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

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.

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 :

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

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

=542525

=125100

=125 %

Now,
9:25=92544
=36100
=36%

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

#### SOLUTION

Solution :

Steps: 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.

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

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.

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

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.

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

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%

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

Given,

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

So, the new value of n
=n + n×25100=n×54

Now, the value of is again decreased by 20.

So, 20 of the new value
=n×54×20100=n×14

So, the new value
=n×54 - n×14 = n

Net percent change
=amount changedoriginal value×100

Net change = n - n = 0.
So on substituting the values we get:
Percent change = 0n×100 = 0
So, there is no percentage change.

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 %

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

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 priceLoss

Selling price=27541.25

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%.

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

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.

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

Given, Selling price = ₹ 13,500

Loss% = 20%

Let the cost price be x.

Loss = 20% of x

Selling price = Cost price - Loss

13500 = x20100×x

13500 = x15x

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.

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

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.

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

Let the cost price of item sold at loss be  x

x10 % of x=45

x10100×x=45

x110×x=45

10xx10=45

x=50Cost price is   50

Loss =   (50 - 45) =   5

Similarly, 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=32Cost price is    32

Profit =  (40 - 32) =   8

Net gain =   (8 - 5) =  ₹ 3

So, the shopkeeper makes a profit of   3.

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

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.