SUMMATIVE ASSESMENT
MATHEMATICS
CLASS-7
Time allowed: 2 1/2 hours Max. marks:60
1. General Instructions:
a. All questions are compulsory.
b. The question paper consists of 26 questions divided into four sections – A, B,
C and D.
c. Section A contains 8 questions of 1 mark each, which are multiple choice
type questions, section B contains 6 questions of 2 marks each, section C
contains 8 questions of 3 marks each and section D contains 4 questions of 4
marks each.
d. There is no overall choice in the paper.
e. Use of calculators is not permitted.
Section-A
(Questions from 1-8 are of one marks each)
1. For any integer is a, a/0 is
a) 0
b) 1
c) a
d) Not defined
2. Which of the following is true?
a) 2/5<6/15
b) 3/7>7/8
c) 7/8>3/7
d) 2/5>6/15
3. 7.65/0.15
a) 51
b) 510
c) 5.1
d) 0.51
4. A coin is flipped to decide which team starts the game. The probability that your team will
start is
a) 1/3
b) 0
c) 1/2
d) 1/4
5. 3 - p/2 = -1 then p=
a) -8
b) 8
c) -4
d) 4
6. The angle which is equal to its compliment is
a) 30o
b) 90o
c) 60o
d) 45o
7. The line segment joining the vertex to the midpoint of opposite side of a triangle is called
a) Orthocentre
b) Median
c) Angular bisector
d) Perpendicular bisector
8. From the figure, x =
a) 50o
b) 115o
c) 65o
d) 165o
Section-B
(Questions from 9-14 are of two marks each)
9. Write four pairs of integers (a, b) such that a ÷ b = –2
10. Dinesh bought 3kg 750 gm apples and 2 kg 125 gm of grapes. Anish bought 2 kg 500 gm of
apples and 2 kg and 750 gm of grapes. Who bought less fruits?
11. Find the median of the data: 85, 86, 74, 55, 68, 61, 84, 84, 73, 59, 76, 72, 69, 80 and 75.
12. Solve: 2x - 3(7 - x )= -1
13. Define linear pair of angles with example.
14. Find x and y from the following figure.
Section-C
(Questions from 15-22 are of three marks each)
15. 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 1200 m below the sea level. What is the vertical distance
between them?
16. Vijay studies for 23/5 hours daily. He devotes 11/4 hours of his time for English and
Mathematics. How much time does she devote for other subjects?
17. The rainfall (in mm) for one week of a city is recorded as shown. Find
Month
Monday Tuesday Wednes Thursda Friday
Saturda
day
y
y
Rainfall 10.1
5.8
2.3
0.2
1.9
4.5
Sunday
6.7
(i) Range of the rainfall and
(ii) Mean of the rainfall.
18. Laxmi's father is 49 years old. He is 4 years older than three times Laxmi's age. What is
Laxmi's age?
19. The lengths of two sides of a triangle are 6 cm and 8 cm. Between which two numbers can
length of the third side fall?
20. In the figure shown, x:y=4:5 . Find the measure of <AOD
21. In the given figure, AB perpendicular to CD . Find the complement of <QBP ?
22. A tree is broken at a height of 5 m from the ground and its top touches the ground at a
distance of 12 m from the base of the tree. Find the original height of the tree.
Section-D
(Questions from23-26 are of four marks each)
23. A shopkeeper earns a profit of Re 1 by selling one pen and incurs a loss of 40 paise per
pencil while selling pencils of her old stock.
(i) In a particular month she incurs a loss of Rs 5. In this period, she sold 45 pens.
How many pencils did she sell in this period?
(ii) In the next month she earns neither profit nor loss. If she sold 70 pens, how many pencils
did she sell?
24. Each side of a regular polygon is 3.2 cm in length. The perimeter of the polygon is 38.4cm.
How many sides does the polygon have?
25. Number of children in six different classes is given below. Represent the data on a bar
graph.
Class
6th
7th
8th
9th
10th
No. of childrens
155
120
105
99
121
(i) How would you choose a scale?
(ii) Which class has the maximum number of children? And the minimum?
(iii) Find the ratio of students of class sixth to the students of class eight.
26. People of Sundargram planted a total of 102 trees in the village garden. Some of the trees
were fruit trees. The number of non-fruit trees was two more than three times the number of
fruit trees. What was the number of fruit trees planted?
SUMMATIVE ASSESMENT
MATHEMATICS
CLASS-7
BLUE PRINT
Time allowed: 3 hours Max. marks:60
1. General Instructions:
a. All questions are compulsory.
b. The question paper consists of 26 questions divided into four sections – A, B,
C and D.
c. Section A contains 8 questions of 1 mark each, which are multiple choice
type questions, section B contains 6 questions of 2 marks each, section C
contains 8 questions of 3 marks each and section D contains 4 questions of 4
marks each.
d. There is no overall choice in the paper.
e. Use of calculators is not permitted.
MARKING SCHEME
Q1. Option (d) is correct.
Q2. Consider 7/8 > 3/7
On cross multiplication we get 49 > 24, which is true
Option (c) is correct.
Q3. Option (a) is correct.
Q4.
When a coin is flipped there are 2 chances
One – my team will start the game
Two – other team will start the game
Also both the teams have equal changes.
Hence probability =1/2
Option (c) is correct.
Q5. Given 3 - p/2= -1
-p/2 = -1 -3
-p/2= -4
Hence p=8
Option (b) is correct.
Q6. Let the angle be x
Compliment of x is 90o –x
Given x = 90 – x
2x= 90
Hence x=45
Option (d) is correct.
Q7.
.
Q8.
The line segment joining the vertex to the midpoint of opposite side of a triangle
is known as median of the triangle.
Option (b) is correct.
In a triangle, exterior angle is equal to sum of interior opposite angles.
X+50=115
X =115-50=65
Option (c) is correct.
Section-B
(Questions from 9-14 are of two marks each)
Q9.
Given condition is a/b=-2
Let a=4, b=-2
Then 4/-2=-2
The first pair of integers is (4,-2 )
a=-4, b=2 then -4/2=-2
The second pair of integers is (-4,2 )
Similarly the other two pairs of integers are ( -6,3) and (6,-3 )
Q10. Given Dinesh bought 3kg 750 gm apples and 2 kg 125 gm of grapes
Total weight of fruits = 3kg 750gm + 2kg 125 gm
Total weight of fruits bought by Dinesh =5 kg 875g
Given Anish bought 2 kg 500 gm of apples and 2 kg and 750 gm of grapes
Total weight of fruits = 2 kg 500 gm + 2 kg 750 gm
Total weight of fruits bought by Anish =5 kg 225g
5kg 875g>5kg 225g
Hence Anish bought less fruits.
Q11. Given observations are: 85, 86, 74, 55, 68, 61, 84, 84, 73, 59, 76, 72, 69, 80 and 75.
Increasing order of the given observations: 55, 59, 61, 68, 69, 72, 73, 74, 75, 76, 80, 84,
84, 85, 86
Since there are 15 observations, the middle most observation is the median of the given data.
Hence median = 8th observation =74
Q12. Given 2x -3(7-x )=-1
2x-21+3x=-1
5x=-1+21
5x=20
Hence x=4.
Q13. Linear pair of angles: A pair of adjacent angles whose non-common sides are opposite rays.
Example: 120o and 60oform a linear pair.
Q14. In a triangle, exterior angle is equal to sum of interior opposite angles.
Hence x+50o=120o
.: X=70o
y+120o=180o {Linear pair}
.: y=60o
Section-C
(Questions from 15-22 are of three marks each)
Q15. Given height of plane above sea level = 5000 m
Position of submarine floating below the seal level = 1200 m
The distance between the plane and the submarine
=5000m+1200m=6200m
Q16. Total hours of study in a day=23/5 hours
Time spent to study English and Mathematics =11/4 hours
Time spent to study other subjects
=23/5 - 11/4=37/20 hours
Q17.
(i)
Range = Highest score – lowest score
=10.1-0.2=9.9
Range of the given data is 9.9 mm.
(ii)
Mean =Sum of scores /No. of scores
=10.1+5.8+2.3+0.2+1.9+4.5+6.7 is divided by 7
=31.5/7=4.5mm
Hence the mean of the given data is 4.5 mm
Q18.
Given Laxmi’s Father age=49 years
Let Laxmi’s age be x years
3 times of Laxmi’s age=3x years
Given that her fathers age is 4 more than 3 times of Laxmi’s age
Hence 49=3x+4
x=15 years
Hence Laxmi’s age is 15 years.
Q19.
We know that the sum of two sides of a triangle is always greater than the third side.
Therefore, one-third side has to be less than the sum of the two sides. The third side is
thus less than 8 + 6 = 14 cm.
The side cannot be less than the difference of the two sides. Thus the third side has to be more
than 8 – 6 = 2 cm.
The length of the third side could be any length greater than 2 and less than 14 cm.
Q20.
Given x:y=4:5
Let the angles be x=4k ,y=5k
x+y=180o [Linear pair]
4k+5k=180o
9k=180o
K=20o
.:x=4k=4*20o=80o
<AOD=x [Vertically opposite angles]
<AOD=80o
Q21.
Given AB perpendicular to CD
<ABD =90o
<ABQ + <DBQ=90o
.:x+2x=90o
3x=90o
x=90/3=30o
From the Fig. <ABC=90o
<ABP + <PBC=90o
<ABP + 45o=90o
.: <ABP=45o
<QBP= <QBA + <ABP= 30o+45o
<QBP=75o
Compliment of <QBP= 90o - <QBP
Q22.
Given height of the tree till the broken part = 5 m
The broken part of the tree touches the ground at a distance of 12 m.
Let the height of broken part of the tree where it touches the ground be AC=x
In Right triangle ABC, AC2=AB2=BC2=52+122=25+144=169
AC=1691/2=13m
Hence the height of the broken part of the tree = 13 m
Original height of the tree=AB+AC=5m+13m=18m
Section-D
(Questions from23-26 are of four marks each)
Q23.
(i) Profit earned by selling one pen = Re 1
Profit earned by selling 45 pens = Rs 45, which we denote by + Rs 45
Total loss given = Rs 5, which we denote by – Rs 5
Profit earned + Loss incurred = Total loss
Therefore, Loss incurred = Total Loss – Profit earned
= Rs (– 5 – 45) = Rs (–50) = –5000 paise
Loss incurred by selling one pencil = 40 paise which we write as – 40 paise
So, number of pencils sold = (–5000) ÷ (– 40) = 125 pencils.
(ii) In the next month there is neither profit nor loss.
So, Profit earned + Loss incurred = 0
i.e., Profit earned = – Loss incurred
Now, profit earned by selling 70 pens = Rs 70
Hence, loss incurred by selling pencils = Rs 70 which we indicate by – Rs 70 or-7000 paise.
Total number of pencils sold = (–7000) ÷ (– 40) = 175 pencils.
Q24. Side of regular polygon, s = 3.2 cm
Perimeter of the regular polygon = 38.4 cm
Let the number of sides of the regular polygon be n
Perimeter of regular polygon of ‘n’ sides=n x s
n x 3.2= 38.4
n=38.4/3.2=12
Hence the number of sides of the regular polygon is 12.
Q25.
Class
6th
No. of childrens
7th
8th
9th
10th
155
120
105
99
121
(i) Scale=1cm=20 Children
(ii) Class 5 has the maximum number of children and class 10 has the minimum number of
children.
(iii) Ratio of number of students of class 6 to the number of students of class 8
=155:105=31:21
Q26.
Given total number of trees in Sundargram = 102
Let the number of fruit trees planted be x
Three times the number of fruit trees planted 3x
Given number of non-fruit trees planted is two more than three times the number of fruit trees.
Hence the number of non-fruit trees planted=3x+2
Total number of trees=x+ (3x+2)=4x+2
.:4x+2=102
4x=102-2=100
.: x=100/4=25
25 fruits trees are planted in garden.