Open App

Class 6 Exam > Class 6 Notes > Class 6 Math: CBSE Past Year Paper - 5

Class 6 Math: CBSE Past Year Paper - 5 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


G e n e r a l 	 I n s t r u c t i o n s :
• 	 T h i s 	 q u e s t i o n 	 p a p e r 	 h a s 	 3 1 	 q u e s t i o n s 	 .
• 	 S e c t i o n 	 A 	 c o n s i s t s 	 o f 	 4 	 q u e s t i o n s 	 o f 	 1 	 m a r k 	 e a c h .
• 	 S e c t i o n 	 B 	 c o n s i s t s 	 o f 	 6 	 q u e s t i o n s 	 o f 	 2 	 m a r k s 	 e a c h .
• 	 S e c t i o n 	 C 	 c o n s i s t s 	 o f 	 1 0 	 q u e s t i o n s 	 o f 	 3 	 m a r k s 	 e a c h .
• 	 S e c t i o n 	 D 	 c o n s i s t s 	 o f 	 1 1 	 q u e s t i o n s 	 o f 	 4 	 m a r k s 	 e a c h .
• 	 A l l 	 q u e s t i o n s 	 a r e 	 c o m p u l s o r y 	 t o 	 a t t e m p t .
• 	 A t t e m p t 	 t h e 	 p a p e r 	 n e a t l y 	 a n d 	 l e g i b l y .
S E C T I O N - A
1 . 	 F i n d 	 t h e 	 e s t i m a t e d 	 d i f f e r e n c e 	 o f 	 4 8 9 3 4 8 	 a n d 	 4 8 3 6 5 .
A n s . 	 4 8 9 3 4 8 	 r o u n d e d 	 o f f 	 t o 	 n e a r e s t 	 t e n 	 t h o u s a n d s 	 = 	 4 9 0 0 0 0
4 8 3 6 5 	 r o u n d e d 	 o f f 	 t o 	 n e a r e s t 	 t e n 	 t h o u s a n d s 	 = 	 5 0 0 0 0
E s t i m a t e d 	 d i f f e r e n c e 	 = 	 4 4 0 0 0 0
2 . 	 I n 	 a 	 p i c t o g r a p h , 	 i f 	 t h e 	 s y m b o l 	 	 r e p r e s e n t s 	 2 0 	 s t u d e n t s , 	 t h e n 	 h o w 	 m a n y 	 s t u d e n t s
w i l l 	 b e 	 r e p r e s e n t e d 	 b y 	 t h e 	 	 s y m b o l ?
A n s . 	 3 5 .
3 . 	 S o l v e 	 a n d 	 w r i t e 	 t h e 	 a n s w e r 	 i n 	 t h e 	 H i n d u 	 A r a b i c 	 N u m e r a l : 	 D 	 - 	 C D X L 	 = 	 _ _ _ _ _ _
A n s . 	 D 	 - 	 C D X L 	 = 	 5 0 0 	 – 	 4 4 0 	 = 	 6 0 	 = 	 L X
4 . 	 H o w 	 m a n y 	 d e g r e e s 	 a r e 	 t h e r e 	 i n 	 t h e 	 a n g l e 	 b e t w e e n 	 t h e 	 h a n d s 	 o f 	 a 	 c l o c k 	 w h e n 	 i t 	 i s
6 	 o ’ c l o c k ?
A n s . 	 1 8 0 	 d e g r e e s 	 = 	 s t r a i g h t 	 a n g l e
Page 2


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


G e n e r a l 	 I n s t r u c t i o n s :
• 	 T h i s 	 q u e s t i o n 	 p a p e r 	 h a s 	 3 1 	 q u e s t i o n s 	 .
• 	 S e c t i o n 	 A 	 c o n s i s t s 	 o f 	 4 	 q u e s t i o n s 	 o f 	 1 	 m a r k 	 e a c h .
• 	 S e c t i o n 	 B 	 c o n s i s t s 	 o f 	 6 	 q u e s t i o n s 	 o f 	 2 	 m a r k s 	 e a c h .
• 	 S e c t i o n 	 C 	 c o n s i s t s 	 o f 	 1 0 	 q u e s t i o n s 	 o f 	 3 	 m a r k s 	 e a c h .
• 	 S e c t i o n 	 D 	 c o n s i s t s 	 o f 	 1 1 	 q u e s t i o n s 	 o f 	 4 	 m a r k s 	 e a c h .
• 	 A l l 	 q u e s t i o n s 	 a r e 	 c o m p u l s o r y 	 t o 	 a t t e m p t .
• 	 A t t e m p t 	 t h e 	 p a p e r 	 n e a t l y 	 a n d 	 l e g i b l y .
S E C T I O N - A
1 . 	 F i n d 	 t h e 	 e s t i m a t e d 	 d i f f e r e n c e 	 o f 	 4 8 9 3 4 8 	 a n d 	 4 8 3 6 5 .
A n s . 	 4 8 9 3 4 8 	 r o u n d e d 	 o f f 	 t o 	 n e a r e s t 	 t e n 	 t h o u s a n d s 	 = 	 4 9 0 0 0 0
4 8 3 6 5 	 r o u n d e d 	 o f f 	 t o 	 n e a r e s t 	 t e n 	 t h o u s a n d s 	 = 	 5 0 0 0 0
E s t i m a t e d 	 d i f f e r e n c e 	 = 	 4 4 0 0 0 0
2 . 	 I n 	 a 	 p i c t o g r a p h , 	 i f 	 t h e 	 s y m b o l 	 	 r e p r e s e n t s 	 2 0 	 s t u d e n t s , 	 t h e n 	 h o w 	 m a n y 	 s t u d e n t s
w i l l 	 b e 	 r e p r e s e n t e d 	 b y 	 t h e 	 	 s y m b o l ?
A n s . 	 3 5 .
3 . 	 S o l v e 	 a n d 	 w r i t e 	 t h e 	 a n s w e r 	 i n 	 t h e 	 H i n d u 	 A r a b i c 	 N u m e r a l : 	 D 	 - 	 C D X L 	 = 	 _ _ _ _ _ _
A n s . 	 D 	 - 	 C D X L 	 = 	 5 0 0 	 – 	 4 4 0 	 = 	 6 0 	 = 	 L X
4 . 	 H o w 	 m a n y 	 d e g r e e s 	 a r e 	 t h e r e 	 i n 	 t h e 	 a n g l e 	 b e t w e e n 	 t h e 	 h a n d s 	 o f 	 a 	 c l o c k 	 w h e n 	 i t 	 i s
6 	 o ’ c l o c k ?
A n s . 	 1 8 0 	 d e g r e e s 	 = 	 s t r a i g h t 	 a n g l e
S E C T I O N - B
5 . 	 F i n d 	 t h e 	 p r o d u c t 	 o f 	 t h e 	 p l a c e 	 v a l u e s 	 o f 	 t w o 	 7 s 	 i n 	 1 9 7 0 4 7 .
A n s . 	 P l a c e 	 v a l u e 	 o f 	 7 	 i n 	 1 9 7 0 4 7 	 i s 	 7 0 0 0 	 a n d 	 7
P r o d u c t 	 = 	 4 9 0 0 0
6 . 	 S i m p l i f y 	 u s i n g 	 d i s t r i b u t i v e 	 p r o p e r t y 	 o f 	 w h o l e 	 n u m b e r s : 	 8 3 5 	 x 	 1 0 5
A n s . 	 8 3 5 	 × 	 1 0 5
= 	 8 3 5 	 × 	 ( 1 0 0 	 + 	 5 )
= 	 8 3 5 	 × 	 1 0 0 	 + 	 8 3 5 	 × 	 5 	 ( u s i n g 	 d i s t r i b u t i v e 	 p r o p e r t y )
= 	 8 3 5 0 0 	 + 	 4 1 7 5 	 = 	 8 7 6 7 5
7 . 	 I s 	 2 8 	 a 	 p e r f e c t 	 n u m b e r ? 	 S h o w 	 t h e 	 w o r k i n g .
A n s . 	 F a c t o r s 	 o f 	 2 8 	 a r e 	 1 , 2 , 4 , 7 , 1 4 	 a n d 	 2 8
S u m 	 o f 	 f a c t o r s 	 = 	 1 	 + 	 2 	 + 	 4 	 + 	 7 	 + 	 1 4 	 + 	 2 8 	 = 	 5 6 	 w h i c h 	 i s 	 2 	 × 	 2 8
T h e r e f o r e , 	 Y e s 	 2 8 	 i s 	 a 	 p e r f e c t 	 n u m b e r
8 . 	 F i n d 	 t h e 	 d i a m e t e r 	 o f 	 a 	 c i r c l e 	 i f 	 i t s 	 r a d i u s 	 i s 	 4 . 5 	 c m .
A n s . 	 D i a m e t e r 	 = 	 2 	 x 	 4 . 5 	 c m 	 = 	 9 	 c m
9 . 	 T h e 	 f o l l o w i n g 	 a r e 	 t h e 	 m a r k s 	 s c o r e d 	 b y 	 2 0 	 s t u d e n t s 	 o f 	 V I 	 A 	 i n 	 M a t h e m a t i c s 	 e x a m :
5 7 , 	 6 5 , 	 8 0 , 	 5 7 , 	 8 0 , 	 6 5 , 	 8 5 , 	 3 4 , 	 5 7 , 	 4 1 , 	 5 4 , 	 8 0 , 	 4 1 , 	 5 4 , 	 3 4 , 	 4 1 , 	 8 5 , 	 5 7 , 	 8 0 , 	 6 5
P r e p a r e 	 a 	 f r e q u e n c y 	 d i s t r i b u t i o n 	 t a b l e .
A n s . 	 F r e q u e n c y 	 d i s t r i b u t i o n 	 t a b l e 	 n e a t l y 	 d r a w n
1 0 . 	 S t a t e 	 t h e 	 c o m m u t a t i v e 	 p r o p e r t y 	 o f 	 a d d i t i o n 	 o f 	 w h o l e 	 n u m b e r s . 	 G i v e 	 a n 	 e x a m p l e .
A n s . 	 a 	 + 	 b 	 = 	 b 	 + 	 a
F o r 	 e g : 	 3 	 + 5 	 = 	 5 	 + 	 3 	 = 	 8
S E C T I O N - C
1 1 . 	 S u b t r a c t 	 2 5 2 7 0 5 3 	 f r o m 	 2 7 2 5 3 5 0 	 a n d 	 w r i t e 	 t h e 	 n u m b e r 	 s e n t e n c e .
A n s . 	 S u b t r a c t i o n 	 s h o w n
N u m b e r 	 s e n t e n c e : 	 2 7 2 5 3 5 0 	 – 	 2 5 2 7 0 5 3 	 = 	 1 9 8 2 9
1 2 . 	 F i n d 	 t h e 	 g r e a t e s t 	 4 - d i g i t 	 n u m b e r 	 w h i c h 	 i s 	 e x a c t l y 	 d i v i s i b l e 	 b y 	 3 7 .
A n s . 	 G r e a t e s t 	 4 - d i g i t 	 n u m b e r = 	 9 9 9 9
D i v i d e 	 9 9 9 9 	 b y 	 3 7 	 R e m a i n d e r = 	 9
N o w , 	 r e q u i r e d 	 n u m b e r 	 = 	 9 9 9 9 	 – 	 9 	 = 	 9 9 9 0
1 3 . 	 I n 	 5 6 * 8 9 1 	 r e p l a c e 	 * 	 b y 	 ( a ) 	 b y 	 3 .
A n s . 	 A c c o r d i n g 	 t o 	 d i v i s i b i l i t y 	 t e s t 	 o f 	 3 , 	 S u m 	 o f 	 d i g i t s 	 = 	 5 	 + 	 6 	 + 	 * 	 + 	 8 	 + 	 9 	 + 	 1
= 	 2 9 	 + 	 *
( a ) 	 S m a l l e s t 	 d i g i t 	 = 	 1 	 ( b ) 	 L a r g e s t 	 d i g i t 	 = 	 7
1 4 . 	 F i n d 	 t h e 	 H C F 	 o f 	 3 8 5 	 a n d 	 6 2 1 . 	 A r e 	 t h e y 	 c o p r i m e
A n s . 	 H C F 	 o f 	 3 8 5 	 a n d 	 6 2 1 	 = 	 1
Y e s 	 t h e y 	 a r e 	 C o p r i m e
1 5 . 	 N a m e 	 a n y 	 s i x 	 a n g l e s 	 i n 	 t h e 	 g i v e n 	 f i g u r e :
A n s . 	 A B C , 	 B A C , 	 A C B , 	 B D C , 	 D C B , 	 	 D B C , 	 A B D 	 a n d 	 A C D 	 ( a t 	 l e a s t 	 6 )
1 6 . 	 D r a w 	 a 	 r o u g h 	 s k e t c h 	 o f 	 a 	 q u a d r i l a t e r a l 	 E F G H 	 a n d 	 n a m e
a . 	 a 	 p a i r 	 o f 	 o p p o s i t e 	 s i d e s .
b . 	 a 	 p a i r 	 o f 	 a d j a c e n t 	 a n g l e s
c . 	 t w o 	 d i a g o n a l s .
Page 4


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


G e n e r a l 	 I n s t r u c t i o n s :
• 	 T h i s 	 q u e s t i o n 	 p a p e r 	 h a s 	 3 1 	 q u e s t i o n s 	 .
• 	 S e c t i o n 	 A 	 c o n s i s t s 	 o f 	 4 	 q u e s t i o n s 	 o f 	 1 	 m a r k 	 e a c h .
• 	 S e c t i o n 	 B 	 c o n s i s t s 	 o f 	 6 	 q u e s t i o n s 	 o f 	 2 	 m a r k s 	 e a c h .
• 	 S e c t i o n 	 C 	 c o n s i s t s 	 o f 	 1 0 	 q u e s t i o n s 	 o f 	 3 	 m a r k s 	 e a c h .
• 	 S e c t i o n 	 D 	 c o n s i s t s 	 o f 	 1 1 	 q u e s t i o n s 	 o f 	 4 	 m a r k s 	 e a c h .
• 	 A l l 	 q u e s t i o n s 	 a r e 	 c o m p u l s o r y 	 t o 	 a t t e m p t .
• 	 A t t e m p t 	 t h e 	 p a p e r 	 n e a t l y 	 a n d 	 l e g i b l y .
S E C T I O N - A
1 . 	 F i n d 	 t h e 	 e s t i m a t e d 	 d i f f e r e n c e 	 o f 	 4 8 9 3 4 8 	 a n d 	 4 8 3 6 5 .
A n s . 	 4 8 9 3 4 8 	 r o u n d e d 	 o f f 	 t o 	 n e a r e s t 	 t e n 	 t h o u s a n d s 	 = 	 4 9 0 0 0 0
4 8 3 6 5 	 r o u n d e d 	 o f f 	 t o 	 n e a r e s t 	 t e n 	 t h o u s a n d s 	 = 	 5 0 0 0 0
E s t i m a t e d 	 d i f f e r e n c e 	 = 	 4 4 0 0 0 0
2 . 	 I n 	 a 	 p i c t o g r a p h , 	 i f 	 t h e 	 s y m b o l 	 	 r e p r e s e n t s 	 2 0 	 s t u d e n t s , 	 t h e n 	 h o w 	 m a n y 	 s t u d e n t s
w i l l 	 b e 	 r e p r e s e n t e d 	 b y 	 t h e 	 	 s y m b o l ?
A n s . 	 3 5 .
3 . 	 S o l v e 	 a n d 	 w r i t e 	 t h e 	 a n s w e r 	 i n 	 t h e 	 H i n d u 	 A r a b i c 	 N u m e r a l : 	 D 	 - 	 C D X L 	 = 	 _ _ _ _ _ _
A n s . 	 D 	 - 	 C D X L 	 = 	 5 0 0 	 – 	 4 4 0 	 = 	 6 0 	 = 	 L X
4 . 	 H o w 	 m a n y 	 d e g r e e s 	 a r e 	 t h e r e 	 i n 	 t h e 	 a n g l e 	 b e t w e e n 	 t h e 	 h a n d s 	 o f 	 a 	 c l o c k 	 w h e n 	 i t 	 i s
6 	 o ’ c l o c k ?
A n s . 	 1 8 0 	 d e g r e e s 	 = 	 s t r a i g h t 	 a n g l e
S E C T I O N - B
5 . 	 F i n d 	 t h e 	 p r o d u c t 	 o f 	 t h e 	 p l a c e 	 v a l u e s 	 o f 	 t w o 	 7 s 	 i n 	 1 9 7 0 4 7 .
A n s . 	 P l a c e 	 v a l u e 	 o f 	 7 	 i n 	 1 9 7 0 4 7 	 i s 	 7 0 0 0 	 a n d 	 7
P r o d u c t 	 = 	 4 9 0 0 0
6 . 	 S i m p l i f y 	 u s i n g 	 d i s t r i b u t i v e 	 p r o p e r t y 	 o f 	 w h o l e 	 n u m b e r s : 	 8 3 5 	 x 	 1 0 5
A n s . 	 8 3 5 	 × 	 1 0 5
= 	 8 3 5 	 × 	 ( 1 0 0 	 + 	 5 )
= 	 8 3 5 	 × 	 1 0 0 	 + 	 8 3 5 	 × 	 5 	 ( u s i n g 	 d i s t r i b u t i v e 	 p r o p e r t y )
= 	 8 3 5 0 0 	 + 	 4 1 7 5 	 = 	 8 7 6 7 5
7 . 	 I s 	 2 8 	 a 	 p e r f e c t 	 n u m b e r ? 	 S h o w 	 t h e 	 w o r k i n g .
A n s . 	 F a c t o r s 	 o f 	 2 8 	 a r e 	 1 , 2 , 4 , 7 , 1 4 	 a n d 	 2 8
S u m 	 o f 	 f a c t o r s 	 = 	 1 	 + 	 2 	 + 	 4 	 + 	 7 	 + 	 1 4 	 + 	 2 8 	 = 	 5 6 	 w h i c h 	 i s 	 2 	 × 	 2 8
T h e r e f o r e , 	 Y e s 	 2 8 	 i s 	 a 	 p e r f e c t 	 n u m b e r
8 . 	 F i n d 	 t h e 	 d i a m e t e r 	 o f 	 a 	 c i r c l e 	 i f 	 i t s 	 r a d i u s 	 i s 	 4 . 5 	 c m .
A n s . 	 D i a m e t e r 	 = 	 2 	 x 	 4 . 5 	 c m 	 = 	 9 	 c m
9 . 	 T h e 	 f o l l o w i n g 	 a r e 	 t h e 	 m a r k s 	 s c o r e d 	 b y 	 2 0 	 s t u d e n t s 	 o f 	 V I 	 A 	 i n 	 M a t h e m a t i c s 	 e x a m :
5 7 , 	 6 5 , 	 8 0 , 	 5 7 , 	 8 0 , 	 6 5 , 	 8 5 , 	 3 4 , 	 5 7 , 	 4 1 , 	 5 4 , 	 8 0 , 	 4 1 , 	 5 4 , 	 3 4 , 	 4 1 , 	 8 5 , 	 5 7 , 	 8 0 , 	 6 5
P r e p a r e 	 a 	 f r e q u e n c y 	 d i s t r i b u t i o n 	 t a b l e .
A n s . 	 F r e q u e n c y 	 d i s t r i b u t i o n 	 t a b l e 	 n e a t l y 	 d r a w n
1 0 . 	 S t a t e 	 t h e 	 c o m m u t a t i v e 	 p r o p e r t y 	 o f 	 a d d i t i o n 	 o f 	 w h o l e 	 n u m b e r s . 	 G i v e 	 a n 	 e x a m p l e .
A n s . 	 a 	 + 	 b 	 = 	 b 	 + 	 a
F o r 	 e g : 	 3 	 + 5 	 = 	 5 	 + 	 3 	 = 	 8
S E C T I O N - C
1 1 . 	 S u b t r a c t 	 2 5 2 7 0 5 3 	 f r o m 	 2 7 2 5 3 5 0 	 a n d 	 w r i t e 	 t h e 	 n u m b e r 	 s e n t e n c e .
A n s . 	 S u b t r a c t i o n 	 s h o w n
N u m b e r 	 s e n t e n c e : 	 2 7 2 5 3 5 0 	 – 	 2 5 2 7 0 5 3 	 = 	 1 9 8 2 9
1 2 . 	 F i n d 	 t h e 	 g r e a t e s t 	 4 - d i g i t 	 n u m b e r 	 w h i c h 	 i s 	 e x a c t l y 	 d i v i s i b l e 	 b y 	 3 7 .
A n s . 	 G r e a t e s t 	 4 - d i g i t 	 n u m b e r = 	 9 9 9 9
D i v i d e 	 9 9 9 9 	 b y 	 3 7 	 R e m a i n d e r = 	 9
N o w , 	 r e q u i r e d 	 n u m b e r 	 = 	 9 9 9 9 	 – 	 9 	 = 	 9 9 9 0
1 3 . 	 I n 	 5 6 * 8 9 1 	 r e p l a c e 	 * 	 b y 	 ( a ) 	 b y 	 3 .
A n s . 	 A c c o r d i n g 	 t o 	 d i v i s i b i l i t y 	 t e s t 	 o f 	 3 , 	 S u m 	 o f 	 d i g i t s 	 = 	 5 	 + 	 6 	 + 	 * 	 + 	 8 	 + 	 9 	 + 	 1
= 	 2 9 	 + 	 *
( a ) 	 S m a l l e s t 	 d i g i t 	 = 	 1 	 ( b ) 	 L a r g e s t 	 d i g i t 	 = 	 7
1 4 . 	 F i n d 	 t h e 	 H C F 	 o f 	 3 8 5 	 a n d 	 6 2 1 . 	 A r e 	 t h e y 	 c o p r i m e
A n s . 	 H C F 	 o f 	 3 8 5 	 a n d 	 6 2 1 	 = 	 1
Y e s 	 t h e y 	 a r e 	 C o p r i m e
1 5 . 	 N a m e 	 a n y 	 s i x 	 a n g l e s 	 i n 	 t h e 	 g i v e n 	 f i g u r e :
A n s . 	 A B C , 	 B A C , 	 A C B , 	 B D C , 	 D C B , 	 	 D B C , 	 A B D 	 a n d 	 A C D 	 ( a t 	 l e a s t 	 6 )
1 6 . 	 D r a w 	 a 	 r o u g h 	 s k e t c h 	 o f 	 a 	 q u a d r i l a t e r a l 	 E F G H 	 a n d 	 n a m e
a . 	 a 	 p a i r 	 o f 	 o p p o s i t e 	 s i d e s .
b . 	 a 	 p a i r 	 o f 	 a d j a c e n t 	 a n g l e s
c . 	 t w o 	 d i a g o n a l s .
A n s . 	 a . 	 A 	 p a i r 	 o f 	 o p p o s i t e 	 s i d e s 	 – 	 E F 	 a n d 	 G H
b . 	 A 	 p a i r 	 o f 	 a d j a c e n t 	 a n g l e s 	 - 	 E 	 a n d 	 G
c . 	 T w o 	 d i a g o n a l s 	 – 	 E G 	 a n d 	 F H
1 7 . 	 C l a s s i f y 	 t h e 	 t r i a n g l e s 	 o n 	 t h e 	 b a s i s 	 o f 	 a n g l e s .
A n s . 	 A c u t e 	 a n g l e d 	 t r i a n g l e , 	 O b t u s e 	 a n g l e d 	 t r i a n g l e 	 a n d 	 R i g h t 	 a n g l e d 	 t r i a n g l e
1 8 . 	 L o o k 	 a t 	 t h e 	 f i g u r e 	 b e l o w
a . 	 W r i t e 	 a n o t h e r 	 w a y 	 o f 	 n a m i n g 	 l i n e 	 b .
b . 	 W r i t e 	 a n o t h e r 	 w a y 	 o f 	 n a m i n g 	 l i n e 	 a .
c . 	 W r i t e 	 a 	 s e t 	 o f 	 c o l l i n e a r 	 p o i n t s .
A n s . 	 a . 	 A n o t h e r 	 w a y 	 o f 	 n a m i n g 	 l i n e 	 b . - 	 Q S
b . 	 A n o t h e r 	 w a y 	 o f 	 n a m i n g 	 l i n e 	 a . 	 – 	 P R
c . 	 A 	 s e t 	 o f 	 c o l l i n e a r 	 p o i n t s . 	 – 	 P , 	 Q 	 a n d 	 T 	 o r 	 R , 	 S 	 a n d 	 T
1 9 . 	 T h e 	 d a t a 	 g i v e n 	 b e l o w 	 p r o v i d e s 	 i n f o r m a t i o n 	 a b o u t 	 t h e 	 n u m b e r 	 o f 	 b o o k s 	 o n 	 d i f f e r e n t
s u b j e c t s 	 i n 	 a 	 l i b r a r y . 	 R e p r e s e n t 	 t h e 	 d a t a 	 u s i n g 	 a 	 p i c t o g r a p h 	 u s i n g 	 a 	 s u i t a b l e 	 s c a l e
	
S u b j e c t E n g l i s h H i n d i M a t h e m a t i c s S c i e n c e
N o . 	 o f 	 b o o k s 1 0 0 0 6 5 0 8 0 0 4 5 0
A n s . 	 P i c t o g r a p h 	 n e a t l y 	 d r a w n
S c a l e 	 s h o w n
2 0 . 	 T h e 	 c o l o u r s 	 o f 	 c a r s 	 p r e f e r r e d 	 b y 	 p e o p l e 	 l i v i n g 	 i n 	 a 	 l o c a l i t y 	 a r e 	 s h o w n 	 b y 	 t h e
f o l l o w i n g 	 p i c t o g r a p h . 	 A n s w e r 	 t h e 	 q u e s t i o n s 	 t h a t 	 f o l l o w :
C o l o u r 	 o f 	 c a r N o . 	 o f 	 p e o p l e 	 ( 1 	 $ 	 = 	 1 5 	 p e o p l e )
B l a c k 	 	 $
S i l v e r 	
R e d 	 $
W h i t e 	 	
a . 	 F i n d 	 t h e 	 n u m b e r 	 o f 	 p e o p l e 	 p r e f e r r i n g 	 s i l v e r 	 c a r s .
b . 	 H o w 	 m a n y 	 l e s s 	 p e o p l e 	 p r e f e r 	 r e d 	 c a r s 	 t h a n 	 w h i t e 	 c a r s ?
c . 	 W h i c h 	 c o l o u r 	 i s 	 m o s t 	 l i k e d 	 b y 	 t h e 	 p e o p l e ?
A n s . 	 a . 	 - 	 6 0
b . 	 - 4 5
c . 	 - 	 W h i t e
S E C T I O N 	 D
2 1 . 	 U s i n g 	 t h e 	 d i g i t s 	 7 , 0 , 3 , 5 	 a n d 	 2 , 	 m a k e 	 a n y 	 f o u r 	 7 	 d i g i t 	 n u m b e r s . 	 I n s e r t 	 c o m m a s 	 u s i n g
I n t e r n a t i o n a l 	 s y s t e m 	 o f 	 n u m e r a t i o n . 	 A r r a n g e 	 t h e m 	 i n 	 a s c e n d i n g 	 o r d e r .
A n s . 	 A n y 	 f o u r 	 7 - 	 d i g i t 	 n u m b e r s 	 u s i n g 	 7 , 0 , 3 , 5 	 a n d 	 2
I n s e r t 	 c o m m a s 	 u s i n g 	 I n t e r n a t i o n a l 	 s y s t e m 	 o f 	 n u m e r a t i o n .
A s c e n d i n g 	 o r d e r
2 2 . 	 T h e 	 f o l l o w i n g 	 t a b l e 	 s h o w s 	 t h e 	 y e a r l y 	 p r o d u c t i o n 	 o f 	 c y c l e s 	 b y 	 a 	 c o m p a n y . 	 D r a w 	 a 	 b a r
g r a p h 	 f o r 	 t h e 	 g i v e n 	 d a t a .
A n s . 	 B a r 	 g r a p h 	 f o r 	 t h e 	 g i v e n 	 d a t a 	 n e a t l y 	 d r a w n .
T i t l e , 	 S c a l e , 	 L a b e l s
2 3 . 	 A 	 V o l v o 	 b u s 	 d r i v e r 	 o f f e r s 	 f r e e 	 s e r v i c e 	 t o 	 a 	 p i l g r i m a g e 	 ( h o l y 	 p l a c e ) 	 w h i c h 	 i s 	 2 1 k m
4 5 m 	 a w a y 	 f r o m 	 a 	 c i t y . 	 T h e 	 b u s 	 m a k e s 	 5 	 r o u n d 	 t r i p s 	 e v e r y 	 d a y . 	 H o w 	 m u c h 	 d i s t a n c e 	 d o e s
i t 	 c o v e r 	 i n 	 t h e 	 m o n t h 	 o f 	 A p r i l ? 	 W h a t 	 v a l u e 	 d o 	 w e 	 l e a r n 	 f r o m 	 t h i s ?

Top Courses for Class 6

View all

FAQs on Class 6 Math: CBSE Past Year Paper - 5

1. What is the CBSE Class 6 Math Past Year Paper - 5?

Ans. The CBSE Class 6 Math Past Year Paper - 5 is a previous year's question paper for the Class 6 Math examination conducted by the Central Board of Secondary Education (CBSE) in India. It is used for practice and revision purposes by students preparing for their exams.

2. How can solving the CBSE Class 6 Math Past Year Paper - 5 benefit students?

Ans. Solving the CBSE Class 6 Math Past Year Paper - 5 can benefit students in several ways. It helps them understand the exam pattern, familiarize themselves with the types of questions asked, and assess their preparation level. By solving the paper, students can identify their strengths and weaknesses and focus on areas that need improvement.

3. Are the questions in the CBSE Class 6 Math Past Year Paper - 5 similar to the actual exam?

Ans. Yes, the questions in the CBSE Class 6 Math Past Year Paper - 5 are designed to be similar to the ones asked in the actual exam. These papers are created by experts who have a good understanding of the exam pattern and syllabus. However, it is important to note that the actual exam may have variations in terms of difficulty level and question distribution.

4. Can solving the CBSE Class 6 Math Past Year Paper - 5 guarantee good marks in the exam?

Ans. Solving the CBSE Class 6 Math Past Year Paper - 5 can certainly improve the chances of scoring well in the exam. It helps students become familiar with the question format, manage time effectively, and gain confidence. However, achieving good marks also depends on regular study, understanding concepts, and practicing additional questions beyond the past year paper.

5. Where can I find the CBSE Class 6 Math Past Year Paper - 5?

Ans. The CBSE Class 6 Math Past Year Paper - 5 can be found on various educational websites, online platforms, or in preparation books specifically designed for CBSE exams. These resources are easily accessible and often provide solutions or answer keys to help students evaluate their performance. Additionally, schools or coaching centers may also provide students with the past year papers for practice.

Related Exams

Class 6

About this Document

	4.85/5 Rating
	Dec 19, 2024 Last updated

Document Description: Class 6 Math: CBSE Past Year Paper - 5 for Class 6 2024 is part of Class 6 preparation. The notes and questions for Class 6 Math: CBSE Past Year Paper - 5 have been prepared according to the Class 6 exam syllabus. Information about Class 6 Math: CBSE Past Year Paper - 5 covers topics like and Class 6 Math: CBSE Past Year Paper - 5 Example, for Class 6 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Class 6 Math: CBSE Past Year Paper - 5.

Introduction of Class 6 Math: CBSE Past Year Paper - 5 in English is available as part of our Class 6 preparation & Class 6 Math: CBSE Past Year Paper - 5 in Hindi for Class 6 courses. Download more important topics, notes, lectures and mock test series for Class 6 Exam by signing up for free. Class 6: Class 6 Math: CBSE Past Year Paper - 5

Description

Full syllabus notes, lecture & questions for Class 6 Math: CBSE Past Year Paper - 5 - Class 6 | Plus excerises question with solution to help you revise complete syllabus | Best notes, free PDF download

Information about Class 6 Math: CBSE Past Year Paper - 5

In this doc you can find the meaning of Class 6 Math: CBSE Past Year Paper - 5 defined & explained in the simplest way possible. Besides explaining types of Class 6 Math: CBSE Past Year Paper - 5 theory, EduRev gives you an ample number of questions to practice Class 6 Math: CBSE Past Year Paper - 5 tests, examples and also practice Class 6 tests.

Download as PDF

Explore Courses for Class 6 exam

Top Courses for Class 6

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Class 6 Math: CBSE Past Year Paper - 5

MCQs

Objective type Questions

Viva Questions

ppt

shortcuts and tricks

Exam

Extra Questions

study material

Sample Paper

Important questions

Semester Notes

video lectures

Free

Class 6 Math: CBSE Past Year Paper - 5

Summary

Previous Year Questions with Solutions

past year papers

pdf

mock tests for examination

practice quizzes

Class 6 Math: CBSE Past Year Paper - 5

;

Additional Information about Class 6 Math: CBSE Past Year Paper - 5 for Class 6 Preparation

Class 6 Math: CBSE Past Year Paper - 5 Free PDF Download

The Class 6 Math: CBSE Past Year Paper - 5 is an invaluable resource that delves deep into the core of the Class 6 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the Class 6 Math: CBSE Past Year Paper - 5 now and kickstart your journey towards success in the Class 6 exam.

Importance of Class 6 Math: CBSE Past Year Paper - 5

The importance of Class 6 Math: CBSE Past Year Paper - 5 cannot be overstated, especially for Class 6 aspirants. This document holds the key to success in the Class 6 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

Class 6 Math: CBSE Past Year Paper - 5 Notes

Class 6 Math: CBSE Past Year Paper - 5 Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to Class 6 Math: CBSE Past Year Paper - 5. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, Class 6 Math: CBSE Past Year Paper - 5 Notes on EduRev are your ultimate resource for success.

Class 6 Math: CBSE Past Year Paper - 5 Class 6 Questions

The "Class 6 Math: CBSE Past Year Paper - 5 Class 6 Questions" guide is a valuable resource for all aspiring students preparing for the Class 6 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study Class 6 Math: CBSE Past Year Paper - 5 on the App

Students of Class 6 can study Class 6 Math: CBSE Past Year Paper - 5 alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the Class 6 Math: CBSE Past Year Paper - 5, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of Class 6 Math: CBSE Past Year Paper - 5 is prepared as per the latest Class 6 syllabus.

Education Revolution