit = 52,000 × \(\frac{9}{13}\)
= ₹ 36000
Vani’s profit = 52,000 – 36,000 = ₹ 16,000

Question 4.
If 3 : 4 and 9 : x are in direct proportion, then what is the value of x?
Answer:
If 3 : 4 and 9 : x are in direct proportion, then their ratio is constant.
∴ \(\frac{3}{4}=\frac{9}{x}\)
⇒ 3 × x = 4 × 9
⇒ x = \(\frac{4 \times 9}{3}\) = 12

Question 5.
If the cost of 4 note books is ₹ 80. What would be the cost of 7 note books?
Answer:
We know that as number of note books increases, the cost also increases such that the ratio of number of note books and the ratio of their costs will remain the same. That means here number of note books and the cost are in direct proportion.

Let the cost of 7 note books be ‘x’.
Then, 4 : 80 = 7 : x
If the ratios are equal, the product of means = The product of the extremes
4 × x = 80 × 7
⇒ x = \(\frac{80 \times 7}{4}\) = ₹ 140
Thus, the cost of 7 note book is equal to ₹ 140.

Question 6.
The scale of a map is given as 1 : 30000. If two cities are 20 cm apart on the map, then o-j, . find the actual distance between them.
Answer:
Let the actual distance be ‘x’ cm. Since the distance on the map is directly proportional to the actual distance.
1 : 30000 = 20 : x
If the ratios are equal, the product of means = The product of the extremes
∴ 1 × x = 30,000 × 20
⇒ x = 6,00,000 cm = 6 km. [1,00,000 cm = 1000 m = 1 km.]
Thus, two cities which are 20cm apart on the map are actually 6 km away from each other.

Question 7.
If 4, 7 and 2, x are inverse proportion, then what is the value of x?
Answer:
4, 7 and 2, x are in inverse proportion . Therefore, 4 × 7 = 2 × x
⇒ x = \(\frac{4 \times 7}{2}\) = 14

Question 8.
If 18 workers can build a wall in 12 days, how many days will eight workers take to build the same wall?

Answer:
If the number of workers decreases, the time taken to build the wall increases in the same proportion.
So, number of workers and the number of days to complete work are in inverse proportion.

Let the number of days to complete the work be ‘x’.

Number of workers	Number of days
18	12
8	x

By taking inverse proportion,
18 : 8 = x : 12
Then,
∴ 18 × 12 = 8 × x
⇒ 8 × x = 18 × 12
⇒ x = \(\frac{18 \times 12}{8}\) = 27days
∴ Eight workers can complete the will in 27 days.

Observe this:
In inverse Proportion, product is always constant.
∴ 18 × 12 = 8 × x
⇒ 8 × x = 18 × 12
⇒ x = \(\frac{18 \times 12}{8}\) = 27days

Question 9.
4 Pumps are required to fill a tank in 1 hr 30 min. How long will it take if only 3 pumps of the same type are used?

Answer:
Let the time be ‘x’.
1 hr 30min = 60 + 30 = 90 minutes ,
If the number of pumps are decrease, the time taken to fill the tank increases.
So, the number of pumps and time taken to fill the tank are in inverse proportion.

Number of pipes	Time taken to fill (min)
4	90
3	x

By taking inverse proportion,
4 : 3 = x : 90
Thus,
∴ 4 × 90 = 3 × x
⇒ 3 × x = 4 × 90
⇒ x = \(\frac{4 \times 90}{3}\) = 120 min
∴ 3 pumps will fill the tank in 120 min or 2 hrs.

Observe this:
In inverse proportion, product is always constant
∴ 4 × 90 = 3 × x
⇒ 3 × x = 4 × 90
⇒ x = \(\frac{4 \times 90}{3}\) = 120 min

Question 10.
If 30 persons use 40 kg. of sugar in 10 days, find in how many days 80 persons will use 320 kg. of sugar? How can we solve?
Answer:
Here, the 3 quantities are persons, weight and number of days.

From the above comparison mustbe in between unknown form to know form.

• Here days and persons are in inverse we denote it with Tmark.
• Here days and sugar are in direct proportion we denote it with 4r mark.

Here number of days is depends on both persons and weight of sugar. So we have to take compound ratio of 80 : 30 and 40 : 320.
∴ 10 : x = 80 × 40 : 30 × 320 = 3200 : 9600
Since the ratios are equal, the product of extremes is equal to product of means.
10 : x = 3200 : 9600 ⇒ 10 × 9600 = x × 3200
⇒ x × 3200 = 10 × 9600 ⇒ x = \(\frac{10 \times 9600}{3200}\) = 30

Question 11.
8 painters can paint a wall of 160m long in 5 days. How many painters are required to paint 240kn. wall in 10 days?
Answer:
Here we have three quantities – number of painters, length of wall and number of days.

Number of painters, is directly proportional to length of the wall.
Number of painters is inversely proportional to the number of days.

Since the number of painters depends both on length of wall and number of days. We will take the compound ratio of 160 : 240 and 10 : 5 1
∴ 8 : x = 160 × 10 : 240 × 5
Since the ratios are equal, product of means is equal to product of extremes
8 : x = 160 × 10 : 240 × 5
⇒ x × 160 × 10 = 8 × 240 × 5
⇒ x = \(\frac{8 \times 240 \times 5}{160 \times 10}\) = 6
Hence the required number of painters = 6

Question 12.
195 men working 10 hours a day can finish a job in 20 days. How many men are employed to finish the job in 15 days if they work 13 hours a day?
Answer:
Here we have three quantities – number of workers, number of hours and number of days.

Here the number of workers inversely proportional to number of hours per day.
Here the number of workers inversely proportional to number of days.
Since number of workers depends on both number of hours and number of days, we will take compound ratio of 13 : 10 and 15 : 20.
Since the ratios are equal, the product of means is equal to product of extremes.
195 : x = 13 × 15 : 20 × 10
⇒ x × 13 × 15 = 195 × 20 × 10
⇒ x = \(\frac{195 \times 20 \times 10}{13 \times 15}\) = 200
Hence the required number of workers = 200

Question 13.
Express the following percentages as fraction, decimal and In ratio.
(i) 45%
Answer:
45% = \(\frac{45}{100}=\frac{9}{20}\)(Fraction)
= 0.45 (decimal form)
= 9:20 (ratio)

(ii) 62%
Answer:
62% = \(\frac{62}{100}=\frac{31}{50}\) = (Fraction)
= 0.62 (decimal form)
= 31:50 (ratio)

Question 14.
Find 24% of 150 and àlso the remaining of that number.
Answer:
24% of 150 = \(\frac{24}{100}\) × 150 = 36
The remaining of that number
= 150 – 36 = 114

Question 15.
Raghu bought pens for 400 and he sold them for ₹ 480 what Is his profit or loss percent?
Answer:
Jyothi solved It this way:
Pen C.P of = ₹ 400, S.P = ₹ 480
S.P > C.P.
So, Raghu gets a profit P = 480 – 400 = ₹ 80
Profit Percentage = \(\frac{80}{400}\) × 100 = 20%

Anwar solved it this way :
Raghu’s Profit = S.P. – C.P = 480 – 400 = ₹ 80
The ratio of profit and cost or the fraction is \(\frac{80}{400}\)
∴ Profit Percentage = \(\frac{80}{400}\) × 100 = 20%

Suresh solved it using proportion :
When C.P is ₹ 400, the profit is ₹ 80 C.P. is ₹ 100, let the profit be ₹ x.
Here C.P and profit are in direct proportion.
x : 80 = 100 : 400 ⇒ \(\frac{x}{80}=\frac{100}{400}\)
⇒ x × 400 = 100 × 80
⇒ x = \(\frac{100 \times 80}{400}\) = 20%
Profit = 20 per 100
Profit Percentage = 20%

Question 16.
Ramana bought a cycle for ₹ 1200 and sold it to his friend Rehman for ₹ 900, then what is Ramanas profit or loss percentage?
Answer:
Here Ramana’s Cost price = ₹ 1200
Selling price = ₹ 900
S.P < C.P, then Ramana got loss
Loss = C.P – S.P = 1200 – 900 = ₹ 300
Loss percentage = \(\frac{300}{1200}\) × 100 = 25%

Question 17.
If John buys a ear for ₹ 1,50,000 and gains 10% on selling it. Then find the selling price.

Answer:
Cost Price = ₹ 1,50,000
Gain % = 10%
Profit = 10% of ₹ 1,50,000
= \(\frac{10}{100}\) × 150000 = ₹ 15000
Selling Price = C.P. + Profit
= ₹ 1,50,000 + ₹ 15,000
= ₹ 1,65,000
This can be solved using proportion:
Gain 10% means
If CP is ₹ 100,the gain is ₹ 10
Thus S.P = 100 + 10 = 110
Now, here C.P = ₹ 1,50,000

Let S.P = x
CP and SP are directly proportional.
\(\frac{110}{100}=\frac{x}{1,50,000}\)
\(\frac{x}{1,50,000}=\frac{110}{100}\)
x = \(\frac{1,50,000 \times 110}{100}\) = ₹ 1,65,000

Question 18.
Kiran sold a refrigerator for ₹ 16800 at a gain of 12%. Then what is the cost price of it?

Answer:
Roopa did the problem uslnj unitary method:
S.P = ₹ 16800 ; Gain% = 12%
If CP is ₹ 100, then Profit is ₹ 12 and thus S.P = ₹ 112
So when S.P is ₹ 112, then C.P is ₹ 100
When S.P is One Rùpee, C.P is \(\frac{110}{112}\) .
Here S.P is ₹ 16800
So, C.P = \(\frac{110}{112}\) × 16800 = ₹ 15000

Sneha solve did the prot1em using proportion as follows:
Gain % = 12% ;S.P = ₹ 16800
If C.P is ₹ 100 then profit is ₹ 12, thus SP = ₹ 12
here S.P = ₹ 16800
C.P be x
C.P and S.P are directly proportional.
∴ \(\frac{x}{16800}=\frac{100}{112}\)
⇒ x = \(\frac{100 \times 16800}{112}\) = ₹ 15000

Question 19.
The cost of an article goes down every year by 10% of its previous value. Find its original cost, if its cost after 2 years is ₹ 32400.
Answer:
Let the cost at the beginning of 1st year be ₹ 100. At the beginning of,2nd year i.e at the end of 1st year it will be decreased by 10% means cost will be ₹ 90.
At the end of 2nd year i.e at the beginning of 3rd year it will be reduced by 10%
i. e. ₹ 90 is reduced by 10%.
90 – 9 = ₹ 81
If the cost of object is ₹ 100 at the beginning, then after 2 years it’s cost will be ₹ 81.
Let the cost of the object ₹ ‘x’ at the beginning, after 2 years it’s cost is ₹ 32,400 .
Thus ratio of original costs = ratio of costs after 2 years
⇒ x : 100 = 32400 : 81
⇒ \(\frac{x}{100}=\frac{32400}{81}\)
⇒ x = \(\frac{32400 \times 100}{81}\) = ₹ 40000

Question 20.
Find discount if,
(i) marked price is ₹ 450, selling price = ₹ 415
Answer:
Discount = marked price – selling price ⇒ 450 – 415 = ₹ 35.

(ii) marked price = ₹ 810, selling price = ₹ 765.
Answer:
Discount = marked price – selling price ⇒ 810 – 765 = ₹ 45.

Question 21.
If discount is ₹ 40, marked price is ₹ 400, then find discount percentage.
Answer:
Discount = ₹ 40, marked price = ₹ 400,
then discount percentage = \(\frac{40}{400}\) × 100 = 10

Question 22.
A shopkeeper marks his goods 20% above the cost price and allows a discount of 10% on them. What percent does he gain?
Answer:
Let the cost price be ₹ 100.
Then the marked price = 100 + 20 = ₹ 120
Discount = 10%, so discount = x 120 = 12% /
SP = Marked price – Discount = ₹ 120 – ₹ 12 = ₹ 108
Gain = \(\frac{8}{100}\) × 100 = 8%
The shopkeeper gains 8% after discount.

Question 23.
Calculate simple interest and the total amount, if
(i) principal = ₹ 5000, time = 2 years, rate = 10%
Answer:
Simple interest
I = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{5000 \times 2 \times 10}{100}\) = ₹ 1000
Total amount = principal + interest = 5000 + 1000 = ₹ 6000

(ii) principal = ₹ 25000, time = 3 years, rate = 12%
Answer:
Simple interest = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{25000 \times 3 \times 12}{100}\) = ₹ 9000
Total amount = principal + interest = 25000 + 9000 = ₹ 34000

Question 24.
If Raheem borrowed a sum of ₹ 25000 at a rate 10% per annum, what is the simple interest and total amount he has to pay for 3 years?
Answer:
Rajesh did like this :
Principal = ₹ 25000 ; Time = 3 years ; Rate of interest = 10%
Simple interest = I = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{25000 \times 3 \times 10}{100}\) = ₹ 7500
Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Sangeetha did like this :
For 1 year we have to pay 10%,
for 3 years we have to pay 3 × 10 = 30% as interest.
Interest = \(\frac{30}{100}\) × 25000 = ₹ 7500
Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Question 25.
What sum will yield an interest of ₹ 6000 at 9% per annum in 3 years 4 months?
Answer:
S.I = ₹ 6000 ; R = 9%

Question 26.
At what rate per annum will ₹ 70000 yield an interest of ₹ 14000 in 2\(\frac{1}{2}\) years?
Answer:
Principal = ₹ 70000 ; Time = 2\(\frac{1}{2}\)years = \(\frac{5}{2}\) years ; Simple interest = ₹ 14000

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Unit Exercise

Question 1.
If the cost of 7 toys is ₹ 1575, then what would be the cost of 6 such toys?
Answer:
We know that as number of toys decreases, the cost also decreases such that the ratio of number of toys and the ratio of their costs will remain the same. That means here number of toys and the cost are in direct proportion.
Let the cost of 6 toys be x.
Then 7 : 1575 = 6 : x
If the ratios are equal, the product of extremes = Product of means
⇒ 7 × x = 1575 x 6

⇒ x = ₹ 1350
∴ Cost of 6 toys is ₹ 1350.

Question 2.
A boy went to a hotel and wants to buy 5 plates of Idly worth ₹ 24 each. After, going to Hotel he observed that rates of Idly were increased to ₹ 30. Now if the boy wants to buy idlies, how many plates of idlies he can buy with the same amount?
Answer:
If the cost of idly increases number of plates decreases.
So, cost of idly and number of plates are in inverse proportion.
Let number of plates at ₹ 30 for the same amount is x.

Cost of each Idly Plate	Number of Plates
24	5
30	x

By taking inverse proportion, 24 : 30 = x : 5
Then, Product of means = Product of extremes

Number of Plates at ₹ 30 for the same amount is 4.

Question 3.
Raju covers a distance of 28 kilometers in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Answer:
We know that as number of kilometers (distance) increases the time taken is also increases with the same speed. That means distance and the time taken are in direct proportion.

Let the time taken to cover the 56 km distance is x hours.
Then 28 : 2 = 56 : x

If the ratios are equal, the product of extremes = the product of means

The time taken to cover the 56 km distance is 4 hours.

Question 4.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days 20 men working 9 hours per day do the same work?
Answer:
Here, we have three quantities number of workers, number of hours and number of days.

Number of Men	Hours	Number of Days
24	8	15
20	9	x

Here the number of worker’s inversely proportional to number of hours perday.
Here the number of workers inversely proportional to number of days.

Let the number of days be ‘x’.
Since number of workers depends on both number of hours and number of days, we will take compound ratio of
24 : 20 and 8 : 9 is 24 × 8 : 20 × 9

Since the ratios are equal,
The product of means = The product of extemes

∴ Number of days required to complete 20 men at 9 hours per day is 16.

Question 5.
Out of 15000 voters in a constituency, 60% of the voters voted. Find the number of people not voted in the constituency.
Answer:
Given number of voters in the constituency = 15000 Percentage of voters voted = 60 %
Number of people voted = 60% of 15000

Number of people voted in the constituency = 9000
Number of people not voted = 15000 – 9000
∴ Number of people not voted = 6000

Question 6.
A shopkeeper bought a suitcase for ₹ 950 and sold it for ₹ 1200. Find its profit or loss percentage.
Answer:
Given cost price of suitcase = ₹ 950
Selling price of suitcase = ₹ 1200
S.P > C.P
So, shopkeeper got profit.
Profit = S.P – C.P = 1200 – 950 = ₹ 250

∴ Profit percent = 26\(\frac{6}{19}\)%.

Question 7.
On selling a mobile for ₹ 4500, a shop-keeper losses 10%. For what amount should be sell it to gain of 15%?
Answer:
Method 1 :
Given selling of mobile = ₹ 4500
Loss = 10%

S.P = C.P – Loss

∴ x = 5000

Method 2:
Let the C.P of mobile = 100%
S.P at a loss of 10% = (100 – 10)%
= ₹ 4500
90% of C.P = ₹ 450O
S.P at a gain of 15% = (100 + 15)%.
= 115%
So, if 90% of CP = ₹ 4500
115% of CP = ?
s.p= \(\frac{115}{90}\) × 4500= ₹ 5750
∴ Cost price of mobile = ₹ 5000
If he gain 15%.

Gain = 15% of 5000

S.P = C.P + Gain = 5000 + 750
∴ Selling price of Mobile = ₹ 5750

Question 8.
A carpenter allows 15% discount on his goods. Find the marked. price of a chair which is sold by him for ₹ 680.
Answer:
Given selling price = ₹ 680
Discount percent = 15%
Let marked price = x

Note:
85% of M.P = ₹ 680
100% of M.P = \(\frac{100}{85}\) × 680
= ₹ 800.

Question 9.
What Is the simple Interest accrued on asum of ₹ 75000 at the rate of 11% for 3 years? Find the total amount.
Answer:
Given Principle = ₹ 7500
Rate of Interest = 11%
Time = 3 years

Total amount = Principle + Interest = ₹ 75000 + ₹ 24750
∴ Total amount = ₹ 99,150

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.7 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.7

Question 1.
Calculate the simple interest accrued on a sum oR 12600, at the rate of 9% per annum for 2 years.
Answer:
Given principle = ₹ 12600
Time = 2 years
Rate of Interest = 9%

∴ Simple Interest = ₹ 2268

Question 2.
Calculate the simple interest accrued for 3 years and the total amount on a sum of ₹ 85000, at the rate of 11% per annum.
Answer:
Given Principle = ₹ 85000
Time = 3 years
Rate of Interest = 11%

Simple Interest = ₹ 28,050
Total amount = Principle + Interest
= 85000 + 28050
∴ Total amount = ₹ 1,13,050

Question 3.
In what time will ₹ 45000 amounts to ₹ 63000, If the simple Interest Is calculated 10% per annum?
Answer:
Method 1:
Given principle = ₹ 45000
Rate of Interest = 10%
Amount = ₹ 63000
Let the time = x years

Interest = ₹ 4500 x
Amount = Principle + Interest
= 45000 + 4500x = 63000
= 4500x = 63000 – 45000 = 18000
⇒ x = \(\frac{18000}{4500}\) = 4
∴ Time = 4 years

Method 2:
Given A = ₹ 63000;
P = ₹ 45000; R = 10%; T=?
So I = A – P = 63000 – 45000 = ₹ 18000

But I = \(\frac{\text { PTR }}{100}\)
⇒ 18000 = \(\frac{45000 \times T \times 10}{100}\)
∴ T = \(\frac{18000 \times 100}{45000 \times 10}\) = 4 years

Question 4.
On a certain amount at the rate of 12% per annum for time 3 years the toti1 interest becomes ₹ 18000. What is the principal amount?
Answer:
Given Rate of Interest = 12%
Time.= 3 years
Interest = ₹ 18000
Let Principle = ₹ x

∴ Principle = ₹ 50,000

Question 5.
In what time the simple interest accrued on a sum of?35000 at the rate of 13% per annum becomes?27300?
Answer:
Given Principle = ₹ 35000 ;
Rate of Interest = 13% ;
Time = x years
Interest = ₹ 27300 .

Time = 6 years

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.6 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.6

Question 1.
A shop selling sewing machines offers a 3% discount on purchases. If the marked price is ₹ 6500. then what is the selling price?
Answer:
Method 1:
Given the marked price = ₹ 6500
bis count = 3%
Disount calculated on marked price.
Discount = 3% of ₹ 6500

∴ Discount = ₹ 195

Selling price
= Marked price – Discount = 6500 – 195
∴ Selling price of sewing machine
= ₹ 6305

Method 2 :
M.P = 100% = ₹ 6500
S.P at a discount of 3%
= (100 – 3)% =?
So, if 100% = 6500
97% = ?
S.P = \(\frac{97}{100}\) × 6500 = ₹ 6305

Question 2.
The marked price of a ceiling fan is ₹ 720 during off season, it is sold for ₹ 684. Determine the discount percentage.

Answer:
Given marked price = ₹ 720
Selling price = ₹ 684
Discount = M.P – S.P = 720 – 684
= ₹ 36
Discount percent = \(\frac{\text { Discount }}{\mathrm{MP}}\) × 100

= 5%
∴ Discount percent = 5%

Question 3.
A publisher gives 32% discount on the printed price of books to book sellers. If the printed price is ₹ 275, then what amount does the seller has to pay to publisher?
Sol. Given printed (marked) price = ₹ 275
Discount percent = 32%

Selling price = M.P – Discount = 275 – 88
∴ Selling price of books = ₹ 187

Question 4.
Rohit buys an item at 25% discount on the marked price. If he bought it for ₹ 660, what is the marked price?
Answer:
Method 1 :
Given selling price = ₹ 660
Discount percentage = 25%
Let the marked price = ₹ x

Marked price – Selling price = Discount
x – 660 = \(\frac{x}{4}\)

∴ Marked price = ₹ 880

Method 2 : M.P = 100% (say)
S.P at a discount of 25% means
S.P = (100 – 25) = 75%
∴ If 75% = ₹ 660 then 100% =?
⇒ SP = \(\frac{100}{75}\) × 660 = ₹ 880

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.5 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.5

Question 1.
Rekha bought a wristwatch for ₹ 2250 and sold it for ₹ 1890. Then find her loss or gain percentage?

Answer:
Given cost price of wristwatch = ₹ 2250
Selling price of wrist watch = ₹ 1890
C.P > S.P, then Rekha got loss.
Loss = C.P – S.P = 2250 – 1890 = ₹ 360

We know loss percent = \(\frac{\text { Loss }}{\text { C.P }}\) × 100

∴ Loss percentage 16%.

Question 2.
A shopkeeper buys a toy for ₹ 250 and sells it for ₹ 300. Find his gain or loss percentage?

Answer:
Given cost price of a toy = ₹ 250
Selling price of a toy = ₹ 300
C.P < S.P, then Shopkeeper got profit.
Profit = S.P – C.P
= 300 – 250 = ₹ 50

We know profit percent

∴ Profit percentage = 20%

Question 3.
The cost price of a chair ₹ 480. 1f he sold at a profit of 10%. what would be selling price of it’?
Answer:
Method: 1
Given cost price of a chair = ₹ 480
Profit percent = 10%
Profit = 10% of ₹ 480

Selling price = C.P + Profit
= ₹ 480 + ₹ 48
∴ Selling price = ₹ 528

Method: 2
C.P = 100% = ₹ 480
Selling it 10% gain.
= S.P = 100% + 10% =?
z 100% = ₹ 480
110% =?
S.P = \(\frac{110}{100}\) × 480= ₹ 528
∴ Selling price = ₹ 528

Question 4.
If Sharma purchased a car for ₹ 350000. After two years he sold at a loss of 12%. Find its selling price’?
Answer:
Given cost price of a car = ₹ 350000
Loss = 12 %
Loss 12% of ₹ 350000

Loss = ₹ 42000
Selling price = CP – Loss
= 350000 – 42000
∴ Selling price of a car = ₹ 3,08,000

Question 5.
A shopkeeper buys wooden tables each at ₹ 2800 and expends ₹ 400 on each table for painting. If he sells it at a cost of ₹ 4000. Find his profit percentage?

Answer:
Given cost price of wooden table = ₹ 2800
Amount spent on painting = ₹ 400
Effective cost price of wooden table
= ₹ 2800 + ₹ 400 = ₹ 3200

Selling price wooden table = 4000
C.P < SP, then shopkeeper got profit.
Profit = S.P – C.P
= ₹ 4000 – ₹ 3200 = ₹ 800

We know profit percentage

∴ Profit percentage = 25%

Question 6.
In a garment shop a saree is sold at a cost of ₹ 1800 after a profit of ₹ 600. Find the profit percentage and cost price?
Answer:
Given selling price of a saree = ₹ 1800
Profit = ₹ 600
Cost price of a saree = S.P – Profit
= 1800 – 600
= ₹ 1200.
We know that profit percentage

∴ Profit percentage = 50%

Question 7.
After incurring a loss of ₹ 258, Jean pant was sold at ₹ 1750. Then find the cost price and loss percentage?
Answer:
Given selling price of Jean pant = ₹ 1750
Loss = ₹ 258
Cost price of Jean pant = S.P + Loss
= 1750 + 258
= ₹ 2008
Loss percent = \(\frac{\text { Loss }}{\text { C.P }}\) × 100
⇒ \(\frac{258}{2008}\) × 100
∴ Loss percentage = 12.85%

Question 8.
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percentage?
Answer:
Method 1:
Let Cost price of 10 articles be ₹ x.
Cost price of 10 articles = 10 x
Selling price of 9 articles = 10 x
Cost price of 9 articles 9x
Profit = S.P — C.P = 10x – 9x
= ₹ x
Profit percentage = \(\frac{\text { Profit }}{\text { C.P }}\) × 100

∴ Profit percentage = \(\frac{100}{9}\)% (or)
11.11% (or) 11\(\frac{1}{9}\)%

Method 2:
By selling 9 articles, there is a gain of 1 article.
∴ Profit Percentage = \(\frac{1}{9}\) × 100
= 11\(\frac{1}{9}\)%

Question 9.
By selling a book for ₹ 258, a bookseller gains 20%. For how much should he sell it to gain 30%?
Answer:
Method 1:
Given selling priceof a book = ₹ 258
Gain = 20%
Cost price of book = ₹ x

∴ Cost price of a book = ₹ 215

If profit percent = 30%

Profit = ₹ 64.50
∴ S.P = C.P + Profit
= 215 + 64.50 = 279.50
To get 30% profit he has to sold at ₹ 279.50.

Method 2:
Let the C.P of the book = 100%
Selling at a gain of 20% means (100 + 20)% = ₹ 258
120% = ₹ 258
Selling at a gain of 30% means (100+30)%=?
If 120% = ₹ 258
then 130% =?
∴ SP = \(\frac{130}{120}\) × 258 = ₹ 279.5

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.3

Question 1.
Find out whether the given quantities vary directly or inversely.
(i) Time taken to cover a distance, speed.
Answer:
Inversely proportional.

(ii) Area of land and its cost.
Answer:
Direct proportion.

(iii) Number of men for work, time taken to complete the work.
Answer:
Inverselv proportional.

(iv) Number of people, quantity of food grains each one gets (total quantity remains same).
Answer:
Inversely proportional.

(v) The length of a journey by bus and price of the ticket.
Answer:
Direct proportion.

Question 2.
If 24 men can construct a wall in. 10 days. In how many days will 15 men do it?
Answer:
If the number of men decreases working days will increase in the same proportion.
So, number of men and the working days are in inverse proportion.

Let the number of days to complete the work be ’x.

Number of mens	Number of days
24	10
15	x

By taking inverse proportion,
24 : 15 = x : 10

Then, product of means = Product of the extremes
⇒ 15 × x = 24 × 10

⇒ x = 16
∴ 15 men will complete the wall in 16 days.

Question 3.
In a hostel there are food provisions for 50 girls for 40 days. If 30 more girls join the hostel, how long will the provisions last?
Answer:
If the number of girls increases, then number of days will decrease in the same proportion.
So, number of girls and the days for food provisions lost are in inverse proportion.
Let the number of food provisions days be ’x’.

Number of girls	Number of days
50	40
50 + 30 = 80	x

By taking inverse proportion,
50 : 80 = x : 40
Then, product of means = Product of extremes
⇒ 80 × x = 50 × 40

⇒ x = 25
∴ Number of days required to com-plete the food provisions for 80 girls is 25 days.

Question 4.
Suman travels a distance for 5 hours with a speed of 48 kilometres per hour. If he wants to travel the same in 4 hours at what speed he should travel?
Answer:
If the time decreases, then the speed will increase in the same proportion. Let the speed to cover the distance in 4 hours is ’x’.

Time	Speed (kmph)
5	48
4	x

By taking inverse proportion,
5 : 4 = x : 48
Then, Product of means = Product of extremes
⇒ 4 × x = 48 × 5

⇒ x = 60 kmph.
∴ Distance covered in 4 hours with the speed of 60 kmph.

Question 5.
A person has money to buy 8 bicycles of worth ₹ 4500 each. If the cost of the bicycle is decreased by ₹ 500, then how many bicycles cap he buy with the amount he has?

Answer:
If the cost of the bicyble decreases, then the number of cycles will increase in the same proportion.
Let the number of bicycles he can buy each with ₹ 4000 (4500 – 500) be ‘x’.

Cost of the bicycle	Number of bicycles
4500	8
(4500-500) = 4000	x

By taking inverse proportion,
4500 : 4000 = x : 8
Then, Product of means = Product of extremes
⇒ 4000 × x = 4500 × 8
⇒ \(\frac{4000 x}{4000}=\frac{4500 \times 8}{4000}\) = 9
⇒ x = 9
∴ Number of bicycles bought with the same amount is 9.

Question 6.
2 pumps are required to fill a tank in 1 hour. How many pumps of the same type are used to fill the tank in 24 minutes?
Answer:
If the filling time decreases, then the number of pumps will increase in the same proportion.
Let the number of pumps to fill the tank be x.

Time taken to fill the tank	Number of Pumps
1 hour (60 minutes)	2
24 minutes	x

By taking inverse proportion,
60 : 24 = x : 2

Then, Product of means = Product of extremes
⇒ 24 × x = 60 × 2

⇒ x = 5
∴ Number of pumps required to fill the tank in 24 minutes is 5.

Question 7.
18 men can reap a field in 10 days. For reaping the same field in 6 days, how many more men are required?

Answer:
If the reaping days decreases, then the number of men increases in the same proportion.
Let the number of men required to reap the field in 6 days is x.

Number of mens	Number of reaping days
18	10
x	6

By taking inverse proportion, 18 : x = 6 : 10
Then, Product of means = Product of extremes

∴ Number of men required to reap the field in 6 days = 30
Number of more men = 30 – 18 = 12
Therefore 12 more men required to reap the field in 6 days.

Question 8.
1200 soldiers in a checkpost had enough food for 28 days. After 4 days some soldiers were transferred to another checkpost and thus remaining food is sufficient for 32 more days. How many soldiers left the checkpost?

Answer:
If the number of soldiers decreases, then the number of days food lost will increases in the same proportion.
Let the number of soldiers after transfer is x.

After 4 days,

Number of Soldiers	Enough Food days
1200	28 – 4 = 24
x	32

By taking inverse proportion, 1200 : x = 32 : 24
Then, Product of means = Product of extremes

∴ Number of Soldiers remained in the checkpost = 900
Number of Soldiers transferred = 1200 – 900 = 300
So, 300 soldiers left the checkpost.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.2

Question 1.
Find out whether the given quantities are in direct proportion or not.
(i) Cost of pens, number of pens.
Answer:
Direct proportion. (As the number of pens increases, their total cost also increases)

(ii) Number of people, food required to them.
Answer:
Direct proportion. (As the number of people increases, food required for them also increases)

(iii) Speed of a car, time taken to reach destination.
Answer:
Not in direct proportion. (As the speed increases, time taken by the car reduces)

(iv) Time taken, distance covered.
Answer:
Direct proportion. (As the time increases, distance covered will be increased)

(v) Cost of the vegetables, number of vegetable bags.
Answer:
There is no relation between cost of vegetables and number of vegetable bags.

Question 2.
Five people went to a Park and paid ₹ 580 for tickets. If three people went to the Park, how much money did they have to pay?

Answer:
We know that if the number of people decreases the amount of the tickets also decreases such that the ratio of number of people and the ratio of their tickets amount will remain the same. That means here number of people and the amount of their tickets are in proportion.

Amount paid for tickets by 5 people = ₹ 580
Let the amount paid for tickets by 3 people be x.
Then, 5 : 3 = 580 : x

If the ratios are equal, the product of means = the product of the extremes.
⇒ 5 × x = 3 × 580
⇒ \(\frac{5 x}{5}=\frac{3 \times 580}{5}\)
⇒ x = 3 × 116
⇒ x = 348
∴ Money paid by three persons for tickets = ₹ 348.

Question 3.
A map drawn with a scale of 1 cm re-presents 26 km. If the original distance between two stations 1404 km., then what would be the distance between the stations in the map?
Answer:
We know that if the distance in the map increases the original distance also increases such that the ratio of distance in the map end ratio of their original distance will remain the same. That means here the distance in map and original distance are in proportion.

Distance 26 km scale represented in the map = 1 cm.
Let the distance 1404 km in the map be equal to x cm.
Then, 26 : 1404 = 1 : x
If the ratio are equal, the product of means = the product of the extremes
⇒ 26 × x = 1404 × 1
⇒ \(\frac{26 x}{26}=\frac{1404}{26}\)
⇒ x = 54 cm
∴ Distance 1404 km between the stations in the map = 54 cm.

Question 4.
The weight of 72 pipes is 180 kg. Then what is the weight of 90 such pipes?

Answer:
We know that if the number of pipes increases their weight also increases, such that the ratio of number of pipes and the ratio of their weights will remain the same. That means here number of pipes and their weights are in proportion.
Given weight of 72 pipes 180 kg
Let the weight of 90 pipes = x kg ,
Then 72 : 90 = 180 : x
If the ratios are equal, the product of means = The product of extremes
⇒ 72 × x = 90 × 180

⇒ x = 225 kg
∴ weight of 90 pipes = 225 kg

Question 5.
A motorbike requires 3 liters of petrol to cover 135 km. on average. How many litres of petrol will be required to cover 495km?
Answer:
We know that if the distance increases quantity of petrol also increases, such that the ratio of distances and the ratio of the required quantity of petrol will remain the same. That means here distance and required quantity of petrol are in proportion.
Given quantity of petrol required to cover 135 km = 3l
Let the quantity of petrol required to cover 495 km = xl
Then 135 : 495 = 3 : x

If the ratios are equal, the product of means = The product of the extremes
⇒ 135 × x = 495 × 3

⇒ x = 11 l
∴ The quantity of petrol required to cover 495 km = 11 l.

Question 6.
The shadow of a pole with the height of 10 m. is 6 m. Then find the height of another pole whose shadow is 9 m. at the same time.
Answer:
We know that if the length of shadow increases the height of the pole also increases, such that the ratio of lengths of shadows and the ratio of heights of poles will remain the same.
That means here lengths of shadows and heights of poles are in proportion.

Given height of the pole when its shadow’s length is 6 m = 10 m
Let the height of the pole when its shadow’s length is 9 m = x m
Then 6 : 9 = 10 : x If the ratios are equal,
The product of means = The product of the extremes
⇒ 6 × x = 9 × 10

⇒ x = 3 × 5 = 15m
∴ The height of the pole when its shadows length is 9 m = 15 m.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Review Exercise

Question 1.
Find the ratio of the following:
(i) 5,8
Answer:
5 : 8

(ii) ₹ 10, ₹ 15
Answer:
10 : 15
2 : 3
∴ ₹ 10 : ₹15 = 2 : 3

(iii) 25 kg., 20. kg.
Answer:
25 kg : 20 kg
25 : 20
5 : 4
∴ 25 kg : 20 kg = 5 : 4

(iv) 5l, 500ml
Answer:
5l : 500 ml
5000 ml: 500 ml (∵ 1l = 1000 ml, 5l = 5000 ml)
5000 : 500
10 : 1
∴ 5l : 500 ml = 10 : 1

(v) 2 km. 500 m, 1 km. 750 m
Answer:
2 km. 500 m : 1 km. 750 m
2000 m + 500 m : 1000 m + 750 m (∵ 1 km = 1000m; 2 km = 2000 m)
2500 : 1750
10 : 7 (∵ HCF = 25)
∴ 2 km. 500 m : 1 km. 750 m = 10 : 7

(vi) 3 hrs, 1 hr. 30 min.
Answer:
3 hrs : 1 hr. 30 min
3 × 60 : (1 × 60) + 30 (∵ 1 hr = 60 min; 3h = 3 × 60 = 180 min)
180 min : 60 + 30 min

2 : 1
∴ 3 hrs : 1 hr. 30 min = 2:1

(vii) 40 days, 1 year
Answer:
40 days : 1 year
40 days : 365 days (∵ 1 year = 365 days)
40 : 365 (∵ HCF = 5)
8 : 73

∴ 40 days : 1 year = 8 : 73
(Or)
40 days : 366 days (∵ 1 year = 366 days for leap year)
20 : 183 (∵ HCF = 2)

Question 2.
Express the following ratios in the simplest form:
(i) 120 : 130
Answer:
Given 120 : 130

12 : 13
∴ 120 : 130 = 12 : 13

(ii) 135:90
Answer:
Given 135 : 90
27 : 18
3 : 2
∴ 135 : 90 = 3 : 2

(iii) 48 : 144
Answer:
Given 48: 144
12 : 36
1 : 3
∴ 48 : 144 = 1:3

(iv) 81 : 54
Answer:
Given 81 : 54
9 : 6
3 : 2
∴ 81 : 54 = 3 : 2

(v) 432 : 378
Answer:
Given 432 : 378
216 : 189
24 : 21
8 : 7
∴ 432 : 378 = 8 : 7

Question 3.
Check whether the two ratios given below are in proportion.

(i) 10 : 20, 25 : 50
Answer:
Given 10 : 20, 25 : 50
a :b = 10 : 20
So, a : b = 1 : 2

c : d = 25 : 50
also, c : d = 1 : 2

If a : b = c: d, then a, b, c, d are in proportion.
a : b = c : d = 1 : 2
Therefore, 10 : 20, 25 : 50 are in proportion.

(ii) 18 : 12, 15 : 10
Answer:
Given 18 : 12, 15 : 10
a : b = 18 : 12
= 6 : 4

c : d = 15 : 10
also c : d = 3 : 2
So, a : b = 3 : 2

If a : b = c : d, then a, b, c, d are in proportion.
a : b = c : d = 3 : 2
Therefore, 18 : 12, 15 : 10 are in proportion.

(iii) 25 : 20, 16 : 14
Answer:
Given 25 : 20, 16 : 14
a : b = 25 : 20
So, a : b = 5 : 4
5 : 4 ≠ 8 : 7
∴ a : b ≠ c : d

c : d = 16 : 14
c : d = 8 : 7

If a : b = c : d, then a, b, c, dare in propottion.
Here a : b ≠ c : d
So, a, b, c, d are not in proportion.

(iv) 54 : 27, 18 : 9
Answer:
Given 54 : 27, 18 : 9
a : b = 54 : 27.
= 6 : 3
So, a : b = 2 : 1

c : d = 18 : 9
also, c : d = 2 : 1

If a : b = c : d, then a, b, c, d are in proportion.
a : b = c : d = 2 : 1

Therefore, 54: 27, 18 : 9 are in proportion.

Question 4.
Find the missing number in each of the following problems:
(i) 15 : 19 = 45 : _______
Answer:
Given 15 : 19 = 45 : x
Let the missing term be ‘x’.
If a : b = c : d, then a, b, c and d are in proportion, then the product of extremes is equal to the product of means that is a . d = b . c
⇒ 15 × x = 19 × 45

⇒ x = 19 × 3
∴ x = 57

(ii) 9 : 13 = ____ : 65
Answer:
Given 9 : 13 = x : 65
Let the missing term be ‘x’.
If a : b = c : d, then a, b, c and d are in proportion, then the product of extremes is equal to the product of means that is a . d = b . c
⇒ 9 × 65 = 13 × x

⇒ 45 = x
∴ x = 45

(iii) 8 : ______ = 72 : 63
Answer:
Given 8 : x = 72 : 63
Let the missing term be ‘x’.
If a : b = c : d, then a, b, c and d are in proportion, then the product of extre¬mes is equal to the product of means that is a . d = b . c
⇒ 8 × 63 = x × 72

⇒ 7 = x
∴ x = 7

Question 5.
Fill in the boxes with equivalent ratio of 3 : 4.

Answer:
Given a : b = 3 : 4 (or) \(\frac{3}{4}\)
\(\frac{a}{b}=\frac{3}{4}\)

Equivalent ratios

Therefore, a : b = 3 : 4 = 6 : 8
= 9 : 12 = 12 : 16 = 15 : 20 = 18 : 24

Question 6.
Pick out any four numbers from below and arrange them so that they are in proportion 2, 3, 10, 12> 15, 18.
Ex : 2 : 10 = 3 : 15
(i) : __________
Answer:
2 : 3; 12 : 18
a : b = 2 : 3, c : d = 12: 18
If a, b, C, d are in proportion, then the product of extremes is equal to the pro¬duct of means.

a . d = b . c
2 × 18 = 3 × 12
36 = 36
So, 2 : 3 = 12 : 18

(ii) : __________
Answer:
10 : 12 = 15 : 18
a : b = 10 : 12
c : d = 15 : 18
If a, b, c, d are in proportion, then the product of extremes is equal to the product of means.

a . d = b . c
10 × 18 = 15 × 12
180 = 180
So, 10 : 12 = 15 : 18

Question 7.
Divide ₹ 1500 into two parts such that they are in the ratio of 7 : 3.
Answer:
Given amount to be divide = ₹ 1500 .
Ratio to be divide = 7:3
Sum of the terms in the ratio = 7 + 3 = 10

[Or 1500 – 1050 = 450]

Question 8.
If there are 20 chocolates in a packet, Rajani and Ragini share them and Rajani takes 12 chocolates, then what is the ratio of chocolates taken by Rajani to Ragini?
Answer:
Total number of Chocolates = 20
Number of chocolates Rajani taken = 12
Number of chocolates Ragini taken = 20 – 12 = 8
Ratio of Rajani to Ragini = 12 : 8
Ratio of chocolates taken by Rajani to Ragini = 3 : 2

Question 9.
A pipe is cut into two parts in such a ratio that the first part to second part is 7:8. If the length of the 2nd part is 48cm, then what is the length of the first part? What is the total length of the pipe before cutting?
Answer:
Given ratio of two parts of pipe = 7:8
Length of the second part = 48 cm
Length of the first part = x cm
Ratio of the two parts = x : 48

Product of extremes = Product of means
⇒ 8 × x = 48 × 7

⇒ x = 42
∴ Length of first part of the pipe = x = 42 cm
Length of the pipe = 42 + 48 = 90 cm

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling InText Questions

[Page No. 108]

Where does the Arithmetic Mean lie ?
The marks obtained by Sarala, Bindu, Geeta and Rekha in Telugu, Hindi and English are given below.

Answer:
Now, let us calculate the average marks obtained by the students in each subject.

Answer:

(i) What do you observe from the above table ?
Answer:
Mean always lie between the maximum and minimum values/observations of the data.

(ii) Does the mean lies between maximum and minimum values in each case ?
Answer:
Yes, Its always true.
i.e., Mean always lies between the maximum and minimum values of the data.

Check Your Progress [Page No: 109]

Question 1.
Find the Arithmetic Mean of first three multiples of 5 ?
Answer:
Given, first three multiples of 5 are: 5, 10, 15.
Sum of observations = 5 + 10 + 15 = 30
Number of observations = 3
Arithmetic mean =\(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{30}{3}\) = 10
∴ Arithmetic mean of first three multiples of 5 is 10.

Let’s Explore [Page No. 109]

Collect the information about the weights of any ten students of your class in kilograms arid answer the following:

Answer:

Name of the student	Weight in kilograms
Aditya	38
Kishore	39
Balu	40
Sreeja	36
Sreekari	35
Kairavi	3.7
Swathi	36
Kirshna	41
Ram	39
Prasad	39

Question 1.
What are the greatest and smallest weights ?
Answer:
Krishna = 41 kg – Greatest weight
Sreekari = 35 kg – Smallest weight

Question 2.
Find Arithmetic mean of collected data.
Answer:
Given observations: 38, 39, 40, 36, 35, 37, 36, 41, 39, 39.
Sum of observations = 38 + 39 + 40 + 36 + 35 + 37 + 36 + 41 + 39 + 39 = 380
Number of observations = 10
Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{380}{10}\) = 38 kg.
Number of observations 10
∴ Arithmetic mean weight of 10 students of our class is 38 kg.

Question 3.
Verify whether Arithmetic Mean lies between greatest and smallest observations or not.
Answer:
Yes, Arithmetic mean 38 kg lies between greatest weight 41 kg and smallest weight 35 kg.

Check Your Progress [Page No. 110]

Question 1.
What is the range of first ten whole numbers?
Answer:
First ten wh4’e numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Maximum value = 9
Minimum value = 0
Range = Maximum Value – Minimum Value = 9 – 0 = 9
∴ Range = 9

Check Your Progress [Page No. 111]

Find the mode of 10, 9, 12, 10, 8, 7, 6, 10, 9, 7, 8, 5 and 2.
Answer:
Given data : 10, 9, 12, 10, 8, 7, 6, 10, 9, 7, 8, 5 and 2.
By arranging the numbers with same values together.
2, 5, 6, 7, 7, 8, 8, 9, 9, 10, 10, 10, 12
As ’10’ occurs more frequently than other observations.
∴ Mode = 10

Let’s Explore [Page No. 111]

Take a dice, roll it 20 times and record the numbers you got on its top face. Find the ‘Mode’ of resulting numbers.
Answer:
A dice rolled 20 times and the numbers are
2, 4, 5, 3, 1, 6, 5, 4, 2, 1, 3, 5, 4, 2, 6, 2, 2, 5, 1, 3.
By arranging the numbers with same values together
1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6.
As 2 occurs more frequently than other observations.
Mode = 2.

Let’s Think [Page No: 111]

The following data shows that the number of hours spent by students for study. Find the mode:

Answer:
In the given data more number of students spend 1 hour for study.
So, mode of the data = 4.

Check Your Progress [Page No: 114]

What is the median of first 7 prime numbers ?
Answer:
First 7 prime numbers are 2, 3, 5, 7, 11,13, 17.
Arrange the given observations in ascending order

In seven observations the 4th observation 7 is the middle most value.
∴ Median = 7.

Let’s Explore [Page Mo. 114]

Question 1.
Collect 10th class pass percentage for last six years of your school (or) your near by school. Find median of the data.
Answer:
10th Class pass percentage for last six years of our school are: 100%, 98%, 93%, 95%, 96%, 97%.
Arrange the given observations in ascending order.

In six observations the 3rd and 4th observations are 96% and 97%.
Here, we have two middle most values 96% and 97%.
Median = Average of two middle most values
= \(\frac{96+97}{2}\) = 96.5%
∴ Median of the data = 96.5%

Project Work [Page No: 114]

Visit any vegetable market along with your parent. Collect the information about the costs of different vegetables. By using this data fill the following table.

Answer:

[Page No. 115]

Prasanna wanted to buy a mobile phone. He selected two mobile phones of different companies having same features. Now he would like to know which mobile phone has good performance. He collected the information of star rating from different magazines and news papers.

Question 1.
What information is given in the table ?
Answer:
The table is showing information about the mobile phones.

Question 2.
Is the above information being useful to Prasanna ?
Answer:
Yes.

Question 3.
Which mobile phone you will suggest to Prasanna ?
Answer:
Mobile A. (21 + \(\frac{1}{2}\) + \(\frac{1}{2}\) rating)

Check Your Progress [Page No. 116]

Observe the adjacent bar graph and answer the following.

Question 1.
Which fruit most of the people like ?
Answer:
Most of the people liked Apple.

Question 2.
How many people likes banana?
Answer:
10.

Observe the following graph. [Page No. 116]

Observe the above double bar graph and answer the following:

Question 1.
In which year the sales of both mobile phone companies are equal ?
Answer:
2018.

Question 2.
In 2017, which mobile phone company has more sales ?
Answer:
Mobile phone – A.

Examples

Question 1.
Mid Day Meal (MDM) taken by the students in six days are 132, 164, 145, 182, 163 and 114, Find Arithmetic Mean of students who took MDM per day.
Answer:
Given, number of students who took MDM 132, 164, 145, 182, 163, 114.
Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{132+164+145+182+163+114}{6}\)
= \(\frac{900}{6}\)
= 150

Question 2.
Ages of students (in years) are 8, 5, 6, 6, 5, 7, 5, 6, 5, 4 7, 6, 7, 6, 5 8 and 6 find the mode of given data.
Answer:
Given, ages of students are 8, 5, 6, 6, 5, 7, 5, 6, 5, 4, 7, 6, 7, 6, 5, 8, 6.
By arranging the numbers with same values together
4, 5, 5, 5, 5,5,6, 6,6,6, 6, 6,7,7, 7, 8, 8.
As 6’ occurs more frequently than other observations.
∴ Mode = 6

Question 3.
Find the mode of the data of Grades A, B, E, A, C, E, B, C, D, A, D, C, F, A and C.
Answer:
Given, A, B, E, A, C, E, B, C, D, A, D, C, F, A, C.
By arranging the letters of same type together
A, A, A, A, B, B, C, C, C, C, D, D, E, E, F
As ‘A’ and ‘C’ occurs most frequently in the data.
∴ Mode = A and C

Question 4.
Find the median of data 32, 43, 25, 67, 46, 71 and 182.
Answer:
Given 32, 43, 25, 67, 46, 71, 182
Arrange the given observations in ascending order.

In ‘7’ observations the 4th observation is the middle most value.
∴ Median = 46.

Question 5.
The monthly incomes of 8 members are ₹8000, ₹9000, f8200, ₹7900, ₹8500, ₹8600, ₹7700 and ₹60000. Find the median income.
Answer:
Given that, the monthly incomes of 8 members are:
₹8000, ₹9000, ₹8200, ₹7900, ₹8500, ₹8600, ₹7700, ₹60000
Arrange the given observations in ascending order.

Here, we have two middle most values 8200 and 8500.
In this case, Median = Average of two middle most values
8200 + 8500 16700
= \(\frac{8200+8500}{2}\) = \(\frac{16700}{2}\) = ₹8,350

Question 6.
The number of CFL bulbs and LED bulbs by a seller every month from March to August are given below. Draw by vertical double bar graph.

Month	CFL Bulbs	LED Bulbs
March	70	75
April	35	30
May	65	75
June	90	100
July	22	35
August	50	50

Answer:
Steps in drawing a Double bar graph:

1. Draw X-axis (horizontal line) and Y – axis (vertical line) on the graph paper and mark their intersection point as ‘O’.
2. Take ‘months’ on X – axis.
3. Take number of CFL and LED bulbs on Y ~ axis.
4. Take an appropriate scale on Y-axis so that number of both the bulbs can be shown easily. Here the maximum value to be plotted on Y – axis as 100, so let us take 1 cm =10 bulbs on Y – axis.
5. Find the length of each bar by dividing the value by 10 (as scale is 1 cm = 10 bulbs).

Eg: Length of bar represents 70 CFL bulbs = \(\frac{70}{10}\) = 7 cm
Length of bar represents 75 LED bulbs = \(\frac{75}{10}\) = 7.5 cm

Question 6.
Draw the bars of uniform width representing ‘CFL bulb’ and ‘LED bulb’ side by side of every month.

Observe the adjacent figure:

Question 1.
Largest portion of the circle is shaded by which colour ?
Answer:
Red.

Question 2.
Are the portions of circle shaded by blue and pink are in same size ?
Answer:
No.

Question 3.
Smallest portion of the circle is shaded by which colour ?
Answer:
Yellow.

Question 2.
Observe the below picture which shows various expenses of Manasa’s family:

(i) For which item highest amount spent?
Answer:
Food.

(ii) For which items same amount spent?
Answer:
Charity and Education.

(iii) For which item least amount spent?
Answer:
Others.

Question 7.
In a school, there are 100 students in class VII and every student is a member of any one of the club. The following table shows the number of students in various clubs, then Construct the pie chart to following data:

Club	Number of members
Mathematics	50
Science	30
Social Studies	40
English	40
Arts	20

Answer:
The angle of each sector will depend on the ratio between the number of students in club and total number of students.
Angle of sector = \(\frac{\text { Value of the item }}{\text { Sum of the values of all items }}\) × 360°

Steps of construction:

1. Draw a circle with any convenient radius and mark it’s centre as ‘O’.
2. Mark a point A, somewhere on the circumference and join OA.
3. Construct ∠AOB = 100° to represent angle of the sector for Maths club.
4. Construct ∠BOC = 60° to represent angle of the sector for Science club.
5. Construct ∠COD = 80° to represent angle of the sector for Social Studies club.
6. Construct ∠DOE = 80° to represent angle of the sector for English club.
7. Now ∠EOA = 40° to represent angle of the sector for Arts club.

Practice Questions [Page No: 124]

In the following alphabetical series, a term (next term) is missing. Choose the missing term from the options.

Question 1.
B, F, J, N, R,’V, …..
(a) Z
(b) W
(c) X
(d) Y
Answer:
(a) Z

Explaination:

So, answer is Z.

Question 2.
A, C, E, G, I, K, …..
(a) P
(b) O
(c) N
(d) M
Answer:
(d) M

Explaination:

So, answer is M.

Question 3.
M, O, R, T, …….
(a) W
(b) U
(c) V
(d) Q
Answer:
(a) W

Explaination:

So, answer is W.

Question 4.
U, S, P, L, …….
(a) F
(b) G
(c) H
(d) I
Answer:
(b) G

Explaination:

So, answer is G.

Question 5.
ZA, YB, XC, WD,
(a) UE
(b) EV
(c) VE
(d) SH
Answer:
(c) VE

Explaination:

So, answer is VE.

Question 6.
AM, BO, CQ, DS, EU, …………
(a) WF
(b) FU
(c) GV
(d) FW
Answer:
(d) FW

Explaination:

In the series next word is FW.

Question 7.
ZY, XV, UR, QM, …..
(a) LG
(b) LI
(c) LH
(d) KJ
Answer:
(a) LG

Explaination:

So, in the series next word is LG.

Question 8.
AC, DF, GI, JL, …..
(a) NO
(b) MO
(c) MN
(d) NP
Answer:
(b) MO

Explaination:

So, in the series next word is MO.

Question 9.
DN, EM, FL, GK, HJ, …..
(a) IK
(b) GI
(c) IJ
(d) II
Answer:
(d) II

Explaination:

So, in the series next word is II.

Question 10.
CBA, STU, FED, VWX,
(a) IHG
(b) GHI
(c) IJK
(d) YZA
Answer:
(a) IHG

Explaination:

So, in the series next word is IHG.

Question 11.
AZC, DYF, GXI, JWL
(a) OVM
(b) UNV
(c) MVO
(d) MNO
Answer:
(c) MVO

Explaination:

So, in the series next word is MVO.

Question 12.
ABK, CDL, EFM, GHN,
(a) JIO
(b) IJO
(c) MNO
(d) ONM
Answer:
(b) IJO

Explaination:

So, in the series next word is IJO.

Question 13.
A2C, D5F, G8I, J11L, …..
(a) Ml40
(b) Ml20
(c) N15P
(d) N12P
Answer:
(a) Ml40

Explaination:

So, in the series next word is M140.

Question 14.
A, CD, HIJ, PQRS, …………..
(a) ZABCD
(b) ZYXW
(c) ABCDE
(d) RSTUV
Answer:
(c) ABCDE

Explaination:

So, in the series next word is ABCDE.

Question 15.
A, BC, DEF, GHIJ,
(a) KLMNP
(b) LMNOP
(c) KLMNO
(d) JKLMN
Answer:
(c) KLMNO

Explaination:
A BC DEF GHIJ KLMNO
So, in the series next word is KLMNO.

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Review Exercise

Question 1.
Mark any three non-collinear points A, B, and C in your notebook, join them to make a triangle, and name it.
Answer:

Question 2.
Observe the given triangle and answer the following :

(i) Write the interior points of the triangle.
Answer:
A, C, D and J.

(ii) Write the points marked on the triangle.
Answer:
B, E, G, P, Q and R.

(iii) Write the exterior points of the triangle.
Answer:
F, H and I.

Question 3.
Observe the given triangle and answer the following:

(i) The opposite side to vertex L is __________
Answer:
\(\overline{\mathrm{KM}}\)

(ii) The opposite side to ZK is __________
Answer:
\(\overline{\mathrm{LM}}\)

(iii) The opposite angle to \(\overline{\mathrm{KL}}\) is __________
Answer:
∠M.

(iv) The opposite vertex to \(\overline{\mathrm{LM}}\) is __________
Answer:
K.

Question 4.
Classify the following angles into acute, obtuse and right angles :
20°, 50°, 102°, 47°, 125°, 65°, 36°, 90°, 95° and 110°.
Answer:
Acute angles : 20°, 50°, 47°, 65°, 36°.
Right angle : 90°.
Obtuse angles : 102°, 125°, 95°, 110°.

Question 5.
Write the intersecting point and concurrent point in the adjust figure.

Answer:
Intersecting point: P
Concurrent point: Q

SCERT AP 7th Class Maths Solutions Pdf Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 2.3

Question 1.
Fill in the blanks in the table. One is done for you.
Answer:

Division	Quotient
1. 362.21 ÷ 10	36.221
2. 5636.1 ÷ 100	56.361
3. 374.9 ÷ 1000	0.3749
4. 2016.4 ÷ 1000	2.0164
5. 123.0 ÷100	1.23
6. 1300.7 ÷ 1000	1.3007
7. 590.01 ÷ 10	59.001

Question 2.
Solve the following,
(i) 5.51 ÷ 2
Answer:
5.51 ÷ 2
= \(\frac{551}{100} \div \frac{2}{1}\)
= \(\frac{551}{100} \times \frac{1}{2}\)
= \(\frac{551 \times 1}{100 \times 2}\)
= \(\frac{551}{200}\)
∴ 5.51 ÷ 2 = 2.755

(ii) 38.4 ÷ 3
Answer:
38.4 ÷ 3
= \(\frac{384}{10} \div \frac{3}{1}\)
= \(\frac{384}{10} \times \frac{1}{3}\)
= \(\frac{384 \times 1}{10 \times 3}\)
= \(\frac{384}{30}\)
∴ 38.4 ÷ 3 = 12.8

(iii) 57.39 ÷ 6
Answer:
57.39 ÷ 6
= \(\frac{5739}{100} \div \frac{6}{1}\)
= \(\frac{5739}{100} \times \frac{1}{6}\)
= \(\frac{5739 \times 1}{100 \times 6}\)
= \(\frac{5739}{600}\)
∴ 57.39 ÷ 6 = 9.565

(iv) 562.1 ÷ 11
Answer:
562.1 ÷ 11
= \(\frac{5621}{10} \div \frac{11}{1}\)

= \(\frac{511}{10}\)
∴ 562.1 ÷ 11 = 51.1

(v) 0.7005 ÷ 5
Answer:
0.7005 ÷ 5
= \(\frac{7005}{10000} \div \frac{5}{1}\)

= \(\frac{1401}{10000}\)
∴ 0.7005 ÷ 5 = 0.1401

(vi) 9.99 ÷ 3
Answer:
9.99 ÷ 3
= \(\frac{999}{100} \div \frac{3}{1}\)

= \(\frac{333}{100}\)
∴ 9.99 ÷ 3 = 3.33

(vii) 13 ÷ 6.5
Answer:
13 ÷ 6.5
= \(\frac{13}{1} \div \frac{65}{10}\)

= \(\frac{2}{1}\)
∴ 13 ÷ 6.5 = 2

(viii) 10.01 ÷ 11
Answer:
10.01 ÷ 11
= \(\frac{1001}{100} \div \frac{11}{1}\)

= \(\frac{91}{100}\)
∴ 10.01 ÷ 11 = 0.91

(ix) 8 ÷ 0.32
Answer:
8 ÷ 0.32

∴ 8 ÷ 0.32 = 25

(x) 320.1 ÷ 33
Answer:
320.1 ÷ 33

∴ 320.1 ÷ 33 = 9.7

Question 3.
Solve the following divisions,
(i) 78.24 ÷ 0.2
Answer:
78.24 ÷ 0.2

∴ 78.24 ÷ 0.2 = 391.2

(ii) 4.845 ÷ 1.5
Answer:
4.845 ÷ 1.5
= \(\frac{4845}{1000} \div \frac{15}{10}\)

(iii) 0.246 ÷ 0.6
Answer:

(iv) 563.2 ÷ 2.2
Answer:
563.2 ÷ 2.2

∴ 563.2 ÷ 2.2 = 256

(v) 0.026 ÷ 0.13
Answer:
0.026 ÷ 0.13

∴ 0.026 ÷ 0.13 = 0.2

(vi) 4.347 ÷ 0.09
Answer:
4.347 ÷ 0.09
\(\frac{4347}{1000} \div \frac{9}{100}\)

∴ 4.347 ÷ 0.09 = 48.3

(vii) 3.9 ÷ 0.13
Answer:
3.9 ÷ 0.13

∴ 3.9 ÷ 0.13 = 30

(viii) 2032 ÷ 0.8
Answer:
20.32 ÷ 0.8

∴ 20.32 ÷ 0.8 = 25.4

(ix) 24.4 ÷ 6.1
Answer:
24.4 ÷ 6.1

∴ 24.4 ÷ 6.1 = 4

(x) 2.164 ÷ 0.008
Answer:
2.164 ÷ 0.008
\(\frac{2164}{1000} \div \frac{8}{1000}\)

∴ 2.164 ÷ 0.008 = 270.5

Question 4.
Solve the following.
(i) Divide 39.54 by 6
Answer:
39.54 by 6
= 39.54 ÷ 6

∴ 39.54 ÷ 6 = 6.59

(ii) Divide 72 by 10
Answer:
7.2 by 10
= 7.2 ÷ 10
= \(\frac{72}{10} \div \frac{10}{1}\)
= \(\frac{72}{10} \times \frac{1}{10}\)
= \(\frac{72 \times 1}{10 \times 10}=\frac{72}{100}\)
∴ 7.2 ÷ 10 = 0.72

(iii) Divide 5.2 by 1.3
Answer:
5.2 by 1.3
= 5.2 ÷ 1.3

∴ 5.2 ÷ 1.3 = 4

Question 5.
‘Sekhar travelled 154.5 km. in 5 hours with uniform speed on his bike. How much distance does he travel in one hour ?
Answer:
Given, distance travelled in 5 hours = 154.5 km

∴ Distance travelled by Sekhar in 1 hour = 30.9 km

Question 6.
If a Mason worked 100 hours in 12.5 days to construct a wall, then how many hours he totally worked in a day?

Answer:
Given number of hours worked in 12.5 days to.construct a wall = 100 hours.
Number of hours worked in one day
= 100 ÷ 12.5
= \(\frac{100}{1} \div \frac{125}{10}\)
= \(\frac{100}{1} \times \frac{10}{125}\)
= \(\frac{1000}{125}\)
∴ Number of hours worked by Mason in one day = 8 hours.

Question 7.
If the cost of dozen eggs is ₹ 61.80, then find the cost of an egg.
Answer:
Given, cost of dozen eggs = ₹ 61.80
We know 1 dozen =12
Cost of one egg = 61.80 ÷ 12

= \(\frac{515}{100}\)
∴ Cost of one egg = ₹ 5.15

Question 8.
If the price of a tablet strip containing 10 tablets is ₹ 26.5, then find the price of each tablet.
Answer:
Given, cost of 10 tablets = ₹ 26.5
Cost of each tablet = 26.5 ÷ 10
= \(\frac{265}{10} \div \frac{10}{1}\)
= \(\frac{265}{10} \times \frac{1}{10}\)
= \(\frac{265}{100}\)

∴ Cost of each tablet = ₹ 2.65.

SCERT AP 7th Class Maths Solutions Pdf Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 2.2

Question 1.
Find the product of the following.
(i) 23.4 × 6
Answer:
= \(\frac{234}{10} \times \frac{6}{1}\)
= \(\frac{1404}{10}\)
∴ 23.4 × 6 = 140.4

(ii) 681.25 × 9
Answer:
681.25 × 9
= \(\frac{68125}{100} \times \frac{9}{1}\)
= \(\frac{68125 \times 9}{100}\)
= \(\frac{613125}{100}\)
∴ 681.25 × 9
= 6131.25

(iii) 53.29 × 14
Answer:
53.29 × 14
= \(\frac{5329}{100} \times \frac{14}{1}\)
= \(\frac{74606}{100}\)
∴ 53.29 × 14
= 746.06

(iv) 8 × 2.52
Answer:
8 × 2.52
= 8 × \(\frac{252}{100}\)
= \(\frac{8 \times 252}{100}\)
= \(\frac{2016}{100}\)
∴ 8 × 2.52 = 20.16

(v) 25 × 2.013
Answer:
25 × 2.013
= 25 × \(\frac{2013}{1000}\)
= \(\frac{25 \times 2013}{1000}\)
= \(\frac{50325}{1000}\)
∴ 25 × 2.013 = 50.325

Question 2.
Fill the blanks in the table.
Answer:

Multiplications	Product
36.21 × 10	362.1
23.104 × 100	2310.4
6.24 × 1000	6240.0
0.02105 × 1000	21.05
9.234 × 100	923.4
1.3004 × 1000	1300.4
5.9001 × 10	59.001

Question 3.
Find the product.
(i) 5.1 × 8.1
Answer:
= \(\frac{51}{10} \times \frac{81}{10}\)
= \(\frac{51 \times 81}{10 \times 10}\)
= \(\frac{4131}{100}\)
∴ 5.1 × 8.1 = 41.31

(ii) 63.205 × 0.27
Answer:
63.205 × 0.27
= \(\frac{63205}{1000} \times \frac{27}{100}\)
= \(\frac{63205 \times 27}{1000 \times 100}\)
= \(\frac{1706535}{100000}\)
∴ 63.205 × 0.27 = 17.06535

(iii) 1.321 × 0.9
Answer:
1.321 × 0.9
= \(\frac{1321}{1000} \times \frac{9}{10}\)
= \(\frac{1321 \times 9}{1000 \times 10}\)
= \(\frac{11889}{10000}\)
∴ 1.321 × 0.9 = 1.1889

(iv) 6.51 × 0.99
Answer:
6.51 × 0.99
= \(\frac{651}{100} \times \frac{99}{100}\)
= \(\frac{651 \times 99}{100 \times 100}\)
= \(\frac{64449}{10000}\)
∴ 6.51 × 0.99 = 6.4449

(v) 837.6 × 0.006
Answer:
837.6 × 0.006
= \(\frac{8376}{10} \times \frac{6}{1000}\)
= \(\frac{8376 \times 6}{10 \times 1000}\)
= \(\frac{50256}{10000}\)
∴ 837.6 × 0.006 = 5.0256

Question 4.
Rithesh reads a book for 2.5 hours everyday. If he reads the entire book in a week, then how many hours all together are required for him to read the book ?

Answer:
Given, time taken to read a book per day = 2.5 hours.

Time taken to read entire book in a week (one week = 7 days)
= 2.5 × 7 days
= \(\frac{25}{10}\) × 7
= \(\frac{175}{10}\)
∴ Time taken to read entire book = 17.5 hours.

Question 5.
Find the area of the rectangle whose length and breadth are 5.3 cm, 2.7 cm respectively.
Answer:
Given length of the rectangle l = 5.3 cm
Breadth of the rectangle b = 2.7 cm
Area of the rectangle = l × b
= 5.3 × 2.7
= \(\frac{53}{10} \times \frac{27}{10}\)
= \(\frac{53 \times 27}{10 \times 10}\)
∴ Area of the rectangle = \(\frac{1431}{100}\)
= 14.31 sq.cm

Question 6.
If the cost of each cement bag is ₹ 326.50, then find the cost of 24 bags of cement.
Answer:
Given, cost of each cement bag
= ₹ 326.50

Cost of 24 cement bags
= 326.50 × 24
= \(\frac{32650}{100}\) × 24
= \(\frac{32650 \times 24}{100}\)
= \(\frac{783600}{100}\)

∴ Cost of 24 cement bags = ₹ 7,836.

Question 7.
Dharmika purchased chudidhar material of 1.40 m at the rate of 7152.5 pet irtetre. Find the amount to be paid.
Answer:
Given cost of one meter chudidhar material = 7152.50
Cost of 1.40 meter chudidhar material = 152.50 × 1.40
= \(\frac{15250}{100} \times \frac{140}{100}\)
= \(\frac{15250 \times 140}{100 \times 100}\)
= \(\frac{2135000}{10000}\)
∴ Cost of 1.40 meter chudidhar material = ₹ 213.50

Question 8.
If a picture chart costs 74.25. Amrutha wants to buy 16 charts to make an album. How much money does she has to pay ?
Answer:
Given, cost of one picture chart = ₹ 4.25
Cost of 16 charts = 4.25 × 16
= \(\frac{425}{100}\) × 16
= \(\frac{425 \times 16}{100}\)
∴ Cost of 16 charts to make album = \(\frac{6800}{100}\) = ₹ 68