5TH GRADE MATH QUESTIONS

Problem 1 :

Divide :

a)

b)

48 ÷ 1.92

Solution :

a)

b) 

Given, 48 ÷ 1.92

= 25

Problem 2 :

Find the following 

a) The product of two numbers is 2160 if HCF is 12 find their LCM.

b) Find the product : 895.41 × 1.01

Solution :

a) Given, the product of two numbers is 2160.

HCF = 12

LCM = ?

The product of two number = HCF × LCM

2160 = 12 × LCM

LCM = 2160/12

LCM = 180

Hence, the LCM is 180.

b) 

Given, 895.41 × 1.01

In 895.41 and 1.01 , ignore the decimal points and consider them as if they were integers. That is  89541 and 101.

Multiply  89541 × 101

In both 895.41 and 1.01, there is 2 digit to the right of the decimal point

So, in the result  9043641, count 4 digits from the right and put a decimal point.

Therefore, 895.41 × 1.01 = 904.3641.

Problem 3 :

a) Simplify using BODMAS : 8.5 - {4.07 - (1.2 - 0.9)} of 1.6.

b) Simplify : 25  ÷ [1 + {8 + (7 1/3 + 8 2/3)}]

Solution :

a) 

= 8.5 - {4.07 - (1.2 - 0.9)} × 1.6 -----> Brackets

=  8.5 - {4.07 - 0.3} × 1.6 -----> Set brackets

= 8.5 - 3.77 × 1.6 -----> Multiplication

= 8.5 - 6.032 -----> Subtraction

= 2.468 -----> Answer

b)

= 25 ÷ [1 + {8 + (7 1/3 + 8 2/3)}]

= 25 ÷ [1 + {8 + (22/3 + 26/3)}]

= 25 ÷ [1 + {8 + (48/3)}]

= 25 ÷ [1 + {8 + 16}]

= 25 ÷ [1 + 8 + 16]

= 25 ÷ 25

= 1

Problem 4 :

Answer the following :

1. Arrange in column and add 769951; 3727690; 85975671 and 562395676.

2. Simplify : 20 ÷ 4 + 5

3. Write the common factor for 120 and 156.

4. 13/12 = 39/? Find the missing one.

5. Write the expanded form for decimal 163.023

Solution :

1.

2. 

= 20 ÷ 4 + 5

= 5 + 5

= 10

3.

The common factor for 120 and 156 is

120 = 2 × 2 × 2 × 3 × 5

156 = 2 × 2 × 3 × 13

∴  HCF of 120 and 156  = 2 × 2 × 3

= 12

4. Let x be the missing number.

x = 36

So, the missing number is 36.

5.

= 163.023

= 100 + 60 + 3 + 0.02 + 0.003 

Problem 5 :

Multiply.

Solution :

1. a)

b)

2. a)

= (8/3) x (1/7) x (4/5)

It is not possible to simplify, so we have to multiply the numerators by the numerator and denominators by the denominator.

= 32/105

b)

= (6/8) x (12/13) x 7

Here 8 and 12 can be simplified using 4 times table. 

= (6/2) x (4/13) x 7

= 3 x (4/13) x 7

= 84/13

So, the answer is 84/13.

Problem 6 :

Find the following :

1. Find the HCF for 136 and 170 using Division method.

2. Find the LCM for 36, 60 and 72 using prime factorization method.

Solution :

1. 

HCF of 136 and 170 is 34.

2.

36 = 2 × 2 × 3 × 3

60 = 2 × 2 × 3 × 5

72 = 2 × 2 × 2 × 3 × 3

∴  LCM of 36, 60 and 72  = 2 × 2 × 2 × 3 × 3 × 5

= 360