Practicing with 6th Standard Maths Question Paper with Answers Kerala Syllabus and 6th Standard Maths Mid Term Model Question Paper will help students prepare effectively for their upcoming exams.
Class 6 Maths Mid Term Model Question Paper Kerala Syllabus
Time : 2 Hours
Total Score : 60
Activity – 1
When two lines are crossed over one of the angle between them is given.
a) Find out the other angles.
Answer:
∠BOD = 80
∠AOC = 80(∵ opposite angles)
∠AOD = 180° – 80°= 100°
∠COB =100°
(∵ ∠AOD, ∠DOB are linear pair)
b) Name one pair of opposite angle.
Answer:
∠AOD, ∠COB
c) Give one pair of linear pair.
Answer:
∠AOC, ∠COB
Activity – 2
A rectangular container with length 50 cm breadth 40 cm and height 10 cm is filled with water.
a) Calculate the volume of water.
Answer:
Volume of the container = lbh
= 50cm × 40cm × 10cm
= 20000 cm3
= 20 litre
b) If 10 litre of this water is transferred to another container what would be the height of the remaining water.
Answer:
Transferred water = 10 litre
remaining water = 10 litre
Volume = 10 litre = 10000 cm3
height of water = \(\frac{\text { Volume }}{l \times \mathrm{b}}=\frac{10000}{50 \times 40}\)
= 5 cm
Activity – 3
a) 423 × 46 = 19458 is so answer the following
i) 42.3× 4.6
ii) 4.23 × 4.6
iii) 4230 × 0.46
iv) 0.423 × 46
v) 4.23 × 0.46
Answer:
423 × 46 = 19458
i) 42.3 × 4.6 = 194.58
ii) 4.23 × 4.6 = 19.458
iii) 4230 × 0. 46 = 1945.8
iv) 0.423 × 46 = 19.458
v) 4.23 × 0.46 = 1.9458
b) A gold smith makes 4.5g weighted bangles from 20 g of gold. How many bangles are made what would be the remaining gold.
Answer:
Total weight of gold = 20g
Weight of one bangle = 4.5g
Number of bangles = \(\frac{20}{4.5}\) = 20 ÷ 4.5
= 20 ÷ \(\frac{45}{10}\)
= 20 × \(\frac{10}{45}\)
= 4\(\frac{4}{9}\)
4 bangles are formed.
Remaining gold = \(\frac{4}{9}\)g
Activity – 4
a) 5. 356 ÷ 0.13
Answer:
5. 356 ÷ 0.13 = \(\frac{5356}{1000} \div \frac{13}{100}\)
= \(\frac{5356}{1000} \times \frac{100}{13}\)
= \(\frac{5356}{13 \times 10}=\frac{412}{10}\)
= 41.2
b) \(\frac{0.01 \times 0.001}{0.01 \times 0.1}\)
Answer:
\(\frac{0.01 \times 0.001}{0.01 \times 0.1}\) = (0.1 × 0.001) ÷ (0.01 × 0.1)
c) By what number 0.001 is multiplied to get 0.0001.
Answer:
0.001 × number = 0.0001
number = 0.0001 ÷ 0.001
= \(\frac{1}{10000} \div \frac{1}{1000}\)
= \(\frac{1}{10000}\) × 1000
= \(\frac{1}{10}\)
= 0.1
d) 33\(\frac{1}{3}\)% is equal to which fraction.
Answer:
33\(\frac{1}{3}\)% = \(\frac{100}{3} \%=\frac{100}{3} \times \frac{1}{100}\)
= \(\frac{1}{3}\)
e) Equivalent percent for 5/8.
Answer:
\(\frac{5}{8}\) = (\(\frac{5}{8}\) × 100)%
= \(\frac{5}{8}\) × 100
= \(\frac{125}{2}\)%
= 62 ½ %
f) Cost of 22 pencil is 79.20 find the cost for 10 pencils.
Answer:
Cost of 22 pencil = Rs. 79.20
Cost of 1 pencil = \(\frac{79.20}{22}\) = Rs. 3.6
Cost of 10 pencils = 3.6 × 10 = Rs. 36
Activity – 5
a) Prepare factor table for 960.
Answer:
960 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5
b) 1) Give the number with only two factors.
2) Give the number with three factors.
3) Give number with only four factors.
4) Give the number with odd number of factors.
Answer:
1) For all prime numbers factors are only 2 Eg. 2,3
2) The square number of prime numbers have 3 factors.
Eg.-4, 9, 25
3) The product of two distinct prime number have only 4 factors Eg. 6,15, 10
4) All square numbers have factors of odd number
Eg. 64, 36 etc.
Activity – 6
a) Weight of 3 childrens are 30.6 kg, 28.9 kg & 29.8 kg. Calculate the average weight.
Answer:
Average weight = \(\frac{\text { Total Weight }}{\text { No.of values }}\)
= \(\frac{30.6+28.9+29.8}{3}\)
= \(\frac{89.3}{3}\)
= 29.76 kg
b) Salim Rent 45% of his salary for family and 1/3 of the rest is kept for education. What would be the remaining percent out of his whole?
Answer:
Percentage of salary kept for family = 45%
Rent = 100 – 45 = 55%
= 55 × \(\frac{1}{100}=\frac{11}{20}\)
For education = 1/3 of \(\frac{11}{20}=\frac{11}{60}\)
Remaining = 1 – \(\frac{38}{60}=\frac{22}{60}=\frac{11}{30}\)
(45 × \(\frac{1}{100}=\frac{9}{20}\))
= 1 – \(\frac{38}{60}=\frac{22}{60}=\frac{11}{30}\)
= \(\frac{11}{30} \times \frac{100}{1}=\frac{110}{3}\)%
= 36 \(\frac{2}{3}\)%
Activity – 7
a) Length of a rectangle is increased by 10% and breadth decreased by 5%. What would be the change in area.
Answer:
Length of rectangle = \(\frac{110}{100}\)
Breath = \(\frac{95}{100}\)
Area = \(\frac{110}{100} \times \frac{95}{100}=\frac{11 \times 19}{2 \times 100}\)
= \(\frac{104.5}{100}\)
Area increased by 4.5 %
b) Convert 2\(\frac{1}{3}\) into percent.
Answer:
(2\(\frac{1}{3}\) × 100)% = \(\frac{7}{3}\) × 100% = \(\frac{700}{3}\)%
c) Calculate 20% of 30% of 500.
Answer:
500 × \(\frac{30}{100} \times \frac{20}{100}\) = 30
Activity – 8
a) Draw an equal sided polygon of 5 sides in a circle.
Answer:
Circle = 360
5 sides in a circle = \(\frac{360}{5}\) = 72°
b) Shade 3/8 portion of a circle and convert this into percent.
Answer:
\(\frac{360}{8}\) = 45°
Angle for 3 parts = 45 × 3 = 135°
Percentage of shaded portion = (\(\frac{135}{360}\) x 100)%
= 37\(\frac{1}{2}\) %