The DAV Class 6 Maths Solutions and **DAV Class 6 Maths Chapter 5 Brain Teasers **Solutions of Percentage and its Applications offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 5 Brain Teasers Solutions

Question 1A.

Tick (✓) the correct answer.

(a) 2.6% expressed as a decimal is _________

(i) 0.26

(ii) 0.026

(iii) 26

(iv) 2.6

Solution:

(ii) 0.026

∴ 2.6% = \(\frac{2.6}{100}\) = 0.026

Hence, (ii) is the correct option.

(b) What per cent is \(\frac{1}{2}\) dozen of one score?

(i) 10%

(ii) 60%

(iii) 30%

(iv) 25%

Solution:

(iii) 30%

x% of 20 = \(\frac{1}{2}\) × 12

\(\frac{x}{100}\) × 20 = 6

x = \(\frac{6 \times 100}{20}\) = 30

Hence, (iii) is the correct option.

(c) If C.P. = ₹ 75, S.P. = ₹ 100, then gain or loss per cent is

(i) Gain of 33%

(ii) Loss of 33%

(iii) Loss of 25%

(iv) Gain of 25%

Solution:

When S.P. > C.P., there is a gain

Gain = S.P. – C.P.

= 100 – 75 = 25

Gain% = \(\frac{\text { Gain }}{\text { C.P. }}\) × 100

= \(\frac{25}{75}\) × 100

= \(\frac{100}{3}\)

= 33 \(\frac{1}{3}\) %

Hence, (i) is the correct option.

(d) Gain or Loss percent is always calculated on

(i) SP

(ii) Loss

(iii) Gain

(iv) CP

Solution:

(iv) CP

Gain or loss per cent is always calculated in C.P.

(e) 33 \(\frac{1}{3}\) % of 1\(\frac{1}{2}\) minute is equal to-

(i) 90 seconds

(ii) 30 seconds

(iii) 50 seconds

(iv) 60 seconds

Solution:

(ii) 30 seconds

33 \(\frac{1}{3}\) % of 1\(\frac{1}{2}\)

= \(\frac{100}{3} \times \frac{1}{100} \times \frac{3}{2}\) minute

= \(\frac{1}{2}\) × 60 sec

= 30 seconds

Hence, (ii) is the correct option.

B. Answer the following questions.

(a) What is the number which is 5% more than 600?

Solution:

600 + 5% of 600 = 600 + \(\frac{5}{100}\) × 600

= 600 + 30 = 630

(b) Find the simple interest on ₹ 200 at 5% per annum for 6 months.

Solution:

Here. P = ₹ 200,

R = 5%,

T = 6 months

= \(\frac{1}{2}\) year

∴ Simple interest S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{200 \times 5 \times 1}{100 \times 2}\)

= ₹ 5

(e) If 140 is reduced to 112, what per cent is the reduction?

Solution:

Reduction = 140 – 112 = 28

∴ Reduction % = \(\frac{\text { Reduction }}{\text { Original number }}\) × 100

= \(\frac{28}{140}\) × 100

= 20 %

(d) What is 12% of 0.5 metre?

Solution:

12% of 0.5 metre

= \(\frac{12}{100}\) × 0.5 × 100 cm

= 12 × 0.5

= 6 cm

(e) If SP = ₹ 260, Profit = ₹ 60, find profit per cent.

Solution:

We know that C.P. = S.P. – Profit

∴ C.P. = ₹ 260 – ₹ 60 = ₹ 200

∴ Profit % = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100

= \(\frac{60}{100}\) × 100

= 30%

Question 2.

Find the value of:

(a) 33 \(\frac{1}{2}\) % of 9012

Solution:

33 \(\frac{1}{3}\) % of 9012

=

= 3004.

(b) 25% of 10% of 1 kg

Solution:

25% of 10% of 1 kg

=

= 25 gm

[∵ 1 kg = 1000 gm]

Question 3.

Find 12 \(\frac{1}{2}\) % less than 16 hours.

Solution:

12 \(\frac{1}{2}\) % of 16 hours

=

= 2 hours

∴ less = 16 hours – 2 hours = 14 hours

Question 4.

John bought 100 eggs for ₹ 40. Out of these, 4 eggs were found to be broken and he sold the remaining eggs at the rate of ₹ 7.50 per dozen. Find his gain or loss percent.

Solution:

∴ Cost price of 100 eggs = ₹ 40

4 eggs were found broken

∴ Remaining eggs = 100 – 4 = 96

S.P. of 96 eggs = \(\frac{7.50 \times 96}{12}\)

= ₹ 60.00

Here, S.P. > C.P.

∴ Gain = ₹ 60 – ₹ 40 = ₹ 20

Gain percent = \(\frac{\text { gain }}{\text { C.P. }}\) × 100

=

= 50%

Hence, the required gain = 50%.

Question 5.

Articles are bought at ₹ 45 per dozen and sold at ₹ 85 per score. Find the gain or loss per cent.

Solution:

C.P. = \(₹ \frac{45}{12}=₹ \frac{15}{4}\)

S.P. = \(₹ \frac{85}{20}=₹ \frac{17}{4}\)

Gain = S.P. – C.P.

= \(\frac{17}{4}-\frac{15}{4}\)

= \(\frac{17-15}{4}\)

= \(\frac{2}{4}=₹ \frac{1}{2}\)

Gain % = \(\frac{\text { Gain }}{\text { C.P. }}\) × 100

= \(\frac{\frac{1}{2}}{\frac{15}{4}}\) × 100

= \(\frac{4 \times 100}{2 \times 15}\)

= \(\frac{40}{3}\)

= 13.33%

Question 6.

Salim deposited ₹ 12000 in a finance company which pays 15% interest per year. Find the amount he is expected

to get after 4\(\frac{1}{2}\) years.

Solution:

Here, P = ₹ 12000,

R = 15% p.a.,

T = 4\(\frac{1}{2}\)

= \(\frac{9}{2}\) years

Simple interest S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 8100

Amount = P + S.I.

= ₹ 12000 + ₹ 8100

= ₹ 20,100

Hence, the required amount = ₹ 20,100.

Question 7.

Ram Lai bought oranges at ₹ 30 per dozen. He had to sell them at a loss of 5%. Find the selling price.

Solution:

C.P. of 1 dozen oranges = ₹ 30

Loss = 5% of ₹ 30

=

= ₹ 1.50

∴ S.P. = C.P. – Loss

= ₹ 30 – ₹ 1.50

= ₹ 28.50

Hence, the required S.P. = ₹ 28.50.

Question 8.

Out of 1200 people, 800 know only English, 50 know only Punjabi and the rest know both languages. Find the percentage of:

(a) People who know only English

(b) People who know both English and Punjabi.

Solution:

Number of people = 1200

Number of people who know English only = 800

Number of people who know Punjabi only = 50

Number of people who know both languages = 800 + 50 = 850

(a) Percentage of the people who know only English

=

= \(\frac{200}{3}\)

= 66 \(\frac{2}{3}\) %

(b) Percentage of the people who know both the languages

=

= \(\frac{425}{6}\)

= 70 \(\frac{5}{6}\) %

Question 9.

Prabal deposited ₹ 5000 in a bank which pays him 5\(\frac{1}{2}\) % interest. After 3 years, he withdraws the money and buys an almirah for ₹ 4700. How much money is left with him?

Solution:

Here, P = ₹ 5000,

R = 5\(\frac{1}{2}\) %

= \(\frac{1}{2}\) %

T = 3 years

Simple interest S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= ₹ 825

Amount = P + S.I.

= ₹ 5000 + ₹ 825

= ₹ 5825

Cost of almirah = ₹ 4700

∴ Money left with him =₹ 5825 – ₹ 4700 = ₹ 1125.

Question 10.

Find the Simple Interest on ₹ 6500 at 8% from 5 January 2015 to 19 March 2015.

Solution:

P = ₹ 6500,

R = 8%

T = 27 days of Jan. + 28 days of Feb. + 18 days of March

= 73 days

= \(\frac{73}{365}\)

= \(\frac{1}{5}\) year

S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= \(\frac{6500 \times 8 \times 1}{100 \times 5}\)

= ₹ 104

### DAV Class 6 Maths Chapter 5 HOTS

Question 1.

The selling price of an article is of its cost price. Find the profit per cent.

Solution:

S.P. = \(\frac{4}{3}\) × C.P. (given)

Profit = S.P. – C.P.

= \(\frac{4}{3}\) C.P. – C.P.

= \(\frac{1}{3}\) C.P.

∴ Profit % = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100

= \(\frac{\frac{1}{3} \text { C.P. }}{\text { C.P. }}\) × 100

= \(\frac{100}{3}\)

= 33 \(\frac{1}{3}\) %

Question 2.

In a school there are 50 students in class VIA. 88% of the students passed a Mathematics test. The same number of boys and girls failed in the exam. If 42% of the total student are girls, how many boys passed the test?

Solution:

No. of students passed in maths 88% of 50 = \(\frac{88}{100}\) × 50 = 44

Total no. of students = 50

∴ Total no. of students failed in exam = 50 – 44 = 6

∴ No. of boys failed in exam = \(\frac{1}{2}\) × 6 = 3

No. of girls = \(\frac{42}{100}\) × 50 = 21

∴ No. of boys = 50 – 21 = 29

∴ No. of boys passed in exam = 29 – 3 = 26.

Question 3.

The cost price of 10 articles is equal to the selling price of 8, find the Profit or loss per cent.

Solution:

C.P. of 10 articles = S.P. of 8 articles = ₹ x

∴ C.P. of 1 article = ₹ \(\frac{x}{10}\)

and S.P. of 1 article = ₹ \(\frac{x}{8}\)

∵ S.P. > C.P., there is a gain

Gain = S.P. – C.P.

= \(\frac{x}{8}-\frac{x}{10}\)

= \(\frac{5 x-4 x}{40}\)

= \(\frac{1}{40}\) x

Profit % = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100

= \(\frac{\frac{1}{40} x}{\frac{x}{10}}\) × 100

= \(\frac{1}{4}\) × 100

= 25%

Question 4.

Find the number by adding 33\(\frac{1}{3}\) % of \(\frac{1}{3}\) of the predecessor of the smallest 5-digit number to the successor of the smallest number formed by the digits of first four composite numbers.

Solution:

Smallest 5-digit number = 10000

Predecessor of smallest 5-digit number = (10000 – 1) = 9999

Composite number are those, which has more than two different factors 4, 6, 8, 9 are the first four composite numbers

∴ Smallest numbers formed by the digits of first four composite numbers = 4689

Successor of 4689 = 4689 + 1 = 4690

As per the condition given in the question, we have

33\(\frac{1}{3}\) % of \(\frac{1}{3}\) of (9999) + 4690 = \(\frac{1}{3}\) × \(\frac{1}{3}\) × 9999 + 4690

= 1111 + 4690

= 5801

### DAV Class 6 Maths Chapter 5 Enrichment Questions

Question 1.

Complete the given magic square.

Solution:

Additional Questions:

Question 1.

Out of 4 dozen bananas, Ramu gave 30 of them to his younger brother. What percent of bananas he has given to him?

Solution:

Number of bananas, Ramu has = 4 × 12 = 48

Number of bananas given to his brother = 30

∴ Percentage =

= \(\frac{125}{2}\)

= 62 \(\frac{1}{2}\) %

Hence, the required percentage = 62 \(\frac{1}{2}\)%.

Question 2.

Find the amount 8% more than 500.

Solution:

8% of 500 = = 40

∴ Amount more than 500 = 500 + 40 = 540.

Question 3.

What percent is 63 of 105?

Solution:

Percent of 63 of 105 =

= 60 %

Question 4.

Rani got 66 marks out of 150 in Hindi. What percent marks she has got in Hindi? To get 80% in the same subject, how many marks should she get?

Solution:

Percentage of marks obtained by Rani in Hindi =

= 44 %

Now 80 % of 150 =

= 120 marks

Hence, required marks obtained = 120.

Question 5.

Mrs. Gupta purchased a house for 4,50,000 and due to sorne reason she had to sell it for 6,00,000. Find her profit percent.

Solution:

C.P. of house = ₹ 4,50,000

S.P. of house = ₹ 6,00,000

Profit. = S.P. – C.P.

= ₹ 6,00,000 – ₹ 4,50,000 = ₹ 1,50,000

Profit percent = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100

= \(\frac{1,50,000}{4,50,000}\)

= \(\frac{100}{3}\) %

= 33 \(\frac{1}{3}\) %

Hence, the required profit 33 \(\frac{1}{2}\) %.

Question 6.

Mr. Deewan deposit ₹ 50,000 in a Nationalised Bank at 5% p.a. simple interest for 2 years. Find the amount that he will get after 2 years.

Solution:

Here, P = ₹ 50,000,

R = 5% p.a.,

T = 2 \(\frac{1}{2}\) = \(\frac{5}{2}\) years

∴ S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 6250

Amount = P + S.I.

= ₹ 50,000 + ₹ 6250

= ₹ 56,250

Hence, the required amount = ₹ 56,250.

Question 7.

Rahul bought 50 eggs for 150. Five eggs were rotten and he sold the remaining eggs for 165. Find his loss or profit percent.

Solution:

Cost of 50 eggs = ₹ 150

Rotten eggs = 50 – 5 = 45

S.R of 45 eggs = ₹ 165

Profit = S.P. – CF.

= ₹ 165 – ₹ 150

= ₹ 15

Profit percent = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100

=

Hence, the required profit percent = 10%.

Question 8.

Find the interest on ₹ 2500 at 2.5% p.a. for 146 days.

Solution:

Here, P = ₹ 2500,

R = 2.5 %

T = \(\frac{146}{365}=\frac{2}{5}\) years

∴ Simple interest S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 25

Hence, the required interest = ₹ 25.

Question 9.

What percent of the multiple of 3 are in the numbers from 1 to 50?

Solution:

Multiples of 3 from 1 to 50 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48 i.e., 16

Required percent = \(\frac{16}{50}\) × 100 = 32 %.

Question 10.

Find the amount if ₹ 30,000 are deposited for 3 years at 10.5 percent p.a.

Solution:

Here, P = ₹ 30,000,

R = 10.5% p.a.

T = 3\(\frac{1}{2}\)

= \(\frac{7}{2}\) years

∴ Simple interest S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 11025

Amount = P + S.I.

= ₹ 30,000 + ₹ 11025 = ₹ 41025

Hence, the required amount = ₹ 41025.

Question 11.

Find the value of:

(a) 20% of ₹ 80

(b) 75% of ₹ 350

(c) 3 % of 180 km

(d) 100% of ₹ 1

(e) 25% of 1 score pencils

(f) 12% of kg

Solution:

(a) 20% of ₹ 80 = \(\frac{20}{100}\) × 80 = ₹ 16

(b) 75% of ₹ 350 = \(\frac{75}{100}\) × 350

= \(\frac{525}{2}\)

= ₹ 262.50

(c) 3\(\frac{1}{3}\) % of 180 km

=

= 6 km

(d) 100 % of ₹ 1

= \(\frac{100}{100}\) × 1

= ₹ 1

(e) 25% of 1 score pencils

=

= 15 pencil

(f) 12% of \(\frac{1}{2}\) kg

=

= \(\frac{1}{16}\) kg

Question 12.

Rohan’s monthly earning is ₹ 8000. He spent 75% of his income. How much money did he spend?

Solution:

Monthly income = ₹ 8000

Spent money = \(\frac{75}{100}\) × 8000 = ₹ 6000

Hence, the money spent = ₹ 6000.

Question 13.

Ritu scored 68% in her Mathematics test. 1f maximum marks were 50, how many marks did she score?

Solution:

Maximum marks = 50

Marks scored = 68% of 50

= \(\frac{68}{100\) × 50 = 34

Hence the required marks obtained 34.

Question 14.

I purchased 5 1 milk and used 80% of it for making sweets. How much milk is Ieft?

Solution:

Quantity of milk purchased = 5 l

Quantity of milk used = 80% of 5 l

= \(\frac{80}{100}\) × 5 l

= 4 l

Quantity of milk left = 5 1 – 4 l = 1 l.

Question 15.

Rakesh covered a distance of 350 km. He travelled 70% by train and 28% by bus and the rest by autorickshaw. Find the distance travelled by autorickshaw.

Solution:

Distance covered by Rakesh = 350 km

Distance travelled by train = 70% of 350

= \(\frac{70}{100}\) × 245 km

Distance travelled by bus = 28% of 350

= \(\frac{28}{100}\) × 350 = 98 km.

Distance travelled by autorickshaw = 350 km – (245 + 98) km

= 350 km – 343 km = 7 km.

Hence the required distance = 7 km.

Question 16.

Find the amount Meena gets on depositing ₹ 1600 at 15% rate of interest for 73 days.

Solution:

Here, P = ₹ 1600,

R = 15% P.a.,

T = \(\frac{73}{365}\) years

= \(\frac{1}{5}\) years

∴ Simple interest S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 48

Amount = P + S.I.

= ₹ 1600 + ₹ 48 = ₹ 1648

Hence the required amount deposited = ₹ 1648.

Question 17.

What percent is:

(a) 5 seconds of 1 minute?

(b) 17 of 68?

Solution:

(a) 1 minute = 60 seconds

∴ 5 seconds of 60 seconds

=

= \(\frac{25}{3}\) %

= 8 \(\frac{1}{3}\) %

(b) 17 of 68

=

= 25 %