# Free Comparing Quantities Subjective Test 01 Practice Test - 7th grade

### Question 1

A group of 36 students went for a picnic. If 6 students can play with 3 balls. How many balls are required for the group of 36 students?

[1 MARK]

#### SOLUTION

Solution :

For 6 students we need 3 balls.

⇒ For 1 student, we need 36=12 balls.

Therefore, 36 students would need:

36×12= 18 balls.

### Question 2

One day Rashi travelled 6 km whereas Soham travelled only 5000 m. What will be the ratio of their distances travelled?

[2 MARKS]

#### SOLUTION

Solution :Units: 1 Mark

Answer: 1 Mark

Given that:

Rashi travelled 6 km.

Soham travelled 5000 m.

We have to find the ratio of their distance travelled.

We can only find the ratio if they have the same units.

First, we will convert both the distances to the same unit.

So, 6 km=6×1000 m=6000 m.

Thus, the required ratio is:

6000 m : 5000 m

⇒ 6:5

### Question 3

In a city out of 15000 voters, 60% voted. Find the number of voters who did not vote. [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Answer: 1 Mark

Given that:

In a city, there are 15000 voters.

60% of the voters voted.

Total percentage of voters = 100%

Voters who voted in % = 60%

⇒ Percentage of voters who did not vote

= 100 - 60 = 40%

Total Voters = 15000

40% of 15000 did not vote.

Then, the number of voters who did not vote is:

= 40% of 15000

⇒40100×15000=6000

∴ 6000 voters did not vote.

### Question 4

Rashi can travel 4 hours after having a Pizza, while Rashid can travel for 6 hours after having a pizza. How many more hours can Rashid travel than Rashi after having a Pizza? Express in terms of percentage.

[2 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Answer: 1 MarkGiven that:

Rashi can travel for 4 hours after having a pizza.

Rashid can travel for 6 hours after having a pizza.

Number of hours Rashid travels more after having a pizza = 6 - 4 = 2 hours

In terms of percentage, it is given by,Extra hoursNumber of hours Rashi can travel after having a pizza×100

On substituting the values we get,

24×100=50

### Question 5

Convert the given decimals to percentage. [2 MARKS]

a) 0.65 b) 2.1

#### SOLUTION

Solution :Each option: 1 Mark

Given decimal is,

a) 0.65

=0.65×100 %

=65×100100 %=65 %

b) 2.1

=2.1×100 %

=21×10010%=210 %

### Question 6

In a shop, there were Fanta, coca cola and sprite bottles in a ratio of 8 : 15 : 12 respectively. If the total number of Fanta bottles is 120, then find the total number bottles. [3 MARKS]

#### SOLUTION

Solution :Calculations: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Let the Fanta, Coca-Cola and Sprite bottles be 8x, 15x, 12x.

The total number of Fanta bottle is 120.

So, 8x = 120

x = 15

Then coca cola bottles =15×15=225

Sprite bottles =12×15=180.

So, the total number of bottle is 120 + 225 + 180 = 525.

Alternative Method:

Let 'x' be the total number of bottles.

Now, the bottles are in the ratio 8: 15: 12

The sum of these numbers = 8 + 15 + 12 = 35

The number of Fanta bottles = 120

Also number of Fanta bottles=835×x = 120

So, x=358×120=525

Hence the total number of bottles is 525.

### Question 7

Aarush used to solve 4 questions in 16 minutes. After much practice, he is now able to solve 6 questions in 18 minutes. What is the percentage decrease in the time he has achieved to solve each question?

[3 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Answer: 1 MarkNumber of questions Aarush used to solve initially = 4 questions in 16 minutes

Time needed to solve a question = 164 = 4 minutesNumber of questions Aarush can solve after practising = 6 questions in 18 minutes

Time needed to solve a question = 186 = 3 minutesDecrease in time to solve a single question = 4 - 3 = 1 minutes.

Therefore, the percentage decrease is given by

Decrease Percent = amount decreasedOriginal or base×100

On substituting the values, we get:

Decrease in percentage = 14×100 = 25%

Hence, after practising Aarush reduced 25% of the time to solve a single question.

### Question 8

Ravi purchased an old scooter for ₹ 12000 and spent ₹ 2850 on its overhauling. Then he sold it to his friend for ₹ 13860. How much percent did he gain or lose?

[3 MARKS]

#### SOLUTION

Solution :Calculations: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Cost price = ₹ 12000, overhead charges = ₹ 2850

∴ Total cost price = ₹ (12000 + 2850) = 14850

Selling Price = ₹ 13860

Selling Price < Cost Price ⇒ Loss

∴ Loss = Cost price - Selling price

⇒ Loss = ₹ (14850-13860) = ₹ 990

Loss percentage = LossCost Price×100

Loss percentage = 99014850×100 = 6.7 %

So, Ravi incurred a loss of 6.7%.

### Question 9

Soham bought a sofa for Rs 6500. If he pays an interest of Rs 975 in 3 years, find the rate of interest. [3 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Here,

It is given that Soham bought a sofa for Rs6500.

He paid Rs 975 as interest in three years.

Let, Principal amount be 'P' = Rs 6500

Interest be 'I' = Rs 975

Time be 'T' = 3 years

The rate of interest =?

Let the rate of interest be 'R'.

We know that,

I=P×T×R100

⇒R=I×100P×T

⇒R=975×1006500×3

⇒R=5

Hence the rate of interest is 5%.

### Question 10

Radhika bought a new mobile phone and saved **₹** 1000 when a discount of 10% was given. What was the price of the mobile phone before the discount? What was the selling price of the phone?

[4 MARKS]

#### SOLUTION

Solution :Steps: 2 Mark

Price before discount: 1 Mark

Selling price: 1 Mark

Given that,

Radhika bought a new mobile phone and saved₹1000 when a discount of 10% was given.

We can infer from the above statement that 10% of the original price =₹1000

10% of the original price =₹1000

Let the price be P.

⇒10% of P = 1000

⇒10100×P = 1000

⇒P=1000×10010

⇒P =₹10000

Hence the original price of the mobile phone is₹10000.

Now, the original price of the mobile phone is₹10000.

Discount given =₹1000

∴ The Selling price or the price at which the phone was sold = Original Price - Discount

On substituting the values we get:

Selling Price = 10000 - 1000 =₹9000

Hence, the mobile phone was sold at₹9000.

### Question 11

Rashi went toy shopping with her mother and saw a toy whose marked price was ₹ 200. The shopkeeper was selling it at a discount of 20%. The shopkeeper still makes a profit of 60%. Find the cost price of the toy. [4 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given that:

Marked price of the toy = ₹ 200

Discount = 20%.Discount = Discount % of Marked Price

= 20100×200

= ₹ 40

Selling price = Marked Price - Discount

= ₹( 200- 40 )= ₹ 160

It is given that even after the discount the shopkeeper was making a profit of 60%.

So, the cost price of the toy

=100160×160=₹100

Hence, the actual price of the toy is ₹ 100.

### Question 12

Meet saves ₹ 4000 from her salary. If this is 10% of her salary. What is her salary and also find her net expenditure? However, Meet wanted to buy a bike, so she started to save 30% of her salary. How much is she saving now? [4 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Answer: 1 Mark each

Given that:

Meet saves ₹ 4000 from her salary, which is 10% of her salary.

Let Meet’s salary be ₹ x

Given that,

10% of x=4000

10100×x=4000

x10=4000

x=4000×10=₹40000

Therefore, Meet’s salary is ₹ 40,000.

Her net expenditure = Salary - savings = 40000 - 4000 = ₹ 36,000

Now in order to buy a bike Meet starts saving 30% of her salary.

Her salary is ₹ 40,000.

The total amount she starts saving monthly = 30100×40,000 = ₹ 12,000

Her net saving monthly is ₹ 12,000.

### Question 13

If Meena gives an interest of **₹ **45 for one year at 9 % rate p.a what is the sum she has borrowed? What is the net amount she pays at the end of one year?

[4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given,

Meena pays an interest of₹45 for one year.

The rate of interest is 9 %

Interest =₹45

Time = 1 year

Rate of interest = 9%

Let the principal be P

We know that,

S.I=P×R×T100

45=P×9×1100

⇒P =45×1009

∴P =₹500

The sum she borrowed is₹500.

The net amount she will pay at the end of one year = 500 + 45 =₹545.

### Question 14

A cupboard bought for ₹ 2,500 and sold at ₹ 3,000. What is the profit or loss in the transaction? Also, find the profit percent or loss percent if it was sold for ₹ 2000.

[4 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Steps: 1 Mark

Answer: 1 Mark Each

Given that,

Cost price = ₹ 2500

Selling price = ₹ 3000

Selling price > Cost price ⇒ Profit

∴ Profit = 3000 - 2500 = ₹ 500

Profit %=ProfitCP×100

Profit %=5002500×100

Profit %=20%

Now it was sold for ₹ 2000.

The selling price, SP = ₹ 2000

SP<CP

Therefore it is a case of loss.

Loss = CP - SP = 2500 - 2000 = ₹ 500

So, the loss % is given by,

Loss % = 5002500×100 = 20 %

### Question 15

Harish borrowed ₹ 7500 from a bank.

a) Find the amount to be paid at the end of 3 years for ₹ 7500 at 5 % p.a.

b) Harish was only able to pay ₹ 5625. Find out the amount which Harish needs to pay to the bank at the end of the 4th year if the rate of interest remains same.

[4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Each answer: 1 Mark

Given,

Harish borrowed ₹ 7500 from a bank.

a) The rate of interest is 5 % p.a.

The principal amount, P =₹7500

Rate of Interest, R = 5 % p.a.

Time, T = 3 years

S.I.=P×R×T100

S.I.=7500×5×3100

S.I.=₹1125

Amount = Principal + Interest

⇒ Amount = 7500 + 1125

Amount =₹8625

Hence, the total amount to be paid at the end of three years is₹8625

b) Now, Harish was able to pay only₹5625.

So, the remaining amount

=₹(8625 - 5625) =₹3000

The new principal amount =₹3000

The rate of interest is same = 5 % p.a.

The interest which he has to pay at the end of the fourth year

= 5100×3000 =₹150

So, the amount which Harish has to pay at the end of the fourth year

= 3000 + 150 =₹3150