# DAV Class 7 Maths Chapter 5 Brain Teasers Solutions

The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 5 Brain Teasers Solutions of Application of Percentage offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 5 Brain Teasers Solutions

Question 1.
A. Tick (✓) the correct option:
(i) 12% of 50 + 5% of 120 is equal to—
(a) 10
(b) 15
(c) 12
(d) 20
(c) 12

12% of 50 + 5% of 120
= (12% of 50) +(5% of 120)
= ($$\frac{12}{100}$$ x 50) + ($$\frac{5}{100}$$ x 120)
= $$\left(\frac{12}{2}\right)+\left(\frac{12}{2}\right)$$
= 6 + 6
= 12
Hence, (c) is the correct answer.

(ii) What is 50% of a number whose 200% is 20?
(a) 15
(b) 5
(c) 20
(d) 10
(b) 5

Let the number be x
∴ 200% of x = 20
⇒ $$\frac{200}{100}$$ × x = 20
⇒ x = $$\frac{20 \times 100}{200}$$
⇒ x = 10
Now, 50% of x i.e., 50% of 10
= $$\frac{50}{100}$$ × 10 = 5
Hence, (b) is the correct answer.

(iii) At what rate of interest per annum should Rita invest if she wants to grow her ? 2500 to ? 4000 in five years?
(a) 10%
(b) 15%
(c) 12%
(d) 13%
(c) 12%

Here, Principal = ₹ 2500
Amount = ₹ 4000
So, simple intrest = Amount – Principal
= ₹ 4000 – ₹ 2500
= ₹ 1500
Time = 5 years
Now, S.I = $$\frac{P \times R \times T}{100}$$
1500 = $$\frac{2500 \times \mathrm{R} \times 5}{100}$$
R = $$\frac{1500 \times 100}{2500 \times 5}$$
R = $$\frac{15 \times 20}{25}$$ = 3 × 4
= 12%
Hence, (c) is the correct answer.

(iv) Rohan gave the pizza delivery person a tip of ₹ 30, which was 20% of his total bill for the pizza ordered. How much did pizza cost him?
(a) ₹ 150
(b) ₹180
(c) ₹ 140
(d) ₹ 200
Let the cost price of pizza be ₹ x.
Now, tip given to the pizza delivery person = 20% of total bill
= $$\frac{20}{100}$$ × x = ₹ $$\frac{x}{5}$$
Also, tip given to the pizza delivery person = ₹ 30
∴ $$\frac{x}{5}$$ = 30
⇒ x = 5 × 30 = ₹ 150.
Hence, (a) is the correct answer.

(v) In a country, there are 215 highway accidents associated with drinking alcohol. Out of these, 113 are caused by excessive speed. Approximately what percent of accidents are speed related?
(a) 47%
(b) 49%
(c) 51%
(d) 53%
Total number of highway accidents = 215
Number of highway accidents caused by excessive speed = 113
∴ Percentage of accidents caused by excessive speed = ($$\frac{113}{215}$$ × 100)%
= ($$\frac{113}{43}$$ × 20)%
= $$\left(\frac{2260}{43}\right)$$%
= 52.558%
= 53% (approx.)
Hence, (d) is the correct answer.

B. Answer the following questions:
(i) What percent of numbers from 1 to 20 are divisible by 4?
Total number of numbers from 1 to 20 = 20
Total number of numbers from 1 to 20 which are divisible by 4 = 5
Percentage of numbers from 1 to 20
which are divisible by 4 = $$\left(\frac{5 \times 100}{20}\right)$$%
= (5 × 5)% = 25%
Hence, 25% of numbers from 1 to 20 are divisible by 4.

(ii) A child had certain amount of chocolates. He ate 35% of the chocolates but still had 13 chocolates. What is the number of chocolates the child originally had?
Let the number of chocolates the child originally had be x.
Chocolates eaten by child = 35% of x
= $$\frac{35}{100}$$ × x = $$\frac{35}{100}$$x
Now, chocolates not eaten by child (which he still had) = 13
∴ $$\frac{35 x}{100}$$ + 13 = x
⇒ x – $$\frac{35 x}{100}$$ = 13
⇒ $$\frac{100 x-35 x}{100}$$ = 13
⇒ $$\frac{65 x}{100}$$ = 13
⇒ x = $$\frac{100 \times 13}{65}=\frac{100}{5}$$ = 20
Therefore, the number of chocolates the child originally had are 20.

(iii) How many more square must be shaded so that only 30% of the figure is left unshaded?

Total number of squares = 20
Total number of squares that must be shaded = (100 – 30)% of 20
= 70% of 20 = $$\frac{70}{100}$$ × 20 = $$\frac{70}{5}$$ = 14
But, number of squares that are already shaded in the figure = 8
Number of squares that must be shaded in the figure = 14 – 8 = 6
Therefore, 6 more squares must be shaded so that only 30% of the figure is unshaded.

(iv) A bucket of 20 litres is 35% full. How many litres of water must be put into it so that it is 65% full?
Total capacity of bucket = 20 litres
Quantity of water in the bucket = 35% of 20 litres
= ($$\frac{35}{100}$$ × 20)litres = ($$\frac{35}{100}$$) litres = 7 litres
Now, quantity of water in the bucket so that it is 65% full
= 65% of 20 litres
= ($$\frac{65}{100}$$ × 20 ) litres
= ($$\frac{65}{100}$$litres
= (($$\frac{65}{5}$$) litres = 13 litres
∴ Quantity of water that must be put into bucket so that it is 65% full = 13 litres – 7 litres = 6 litres.
So, 6 litres of water must be put into the bucket so that it is 65% full.

(v) Mr Sanjay deposited ₹ 15,000 in the bank for his five-year old daughter as he wishes to give his daughter the amount of ₹ 21000 on her thirteenth birthday. At what rate of interest should the money be invested?
Here, Principal (P) = ₹ 15,000
Amount = ₹ 21,000
∴ Simple Interest (S.I.) = ₹ 21,000 – ₹ 15,000
= ₹ 6,000
Time = (13 – 5) = 8 years
Now, S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
6000 = $$\frac{15,000 \times \mathrm{R} \times 8}{100}$$
⇒ R = $$\frac{6000 \times 100}{15,000 \times 8}$$
⇒ R = $$\frac{6000 \times 100}{15,000 \times 8}$$
⇒ R = $$\frac{200}{40}$$
⇒ R = 5%
So, the money should be invested at 5% rate of interest.

Question 2.
Rahul sold a watch to Sohan at a gain of 10% and Sohan sold it to Mohan at a loss of 10%. If Mohan Paid ₹ 990 for it, find the price paid by Rahul.
S.P. of watch for Sohan = ₹ 990
Loss = 10%
S.P = C.P(1 – $$\frac{\text { Loss }}{100}$$)
990 = C.P.(1 – $$\frac{10}{100}$$)
990 = C.P. × $$\frac{9}{10}$$
C.P. = 990 × $$\frac{10}{9}$$ = ₹ 1100
C.P. of watch for Sohan = S.P. of watch for Rahul

∴ S.P = ₹ 1100, gain = ₹ 10%
S.P. = C.P. (1 + $$\frac{\text { gain }}{100}$$)
1100 = C.P(1 + $$\frac{10}{100}$$)
⇒ 1100 = C.P.($$\frac{11}{10}$$)
∴ C.P. = $$\frac{1100 \times 10}{11}$$ = ₹ 1000
Hence the price paid by Rahul = ₹ 1000

Question 3.
In a school there are 50 teachers, 30% of them are men and the rest are women. If 60% of the male teachers are married, find the number of married male teachers.
Total number of teachers = 50
Number of married male teachers = $$\frac{30}{100}$$ × 50 = 15
Number of married male teachers = $$\frac{60}{100}$$ × 15 = 9
Hence the number of married male teachers = 9

Question 4.
(i) If x % of y is 13x, then find the value of y.
$$\frac{x}{100}$$ × y = 13x
y = $$\frac{13 x \times 100}{x}$$ = 1300
Hence the value of y = ₹ 1300

(ii) Find 12$$\frac{1}{2}$$% 3$$\frac{1}{2}$$% of 256.
= $$\frac{25}{2 \times 100} \times \frac{7}{2 \times 100}$$ × 256
= $$\frac{28}{25}$$
= 1.12
Hence the value of y = ₹ 1300

Question 5.
In an exam, 14% students failed and 559 students passed. Determine the number of students who failed.
Let the total number of students be x
Number of failed students = $$\frac{14}{100}$$ × x = $$\frac{7 x}{50}$$
Number of passed student = x – $$\frac{7 x}{50}=\frac{43 x}{50}$$
$$\frac{43 \times x}{50}$$ = 559
x = $$\frac{559 \times 50}{43}$$ = 650
∴ Number of failed students = 650 – 559 = 91

Question 6.
30% of the maximum marks are required to pass a test. A student gets 135 marks and is declared failed by 15 marks. Find the maximum marks.
Number of marks required to pass a test = 135 + 15 = 150
Let the maximum marks be x
∴ $$\frac{30}{100}$$ × x = 150
⇒ x = $$\frac{150 \times 100}{30}$$
⇒ x = 500
Hence the maximum marks = 500

Question 7.
A man bought cardbord sheet for ₹ 3600 and spent ₹ 100 on transport. Paying ₹ 300 for labour, he had 330 boxes made, which he sold at ₹ 14 each. Find the profit%.
Cost price of the sheet = ₹ 3600
Transportation cost = ₹ 100
Money paid to labours = ₹ 300
Total C.P. = ₹ 3600 + ₹ 100 + ₹ 300
= ₹ 4000
S.P. of 330 boxes = 14 × 330 = ₹ 4620
Profit = S.P – C.P
= ₹ 4620 – ₹ 4000
= ₹ 620
∴ Profit% = $$\frac{\text { Profit } \times 100}{\text { C.P. }}$$
= $$\frac{620 \times 100}{4000}$$ = 15.5%
Hence, profit % = 15.5

Question 8.
The S.P. of an article is three fourth of its C.P. What is the loss %?
Let the principal be ₹ x
S.I. = $$\frac{3}{4}$$x
S.P = C.P.(1 – $$\frac{\text { Loss }}{100}$$)
$$\frac{3}{4}$$x = x(1 – $$\frac{\text { Loss }}{100}$$)
1 – $$\frac{\text { Loss }}{100}=\frac{3}{4}$$
$$\frac{\text { Loss }}{100}$$ = 1 – $$\frac{3}{4}=\frac{1}{4}$$
Loss = $$\frac{100}{4}$$ = 25%
Hence Loss% = 25

Question 9.
In what time will the simple interest on a certain sum be $$\frac{3}{4}$$ th of the principal at 6% per annum?
Let the principal be ₹ x
∴ S.I = $$\frac{3}{4}$$x
S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
$$\frac{3}{4}$$x = $$\frac{x \times 6 \times T}{100}$$
T = $$\frac{3 \times 100}{4 \times 6}=\frac{25}{2}$$
= 12$$\frac{1}{2}$$ years
Hence the time = 12$$\frac{1}{2}$$

Question 10.
At what rate of interest per annum will a sum of? 8000 amount to ? 9260 in 3 y years?
P = ₹ 8000, A = ₹ 9260
∴ S.I. = A – P = ₹ 9260 – ₹ 8000
= ₹ 1260

S.I = $$\frac{P \times R \times T}{100}$$
1260 = $$\frac{8000 \times \mathrm{R} \times 7}{100 \times 2}$$
R = $$\frac{1260 \times 100 \times 2}{8000 \times 7}$$
R = $$\frac{9}{2}$$ = 4$$\frac{1}{2}$$% p.a

Hence the rate of interest = 4$$\frac{1}{2}$$% p.a.

Question 11.
If 40% of 70 is x more than 30% of 80, then find x.
According to the question, we have
40% of 70 = (30% of 80) + x
⇒ $$\frac{40}{100}$$ × 70 = ($$\frac{30}{100}$$ × 80) + x
⇒ 28 = 24 + x
⇒ x = 28 – 24
⇒ x = 4

Question 12.
A manufacturer sells three products A, B, C.
Product A costs ₹ 200 and is sold for ₹ 260.
Product B costs ₹ 150 and is sold for ₹ 180.
Product C costs ₹ 100 and is sold for ₹ 140.
Which product should be manufactured more in order to have maximum profit percentage?
Cost price of product A = ₹ 200
Selling price of product A = ₹ 260
∴ Profit on product A = ₹ 260 – ₹ 200
= ₹ 60
Now Profit % on product A = $$\frac{60}{200}$$ × 100 = 30%
Cost price of product B = ₹ 150
Selling price of product B = ₹ 180
Profit on product B = ₹ 180 – ₹ 150 = ₹ 30
Profit on product B = $$\frac{30}{150}$$ × 100 = 20%
Cost price of product C = ₹ 100
Selling price of product C = ₹ 140
Profit on product C = ₹ 140 – ₹ 100 = ₹ 40
Profit % on product C = $$\frac{40}{100}$$ × 100 = 40%
As profit percentage is maximum in manufacturing product C i.e., 40% so, product C should be manufactured more.

Question 13.
Find x:
48% of 480 + 25% of 250 – x = 200.
48% of 480 + 25% of 250 – x = 200

## DAV Class 7 Maths Chapter 5 HOTS

Question 1.
There are 40 papers of students of a class to be checked by two teachers. Mr. Ashok can check five papers in an hour and Mrs. Meena can check four papers in an hour. If Mr. Ashok spends three hours in checking the papers and Mrs. Meena works for two hours, what percentage of the papers will be checked in all?
Total number of papers = 40
Number of papers checked by Mr. Ashok in an hour = 5
Number of papers checked by Mrs. Ashok in 3 hours = 15 papers
Now, number of papers checked by Mrs. Meena in an hour = 4
Number of papers checked by Mrs. Meena in 2 hours = 8 papers
So, total number of papers checked by both Mr. Ashok and Mrs. Meena in all = (15 + 8) papers
∴ Percentage of papers that will be checked in all = ($$\frac{23 \times 5}{2}$$ x 100)%
= ($$\frac{23 \times 5}{2}$$)%
= ($$\frac{115}{2}$$)%
= 57.5%
Hence, 57.5% of the papers will be checked in all.

Question 1.
Find C.P. when;
(i) a cycle is sold for ₹ 1485 at a profit of 8%
C.P. = ?, S.P. = ₹ 1485, Profit = 8%
⇒ 1485 = C.P.(1 + $$\frac{\text { Profit }}{100}$$)
⇒ 1485 = C.P.(1 + $$\frac{8}{100}$$)
⇒ 1485 = C.P.($$\frac{27}{25}$$)
⇒ C.P. = $$\frac{1485 \times 25}{27}$$ = 1375

(ii) a fan is sold for ₹ 657.60 at a loss of 4%
C.P. = ?, S.P. = ₹ 657.60, Loss = 4%
⇒ 657.60 = C.P.(1 – $$\frac{\text { loss }}{100}$$
⇒ 657.60 = C.P.(1 – $$\frac{4}{100}$$)
⇒ 657.60 = C.P. x $$\frac{24}{25}$$
⇒ C.P = $$\frac{657.60 \times 25}{24}$$
⇒ C.P = ₹ 685

Question 2.
Find S.P. when CP = ₹ 435 and Loss = 16%
Solution:
S.P. = C.P.(1 – $$\frac{\text { Loss }}{100}$$)
= 435(1 – $$\frac{16}{100}$$)
= 435 × $$\frac{21}{25}$$
= ₹ 360.40

Question 3.
In a family of 25 persons, 72% read English newspaper, 16% read Hindi newspaper and the rest do not read any newspaper. Find the number of people who read no newspaper.
Number of persons who read English newspaper = 25 × $$\frac{72}{100}$$
Number of persons who read Hindi newspaper = 25 × $$\frac{16}{100}$$ = 4
Remaining people who do not read any newspaper = 25 – (18 + 4)
= 25 – 22
= 3 persons

Question 4.
What is 30% of 40% of 50% of 2400.
30% of 40% of 50% of 2400 = $$\frac{30}{100} \times \frac{40}{100} \times \frac{50}{100}$$ × 2400
= $$\frac{3}{10} \times \frac{2}{5} \times \frac{1}{2}$$ × 2400
= $$\frac{3}{50}$$ × 2400
= 3 × 48
= 144

Question 5.
A man bought 10 pencils for ₹ 11 and sold 11 pencils for ₹ 10. Find his profit or loss percent.
Solution:
C.P. of 10 pencils = ₹ 11
S.P. of 11 pencils = ₹ 10
S.P. of 1 pencils = ₹ $$\frac{10}{11}$$

Here C.P. > S.P.
it is a loss
Loss = C.P – S.P
= $$\frac{11}{10}-\frac{10}{11}=\frac{121-100}{110}=\frac{21}{110}$$

∴ Loss % = $$\frac{\text { loss }}{\text { C.P. }}$$ × 100
= $$\frac{\frac{21}{110}}{\frac{11}{10}}$$ × 100
= $$\frac{21}{110} \times \frac{10}{11}$$ × 100
= $$\frac{21}{110} \times \frac{10}{11}$$ %
= 17.35 %

Question 6.
Find the simple interest on ₹ 35000 at 5% p.a. for 2y years.
Here P = ₹ 35000, R = 5% p.a., T = 2$$\frac{1}{2}$$ = $$\frac{5}{2}$$ years.
S.I = $$\frac{P \times R \times T}{100}$$
= $$\frac{35000 \times 5 \times 5}{100 \times 2}$$
= ₹ 4375

Question 7.
At what rate percent will ₹ 1500 amount to ₹ 2400 in 4 years?
Here P = ₹ 1500, A = ₹ 2400 , T = 4 years, R = ?
S.I. = A – P
= ₹ 2400 – ₹ 1500
= ₹ 900

∴ S.I = $$\frac{P \times R \times T}{100}$$
⇒ 900 = $$\frac{1500 \times R \times 4}{100}$$
⇒ R = $$\frac{900 \times 100}{1500 \times 4}$$ = 15%
Hence rate of interest = 15% p.a.

Question 8.
Find the simple interest on ₹ 4800 in 2 y years at 3% p.a.
Here S.I. = ?, P = ₹ 4800, T = y years, R = 3% p.a.
S.I = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{4800 \times 3 \times 5}{100 \times 2}$$
= ₹ 360

Question 9.
A man borrowed a loan from a bank. After 2 years, he paid ₹ 5434. If the rate of interest is 2$$\frac{1}{4}$$% p.a, find sum of money.
Here A = ₹ 5434
R = 2$$\frac{1}{4}$$% = $$\frac{9}{4}$$% p.a.
T = 2 years
P =?

∴ The money borrowed = ₹ 5200

Question 10.
A sells a watch to B at a gain of 20% and B sells it to C at a loss of 10% and C sells it for ₹ 1404 gaining 4%. How much did A pay for it?
Let C.P. for A be ? x
∴ S.P. for A = C.P. for B = x(1 + $$\frac{20}{100}$$) = $$\frac{6 x}{5}$$
S.P for B = C.P for C
⇒ $$\frac{6 x}{5}$$(1 – $$\frac{10}{100}$$)
= $$\frac{6 x}{5} \times \frac{9}{10}$$
= ₹ $$\frac{54}{50}$$x

S.P for C = C.P for D (Say)
= $$\frac{54 x}{100}$$(1 + $$\frac{4}{100}$$)
⇒ $$\frac{54 x}{100} \times \frac{104}{100}$$

But S.P made by C = 1404
∴ $$\frac{54 x}{100} \times \frac{104}{100}$$
∴ x = $$\frac{1404 \times 100 \times 100}{54 \times 104}$$ = 1250
Hence, A paid for the watch = 1250