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These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12\(\frac{1}{2}\)% per annum compounded annually.
(b) ₹ 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.
(c) ₹ 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.
(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.
(You could use the year by year calculation using SI formula to verify)
(e) ₹ 10,000 for 1 year at 8% per annum compounded half-yearly.
Solution:
(a) Here P = ₹ 10800, n = 3 years

Compound interest = ₹ 15377.34 – ₹ 10,800 = ₹ 4577.34
Amount = ₹ 15377.34
Compound interest = ₹ 4577.34

(b) HereP = ₹ 18,000, n = 2\(\frac{1}{2}\), R = 10% p.a
For first 2 years

CI = A – P = 21,780 – 18000 = ₹ 3780
For the last \(\frac{1}{2}\) year, principal become ₹ 21,780
So, Interest = \(\frac{21780 \times 10 \times 1}{2 \times 100}\) = ₹ 1089
∴ Total interest = 3780 + 1089 = ₹ 4869

(c) Here P = ₹ 62,500, R = \(\frac{8}{2}\) = 4% per half year
n = \(\frac{3}{2}\) × 2 = 3 years

A = ₹ 4 × 26 × 26 × 26 = ₹ 70304
Compound interest = Amount – Principal
= ₹ 70304 – ₹ 62500
= ₹ 7804

(d) Here P = ₹ 8000, n = 2 (2 half year),
R = \(\frac{9}{2}\) % (half yearly)

Compound interest = ₹ 8736.2 – ₹ 8000 = ₹ 736.20

(e) Here P = ₹ 10,000; n = 2 (2 half year);
R = \(\frac{8}{2}\) % = 4% (half yearly)

Compound Interest = ₹ 10816 – ₹ 10,000 = ₹ 816

Question 2.
Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\) years).
Solution:
Here P = ₹ 26,400, R% = 15% p.a, n = 2 years

Simple interest on ₹ 34914 at 15% p.a for 4 months (\(\frac{1}{3}\) years)
= ₹ \(\frac{34914 \times 15 \times 1}{100 \times 3}\)
= ₹ 1745.70
Requiredamount = ₹ 34914 + ₹ 1745.70 = ₹ 36,659.70

Question 3.
Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
For Fabina,
P = ₹ 12,500, R% = 12%, T = 3 years
Simple interest = \(\frac{\text { PRT }}{100}\)
= ₹ \(\frac{12500 \times 12 \times 3}{100}\)
= ₹ 4500
For Radha,
P = ₹ 12500, R = 10% p.a. n = 3 years

Compound interest = Amount – Principal
= ₹ 16,637.50 – ₹ 12500
= ₹ 4137.50
Difference between C.I and S.I = ₹ 4500 – ₹ 4137.50 = ₹ 362.5
Fabina pays more by ₹ 362.50

Question 4.
I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
Simple interest
Here P = ₹ 12000, R% = 6%, T = 2 years
I = \(\frac{\text { PRT }}{100}\)
= ₹ \(\frac{12000 \times 6 \times 2}{100}\)
= ₹ 1440
Compound interest
Here P = ₹ 12000, R = 6%, n = 2 years

C.I. = ₹ 13,483.20 – ₹ 12000 = ₹ 1483.20
∴ Excess amount = ₹ 1483.20 – ₹ 1440 = ₹ 43.20
I would have to pay to him an excess amount of ₹ 43.20.

Question 5.
Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Solution:
(i) After 6 months
Here, P = ₹ 60,000, R = 12% per annum = \(\frac {1}{2}\) × 12 = 6%, n = 1 (one half year)

He would get ₹ 63,600 after 6 months

(ii) After one year
Here P = ₹ 60,000, R = \(\frac{12}{2}\) = 6% per half year
n = (1 × 2) = 2 half years

Hence, he would get ₹ 67,416 after one year

Question 6.
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1\(\frac{1}{2}\) years if the interest is
(i) compounded annually
(ii) compounded half-yearly
Solution:
(i) Compounded annually
Here P = ₹ 80,000, R = 10% p.a, n = 1 year

Total interest = ₹ 8000 + ₹ 4400 = ₹ 12400
∴ Required amount = ₹ 80000 + ₹ 12400 = ₹ 92400

(ii) Compound half yearly
Here P = ₹ 80,000
R = \(\frac{10}{2}\)% = 5% per half year
n = 1\(\frac{1}{2}\) × 2 years = 3 half years

= \(\frac{80000 \times 105 \times 105 \times 105}{100 \times 100 \times 100}\)
= ₹ 92610
∴ The required amount = ₹ 92610
Difference in amount = ₹ 92610 – ₹ 92400 = ₹ 210
Hence, difference in amount is ₹ 210.

Question 7.
Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Solution:
(i) Here P = ₹ 8000, R = 5%, n = 2 years

The amount credited at the end of 2nd year = ₹ 8820

(ii) Here P = ₹ 8820, R = 5%, n = 1 year
Note: To find the interest for the 3rd year
(Find the S.I. for one year)
I = \(\frac{\text { PRT }}{100}\)
= \(\frac{8820 \times 5 \times 1}{100}\)
= ₹ 441
∴ Interest for the 3rd year is ₹ 441

Question 8.
Find the amount and the compound interest on ₹ 10,000 for 1\(\frac{1}{2}\) years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?
Solution:
If compounded half-yearly
Here P = ₹ 10,000
R = \(\frac{10}{2}\) = 5% per half year

C.I = A – P
= ₹ 11576.25 – ₹ 10,000
= ₹ 1576.25
₹ 1576.25 is the required C.I, when compounded half-yearly.

If compounded annually
Here, P = ₹ 10,000, R = 10%, n = 1 year

Total compound Interest = ₹ (11000 – 10,000) + 550
= ₹ 1000 + ₹ 550
= ₹ 1550
When compounded half-yearly, the compound interest = ₹ 1576.25
When compounded annually, the compound interest = ₹ 1550
Hence, the interest when compounded half-yearly would be more than the interest when compounded annually.

Question 9.
Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12\(\frac{1}{2}\)% per annum, interest being compounded half-yearly.
Solution:

Hence, the required amount ₹ 4913.

Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?
Solution:
Let the population in 2001 be ‘x’
R = 5%, n = 2years (2003 – 2001)
Population in 2003 is 54,000

= 48979.59
= 48980 (approx).
Hence, the population in 2001 was about 48980

(ii) Here P = 54,000, R = 5% p.a., n = 2 years (2003 – 2005 = 2)

Hence, the population in 2005 would be 59,535.

Question 11.
In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Here P = 5,06,000, R = 2.5% per hour, n = 2 hours

= 531,616.25
= 531616 (approx).
Number of bacteria at the end of 2 hours = 531616 (approx.)

Question 12.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Here P = ₹ 42,000 R = 8% p.a, n = 1 year
When the value depreciated

Value of the scooter after one year = ₹ 38,640

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.2

Question 1.
A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.
Solution:
Let the original salary of a man be ₹ x.
Increase in salary = 10% = \(\frac{10}{100}\) × x = \(\frac{x}{10}\)
New salary = ₹ 1,54,000
x + \(\frac{x}{10}\) = 154000
\(\frac{10 x+x}{10}\) = 154000
11x = 154000 × 10
x = \(\frac{154000 \times 10}{11}\) = ₹ 1,40,000
∴ Original salary = ₹ 1,40,000

Question 2.
On Sunday 845 people went to the zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the zoo on Monday?
Solution:
Decrease Value = 845 – 169 = 676
Decrease % = \(\frac{\text { Decrease Value }}{\text { Original Value }} \times 100\)
= \(\frac{676}{845}\) × 100 = 80%
Decrease % = 80%

Question 3.
A shopkeeper buys 80 articles for ₹ 2400 and sells them for a profit of 16%. Find the selling price of one article.
Solution:
Cost price of 80 articles = ₹ 2400
Profit = \(\frac{16}{100}\) × 2400 = ₹ 384
Selling price of 80 articles = ₹ 2400 + 384 = ₹ 2784
Selling price of one article = ₹ \(\frac{2784}{80}\) = ₹ 34.80

Question 4.
The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.
Solution:
Total cost price of the article = ₹ 15,500 + ₹ 450 = ₹ 15,950
Profit = 15% of C.P.
= \(\frac{15}{100}\) × 15950
= ₹ 2392.50
Selling Price of the article = ₹ 15950 + ₹ 2392.50 = ₹ 18342.50
∴ S.P of the article = ₹ 18342.50

Question 5.
A VCR and TV were bought for ₹ 8000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the T. V. Find the gain or loss per cent on the whole transaction.
Solution:
Total C.P. of VCR and T.V. = ₹ 2 × 8000 = ₹ 16000
Loss on VCR = \(\frac{4}{100}\) × 8000 = ₹ 320
Selling price of the VCR = ₹ 8000 – 320 = ₹ 7680
Profit on T.V. = \(\frac{8}{100}\) × 8000 = ₹ 640
Selling price of the T.V. = ₹ 8000 + ₹ 640 = ₹ 8640
Total selling price = ₹ 7680 + ₹ 8640 = ₹ 16,320
Combined Profit = ₹ 16320 – ₹ 16000 = ₹ 320
Gain % = \(\frac{\text { Gain }}{\text { Total C.P. }} \times 100\)
= \(\frac{320}{16000}\) × 100
= 2%
Profit % on the whole = 2%

Question 6.
During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?
Solution:
Marked price of a pairs of jeans = ₹ 1450
Discount = 10% of M.P
= \(\frac{10}{100}\) × 1450
= ₹ 145
Sale price of a pair of jeans = ₹ 1450 – ₹ 145 = ₹ 1305
Marked price of two shirts = ₹ 850 × 2 = ₹ 1700
Discount = 10% of M.P
= \(\frac{10}{100}\) × 1700
= ₹ 170
Sale price of two shirts = ₹ 1700 – 170 = ₹ 1530
∴ Total amount paid by the customer = ₹ 1305 + ₹ 1530 = ₹ 2835
Amount paid by the customer = ₹ 2835

Question 7.
A milkman sold two of his buffaloes for ₹ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Find CP of each)
Solution:
Selling price of 1st buffalo = ₹ 20,000
gain% = 5%
Cost price = \(\left(\frac{100}{100+\text { Profit } \%}\right) \times \mathrm{S} . \mathrm{P}\)

Total selling price = ₹ 2 × 20,000 = ₹ 40,000
Loss = Total C.P. – Total S.P
= ₹ 41269.84 – ₹ 40,000
= ₹ 1269.84

Question 8.
The price of a TV is ₹ 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.
Solution:
Cost price of the T.V. = ₹ 13,000
Sales tax = 12% of C.P
= \(\frac{12}{100}\) × 13000
= ₹ 1560
Bill amount = ₹ 1560 + ₹ 13,000 = ₹ 14,560
∴ Vinod has to pay ₹ 14,560 to buy a TV.

Question 9.
Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600. Find the marked price.
Solution:
Let the marked price be ₹ ‘x’
Discount = 20% of M.P
= \(\frac{20}{100}\) × x
= \(\frac{\mathrm{x}}{5}\)
Amount paid by Arun = ₹ 1600
x – \(\frac{\mathrm{x}}{5}\) = 1600
\(\frac{5 x-x}{5}=1600\)
x = \(\frac{1600 \times 5}{4}\) = ₹ 2000
∴ Amount paid by Arun = ₹ 2,000

Question 10.
I purchased a hair-dryer for ₹ 5400 including 8% VAT. Find the price before VAT was added.
Solution:
Let the original price of a hair-dryer be ‘x’
VAT = 8% of original price = \(\frac{8}{100}\) × x = \(\frac{2 \mathrm{x}}{25}\)
Bill amount = ₹ 5400
x + \(\frac{2 \mathrm{x}}{25}\) = 5400
\(\frac{25 x+2 x}{25}\) = 5400
\(\frac{27 \mathrm{x}}{25}\) = 5400
x = \(\frac{5400 \times 25}{27}\) = ₹ 5000
∴ Original price of the hair dryer = ₹ 5000

Question 11.
An article was purchased for ₹ 1239 including GST of 18%. Find the price of the article before GST was added?
Solution:
Cost with GST included = ₹ 1239
Cost with out GST = x rupees
GST = 18%
If we take this into equation
x + ((18 ÷ 100 ) × x) = 1239
Cost before GST + GST = Cost with GST
⇒ x + ( 9x ÷ 50 ) = 1239
⇒ (50x + 9x) ÷ 50 = 1239
⇒ 59x ÷ 50 = 1239
⇒ 59x = 1239 × 50
⇒ 59x = 61950
⇒ x = 61950 ÷ 59
⇒ x = 1050
Price before GST = ₹ 1050

These NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Exercise 9.1

Question 1.
List five rational numbers between
(i) -1 and 0
(ii) -2 and -1
(iii) \(\frac{-4}{5}\) and \(\frac{-2}{3}\)
(iv) \(\frac{-1}{2}\) and \(\frac{2}{3}\)
Answer:
(i) -1 and 0

(ii) -2 and -1

are the five rational numbers between -2 and -1

(iii) \(\frac{-4}{5}\) and \(\frac{-2}{3}\)

The five rational numbers are
\(\frac{-47}{60}<\frac{-23}{30}<\frac{-3}{4}<\frac{-11}{15} \text { and } \frac{-43}{60}\)

(iv) \(\frac{-1}{2}\) and \(\frac{2}{3}\)
L. C. M of 2 and 3 is 6

Question 2.
Write four more rational numbers in each of the following patterns.
(i) \(\frac{-3}{5}, \frac{-6}{10}, \frac{-9}{15}, \frac{-12}{20}\) ………………
(ii) \(\frac{-1}{4}, \frac{-2}{8}, \frac{-3}{12}\) ………………
(iii) \(\frac{-1}{6}, \frac{2}{-12}, \frac{3}{-18}, \frac{4}{-24}\) ………………
(iv) \(\frac{-2}{3}, \frac{2}{-3}, \frac{4}{-6}, \frac{6}{-9}\) ………………
Answer:

Thus, we observe a pattern in these numbers.
The next four more rational numbers are

The four required rational numbers are \(\frac{-15}{25}, \frac{-18}{30}, \frac{-21}{35}\) and \(\frac{-24}{40}\)

Thus, we observe a pattern in these numbers.
Next Four rational numbers are

The next four required rational numbers are \(\frac{-4}{16}, \frac{-5}{20}, \frac{-6}{24}\) and \(\frac{-7}{28}\)

Thus, we observe a pattern in these numbers
The next four rational numbers would be

The required four rational numbers are \(\frac{5}{-30}, \frac{6}{-36} ; \frac{7}{-42}\) and \(\frac{8}{-48}\)

Thus, we observe a pattern in these numbers
Four more rational numbers would be

The required four rational numbers are \(\frac{8}{-12}, \frac{10}{-15} ; \frac{12}{-18}\) and \(\frac{14}{-21}\)

Question 3.
Give four rational numbers equivalent to:
(i) \(\frac{-2}{7}\)
(ii) \(\frac{5}{-3}\)
(iii) \(\frac{4}{9}\)
Answer:
(i) Four rational numbers equivalent to

(ii) Four rational numbers equivalent to

(iii) Four rational numbers equivalent to

Question 4.
Draw the number line and represent the following rational numbers on it:
(i) \(\frac{3}{4}\)
(ii) \(\frac{-5}{8}\)
(iii) \(\frac{-7}{4}\)
(iv) \(\frac{7}{8}\)
Answer:

Question 5.
ThepointsP,Q,RS,T,U,AandBonthe number line are such that, TR = RS = SU
and AP = PQ = QB. Name the rational numbers represented by P, Q, R, and S.

Answer:
Since AP = PQ = QB
Distance between 2 and 3 is divided into 3 equal parts.
Similarly, distance between -2 and -1 is also divided into three equal parts.
P represents the rational number
\(\left(2+\frac{1}{3}\right) \text { i.e. } \frac{7}{3}\)
Q represents the rational number
\(\left(2+\frac{2}{3}\right) \text { i.e. } \frac{8}{3}\)
R represents the rational number
\(\left(-1-\frac{1}{3}\right) \text { i.e. } \frac{-4}{3}\)
S represents the rational number
\(\left(-1-\frac{2}{3}\right) \text { ie } \frac{-5}{3}\)

Question 6.
Which of the following pairs represent the same rational number?
(i) \(\frac{-7}{21}\) and \(\frac{3}{9}\)
(ii) \(\frac{-16}{20}\) and \(\frac{20}{-25}\)
(iii) \(\frac{-2}{-3}\) and \(\frac{2}{3}\)
(iv) \(\frac{-3}{5}\) and \(\frac{-12}{20}\)
(v) \(\frac{8}{-5}\) and \(\frac{-24}{15}\)
(vi) \(\frac{1}{3}\) and \(\frac{-1}{9}\)
(vii) \(\frac{-5}{-9}\) and \(\frac{5}{-9}\)
Answer:
(i) \(\frac{-7}{21}\) and \(\frac{3}{9}\)
Here \(\frac{-7}{21}\) is a negative rational number and \(\frac{3}{9}\) is a positive rational number.
\(\frac{-7}{21}=\frac{-1}{3} ; \frac{3}{9}=\frac{1}{3} ; \frac{-7}{21} \neq \frac{3}{9}\)
Thus, \(\frac{-7}{21}\) and \(\frac{3}{9}\) does not represent the same rational number.

(ii) \(\frac{-16}{20}\) and \(\frac{20}{-25}\)
We have \(\frac{-16}{20}\)= \(\frac{-4}{5}\)

Thus, \(\frac{-16}{20}\) and \(\frac{20}{-25}\) represent the same rational number.

(iii) \(\frac{-2}{-3}\) and \(\frac{2}{3}\)
\(\frac{-2}{-3}=\frac{+2}{3} ; \frac{2}{3}=\frac{2}{3}\)
Thus, \(\frac{-2}{-3}\) and \(\frac{2}{3}\) represent the same rational number.

(iv) \(\frac{-3}{5}\) and \(\frac{-12}{20}\)
We have, \(\frac{-3}{5}=\frac{-3}{5} ; \frac{-12}{20}=\frac{-3}{5}\)
So, \(\frac{-3}{5}=\frac{-12}{20}\)
Thus \(\frac{-3}{5}\) and \(\frac{-12}{20}\) represent the same rational number.

(v) \(\frac{8}{-5}\) and \(\frac{-24}{15}\)
We have, \(\frac{8}{-5}=\frac{-8}{5} ; \frac{-24}{15}=\frac{-8}{5}\)
So, \(\frac{8}{-5}\) = \(\frac{-24}{15}\)
Thus \(\frac{8}{-5}\) and \(\frac{-24}{15}\) represent the same rational number.

(vi) \(\frac{1}{3}\) and \(\frac{-1}{9}\)
Here \(\frac{1}{3}\) is positive integer and \(\frac{-1}{9}\) is a negative integer.
∴ \(\frac{1}{3} \neq \frac{-1}{9}\)
Thus \(\frac{1}{3}\) and \(\frac{-1}{9}\) does not represent the same rational number.

(vii) \(\frac{-5}{-9}\) and \(\frac{5}{-9}\)
\(\frac{-5}{-9}\) and \(\frac{5}{-9}\) is a positive integer.
and \(\frac{-5}{-9}\) = \(\frac{5}{-9}\) is a negative integer
∴ \(\frac{-5}{-9} \neq \frac{5}{-9}\)
Thus \(\frac{-5}{-9}\) and \(\frac{5}{-9}\) do not represent tha same rational number.

Question 7.
Rewrite the following rational numbers in the simplest form:
(i) \(\frac{-8}{6}\)
(ii) \(\frac{25}{45}\)
(iii) \(\frac{-44}{72}\)
(iv) \(\frac{-8}{10}\)
Answer:
(i) \(\frac{-8}{6}\)
\(\frac{-8}{6}=\frac{-4}{3}\) (Divide both sides by 2)
The simplest form is \(\frac{-4}{3}\)

(ii) \(\frac{25}{45}\)
\(\frac{25}{45}=\frac{5}{9}\) (Divide both sides by 5)
The simplest form is \(\frac{5}{9}\)

(iii) \(\frac{-44}{72}\)
\(\frac{-44}{72}=\frac{-11}{18}\)
Dividing both sides by 4)
The simplest \(\frac{-11}{18}\)

(iv) \(\frac{-8}{10}\)
\(\frac{-8}{10}=\frac{-4}{5}\) (Dividing both sides by 2)
The simplest form is \(\frac{-4}{5}\)

Question 8.
Fill in the boxes with the correct symbol out of > , <, and =


Answer:
\(\frac{-7}{6}\) is a negative rational number. 6

(ii) \(\frac{-4}{5} \text { and } \frac{-5}{7}\)
are negative rational numbers.
L.C.M of 5 and 7 is 35.

are negative rational numbers.
L.C.M of 8 and 16 is 16.

(iv) \(\frac{-8}{5}\) and \(\frac{-7}{4}\) are negative rational numbers.
LCM of 5 and 4 = 20

(v) \(\frac{1}{-3}\) and \(\frac{-1}{4}\) negative rational numbers.
L.C.M of 3 and 4 is 12.

(vi) \(\frac{5}{-11}\) and \(\frac{-5}{11}\) are negative rational numbers.

(vii) \(\frac{-7}{6}\) is a negative integer since 0 is 6
greater than every negative number.

Question 9.
Which is greater in each of the following?
(i) \(\frac{2}{3}, \frac{5}{2}\)
(ii) \(\frac{-5}{6}, \frac{-4}{3}\)
(iii) \(\frac{-3}{4}, \frac{2}{-3}\)
(iv) \(\frac{-1}{4}, \frac{1}{4}\)
(v) \(-3 \frac{2}{7} ;-3 \frac{4}{5}\)
Answer:
(i) \(\frac{2}{3}, \frac{5}{2}\)
L.C.M of 2 and 3 is 6

Thus \(\frac{5}{2}\) is greater rational number.

(ii) \(\frac{-5}{6}, \frac{-4}{3}\)
L.C.M of 6 and 3 is 6

Thus \(\frac{-5}{6}\) is greater rational number.

(iii) \(\frac{-3}{4}, \frac{2}{-3}\)
L.C.M of 4 and 3 is 12

\(\frac{2}{-3}>\frac{-3}{4}\)
Thus, the rational number \(\frac{2}{-3}\) is greater.

(iv) \(\frac{-1}{4}, \frac{1}{4}\)
since a positive rational number is always greater than a negative rational number.
\(\frac{1}{4}>\frac{-1}{4}\)
Thus, greater rational number is \(\frac{1}{4}\)

(v) \(-3 \frac{2}{7} ;-3 \frac{4}{5}\)

Thus, rational number -3\(\frac{2}{7}\) is greater.

Question 10.
Write the following rational numbers in ascending order.
(i) \(\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}\)
(ii) \(\frac{-1}{3}, \frac{-2}{9}, \frac{-4}{3}\)
(iii) \(\frac{-3}{7}, \frac{-3}{2}, \frac{-3}{4}\)
Answer:
(i) \(\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}\)
\(-\frac{3}{5}<-\frac{2}{5}<-\frac{1}{5}\)
The ascending order is \(-\frac{3}{5},-\frac{2}{5},-\frac{1}{5}\)

(ii) \(\frac{-1}{3}, \frac{-2}{9}, \frac{-4}{3}\)
L.C.M of 3 and 9 is 9

Thus, the ascending order is
\(\frac{-4}{3}, \frac{-1}{3}, \frac{-2}{9}\)

(iii) \(\frac{-3}{7}, \frac{-3}{2}, \frac{-3}{4}\)
L.C.M of 7, 2 and 4 is 28

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.5

Question 1.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR. Ans: In the right APQR
Answer:
In the right ΔPQR

QR² = PQ² + PR²
(pythagoras property)
= 10² + 24²
= 100 + 576
= 676
QR² = 26²
∴ QR = 26 cm

Question 2.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Answer:
In the right ΔABC

AB² = AC² + BC²
(using pythagoras property)
25² = 7² + BC²
625 = 49 + BC²
625 – 49 = BC²
576 = BC²
24² = BC²
∴ BC = 24 cm

Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Answer:
Let the distance of the foot of a ladder from the wall be ‘a’m

a2² + 12² = 15²
(using pythagoras property)
a² + 144 = 225
a² = 225 – 144
a² = 81
a² = 9²
a = 9 m
The distance of the foot of the ladder from the wall = 9 m.

Question 4.
Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.
Answer:
(i) 2.5 cm, 6.5 cm, 6 cm².
The longest side is 6.5 cm
2.5² + 6² = 6.25 + 36
= 42.25
6.5² = 42.25
∴ 6.5² = 2.5² + 6²
(pythagoras property)
The given lengths can be the sides of a right triangle.
The right angle is the angle between the sides 2.5 cm and 6 cm

(ii) 2 cm, 2 cm, 5 cm
The longest side is 5 cm
2² + 2² = 4 + 4 = 8
5² = 25
5² ≠ 2² + 2²
∴ The given length cannot be the sides of a right triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm
The longest side is 2.5 cm
1.5² + 2² = 2.25 + 4
= 6.25
2.5² = 6.25
1.5² + 2² = 2.5²
(pythagoras property)
The given lengths can be sides of a right triangle.
The right angle is the angle between the sides 1.5 cm and 2 cm.

Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Answer:
Let the tree BD be broken at the point C, such that CD = CA
In the right triangle ABC, using the pythagoras property; we get
AC² = AB² + BC²
AC² = 12² + 5²
= 144 + 25
AC² = 169
AC² = 13²
∴ AC = 13
Now, height of the tree
BD = BC + CD (CD = AC)
= 5 m + 13 m
= 18 m
The height of the tree is 18 m.

Question 6.
Angles Q and R of PQR are 25° and 65°. Write which of the following is true:
(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²
Answer:
In ΔPQR
∠P + ∠Q +∠R = 180°
∠P + 25° + 65° = 180°

∠P + 90° = 180°
∠P = 180° – 90°
= 90°
So, ΔPQR is a triangle right angled at P.
∴ QR is the hypotenuse.
∴ using the pythagoras property
QR² = PQ² + PR² So,

(ii) PQ² + RP² = QR² is true

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer:
Let the breadth AD be x cm.
In the right triangle ABD,

BD² = AD² + AB²
41²= x² + 40²
x² = 41² – 40²
= 1681 – 1600
x² = 81
x² = 9²
x = 9 cm
Perimeter of the rectangle = 2 (l + b)
= 2 (40 + 9)
= 2 × 49
= 98 cm
Thus, the perimeter of the rectangle
= 98 cm

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Answer:
Since the diagonals of a rhombus bisect each other at right angles.

∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
AO = \(\frac { 1 }{ 2 }\) AC; AO = \(\frac { 1 }{ 2 }\) × 30 = 15
∵ AO = OC = 15 cm and BO = OD = 8 cm
In the right ΔAOB,
AB² = AO² + BO²
AB² = 15² + 8²
AB² = 225 + 64
AB² = 289
AB² = 17²
AB = 17 cm
∵ Side of the rhombus is 17 cm.
∴ Perimeter of the rhombus ABCD = 4 × 17
(all the four sides are equal) = 68 cm

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.1

Question 1.
Find the ratio of the following:
(a) Speed of a cycle 15km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise to ₹ 5
Solution:
(a) Speed of the cycle = 15 km per hour.
Speed of the scooter = 30 km per hour.
∴ Ratio = Speed of the cycle : Speed of the scooter
= 15 : 30
= 1 : 2 (Divided by 15)

(b) 10 km = 10 × 1000 m = 10000 m
Ratio = 5 m : 10 km
= 5 : 10000
= 1 : 2000

(c) ₹ 5 = 5 × 100 paise = 500 paise
Ratio = 50 paise : ₹ 5
= 50 : 500
= 1 : 10

Question 2.
Convert the following ratios to percentages.
(a) 3 : 4
(b) 2 : 3
Solution:
(a) 3 : 4 = \(\frac{3}{4}\)
= \(\frac{3}{4}\) × 100%
= 75%

(b) 2 : 3 = \(\frac{2}{3}\)
= \(\frac{2}{3}\) × 100
= \(\frac{200}{3}\)
= 66\(\frac{2}{3}\) %

Question 3.
72% of 25 students are good at mathematics. How many are not good in mathematics.
Solution:
Total number of students = 25
Number of students good in mathematics = 72% of 25
= \(\frac{72}{100}\) × 25
= 18
Number of students who are not good in mathematics = 25 – 18 = 7
Percentage of students not good in mathematics = \(\frac{7}{25}\) × 100
= 7 × 4
= 28%
or = \(\frac{28}{100}\) × 25 = 7 students are not good in Mathematics.

Question 4.
A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Solution:
Number of matches won by the team = 10
Let the total number of matches be ‘x’
The team won 40% the total number matches.
∴ \(\frac{40}{100}\) × x = 10
x = \(\frac{10 \times 100}{40}\) = 25
∴ Total number of matches played = 25.

Question 5.
If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Amount left with Chameli = (100 – 75)% = 25%
Let the total amount with Chameli be ₹ ‘x’
∴ 25% of total money = ₹ 600
\(\frac{25}{100}\) × x = 600
x = \(\frac{600 \times 100}{25}\) = ₹ 2400
Amount in the beginning = ₹ 2400

Question 6.
If 60% of people in a city like a cricket 30% like football and the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.
Solution:
People who like cricket = 60%
People who like football = 30%
People who like other games = 100 – (60 + 30)
= 100 – 90
= 10%
Total number of people = 50,00,000
No. of people who like cricket = \(\frac{60}{100}\) × 50,00,000 = 30,00,000
No. of people who like football = \(\frac{30}{100}\) × 50,00,000 = 15,00,000
No. of people who like other games = \(\frac{10}{100}\) × 50,00,000 = 5,00,000

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.4

Question 1.
Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
Answer:
(i) 2 cm, 3 cm, 5 cm
2 cm + 3 cm = 5 cm
third side = 5 cm
∵ Sum of the measures of length of two sides = length of the third side
∴ It is not possible with these sides to form a triangle.

(ii) 3 cm, 6 cm, 7 cm
3 cm + 6 cm = 9 cm
9 cm > 7 cm
3 cm + 7 cm = 10 cm and 10 cm > 6 cm.
6 cm + 7 cm = 13 cm and 13 cm > 3 cm.
Thus, a triangle can be possible with these sides.

(iii) 6 cm, 3 cm and 2 cm.
6 cm + 3 cm = 9 cm and 9 cm > 2 cm
3 cm + 2 cm = 5 cm and 5 cm < 6 cm 2 cm + 6 cm = 8 cm and 8 cm > 5 cm
Thus, a triangle cannot be possible with these sides.

Question 2.
Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?

Answer:
(i) Yes OP + QQ> PQ
( The sum of lengths of any two sides of a triangle is greater than the length of the third side)
(ii) Yes, OQ + OR > QR
(iii) Yes, OR + OP > RP

Question 3.
AM is a median of a triangle ABC. IsAB + BC + CA > 2 AM?
(Consider the sides of triangles ΔABM and ΔAMC.)
Answer:
Since, the sum of the length of any two sides of a triangle is greater than the length
of the third side
∴ In ΔABM, we have
(AB + BM) > AM …………. (i)
Similarly in ΔACM
(CA + CM) > AM ………. (ii)

Adding (1) and (2), we have
[ (AB + BM) + (CA + CM)] > AM + AM
[ AB + (BM + CM) + CA] > 2 AM
[ AB + BC + CA]> 2AM
Thus, (AB + BC + CA) > 2 AM

Question 4.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
Answer:
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

In ABC, we have
AB + BC > AC …………(i)
Similarly, in ΔACD, we have
CD + DA > AC …………(ii)
Adding (i) and (ii), we get
[(AB + BC) + (CD + DA)] > 2 AC …………(iii)
Again In ΔABD, we have
AB + DA > BD …………….(iv)
In ΔBCD, we have
BC + CD > BD …(v)
Adding (iv) and (v), we get
[(AB + DA) + (BC + CD)] > 2 BD …………..(vi)
Now Adding (iii) and (vi), we get
2[(AB + BC) + (CD+ DA)] > 2 (AC + BD) (AB + BC + CD + DA) > (AC + BD)

Question 5.
ABCD is a quadrilateral.
Is AB + BC + CD + DA < 2(AC + BD)?
Answer:
Since the sum of the length of any two sides of a triangle is greater than the length of the third side.

∴ In ΔAOB, we have (OA + OB) > AB …………(i)
Similarly,
In ΔOBC, we have
(OB + OC) > BC …………..(ii)
In ΔOCD, we have
(OC + OD) > CD ………….(iii)
In ΔOAD we have
(OA + OD) > AD …………..(iv)
Adding (i), (ii), (iii) and (iv), we have
2] OA + OB + OC + OD] > (AB + BC + CD + DA)]
⇒ AB + BC + CD + DA < 2 (OA + OB + OC + OD)
⇒ AB + BC + CD + DA < 2 [(OA + OC) + (OB + OD)]
⇒ AB + BC + CD + DA < 2 (AC + BD)

Question 6.
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Answer:
Since, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Sum of 12 cm and 15 cm is greater than the length of the third side,
i.e. (12 cm + 15 cm) > Third side 27 cm > Third side (or) Third side < 27 cm.
Also, the difference of the lengths of any two sides is less than the length of the third side.
(15 cm – 12 cm) < Third side 3 cm < Third side
Thus, we have 3 cm < Third side < 27 cm.
∴ The third side should be of any length between 3 cm and 27 cm.