29 + Interview Questions in PROPORTION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	The ratio of the length and the width of a school ground is 5 : 2. Find the length, if the width is 40 meters.
Answer» Let the length = x m, width = 40 m The ratio of length to width = x : 40 as per given statement 5 : 2 = x : 40 => 2 x X = 40 x 5 x =(40 x 5)/2 = 20 x 5 = 100 m

1.

The ratio of the length and the width of a school ground is 5 : 2. Find the length, if the width is 40 meters.

Answer»

Let the length = x m,

width = 40 m

The ratio of length to width = x : 40

as per given statement 5 : 2 = x : 40

=> 2 x X = 40 x 5

x =(40 x 5)/2 = 20 x 5 = 100 m

Discussion

2.	Sathyan got Rs. 500 after working for 6 hours. Gopi got Rs. 400 after working for 4 \(\frac{1}{2}\) hours. Are the wages obtained proportional to the working time ?
Answer» Ratio of working hours = 6 : 4 \(\frac{1}{2}\) = 12 : 9 = 4 : 3 Ratio of wages = 800 : 600 = 8 : 6 = 4 : 3 Since the ratios are equal, the working hours are proportional to the wages.

2.

Sathyan got Rs. 500 after working for 6 hours. Gopi got Rs. 400 after working for 4 \(\frac{1}{2}\) hours. Are the wages obtained proportional to the working time ?

Answer»

Ratio of working hours = 6 : 4 \(\frac{1}{2}\)

= 12 : 9 = 4 : 3

Ratio of wages = 800 : 600 = 8 : 6 = 4 : 3

Since the ratios are equal, the working hours are proportional to the wages.

Discussion

3.	Find the ratio of the following (i) Smita works in office for 6 hours and Kajal works for 8 hours in her office. Find the ratio of their working hours.(ii) speed of a cycle is 15km/h and speed of the scooter is 30km/h.
Answer» i) The ratio of working hours of smita and kajal = 6:8 = (2 × 3 ) : (2 × 4) = 3 : 4 ii) The ratio of speeds of a cycle and a sector = 15 : 30 = (15 × 1) : 15 × 2 = 1 : 2

3.

Find the ratio of the following (i) Smita works in office for 6 hours and Kajal works for 8 hours in her office. Find the ratio of their working hours.(ii) speed of a cycle is 15km/h and speed of the scooter is 30km/h.

Answer»

i) The ratio of working hours of smita and kajal = 6:8

= (2 × 3 ) : (2 × 4) = 3 : 4

ii) The ratio of speeds of a cycle and a sector

= 15 : 30 = (15 × 1) : 15 × 2 = 1 : 2

Discussion

4.	In a primary school there shall be 3 teachers to 60 students. If there are 400 students enrolled in the school, how many teachers should be there in the school in the same ratio?
Answer» No. of teachers are required for 400 students at the rate of 3 teachers to 60 students are ⇒ 60 : 3 = 400 : x ⇒ 60/3 = 400/x ⇒ x = (400 x 3)/60 ∴ x = 20

4.

In a primary school there shall be 3 teachers to 60 students. If there are 400 students enrolled in the school, how many teachers should be there in the school in the same ratio?

Answer»

No. of teachers are required for 400 students at the rate of 3 teachers to 60 students are

⇒ 60 : 3 = 400 : x

⇒ 60/3 = 400/x

⇒ x = (400 x 3)/60

∴ x = 20

Discussion

5.	The ratio of carbon, sulphur and potassium nitrate to make gun powder is 3 : 2: 1. How much quantity of each is required to make 1.2 kg of gun powder ?
Answer» Quantity of gun powder = 1.2 kg Quantity of carbon = \(1.2\times\frac{3}{6}\) = 0.2 x 3 = 0.6 kg Quantity of sulphur = \(1.2\times\frac{2}{6}=0.2\times2\) = 0.4 kg Quantity of potassium nitrate \(=1.2\times\frac{1}{6}=0.2\times1=0.2\) kg

5.

The ratio of carbon, sulphur and potassium nitrate to make gun powder is 3 : 2: 1. How much quantity of each is required to make 1.2 kg of gun powder ?

Answer»

Quantity of gun powder = 1.2 kg

Quantity of carbon = \(1.2\times\frac{3}{6}\)

= 0.2 x 3 = 0.6 kg

Quantity of sulphur = \(1.2\times\frac{2}{6}=0.2\times2\)

= 0.4 kg

Quantity of potassium nitrate

\(=1.2\times\frac{1}{6}=0.2\times1=0.2\) kg

Discussion

6.	Find the buying price of each of the following items when a sales tax of 5% is added on them. (1) a towel of ₹50 (ii) Two bars of soap at ₹35 each.
Answer» Given that Sales tax = 5% (i) Cost of a towel = ₹ 50 Sales Tax = 5% of 50 = \(\frac{5}{100}\) x 50 = \(\frac{5}{2}\) = ₹ 2.50 ∴ C.P. = Net Price + Sales Tax = 50 + 2.50 = ₹ 52.50 (ii) The cost of two soaps at the rate of ₹ 35 each = 2 × 35 = ₹ 70 Sales Tax = 5% of 70 = \(\frac{5}{100}\) × 70 = \(\frac{7}{2}\) = ₹ 3.50 ∴ C.P. = Net Price + Sales Tax = 70 + 3.50 = ₹ 73.50

6.

Find the buying price of each of the following items when a sales tax of 5% is added on them. (1) a towel of ₹50 (ii) Two bars of soap at ₹35 each.

Answer»

Given that Sales tax = 5%

(i) Cost of a towel = ₹ 50

Sales Tax = 5% of 50

= \(\frac{5}{100}\) x 50 = \(\frac{5}{2}\) = ₹ 2.50

∴ C.P. = Net Price + Sales

Tax = 50 + 2.50 = ₹ 52.50

(ii) The cost of two soaps at the rate of ₹ 35 each = 2 × 35 = ₹ 70

Sales Tax = 5% of 70

= \(\frac{5}{100}\) × 70 = \(\frac{7}{2}\) = ₹ 3.50

∴ C.P. = Net Price + Sales

Tax = 70 + 3.50 = ₹ 73.50

Discussion

7.	A Super-Bazar prices an item in rupees and paise so that when 4% sales tax is added, no rounding is necessary because the result is exactly in ‘n’ rupees, where ‘n’ is a positive integer. Find the smallest value of ’n’.
Answer» Let the cost price = x say ∴ If x is increased 4% sales tax is added then x + 4% of x = n x + 4/100 × x = n = \(\frac{104x}{100}\) = n = x = n × \(\frac{100}{104}\) = \(\frac{25\,\times\,n}{26}\) ∴ n should be a least multiple of 26, then only the value of the article should be represented in only rupees. ∴ x = \(\frac{25\,\times\,13}{26} = \frac{25}2\) = 12.5 [∵n = 13, 26, 39. from them 13 should betaken] :. Required value of the article =12.50 + \(\frac{4}{100}\) x 12.5 = 12.50 + 0.5 = ₹13

7.

A Super-Bazar prices an item in rupees and paise so that when 4% sales tax is added, no rounding is necessary because the result is exactly in ‘n’ rupees, where ‘n’ is a positive integer. Find the smallest value of ’n’.

Answer»

Let the cost price = x say

∴ If x is increased 4% sales tax is added then

x + 4% of x = n

x + 4/100 × x = n

= \(\frac{104x}{100}\) = n

= x = n × \(\frac{100}{104}\) = \(\frac{25\,\times\,n}{26}\)

∴ n should be a least multiple of 26, then only the value of the article should be represented in only rupees.

∴ x = \(\frac{25\,\times\,13}{26} = \frac{25}2\) = 12.5

[∵n = 13, 26, 39. from them 13 should betaken]

:. Required value of the article =12.50 + \(\frac{4}{100}\) x 12.5

= 12.50 + 0.5 = ₹13

Discussion

8.	By what percent is Rs.2000 less than Rs.2400? Is it the same as the percent by which Rs.2400 is more than Rs.2000?
Answer» Decrease in percent of Rs.2000 less than Rs.2400 = \(\frac{(2400\,-\,2000)}{2400} \,\times\,100\) = \(\frac{400}{2400} \,\times\,100\) = \(\frac{400} {24} = \frac{50}3\) = 16\(\frac23\)% Increase in percent of Rs.2400 more than Rs.2000 = \(\frac{(2400\,-\,2000)}{2000} \,\times\,100\) = \(\frac{400}{2000} \,\times\,100\) = \(\frac{400} {20} \) = 20%

8.

By what percent is Rs.2000 less than Rs.2400? Is it the same as the percent by which Rs.2400 is more than Rs.2000?

Answer»

Decrease in percent of Rs.2000 less than Rs.2400

= \(\frac{(2400\,-\,2000)}{2400} \,\times\,100\) = \(\frac{400}{2400} \,\times\,100\)

= \(\frac{400} {24} = \frac{50}3\) = 16\(\frac23\)%

Increase in percent of Rs.2400 more than Rs.2000

= \(\frac{(2400\,-\,2000)}{2000} \,\times\,100\) = \(\frac{400}{2000} \,\times\,100\)

= \(\frac{400} {20} \) = 20%

Discussion

9.	With most of the Xerox machines. you can reduce or enlarge your original by entering a percentage for the copy. Reshma wanted to enlarge a 2 cm by 4 cm drawing. She set the Xerox machine for 150% and copied her drawing. What will be the dimensions of the copy of the drawing be’?
Answer» Length of the copy = 2 cm breadth = 4 cm If the length is increase in 150% then its measure = 150 % of 2 cm = \(\frac{150}{100}\) × 2 = 1.5 × 2 = 3 cm If the breadth is increase in 150% then its measure = 150 % of 4 cm 150 = \(\frac{150}{100}\) × 4 = 1.5 × 4 = 6 cm 100 ∴ New length = 3 cm breadth = 6 cm

9.

With most of the Xerox machines. you can reduce or enlarge your original by entering a percentage for the copy. Reshma wanted to enlarge a 2 cm by 4 cm drawing. She set the Xerox machine for 150% and copied her drawing. What will be the dimensions of the copy of the drawing be’?

Answer»

Length of the copy = 2 cm

breadth = 4 cm

If the length is increase in 150% then its measure = 150 % of 2 cm

= \(\frac{150}{100}\) × 2 = 1.5 × 2 = 3 cm

If the breadth is increase in 150% then its measure = 150 % of 4 cm 150

= \(\frac{150}{100}\) × 4 = 1.5 × 4 = 6 cm 100

∴ New length = 3 cm

breadth = 6 cm

Discussion

10.	Find the amount and the compound interest on ₹ 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.
Answer» P = ₹ 6500 R = 5% T = 1 years ∴ \(\frac{PTR}{100}\) = \(\frac{6500\,\times\,5\,\times1}{100}\) = 325 ∴ A = P + I = 6500 + 325 = 6825 ∴ P = 6825 (At the beginning of 2,id year A=P) R = 6% T = 1 year ∴ \(\frac{PTR}{100}\) = \(\frac{6825\,\times\,6\,\times1}{100}\) = 409.5 ∴ A = P + I = 6825 + 409.5 ∴ Amount = ₹ 7234.50 C.I. = A – P = ₹ 7234.50 – 6500 = ₹734.50

10.

Find the amount and the compound interest on ₹ 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.

Answer»

P = ₹ 6500

R = 5%

T = 1 years

∴ \(\frac{PTR}{100}\) = \(\frac{6500\,\times\,5\,\times1}{100}\) = 325

∴ A = P + I = 6500 + 325 = 6825

∴ P = 6825

(At the beginning of 2,id year A=P)

R = 6%

T = 1 year

∴ \(\frac{PTR}{100}\) = \(\frac{6825\,\times\,6\,\times1}{100}\) = 409.5

∴ A = P + I = 6825 + 409.5

∴ Amount = ₹ 7234.50

C.I. = A – P

= ₹ 7234.50 – 6500

= ₹734.50

Discussion

11.	A person invests 10000 rupees and 15000 rupees in two different schemes. After one year, he got 900 rupees as interest for the first amount and 1200 rupees as interest for the second amount.i. Are the interests proportional to the investments?ii. What is the ratio of the interest to the amount invested in the first scheme? What about the second?.iii. What is the annual rate of interest in the first scheme? And in the second?
Answer» i. The ratio between the amounts invested = 10000:15000 = 2 : 3 The ratio between the interests = 900 : 1200 = 3 : 4 Since the ratios are different. So, interests will be not proportional to the investments. ii. The ratio between the amount invested and interest in scheme 1 = 10000 : 900= 100 : 9 The ratio between the amount invested and interest in scheme 2 = 15000 : 1200 = 25 : 2 iii. Rate of interest in the first scheme = 900/10000 × 100 = 9% Rate of interest in the second scheme = 1200/15000 × 100 = 8%

11.

A person invests 10000 rupees and 15000 rupees in two different schemes. After one year, he got 900 rupees as interest for the first amount and 1200 rupees as interest for the second amount.i. Are the interests proportional to the investments?ii. What is the ratio of the interest to the amount invested in the first scheme? What about the second?.iii. What is the annual rate of interest in the first scheme? And in the second?

Answer»

i. The ratio between the amounts invested = 10000:15000 = 2 : 3

The ratio between the interests = 900 : 1200 = 3 : 4

Since the ratios are different. So, interests will be not proportional to the investments.

ii. The ratio between the amount invested and interest in scheme 1

= 10000 : 900= 100 : 9

The ratio between the amount invested and interest in scheme 2

= 15000 : 1200 = 25 : 2

iii. Rate of interest in the first scheme

= 900/10000 × 100 = 9%

Rate of interest in the second scheme = 1200/15000 × 100 = 8%

Discussion

12.	Find compound interest on Rs.12600 for 2 years at 10% per annum compounded annually.
Answer» P = Rs.12,600; R = 10%; n = 2 years A = P\([1\,+\,\frac R{100}]^n\) = 12600 \([1\,+\,\frac {10}{100}]^2\) = 12600\([1\,+\,\frac 1{10}]^2\) = 12600 x \(\frac{11}{10}\) x \(\frac{11}{10}\) = 126 x 121 A = Rs.15246 ∴ Compound Interest = Amount – Principal = 15,246 – 12,600 ∴ C.I. = Rs.2646 /-

12.

Find compound interest on Rs.12600 for 2 years at 10% per annum compounded annually.

Answer»

P = Rs.12,600; R = 10%; n = 2 years

A = P\([1\,+\,\frac R{100}]^n\) = 12600 \([1\,+\,\frac {10}{100}]^2\) = 12600\([1\,+\,\frac 1{10}]^2\)

= 12600 x \(\frac{11}{10}\) x \(\frac{11}{10}\) = 126 x 121

A = Rs.15246

∴ Compound Interest = Amount – Principal

= 15,246 – 12,600

∴ C.I. = Rs.2646 /-

Discussion

13.	What will happen if interest is compounded quarterly? How many conversion periods will be there? What about the quarter year rate . how much will it be of the annual rate?
Answer» Here C.I will be calculated for every 3 months. So, 4 time periods will be occurred in 1 year. Rate of Interest (R) = R/4 [ ∵ 12/3 = 4 ] Amount = \(P[1\,+\,(\frac{\frac R4}{100})]^4\) A = \(P[1\,+\,(\frac{R}{400})]^4\)

13.

What will happen if interest is compounded quarterly? How many conversion periods will be there? What about the quarter year rate . how much will it be of the annual rate?

Answer»

Here C.I will be calculated for every 3 months.

So, 4 time periods will be occurred in 1 year.

Rate of Interest (R) = R/4 [ ∵ 12/3 = 4 ]

Amount = \(P[1\,+\,(\frac{\frac R4}{100})]^4\)

A = \(P[1\,+\,(\frac{R}{400})]^4\)

Discussion

14.	When a weight is suspended by a spring, the extension is proportional to the weight.Explain how this can be used to mark weights on a spring balance.
Answer» Mark the pointer when no weight is hanging on it. Then mark the point when a constant weight is hanging on it. For example when we hang 1 kg on it a 2 cm extension is made. So we mark 1, 2, 3, 4 from the points marked first, i.e., 2, 4, 6 and 8. These distances are divided into 10 equal parts, then we can mark the points 1.1 kg and 2.4 kg etc.

14.

When a weight is suspended by a spring, the extension is proportional to the weight.Explain how this can be used to mark weights on a spring balance.

Answer»

Mark the pointer when no weight is hanging on it. Then mark the point when a constant weight is hanging on it. For example when we hang 1 kg on it a 2 cm extension is made. So we mark 1, 2, 3, 4 from the points marked first, i.e.,

2, 4, 6 and 8.

These distances are divided into 10 equal parts, then we can mark the points 1.1 kg and 2.4 kg etc.

Discussion

15.	The ratio of copper and zinc in an alloy is 9 : 8. If the weight of zinc, in the alloy, is 9.6 kg ; find the weight of copper in the alloy.
Answer» Let the weight of copper = x kg Weight of zinc = 9.6 kg. According to question, 9 : 8 = x : 9.6 => 8 x X = 9 x 9.6 => x = (9 x 9.6)/8 = 9 x 1.2 = 10.8 kg. Weight of copper in alloy = 10.8

15.

The ratio of copper and zinc in an alloy is 9 : 8. If the weight of zinc, in the alloy, is 9.6 kg ; find the weight of copper in the alloy.

Answer»

Let the weight of copper = x kg

Weight of zinc = 9.6 kg.

According to question,

9 : 8 = x : 9.6

=> 8 x X = 9 x 9.6

=> x = (9 x 9.6)/8 = 9 x 1.2 = 10.8 kg.

Weight of copper in alloy = 10.8

Discussion

16.	In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10 : 3: 12 . When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is it calcium carbonate?
Answer» The ratio between calcium, carbon and oxygen in calcium carbonate = 10 : 3: 12 The ratio of calcium, carbon and oxygen in the given compound = 60 : 20: 70 = 6 : 2: 7 Since this ratio is not equal to the ratio of calcium, carbon and oxygen in calcium carbonate, it is not calcium carbonate.

16.

In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10 : 3: 12 . When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is it calcium carbonate?

Answer»

The ratio between calcium, carbon and oxygen in calcium carbonate = 10 : 3: 12

The ratio of calcium, carbon and oxygen in the given compound = 60 : 20: 70 = 6 : 2: 7

Since this ratio is not equal to the ratio of calcium, carbon and oxygen in calcium carbonate, it is not calcium carbonate.

Discussion

17.	During rainfall, the volume of water falling in each square metre may be considered equal.i. Prove that the volume of water falling in a region is proportional to the area of the region.ii. Explain why the heights of rainwater collected in different sized hollow prisms kept near one another are equal.
Answer» i. The volume of water falling in each square metre are equal. Let the volume of the rain falling on the 1 square metre be k The volume of the rain falling on the 2 square metre = 2k The volume of the rain falling on the 3 square metre = 3k The volume of the rain falling on the x square metre y = kx Here x and y are proportional. ii. Let x be the base area of the vessel and h be the height of the water collected, then the volume of water y = xh. y/x =h The volume of water collected is different in hollow prisms having different base area. But \(\frac{volume}{Area}\) is always equal to the height of water level. So the height at which rainwater collected is same.

17.

During rainfall, the volume of water falling in each square metre may be considered equal.i. Prove that the volume of water falling in a region is proportional to the area of the region.ii. Explain why the heights of rainwater collected in different sized hollow prisms kept near one another are equal.

Answer»

i. The volume of water falling in each square metre are equal.

Let the volume of the rain falling on the 1 square metre be k The volume of the rain falling on the 2 square metre = 2k The volume of the rain falling on the 3

square metre = 3k

The volume of the rain falling on the x square metre y = kx Here x and y are proportional.

ii. Let x be the base area of the vessel and h be the height of the water collected, then the volume of water y = xh.

y/x =h

The volume of water collected is different in hollow prisms having different base area.

But \(\frac{volume}{Area}\) is always equal to the height of water level. So the height at which rainwater collected is same.

Discussion

18.	For each pair of quantities given below, check whether the first is proportional to the second. For proportional quantities, calculate the constant of proportionality.i. Perimeter and radius of circles.ii. Area and radius of circles.iii. The distance travelled and the number of rotations of a circular ring moving along a line.iv. The interest got in a year and the amount deposited in a scheme in which interest is compounded annually.v. The volume of water poured into a hollow prism and the height of the water level.
Answer» i. The perimeter of a circle is n times its diameter. That is 2n times the radius. ∴ Perimeter = 2πr ∴ The perimeter and radius are proportional. The constant of proportionality is 2π ii. The area of a circle is π times the square of the radius. ∴ Area = πr² ∴ Area and radius are not proportional. iii. When the ring rotates once, the distance travelled is equal to its perimeter. When it rotates twice the distance travelled is twice its perimeter. When it rotates ‘n’ times, the distance travelled is ‘n ‘ times the perimeter of the ring. iv. If the amount deposited is P and the rate of interest is R. Annual interest = I = PNR, I = P × R (N = 1) The amount and interest are proportional. Constant of proportionality is rate of interest, R. v. Volume of a hollow prism = base × height Volume, of water and height of water level, are proportional. Constant of proportionality is the base area.

18.

For each pair of quantities given below, check whether the first is proportional to the second. For proportional quantities, calculate the constant of proportionality.i. Perimeter and radius of circles.ii. Area and radius of circles.iii. The distance travelled and the number of rotations of a circular ring moving along a line.iv. The interest got in a year and the amount deposited in a scheme in which interest is compounded annually.v. The volume of water poured into a hollow prism and the height of the water level.

Answer»

i. The perimeter of a circle is n times its diameter.

That is 2n times the radius.

∴ Perimeter = 2πr

∴ The perimeter and radius are proportional.

The constant of proportionality is 2π

ii. The area of a circle is π times the square of the radius.

∴ Area = πr²

∴ Area and radius are not proportional.

iii. When the ring rotates once, the distance travelled is equal to its perimeter. When it rotates twice the distance travelled is twice its perimeter. When it rotates ‘n’ times, the distance travelled is ‘n ‘ times the perimeter of the ring.

iv. If the amount deposited is P and the rate of interest is R.

Annual interest = I = PNR,

I = P × R (N = 1)

The amount and interest are proportional. Constant of proportionality is rate of interest, R.

v. Volume of a hollow prism = base × height

Volume, of water and height of water level, are proportional. Constant of proportionality is the base area.

Discussion

19.	In the year 2012, it was estimated that there were 36.4 crore Internet users worldwide. In the next ten years, that number will be increased by 125%. Estimate the number of Internet users worldwide in 2022.
Answer» Internet users in the year 2012 = 36.4 crores. The number will be increased by next 10 years = 125% ∴ The no. of internet users in the year 2022 = 36.4 + 125% of 36.4 = 36.4 + 125/100 × 36.4 = 36.4 + 45.5 = 81.9 crores.

19.

In the year 2012, it was estimated that there were 36.4 crore Internet users worldwide. In the next ten years, that number will be increased by 125%. Estimate the number of Internet users worldwide in 2022.

Answer»

Internet users in the year 2012 = 36.4 crores.

The number will be increased by next 10 years = 125%

∴ The no. of internet users in the year 2022

= 36.4 + 125% of 36.4

= 36.4 + 125/100 × 36.4

= 36.4 + 45.5

= 81.9 crores.

Discussion

20.	A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.i. The rate of water flow and the height of the water level.ii The rate of water flow and the time taken to fill the tank.
Answer» i. Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then x = Ay Height of the water level in the tank is proportional to the rate of the water flowing. ii. If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in ‘t’ second is given by C = V × t \(V=\frac{C}{t}=C\times\frac{1}{t}\) That is the rate of water flow and the time taken for filling the tank are inversely proportional. C is the constant of proportionality.

20.

A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.i. The rate of water flow and the height of the water level.ii The rate of water flow and the time taken to fill the tank.

Answer»

i. Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then

x = Ay

Height of the water level in the tank is proportional to the rate of the water flowing.

ii. If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in ‘t’ second is given by C = V × t

\(V=\frac{C}{t}=C\times\frac{1}{t}\)

That is the rate of water flow and the time taken for filling the tank are inversely proportional. C is the constant of proportionality.

Discussion

21.	In triangles of the same area, how do we say the relation between the length of the longest side and the length of the perpendicular from the opposite vertex? What if we take the length of the shortest side instead?
Answer» Let ‘a’ be the large side, h be the length of perpendicular from opposite vertices, ‘A’ be the area, then A = \(\frac{1}{2}\)ah, a = \(\frac{2A}{h}\) Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex. i.e., The length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

21.

In triangles of the same area, how do we say the relation between the length of the longest side and the length of the perpendicular from the opposite vertex? What if we take the length of the shortest side instead?

Answer»

Let ‘a’ be the large side, h be the length of perpendicular from opposite vertices, ‘A’ be the area, then

A = \(\frac{1}{2}\)ah, a = \(\frac{2A}{h}\)

Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.

i.e., The length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

Discussion

22.	Will the dye which contains 10 L blue colour and 15 L white colour and another dye which contain 12L blue colour and 17 L white colour have j the same colour? Why?
Answer» In the first dye , blue colour: white colour = 10 : 15 = 2 : 3 In the second dye, blue colour: white colour = 12: 17 The ratios are not same. Hence the both will not have same colour.

22.

Will the dye which contains 10 L blue colour and 15 L white colour and another dye which contain 12L blue colour and 17 L white colour have j the same colour? Why?

Answer»

In the first dye , blue colour: white colour = 10 : 15 = 2 : 3

In the second dye, blue colour: white colour = 12: 17

The ratios are not same. Hence the both will not have same colour.

Discussion

23.	A farmer obtained a yielding of 1720 bags of cotton last year. This year she expects her crop to be 20% more. How many bags of cotton does she expect this year?
Answer» During the last year yielding the bags of cotton = 1720 If she expects 20% crop to be more then = 20% of 1720 = 20/100 × 1720 = 2 × 172 = 344 bags Her expectation of total bags = 1720 + 344 = 2064

23.

A farmer obtained a yielding of 1720 bags of cotton last year. This year she expects her crop to be 20% more. How many bags of cotton does she expect this year?

Answer»

During the last year yielding the bags of cotton = 1720

If she expects 20% crop to be more then

= 20% of 1720

= 20/100 × 1720

= 2 × 172

= 344 bags

Her expectation of total bags

= 1720 + 344

= 2064

Discussion

24.	In regular polygons, what is the relation between the number of sides and the degree measure of an outer angle? Can it be stated in terms of proportion?
Answer» The sum of the exterior angles of all polygon is 360°. If ‘n’ is the number of sides. Measure of an exterior angle = \(\frac{Sum\,of\,exterior\,angles}{No.of\,sides}\) One outer angle = 360/n (n = number of sides) If the measure of an exterior angle is ‘x’. x = \(\frac{1}{n}\times360 ^\circ\) One outer angle and number of sides are inversely proportional. The constant of proportionality is 1/n.

24.

In regular polygons, what is the relation between the number of sides and the degree measure of an outer angle? Can it be stated in terms of proportion?

Answer»

The sum of the exterior angles of all polygon is 360°. If ‘n’ is the number of sides.

Measure of an exterior angle = \(\frac{Sum\,of\,exterior\,angles}{No.of\,sides}\)

One outer angle = 360/n

(n = number of sides)

If the measure of an exterior angle is ‘x’.

x = \(\frac{1}{n}\times360 ^\circ\)

One outer angle and number of sides are inversely proportional. The constant of proportionality is 1/n.

Discussion

25.	150 litres of water is flowing through a pipe in 6 minutes. If 200 litres of water flows through it in 8 minutes check whether the quantity of water and time of flow are proportional ?
Answer» Quantity of water flowing in 6 minutes = 150 litres Ratio between the quantity of water and time of flow= 150: 6 = 25: 1 Quantity of water flowing in 8 minutes = 200 litres Ratio between quantity of water and times of flow = 200 : 8 = 25: 1 Since the radios are equal, the amount of water flowing and the time of flow are proportional.

25.

150 litres of water is flowing through a pipe in 6 minutes. If 200 litres of water flows through it in 8 minutes check whether the quantity of water and time of flow are proportional ?

Answer»

Quantity of water flowing in 6 minutes = 150 litres

Ratio between the quantity of water and time of flow= 150: 6 = 25: 1

Quantity of water flowing in 8 minutes = 200 litres

Ratio between quantity of water and times of flow = 200 : 8 = 25: 1

Since the radios are equal, the amount of water flowing and the time of flow are proportional.

Discussion

26.	Rajendra and Rehana own a business. Rehana receives 25% of the profit in each month. If Rehana received ₹ 2080 in particular month, what is the total profit in that month?
Answer» Total Profit = x say 25% of x = 2080 ⇒ 25/100 × x = 2080 ⇒ x/4 = 2080 ⇒ x = 2080 × 4 ∴ x = ₹ 8320

26.

Rajendra and Rehana own a business. Rehana receives 25% of the profit in each month. If Rehana received ₹ 2080 in particular month, what is the total profit in that month?

Answer»

Total Profit = x say

25% of x = 2080

⇒ 25/100 × x = 2080

⇒ x/4 = 2080

⇒ x = 2080 × 4

∴ x = ₹ 8320

Discussion

27.	The two sides of a triangle having perimeter 10 m are \(2\frac{1}{2}\) m and \(3\frac{1}{2}\) m.What is the ratio of the length of the three sides of triangles?
Answer» Perimeter of triangle = 10 m First side = \(3\frac{1}{2}\) m = \(\frac{5}{2}\) m Second side = \(3\frac{1}{2}\) m = \(\frac{7}{2}\) m Third side = 4 m Ratio of three sides of a triangle \(=\frac{5}{2}:\frac{7}{2}:4=5:7:8\)

27.

The two sides of a triangle having perimeter 10 m are \(2\frac{1}{2}\) m and \(3\frac{1}{2}\) m.What is the ratio of the length of the three sides of triangles?

Answer»

Perimeter of triangle = 10 m

First side = \(3\frac{1}{2}\) m = \(\frac{5}{2}\) m

Second side = \(3\frac{1}{2}\) m = \(\frac{7}{2}\) m

Third side = 4 m Ratio of three sides of a triangle

\(=\frac{5}{2}:\frac{7}{2}:4=5:7:8\)

Discussion

28.	Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Answer» Ratio of investments = 60000 : 100000 = 6 : 10 = 3 : 5 Ratio of profit divided 1800 : 3000 = 18 : 30 = 3 : 5 Ratio of investments and Ratio of profit divided are equal. Hence they are proportional.

28.

Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?

Answer»

Ratio of investments = 60000 : 100000 = 6 : 10 = 3 : 5

Ratio of profit divided

1800 : 3000 = 18 : 30 = 3 : 5

Ratio of investments and Ratio of profit divided are equal. Hence they are proportional.

Discussion

29.	A pre-owned car show-room owner bought a second hand car for ₹ 1,50,000. He spent ₹20,000 on repairs and painting, then sold it for ₹ 2,00,000. Find whether he gets profit or loss. If so, what percent?
Answer» After repair, the C.P of a car = 1,50,000 + 20,000 = 1,70,000 S.P. of a ear = ₹ 2,00,000 ∴ S.P. > C.P. ∴ Profit = S.P.-C.P. = 2,00,000-1,70,000= 30,000 Profit = Profit / C.P. × 100 = \(\frac{30000}{170000} \) × 100 = Profit = \(\frac{300}{17} \) Profit% = 17.64%

29.

A pre-owned car show-room owner bought a second hand car for ₹ 1,50,000. He spent ₹20,000 on repairs and painting, then sold it for ₹ 2,00,000. Find whether he gets profit or loss. If so, what percent?

Answer»

After repair, the C.P of a car = 1,50,000 + 20,000 = 1,70,000

S.P. of a ear = ₹ 2,00,000

∴ S.P. > C.P.

∴ Profit = S.P.-C.P.

= 2,00,000-1,70,000= 30,000

Profit = Profit / C.P. × 100

= \(\frac{30000}{170000} \) × 100

= Profit = \(\frac{300}{17} \)

Profit% = 17.64%

Discussion

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