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Model Question Paper-3 Class-6 ML Aggarwal ICSE Maths Solutions

Model Question Paper-3 Class-6 ML Aggarwal ICSE Maths Solutions . APC Understanding Mathematics for ICSE Class-6 Model Question Paper-3 Solutions based on Chapter-1 to 8. Solutions of Section-A , B, C and D of Model Question Paper-3 for Class-6 ML Aggarwal. Visit official Website CISCE for detail information about ICSE Board Class-6 Mathematics.

Model Question Paper-3 Class-6 ML Aggarwal ICSE Maths Solutions


-: Select Topic:-

Section-A

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Section-B

Section-C

Section-D

 


 Model Question Paper 3

 

( based on chapter-1 to 8)

Time Allowed-2 hour 30 minutes

max mark-90


Note :

(i) All Questions are compulsory

(ii) Questions 1-2 carry 1 mark each, Questions 3-5 carry 2 marks each, Questions 6-8 carry 3 marks each and Questions 9-10 carry 4 marks each.


Section-A

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Questions 1 to 8 are of 1 mark each.
Choose the correct answer from the given four options (1 to 8):

Question 1.
The difference between the place value and face value of 5 in the numeral 70542 is:
(a) 0
(b) 42
(c) 495
(d) 537

Solution:

Place value of 5
70540
500
Face value of 5
70542
5
Difference = 500 – 5 = 495 (c)

Question 2.
he sum of the successor of 99 and the predecessor of 101 is
(a) 198
(b) 200
(c) 201
(d) 199

Solution:

Successor of 99 = 100
Predecessor of 101 = 100
∴ Sum = 100 + 100 = 200 (b)

Question 3

If x and y are two co-prime numbers, then their LCM is
ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 1

Solution:

ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 2

Question 4.
Which of the following statement is true?
(a) |15 – 6| = |15| + |—6|
(b) Additive inverse of-3 is 3
(c) -1 lies on the right of 0 on the number line
(d) -4 is greater than -3

Solution:

Additive inverse of -3 is 3 (b)

Question 5.
The fraction equivalent to \frac{360}{540} is
ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 3

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Solution:

\frac{360}{540} when divide by 90 = \frac{360 \div 90}{540 \div 90}=\frac{4}{6} (a)

Question 6.
Which of the following numbers is divisible by 6?
(a) 5372
(b) 6495
(c) 7632
(d) 7568

Solution:

The given number = 7632
It’s unit digit = 2
So, it is divisible by 2
Sum of its digits = 7 + 6 + 3 + 2 = 18
Which is divisible by 3
∴ 7632 is divisible by both 2 and 3.
Hence it is divisible by 6 (c)

Question 7.
5 kg 5g is equal to
(a) 5.5 kg
(b) 5.05 kg
(c) 5.005 kg
(d) 5.0005 kg

Solution:

5 kg + \frac{5}{1000} kg = 5 kg + 0.005 kg = 5.005 kg

Question 8.
The ratio of number of girls to the number of boys in a class is 7 : 5. If there are 15 boys in the class, then the number of girls in the class is
(a) 14
(b) 21
(c) 28
(d) 36

Solution:

Let number of girls = x
As per question,
7 : 5 = x : 15
\frac{7}{5}=\frac{x}{15}
x = \frac{7}{5} × 15 = 21 girls


Section – B

Questions 9 to 14 are of 2 marks each.

Question 9.
Write the greatest and the smallest 4-digit numbers using four different digits with the condition that the digit 4 occurs at tens place.

Solution:

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9847; 1042

Question 10.
Find the prime factorisation of 980.

Solution:

980
ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 4
∴ 980 = 2 × 2 × 5 × 7 × 7

 

Question 11.
Find the value of-15 – (-2) – 71 – (-8) + 6.

Solution:

-15 + 2 – 71 + 8 + 6
8 + 6 + 2 – 15 – 71 = -70

Question 12.
What fraction of the given figure is the shaded part?
ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 5

Solution:

Total parts in the given figure = 16
Shaded parts = 53
∴ Fraction = \frac{5}{16}

Question 13.
Write the mixed fraction 7 \frac{3}{40} as a decimal number.

Solution:

Decimal number of 7 \frac{3}{40}=\frac{283}{40} = 7.075

Question 14.
The length of a pencil is 14 cm and its diameter is 7 mm. What is the ratio of the diameter of the pencil to that of its length?

Solution:

Length of a pencil = 14 cm = 140 mm
Diameter of a pencil = 7 mm

Ans 14 Model Question Paper 3 Ratio and Proportion Class-6 ML Aggarwal ICSE Maths APC Solutions


Section – C

 Questions 15 to 24 are of 4 marks each.

Question 15.
Estimate the product 2459 x 653 by rounding off each factor to its
(i) greatest place
(ii) nearest hundreds.

Solution:

(i) Rounding off each factor to its greatest place
∴ Product = 2400 × 600 = 1440000
(ii) nearest hundreds of 2459 is 2500 and 653 is 700
∴ Product = 2500 × 700 = 1750000

Question 16.
By using distributive laws, find 257 × 1007.

Solution:

257 × 1007
= 257 × (1000 + 7)
= 257000 + 1799
= 258799

Question 17.
Using number line, subtract (-3) from (-8).

Solution:

Start from -8 on the number line.
Move 3 units to the right. We reach at -5
∴ -8 – (-3) = -5
ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 7

Question 18.
Find the greatest number that will divide 76, 113 and 186 leaving remainders 4, 5 and 6 respectively.

Solution:

76 – 4 = 72
113 – 5 = 108
186 – 6 = 180
HCF of 72, 108 and 180 is 36
∴ The greatest number that will divide 72,108 and 180 is 36.

Question 19.
Arrange the following integers in descending order:
-506, 2376, 2367, -311, -509, 245.

Solution:

-506, 2376, 2367, -311, -509, 245
Descending order:
2376, 2367, 245, -311, -506, -509

Question 20.
Arrange the following fractions in ascending order \frac{5}{12}, \frac{1}{4}, \frac{7}{8}, \frac{5}{6}.

Solution:

\frac{5}{12}, \frac{1}{4}, \frac{7}{8}, \frac{5}{6}
Taking LCM of 12, 4, 8 and 6
ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 8
= 2 × 2 × 2 × 3 = 24
Making all the fractions with denominator = 24
ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 3 9

Question 21.

Simplify: 3 \frac{5}{6}+4 \frac{3}{4}-5-1 \frac{3}{8}.

Solution:

Ans 21 Model Question Paper 3 Ratio and Proportion Class-6 ML Aggarwal ICSE Maths APC Solutions

Question 22

Write all possible natural numbers using the digits 3, 0, 8. Repetition of digits is not allowed. Also find their sum.

Solution:

The given digits are 3, 0, 8 and repetition of digits is not allowed.
The one- digit numbers that can be formed are 0, 3 and 8.
We are required to write 2-digit numbers.
Out of the given digits, the possible ways of choosing the two digits are
3, 0, 3, 8; 8, 0
Using the digits 0 and 3, the numbers are 30.

Similarily, Using the digits 3 and 8, the numbers are 38 and 83.
Using the digits 8 and 0, the numbers are 80.
Hence, all possible 2-digit numbers are 30, 38, 83, 80
Now, We are required to write 3-digit numbers using the digits 0, 3, 8
and the repetition of the digits is not allowed.
Keeping 0 at unit’s place, the 3-digit number obtained are 380 and 830.
Keeping 3 at unit’s place, the 3-digit number obtained are 803.
Keeping 8 at unit’s place, the 3-digit number obtained is 308.
Hence, all possible 3-digit numbers are : 380, 830, 803, 308
All possible numbers using the digits 3, 0 and 8 are :
0, 3, 8, 30, 80, 38, 83, 380, 830, 803, 308.
∴ Sum of all possible numbers obtained using the digits 3, 0, 8
= 0 + 3 + 8 + 30 + 80 + 38 + 83 + 380 + 830 + 803 + 308 = 2563

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