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G e n e r a l 	 I n s t r u c t i o n :
T h i s 	 q u e s t i o n 	 p a p e r 	 c o n t a i n s 	 8 	 q u e s t i o n s
A l l 	 t h e 	 q u e s t i o n s 	 a r e 	 c o m p u l s o r y .
S e c t i o n 	 A
1 . 	 I f 	 t w o 	 i n t e g e r s 	 a r e 	 o f 	 u n l i k e 	 s i g n s , 	 t h e n 	 w h a t 	 i s 	 t h e 	 s i g n 	 o f 	 t h e i r 	 p r o d u c t ?
S o l . 	 N e g a t i v e
2 . 	 W h a t 	 i s 	 t h e 	 c h a n c e 	 o f 	 y o u r 	 b e i n g 	 y o u n g e r 	 t o m o r r o w 	 t h a n 	 t o d a y ?
S o l . 	 N o 	 c h a n g e
3 . 	 S i m p l i f y : 	 ( - 8 0 3 ) 	 × 	 1 0 1 	 + 	 8 0 3 	 × 	 ( - 1 ) .
S o l . 	 ( - 8 0 3 ) 	 × 	 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 ( 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 ( 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 1 0 2
- 8 1 9 0 6
4 . 	 A 	 d i e 	 i s 	 t h r o w n 	 2 0 	 t i m e s . 	 T h e 	 n u m b e r 	 4 	 a p p e a r s 	 6 	 t i m e s . 	 W h a t 	 i s 	 t h e 	 p r o b a b i l i t y
o f 	 n o t 	 g e t t i n g 	 n u m b e r 	 4 ?
S o l . 	 P r o b a b i l i t y
P r o b a b i l i t y 	 o f 	 g e t t i n g
P r o b a b i l i t y 	 o f 	 n o t 	 g e t t i n g 	 4 	 	
= 	 0 . 7
Page 2


G e n e r a l 	 I n s t r u c t i o n :
T h i s 	 q u e s t i o n 	 p a p e r 	 c o n t a i n s 	 8 	 q u e s t i o n s
A l l 	 t h e 	 q u e s t i o n s 	 a r e 	 c o m p u l s o r y .
S e c t i o n 	 A
1 . 	 I f 	 t w o 	 i n t e g e r s 	 a r e 	 o f 	 u n l i k e 	 s i g n s , 	 t h e n 	 w h a t 	 i s 	 t h e 	 s i g n 	 o f 	 t h e i r 	 p r o d u c t ?
S o l . 	 N e g a t i v e
2 . 	 W h a t 	 i s 	 t h e 	 c h a n c e 	 o f 	 y o u r 	 b e i n g 	 y o u n g e r 	 t o m o r r o w 	 t h a n 	 t o d a y ?
S o l . 	 N o 	 c h a n g e
3 . 	 S i m p l i f y : 	 ( - 8 0 3 ) 	 × 	 1 0 1 	 + 	 8 0 3 	 × 	 ( - 1 ) .
S o l . 	 ( - 8 0 3 ) 	 × 	 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 ( 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 ( 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 1 0 2
- 8 1 9 0 6
4 . 	 A 	 d i e 	 i s 	 t h r o w n 	 2 0 	 t i m e s . 	 T h e 	 n u m b e r 	 4 	 a p p e a r s 	 6 	 t i m e s . 	 W h a t 	 i s 	 t h e 	 p r o b a b i l i t y
o f 	 n o t 	 g e t t i n g 	 n u m b e r 	 4 ?
S o l . 	 P r o b a b i l i t y
P r o b a b i l i t y 	 o f 	 g e t t i n g
P r o b a b i l i t y 	 o f 	 n o t 	 g e t t i n g 	 4 	 	
= 	 0 . 7
5 . 	 a . 	 T h e 	 m e a n 	 o f 	 1 5 	 o b s e r v a t i o n s 	 i s 	 3 8 . 2 . 	 I t 	 w a s 	 f o u n d 	 l a t e r 	 o n , 	 t h a t 	 t h e 	 n u m b e r 	 2 5 	 w a s
m i s r e a d 	 a s 	 5 2 . 	 F i n d 	 t h e 	 a c t u a l 	 m e a n .
b . 	 1 3 	 s t u d e n t s 	 w e r e 	 g i v e n 	 a 	 c r o s s w o r d 	 p u z z l e . 	 T h e 	 t i m e 	 t a k e n 	 ( i n 	 m i n u t e s ) 	 b y 	 t h e m 	 t o
c o m p l e t e 	 t h e 	 c r o s s w o r d 	 i s 	 g i v e n 	 b e l o w :
1 0 , 	 1 2 , 	 1 5 , 	 8 , 	 1 0 , 	 1 2 , 	 1 2 , 	 8 , 	 1 0 , 	 8 , 	 1 2 , 	 1 1 , 	 1 1
F i n d 	 t h e 	 m e d i a n 	 a n d 	 m o d e 	 f o r 	 t h e 	 a b o v e 	 d a t a .
S o l . 	 ( a ) 	 M e a n 	 = 	 3 8 . 2
N o . 	 o f 	 o b s e r v a t i o n 	 = 	 1 5
C o r r e c t 	 m e a n
B u t 	
C o r r e c t 	 m e a n 	 	
	 	 = 3 6 . 4
( b ) 	 A r r a n g e 	 o b s . 	 I n 	 a s c e n d i n g 	 o r d e r
8 , 	 8 , 	 8 , 	 1 0 , 	 1 0 , 	 1 0 , 	 1 1 , 	 1 1 , 	 1 2 , 	 1 2 , 	 1 2 , 	 1 5
N o . 	 o f 	 o b s e r v a t i o n 	 = 	 1 3 	 ( o d d )
M e a d i a n
= 	 1 1 	 	 M o d e 	 = 	 1 2
6 . 	 S i m p l i f y :
[ 	 - 	 4 2 	 – 	 { 	 2 4 	 ÷ ( 	 - 	 1 5 	 + 	 3 	 ) 	 + 	 ( 7 ) 	 } 	 – 	 ( 	 - 	 3 0 ) 	 ÷ 	 ( - 	 6 ) ]
S o l . 	 [ - 4 2 	 – 	 { 2 4 	 ÷ 	 ( - 1 5 	 + 	 3 ) 	 + 	 7 } 	 – ( - 3 0 ) 	 ÷ 	 ( - 	 6 ) ]
[ - 4 2 	 – 	 { 2 4 	 ÷ 	 ( - 1 2 ) 	 + 	 7 } 	 – 5 ]
[ - 4 2 	 – 	 { - 2 	 + 	 7 } 	 – 5 ]
[ - 4 2 	 – 	 { 5 } 	 – 5 ]
Page 3


G e n e r a l 	 I n s t r u c t i o n :
T h i s 	 q u e s t i o n 	 p a p e r 	 c o n t a i n s 	 8 	 q u e s t i o n s
A l l 	 t h e 	 q u e s t i o n s 	 a r e 	 c o m p u l s o r y .
S e c t i o n 	 A
1 . 	 I f 	 t w o 	 i n t e g e r s 	 a r e 	 o f 	 u n l i k e 	 s i g n s , 	 t h e n 	 w h a t 	 i s 	 t h e 	 s i g n 	 o f 	 t h e i r 	 p r o d u c t ?
S o l . 	 N e g a t i v e
2 . 	 W h a t 	 i s 	 t h e 	 c h a n c e 	 o f 	 y o u r 	 b e i n g 	 y o u n g e r 	 t o m o r r o w 	 t h a n 	 t o d a y ?
S o l . 	 N o 	 c h a n g e
3 . 	 S i m p l i f y : 	 ( - 8 0 3 ) 	 × 	 1 0 1 	 + 	 8 0 3 	 × 	 ( - 1 ) .
S o l . 	 ( - 8 0 3 ) 	 × 	 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 ( 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 ( 1 0 1 	 + 	 1 )
- 8 0 3 	 × 	 1 0 2
- 8 1 9 0 6
4 . 	 A 	 d i e 	 i s 	 t h r o w n 	 2 0 	 t i m e s . 	 T h e 	 n u m b e r 	 4 	 a p p e a r s 	 6 	 t i m e s . 	 W h a t 	 i s 	 t h e 	 p r o b a b i l i t y
o f 	 n o t 	 g e t t i n g 	 n u m b e r 	 4 ?
S o l . 	 P r o b a b i l i t y
P r o b a b i l i t y 	 o f 	 g e t t i n g
P r o b a b i l i t y 	 o f 	 n o t 	 g e t t i n g 	 4 	 	
= 	 0 . 7
5 . 	 a . 	 T h e 	 m e a n 	 o f 	 1 5 	 o b s e r v a t i o n s 	 i s 	 3 8 . 2 . 	 I t 	 w a s 	 f o u n d 	 l a t e r 	 o n , 	 t h a t 	 t h e 	 n u m b e r 	 2 5 	 w a s
m i s r e a d 	 a s 	 5 2 . 	 F i n d 	 t h e 	 a c t u a l 	 m e a n .
b . 	 1 3 	 s t u d e n t s 	 w e r e 	 g i v e n 	 a 	 c r o s s w o r d 	 p u z z l e . 	 T h e 	 t i m e 	 t a k e n 	 ( i n 	 m i n u t e s ) 	 b y 	 t h e m 	 t o
c o m p l e t e 	 t h e 	 c r o s s w o r d 	 i s 	 g i v e n 	 b e l o w :
1 0 , 	 1 2 , 	 1 5 , 	 8 , 	 1 0 , 	 1 2 , 	 1 2 , 	 8 , 	 1 0 , 	 8 , 	 1 2 , 	 1 1 , 	 1 1
F i n d 	 t h e 	 m e d i a n 	 a n d 	 m o d e 	 f o r 	 t h e 	 a b o v e 	 d a t a .
S o l . 	 ( a ) 	 M e a n 	 = 	 3 8 . 2
N o . 	 o f 	 o b s e r v a t i o n 	 = 	 1 5
C o r r e c t 	 m e a n
B u t 	
C o r r e c t 	 m e a n 	 	
	 	 = 3 6 . 4
( b ) 	 A r r a n g e 	 o b s . 	 I n 	 a s c e n d i n g 	 o r d e r
8 , 	 8 , 	 8 , 	 1 0 , 	 1 0 , 	 1 0 , 	 1 1 , 	 1 1 , 	 1 2 , 	 1 2 , 	 1 2 , 	 1 5
N o . 	 o f 	 o b s e r v a t i o n 	 = 	 1 3 	 ( o d d )
M e a d i a n
= 	 1 1 	 	 M o d e 	 = 	 1 2
6 . 	 S i m p l i f y :
[ 	 - 	 4 2 	 – 	 { 	 2 4 	 ÷ ( 	 - 	 1 5 	 + 	 3 	 ) 	 + 	 ( 7 ) 	 } 	 – 	 ( 	 - 	 3 0 ) 	 ÷ 	 ( - 	 6 ) ]
S o l . 	 [ - 4 2 	 – 	 { 2 4 	 ÷ 	 ( - 1 5 	 + 	 3 ) 	 + 	 7 } 	 – ( - 3 0 ) 	 ÷ 	 ( - 	 6 ) ]
[ - 4 2 	 – 	 { 2 4 	 ÷ 	 ( - 1 2 ) 	 + 	 7 } 	 – 5 ]
[ - 4 2 	 – 	 { - 2 	 + 	 7 } 	 – 5 ]
[ - 4 2 	 – 	 { 5 } 	 – 5 ]
[ - 4 2 	 – 	 5 	 – 	 5 ]
- 5
7 . 	 S e e i n g 	 t h e 	 p o o r 	 a n d 	 u n d e r p r i v i l e g e d 	 c h i l d r e n 	 i n 	 t h e 	 n e i g h b o u r h o o d , 	 t h e 	 r e s i d e n t s 	 o f
D w a r k a 	 d e c i d e 	 t o 	 d i s t r i b u t e 	 8 0 	 t a b l e 	 f a n s 	 d u r i n g 	 s u m m e r . 	 T h e 	 c o s t 	 o f 	 o n e 	 f a n 	 i s 	 R s 	 6 5 0 .
F i n d 	 t h e 	 t o t a l 	 c o s t 	 o f 	 f a n s 	 a n d 	 w r i t e 	 t w o 	 v a l u e s 	 t h a t 	 y o u 	 l e a r n 	 f r o m 	 i t .
S o l . 	 C o s t 	 o f 	 1 	 t a b l e 	 f a n 	 = 	 R s 	 6 5 0
C o s t 	 o f 	 8 0 	 t a b l e 	 f a n 	 = R s 	 6 5 0 	 × 	 8 0
T o t a l 	 c o s t 	 o f 	 t a b l e 	 f a n 	 = 	 	 R s 	 5 2 0 0 0
8 . 	 T h e 	 f o l l o w i n g 	 d a t a 	 s h o w s 	 t h e 	 n u m b e r 	 o f 	 c a r s 	 p r o d u c e d 	 i n 	 a 	 f a c t o r y 	 f o r 	 t h e 	 f i r s t 	 s i x
m o n t h s 	 o f 	 t h e 	 y e a r s 	 2 0 1 4 	 a n d 	 2 0 1 5 .
J a n F e b M a r c h A p r i l M a y J u n 	
2 0 1 4 8 6 0 0 9 6 0 0 7 6 0 0 4 4 0 0 3 2 0 0 8 0 0
2 0 1 5 8 6 0 0 9 4 0 0 8 0 0 0 6 4 0 0 4 0 0 0 1 2 0 0
P l o t 	 a 	 d o u b l e 	 b a r 	 g r a p h 	 u s i n g 	 t h e 	 d a t a 	 g i v e n 	 a b o v e .
S o l . 	 C o r r e c t 	 B a r 	 G r a p h
W i t h 	 s c a l e , 	 a x i s ,
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FAQs on Class 7 Math: CBSE Past Year Paper - 7 - Mathematics (Maths) Class 7

1. What are the important topics to study in Class 7 Math for the CBSE exam?
Ans. Some important topics to study in Class 7 Math for the CBSE exam are integers, fractions and decimals, algebraic expressions, geometry, data handling, and practical geometry.
2. How can I prepare for the Class 7 Math CBSE exam effectively?
Ans. To prepare effectively for the Class 7 Math CBSE exam, you can follow these strategies: - Understand the concepts thoroughly by referring to the textbook and taking notes. - Practice solving a variety of problems from different topics. - Solve previous year question papers and sample papers to get familiar with the exam pattern and types of questions. - Seek help from your teacher or classmates if you have any doubts. - Create a study schedule and allocate dedicated time for Math practice every day.
3. Are there any online resources available for Class 7 Math CBSE exam preparation?
Ans. Yes, there are several online resources available for Class 7 Math CBSE exam preparation. You can find video lessons, interactive quizzes, practice tests, and study materials on various educational websites and learning platforms. Some popular online resources include Khan Academy, BYJU'S, Meritnation, and TopperLearning.
4. What is the marking scheme for the Class 7 Math CBSE exam?
Ans. The marking scheme for the Class 7 Math CBSE exam usually consists of two sections - theory and practical. The theory section carries a maximum of 80 marks, while the practical section carries a maximum of 20 marks. The theory section may have multiple-choice questions, short answer questions, and long answer questions, and the practical section may involve solving problems, drawing geometrical figures, or interpreting data.
5. How can I improve my problem-solving skills for the Class 7 Math CBSE exam?
Ans. To improve your problem-solving skills for the Class 7 Math CBSE exam, you can follow these tips: - Understand the problem statement carefully by reading it multiple times. - Identify the given information and what needs to be found. - Break down complex problems into smaller, manageable steps. - Use appropriate mathematical concepts and formulas to solve the problem. - Practice solving different types of problems regularly to enhance your problem-solving abilities. - Review your solutions and learn from any mistakes or errors made.
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