Samacheer Kalvi Class 5 Maths Solutions Term 1 Chapter 2 Numbers Exercise 2.8

Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 1 Chapter 2 Numbers Ex 2.8

Question 1. Find quotient and remainder.
(a) \( 5732 \div 9 \)
(b) \( 4735 \div 5 \)
(c) \( 3032 \div 7 \)
(d) \( 43251 \div 10 \)
(e) \( 2532 \div 4 \)
Answer:
(a) For \( 5732 \div 9 \):
\( \begin{array}{r} 636 \\ 9 \overline{) 5732} \\ -54 \downarrow \\ \hline 33 \\ -27 \downarrow \\ \hline 62 \\ -54 \\ \hline 8 \end{array} \)
Quotient = 636, Remainder = 8. This process helps us distribute 5732 items equally among 9 groups.
(b) For \( 4735 \div 5 \):
\( \begin{array}{r} 947 \\ 5 \overline{) 4735} \\ -45 \downarrow \\ \hline 23 \\ -20 \downarrow \\ \hline 35 \\ -35 \\ \hline 0 \end{array} \)
Quotient = 947, Remainder = 0. When the remainder is zero, it means the number is perfectly divisible.
(c) For \( 3032 \div 7 \):
\( \begin{array}{r} 433 \\ 7 \overline{) 3032} \\ -28 \downarrow \\ \hline 23 \\ -21 \downarrow \\ \hline 22 \\ -21 \\ \hline 1 \end{array} \)
Quotient = 433, Remainder = 1. A remainder is what is left over when one number cannot be divided exactly by another.
(d) For \( 43251 \div 10 \):
\( \begin{array}{r} 4325 \\ 10 \overline{) 43251} \\ -40 \downarrow \\ \hline 32 \\ -30 \downarrow \\ \hline 25 \\ -20 \downarrow \\ \hline 51 \\ -50 \\ \hline 1 \end{array} \)
Quotient = 4325, Remainder = 1. Dividing by 10 is easy; the remainder is simply the last digit of the number.
(e) For \( 2532 \div 4 \):
\( \begin{array}{r} 633 \\ 4 \overline{) 2532} \\ -24 \downarrow \\ \hline 13 \\ -12 \downarrow \\ \hline 12 \\ -12 \\ \hline 0 \end{array} \)
Quotient = 633, Remainder = 0. Division is the process of splitting a number into equal parts or finding how many times one number fits into another.
In simple words: For each division problem, we find how many times the divisor fits into the dividend (the quotient) and how much is left over (the remainder).

🎯 Exam Tip: Always double-check your division by multiplying the quotient by the divisor and adding the remainder; it should equal the original dividend.

Question 2. Answer the following:
(a) 3057 families are living in a town. The town panchayat decided to split the town into 3 wards. How many families will be there in each panchayat?
(b) A water board distributes 28,049 litres daily to a town in 7 lorries. How much water will each lorry carry?
(c) A company gives Rs 93,300 as salary for 6 workers equally. How much salary will each worker get?
Answer:
(a) Total number of families in the town = 3057
Number of wards = 3
Number of families in each ward = Total families \( \div \) Number of wards
\( = 3057 \div 3 \)
\( = 1019 \)
Each ward will have 1019 families. Equal distribution is a key concept in fair sharing.

(b) Total litres of water distributed daily = 28049 litres
Number of lorries = 7
Litres of water each lorry will carry = Total litres \( \div \) Number of lorries
\( = 28049 \div 7 \)
\( = 4007 \) litres
Each lorry will carry 4007 litres of water. Division helps us figure out how much each part gets when a total is shared.

(c) Total salary given to workers = Rs 93,300
Number of workers = 6
Salary each worker will get = Total salary \( \div \) Number of workers
\( = 93300 \div 6 \)
\( = 15550 \)
Each worker will get Rs 15,550 as salary. This calculation ensures everyone receives a fair share of the total amount.
In simple words: We used division to find out how many families are in each ward, how much water each lorry carries, and how much salary each worker gets when things are shared equally.

🎯 Exam Tip: For word problems, always identify the total amount and the number of parts to divide it into. Clearly state your steps and final answer with units.