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Saturday, April 4, 2015
Wednesday, April 1, 2015
CBSE Class 9  Maths  Number Systems (Very Short Questions and Answers)
NUMBER SYSTEMS
Very Short Q & A
Q1: Categorize the following numbers as
(i) Natural Numbers
(ii) Whole Numbers
(iii) Integers
(i) Natural Numbers
(ii) Whole Numbers
(iii) Integers
(iv) Rational Numbers
7, 3/2, 0, 2, 9/5, 1/4, 8
Answer:
(i) Natural Numbers {2, 7}
(i) Natural Numbers {2, 7}
(ii) Whole Numbers {0, 2, 7}
(iii) Integers {8, 0, 2, 7}
(iv) Rational Numbers {8, 9/5, 0, 1/4, 3/2, 2, 7}
Q2: What is the rational number which does not have reciprocal?
Answer: 0 (zero)
Q3: Write the greatest negative integer.
Answer: 1
Q4: Name the smallest natural number.
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class9maths
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Sunday, March 29, 2015
CBSE Class 10  Maths  CH1  Real Numbers (Euclid's Division Lemma)
Euclid's Division Lemma
Important Questions asked in Examination...
Euclid (4th BC)
credits: wikipedia
credits: wikipedia
Q1: State Euclid's Division Lemma.
Answer: Given positive integers a and b ( b ≠ 0 ), there exists unique integer numbers q and r satisfying a = bq + r, 0 ≤ r < b. where
a is called dividend
b is called divisor
q is called quotient
r is called remainder.
e.g. 17 = 5 × 3 + 2
Q2: Prove Euclid's Division Lemma.
Answer: According to Euclid's Division lemma, for a positive pair of integers there exists unique integers q and r, such that
a = bq + r, where 0 ≤ r < b
Let us assume q and r are not unique i.e. let there exists another pair q0 and r0 i.e. a = bq0 + r0, where 0 ≤ r0 < b
⇒ bq + r = bq0 + r0
⇒ b(q  q0) = r  r0 ................ (I)
Since 0 ≤ r < b and 0 ≤ r0 < b, thus 0 ≤ r  r0 < b ......... (II)
The above eq (I) tells that b divides (r  r0) and (r  r0) is an integer less than b. This means (r  r0) must be 0.
⇒ r  r0 = 0 ⇒ r = r0
Eq (I) will be, b(q  q0) = 0
Since b ≠ 0, ⇒ (q  q0) = 0 ⇒ q = q0
Since r = r0 and q = q0, ∴ q and r are unique.
Q3: Prove that the product of two consecutive positive integers is divisible by 2?
Answer: Given positive integers a and b ( b ≠ 0 ), there exists unique integer numbers q and r satisfying a = bq + r, 0 ≤ r < b. where
a is called dividend
b is called divisor
q is called quotient
r is called remainder.
e.g. 17 = 5 × 3 + 2
Q2: Prove Euclid's Division Lemma.
Answer: According to Euclid's Division lemma, for a positive pair of integers there exists unique integers q and r, such that
a = bq + r, where 0 ≤ r < b
Let us assume q and r are not unique i.e. let there exists another pair q0 and r0 i.e. a = bq0 + r0, where 0 ≤ r0 < b
⇒ bq + r = bq0 + r0
⇒ b(q  q0) = r  r0 ................ (I)
Since 0 ≤ r < b and 0 ≤ r0 < b, thus 0 ≤ r  r0 < b ......... (II)
The above eq (I) tells that b divides (r  r0) and (r  r0) is an integer less than b. This means (r  r0) must be 0.
⇒ r  r0 = 0 ⇒ r = r0
Eq (I) will be, b(q  q0) = 0
Since b ≠ 0, ⇒ (q  q0) = 0 ⇒ q = q0
Since r = r0 and q = q0, ∴ q and r are unique.
Q3: Prove that the product of two consecutive positive integers is divisible by 2?
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class10maths
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Saturday, January 17, 2015
CBSE Class 9  CH 12  Sound (Study Points)
Sound
Visual Representation of Sound Waveform (image courtsey: openclipart) 
Study Points
1. Sound travels from one point to another in the form of waves.
2. Sound is produced due to vibration of different objects.
3. Sound waves are longitudinal in nature.
4. Sound travels as a longitudinal wave through a material medium.
5. Sound travels as successive compressions and rarefactions in the medium.
6. In sound propagation, it is the energy of the sound that travels and not the particles of the medium.
7. Sound cannot travel in vacuum.
8. Sound waves cannot be polarised.
9. The change in density from one maximum value to the minimum value and again to the maximum value makes one complete oscillation.
10. The part or region of a longitudinal wave in which the density of the particles of the medium is higher than the normal density is known as compression.
11. The part or region of a longitudinal wave in which the density of the particles of the medium is lesser than the normal density is called a rarefaction.
12. The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength, λ.
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class9science
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CBSE Class 12  Chemistry  Question Paper (2014)
CHEMISTRY (Theory)
2014


[Time allowed: 3hrs]  [Maximum Marks:70] 
General Instructions :
(i) All questions are compulsory.
(ii) Questions numbers 1 to 8 are very shortanswer questions and carry 1 mark each.
(iii) Questions numbers 9 to 18 are also shortanswer questions and carry 2 marks each.
(iv) Questions numbers 19 to 27 are longanswer questions and carry 3 marks each.
(v) Questions numbers 28 to 30 are longanswer questions and carry 5 marks each.
(vi) Use Log Tables,if neccessary.Use of calculators is not allowed.
(i) All questions are compulsory.
(ii) Questions numbers 1 to 8 are very shortanswer questions and carry 1 mark each.
(iii) Questions numbers 9 to 18 are also shortanswer questions and carry 2 marks each.
(iv) Questions numbers 19 to 27 are longanswer questions and carry 3 marks each.
(v) Questions numbers 28 to 30 are longanswer questions and carry 5 marks each.
(vi) Use Log Tables,if neccessary.Use of calculators is not allowed.
1.  Give one example of sol and gel?  1 
2.  Which reducing agent is employed to get copper from the leached low grade copper ore?  1 
3.  Write the IUPAC name of the compound  1 
CH_{3}  CH  CH_{2}  CHO
NH_{2} 

4.  Which of the following isomers is more volatile :  1 
onitrophenol or pnitrophenol ?  
5.  Some liquids on mixing form 'azeotropes'. What are 'azeotropes' ?  1 
6.  Arrange the following in increasing order of basic strength :  1 
C_{6}H_{5}NH_{2}, C_{6}H_{5}NHCH_{3}, C_{6}H_{5}CH_{2}NH_{2} 
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class12chemistry
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Saturday, January 10, 2015
CBSE Class 9  Maths  CH 13  Surface Area and Volume (Study Points)
Surface Area & Volume
Key Concepts
S.no.  Name  Figure  Lateral/ Curved Surface Area 
Total Surface Area(TSA) 
Volume(V)  Symbols use for 

1  cuboid  2(l + b)xh  2(lb + bh + hl)  lbh  l=length b=breadth h=height 

2  cube  4s^{2}  6s^{2}  s^{3}  s = side  
3 
Right circular cylinder 
2Πrh  2Πr(h + r)  Πr^{2}h  h = height r = radius of base 

4 
Right circular cone 
Πrl  Πr(l + r) 
1 3
Πr^{2}h

r = radius of base
h = heightl = Slant height


5  Sphere  4Πr^{2}  4Πr^{2} 
4 3
Πr^{3} 
r = OA = radius 
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class9maths
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Sunday, January 4, 2015
CBSE Class 8/9/10  Reading Comprehension (Set5)
Reading Comprehension
Read the following passages carefully and mark out the correct answers from among the alternatives given below each question in every passage:
PASSAGE  1
The load was very heavy for the old camel. He tried his best to carry it as far as the next village. There his master would be able to get another camel to carry this heavy load. But he was so tried and exhausted that he could not go any further. His master took off the load from the camel’s back and put it on the back of another camel and went his way. A tiger was passing by at this time. He had been hurt by the tusk of an elephant. As the tiger was in pain, he found it difficult to walk. So he lay down by the side of the camel. The camel began licking the tiger’s wound with his long tongue and offered him food that his master had left behind. In a few days, the tiger and the camel recovered. As the tiger was very hungry the old camel told him to kill him and eat his meat. The tiger could not think of killing his friend.
At that time a deer came running towards them. A hunter had shot at him with an arrow. He lay down by the side of the tiger. He told the tiger to kill him and eat his flesh as he did not want the wicked hunter to take him away. The tiger did as he was told. Just then the hunter came on the spot. The angry tiger jumped on him and killed him also. Another young deer was in the bag that the hunter was carrying. The tiger set it free. The tiger and the camel lived happily ever after in the forest.
Q1: The camel began to lick the tiger’s wound because the
(a) tiger was in pain
(b) tiger was tired
(c) tiger could not walk
(d) camel was hungry
(e) camel’s tongue is long.
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class10english,
class8english,
class9english,
CTET
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