Class 7 Exam  >  Class 7 Notes  >  Mathematics (Maths) Class 7  >  Class 7 Math: CBSE Past Year Paper - 6

Class 7 Math: CBSE Past Year Paper - 6 | Mathematics (Maths) Class 7 PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


S E C T I O N 	 – 	 A 	 	 	 	 [ 1 	 m a r k 	 e a c h ]
1 . 	 W h a t 	 i s 	 t h e 	 d e g r e e 	 o f 	 t h e 	 p o l y n o m i a l 	 x 3 	 y 	 – 	 2 x y 4 	 + 	 1 ?
A n s . 	 D e g r e e 	 = 	 5 	
2 . 	 A 	 d i c e 	 i s 	 t h r o w n 	 a t 	 r a n d o m , 	 f i n d 	 t h e 	 p r o b a b i l i t y 	 o f 	 g e t t i n g 	 a 	 n u m b e r 	 d i v i s i b l e 	 b y 	 3 .
A n s . 	
3 . 	 W r i t e 	 t h e 	 n u m b e r 	 o f 	 s i d e s 	 o f 	 t h e 	 b a s e 	 i n 	 a 	 p e n t a g o n a l 	 p r i s m .
A n s . 	 5
4 . 	 C o m p u t e 	 t h e 	 r a n g e 	 o f 	 4 . 9 , 	 5 . 5 , 	 2 5 . 6 , 	 2 1 . 3 , 	 0 . 6 , 	 2 . 2 , 	 0 . 5 .
A n s . 	 R a n g e 	 = 	 2 5 . 6 	 - 0 . 5 	 = 	 2 5 . 1
S E C T I O N 	 – 	 B 	 [ 2 	 m a r k s 	 e a c h ]
5 . 	 A 	 m a n 	 t r a v e l l e d 	 9 0 	 k m 	 t o 	 t h e 	 n o r t h 	 o f 	 J a i p u r 	 a n d 	 t h e n 	 1 5 0 	 k m 	 t o 	 t h e 	 s o u t h 	 o f 	 i t . 	 H o w
f a r 	 f r o m 	 J a i p u r 	 w a s 	 h e 	 f i n a l l y ?
A n s . 	 N o r t h 	 = 	 + 9 0 	 k m
S o u t h 	 = 	 - 1 5 0 	 k m 	 - - - - - - - ( 0 . 5 )
F i n a l 	 = 	 + 9 0 	 - 1 5 0 	 = 	 - 	 6 0 	 k m 	 - - - - - - - ( 1 )
H e n c e 	 h e 	 i s 	 6 0 	 k m 	 s o u t h 	 f r o m 	 J a i p u r 	 - - - - - - - ( 0 . 5 )
6 . 	 C o m p l e t e 	 t h e 	 f o l l o w i n g 	 t a b l e :
A n s .
( 0 . 5 	 m a r k s 	 e a c h 	 b o x )
Page 2


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


S E C T I O N 	 – 	 A 	 	 	 	 [ 1 	 m a r k 	 e a c h ]
1 . 	 W h a t 	 i s 	 t h e 	 d e g r e e 	 o f 	 t h e 	 p o l y n o m i a l 	 x 3 	 y 	 – 	 2 x y 4 	 + 	 1 ?
A n s . 	 D e g r e e 	 = 	 5 	
2 . 	 A 	 d i c e 	 i s 	 t h r o w n 	 a t 	 r a n d o m , 	 f i n d 	 t h e 	 p r o b a b i l i t y 	 o f 	 g e t t i n g 	 a 	 n u m b e r 	 d i v i s i b l e 	 b y 	 3 .
A n s . 	
3 . 	 W r i t e 	 t h e 	 n u m b e r 	 o f 	 s i d e s 	 o f 	 t h e 	 b a s e 	 i n 	 a 	 p e n t a g o n a l 	 p r i s m .
A n s . 	 5
4 . 	 C o m p u t e 	 t h e 	 r a n g e 	 o f 	 4 . 9 , 	 5 . 5 , 	 2 5 . 6 , 	 2 1 . 3 , 	 0 . 6 , 	 2 . 2 , 	 0 . 5 .
A n s . 	 R a n g e 	 = 	 2 5 . 6 	 - 0 . 5 	 = 	 2 5 . 1
S E C T I O N 	 – 	 B 	 [ 2 	 m a r k s 	 e a c h ]
5 . 	 A 	 m a n 	 t r a v e l l e d 	 9 0 	 k m 	 t o 	 t h e 	 n o r t h 	 o f 	 J a i p u r 	 a n d 	 t h e n 	 1 5 0 	 k m 	 t o 	 t h e 	 s o u t h 	 o f 	 i t . 	 H o w
f a r 	 f r o m 	 J a i p u r 	 w a s 	 h e 	 f i n a l l y ?
A n s . 	 N o r t h 	 = 	 + 9 0 	 k m
S o u t h 	 = 	 - 1 5 0 	 k m 	 - - - - - - - ( 0 . 5 )
F i n a l 	 = 	 + 9 0 	 - 1 5 0 	 = 	 - 	 6 0 	 k m 	 - - - - - - - ( 1 )
H e n c e 	 h e 	 i s 	 6 0 	 k m 	 s o u t h 	 f r o m 	 J a i p u r 	 - - - - - - - ( 0 . 5 )
6 . 	 C o m p l e t e 	 t h e 	 f o l l o w i n g 	 t a b l e :
A n s .
( 0 . 5 	 m a r k s 	 e a c h 	 b o x )
7 . 	 I s 	 e v e r y 	 r a t i o n a l 	 n u m b e r 	 a 	 f r a c t i o n ? 	 G i v e 	 a 	 r e a s o n 	 i n 	 s u p p o r t 	 o f 	 y o u r 	 a n s w e r .
A n s . 	 E v e r y 	 r a t i o n a l 	 n u m b e r 	 n e e d 	 n o t 	 b e 	 a 	 f r a c t i o n 	 a s 	 r a t i o n a l 	 n u m b e r s 	 c a n 	 b e 	 m a d e 	 u p 	 o f
i n t e g e r s 	 w h e r e 	 a s 	 f r a c t i o n s 	 a r e 	 m a d e 	 u p 	 o f 	 w h o l e 	 n u m b e r s .
8 . 	 E x p r e s s 	 i n 	 d e c i m a l 	 f o r m . 	 I s 	 i t 	 t e r m i n a t i n g 	 o r 	 n o n - t e r m i n a t i n g ?
A n s . 	 	
I t 	 i s 	 n o n - 	 t e r m i n a t i n g 	 r e p e a t i n g 	 d e c i m a l 	 r e p r e s e n t a t i o n . 	 - - - ( 0 . 5 )
9 . 	 T h i s 	 i s 	 a 	 m a g i c 	 s q u a r e 	 a s 	 t h e 	 s u m 	 i n 	 e a c h 	 r o w , 	 c o l u m n 	 a n d 	 d i a g o n a l 	 i s 	 t h e 	 s a m e . 	 F i n d
t h e 	 s u m 	 o f 	 e a c h 	 r o w , 	 c o l u m n 	 , 	 d i a g o n a l 	 w h e n 	 a = 	 5 . 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	
A n s . 	 S u m 	 = 	 3 a 2 	 - - - - - - - - - - - - - ( 1 )
V a l u e 	 f o r 	 a 	 = 	 5 	 i s 	 7 5 	 - - - - - - - - ( 1 )
1 0 . 	 T h e 	 e x t e r i o r 	 a n g l e 	
A n s .
	
S E C T I O N 	 – 	 C 	 [ 3 	 m a r k s 	 e a c h 	 ]
1 1 . 	 S i m p l i f y :
1 2 	 ÷ 	 [ 1 5 	 – 	 { 9 	 ÷ 	 ( 1 7 	 + 	 3 	 x 	 2 	 – 	 2 0 ) } ]
A n s . 	 1 2 	 ÷ 	 [ 1 5 	 – 	 { 9 	 ÷ 	 ( 1 7 	 + 	 3 	 x 	 2 	 – 	 2 0 ) } ]
S o l v i n g 	 u s i n g 	 B O D M A S 	 a n d 	 g e t t i n g 	 t h e 	 a n s w e r 	 1
1 2 . 	 I f 	 t h e 	 c o s t 	 o f 	 2 0 	 M a t h e m a t i c s 	 b o o k s 	 o f 	 c l a s s 	 V I I 	 i s 	 R s . 3 9 0 . 5 0 , 	 w h a t 	 i s 	 t h e 	 c o s t 	 o f 	 2 	 s u c h
M a t h e m a t i c s 	 b o o k s ?
A n s . 	 C o s t 	 o f 	 2 0 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 3 9 0 . 5 0
C o s t 	 o f 	 1 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 3 9 0 . 5 0 	 ÷ 	 2 0 	 = 	 R s . 	 1 9 . 5 2 5 	 - - - - - - - - - - - - - - ( 1 . 5 )
C o s t 	 o f 	 2 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 1 9 . 5 2 5 	 x 	 2 	 = 	 R s . 	 3 9 . 0 5 	 - - - - - - - - - - - - - ( 1 . 5 )
1 3 . 	 W h a t 	 s h o u l d 	 b e 	 s u b t r a c t e d 	 f r o m 	 x
2
	 + 	 x y 	 + 	 y
2
	 + 	 x 	 + 	 2 y 	 – 	 3 	 t o 	 o b t a i n 	 x 2 	 – 	 3 y 2 	 + 	 4 x y 	 – 	 7 ?
A n s . 	 R e q u i r e d 	 e x p r e s s i o n 	 = 	 x 2 	 + 	 x y 	 + 	 y 2 	 + 	 x 	 + 	 2 y 	 – 	 3 	 – 	 ( x 2 	 – 	 3 y 2 	 + 	 4 x y 	 – 	 7 ) 	 - - - - - - - - - - - - ( 1 )
= 	 x 	 2 	 + 	 x y 	 + 	 y 2 	 + 	 x 	 + 	 2 y 	 – 	 3 	 – x 2 	 + 	 3 y 2 	 - 	 4 x y 	 + 	 7 	 - - - - - - - - - - - - ( 1 )
= 	 - 3 x y 	 + 	 4 y 2 	 + 	 x 	 + 	 2 y 	 + 	 4 	 - - - - - - - - - - - - ( 1 )
1 4 . 	 I n 	 a 	 c l a s s 	 t e s t 	 3 	 m a r k s 	 a r e 	 a w a r d e d 	 f o r 	 e v e r y 	 c o r r e c t 	 a n s w e r 	 a n d 	 ( - 2 ) 	 m a r k s 	 a r e
g i v e n 	 f o r 	 e v e r y 	 i n c o r r e c t 	 a n s w e r 	 a n d 	 0 	 f o r 	 q u e s t i o n s 	 n o t 	 a t t e m p t e d . 	
a . 	 M o h a n 	 g e t s 	 f o u r 	 c o r r e c t 	 a n d 	 s i x 	 i n c o r r e c t 	 a n s w e r s . 	 W h a t 	 i s 	 h i s 	 s c o r e ? 	
b . 	 R a m 	 s c o r e s 	 2 0 	 m a r k s , 	 i f 	 h e 	 g e t s 1 2 	 c o r r e c t 	 a n s w e r s 	 t h e n , 	 f i n d 	 t h e 	 n u m b e r 	 o f 	 i n c o r r e c t
a n s w e r s ? 	
c . 	 H e e n a 	 s c o r e s 	 - 5 	 m a r k s , 	 i f 	 s h e 	 g e t s 	 7 	 c o r r e c t 	 a n s w e r s 	 t h e n , 	 f i n d 	 t h e 	 n u m b e r 	 o f
i n c o r r e c t 	 a n s w e r s ?
A n s . 	 a . 	 M o h a n ’ s 	 s c o r e 	 = 	 4 	 x 	 3 	 + 	 6 	 x 	 ( - 2 ) 	 = 	 1 2 	 - 1 2 	 = 	 0 	 - - - - - - - - - - - - ( 1 )
b . 	 2 0 	 = 	 1 2 	 x 	 3 	 + 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 2 0 	 – 	 3 6 	 = 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 i n c o r r e c t 	 = 	 8 	 - - - - - - - - - - - - ( 1 )
c . - 5 	 = 	 7 	 x 	 3 	 + 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 - 5 	 - 2 1 	 = 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 i n c o r r e c t 	 = 	 1 3 	 - - - - - - - - - - - - ( 1 ) 	
1 5 . 	 T h e 	 p r o d u c t 	 o f 	 t w o 	 r a t i o n a l 	 n u m b e r s 	 i s . 	 I f 	 o n e 	 o f 	 t h e m 	 i s , 	 f i n d 	 t h e 	 o t h e r .
A n s .
O t h e r 	 n o . 	 	 	 	 - - ( 1 )
1 6 . 	 I d e n t i f y 	 t h e 	 s o l i d 	 s h a p e : 	
a . 	 I 	 h a v e 	 o n l y 	 o n e 	 c u r v e d 	 f a c e 	 w i t h o u t 	 a n y 	 e d g e s 	 o r 	 v e r t i c e s . 	
b . 	 I 	 h a v e 	 f i v e 	 f a c e s 	 o f 	 t h e 	 s a m e 	 s h a p e 	 a n d 	 o n e 	 f a c e 	 i s 	 d i f f e r e n t . 	
c . 	 M y 	 t w o 	 f a c e s 	 a r e 	 c o n g r u e n t 	 t r i a n g l e s 	 a n d 	 o t h e r s 	 a r e 	 a l l 	 r e c t a n g l e s .
A n s . 	 a . 	 S p h e r e 	 - - - - - ( 1 )
b . 	 P e n t a g o n a l 	 p y r a m i d 	 - - - - - ( 1 )
c . 	 T r i a n g u l a r 	 p r i s m 	 - - - - - ( 1 )
Page 4


S E C T I O N 	 – 	 A 	 	 	 	 [ 1 	 m a r k 	 e a c h ]
1 . 	 W h a t 	 i s 	 t h e 	 d e g r e e 	 o f 	 t h e 	 p o l y n o m i a l 	 x 3 	 y 	 – 	 2 x y 4 	 + 	 1 ?
A n s . 	 D e g r e e 	 = 	 5 	
2 . 	 A 	 d i c e 	 i s 	 t h r o w n 	 a t 	 r a n d o m , 	 f i n d 	 t h e 	 p r o b a b i l i t y 	 o f 	 g e t t i n g 	 a 	 n u m b e r 	 d i v i s i b l e 	 b y 	 3 .
A n s . 	
3 . 	 W r i t e 	 t h e 	 n u m b e r 	 o f 	 s i d e s 	 o f 	 t h e 	 b a s e 	 i n 	 a 	 p e n t a g o n a l 	 p r i s m .
A n s . 	 5
4 . 	 C o m p u t e 	 t h e 	 r a n g e 	 o f 	 4 . 9 , 	 5 . 5 , 	 2 5 . 6 , 	 2 1 . 3 , 	 0 . 6 , 	 2 . 2 , 	 0 . 5 .
A n s . 	 R a n g e 	 = 	 2 5 . 6 	 - 0 . 5 	 = 	 2 5 . 1
S E C T I O N 	 – 	 B 	 [ 2 	 m a r k s 	 e a c h ]
5 . 	 A 	 m a n 	 t r a v e l l e d 	 9 0 	 k m 	 t o 	 t h e 	 n o r t h 	 o f 	 J a i p u r 	 a n d 	 t h e n 	 1 5 0 	 k m 	 t o 	 t h e 	 s o u t h 	 o f 	 i t . 	 H o w
f a r 	 f r o m 	 J a i p u r 	 w a s 	 h e 	 f i n a l l y ?
A n s . 	 N o r t h 	 = 	 + 9 0 	 k m
S o u t h 	 = 	 - 1 5 0 	 k m 	 - - - - - - - ( 0 . 5 )
F i n a l 	 = 	 + 9 0 	 - 1 5 0 	 = 	 - 	 6 0 	 k m 	 - - - - - - - ( 1 )
H e n c e 	 h e 	 i s 	 6 0 	 k m 	 s o u t h 	 f r o m 	 J a i p u r 	 - - - - - - - ( 0 . 5 )
6 . 	 C o m p l e t e 	 t h e 	 f o l l o w i n g 	 t a b l e :
A n s .
( 0 . 5 	 m a r k s 	 e a c h 	 b o x )
7 . 	 I s 	 e v e r y 	 r a t i o n a l 	 n u m b e r 	 a 	 f r a c t i o n ? 	 G i v e 	 a 	 r e a s o n 	 i n 	 s u p p o r t 	 o f 	 y o u r 	 a n s w e r .
A n s . 	 E v e r y 	 r a t i o n a l 	 n u m b e r 	 n e e d 	 n o t 	 b e 	 a 	 f r a c t i o n 	 a s 	 r a t i o n a l 	 n u m b e r s 	 c a n 	 b e 	 m a d e 	 u p 	 o f
i n t e g e r s 	 w h e r e 	 a s 	 f r a c t i o n s 	 a r e 	 m a d e 	 u p 	 o f 	 w h o l e 	 n u m b e r s .
8 . 	 E x p r e s s 	 i n 	 d e c i m a l 	 f o r m . 	 I s 	 i t 	 t e r m i n a t i n g 	 o r 	 n o n - t e r m i n a t i n g ?
A n s . 	 	
I t 	 i s 	 n o n - 	 t e r m i n a t i n g 	 r e p e a t i n g 	 d e c i m a l 	 r e p r e s e n t a t i o n . 	 - - - ( 0 . 5 )
9 . 	 T h i s 	 i s 	 a 	 m a g i c 	 s q u a r e 	 a s 	 t h e 	 s u m 	 i n 	 e a c h 	 r o w , 	 c o l u m n 	 a n d 	 d i a g o n a l 	 i s 	 t h e 	 s a m e . 	 F i n d
t h e 	 s u m 	 o f 	 e a c h 	 r o w , 	 c o l u m n 	 , 	 d i a g o n a l 	 w h e n 	 a = 	 5 . 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	
A n s . 	 S u m 	 = 	 3 a 2 	 - - - - - - - - - - - - - ( 1 )
V a l u e 	 f o r 	 a 	 = 	 5 	 i s 	 7 5 	 - - - - - - - - ( 1 )
1 0 . 	 T h e 	 e x t e r i o r 	 a n g l e 	
A n s .
	
S E C T I O N 	 – 	 C 	 [ 3 	 m a r k s 	 e a c h 	 ]
1 1 . 	 S i m p l i f y :
1 2 	 ÷ 	 [ 1 5 	 – 	 { 9 	 ÷ 	 ( 1 7 	 + 	 3 	 x 	 2 	 – 	 2 0 ) } ]
A n s . 	 1 2 	 ÷ 	 [ 1 5 	 – 	 { 9 	 ÷ 	 ( 1 7 	 + 	 3 	 x 	 2 	 – 	 2 0 ) } ]
S o l v i n g 	 u s i n g 	 B O D M A S 	 a n d 	 g e t t i n g 	 t h e 	 a n s w e r 	 1
1 2 . 	 I f 	 t h e 	 c o s t 	 o f 	 2 0 	 M a t h e m a t i c s 	 b o o k s 	 o f 	 c l a s s 	 V I I 	 i s 	 R s . 3 9 0 . 5 0 , 	 w h a t 	 i s 	 t h e 	 c o s t 	 o f 	 2 	 s u c h
M a t h e m a t i c s 	 b o o k s ?
A n s . 	 C o s t 	 o f 	 2 0 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 3 9 0 . 5 0
C o s t 	 o f 	 1 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 3 9 0 . 5 0 	 ÷ 	 2 0 	 = 	 R s . 	 1 9 . 5 2 5 	 - - - - - - - - - - - - - - ( 1 . 5 )
C o s t 	 o f 	 2 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 1 9 . 5 2 5 	 x 	 2 	 = 	 R s . 	 3 9 . 0 5 	 - - - - - - - - - - - - - ( 1 . 5 )
1 3 . 	 W h a t 	 s h o u l d 	 b e 	 s u b t r a c t e d 	 f r o m 	 x
2
	 + 	 x y 	 + 	 y
2
	 + 	 x 	 + 	 2 y 	 – 	 3 	 t o 	 o b t a i n 	 x 2 	 – 	 3 y 2 	 + 	 4 x y 	 – 	 7 ?
A n s . 	 R e q u i r e d 	 e x p r e s s i o n 	 = 	 x 2 	 + 	 x y 	 + 	 y 2 	 + 	 x 	 + 	 2 y 	 – 	 3 	 – 	 ( x 2 	 – 	 3 y 2 	 + 	 4 x y 	 – 	 7 ) 	 - - - - - - - - - - - - ( 1 )
= 	 x 	 2 	 + 	 x y 	 + 	 y 2 	 + 	 x 	 + 	 2 y 	 – 	 3 	 – x 2 	 + 	 3 y 2 	 - 	 4 x y 	 + 	 7 	 - - - - - - - - - - - - ( 1 )
= 	 - 3 x y 	 + 	 4 y 2 	 + 	 x 	 + 	 2 y 	 + 	 4 	 - - - - - - - - - - - - ( 1 )
1 4 . 	 I n 	 a 	 c l a s s 	 t e s t 	 3 	 m a r k s 	 a r e 	 a w a r d e d 	 f o r 	 e v e r y 	 c o r r e c t 	 a n s w e r 	 a n d 	 ( - 2 ) 	 m a r k s 	 a r e
g i v e n 	 f o r 	 e v e r y 	 i n c o r r e c t 	 a n s w e r 	 a n d 	 0 	 f o r 	 q u e s t i o n s 	 n o t 	 a t t e m p t e d . 	
a . 	 M o h a n 	 g e t s 	 f o u r 	 c o r r e c t 	 a n d 	 s i x 	 i n c o r r e c t 	 a n s w e r s . 	 W h a t 	 i s 	 h i s 	 s c o r e ? 	
b . 	 R a m 	 s c o r e s 	 2 0 	 m a r k s , 	 i f 	 h e 	 g e t s 1 2 	 c o r r e c t 	 a n s w e r s 	 t h e n , 	 f i n d 	 t h e 	 n u m b e r 	 o f 	 i n c o r r e c t
a n s w e r s ? 	
c . 	 H e e n a 	 s c o r e s 	 - 5 	 m a r k s , 	 i f 	 s h e 	 g e t s 	 7 	 c o r r e c t 	 a n s w e r s 	 t h e n , 	 f i n d 	 t h e 	 n u m b e r 	 o f
i n c o r r e c t 	 a n s w e r s ?
A n s . 	 a . 	 M o h a n ’ s 	 s c o r e 	 = 	 4 	 x 	 3 	 + 	 6 	 x 	 ( - 2 ) 	 = 	 1 2 	 - 1 2 	 = 	 0 	 - - - - - - - - - - - - ( 1 )
b . 	 2 0 	 = 	 1 2 	 x 	 3 	 + 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 2 0 	 – 	 3 6 	 = 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 i n c o r r e c t 	 = 	 8 	 - - - - - - - - - - - - ( 1 )
c . - 5 	 = 	 7 	 x 	 3 	 + 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 - 5 	 - 2 1 	 = 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 i n c o r r e c t 	 = 	 1 3 	 - - - - - - - - - - - - ( 1 ) 	
1 5 . 	 T h e 	 p r o d u c t 	 o f 	 t w o 	 r a t i o n a l 	 n u m b e r s 	 i s . 	 I f 	 o n e 	 o f 	 t h e m 	 i s , 	 f i n d 	 t h e 	 o t h e r .
A n s .
O t h e r 	 n o . 	 	 	 	 - - ( 1 )
1 6 . 	 I d e n t i f y 	 t h e 	 s o l i d 	 s h a p e : 	
a . 	 I 	 h a v e 	 o n l y 	 o n e 	 c u r v e d 	 f a c e 	 w i t h o u t 	 a n y 	 e d g e s 	 o r 	 v e r t i c e s . 	
b . 	 I 	 h a v e 	 f i v e 	 f a c e s 	 o f 	 t h e 	 s a m e 	 s h a p e 	 a n d 	 o n e 	 f a c e 	 i s 	 d i f f e r e n t . 	
c . 	 M y 	 t w o 	 f a c e s 	 a r e 	 c o n g r u e n t 	 t r i a n g l e s 	 a n d 	 o t h e r s 	 a r e 	 a l l 	 r e c t a n g l e s .
A n s . 	 a . 	 S p h e r e 	 - - - - - ( 1 )
b . 	 P e n t a g o n a l 	 p y r a m i d 	 - - - - - ( 1 )
c . 	 T r i a n g u l a r 	 p r i s m 	 - - - - - ( 1 )
1 7 . 	 C a l c u l a t e 	 t h e 	 m e a n 	 o f 	 t h e 	 f o l l o w i n g 	 d a t a :
A n s . 	 C a l c u l a t i n g
	
1 8 . 	 F i n d 	 t h e 	 m e d i a n 	 o f 	 a l l 	 t h e 	 p r i m e 	 n u m b e r s 	 l y i n g 	 b e t w e e n 	 1 	 a n d 	 5 0 .
A n s . 	 P r i m e 	 n u m b e r s 	 l y i n g 	 b e t w e e n 	 1 	 a n d 	 5 0 	 = 	 2 , 	 3 , 	 5 , 	 7 , 	 1 1 , 	 1 3 , 	 1 7 , 	 1 9 , 	 2 3 , 	 2 9 , 	 3 1 , 	 3 7 , 	 4 1 , 	 4 3 , 	 4 7
N 	 = 	 1 5 	 ( o d d )
M e d i a n 	 = 	 o b s e r v a t i o n 	 = 	 8 t h 	 o b s e r v a t i o n 	 = 	 1 9
1 9 . 	 I s 	 l ? 	 J u s t i f y 	 y o u r 	 a n s w e r .
A n s . 	 V e r t i c a l l y 	 o p p o s i t e 	 a n g l e 	 o f 	 	 i s 	 t h e 	 c o - 	 i n t e r i o r 	 a n g l e 	 o f 	 	 a n d 	 t h e i r 	 s u m 	 i s 	 1 8 0 0 	 - -
- - ( 2 ) 	 S i n c e 	 t h e 	 s u m 	 o f 	 c o - i n t e r i o r 	 a n g l e s 	 o n 	 t h e 	 s a m e 	 s i d e 	 o f 	 t h e 	 t r a n s v e r s a l 	 i s 	 1 8 0 0 	 h e n c e
l i n e s 	 a r e 	 p a r a l l e l . 	 - - - - ( 1 )
2 0 . 	 A 	 M a t h e m a t i c s 	 t e a c h e r 	 w a n t s 	 t o 	 s e e 	 w h e t h e r 	 t h e 	 n e w 	 t e c h n i q u e 	 o f 	 t e a c h i n g , 	 w h i c h
s h e 	 a p p l i e d 	 a f t e r 	 h a l f 	 y e a r l y 	 w a s 	 e f f e c t i v e 	 o r 	 n o t . 	 S h e 	 t a k e s 	 t h e 	 m a r k s 	 o f 	 5 	 c h i l d r e n 	 i n
t h e 	 h a l f 	 y e a r l y 	 t e s t 	 ( o u t 	 o f 	 5 0 ) 	 a n d 	 t h e n 	 t h e 	 a n n u a l 	 t e s t 	 ( o u t 	 o f 	 5 0 ) 	 a n d 	 r e c o r d s 	 t h e
f o l l o w i n g 	 s c o r e s .
D r a w 	 a 	 d o u b l e 	 b a r 	 g r a p h . 	 D o 	 y o u 	 t h i n k 	 h e r 	 n e w 	 t e c h n i q u e 	 h a s 	 i m p r o v e d 	 t h e 	 r e s u l t 	 o f
t h e 	 s t u d e n t s ?
Page 5


S E C T I O N 	 – 	 A 	 	 	 	 [ 1 	 m a r k 	 e a c h ]
1 . 	 W h a t 	 i s 	 t h e 	 d e g r e e 	 o f 	 t h e 	 p o l y n o m i a l 	 x 3 	 y 	 – 	 2 x y 4 	 + 	 1 ?
A n s . 	 D e g r e e 	 = 	 5 	
2 . 	 A 	 d i c e 	 i s 	 t h r o w n 	 a t 	 r a n d o m , 	 f i n d 	 t h e 	 p r o b a b i l i t y 	 o f 	 g e t t i n g 	 a 	 n u m b e r 	 d i v i s i b l e 	 b y 	 3 .
A n s . 	
3 . 	 W r i t e 	 t h e 	 n u m b e r 	 o f 	 s i d e s 	 o f 	 t h e 	 b a s e 	 i n 	 a 	 p e n t a g o n a l 	 p r i s m .
A n s . 	 5
4 . 	 C o m p u t e 	 t h e 	 r a n g e 	 o f 	 4 . 9 , 	 5 . 5 , 	 2 5 . 6 , 	 2 1 . 3 , 	 0 . 6 , 	 2 . 2 , 	 0 . 5 .
A n s . 	 R a n g e 	 = 	 2 5 . 6 	 - 0 . 5 	 = 	 2 5 . 1
S E C T I O N 	 – 	 B 	 [ 2 	 m a r k s 	 e a c h ]
5 . 	 A 	 m a n 	 t r a v e l l e d 	 9 0 	 k m 	 t o 	 t h e 	 n o r t h 	 o f 	 J a i p u r 	 a n d 	 t h e n 	 1 5 0 	 k m 	 t o 	 t h e 	 s o u t h 	 o f 	 i t . 	 H o w
f a r 	 f r o m 	 J a i p u r 	 w a s 	 h e 	 f i n a l l y ?
A n s . 	 N o r t h 	 = 	 + 9 0 	 k m
S o u t h 	 = 	 - 1 5 0 	 k m 	 - - - - - - - ( 0 . 5 )
F i n a l 	 = 	 + 9 0 	 - 1 5 0 	 = 	 - 	 6 0 	 k m 	 - - - - - - - ( 1 )
H e n c e 	 h e 	 i s 	 6 0 	 k m 	 s o u t h 	 f r o m 	 J a i p u r 	 - - - - - - - ( 0 . 5 )
6 . 	 C o m p l e t e 	 t h e 	 f o l l o w i n g 	 t a b l e :
A n s .
( 0 . 5 	 m a r k s 	 e a c h 	 b o x )
7 . 	 I s 	 e v e r y 	 r a t i o n a l 	 n u m b e r 	 a 	 f r a c t i o n ? 	 G i v e 	 a 	 r e a s o n 	 i n 	 s u p p o r t 	 o f 	 y o u r 	 a n s w e r .
A n s . 	 E v e r y 	 r a t i o n a l 	 n u m b e r 	 n e e d 	 n o t 	 b e 	 a 	 f r a c t i o n 	 a s 	 r a t i o n a l 	 n u m b e r s 	 c a n 	 b e 	 m a d e 	 u p 	 o f
i n t e g e r s 	 w h e r e 	 a s 	 f r a c t i o n s 	 a r e 	 m a d e 	 u p 	 o f 	 w h o l e 	 n u m b e r s .
8 . 	 E x p r e s s 	 i n 	 d e c i m a l 	 f o r m . 	 I s 	 i t 	 t e r m i n a t i n g 	 o r 	 n o n - t e r m i n a t i n g ?
A n s . 	 	
I t 	 i s 	 n o n - 	 t e r m i n a t i n g 	 r e p e a t i n g 	 d e c i m a l 	 r e p r e s e n t a t i o n . 	 - - - ( 0 . 5 )
9 . 	 T h i s 	 i s 	 a 	 m a g i c 	 s q u a r e 	 a s 	 t h e 	 s u m 	 i n 	 e a c h 	 r o w , 	 c o l u m n 	 a n d 	 d i a g o n a l 	 i s 	 t h e 	 s a m e . 	 F i n d
t h e 	 s u m 	 o f 	 e a c h 	 r o w , 	 c o l u m n 	 , 	 d i a g o n a l 	 w h e n 	 a = 	 5 . 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	
A n s . 	 S u m 	 = 	 3 a 2 	 - - - - - - - - - - - - - ( 1 )
V a l u e 	 f o r 	 a 	 = 	 5 	 i s 	 7 5 	 - - - - - - - - ( 1 )
1 0 . 	 T h e 	 e x t e r i o r 	 a n g l e 	
A n s .
	
S E C T I O N 	 – 	 C 	 [ 3 	 m a r k s 	 e a c h 	 ]
1 1 . 	 S i m p l i f y :
1 2 	 ÷ 	 [ 1 5 	 – 	 { 9 	 ÷ 	 ( 1 7 	 + 	 3 	 x 	 2 	 – 	 2 0 ) } ]
A n s . 	 1 2 	 ÷ 	 [ 1 5 	 – 	 { 9 	 ÷ 	 ( 1 7 	 + 	 3 	 x 	 2 	 – 	 2 0 ) } ]
S o l v i n g 	 u s i n g 	 B O D M A S 	 a n d 	 g e t t i n g 	 t h e 	 a n s w e r 	 1
1 2 . 	 I f 	 t h e 	 c o s t 	 o f 	 2 0 	 M a t h e m a t i c s 	 b o o k s 	 o f 	 c l a s s 	 V I I 	 i s 	 R s . 3 9 0 . 5 0 , 	 w h a t 	 i s 	 t h e 	 c o s t 	 o f 	 2 	 s u c h
M a t h e m a t i c s 	 b o o k s ?
A n s . 	 C o s t 	 o f 	 2 0 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 3 9 0 . 5 0
C o s t 	 o f 	 1 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 3 9 0 . 5 0 	 ÷ 	 2 0 	 = 	 R s . 	 1 9 . 5 2 5 	 - - - - - - - - - - - - - - ( 1 . 5 )
C o s t 	 o f 	 2 	 M a t h e m a t i c s 	 b o o k s 	 = 	 R s . 	 1 9 . 5 2 5 	 x 	 2 	 = 	 R s . 	 3 9 . 0 5 	 - - - - - - - - - - - - - ( 1 . 5 )
1 3 . 	 W h a t 	 s h o u l d 	 b e 	 s u b t r a c t e d 	 f r o m 	 x
2
	 + 	 x y 	 + 	 y
2
	 + 	 x 	 + 	 2 y 	 – 	 3 	 t o 	 o b t a i n 	 x 2 	 – 	 3 y 2 	 + 	 4 x y 	 – 	 7 ?
A n s . 	 R e q u i r e d 	 e x p r e s s i o n 	 = 	 x 2 	 + 	 x y 	 + 	 y 2 	 + 	 x 	 + 	 2 y 	 – 	 3 	 – 	 ( x 2 	 – 	 3 y 2 	 + 	 4 x y 	 – 	 7 ) 	 - - - - - - - - - - - - ( 1 )
= 	 x 	 2 	 + 	 x y 	 + 	 y 2 	 + 	 x 	 + 	 2 y 	 – 	 3 	 – x 2 	 + 	 3 y 2 	 - 	 4 x y 	 + 	 7 	 - - - - - - - - - - - - ( 1 )
= 	 - 3 x y 	 + 	 4 y 2 	 + 	 x 	 + 	 2 y 	 + 	 4 	 - - - - - - - - - - - - ( 1 )
1 4 . 	 I n 	 a 	 c l a s s 	 t e s t 	 3 	 m a r k s 	 a r e 	 a w a r d e d 	 f o r 	 e v e r y 	 c o r r e c t 	 a n s w e r 	 a n d 	 ( - 2 ) 	 m a r k s 	 a r e
g i v e n 	 f o r 	 e v e r y 	 i n c o r r e c t 	 a n s w e r 	 a n d 	 0 	 f o r 	 q u e s t i o n s 	 n o t 	 a t t e m p t e d . 	
a . 	 M o h a n 	 g e t s 	 f o u r 	 c o r r e c t 	 a n d 	 s i x 	 i n c o r r e c t 	 a n s w e r s . 	 W h a t 	 i s 	 h i s 	 s c o r e ? 	
b . 	 R a m 	 s c o r e s 	 2 0 	 m a r k s , 	 i f 	 h e 	 g e t s 1 2 	 c o r r e c t 	 a n s w e r s 	 t h e n , 	 f i n d 	 t h e 	 n u m b e r 	 o f 	 i n c o r r e c t
a n s w e r s ? 	
c . 	 H e e n a 	 s c o r e s 	 - 5 	 m a r k s , 	 i f 	 s h e 	 g e t s 	 7 	 c o r r e c t 	 a n s w e r s 	 t h e n , 	 f i n d 	 t h e 	 n u m b e r 	 o f
i n c o r r e c t 	 a n s w e r s ?
A n s . 	 a . 	 M o h a n ’ s 	 s c o r e 	 = 	 4 	 x 	 3 	 + 	 6 	 x 	 ( - 2 ) 	 = 	 1 2 	 - 1 2 	 = 	 0 	 - - - - - - - - - - - - ( 1 )
b . 	 2 0 	 = 	 1 2 	 x 	 3 	 + 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 2 0 	 – 	 3 6 	 = 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 i n c o r r e c t 	 = 	 8 	 - - - - - - - - - - - - ( 1 )
c . - 5 	 = 	 7 	 x 	 3 	 + 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 - 5 	 - 2 1 	 = 	 i n c o r r e c t 	 x 	 ( - 2 ) 	 = > 	 i n c o r r e c t 	 = 	 1 3 	 - - - - - - - - - - - - ( 1 ) 	
1 5 . 	 T h e 	 p r o d u c t 	 o f 	 t w o 	 r a t i o n a l 	 n u m b e r s 	 i s . 	 I f 	 o n e 	 o f 	 t h e m 	 i s , 	 f i n d 	 t h e 	 o t h e r .
A n s .
O t h e r 	 n o . 	 	 	 	 - - ( 1 )
1 6 . 	 I d e n t i f y 	 t h e 	 s o l i d 	 s h a p e : 	
a . 	 I 	 h a v e 	 o n l y 	 o n e 	 c u r v e d 	 f a c e 	 w i t h o u t 	 a n y 	 e d g e s 	 o r 	 v e r t i c e s . 	
b . 	 I 	 h a v e 	 f i v e 	 f a c e s 	 o f 	 t h e 	 s a m e 	 s h a p e 	 a n d 	 o n e 	 f a c e 	 i s 	 d i f f e r e n t . 	
c . 	 M y 	 t w o 	 f a c e s 	 a r e 	 c o n g r u e n t 	 t r i a n g l e s 	 a n d 	 o t h e r s 	 a r e 	 a l l 	 r e c t a n g l e s .
A n s . 	 a . 	 S p h e r e 	 - - - - - ( 1 )
b . 	 P e n t a g o n a l 	 p y r a m i d 	 - - - - - ( 1 )
c . 	 T r i a n g u l a r 	 p r i s m 	 - - - - - ( 1 )
1 7 . 	 C a l c u l a t e 	 t h e 	 m e a n 	 o f 	 t h e 	 f o l l o w i n g 	 d a t a :
A n s . 	 C a l c u l a t i n g
	
1 8 . 	 F i n d 	 t h e 	 m e d i a n 	 o f 	 a l l 	 t h e 	 p r i m e 	 n u m b e r s 	 l y i n g 	 b e t w e e n 	 1 	 a n d 	 5 0 .
A n s . 	 P r i m e 	 n u m b e r s 	 l y i n g 	 b e t w e e n 	 1 	 a n d 	 5 0 	 = 	 2 , 	 3 , 	 5 , 	 7 , 	 1 1 , 	 1 3 , 	 1 7 , 	 1 9 , 	 2 3 , 	 2 9 , 	 3 1 , 	 3 7 , 	 4 1 , 	 4 3 , 	 4 7
N 	 = 	 1 5 	 ( o d d )
M e d i a n 	 = 	 o b s e r v a t i o n 	 = 	 8 t h 	 o b s e r v a t i o n 	 = 	 1 9
1 9 . 	 I s 	 l ? 	 J u s t i f y 	 y o u r 	 a n s w e r .
A n s . 	 V e r t i c a l l y 	 o p p o s i t e 	 a n g l e 	 o f 	 	 i s 	 t h e 	 c o - 	 i n t e r i o r 	 a n g l e 	 o f 	 	 a n d 	 t h e i r 	 s u m 	 i s 	 1 8 0 0 	 - -
- - ( 2 ) 	 S i n c e 	 t h e 	 s u m 	 o f 	 c o - i n t e r i o r 	 a n g l e s 	 o n 	 t h e 	 s a m e 	 s i d e 	 o f 	 t h e 	 t r a n s v e r s a l 	 i s 	 1 8 0 0 	 h e n c e
l i n e s 	 a r e 	 p a r a l l e l . 	 - - - - ( 1 )
2 0 . 	 A 	 M a t h e m a t i c s 	 t e a c h e r 	 w a n t s 	 t o 	 s e e 	 w h e t h e r 	 t h e 	 n e w 	 t e c h n i q u e 	 o f 	 t e a c h i n g , 	 w h i c h
s h e 	 a p p l i e d 	 a f t e r 	 h a l f 	 y e a r l y 	 w a s 	 e f f e c t i v e 	 o r 	 n o t . 	 S h e 	 t a k e s 	 t h e 	 m a r k s 	 o f 	 5 	 c h i l d r e n 	 i n
t h e 	 h a l f 	 y e a r l y 	 t e s t 	 ( o u t 	 o f 	 5 0 ) 	 a n d 	 t h e n 	 t h e 	 a n n u a l 	 t e s t 	 ( o u t 	 o f 	 5 0 ) 	 a n d 	 r e c o r d s 	 t h e
f o l l o w i n g 	 s c o r e s .
D r a w 	 a 	 d o u b l e 	 b a r 	 g r a p h . 	 D o 	 y o u 	 t h i n k 	 h e r 	 n e w 	 t e c h n i q u e 	 h a s 	 i m p r o v e d 	 t h e 	 r e s u l t 	 o f
t h e 	 s t u d e n t s ?
A n s . 	 2 0 . 	 D r a w i n g 	 t h e 	 d o u b l e 	 b a r 	 g r a p h 	 - - - - - - - - - - - ( 2 . 5 )
Y e s , 	 t h e 	 n e w 	 t e c h n i q u e 	 h a s 	 i m p r o v e d 	 t h e 	 r e s u l t 	 o f 	 t h e 	 s t u d e n t s . 	 - - - ( 0 . 5 )
S E C T I O N 	 – 	 D 	 [ 4 	 m a r k s 	 e a c h ]
2 1 . 	 S i m p l i f y : 	
( 4 3 . 7 	 – 	 1 8 . 2 3 	 + 	 9 6 . 3 8 	 – 	 2 7 . 6 2 ) 	 ÷ 	 9
A n s . 	 2 1 . 	 ( 4 3 . 7 	 – 	 1 8 . 2 3 	 + 	 9 6 . 3 8 	 – 	 2 7 . 6 2 ) 	 ÷ 	 9
= 	 ( 1 4 0 . 0 8 	 – 	 4 5 . 8 5 ) 	 ÷ 	 9 	 - - - - - - - - ( 2 )
= 	 9 4 . 2 3 	 ÷ 	 9 	 - - - - - - - - ( 1 )
= 	 1 0 . 4 7 	 - - - - - - - - ( 1 )
2 2 . 	 H o w 	 m a n y 	 i c e 	 c r e a m 	 c o n e s 	 c a n 	 b e 	 m a d e 	 f r o m 	 a 	 4 . 5 1 	 K g 	 b o x 	 o f 	 i c e 	 c r e a m 	 i f 	 e a c h 	 i c e
c r e a m 	 c o n e 	 c o n t a i n s 	 2 2 	 g 	 o f 	 i c e 	 c r e a m ?
A n s . 	 W e i g h t 	 o f 	 b o x 	 o f 	 i c e - c r e a m s 	 = 	 4 . 5 	 1 k g 	 = 	 4 5 1 0 	 g 	 - - - - - - - - - - ( 1 )
W e i g h t 	 o f 	 e a c h 	 i c e 	 c r e a m 	 c o n e 	 = 	 2 2 	 g
N o . 	 o f 	 i c e - c r e a m 	 c o n e s 	 r e q u i r e d 	 = 	 W e i g h t 	 o f 	 b o x 	 o f 	 i c e - c r e a m s 	 ÷ 	 W e i g h t 	 o f 	 e a c h 	 i c e 	 c r e a m
c o n e 	 - ( 1 ) 	 = 	 4 5 1 0 	 ÷ 	 2 2 	 = 	 2 0 5
2 3 . 	 A y u s h i 	 b o u g h t 	 1 5 0 . 2 5 	 k g 	 p u l s e s , 	 5 0 0 . 7 5 	 k g 	 w h e a t 	 a n d 	 2 7 5 . 5 0 	 k g 	 r i c e . 	 S h e 	 s e n t 	 a l l 	 t h e s e
t h i n g s 	 t o 	 t h e 	 f l o o d 	 a f f e c t e d 	 a r e a s 	 i n 	 t h e 	 r a i n y 	 s e a s o n .
a . 	 H o w 	 m u c h 	 t o t a l 	 q u a n t i t y 	 d i d 	 s h e 	 b u y ?
b . 	 H o w 	 m u c h 	 m o r e 	 w h e a t 	 d i d 	 s h e 	 b u y 	 t h a n 	 p u l s e s ?
c . 	 W h a t 	 m o r a l 	 d o 	 y o u 	 l e a r n 	 f r o m 	 i t ?
A n s . 	 a . 	 T o t a l 	 q u a n t i t y 	 s h e 	 b o u g h t 	 = 	 1 5 0 . 2 5 	 + 	 5 0 0 . 7 5 	 + 	 2 7 5 . 5 0 	 k g 	 = 	 9 2 6 . 5 0 	 k g 	 … … … . . ( 2 )
b . 	 M o r e 	 w h e a t 	 s h e 	 b u y 	 t h a n 	 p u l s e s 	 = 	 5 0 0 . 7 5 	 – 	 1 5 0 . 2 5 	 = 	 3 5 0 . 5 	 k g 	 - - - - - - ( 1 )
c . 	 H e l p i n g 	 i n 	 d i s a s t e r 	 … … … . . ( 1 )
2 4 . 	 D i v i d e 	 t h e 	 s u m 	 o f 	 a n d 	 b y 	 t h e i r 	 d i f f e r e n c e .
A n s . 	 4 . 	 S u m 	 = 	 - - - - ( 1 )
D i f f e r e n c e 	 = 	 - - - - ( 1 )
Read More
76 videos|345 docs|39 tests

Top Courses for Class 7

FAQs on Class 7 Math: CBSE Past Year Paper - 6 - Mathematics (Maths) Class 7

1. What are the important topics to study for Class 7 Math CBSE exam?
Ans. The important topics to study for Class 7 Math CBSE exam include integers, fractions and decimals, algebraic expressions, equations and inequalities, geometry, data handling, and practical geometry.
2. How can I score well in the Class 7 Math CBSE exam?
Ans. To score well in the Class 7 Math CBSE exam, you should practice solving a variety of problems from each chapter, understand the concepts thoroughly, revise regularly, and solve previous year papers and sample papers. Additionally, seeking help from your teacher or tutor can also be beneficial.
3. What is the best way to prepare for the Class 7 Math CBSE exam?
Ans. The best way to prepare for the Class 7 Math CBSE exam is to follow a structured study plan. Start by understanding the concepts of each chapter, practice solving problems, revise regularly, and make use of resources such as textbooks, reference books, and online materials. It is also advisable to solve previous year papers and sample papers to get familiar with the exam pattern.
4. 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 tutorials, practice questions, interactive quizzes, and study materials on various educational websites. Some popular online platforms for CBSE exam preparation include Khan Academy, BYJU'S, and TopperLearning.
5. How can I manage my time effectively during the Class 7 Math CBSE exam?
Ans. To manage your time effectively during the Class 7 Math CBSE exam, it is important to practice solving problems within a stipulated time frame. Divide your time equally among the different sections of the exam. Start with the questions you are confident about and then move on to the more challenging ones. Avoid spending too much time on a single question and leave some time for revising your answers.
76 videos|345 docs|39 tests
Download as PDF
Explore Courses for Class 7 exam

Top Courses for Class 7

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

shortcuts and tricks

,

Semester Notes

,

mock tests for examination

,

past year papers

,

Important questions

,

ppt

,

Summary

,

pdf

,

Extra Questions

,

Objective type Questions

,

video lectures

,

practice quizzes

,

Class 7 Math: CBSE Past Year Paper - 6 | Mathematics (Maths) Class 7

,

Sample Paper

,

Free

,

Previous Year Questions with Solutions

,

Viva Questions

,

MCQs

,

Exam

,

Class 7 Math: CBSE Past Year Paper - 6 | Mathematics (Maths) Class 7

,

Class 7 Math: CBSE Past Year Paper - 6 | Mathematics (Maths) Class 7

,

study material

;