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

4 drivers: 1 hour

$\Rightarrow$  4 : 1

20 drivers: 5 hours

$\Rightarrow$  20 : 5 $\Rightarrow$  4 : 1

$\therefore$ Given ratios are equivalent.

Each answer: 1 Mark

Given that,

Number circles = 26

Number of hearts = 1 

Number of stars=1

$\therefore$ Number of circles: number of hearts

$\Rightarrow$  26 : 1


$\therefore$ Number of hearts: number of stars

$\Rightarrow$  1 : 1

Each Ratio: 1 Mark

Given ratio is, 5:4

$\Rightarrow \frac{5}{4}$

$=\frac{5}{4}$ x $\frac{25}{25}$

=$\frac{125}{100}$

=125 $\%$

Now,
$9:25=\frac{9}{25}$ x $\frac{4}{4}$
$=\frac{36}{100}$
$=36\%$

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%

$\therefore$  Number of students not studying Sanskrit,

$=\frac{65}{100}\times 860=559$
So, 559 students do not study Sanskrit.

Formula: 1 Mark
Answer: 1 Mark

Given that,

Marked Price of the watch = ₹ 1250.

6% discount on MP

$\Rightarrow \frac{6}{100}\times 1250=₹75$

SP = MP - Discount.

SP = ₹ 1250 - ₹ 75.

SP = ₹ 1175.

The selling price of the watch is  ₹ 1175.

Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

Let Nishu's pocket money be $n$

Given, she saves 10% of her pocket money. 

$\Rightarrow$ 10% of n = ₹ 120

$\Rightarrow \frac{10}{100}\times n=120$

$\Rightarrow n=\frac{120\times 100}{10}$

$\Rightarrow n=1200$

So, Nishu's pocket money is ₹ 1200.

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.

$\therefore$ Increase in the price = ₹ 20,000

$Percentageincrease=\frac{Increaseintheprice}{originalprice}\times 100$

$Percentageincrease=\frac{20000}{350000}\times 100$

$Percentageincrease=5.714\%$
The price of the car went up by 5.714$\%$

Formula: 1 Mark
Steps: 1 Mark 
Answer: 1 Mark

Given, 

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

So, the new value of n
$=n+n\times \frac{25}{100}=n\times \frac{5}{4}$

Now, the value of is again decreased by $20$.

So, $20$ of the new value
$=n\times \frac{5}{4}\times \frac{20}{100}=n\times \frac{1}{4}$

So, the new value
$=n\times \frac{5}{4}$ - $n\times \frac{1}{4}=n$

Net percent change
$=\frac{\text{amount changed}}{\text{original value}}\times 100$

Net change = n - n = 0.
So on substituting the values we get:
Percent change = $\frac{0}{n}\times 100$ = 0
So, there is no percentage change.

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
$=\left(\frac{Increase}{Originalmarks}\right)\times 100$

Increase in percentage
$=\left(\frac{15}{70}\right)\times 100$ = 21.42 %

Concept: 1 Mark
Application: 1 Mark 
Each Answer: 1 Mark

Given that
Amina bought a book for ₹ 275.
So the Cost price, CP = ₹ 275.

$Loss\%=15\%$

$Loss=15\%of275$

$Loss=\frac{15}{100}\times 275$

$Loss=\frac{4125}{100}$

$Loss=41.25$

$Sellingprice=Costprice-Loss$

$Sellingprice=275-41.25$

$Sellingprice=₹233.75$

Hence she sold it at ₹ 233.75.

If she has sold at ₹ 286,
The selling price, SP = ₹ 286
Profit = SP - CP
$\therefore$ On substituting the values we get, Profit = (286 - 275) = ₹ 11

Now, Profit % $=\frac{profit}{CostPrice}\times 100$

Profit % $=\frac{11}{275}\times 100=4$

$\therefore$ Had she sold it at ₹ 286 her profit% would have been 4%.

Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Given, Selling price = ₹ 13,500

$Loss\%=20\%$

Let the cost price be $x.$

$\therefore Loss=20\%ofx$

Selling price = Cost price - Loss 

$\Rightarrow 13500=x-\frac{20}{100}\times x$

$\Rightarrow 13500=x-\frac{1}{5}x$

$\Rightarrow 13500=\frac{4}{5}x$

$\Rightarrow x=16875$

Therefore, she bought it for ₹ 16875.
So, cost price of the article = ₹ 16875

20% of the cost price
$=\frac{20}{100}\times 16875=₹3375$

Net amount = 16875 + 3375 = ₹ 20250
So, in order to sell the washing machine at 20% profit, she needs to sell it at ₹ 20250.

Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Given,

Principal (P) = ₹ 1200

Rate (R) = $12\%$ p.a.

Time (T) = 3 years

$S.I.=\frac{P\times R\times T}{100}$

$S.I.=\frac{1200\times 12\times 3}{100}$

$S.I.=₹432$

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.

Application: 1 Mark 
Steps: 2 Marks
Answer: 1 Mark

Let the cost price of item sold at loss be ₹ x

$\Rightarrow x-10\%ofx=45$

$\Rightarrow x-\frac{10}{100}\times x=45$

$\Rightarrow x-\frac{1}{10}\times x=45$

$\Rightarrow \frac{10x-x}{10}=45$

$\Rightarrow x=50\Rightarrow \text{Cost price is}$  ₹ 50

$\Rightarrow Loss$ =  ₹ (50 - 45) =  ₹ 5

Similarly, let the cost price of an item sold at profit be ₹ y.

$\Rightarrow y+25\%ofy=40$

$\Rightarrow y+\frac{25}{100}\times y=40$

$\Rightarrow y+\frac{1}{4}\times y=40$

$\Rightarrow \frac{4y+y}{4}=40$

$\Rightarrow y=32\Rightarrow \text{Cost price is}$   ₹ 32

$\Rightarrow \text{Profit}$ =  ₹(40 - 32) =  ₹ 8

Net gain =  ₹ (8 - 5) =  ₹ 3
 
So, the shopkeeper makes a profit of  ₹ 3.

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

$\therefore S.I=\frac{P\times R\times T}{100}$

$\Rightarrow I=\frac{62500\times 4\times 1}{100}$

$\Rightarrow I=2500$

$\Rightarrow$ After 1 year the population will decrease by 2500 people

$\therefore$ Population 1 year ago is = 62500 - 2500 = 60,000

For present population, 

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

$\therefore S.I=\frac{P\times R\times T}{100}$

$\Rightarrow I=\frac{60000\times 4\times 1}{100}$

$\Rightarrow I=2400$

$\Rightarrow$ 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.