Question Set B

CBSE 7th Standard Mathematics Practical Geometry Sample Question Papers Study Material - QB365

Download CBSE Class 7th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 7th Standard CBSE Mathematics

Questions:

Generic 2

9 x 2 = 18

Find the product of \(\frac { 6 }{ 3 } \)and \(\frac { 1 }{ 4 } \)

Find (-31) x (-100)

Arrange the following in descending order \(\frac{1}{5},\frac{3}{7},\frac{7}{10}\)

Solve (-21) - (30)

Solve (-6) - (-16)

Solve 14- (-20)

Solve 10 + (-4)

Add the following integers using number line: -3 and 4

Find the mode of the following data.1 , 4, 6, 5, 6, 2, 3, 4, 5, 3, 2, 4, 5, 2, 3

Generic 3

5 x 3 = 15

10.

Verify a + (b + c) = (a + b) + C for the following values of a, b and c. a =18, b = 15, c = 13.

11.

Write 5 rational numbers between \(\frac { 1 }{ 3 } \) and \(\frac { 1 }{ 4 } \).

12.

The bar graph given below shows the marks of students of a class in a particular subject

If 40 is the pass mark, then how many students have failed?

13.

Find the mode of the following data - 12,14,12,16,15,13,14,18,19,12,14,15,16,15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15,13,15,14

14.

A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200m below the sea level. What is the vertical distance between them?

Multiple Choice Question

5 x 1 = 5

15.

In the given figure, AB = AC = BC, then the value of each angle is equal to

(a)

40°

(b)

70°

(c)

90°

(d)

60°

16.

The value of (-2) x (-1) x (1) is

(a)

(b)

(c)

-4

(d)

17.

The number of trees in different parks of a city are 33, 38, 48, 33, 34, 34, 33 and 24. The mode of this data is

(a)

(b)

(c)

(d)

18.

The mode of the data 22,29, 27, 23, 43, 41, 27 is

(a)

23 and 27

(b)

(c)

23 and 43

(d)

19.

How many rational numbers are there between two rational numbers?

(a)

(b)

(c)

unlimited

(d)

100

Fill up / 1 Marks

5 x 1 = 5

20.

In a right triangle, the longest side is called ______________.

21.

The representation of data with bars of uniform width is called _____

22.

(- 206) \(\div\) ____ = 1

23.

Replace the blank with an integer to make it a true statement. (-3) x _____ =27

24.

\(\frac { -2 }{ 9 } -\frac { 7 }{ 9 } =\_ \_ \_ \_ \_ \_ \)

Match the following

5 x 1 = 5

25.

\(\angle A=45^0,\angle B=45^0,\)AB= 4 cm

26.

\(7 \frac{5}{6}\times \frac{3}{47} \times46\)

27.

(60) \(\div\) 1

28.

Range

29.

\(\frac { b }{ a } \div \left( \frac { d }{ c } \right) \)

Generic 5

3 x 5 = 15

30.

Draw \(\triangle PQR\) with PQ= 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

31.

Look at the following table and fill up the blank spaces.

Hundreds (100)	Tens (10)	Ones (1)	Tenths (\(\frac{1}{10}\))	Hundredths (\(\frac{1}{100}\))	Thousandths (\(\frac{1}{1000}\))	Number
2	5	3	1	4	7	253.147
6	2	9	3	2	1	_______
0	4	3	1	9	2	_______
_____	1	4	2	5	1	514.251
2	______	6	5	1	2	236.512
_____	2	______	5	_______	3	724.503
6	______	4	_______	2	________	614.326
0	1	0	5	3	0	______

32.

Study the double bar graph given below and answer the questions that follow:

(a) What information does the above double graph depict?
(b) Name the fruits for which cost of 1 kg is greater in city I as compared to city II.
(c) What is the difference of rates for apples in both the cities?
(d) Find the ratio of the cost of mangoes per kg in city I to the cost of mangoes per kg in city II.

True or False

1 x 1 = 1

33.

We can draw exactly one triangle whose angles are 70°, 30° and 80°.

(a)

True

(b)

False

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

9 x 2 = 18

\(\frac { 1 }{ 2 } \)

We have, (-31) x (-100) = 31 x 100 =3100

We have\(\frac{1}{5},\frac{3}{7},\frac{7}{10}\)

LCM of 5, 7 and 10 = 5\(\times\)7\(\times\)2 = 70
On converting given fractions into like fractions, we get
\(\frac { 1 }{ 5 } =\frac { 1\times 4 }{ 5\times 14 } =\frac { 14 }{ 70 } \)
\(\frac { 3 }{ 7 } =\frac { 3\times 10 }{ 7\times 10 } =\frac { 30 }{ 70 } \)
and \(\frac { 7 }{ 10 } =\frac { 7\times 7 }{ 10\times 7 } =\frac { 49 }{ 70 } \)
∵ 14 < 30 < 49 [numerators of fractions]
∴ \(\frac { 14 }{ 70 } <\frac { 30 }{ 70 } <\frac { 40 }{ 70 } \)
Thus,
Hence, descending order of given fractions is \(\frac { 7 }{ 10 } ,\frac { 3 }{ 7 } ,\frac { 1 }{ 5 } \)

(-21) - (30) = (-21) + (-30) = -51 [\(\because\) (- )(-) = (+), higher integer sign is (-)]

(-6) - (-16) = -6 + (16)= + 10 or 10 [\(\because\) (-)( -) = (+), difference = 10, higher integer sign is '+']

14-(-20)=14+(20)=34 [\(\because\) (-)(-) = +]

10 + (-4) = 10 - 4 = +6 or 6 [\(\because\) difference = 6, higher integer sign is' +']

We have, -3 and 4
In this case, firstly we go to (-3) and then move (4) step to the right of (-3).
Thus, we reached to (1), i.e. (-3) + (4) = 1

Given data are 1,4, 6, 5, 6, 2, 3, 4, 5, 3, 2, 4, 5,2 and 3.
On the basis of given data, we can arrange the data in tabular form:

Data	Tally Marks	Frequency
1	I	1
2	III	3
3	III	3
4	III	3
5	III	3
6	II	2
Total		15

Since, mode is the observation having highest frequency.
Here, 2, 3, 4 and 5 have frequency as 3, so mode of the given data is 2, 3, 4 and 5, because all occur 3 times

Generic 3

5 x 3 = 15

10.

Given, a=18, b=15, c=13
Taking LHS,
a + (b + c) = 18 + (15 + 13) = 18 + 28 = 46
Now, taking RHS,
(a + b) + c = (18 + 15) + 13 = 33 + 13 = 46
Hence, a + (b + c) = (a + b) + C

11.

Given rational numbers are \(\frac { 1 }{ 3 } \) and \(\frac { 1 }{ 4 } \).
LCM of 3 and 4 is 12,
\(\frac { 1\times 4 }{ 3\times 4 } =\frac { 4 }{ 12 } \) and \(\frac { 1\times 3 }{ 4\times 3 } =\frac { 3 }{ 12 } \)
Now, \(\frac { 4\times 10 }{ 12\times 10 } =\frac { 40 }{ 120 } \) and \(\frac { 3\times 10 }{ 12\times 10 } =\frac { 30 }{ 120 } \)
\(\frac { 30 }{ 120 } <\frac { 31 }{ 120 } <\frac { 32 }{ 120 } <\frac { 33 }{ 120 } <\frac { 34 }{ 120 } <\frac { 35 }{ 120 } <.......<\frac { 40 }{ 120 } \)
Hence, 5 rational numbers between \(\frac { 1 }{ 3 } \) and \(\frac { 1 }{ 4 } \) are
\(\frac { 31 }{ 120 } ,\frac { 32 }{ 120 } ,\frac { 33 }{ 120 } ,\frac { 34 }{ 120 } ,\frac { 35 }{ 120 } .\)

12.

If 40 is the pass mark, then students who getting less than 40 marks have failed. Hence, from the bargraph we can see students getting marks between 30-39 have failed i.e. 4 students have failed

13.

On arranging the data in ascending order, we get 12,12,12,13,13,13,13,14,14,14,14,14,15,15,15, 15,15,15,15,15,15,15,16,16,16,16,16,16,17,17, 18, 19
Now, we will arrange it in a table form

Numbers	Occuring Time(Frequency)
12	Three times
13	Four Times
14	Five Times
15	Elevan times
16	Six times
17	Two Times
18	One times
19	One times

It is clear from the table that, 15 occurs the maximum
number of times i.e 11.
Hence, mode of the data is 15.

14.

Here, the sea level is at 0 m and the plane is 5000m above the sea level.
\(\therefore\) Distance between plane and the sea level = 5000 m
Also, the submarine is floating 1200 m below the sea level.
\(\therefore\) Distance between the submarine and the sea level = 1200 m
Hence, the vertical distance between the plane and the submarine = Distance between the plane and the sea level + Distance between the sea level and submarine
= 5000 + 1200 = 6200 m

Multiple Choice Question

5 x 1 = 5

15.

(d)
60°

16.

(d)
2

17.

18.

(b)
27

19.

Fill up / 1 Marks

5 x 1 = 5

20.

()
the hypotenuse

21.

The representation of data with bars of uniform width is called bar graph

22.

()
We have, (- 206) \(\div\) (- 206) = -206/-206 = 1 [ \(\because\) (-a) \(\div\) (-a) = 1 ]

23.

()
(- 3) x (- 9) = 27

24.

\(\frac { -2 }{ 9 } -\frac { 7 }{ 9 } =\frac { (-2+7) }{ 9 } =\frac { -9 }{ 9 } =-1\)

Match the following

5 x 1 = 5

25.

()

26.

()
23

27.

()
60

28.

()
Highest observation - Lowest observation

29.

()
\(\left( \frac { bc }{ ad } \right) \)

Generic 5

3 x 5 = 15

30.

Given, three sides of \(\triangle PQR\) are PQ = 4 cm, QR = 35 cm and PR = 4cm.
To construct a triangle with given sides, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch with given measures marked on it.

Step II Draw a line segment QR = 35 cm.

Step III With Q as centre and radius 4 cm, draw an arc.

Step IV Now, with R as centre and radius 4 cm, draw another arc intersecting the previous arc at P.

Step V Join PQ and PR.

Thus, \(\triangle PQR\) is the required triangle.
Here, PQ = PR = 4 cm, i.e. two sides of \(\triangle PQR\) are equal.
Therefore, \(\triangle PQR\) is an isosceles triangle.

31.

The complete table is shown below

Hudrends (100)	Tens (10)	Ones(1)	Tenths(\(\frac{1}{10}\))	Hudrendths(\(\frac{1}{100}\))	Thousandths(\(\frac{1}{1000}\))	Number
2	5	3	1	4	7	253.147
6	2	9	3	2	1	629.321
0	4	3	1	9	2	043.192
5	1	4	2	5	1	514.251
2	3	6	5	1	2	236.512
7	2	4	5	0	3	724.503
6	1	4	3	2	6	614.326
0	1	0	5	3	0	010.530

32.

After the study of double bar graph,
Cost of apple in city 1=82
Cost of apple in city II = 75
Cost of banana in city I = 45
Cost of banana in city II = 32
Cost of mango in city I = 75
Cost of mango in city II = 60
Cost of watermelon in city I = 20
Cost of watermelon in city II = 26
Cost of cherry in city I = 38
Cost of cherry in city II = 30
(a) Clearly, the double bar graph compares the cost of different fruits per kg in city I and II.
(b) Cost of apple in city I = 82, where in city II is = 75
Cost of Banana in city I = 45, where in city II = 32
Cost of Mango in city 1=75, where in city II = 60
Cost of Cherry in city I = 38, where in city II = 30
Hence, apple, banana, mango, and cherry
prices/costs in city I are greater than city II.
(c) Difference of rates for apples in cities I and II
= 82 -75 =7
(d) The ratio of the cost of mangoes per kg in city I to
the cost of mangoes per kg in city II
=75: 60 = 5: 4