Bangladesh International School andCollege
Nirjhor, Dhaka Cantonment
CT-2-2025
Class -IV
Chapter-5 (Problems Related to Four Basic Operations )
3. Express the situation to single Mathematical sentence and solve it.
(a) Price of 5 pencils is taka 60. How much is the price of 9 pencils?
(b) Divisor is 3 times the remainder and the quotient is 4 times the divisor. Remainder
is 2. What is the dividend?
(c) Monthly salary of Ms Shampa is taka 7500. She spends taka 7250 for every month. How
much money does she save in a year?
Solution
(a) Mathematical sentence = 60 ÷ 5 × 9
60 ÷ 5 × 9
= 12 × 9
= 108
∴ Price of 9 pencils is Tk. 108 Answer: 108 taka
(b) Given,
Remainder = 2
Divisor
= 3 × 2 = 6, (since divisor is 3 times the remainder)
Quotient = 4 × 6 = 24
we know,
Dividend = Divisor × Quotient + Remainder.
Mathematical sentence = 6 × 24 + 2
= 144 + 2
= 146
Answer: Dividend = 146.
(c) Mathematical sentence = (7500 − 7250) × 12 Ms
Shampa's monthly salary = Tk 7500
Her monthly expenses = Tk. 7250
The amount she saves every month = Tk. (7500 − 7250)
= Tk. 250
The amount she saves in a year = Tk. (250 × 12)
= Tk. 3000
∴ Ms Shampa saves Tk. 3000 every year
Answer: 3000 taka
4.Rupa and Moni have taka 875 together. Moni has taka 125 more than that of
Rupa. How much money do they have separately?
Solution:
Given,
Rupa and Moni altogether have Tk 875
Moni has Tk 125 more than Rupa.
The amount Rupa has = Tk. (875 − 125) ÷ 2
= Tk 750 ÷2
= Tk. 375
∴ The amount, Moni has Tk. (375 + 125) = Tk 500
Answer: Moni and Rupa have Tk. 500 and tk 375 respectively.
5.Sum of the ages of father and son is 55 years. Father's age is 4 times the son's.
How old are they individually?
Solution:
According to the statement,
4 × son's age + son's age = 55 years.
Or, 5 × son's age = 55 years
Or, Son's age
= (55÷5) years
= 11 years.
∴ Father's age = 4 × 11 years
= 44 years.
Answer: Father's age is 44 years and son's age is 11 years.
6.Price of 4 hens and 3 ducks together is taka 639. Price of 1 duck is taka 85.
What is the price of 1 hen?
Solution:
The price of 1 duck = Tk 85
the price of 3 duck = Tk. ( 85 × 3 )
= Tk. 255 .
The price of 4 hens = Tk. (639 - 255) ÷4
= Tk. (384÷4)
= Tk. 96
Answer: The price of a hen is Tk. 96.
Chapter: 7 (Multiples and Factors)
1. There are two bells. Bell A rings every 8 minutes and bell B rings every 6 minutes.
The bells ring together at noon. When is the next time they will ring together?
Solution:
Let, the two bells ring together at 12:00 p.m.
Again, bell A rings every 8 minutes and bell B rings every 6 minutes.
ATQ,
The two bells will ring together at the least common time of ringing the bells.
Multiples of 8: 8, 16, 24, 32, 40, 48........
Multiples of 6: 6, 12, 18, 24, 30, 36,.........
The LCM of 8 and 6 is 24
The two bells will ring together at 12:24 p.m.
Answer : 12:24 p.m
2. I have a piece of paper with 36 cm long and 24 cm wide and I want to spread squares
of the same size all over the surface of the paper. How long is the length of the side of
the biggest square possible?
24 cm
36 cm
Solution:
Length of the sheet of paper is 36 cm and width 24 cm.
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
The common factors of 36 and 24 are 1, 2, 3, 4, 6, 12
HCF of 36 and 24 is: 12
The length of the side of the biggest square is 12 cm.
Answer : 12cm
3. We arrange rectangular tiles 5 cm long and 7 cm wide like the figure on the right to
make the smallest square possible. How many cm are the sides of the smallest square
we can make?
7cm
5cm
Solution:
ATQ,
Area of the tiles having length of 7 cm and breadth of 5 cm is = (7× 5) sq cm.
=35 sq cm.
7cm
5cm
It is evident from the above figure that length and width of the figure is the same, that is 35
cm. So, it is the smallest squares which can be formed with tiles each of length and width of
7 cm and 5 cm respectively.
One side of the smallest square which can be formed with tiles of given size is = 35cm.
Answer : 35cm
4. There are 45 apples and 18 oranges to Rasel. Rasel want to distribute apples and
oranges equally to maximum number of children without any remaining. Find the
number of children. And how many apples and oranges do each children get?
Solution:
Total apples =45 pieces
Total orange= 18 pieces
The highest number of children among which the fruits will be exactly distributed is nothing
but The greatest common factors of 45 and 18.
Factors of 45: 1, 3, 5, 9, 15, 45.
Factors of 18: 1, 2, 3, 6, 9, 18.
HCF of 45 and 18 is 9.
maximum number of children among which the fruits can be distributed is 9.
9 children will get = 45 apples
1 child will get
= (45 ÷ 9 )apples
= 5 apples
9 children will get= 18 oranges
1 child will get
= (18 ÷ 9) oranges
=2 oranges
Answer: Number of children 9,
Each student get 5 apples And 2 oranges.