5TH GRADE MATH PRACTICE QUESTIONS

Question 1 :

If a 6-sided dice is rolled twice, then the probability for a sum of 2 or 12 is.

(A) 1/18      (B) 1/15       (C) 1/16

Solution :

Sample space n(S)  =  36

Let A be the event of getting the sum 2 or 12.

A  =  { (1, 1) (6, 6)}

n(A)  =  2

p(A)  =  n(A) / n(S)

p(A)  =  2/36

p(A)  =  18

Question 2 :

The two angles of a triangle are complementary, Then, the measure of the third angle is

(A) 20     (B) 40       (C) 90

Solution :

If two angles are complementary, then sum of those angles will be 90. Then the other angle is also 90, because the sum of interior angles of a triangle is 180.

Question 3 :

There are 9 soup cans in a row and 8 are stacked on top, then 7, then 6 and so forth until there is only 1 can stacked on the very top. How many soup cans were used to form this arrangement ?

(A) 14        (B) 52        (C) 45

Solution :

By writing the number of soup cans in each row as sequence, we get 

9, 8, 7, .......1

To find the number of soup cans in total, we should find the sum of the above sequence.

Number of terms in the sequence is 9.

  =  n(n + 1) / 2

  =  9(9 + 1) / 2

  =  9 (5)

  =  45

Hence there are 45 soup cans are in the arrangement.

Question 4 :

How many positive single digit prime numbers greater than one is there?

(A) 2      (B) 4        (C) 1

Solution :

Prime numbers are 2, 3,  5, 7.

So, we have 4 one digit prime numbers are there.

Question 5 :

How many positive even prime numbers are there?

(A) 2       (B) 1     (C) 5

Solution :

There is only one even prime number is there. That is 2.

Question 6 :

If the product of 3 and x² is 48, then the value of “x” is

(A) 4       (B) 5      (C) 8

Solution :

3x²  =  48

x²  =  16

x  =  √16

x  =  4

Question 7 :

The sum of the Interior and exterior angles of a triangle is 

(A) 180     (B) 360     (C) 540

Solution :

The sum of interior angle of the triangle is 180 and the sum of exterior angle of triangle is 180.

Question 8 :

If the radius and height of a right circular cylinder are 8 and 5 respectively, then the volume of the cylinder is

(A) 320π       (B) 420π          (C) 360π

Solution :

Radius (r)  =  8 and height (h)  =  5

Volume of cylinder  =  πr²h

  =  π ⋅ 8² ⋅ 5

  =  320π

Volume of cylinder is 320π.

Question 9 :

A group of 2000 bees can make 7 jars of honey in one year. How long will it take for 5000 bees to make 70 jars of honey (in years) ?

(A) 6          (B) 7         (C) 4

Solution :

A group of 2000 bees can make 7 jars of honey in one year.

2000 ⋅ 70 ⋅ 1  =  5000 ⋅ 7 ⋅ x

x  =  (2000 ⋅ 70) / (5000 ⋅ 7)

x  =  4

Then the required number of years is 4.

Question 10 :

The radius and height of a cone is 7/2 and 12 respectively, calculate the volume of the cone?

(A) 18π        (B) 49π         (C) 25π

Solution :

Radius (r)  =  7/2 and height (h)  =  12

Volume of cone  =  (1/3) πr²h

  =  (1/3) π (7/2)²(12)

  =  49π

Question 11 :

What is the value of 30 - (8 + 16)  ÷ 3 x 2 ?

Solution :

= 30 - (8 + 16) ÷ 3 x 2

= 30 - 24 ÷ 3 x 2

= 30 - 8 x 2

= 30 - 16

= 14

So, the answer is 14.

Question 12 :

In the figure EOH, FOJ and GOK are straight lines. ∠EOF = 31° and ∠KOH = 96°. Find ∠GOF

Solution :

∠GOF = ∠KOJ

∠EOF = ∠JOH = 31

∠EOG = ∠KOH 

∠EOF + ∠FOG = 96

31 + ∠FOG = 96

 ∠FOG = 96 - 31

= 65

Question 12 :

The ribbon was cut into two pieces in the ratio 3 : 5. The length of the longer piece was 35 cm. What was the length of the ribbon at first ?

Solution :

3x and 5x are the measures of two pieces.

Length of the ribbon originally = 3x + 5x

= 8x

Length of longer piece = 35 cm

5x = 35

x = 35/5

x = 7

Original length of the ribbon = 8(7)

= 56 cm

Question 12 :

The bar graphs shows the time taken by 5 children to complete a race.

a) Who was the fastest runner ?

b) How much longer did Dan take than Elvin to complete

Solution :

a) By observing the bar graph, it is clear that Ben runs faster.

b) Time taken by Dan = 64

Time taken by Elvin = 48

= 64 - 48

= 16 seconds

Question 13 :

In the figure, rectangle WXYZ is made 8 identical smaller rectangles. What fraction of rectangle WXYZ is shaded ?

Solution :

Let L be the length of the rectangle and W be the width of the one small rectangle.

Area of triangle at left = 1/2 x base x height

= 1/2 x 3L x W

= 3LW/2

Area of trapezium at right = 1/2 x height x sum of parallel sides

= (1/2) x W x (2L + 5L)

= 7LW/2

Area of shaded part = 3LW/2 + 7LW/2

= 10LW/2

= 5 LW

Area of the entire rectangle = l

= 8L x W

= 8LW

Fraction part of the area shaded 

= 5LW/8LW

= 5/8 of the part is shaded.

Question 14 :

Devi had 5/8 kg of sugar. She used 4/5 of it to make some desert. What was the mass of the sugar that she used ?

Solution :

Quantity of sugar = 5/8 kg

Quantity of sugar used = 4/5 of 5/8

= 4/5 x (5/8)

= 4/8

= 1/2 kg

So, the amount of sugar used is 1/2 kg.

Question 15 :

Mrs Tan baked 505 muffins

She sold 125 muffins in the morning and 2/5 of the muffins in the evening. How many muffins did she sell altogether ?

Solution :

Number of muffins baked = 505

Number of muffins sold in the morning = 125

= 2/5 of 505

= (2/5) x 505

= 2 x 101

= 202

Number of muffins together = 202 + 125

= 327

Question 16 :

Mr Lee spent 4/9 of his salary on food and 3/10 of the remainder on transport. He then saved the rest of his salary. Mr. Lee spent $56 more on food than the amount he saved. How much was Mr. Lee's salary ?

Solution :

Let x be the salary.

Amount spent on food = 4/9 of salary

= 4x/9

Amount spent on transport = 3/10 of 5/9 of salary

= 3/10 (5/9) x

= x/6

Amount remaining = x - [(4x/9) + x/6]

= [18x - (8x + 3x)]/18

= (18x - 11x)/18

= 7x/18

4x/9 = (7x/18) + 56

4x/9 - 7x/18 = 56

(8x - 7x)/18 = 56

x = 56(18)

= 1008

So, the required salary is 1008.

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