7th Maths : Term 3 Unit 2 : Percentage and Simple Interest : Exercise 2.5 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

**Exercise
2.5**

** **

__Miscellaneous
Practice problems__

** **

**1.
When Mathi was buying her flat she had to put down a deposit of 1/10 of the value
of the**** flat. What percentage was this?**

**Solution: **

Percentage of 1/10 = 1/10 × 100% = 10%

Mathi has to put down a deposit of 10% of the value of the flat.

** **

**2. Yazhini scored 15 out of 25 in a test.
Express the marks scored by her in percentage.**

**Solution: **

Yazhini's score = 15 out of 25 = 15 / 25

Score in percentage = (15 / 25) × 100% = 60%

** **

**3. Out of total 120 teachers of a school
70 were male. Express the number of male teachers as percentage.**

**Solution: **

Total teachers of the school = 120

Number of male teachers = 70

∴ Percentage of male teacher = (70 / 120) × 100% = 700/12 %

Score in percentage = 58.33%

Percentage of male teachers = 58.33%

** **

**4. A cricket team won 70 matches during
a year and lost 28 matches and no results for two matches. Find the percentage of
matches they won.**

**Solution: **

Number of Matches won = 70

Number of Matches lost = 28

“No result” Matches = 2

Total Matches = 70 + 28 + 2= 100

Percentage of Matches won = 70/100 × 100% = 70%

The won 70% of the matches

** **

**5.
There are 500 students in a rural school. If 370 of them can swim, what percentage
of them can swim and what percentage cannot?**

**Solution:**

Total number of students
= 500

Number of students who can swim = 370

Percentage of students who can swim = 370/500 × 100% = 74%

Number of students who cannot swim = 500 – 370 = 130

Percentage of students who cannot swim = 130/500 × 100% = 26%

i.e. 74% can swim and 26%
cannot swim

** **

**6. The ratio of Saral’s income to her
savings is 4 : 1. What is the percentage of money saved by her?**

**Solution: **

Total parts of money = 4+1=5

Part of money saved = 1

∴ Percentage of money saved = 1/5 × l00% = 20%

∴ 20% of money is saved by Saral

** **

**7. A salesman is on a commission rate
of 5%. How much commission does he make on sales worth ₹ 1,500?**

**Solution: **

Total amount on sale = ₹ 1500

Commission rate = 5%

Commission received = 5% of ₹ 1500 = (5 / 100) × 1500 = ₹ 75

∴ Commission received = ₹ 75

** **

**8. In the year 2015 ticket to the world
cup cricket match was ₹ 1,500. This year the price has been increased by 18%. What
is the price of a ticket this year?**

**Solution: **

Price of a ticket in 2015 = ₹ 1500

Increased price this year = 18% of price in 2015

= 18% of ₹1500 = (18 / 100) × 1500 = ₹ 270

Price of ticket this year = last year price + increased price

= ₹ 1500 + ₹ 270 = ₹ 1770

Price of ticket this year = ₹ 1770

** **

**9. 2 is what percentage of 50?**

**Solution:**

Let the required percentage be *x*

*x *% of 50 = 2

*(x* / 100) × 50 = 2

*x* =( 2 × 100) / 50 = 4%

∴ 4% of 50 is 2

** **

**10. What percentage of 8 is 64?**

**Solution:**

Let the required percentage be *x*

So *x*% of 8 = 64

(*x / *100)* *× 8 = 64

*x *= [64 × 100] / 8 = 800

∴ 800% of 8 is 64

** **

**11. Stephen invested ₹ 10,000 in a savings
bank account that earned 2% simple interest. Find the interest earned if the amount
was kept in the bank for 4 years.**

**Solution: **

Principal (P) = ₹ 10,000

Rate of interest (*r*) =
2%

Time (*n*) = 4 years

∴ Simple Interest I = P*nr */
100 = (10000 × 4 × 2) / 100 = ₹ 800

Stephen will earn ₹800

** **

**12. Riya bought ₹ 15,000 from a bank
to buy a car at 10% simple interest. If she paid ₹ 9,000 as interest while clearing
the loan, find the time for which the loan was given.**

**Solution: ** Here Principal (P) = ₹ 15,000

Rate of interest (*r*) =
10%

Simple Interest (I) = ₹ 9000

I = P*nr */ 100

9000 = (15000 × *n* × 10)
/ 100

*n *= ( 9000 × 100) / ( 15000 × 10)

*n *= 6 years

.’. The loan was given for 6 years.

** **

**13.
In how much time will the simple interest on ₹ 3,000 at the rate of 8% per annum
be the same as simple interest on ₹ 4,000 at 12% per annum for 4 years?**

**Solution: **

Let the required number of years be *x*

Simple Interest I = P*nr*
/ 100

Principal P_{1} = ₹ 3000

Rate of interest (*r*) =
8%

Time (*n*_{1}) =
*n*_{1 }years

Simple Interest I_{1} = (3000 × 8 × *n*_{1 }) / 100 = 240 *n*_{1}

Principal (P_{2}) = ₹ 4000

Rate of interest (*r*) =
12%

Time *n*_{2 }= 4
years

Simple Interest I_{2} = (4000 × 12 × 4) / 100

I_{2} = 1920

If I_{1} = I_{2}

240 *n*_{1} = 1920

*n*_{1 }= 1920 / 240 = 8

∴ The required time = 8 years

** **

__Challenge
Problems__

** **

**14. A man travelled 80 km by car
and 320 km by train to reach his destination. Find what percent of total
journey did he travel by car and what per cent by train?**

**Solution: **

Distance travelled by car = 80 km.

Distance travelled by train = 320 km

Total distance = 80 + 320 km = 400 km

Percentage of distance travelled by car = [80 / 400] × 100% =
20%

Percentage of distance travelled by train = [320 / 800] × 100% =
40%

** **

**15. Lalitha took a math test and got
35 correct and 10 incorrect answers. What was the percentage of correct answers?**

**Solution: **

Number of correct answers = 35

Number of incorrect answers = 10

Total number of answers = 35 + 10 = 45

Percentage of correct answers = [35 / 45] × 100% = 77.777% =
77.78%

** **

**16. Kumaran worked 7 months out of the
year. What percentage of the year did he work?**

**Solution: ** Total number of months in
a year = 12

Number of Months Kumaran worked = 7

Percentage of the year he worked = [7 / 12] × 100%

Percentage of the year worked = 58.33%

** **

**17. The population of a village is 8000.
Out of these, 80% are literate and of these literate people, 40% are women. Find
the percentage of literate women to the total population?**

**Solution: ** Population of the village
= 8000 people

literate people = 80% of population

= 80% of 8000 = ( 80 / 100 ) × 8000

literate people = 6400

Percentage of women = 40%

Number of women = 40% of literate people = (40/100) × 6400 =
2560

∴ literate women : Total population = 8000 : 2560 = 25 : 8

** **

**18. A student earned a grade of 80% on
a math test that had 20 problems. How many problems on this test did the student
answer correctly?**

**Solution: **Total number of
problems in the test = 20

Students score = 80%

Number of problem answered = (80/100) × 20 = 16

** **

**19. A metal bar weighs 8.5 kg.
85% of the bar is silver. How many kilograms of silver are in the bar?**

**Solution: **Total weight of the
metal = 8.5 kg

Percentage of silver in the metal = 85%

Weight of silver in the metal = 85% of total weight = (85/100) ×
8.5 kg = 7.225 kg

7.225 kg of silver are in the bar.

** **

**20. Concession card holders pay ₹ 120
for a train ticket. Full fare is ₹ 230. What is the percentage of discount for concession
card holders?**

**Solution: **

Train ticket fare = ₹ 230

Ticket fare on concession = ₹120

Discount = Ticket fare – concession fare = 230 – 120 = ₹110

Percentage of discount = [ Discount / Original rate ] × 100% = [110
/ 230] × 100 = 47.826% = 47.83%

Percentage of discount = 47.83%

** **

**21. A tank can hold 200 litres of water.
At present, it is only 40 % full. How many litres of water to fill in the tank,
so that it is 75 % full?**

**Solution: **

Capacity of the water tank = 200 litres

Percentage of water in the tank = 40%

Percentage of water to fill = Upto 75%

Difference in percentage = 75% – 40% = 35%

∴ Volume of water to be filled = Percentage of difference × total
capacity

= [35 / 100] × 200 = 70 *l*

70 *l* of water to be
filled.

** **

**22. Which is greater 16 (2/3) or (2/5)
or 0.17 ?**

**Solution: **

16( 2 / 3) = 50 / 30 = (50/3) × 100% = 1666.67%

⇒ 2/5 = (2/5) × 100 = 40%

0.17 = 17 / 100 =17%

∴ 1666.67 is greater. ∴

16(2/3) is greater.

** **

**23. The value of a machine depreciates
at 10% per year. If the present value is ₹ 1,62,000, what is the worth of the machine
after two years.**

**Solution: **

Present value of the machine = ₹ 1,67,000

Rate of depreciation = 10% per annum

Time (*n*) = 2 years

For 1 year depreciation amount = [1,62,000 × 1 × 10] / 100 = ₹
16,200

Worth of the machine after one year = Worth of Machine –
Depreciation

= 1,67,000 – 16,200 = 1,45,800

Depreciation of the machine for 2^{nd} year = 145800 × 1
× 10/100 = 14580

Worth of the machine after 2 years = 1,45,800 – 14,580 = 1,31,220

∴ Worth of the machine after 2 years = ₹ 1,31,220

** **

**24. In simple interest, a sum of money
amounts to ₹ 6,200 in 2 years and ₹ 6,800 in 3 years. Find the principal and rate
of interest.**

**Solution: **

Let the principal P = ₹ 100

If A = ₹ 6200

=> Principal + Interest for 2 years = 6200

A = ₹ 7400

=> Principal + Interest for 3 years = 7400

∴ Difference gives the Interest for 1 year

∴ Interest for 1 year = 7400 – 6200

I = 1200

P*nr* / 100 = 1200 ⇒ [P × 1 × *r *]* */ 100 = 1200

If the Principal = 10,000 then

[ 10,000 × 1 × *r* ] / 100
= 1200 ⇒* r *= 12%

Rate of interest = 12% per annum

** **

**25. A sum of ₹ 46,900 was lent out at
simple interest and at the end of 2 years, the total amount was ₹ 53,466.Find the
rate of interest per year.**

**Solution: **

Here principal P = ₹ 46900

Time *n* = 2 years

Amount A = ₹ 53466

Let *r n* be the rate of
interest per year

Interest I = P*nr* / 100

A = P + I

53466 = 46900 + (46900 × 2 ×*
r ) / *100

53466 – 46900 = ( 46900 × 2 × *r ) / *100

6566 = 469 × 2 ×* r*

*r* = 6566 / (2 × 469) % =
7%

Rate of interest = 7% per year

** **

**26. Arun lent ₹ 5,000 to Balaji for 2
years and ₹ 3,000 to Charles for 4 years on simple interest at the same rate of
interest and received ₹ 2,200 in all from both of them as interest. Find the rate
of interest per year.**

Principal lent to Balaji P_{1} = ₹ 5000

Time *n*_{1 }= 2
years

Let *r *be the rate of
interest per year

Simple interest got from Balaji = P*nr* / 100 ⇒ I_{1 }= [ 5000 × 25 × *r* ] / 100

Again principal let to Charles P_{2} = ₹ 3000

Time (*n*_{2}) =
4 years

Simple interest got from Charles (I_{2}) = [ 3000 × 4 x* r* ] / 100

Altogether Arun got ₹ 2200 as interest.

∴ I_{1} + I_{2} = 2200

[ (5000 × 2 × *r)* / 100
] + [ (3000 × 4 × *r) * / 100 ] = 2200

100*r* +120*r* = 2200

220*r* = 2200 = 2200 /
220

*r *= 10%

Rate of interest per year = 10%

** **

**27.If
a principal is getting doubled after 4 years, then calculate the rate of interest.**** ****(Hint : Let P = ₹ 100).**

**Solution: **

Let the principal P = ₹ 100

Given it is doubled after 4 years

i.e. Time *n* = 4 years

After 4 years A = ₹ 200

∴ A = P + I

A – P = 1

200 – 100 = I

After 4 years interest I = 100

I = P*nr* / 100 ⇒ 100 = [ 100 × 4 × *r *] /
100

4r = 100 ⇒ *r*
= 25%

Rate of interest *r* = 25%

** **

__ANSWERS:__

**Exercise 2.5 **

1. 10%

2. 60%

3. 58.33%

4. 70 %

5. 74% cam swim; 26%
canot swim

6. 20%

7. ₹ 75

8. ₹ 1,770

9. 4%

10. 800%

11. ₹ 800

12. 6 years

13. 8 years

**Challange Problems **

14. 20% ; 80%

15. 77.77%

16. 58.33%

17. 32%

18. 16

19. 7.225* kg *

20. 47.82%

21. 70 *l*

22. 16 (2/3)

23. 1,31,220

24. 12 %

25. 7%

26. 10 %

27. 25 %

Tags : Questions with Answers, Solution | Percentage and Simple Interest | Term 3 Chapter 2 | 7th Maths , 7th Maths : Term 3 Unit 2 : Percentage and Simple Interest

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7th Maths : Term 3 Unit 2 : Percentage and Simple Interest : Exercise 2.5 | Questions with Answers, Solution | Percentage and Simple Interest | Term 3 Chapter 2 | 7th Maths

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