5TH GRADE MATH PRACTICE PROBLEMS

Problem 1 :

Find the value of (100 + 250 ÷ 10) × 3.

Solution :

= (100 + 25)  × 3

= 125 × 3

= 375

So, option (4) is correct.

Problem 2 :

Which of the following numbers when rounded to the nearest hundred becomes 61,400 ?

Solution :

First consider option (1).

61,349 - Nearest Hundreds : 61,300

Consider option (2).

61,449 - Nearest Hundreds : 61,400

Consider option (3).

61,450 - Nearest Hundreds : 61,500

Consider option (4).

61,495 - Nearest Hundreds : 61,500

So, option (2) is correct.

Problem 3 :

How many sixths are there in 4 5/6 ?

Solution :

So, 29 sixths are there in 4 5/6.

So, option (1) is correct.

Problem 4 :

Which of the following is not equal to 1/2 ?

Solution :

Hence, option (2) is correct.

Problem 5 :

What is the value of X in the scale below ?

Solution :

Between 4.8 and 4.9, we have 5 spaces.

0.1 / 5 = 0.02

So, the value of x is 5.04.

Hence, option (1) is correct.

Problem 6 :

A typist can type 160 words in 5 minutes. At this rate, how long does the typist take to type 320 words ? 

Solution :

160 words = 5 minutes

Multiplying by 2 on both sides

320 words = 10 minutes.

So, a typist can type 320 words in 10 minutes.

Problem 7 :

What is the missing number in the x ?

7 : x = 3 : 15

Solution :

By using cross multiplication on we get,

x = 35

So, option (4) is correct.

Problem 8 :

Four letters P, E, A and R are shown below. How many of the letters has/have a line of symmetry ?

Solution :

Two letters has a line of symmetry.

So, option (2) is correct.

Problem 9 :

In the figure below, AF, BD and CE are straight lines. ∠AFE = 43º, ∠AFB is a right angle. Find ∠CFD.

Solution :

By observing the figure,

Given, ∠AFE = 43^º, 

∠AFB is a right angle. 

So, ∠AFB = 90^º

∠CFD = ∠AFE + ∠AFB

∠CFD = 90^º + 43^º

∠CFD = 133^º

So, option (3) is correct.

Problem 10 :

What is the area of triangle ABC ?

Solution :

Area of triangle = 1/2 (b × h)

= 1/2 (24 × 10)

= 1/2 × 240

= 120 cm²

So, option (3) is correct.

Problem 11 :

The ratio of the cost of an eraser to the cost of a pen is 2 : 5. A ruler costs twice as much as the eraser. What is the ratio of the cost of the ruler to the cost of the eraser to the total cost of the three items ?

Solution :

The cost of an eraser = 2

The cost of a pen = 5

A ruler costs twice as much as the eraser. So, the cost of an eraser = 4

Total cost of the three items = 2 + 5 + 4

= 11

So, the ratio of the cost of the ruler to the cost of the eraser = 4 : 2 : 11.

Hence, option (4) is correct.

Problem 12 :

The average mass of four boys is 52 kg. The masses of Cameron and James are shown in the table below. Leon and Zach are of the same mass each. What is the total mass of Zach and Cameron ?

Solution :

Average mass of four boys is 52 kg.

The masses of Cameron = 42 kg

The masses of  James = 56 kg

The Leon and Zach are of the same mass.

So, The masses of Leon = x

The masses of Zach = x

Total mass is

x = 55

Leon and Zach are of the same mass of 55 and 55.

So, option (1) is correct.

Problem 13 :

Which one of the figures below shows that 40% of the figure is shaded ?

Solution :

Total square box is 20.

40% of the figure is shaded is 40/100.

= 0.4

= 0.4 × 20

= 8

So, option (c) is correct.