Class 9 Exam > Class 9 Notes > Class 9 Math: Sample Question Paper- 12 (With Solutions)
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Page 1 7 / 2 8 C B S E C l a s s 9 M a t h e m a t i c s S a m p l e P a p e r S o l u t i o n S e c t i o n - A ( Q u e s t i o n n u m b e r s 1 t o 6 c a r r y 1 m a r k e a c h ) 1 . I f x a / b = 1 , t h e n f i n d t h e v a l u e o f ‘ a ’ . A n s w e r : a = 0 S i n c e a / b = 0 , t h e n x 0 = 1 2 . I f p ( x ) = 2 x 3 + 5 x 2 – 3 x - 2 i s d i v i d e d b y x - 1 , t h e n f i n d t h e r e m a i n d e r . A n s w e r : 2 p ( 1 ) i s t h e r e m a i n d e r . p ( 1 ) = 2 ( 1 ) 3 + 5 ( 1 ) 2 – 3 ( 1 ) - 2 = 2 + 5 - 3 - 2 = 2 3 . T h e d i s t a n c e o f t h e p o i n t ( 0 , - 1 ) f r o m t h e o r i g i n i s _ _ _ _ _ _ _ _ . A n s w e r : 1 d = = = 1 4 . I f t h e v e r t i c a l a n g l e o f a n i s o s c e l e s t r i a n g l e i s 1 0 0 0 , t h e n f i n d t h e m e a s u r e s o f i t s b a s e a n g l e s . A n s w e r : 4 0 ° , 4 0 ° L e t m e a s u r e o f e a c h b a s e a n g l e o f a n i s o s c e l e s t r i a n g l e b e x . T h e r e f o r e , w e h a v e 1 0 0 ° + x + x = 1 8 0 ° 2 x = 1 8 0 ° – 1 0 0 ° x = 4 0 ° . Page 2 7 / 2 8 C B S E C l a s s 9 M a t h e m a t i c s S a m p l e P a p e r S o l u t i o n S e c t i o n - A ( Q u e s t i o n n u m b e r s 1 t o 6 c a r r y 1 m a r k e a c h ) 1 . I f x a / b = 1 , t h e n f i n d t h e v a l u e o f ‘ a ’ . A n s w e r : a = 0 S i n c e a / b = 0 , t h e n x 0 = 1 2 . I f p ( x ) = 2 x 3 + 5 x 2 – 3 x - 2 i s d i v i d e d b y x - 1 , t h e n f i n d t h e r e m a i n d e r . A n s w e r : 2 p ( 1 ) i s t h e r e m a i n d e r . p ( 1 ) = 2 ( 1 ) 3 + 5 ( 1 ) 2 – 3 ( 1 ) - 2 = 2 + 5 - 3 - 2 = 2 3 . T h e d i s t a n c e o f t h e p o i n t ( 0 , - 1 ) f r o m t h e o r i g i n i s _ _ _ _ _ _ _ _ . A n s w e r : 1 d = = = 1 4 . I f t h e v e r t i c a l a n g l e o f a n i s o s c e l e s t r i a n g l e i s 1 0 0 0 , t h e n f i n d t h e m e a s u r e s o f i t s b a s e a n g l e s . A n s w e r : 4 0 ° , 4 0 ° L e t m e a s u r e o f e a c h b a s e a n g l e o f a n i s o s c e l e s t r i a n g l e b e x . T h e r e f o r e , w e h a v e 1 0 0 ° + x + x = 1 8 0 ° 2 x = 1 8 0 ° – 1 0 0 ° x = 4 0 ° . 8 / 2 8 5 . T h e r a t i o o f t h e w h o l e s u r f a c e a r e a o f a s o l i d s p h e r e a n d a s o l i d h e m i s p h e r e i s _ _ _ _ . A n s w e r : 4 : 3 T o t a l S u r f a c e a r e a o f a s p h e r e : T o t a l S u r f a c e a r e a o f a s o l i d h e m i s p h e r e = = 4 : 3 6 . T h e r e a r e 6 0 b o y s a n d 4 0 g i r l s i n a c l a s s . A s t u d e n t i s s e l e c t e d 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 t h a t s t u d e n t i s a g i r l . A n s w e r : T o t a l n o . o f s t u d e n t s i n t h e c l a s s = 1 0 0 N o . o f g i r l s i n t h e c l a s s = 4 0 P ( t h e s t u d e n t i s a g i r l ) = S e c t i o n B ( Q u e s t i o n n u m b e r s 7 t o 1 2 c a r r y 2 m a r k s e a c h ) 7 . I f p = 2 - a , t h e n p r o v e t h a t a 3 + 6 a p + p 3 – 8 = 0 . A n s w e r : S i n c e p = 2 – a , p + a – 2 = 0 p 3 + a 3 + ( - 2 ) 3 = 3 ( p ) ( a ) ( - 2 ) [ S i n c e i f a + b + c = 0 , t h e n a 3 + b 3 + c 3 = 3 a b c ] p 3 + a 3 – 8 = - 6 a p a 3 + 6 a p + p 3 – 8 = 0 . 8 . I n t h e a d j o i n i n g f i g u r e , w e h a v e A B = B C , B X = B Y . S h o w t h a t A X = C Y ( u s i n g a p p r o p r i a t e E u c l i d ’ s a x i o m ) A n s w e r : W e h a v e Page 3 7 / 2 8 C B S E C l a s s 9 M a t h e m a t i c s S a m p l e P a p e r S o l u t i o n S e c t i o n - A ( Q u e s t i o n n u m b e r s 1 t o 6 c a r r y 1 m a r k e a c h ) 1 . I f x a / b = 1 , t h e n f i n d t h e v a l u e o f ‘ a ’ . A n s w e r : a = 0 S i n c e a / b = 0 , t h e n x 0 = 1 2 . I f p ( x ) = 2 x 3 + 5 x 2 – 3 x - 2 i s d i v i d e d b y x - 1 , t h e n f i n d t h e r e m a i n d e r . A n s w e r : 2 p ( 1 ) i s t h e r e m a i n d e r . p ( 1 ) = 2 ( 1 ) 3 + 5 ( 1 ) 2 – 3 ( 1 ) - 2 = 2 + 5 - 3 - 2 = 2 3 . T h e d i s t a n c e o f t h e p o i n t ( 0 , - 1 ) f r o m t h e o r i g i n i s _ _ _ _ _ _ _ _ . A n s w e r : 1 d = = = 1 4 . I f t h e v e r t i c a l a n g l e o f a n i s o s c e l e s t r i a n g l e i s 1 0 0 0 , t h e n f i n d t h e m e a s u r e s o f i t s b a s e a n g l e s . A n s w e r : 4 0 ° , 4 0 ° L e t m e a s u r e o f e a c h b a s e a n g l e o f a n i s o s c e l e s t r i a n g l e b e x . T h e r e f o r e , w e h a v e 1 0 0 ° + x + x = 1 8 0 ° 2 x = 1 8 0 ° – 1 0 0 ° x = 4 0 ° . 8 / 2 8 5 . T h e r a t i o o f t h e w h o l e s u r f a c e a r e a o f a s o l i d s p h e r e a n d a s o l i d h e m i s p h e r e i s _ _ _ _ . A n s w e r : 4 : 3 T o t a l S u r f a c e a r e a o f a s p h e r e : T o t a l S u r f a c e a r e a o f a s o l i d h e m i s p h e r e = = 4 : 3 6 . T h e r e a r e 6 0 b o y s a n d 4 0 g i r l s i n a c l a s s . A s t u d e n t i s s e l e c t e d 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 t h a t s t u d e n t i s a g i r l . A n s w e r : T o t a l n o . o f s t u d e n t s i n t h e c l a s s = 1 0 0 N o . o f g i r l s i n t h e c l a s s = 4 0 P ( t h e s t u d e n t i s a g i r l ) = S e c t i o n B ( Q u e s t i o n n u m b e r s 7 t o 1 2 c a r r y 2 m a r k s e a c h ) 7 . I f p = 2 - a , t h e n p r o v e t h a t a 3 + 6 a p + p 3 – 8 = 0 . A n s w e r : S i n c e p = 2 – a , p + a – 2 = 0 p 3 + a 3 + ( - 2 ) 3 = 3 ( p ) ( a ) ( - 2 ) [ S i n c e i f a + b + c = 0 , t h e n a 3 + b 3 + c 3 = 3 a b c ] p 3 + a 3 – 8 = - 6 a p a 3 + 6 a p + p 3 – 8 = 0 . 8 . I n t h e a d j o i n i n g f i g u r e , w e h a v e A B = B C , B X = B Y . S h o w t h a t A X = C Y ( u s i n g a p p r o p r i a t e E u c l i d ’ s a x i o m ) A n s w e r : W e h a v e 9 / 2 8 A B = B C - - - - - - - - - - ( 1 ) B X = B Y - - - - - - - - - - - ( 2 ) S u b t r a c t ( 2 ) f r o m ( 1 ) A B – B X = B C – B Y N o w , b y E u c l i d ’ s a x i o m 3 , w e h a v e I f e q u a l s a r e s u b t r a c t e d f r o m e q u a l s , t h e r e m a i n d e r s a r e e q u a l . H e n c e , A X = C Y ( S i n c e B X = B Y ) 9 . I f t w o o p p o s i t e a n g l e s o f a p a r a l l e l o g r a m a r e ( 6 3 - 3 x ) ° a n d ( 4 x - 7 ) ° . F i n d a l l t h e a n g l e s o f t h e p a r a l l e l o g r a m . A n s w e r : I n a p a r a l l e l o g r a m , t h e o p p o s i t e a n g l e s a r e e q u a l . ( 6 3 - 3 x ) ° = ( 4 x - 7 ) ° 4 x + 3 x = 6 3 + 7 7 x = 7 0 x = 1 0 ( 6 3 - 3 x ) ° = 3 3 ° ( 4 x - 7 ) ° = 3 3 ° S u m o f a l l i n t e r i o r a n g l e s o f a p a r a l l e l o g r a m = 3 6 0 ° S u m o f t h e o t h e r t w o o p p o s i t e a n g l e s = 3 6 0 ° - ( 3 3 ° + 3 3 ° ) = 3 6 0 ° - 6 6 ° = 2 9 4 ° E a c h o f t h e o t h e r t w o o p p o s i t e a n g l e s = = 1 4 7 ° H e n c e t h e f o u r a n g l e s o f a p a r a l l e l o g r a m a r e 3 3 ° , 1 4 7 ° , 3 3 ° , 1 4 7 ° 1 0 . T h r e e S c h o o l s s i t u a t e d a t P , Q a n d R i n t h e f i g u r e a r e e q u i d i s t a n t f r o m e a c h o t h e r a s s h o w n i n t h e f i g u r e . F i n d Q O R . Page 4 7 / 2 8 C B S E C l a s s 9 M a t h e m a t i c s S a m p l e P a p e r S o l u t i o n S e c t i o n - A ( Q u e s t i o n n u m b e r s 1 t o 6 c a r r y 1 m a r k e a c h ) 1 . I f x a / b = 1 , t h e n f i n d t h e v a l u e o f ‘ a ’ . A n s w e r : a = 0 S i n c e a / b = 0 , t h e n x 0 = 1 2 . I f p ( x ) = 2 x 3 + 5 x 2 – 3 x - 2 i s d i v i d e d b y x - 1 , t h e n f i n d t h e r e m a i n d e r . A n s w e r : 2 p ( 1 ) i s t h e r e m a i n d e r . p ( 1 ) = 2 ( 1 ) 3 + 5 ( 1 ) 2 – 3 ( 1 ) - 2 = 2 + 5 - 3 - 2 = 2 3 . T h e d i s t a n c e o f t h e p o i n t ( 0 , - 1 ) f r o m t h e o r i g i n i s _ _ _ _ _ _ _ _ . A n s w e r : 1 d = = = 1 4 . I f t h e v e r t i c a l a n g l e o f a n i s o s c e l e s t r i a n g l e i s 1 0 0 0 , t h e n f i n d t h e m e a s u r e s o f i t s b a s e a n g l e s . A n s w e r : 4 0 ° , 4 0 ° L e t m e a s u r e o f e a c h b a s e a n g l e o f a n i s o s c e l e s t r i a n g l e b e x . T h e r e f o r e , w e h a v e 1 0 0 ° + x + x = 1 8 0 ° 2 x = 1 8 0 ° – 1 0 0 ° x = 4 0 ° . 8 / 2 8 5 . T h e r a t i o o f t h e w h o l e s u r f a c e a r e a o f a s o l i d s p h e r e a n d a s o l i d h e m i s p h e r e i s _ _ _ _ . A n s w e r : 4 : 3 T o t a l S u r f a c e a r e a o f a s p h e r e : T o t a l S u r f a c e a r e a o f a s o l i d h e m i s p h e r e = = 4 : 3 6 . T h e r e a r e 6 0 b o y s a n d 4 0 g i r l s i n a c l a s s . A s t u d e n t i s s e l e c t e d 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 t h a t s t u d e n t i s a g i r l . A n s w e r : T o t a l n o . o f s t u d e n t s i n t h e c l a s s = 1 0 0 N o . o f g i r l s i n t h e c l a s s = 4 0 P ( t h e s t u d e n t i s a g i r l ) = S e c t i o n B ( Q u e s t i o n n u m b e r s 7 t o 1 2 c a r r y 2 m a r k s e a c h ) 7 . I f p = 2 - a , t h e n p r o v e t h a t a 3 + 6 a p + p 3 – 8 = 0 . A n s w e r : S i n c e p = 2 – a , p + a – 2 = 0 p 3 + a 3 + ( - 2 ) 3 = 3 ( p ) ( a ) ( - 2 ) [ S i n c e i f a + b + c = 0 , t h e n a 3 + b 3 + c 3 = 3 a b c ] p 3 + a 3 – 8 = - 6 a p a 3 + 6 a p + p 3 – 8 = 0 . 8 . I n t h e a d j o i n i n g f i g u r e , w e h a v e A B = B C , B X = B Y . S h o w t h a t A X = C Y ( u s i n g a p p r o p r i a t e E u c l i d ’ s a x i o m ) A n s w e r : W e h a v e 9 / 2 8 A B = B C - - - - - - - - - - ( 1 ) B X = B Y - - - - - - - - - - - ( 2 ) S u b t r a c t ( 2 ) f r o m ( 1 ) A B – B X = B C – B Y N o w , b y E u c l i d ’ s a x i o m 3 , w e h a v e I f e q u a l s a r e s u b t r a c t e d f r o m e q u a l s , t h e r e m a i n d e r s a r e e q u a l . H e n c e , A X = C Y ( S i n c e B X = B Y ) 9 . I f t w o o p p o s i t e a n g l e s o f a p a r a l l e l o g r a m a r e ( 6 3 - 3 x ) ° a n d ( 4 x - 7 ) ° . F i n d a l l t h e a n g l e s o f t h e p a r a l l e l o g r a m . A n s w e r : I n a p a r a l l e l o g r a m , t h e o p p o s i t e a n g l e s a r e e q u a l . ( 6 3 - 3 x ) ° = ( 4 x - 7 ) ° 4 x + 3 x = 6 3 + 7 7 x = 7 0 x = 1 0 ( 6 3 - 3 x ) ° = 3 3 ° ( 4 x - 7 ) ° = 3 3 ° S u m o f a l l i n t e r i o r a n g l e s o f a p a r a l l e l o g r a m = 3 6 0 ° S u m o f t h e o t h e r t w o o p p o s i t e a n g l e s = 3 6 0 ° - ( 3 3 ° + 3 3 ° ) = 3 6 0 ° - 6 6 ° = 2 9 4 ° E a c h o f t h e o t h e r t w o o p p o s i t e a n g l e s = = 1 4 7 ° H e n c e t h e f o u r a n g l e s o f a p a r a l l e l o g r a m a r e 3 3 ° , 1 4 7 ° , 3 3 ° , 1 4 7 ° 1 0 . T h r e e S c h o o l s s i t u a t e d a t P , Q a n d R i n t h e f i g u r e a r e e q u i d i s t a n t f r o m e a c h o t h e r a s s h o w n i n t h e f i g u r e . F i n d Q O R . 1 0 / 2 8 A n s w e r : I n P Q R , w e h a v e P Q = Q R = P R ( S i n c e P , Q a n d R a r e e q u i d i s t a n t ) S o , P Q R i s a n e q u i l a t e r a l t r i a n g l e . Q P R = 6 0 ° S o , Q O R = 2 Q P R = 2 ( 6 0 ° ) = 1 2 0 ° ( S i n c e a n g l e s u b t e n d e d b y a n a r c a t t h e c e n t r e i s d o u b l e t h e a n g l e s u b t e n d e d b y i t a t a n y p o i n t o n t h e r e m a i n i n g p a r t o f t h e c i r c l e . ) 1 1 . T h e d i a m e t e r o f t h e t w o r i g h t c i r c u l a r c o n e s a r e e q u a l i f t h e i r s l a n t h e i g h t s a r e i n t h e r a t i o 3 : 2 , t h e n w h a t i s t h e r a t i o o f t h e i r c u r v e d s u r f a c e a r e a s ? A n s w e r : L e t t h e r a d i i a n d s l a n t h e i g h t s o f t w o r i g h t c i r c u l a r c o n e s a r e r 1 , l 1 a n d r 2 , l 2 r e s p e c t i v e l y . R a t i o o f t h e i r C u r v e d S u r f a c e A r e a = ( S i n c e r 1 = r 2 ) = = 3 : 2 1 2 . A b a t s m a n i n h i s 1 1 t h i n n i n g s m a k e s a s c o r e o f 6 8 r u n s a n d t h e r e b y i n c r e a s e s h i s a v e r a g e s c o r e b y 2 . W h a t i s h i s a v e r a g e s c o r e a f t e r t h e 1 1 t h i n n i n g s . A n s w e r : L e t t h e a v e r a g e s c o r e o f 1 1 i n n i n g s b e x : T h e n t h e a v e r a g e s c o r e o f 1 0 i n n i n g s = x – 2 T o t a l s c o r e o f 1 1 i n n i n g s = 1 1 x Page 5 7 / 2 8 C B S E C l a s s 9 M a t h e m a t i c s S a m p l e P a p e r S o l u t i o n S e c t i o n - A ( Q u e s t i o n n u m b e r s 1 t o 6 c a r r y 1 m a r k e a c h ) 1 . I f x a / b = 1 , t h e n f i n d t h e v a l u e o f ‘ a ’ . A n s w e r : a = 0 S i n c e a / b = 0 , t h e n x 0 = 1 2 . I f p ( x ) = 2 x 3 + 5 x 2 – 3 x - 2 i s d i v i d e d b y x - 1 , t h e n f i n d t h e r e m a i n d e r . A n s w e r : 2 p ( 1 ) i s t h e r e m a i n d e r . p ( 1 ) = 2 ( 1 ) 3 + 5 ( 1 ) 2 – 3 ( 1 ) - 2 = 2 + 5 - 3 - 2 = 2 3 . T h e d i s t a n c e o f t h e p o i n t ( 0 , - 1 ) f r o m t h e o r i g i n i s _ _ _ _ _ _ _ _ . A n s w e r : 1 d = = = 1 4 . I f t h e v e r t i c a l a n g l e o f a n i s o s c e l e s t r i a n g l e i s 1 0 0 0 , t h e n f i n d t h e m e a s u r e s o f i t s b a s e a n g l e s . A n s w e r : 4 0 ° , 4 0 ° L e t m e a s u r e o f e a c h b a s e a n g l e o f a n i s o s c e l e s t r i a n g l e b e x . T h e r e f o r e , w e h a v e 1 0 0 ° + x + x = 1 8 0 ° 2 x = 1 8 0 ° – 1 0 0 ° x = 4 0 ° . 8 / 2 8 5 . T h e r a t i o o f t h e w h o l e s u r f a c e a r e a o f a s o l i d s p h e r e a n d a s o l i d h e m i s p h e r e i s _ _ _ _ . A n s w e r : 4 : 3 T o t a l S u r f a c e a r e a o f a s p h e r e : T o t a l S u r f a c e a r e a o f a s o l i d h e m i s p h e r e = = 4 : 3 6 . T h e r e a r e 6 0 b o y s a n d 4 0 g i r l s i n a c l a s s . A s t u d e n t i s s e l e c t e d 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 t h a t s t u d e n t i s a g i r l . A n s w e r : T o t a l n o . o f s t u d e n t s i n t h e c l a s s = 1 0 0 N o . o f g i r l s i n t h e c l a s s = 4 0 P ( t h e s t u d e n t i s a g i r l ) = S e c t i o n B ( Q u e s t i o n n u m b e r s 7 t o 1 2 c a r r y 2 m a r k s e a c h ) 7 . I f p = 2 - a , t h e n p r o v e t h a t a 3 + 6 a p + p 3 – 8 = 0 . A n s w e r : S i n c e p = 2 – a , p + a – 2 = 0 p 3 + a 3 + ( - 2 ) 3 = 3 ( p ) ( a ) ( - 2 ) [ S i n c e i f a + b + c = 0 , t h e n a 3 + b 3 + c 3 = 3 a b c ] p 3 + a 3 – 8 = - 6 a p a 3 + 6 a p + p 3 – 8 = 0 . 8 . I n t h e a d j o i n i n g f i g u r e , w e h a v e A B = B C , B X = B Y . S h o w t h a t A X = C Y ( u s i n g a p p r o p r i a t e E u c l i d ’ s a x i o m ) A n s w e r : W e h a v e 9 / 2 8 A B = B C - - - - - - - - - - ( 1 ) B X = B Y - - - - - - - - - - - ( 2 ) S u b t r a c t ( 2 ) f r o m ( 1 ) A B – B X = B C – B Y N o w , b y E u c l i d ’ s a x i o m 3 , w e h a v e I f e q u a l s a r e s u b t r a c t e d f r o m e q u a l s , t h e r e m a i n d e r s a r e e q u a l . H e n c e , A X = C Y ( S i n c e B X = B Y ) 9 . I f t w o o p p o s i t e a n g l e s o f a p a r a l l e l o g r a m a r e ( 6 3 - 3 x ) ° a n d ( 4 x - 7 ) ° . F i n d a l l t h e a n g l e s o f t h e p a r a l l e l o g r a m . A n s w e r : I n a p a r a l l e l o g r a m , t h e o p p o s i t e a n g l e s a r e e q u a l . ( 6 3 - 3 x ) ° = ( 4 x - 7 ) ° 4 x + 3 x = 6 3 + 7 7 x = 7 0 x = 1 0 ( 6 3 - 3 x ) ° = 3 3 ° ( 4 x - 7 ) ° = 3 3 ° S u m o f a l l i n t e r i o r a n g l e s o f a p a r a l l e l o g r a m = 3 6 0 ° S u m o f t h e o t h e r t w o o p p o s i t e a n g l e s = 3 6 0 ° - ( 3 3 ° + 3 3 ° ) = 3 6 0 ° - 6 6 ° = 2 9 4 ° E a c h o f t h e o t h e r t w o o p p o s i t e a n g l e s = = 1 4 7 ° H e n c e t h e f o u r a n g l e s o f a p a r a l l e l o g r a m a r e 3 3 ° , 1 4 7 ° , 3 3 ° , 1 4 7 ° 1 0 . T h r e e S c h o o l s s i t u a t e d a t P , Q a n d R i n t h e f i g u r e a r e e q u i d i s t a n t f r o m e a c h o t h e r a s s h o w n i n t h e f i g u r e . F i n d Q O R . 1 0 / 2 8 A n s w e r : I n P Q R , w e h a v e P Q = Q R = P R ( S i n c e P , Q a n d R a r e e q u i d i s t a n t ) S o , P Q R i s a n e q u i l a t e r a l t r i a n g l e . Q P R = 6 0 ° S o , Q O R = 2 Q P R = 2 ( 6 0 ° ) = 1 2 0 ° ( S i n c e a n g l e s u b t e n d e d b y a n a r c a t t h e c e n t r e i s d o u b l e t h e a n g l e s u b t e n d e d b y i t a t a n y p o i n t o n t h e r e m a i n i n g p a r t o f t h e c i r c l e . ) 1 1 . T h e d i a m e t e r o f t h e t w o r i g h t c i r c u l a r c o n e s a r e e q u a l i f t h e i r s l a n t h e i g h t s a r e i n t h e r a t i o 3 : 2 , t h e n w h a t i s t h e r a t i o o f t h e i r c u r v e d s u r f a c e a r e a s ? A n s w e r : L e t t h e r a d i i a n d s l a n t h e i g h t s o f t w o r i g h t c i r c u l a r c o n e s a r e r 1 , l 1 a n d r 2 , l 2 r e s p e c t i v e l y . R a t i o o f t h e i r C u r v e d S u r f a c e A r e a = ( S i n c e r 1 = r 2 ) = = 3 : 2 1 2 . A b a t s m a n i n h i s 1 1 t h i n n i n g s m a k e s a s c o r e o f 6 8 r u n s a n d t h e r e b y i n c r e a s e s h i s a v e r a g e s c o r e b y 2 . W h a t i s h i s a v e r a g e s c o r e a f t e r t h e 1 1 t h i n n i n g s . A n s w e r : L e t t h e a v e r a g e s c o r e o f 1 1 i n n i n g s b e x : T h e n t h e a v e r a g e s c o r e o f 1 0 i n n i n g s = x – 2 T o t a l s c o r e o f 1 1 i n n i n g s = 1 1 x 1 1 / 2 8 T o t a l s c o r e o f 1 0 i n n i n g s = 1 0 ( x - 2 ) = 1 0 x – 2 0 S c o r e o f t h e 1 1 t h i n n i n g s = T o t a l s c o r e o f 1 1 i n n i n g s – T o t a l s c o r e o f 1 0 i n n i n g s = 1 1 x – ( 1 0 x – 2 0 ) = x + 2 0 x + 2 0 = 6 8 ( G i v e n ) x = 4 8 H e n c e , t h e a v e r a g e s c o r e a f t e r t h e 1 1 t h i n n i n g s i s 4 8 . S e c t i o n C ( Q u e s t i o n n u m b e r s 1 3 t o 2 2 c a r r y 3 m a r k s e a c h ) 1 3 . R e p r e s e n t o n t h e n u m b e r l i n e A n s w e r : C o n s t r u c t i o n : 1 . T a k e a l i n e s e g m e n t A O = 3 u n i t o n t h e x a x i s . 2 . D r a w a p e r p e n d i c u l a r o n O a n d d r a w a l i n e O C = 1 u n i t 3 . N o w j o i n A C . 4 . T a k e A a s c e n t r e a n d A C a s r a d i u s , d r a w a n a r c w h i c h c u t s t h e x a x i s a t E . 5 . A E r e p r e s e n t s u n i t sRead More
FAQs on Class 9 Math: Sample Question Paper- 12 (With Solutions)
| 1. What is the format of the Class 9 Math Sample Question Paper-12? |  |
Ans. The Class 9 Math Sample Question Paper-12 follows a specific format that includes a set of questions along with their solutions. This format helps students understand the types of questions they may encounter in their exams and provides them with solutions to check their answers.
| 2. How can the Class 9 Math Sample Question Paper-12 help students in their exam preparation? | |
Ans. The Class 9 Math Sample Question Paper-12 is a valuable resource for students preparing for their exams. By solving this question paper, students can familiarize themselves with the exam pattern, gain confidence in their problem-solving skills, and identify any gaps in their understanding of the subject. The provided solutions also help them understand the correct approach to solving different types of math problems.
| 3. Are the questions in the Class 9 Math Sample Question Paper-12 similar to those asked in the actual exam? | |
Ans. The questions in the Class 9 Math Sample Question Paper-12 are designed to simulate the level of difficulty and the types of questions that students can expect in their actual exam. While the exact questions may not appear in the exam, solving this sample paper will help students prepare effectively by covering various concepts and problem-solving techniques.
| 4. Can students rely solely on the provided solutions in the Class 9 Math Sample Question Paper-12 for exam preparation? | |
Ans. While the solutions provided in the Class 9 Math Sample Question Paper-12 are helpful for understanding the correct approach to solving the questions, it is essential for students to practice solving similar problems on their own. Relying solely on the provided solutions may hinder the development of problem-solving skills and limit their ability to tackle unfamiliar questions in the exam.
| 5. How can students make the most of the Class 9 Math Sample Question Paper-12 in their exam preparation? | |
Ans. To make the most of the Class 9 Math Sample Question Paper-12, students should attempt to solve the questions without referring to the solutions initially. Once they have completed the paper, they can compare their answers with the provided solutions and identify any mistakes or areas where they need improvement. They should also analyze the solutions to understand the logical steps and concepts used in solving each question. Regular practice of such sample papers can significantly enhance their exam preparation.
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