Class 5 math EV chapter 5 creative questions with ans
Question 01
Four bells rang at once in the beginning and then rang after every 5, 7, 12, 15 seconds.
(a) What is the LCM of the time of ringing of first two bells
(b) What is general price factor of time of ringing of 3rd and 4th bell?
(c) Which smallest number can be divided by the ringing times of the bells without remainder?
(d) After what minimum time the bells will ring together again?
Solution:
(a)
5 = 1×5 and 7 =1 x 7
LCM of 5 and 7 is = 1 x 5 x 7 = 35
(b)
12=2x2x3 and 15=3 x 5
The common prime factor of 12 and 15 is 3
(c)
The LCM of the ringing times is required smallest number
3) 5, 7, 12, 15
5) 5, 7, 4, 5
1, 7, 4, 1
LCM of the numbers = 3 x 5 x 7 x 4 = 420
Required smallest number is 420
(d)
The LCM of 5, 7, 12, 15 is the required minimum time
The minimum time is 420 sec and =\[\frac{420}{60}\] minute = 7 minute
After 7 minutes the bell will ring together again
Question 02
In a class the teacher told the students to write the smallest number of 4 digits and biggest number of 3 digits, whose digit of ones place will be 8.
(a) Write the 2 numbers
(b) What is the sum and differences of the two numbers?
(c) Write the prime factors of the biggest number.
(d) What is the LCM of the numbers?
(e) What is the GCF of the numbers?
Solution:
(a)
Smallest number of four digits whose digit in the ones place is 8 is = 1008
And, biggest number of three digits whose digit in the ones place is 8 is = 998
(b)
Sum of the numbers = (1008 + 998) = 2006
Difference of the numbers = (1008 – 998) = 10
(c)
Biggest number is 1008
1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
The prime factors of 1008 are 2,2,2,2,3,3,7
(d)
998 = 2×499
Prime factors of 998 are 2, 499
The common prime factors of 998 and 1008 is 2
GCF of 1008 and 998 is 2
(e)
LCM of 1008 and 998 is
=2 x 2 x 2 x 2 x 3 x 3 x 7 x 499
=16 x 9 x 7 x 499 = 502992
LCM of 1008 and 998 is 502992
Creative math questions and solutions for Class 5 EV Chapter 5 to make learning easier and more fun.
Question 03
Three values are given, 24, 48, 72.
(a) Write down 2 multiples of 24, 48, 72
(b) Write down the common multiples of 24 and 72
(c) Find the L.C.M of 24, 72
(d) Find the L.C.M of 24, 48, 72.
Ans.
(a)
2 multiples of 24 are = 48, 72
2 multiples of 48 are = 96, 144
2 multiples of 72 are = 144, 216
(b)
Multiples of 24 are = 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288
Multiples of 72 are = 144, 216, 288, 360, 432
Common multiples of 24 and 72 are = 144, 216, 288, …..
(c)
2)24, 72
2)12, 36
2)6, 18
3)3, 9
1, 3
L.C.M of 24, 72 is = 2 x 2 x 2 x3 x 1 x 3 = 72
(d)
2)24, 48, 72
2)12, 24, 36
2)6, 12, 18
3) 3, 6, 9
1, 2, 3
L.C.M of 24, 48, 72 is = 2 x 2 x 2 x 3 x 2 x 3 x 1
= 144
Question 04
Mr. Amzad brought about 40 mangoes, 85 apples from the market. He distributed the fruits equally. The price of 4 mangoes is 40 taka.
(a) What is the price of mangoes?
(b) If per piece cost 5 taka how much price will the apples be?
(c) He distributed the fruits how many members?
(d) Each person gave how many fruit?
Answer:
(a)
The price of 4 piece of mango is 40 taka
The price of 1 piece of mango is (40÷4) = 10 taka
The price of 40 pieces of mango is (40×10) = 400 taka
(b)
1 piece apples cost 5 taka
85 pieces of apples cost (5×85) = 425 taka
(c)
40 = 2 x 20
= 2 x 2 x 10
= 2 x 2 x 2 x 5
85 = 5 x17
GCF = 5
He distributed the fruits among 5 person.
(d)
Each person gave (40 ÷ 5) = 8 piece mango
Each person gave (85 ÷ 5) 17 piece apple
Easy and fun questions with answers for Class 5 EV Chapter 5 to help students improve their math skills.
Question 05
A number of saplings is such that when 3, 5, 6, 8, 10 or 15 are planted in each row, every time two saplings are left out.
(a) What is the GCF of last 3 numbers?
(b) What is the LCM of the given numbers?
(c) what is the minimum number of saplings?
(d) If we plant 18 saplings in 16 rows then how many more saplings do we need?
Solution:
(a)
8 = 1 x 2 x 2 x 2
10 = 1 x 2 x 5
15 = 1 x 3 x 5
GCD of 8, 10 and 15 is =1
(b)
2) 3, 5, 6, 8, 10, 15
3) 3, 5, 2, 4, 5, 15
1, 5, 1, 4, 1, 5
LCM = 2 x 3 x 5 x 4 = 120
(c)
The minimum number of sampling is = (120 + 2) =122
(d)
Number of extra saplings ={(16 x 8) – 122}
=(128 – 122)
= 6
Question 06
Tushi set a few no. of bell in their drawing room. After ringing together they rand after every 6, 9, 12, 15 seconds respectively.
(a) Find the GCF
(b) Find the LCM
(c) What is the summation of 4 multiples of 9?
(d) When the bells will ring together again?
Ans.
(a)
6 = 2 x 3
9 = 3 x 3
12 = 2 x 2 x 3
15 = 3 x 5
GCF of 6, 9, 12, 15 is = 3
(b)
3) 6, 9, 12, 15
2) 2, 3, 4, 5
1, 3, 2, 5
LCM of 6, 9, 12, 15 = 3 x 2 x 3 x 2 x 5 x 1 = 180
(c)
4 multiples of 9 are = 9, 18, 27, 36.
Sum of 4 multiples of 9 is = 9 + 18 + 27 + 36=90
(d)
3)6, 9, 12, 15
2)1, 3, 2, 5
1, 3, 2, 5
LCM = 3 x 2 x 3 x 2 x 5 x 1=180
The bells of the drawing room will ring again together after 10 seconds.
Question 07
16, 24, 2 and 40 are four even numbers.
(a) Write the factors of first 2 numbers
(b) Write 2 multiples of last wo numbers.
(c) Which smallest number when divided by the four number gives 6 as remainder?
(d) Find the next number after 4 & 6 which can be divided by 32 without any remainder.
Solution:
(a)
Factors of 16 are = 1, 2, 4, 8, 16
Factors of 24 are = 1, 2, 3, 4, 8, 12, 24
(b)
32 x 1 = 32
32 x 2= 64
The multiples of 32 are 32, 64 and 40 x 2 = 80
(c)
The multiples of 40 are 40, 80.
2) 16, 24, 32, 40
2) 8, 12, 16, 20
2) 4, 6, 8, 10
2) 2, 3, 4, 5
1, 3, 2, 5
LCM of 16, 24, 32, 40 is = 2x2x2x3x2x5 = 480
Required smallest number = (480 + 6) = 486
(d)
Required number = (486 + 32) -6
= 518 – 6 = 512
Question 08
210 mangoes and lychees are divided among some boys and girls. If the number of mangoes is 60 then.
(a) What is the number of lychees?
(b) Write the number of mangoes and lychees in prime factors.
(c) Among how many maximum number of boys and girls the fruits can be divided equally so that no fruit is left?
(d) How many fruits will each one get?
(e) How many mangoes and how many lychees will each get?
Solution:
(a)
Number of lychees = (210-60) = 150
(b)
Number of mangoes and lychees are 60 and 150
60 = 2 x 2 x 3 x 5
150 = 2 x 3 x 5 x 5
(c)
If we divide 60 mangoes and 150 lychees among boys and girls with no fruit being left, then GDS of 60 and 150 is the required maximum number of boys and girls.
The common prime factors of 60 and 150 are = 2,3 and 5
Required maximum number of boys and girls is 30
(d)
Number of fruits each one will get is = (210 ÷ 30) =7
(e)
Number of mangoes each one will get is = (60 ÷ 30)= 2
Number of lychees each one will get is = (150 ÷ 30) = 5
Master Class 5 Math EV Chapter 5 with creative questions and simple answers for better learning.
Question 09
The length and breath of a rectangular house is 7.20 meters and 44 decimeters. The floor of the house has to be fitted with marbles so that no marble has to be broken.
(a) Express the length of the house in decimeter
(b) What is the area of the house in square decimeter?
(c) Express the length and breath of the house in prime factors.
(d) What is the maximum size of the marble stone needed?
(e) How many marble stones are needed for the floor?
Solution:
(a)
Length of the house = (7.20 x 10) decimeter
= 72 decimeters
(b)
Area of the rectangular house =(length x breath)
=(72 x 44) square decimeters
= 3168 square decimeters
(c)
Length of the house is 72 decimeters and breath is 44 decimeters.
72 = 8 x 9 = 2 x 2 x 2 x 3 x 3
44 = 2 x 22 = 2 x 2 x 11
(d)
The length of the required marble will be equal to the GCF of 72 and 44.
GCF of 72 and 44 = 2 x 2 = 4
The maximum length of the required marble stone is 4 decimeters.
(e)
Are of the square marble stones is
=(4 x 4) square decimeters
= 16 square decimeters
Numbers of marble stones needed for the floor is (3168 ÷ 16) = 198
Question 10
The capacity of holding water for 2 drums are 228 liters and 348 liters.
(a) Express the capacity of first drum in prime factors.
(b) What are the common prime factors of the capacity of 1st and 2nnd drum?
(c) What will be the maximum capacity of a pitcher with which we can fill the drums by pouring water integer numbers of times?
(d) How many pitcher of water each drum can hold?
(e)How many pitcher of water is totally needed to fill 2 drums?
Solution:
(a)
Water holding capacity of the 1st drum is 228 liter.
228= 2 x 2 x 3 x 19
(b)
Water holding capacity of the 2nd drum is 348 liters.
348 = 2 x 2 x 3 x29
The common factors of 228 and 348 are 2, 2,3
(c)
The GCF of the capacities of the two drums is the required maximum capacity of the pitcher.
GCF of 228 and 348 is = 2 x 2 x 3 = 12
The maximum water holding capacity of the pitcher is 12 liters.
(d)
Amount of water needed to fill the first drum = (228 ÷ 12) pitchers = 19 pitchers
Amount of water needed to fill the 2nd drum = (348 ÷ 12) pitchers = 29 pitchers
(e)
To fill the two drums total amount of water needed is = (29 + 19) pitchers
= 48 pitchers
More Questions:
Q-11
Three bells having tolled together began to toll after every 9, 12 and 15 minutes.
(a) What is done to find out after what minimum time will the bells toll together again?
(b) After what time will the bells toll together again?
(c) If the bells began tolling after every 6, 9 and 12 minutes, after what minimum times will the bells toll together
again?
Q-12
100 mangoes and 180 leychees are divided among some children.
(a) What is the largest number of children among whom mangoes and lychees are divided without any reminder?
(b) How many mangoes will each of them get?
(c) How many lychees will each of them get?
Q-13
There are 126 mangoes, 231 lychees and 357 jackfruit sapling are distributed for planting a village.
(a) What is the largest number of villagers among whom sapling can be divided equally?
(b) How many mangoes, how many lychees and how many jackfruit will each of them get?
Q-14
There are two bells. One bell rings after each 12 minutes and the other bell rings after 5 minutes.
(a) What is the L.C.M of the numbers representing ringing time of the two bells?
(b) If a bell ringing after every 7 minutes is included with the given two bells, then the 3 bells will ring altogether. after how long time they once ring together?
(c) If the two bells ring together at 3 p.m, then when will they ring again altogether?
Q-15
There are two drums of capacity to contain water of 228 litres and 348 litres.
(a) What is the G.C.F. of 228 and 348?
(b) How many pitchers of water of the highest capacity will be required to fill both the drums separately?
(c) What is the least number exactly divisible by 228 and 348?
Q-16
The length and the breadth of a hall room of rectangular size are 12 metre and 7 metre respectively.
(a) What is the biggest square size tile that can be used to cover the hall room without breaking any of them.
(b) If the length of the hallroom is increased by 3 metre and the breadth is decreased by 2 metre, what will be the change of area of the hallroom?
Q-17
12, 18, 24 and 30 are four even numbers with 2 digits.
(a) Find out the first 4 multiple of 12.
(b) Find out the LCM of the given numbers.
(c) What is the greatest common number among the given numbers?
Explore creative and easy math questions with answers for Class 5 EV Chapter 5 to help students learn effortlessly.
Q-18
Four bells rang at the same time and then rang again at an interval of 5, 7, 12 and 15 seconds respectively.
(a) What is a prime number?
(b) Write the 1st 4 multiples of 15.
(c) After what interval of time will the bell ring again together?
(d) What is the G.C.F. of 5, 7, 12 and 15?
Q-19
147 lichees were distributed among some boys and it was found that each boy got 11 lichees and 4 lichees were in excess.
(a) Express the above information mathematically in the form of an open sentence?
(b) Find out the number of boys among whom the lichees were distributed.
(c) If \[\frac13\] of the liches are distributed among some boys and each boy gets 11 lichees leaving 5 as excess, express the above information in an open sentence.
Q-20
A particular numbers is multiplied by and then the product is divided by 6 to get 21 as quotient.
(a) Write the open sentence with the help of the information given above.
(b) What is the value of the particular number according to the open sentence in (a)?
(c) If 3 times x equals 3 more than 45, what is the value of x?
