Sample Solution Paper 2 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev

 Page 1


  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
CBSE Board 
Class VI Mathematics 
Term I 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
Arrange the numbers in place-value chart: 
Cr T L L T Th Th H T O 
4 7 8 9 6 3 0 4 
4 7 8 9 6 3 4 0 
 
Clearly both numbers have 8 digits. 
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have 
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively. 
  
But at tens place, first number has 0 and second number has 4. 
Clearly, 0 < 4 
Hence  
 47896304 < 47896340 
  
2. Correct answer: B 
To add 0 and 4 on number line, move 4 steps to the right of 0. 
 
3. Correct answer: D 
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest 
hundreds. 
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its 
nearest hundreds as 200. 
So, the required estimation of the product is 50 × 200 = 10000 
 
4. Correct answer: A 
Since, 
  
Therefore, 36 = 2 × 2 × 3 × 3 
Page 2


  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
CBSE Board 
Class VI Mathematics 
Term I 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
Arrange the numbers in place-value chart: 
Cr T L L T Th Th H T O 
4 7 8 9 6 3 0 4 
4 7 8 9 6 3 4 0 
 
Clearly both numbers have 8 digits. 
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have 
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively. 
  
But at tens place, first number has 0 and second number has 4. 
Clearly, 0 < 4 
Hence  
 47896304 < 47896340 
  
2. Correct answer: B 
To add 0 and 4 on number line, move 4 steps to the right of 0. 
 
3. Correct answer: D 
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest 
hundreds. 
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its 
nearest hundreds as 200. 
So, the required estimation of the product is 50 × 200 = 10000 
 
4. Correct answer: A 
Since, 
  
Therefore, 36 = 2 × 2 × 3 × 3 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
5. Correct answer: D 
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7 
 
6. Correct answer: B 
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending, 
it can be seen that they would meet at a point. Hence, they are intersecting lines. 
 
7. Correct answer: B 
The number just before 1000000 is one less than 1000000. 
The required number = 1000000 – 1 = 999999. 
 
8. Correct answer: B 
On a number line, –110 lies next to –111 on the right.  
Therefore, the successor of –111 is –110. 
 
9. Correct answer: A 
The given fraction is  
Dividing the numerator and denominator by 3, we get 
 
Thus,  
 is equivalent to . 
 
10. Correct answer: D 
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6. 
The number 653 is neither divisible by 3 nor by 2 and hence not by 6. 
Now, consider the number 531. 
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3. 
But it is not an even number, so it is not divisible by 2. 
Thus, 531 is divisible by 3 but not by 6. 
 
11. Correct answer: B 
  
 
12. Correct answer: C 
Every quadrilateral has four pairs of adjacent angles. 
Example: For the quadrilateral ABCD, the pairs of adjacent angles are  
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A. 
 
Page 3


  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
CBSE Board 
Class VI Mathematics 
Term I 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
Arrange the numbers in place-value chart: 
Cr T L L T Th Th H T O 
4 7 8 9 6 3 0 4 
4 7 8 9 6 3 4 0 
 
Clearly both numbers have 8 digits. 
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have 
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively. 
  
But at tens place, first number has 0 and second number has 4. 
Clearly, 0 < 4 
Hence  
 47896304 < 47896340 
  
2. Correct answer: B 
To add 0 and 4 on number line, move 4 steps to the right of 0. 
 
3. Correct answer: D 
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest 
hundreds. 
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its 
nearest hundreds as 200. 
So, the required estimation of the product is 50 × 200 = 10000 
 
4. Correct answer: A 
Since, 
  
Therefore, 36 = 2 × 2 × 3 × 3 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
5. Correct answer: D 
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7 
 
6. Correct answer: B 
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending, 
it can be seen that they would meet at a point. Hence, they are intersecting lines. 
 
7. Correct answer: B 
The number just before 1000000 is one less than 1000000. 
The required number = 1000000 – 1 = 999999. 
 
8. Correct answer: B 
On a number line, –110 lies next to –111 on the right.  
Therefore, the successor of –111 is –110. 
 
9. Correct answer: A 
The given fraction is  
Dividing the numerator and denominator by 3, we get 
 
Thus,  
 is equivalent to . 
 
10. Correct answer: D 
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6. 
The number 653 is neither divisible by 3 nor by 2 and hence not by 6. 
Now, consider the number 531. 
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3. 
But it is not an even number, so it is not divisible by 2. 
Thus, 531 is divisible by 3 but not by 6. 
 
11. Correct answer: B 
  
 
12. Correct answer: C 
Every quadrilateral has four pairs of adjacent angles. 
Example: For the quadrilateral ABCD, the pairs of adjacent angles are  
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A. 
 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
Section B  
 
13.    
 
T Cr Cr T L L T Th Th H T O 
(i) 
   
7 0 7 0 7 5 
(ii) 
 
5 3 6 1 8 4 9 3 
 
i. Seven lakh seven thousand seventy five. 
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three. 
 
14.   
 
1. Two lines in the same plane which never intersect are called parallel lines.  
2. Parallel lines remain the same distance apart over their entire length. 
 
15.  
i. Going 6 m to the West 
ii. A withdrawal of Rs 100 
iii. 10 km below sea level  
iv. Spending Rs 500 
 
16. Population of the village = 13295 
Increase in population= Average growth - 1 = 399. 
Population in the successive year = 13295 + 399 = 13694 
 
17.   
  
         Prime factorisation of 455 is 5 × 7 × 13 
         Therefore, the dimensions of the cuboid are 5 cm, 7 cm, 13 cm. 
 
 
 
 
Page 4


  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
CBSE Board 
Class VI Mathematics 
Term I 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
Arrange the numbers in place-value chart: 
Cr T L L T Th Th H T O 
4 7 8 9 6 3 0 4 
4 7 8 9 6 3 4 0 
 
Clearly both numbers have 8 digits. 
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have 
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively. 
  
But at tens place, first number has 0 and second number has 4. 
Clearly, 0 < 4 
Hence  
 47896304 < 47896340 
  
2. Correct answer: B 
To add 0 and 4 on number line, move 4 steps to the right of 0. 
 
3. Correct answer: D 
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest 
hundreds. 
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its 
nearest hundreds as 200. 
So, the required estimation of the product is 50 × 200 = 10000 
 
4. Correct answer: A 
Since, 
  
Therefore, 36 = 2 × 2 × 3 × 3 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
5. Correct answer: D 
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7 
 
6. Correct answer: B 
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending, 
it can be seen that they would meet at a point. Hence, they are intersecting lines. 
 
7. Correct answer: B 
The number just before 1000000 is one less than 1000000. 
The required number = 1000000 – 1 = 999999. 
 
8. Correct answer: B 
On a number line, –110 lies next to –111 on the right.  
Therefore, the successor of –111 is –110. 
 
9. Correct answer: A 
The given fraction is  
Dividing the numerator and denominator by 3, we get 
 
Thus,  
 is equivalent to . 
 
10. Correct answer: D 
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6. 
The number 653 is neither divisible by 3 nor by 2 and hence not by 6. 
Now, consider the number 531. 
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3. 
But it is not an even number, so it is not divisible by 2. 
Thus, 531 is divisible by 3 but not by 6. 
 
11. Correct answer: B 
  
 
12. Correct answer: C 
Every quadrilateral has four pairs of adjacent angles. 
Example: For the quadrilateral ABCD, the pairs of adjacent angles are  
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A. 
 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
Section B  
 
13.    
 
T Cr Cr T L L T Th Th H T O 
(i) 
   
7 0 7 0 7 5 
(ii) 
 
5 3 6 1 8 4 9 3 
 
i. Seven lakh seven thousand seventy five. 
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three. 
 
14.   
 
1. Two lines in the same plane which never intersect are called parallel lines.  
2. Parallel lines remain the same distance apart over their entire length. 
 
15.  
i. Going 6 m to the West 
ii. A withdrawal of Rs 100 
iii. 10 km below sea level  
iv. Spending Rs 500 
 
16. Population of the village = 13295 
Increase in population= Average growth - 1 = 399. 
Population in the successive year = 13295 + 399 = 13694 
 
17.   
  
         Prime factorisation of 455 is 5 × 7 × 13 
         Therefore, the dimensions of the cuboid are 5 cm, 7 cm, 13 cm. 
 
 
 
 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
18. The opposite sides of a parallelogram are parallel and equal. 
Therefore, LM = NO 
 
 
19. 90, 91, 92, 93, 94, 95, 96 are the required numbers. 
 
20. Number of circles in step 1 = 3 = 1 × 2 + 1 
Number of circles in step 2 = 5 = 2 × 2 + 1 
Thus, we can observe that the number of circles is obtained by multiplying the step 
number by 2 and then adding 1. 
Therefore, number of circles in the 100
th
 step = (100 × 2) + 1 = 201 
 
21.   
  
 20570 = 2 × 5 × 11 × 11 × 17 
 
22. Anna is 7 feet above sea level. 
She jumps 3 feet down and walks another 2 feet down. Total distance travelled 
downwards = 3 + 2 = 5 feet. 
 
23. (–13) + (–19) + (+15) + (–10) 
 = –13 – 19 + 15 – 10 
 = –13 – 19 – 10 + 15 
 = –42 + 15 
 = –27 
 
24. 21397 can be estimated as 21000 
27807 can be estimated as 28000 
42305 can be estimated as 42000 
On adding, we get 21000 + 28000 + 42000 = 91000 
Page 5


  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
CBSE Board 
Class VI Mathematics 
Term I 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
Arrange the numbers in place-value chart: 
Cr T L L T Th Th H T O 
4 7 8 9 6 3 0 4 
4 7 8 9 6 3 4 0 
 
Clearly both numbers have 8 digits. 
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have 
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively. 
  
But at tens place, first number has 0 and second number has 4. 
Clearly, 0 < 4 
Hence  
 47896304 < 47896340 
  
2. Correct answer: B 
To add 0 and 4 on number line, move 4 steps to the right of 0. 
 
3. Correct answer: D 
Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest 
hundreds. 
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its 
nearest hundreds as 200. 
So, the required estimation of the product is 50 × 200 = 10000 
 
4. Correct answer: A 
Since, 
  
Therefore, 36 = 2 × 2 × 3 × 3 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
5. Correct answer: D 
13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7 
 
6. Correct answer: B 
NO and PQ can be extended indefinitely on both sides, so they are lines. On extending, 
it can be seen that they would meet at a point. Hence, they are intersecting lines. 
 
7. Correct answer: B 
The number just before 1000000 is one less than 1000000. 
The required number = 1000000 – 1 = 999999. 
 
8. Correct answer: B 
On a number line, –110 lies next to –111 on the right.  
Therefore, the successor of –111 is –110. 
 
9. Correct answer: A 
The given fraction is  
Dividing the numerator and denominator by 3, we get 
 
Thus,  
 is equivalent to . 
 
10. Correct answer: D 
The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6. 
The number 653 is neither divisible by 3 nor by 2 and hence not by 6. 
Now, consider the number 531. 
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3. 
But it is not an even number, so it is not divisible by 2. 
Thus, 531 is divisible by 3 but not by 6. 
 
11. Correct answer: B 
  
 
12. Correct answer: C 
Every quadrilateral has four pairs of adjacent angles. 
Example: For the quadrilateral ABCD, the pairs of adjacent angles are  
(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A. 
 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
Section B  
 
13.    
 
T Cr Cr T L L T Th Th H T O 
(i) 
   
7 0 7 0 7 5 
(ii) 
 
5 3 6 1 8 4 9 3 
 
i. Seven lakh seven thousand seventy five. 
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three. 
 
14.   
 
1. Two lines in the same plane which never intersect are called parallel lines.  
2. Parallel lines remain the same distance apart over their entire length. 
 
15.  
i. Going 6 m to the West 
ii. A withdrawal of Rs 100 
iii. 10 km below sea level  
iv. Spending Rs 500 
 
16. Population of the village = 13295 
Increase in population= Average growth - 1 = 399. 
Population in the successive year = 13295 + 399 = 13694 
 
17.   
  
         Prime factorisation of 455 is 5 × 7 × 13 
         Therefore, the dimensions of the cuboid are 5 cm, 7 cm, 13 cm. 
 
 
 
 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
18. The opposite sides of a parallelogram are parallel and equal. 
Therefore, LM = NO 
 
 
19. 90, 91, 92, 93, 94, 95, 96 are the required numbers. 
 
20. Number of circles in step 1 = 3 = 1 × 2 + 1 
Number of circles in step 2 = 5 = 2 × 2 + 1 
Thus, we can observe that the number of circles is obtained by multiplying the step 
number by 2 and then adding 1. 
Therefore, number of circles in the 100
th
 step = (100 × 2) + 1 = 201 
 
21.   
  
 20570 = 2 × 5 × 11 × 11 × 17 
 
22. Anna is 7 feet above sea level. 
She jumps 3 feet down and walks another 2 feet down. Total distance travelled 
downwards = 3 + 2 = 5 feet. 
 
23. (–13) + (–19) + (+15) + (–10) 
 = –13 – 19 + 15 – 10 
 = –13 – 19 – 10 + 15 
 = –42 + 15 
 = –27 
 
24. 21397 can be estimated as 21000 
27807 can be estimated as 28000 
42305 can be estimated as 42000 
On adding, we get 21000 + 28000 + 42000 = 91000 
  
 
CBSE VI | Mathematics 
Sample Paper 2 – Solution 
 
     
Section C 
25. C stands for 100 
D stands for 500  
V stands for 5 
I stands for 1 
X stands for 10 
M stands for 1000 
In ascending order, the numbers can be arranged as 
1 < 5 < 10 < 100 < 500 < 1000 
Thus, the given roman numerals can be arranged in ascending order as 
I, V, X, C, D, M 
 
26. First we find the LCM of 48, 60, 72. 
48 = 2 × 2 × 2 × 2 × 3 
60 = 2 × 2 × 3 × 5 
72 = 2 × 2 × 2 × 3 × 3 
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720 
Hence, they will meet after = 2 rounds. 
 
27. The given fractions are . 
  
LCM of 2, 3, 6, 9 = (2 x 3 x 3) = 18 
So, we convert each of the given fractions into an equivalent fraction with 18 as the 
denominator. 
Thus, we have: 
 
 Hence, the like fractions are