# Free Comparing Quantities Subjective Test 02 Practice Test - 7th grade

### Question 1

Are these ratios equivalent? [1 MARK]

4 drivers : 1 hour

20 drivers : 5 hours

#### SOLUTION

Solution :

4 drivers: 1 hour

⇒ 4 : 1

20 drivers: 5 hours

⇒ 20 : 5 ⇒ 4 : 1

∴ Given ratios are equivalent.

### Question 2

Dylan drew 1 heart, 1 star, and 26 circles. What is the ratio of circles to hearts? What is the ratio of hearts to star? [2 MARKS]

#### SOLUTION

Solution :Each answer: 1 Mark

Given that,

Number circles = 26

Number of hearts = 1

Number of stars=1

∴ Number of circles: number of hearts

⇒ 26 : 1

∴ Number of hearts: number of stars

⇒ 1 : 1

### Question 3

Convert the ratios 5:4 and 9:25 into a percentage. [2 MARKS]

#### SOLUTION

Solution :Each Ratio: 1 Mark

Given ratio is, 5:4

⇒54

=54 x 2525

=125100

=125 %

Now,

9:25=925 x 44

=36100

=36%

### Question 4

In a school, there were 860 students, 35% study Sanskrit. How many students do not study Sanskrit?

[2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Answer: 1 Mark

Total number of students = 860

Number of students who study Sanskrit = 35%

Therefore, number of students who do not study Sanskrit =100% - 35% = 65%

∴ Number of students not studying Sanskrit,

=65100×860=559

So, 559 students do not study Sanskrit.

### Question 5

The marked price of a watch is ₹ 1250 and the shopkeeper allows 6% of discount on it. Find the selling price of the watch.

[2 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Answer: 1 Mark

Given that,

Marked Price of the watch = ₹ 1250.

6% discount on MP

⇒6100×1250=₹ 75

SP = MP - Discount.

SP = ₹ 1250 - ₹ 75.

SP = ₹ 1175.The selling price of the watch is ₹ 1175.

### Question 6

Nishu saves ₹ 120 from her pocket money. If she saves 10% of her pocket money, what is her pocket money? [3 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Let Nishu's pocket money be n

Given, she saves 10% of her pocket money.

⇒ 10% of n = ₹ 120

⇒10100×n=120

⇒n=120×10010

⇒n=1200

So, Nishu's pocket money is ₹ 1200.

### Question 7

Arun bought a car for ₹ 3,50,000. Next year, the price went up to ₹ 3,70,000. What was the percentage of price increase?

[3 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Given that,

Original price = ₹ 3,50,000.

New price = ₹ 3,70,000.

Increase in the price = ₹ 3,70,000 - ₹ 3,50,000.

∴ Increase in the price = ₹ 20,000

Percentage increase=Increase in the priceoriginal price×100

Percentage increase=20000350000×100

Percentage increase=5.714%

The price of the car went up by 5.714%

### Question 8

If number ‘n’ is increased by 25%, then it is decreased by 20%. What is the net percentage change in the value of n? [3 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Answer: 1 MarkGiven,

The number 'n' is first increased by 25%.

25So, the new value of n

=n + n×25100=n×54Now, the value of is again decreased by 20.

So, 20 of the new value

=n×54×20100=n×14So, the new value

=n×54 - n×14 = nNet percent change

=amount changedoriginal value×100Net change = n - n = 0.

So on substituting the values we get:

Percent change = 0n×100 = 0

So, there is no percentage change.

### Question 9

Aju scored 70 in 1st test and 85 in 2nd test of Mathematics out of 100. What is percentage increase in marks scored by him?

[3 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Application: 1 Mark

Calculation: 1 Mark

Marks scored in the 1st test = 70

Marks scored in the 1st test = 85

Increase in marks = 1st test mark - 2nd test marks

Increase in marks = 85 - 70 = 15

Increase in percentage

=(IncreaseOriginal marks)×100

Increase in percentage

=(1570)×100 = 21.42 %

### Question 10

Amina buys a book for ₹ 275 and sells it at a loss 15%. How much does she sell it for? If she had sold it for ₹ 286 calculate the profit percent.

[4 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Application: 1 Mark

Each Answer: 1 Mark

Given that

Amina bought a book for ₹ 275.

So the Cost price, CP = ₹ 275.

Loss%=15%

Loss=15% of 275

Loss=15100×275

Loss=4125100

Loss=41.25

Selling price=Cost price−Loss

Selling price=275−41.25

Selling price=₹233.75

Hence she sold it at ₹ 233.75.

If she has sold at ₹ 286,

The selling price, SP = ₹ 286

Profit = SP - CP

∴ On substituting the values we get, Profit = (286 - 275) = ₹ 11

Now, Profit % =profitCostPrice×100

Profit % =11275×100=4

∴ Had she sold it at ₹ 286 her profit% would have been 4%.

### Question 11

A sum of **₹** 56000 gives **₹** 280 as interest in 2 years. Find the rate of interest. If the interest rate was 11%, then find the interest in two years. [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given,

Principal =₹56,000

Time = 2 years

Interest =₹280

Let the rate of interest be R % p.a.

S.I=P×R×T100

⇒R=I×100P×T

⇒R=280×10056000×2

⇒R=280560×2

⇒R=14=0.25

∴ The rate of interest is 0.25 % .

We have to find the interest if the rate of interest was 11 %,

Interest = P×R×T100

On substituting the values we get,

Interest = 56000×11×2100 =₹12320

Hence, the interest in two years if the rate was 11% is₹12320.

### Question 12

Juhi sold a washing machine for ₹ 13,500 at the loss of 20%. Find the cost price of the washing machine. At what price she must sell to gain 20%?

[4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given, Selling price = ₹ 13,500

Loss% = 20%

Let the cost price be x.

∴Loss = 20% of x

Selling price = Cost price - Loss

⇒13500 = x−20100×x

⇒13500 = x−15x

⇒13500 = 45x

⇒x = 16875

Therefore, she bought it for ₹ 16875.

So, cost price of the article = ₹ 16875

20% of the cost price

=20100×16875=₹3375

Net amount = 16875 + 3375 = ₹ 20250

So, in order to sell the washing machine at 20% profit, she needs to sell it at ₹ 20250.

### Question 13

Find the amount to be paid at the end of 3 years of the sum of ₹ 1,200 at the rate of 12% p.a.

[4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given,

Principal (P) = ₹ 1200

Rate (R) = 12% p.a.

Time (T) = 3 years

S.I.=P×R×T100

S.I.=1200×12×3100

S.I.=₹432

Total amount to be paid at the end of three years = P + S.I.

Amount = ₹ 1200 + ₹ 432

Amount = ₹ 1632

₹ 1632 has to be paid at the end of three years.

### Question 14

If a shopkeeper incurs 10% loss and 25% profit on items sold for **₹** 45 and **₹** 40 respectively, then find his net profit/loss.

[4 MARKS]

#### SOLUTION

Solution :Application: 1 Mark

Steps: 2 Marks

Answer: 1 MarkLet the cost price of item sold at loss be

₹x⇒x−10 % of x=45

⇒x−10100×x=45

⇒x−110×x=45

⇒10x−x10=45

⇒x=50⇒Cost price is

₹50⇒Loss =

₹(50 - 45) =₹5Similarly, let the cost price of an item sold at profit be

₹y.⇒y+25 % of y=40

⇒y+25100×y=40

⇒y+14×y=40

⇒4y+y4=40

⇒y=32⇒Cost price is

₹32⇒Profit =

₹(40 - 32) =₹8Net gain =

₹(8 - 5) =₹3

So, the shopkeeper makes a profit of₹3.

### Question 15

The population of a town 2 years ago was 62500. Since some people migrate to different cities the number of people decreases every year at the rate of 4% per annum. Find its present population. [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Application: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Given that,

Population 2 years ago = 62500.

Rate of decrease = 4 % p.a.

For population 1 year ago,

P = 62500 , R = 4 %, T = 1 year

∴S.I=P×R×T100

⇒I=62500×4×1100

⇒I=2500

⇒ After 1 year the population will decrease by 2500 people

∴ Population 1 year ago is = 62500 - 2500 = 60,000

For present population,

P = 60,000, R = 4 %, T = 1 year

∴S.I=P×R×T100

⇒I=60000×4×1100

⇒I=2400

⇒ Present population = 60,000 - 2400 = 57,600

[We are subtracting the population because population is decreasing by 4% p.a.]

The present population is 57,600.