Free Fractions and Decimals Subjective Test 01 Practice Test - 7th grade

Question 1

Karan solves a question in 15 seconds. One of his friends challenges him that he can complete the work in one-fourth of the time that Karan takes. What is the time taken by Karan's friend to complete the work? [1 MARK]

SOLUTION

Solution :
Time taken by Karan to solve a question= 15 seconds

As per the question,

Time taken by Karan's friend to solve the question
= (1/4) of the time taken by Karan to solve the question
=(15)(1/4) = 3.75 seconds

Hence, Karan's friend will take 3.75 seconds to solve the question.

Question 2

A bag contains 35 apples and 28 bananas. They are distributed to 7 students such that all have an equal number of apples and bananas. How many fruits does each student get? [2 MARKS]

SOLUTION

Solution :
Concept: 1 Mark
Application: 1 Mark
Given that
Total number of apples = 35

The number of apples each student gets = (35/7) = 5 apples

Total number of bananas = 28

The number of bananas each student gets =(28/7) = 4 bananas

Total number of fruits a student gets = 5 apples + 4 bananas = 9
Hence, each student will get 5 apples and 4 bananas.

Question 3

Ramesh does $\frac{1}{10}$ th of work in an hour while his friend does $\frac{1}{10000}$ th of what Ramesh does in the given time. What part of work did Ramesh's friend complete in an hour ? [2 MARKS]

SOLUTION

Solution :
Steps: 1 Mark
Answer: 1 Mark

Let the total work be W

The work completed by Ramesh = $\frac{1}{10} \times W$

The work completed by Ramesh's friend = $\frac{1}{10} \times \frac{1}{10000} W$

= $\frac{1}{100000} \times W$

Ramesh's friend completes $\frac{1}{100000}$ ^th of the work in the given time,

Question 4

The area of a rectangle is 50 $m^{2}$ . Find the breadth of the rectangle given that its length is $\frac{4}{7}$ m. [2 MARKS]

SOLUTION

Solution :
Formula: 1 Mark
Result: 1 Mark

We know that area of the rectangle = 50 $m^{2}$

Given that the length of the rectangle
= $\frac{4}{7}$ metres

Now, Length $\times$ breadth= Area of the rectangle
Let the breadth be 'b' meters

$\frac{4}{7}$ $\times$ b = 50

b = $\frac{175}{2}$ metres

Hence, the breadth of the rectangle is $\frac{175}{2}$ metres.

Question 5

Damayanti ate $\frac{2}{9}$ th part of a pizza. The remaining was divided equally between Hari and Shankar. What part of pizza did Hari and Shankar get? [2 MARKS]

SOLUTION

Solution : Steps: [1 Mark]
Answer: [1 Mark]

Amount of pizza eaten by Damayanti $= \frac{2}{9}$

Amount of pizza not eaten by Damayanti

$\frac{9}{9} - \frac{2}{9} = \frac{7}{9}$

This portion is to be divided equally amongst Hari and Shankar.

$∴$ Hari and Shankar will get $\frac{7}{9} \div \frac{2}{1}$

$\Rightarrow \frac{7}{9} \times \frac{1}{2}$

$\Rightarrow \frac{7}{18}$ of the pizza each

So Hari and Shankar each get $\frac{7}{18}$ of the pizza

Question 6

A man completes a work in 1 hour in his first attempt and in his second attempt he completes it in (1/4) of 1 hour. In his third attempt, he completes it in (4/5)^thtime of what he took in his second attempt. How long did he take to complete the work in his third attempt? What is the difference in the time taken by him in the first and third attempt? [3 MARKS]

SOLUTION

Solution :
Steps: [1 Marks]
Each answer: [1 Mark]

Time taken by the man in his first attempt = 1 hour

Time taken by the man in his second attempt = $\frac{1}{4} \times 1$

= $\frac{1}{4}$ hour

Time taken by the man in his third attempt = $\frac{1}{4} \times \frac{4}{5}$

= $\frac{1}{5}$

Hence he took $\frac{1}{5}$ hours to complete the work on his third attempt.

The difference in the time taken by him in his first attempt and third attempt
$= 1 - \frac{1}{5} = \frac{4}{5} h o u r s$

Question 7

In a match of 50 overs, the team batting second was given a target of 310.They lost the match by 10 runs. The team batting second, i.e. the team which lost, scored one-half of their runs in 30 overs and in next five overs they scored one-half of the remaining runs. Later in the next 8 overs, they scored $\frac{1}{5}$ ^th of the remaining runs. In last seven overs, they scored the remaining runs. How many runs did they score in the last seven overs?
[3 MARKS]

SOLUTION

Solution :
Steps: [2 Marks]
Result: [1 Mark]

Total runs scored by the team that lost = 310 - 10 = 300

Now, total no. of overs played by them = 50

i.e. in 30 overs, this team scored $= \frac{300}{2}$

=150 runs

The balance 300 - 150 = 150 runs were scored by this team in balance 20 overs.

By the given condition

In the next 5 overs $\frac{1}{2}$ of 150 runs

= 75 runs were scored.

$\Rightarrow$ 150 runs - 75 runs = 75 runs are still in balance.

Again in the next 8 overs.

$\frac{1}{5}$ of 75 = 15 runs were scored.

Now No. of overs left = 20 - (5 + 8)

=20 - 13
=7 overs

In seven overs, runs scored will be 75 - 15 = 60 runs

Question 8

A circle is divided into four equal parts. One of the parts is again divided into two equal parts. If the area of the given circle is 400 metre square. What is the area of the shaded region? [3 MARKS]

SOLUTION

Solution :
Visualizing the question : 1 Mark
Steps : 1 Mark
Result : 1 Mark

Area of the given circle = 400 $m^{2}$

Area of each sector= $\frac{400}{4} = 100 m^{2}$

Area of the shaded portion = $\frac{100}{2}$
= $50 m^{2}$

Question 9

Question 6
Lipika reads a book for $1 \frac{3}{4}$ hours in a day. If she read the entire book in 6 days how many hours did she take to read the book? In how many days will she be able to complete the book if she would have read for $\frac{1}{2}$ hours each day? [3 Marks]

SOLUTION

Solution : Steps: 1 Mark
Calculation: 2 Marks

Time taken by Lipika to read a book $= 1 \frac{3}{4} h o u r s$
She reads the entire book in 6 days.
Now, total number of hours she took to read the entire book

$= 1 \frac{3}{4} \times 6 = \frac{7}{4} \times 6 = \frac{21}{2} = 10 \frac{1}{2} h o u r s$

Now if she reads for $\frac{1}{2}$ hours daily, time required by her to complete the book

$= 10 \frac{1}{2} \div \frac{1}{2} = \frac{21}{2} \div \frac{1}{2} = 21$

Hence, if she reads for half an hour daily she will finish reading the book in 21 days.

Question 10

The length of the shaded rectangle is $(\frac{1}{10})$ of the bigger rectangle and its breadth is 0.001 times the bigger rectangle. If the area of the bigger rectangle is $10000 m^{2}$ . Find the area of the shaded rectangle. [4 MARKS]

SOLUTION

Solution :
Steps: [3 Marks]
Answer: [1 Mark]

Let the length and breadth of the bigger rectangle be $a$ meters and $b$ meters respectively
As per the question.

The length of the shaded rectangle is $(\frac{1}{10})$ of the bigger rectangle.

The length of the shaded rectangle is $0.1 \times a$ meters.

The breadth of the shaded rectangle is 0.001 times the bigger rectangle.

The breadth of the shaded rectangle is $0.001 \times b$ meters.

Area of the given bigger rectangle $= a \times b$ metres $= 10000 m^{2}$

Area of the shaded rectangle $= (0.1 a) \times (0.001 b)$

$= (0.0001) \times (a b)$

$= (0.0001) \times (10000)$
$= 1 m^{2}$
So, the area of the shaded rectangle is $1 m^{2}$ .

Question 11

Ram purchased 1.5 kgs of tomato for 35 rupees. If 5.6 kg of tomato is required to prepare 5 kg tomato pickle. What is the cost of 1 kg tomato pickle, given that only tomatoes are used for the preparation of the pickle? [4 MARKS]

SOLUTION

Solution :
Steps: 3 Marks
Result: 1 Mark

Given that
Ram purchased 1.5 kgs of tomato for 35 rupees.

Also, 5.6 kg of tomato is required to prepare 5 kg of tomato pickle.

The first step is to find per kg cost of tomato.

Now, cost of 1 kg of tomato = $\frac{35}{1.5}$ rupees = $\frac{35 \times 2}{1.5 \times 2}$ = $\frac{70}{3}$ rupees

Given that 5.6 kg of tomato is required to prepare 5 kg tomato pickle

Then tomatoes required for making of 1 kg of pickle = $\frac{5.6}{5}$ kgs = 1.12 kgs

Cost of 1kg of tomato pickle $= (1.12) \times \frac{70}{3}$ = 26.133 rupees

Question 12

A tap fills one half of a bottle in 1 minute. Three fourth of the remaining in next 1 minute. If the remaining volume is 7 $m^{3}$ , then find the volume of the bottle.
[4 MARKS]

SOLUTION

Solution :
Steps: [2 Marks]
Application: [1 Mark]
Answer: [1 Mark]
Let the total volume of the bottle be V $m^{3}$

Volume filled in the first minute is $\frac{V}{2}$

Volume of the bottle remaining= $V - \frac{V}{2} = \frac{V}{2}$
Volume filled in the second minute is $(\frac{3}{4}) \times (\frac{V}{2}) m^{3}$

Volume filled in the third minute is $V - (\frac{V}{2}) - (\frac{3}{4}) \times (\frac{V}{2})$

$= (\frac{V}{8}) m^{3}$

Then the remaining volume $= 7 m^{3}$

$(\frac{V}{8}) = 7$

$V = 56 m^{3}$
The volume of the bottle is $56 m^{3}$

Question 13

A train got derailed and ${\frac{11}{14}}^{t h}$ of the length of the train was found off the track. If the length of the train was 322 m, find out what length of the train was found off the track? What length of the train was still on track? [4 MARKS]

SOLUTION

Solution : Steps: 2 Marks
Answers: 2 Marks

Given the length of the train is 322 m

Let x metres be the length that was found off the track. Then using unitary method we have

Or, $x \times \frac{14}{14} = 322 \times \frac{11}{14}$

Or, $14 x = 322 \times 11$

$∴ x = \frac{322 \times 11}{14}$

$= \frac{322}{14} \times 11$

$= 23 \times 11$

$= 253 m e t r e s$

The total length of the train = 322 m
Total length of the train off the track = 253 m
So, the length of the train which was still on track = 322 - 253 = 69 m.

Question 14

Look at the figure above. This is a staircase consisting of 5 steps. The height of each in-step(green area) is 6 cms.

(i) What is the total length of the pink line if the breadth of each step is 32 cms.

(ii) Write the ratio of the height of the instep to its total length of the pink line in the form of a fraction.

[4 MARKS]

SOLUTION

Solution : (i) Formula: 1 Mark
Result: 1 Mark
(ii) Formula: 1 Mark
Result: 1 Mark

(i) Given: The height of the staircase = 6 cms

The breadth of the staircase = 32 cms

So, length of the pink line will be = (breadth + height) of five steps.
$\Rightarrow$ 5 (6 cms) + 5 (32 cms)

$\Rightarrow 5 [6 + 32] c m s$

$\Rightarrow 5 [38] c m s$

$\Rightarrow 190 c m s$

(ii) $\frac{T h e t o t a l l e n g t h o f t h e i n - s t e p}{T h e t o t a l l e n g t h o f t h e p i n k l i n e}$

$\Rightarrow \frac{5 (6) c m s}{190 c m s}$

$\Rightarrow \frac{30}{190} \Rightarrow \frac{3}{19}$

Question 15

Shivaji galloped with his horse for 21 km. He crossed the river which was $\frac{1}{3}$ the length he covered with the horse. He walked the remaining distance which was 4 times the breadth of the river.
Calculate the total distance covered by him. [4 MARKS]

SOLUTION

Solution : Concept: 1 Mark
Steps: 2 Marks
Result: 1 Mark

The breadth of the river $= \frac{1}{3} \times 21$

$= 7 k m$

Length covered by walking $= 4 (\frac{1}{3} o f 21) k m$

$\Rightarrow 4 (7 k m)$

$\Rightarrow 28 k m$

Hence, the total distance covered by Shivaji = 21km + 7km + 28km

$\Rightarrow$ 28km + 28km
= 56km

The total distance covered by Shivaji is 56km.

Free Fractions and Decimals Subjective Test 01 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsPerimeter and Area 03Rational Numbers Subjective Test 01Simple Equations Subjective Test 01

Related Exams
Perimeter and Area 03
Rational Numbers Subjective Test 01
Simple Equations Subjective Test 01