1. Give the successor and predecessor of a) 60229 b) 50000 c)25890 2. Solve by suitable rearrangement a) 1762 + 553 + 1338 + 547 b) 4 x 245 x 25 3. Find the value using suitable property. a) 2639 x 8 + 2639 x 92 b) 6796 x 141 – 41 x 6796 4. Write the number names in Indian system of numeration. a) 9475340 b) 30245601 5. Write the number names in International system of numeration. a) 9543201 b) 7000023 6. Using divisibility test, determine which of the following numbers are divisible by 11: a) 10600601 b) 169308 c) 434434 7. Find the first three common multiples of 9,12 and 18 8. Find the common factors of 75,100 and 60 9. Find the HCF of 54 and 81 by prime factorization method. 10. State the divisibility rules for 4 and 8 and check if the number 684572 is divisible by 4 and 8. 11. Test the divisibility of the following number by 6 a) 5634 b) 65990 12. Test the divisibility of the following number by 9 a) 94567 b) 91827 13. If a pencil box costs Rs 85 and a packet of sketch pens costs Rs 24, what is the total cost of 120 pencil boxes and 300 packets of sketch pens? 14. Fruit baskets each weighing 4kg 500g are loaded in a van. How many such baskets can be loaded in a van which can carry only 900kg? 15. Draw the lines of symmetry of the following shapes. a) Equilateral triangle b) Square c) Rectangle d) Isosceles triangle e) Rhombus
Answers
Answer:
successor predecessor
1.a)60230 60228
b)50001 49999
c) 25891 25889
2.a) 1762+ 1338+ 553 +547 = 3200
b) 245× 25 × 4=24,500
3.a) 2,639×8+2,639×92 = 2,63,900
b)6,796×141-41×6,796 = 6,79,600
4. a) 9575340 = 95,75,340
b) 30245601 = 3,02,45,601
5.a) 9543201 = 9,543,201
b) 7000023 = 7,000,023
6.
Step-by-step explanation:
a) ) 60229 Successor = 60229 Predecessor = 60228
b) 50000 Successor = 50001 Predecessor = 49999
c) 25890 Successor = 25891 Predecessor = 258889
2. Solve by suitable rearrangement
a) 1762 + 553 + 1338 + 547 b) 4 x 245 x 25
Ans: a) 1762 + 553 + 1338 + 547
= (1762 + 1338) + ( 553 + 547 )
= 3100 + 1076
= 4176
b) 4 x 245 x 25
= 4 x 25 x 245
= 100 x 245
= 24500
3. Find the value using suitable property.
a) 2639 x 8 + 2639 x 92 b) 6796 x 141 – 41 x 6796
Ans: a) 2639 x 8 + 2639 x 92
= 2639 x ( 92 + 8 )
= 2639 x 100
= 263900
b) 6796 x 141 – 41 x 6796
= 6796 x ( 141 – 41 )
= 6796 x 100
= 679600
4. Write the number names in Indian system of numeration.
a) 9475340 b) 30245601
Ans: 94,75,340 = Ninety four lakh, seventy five thousand, three hundred forty.
3,02,45,601 = Three crore, two lakh, forty five thousand, six hundred one.
5. Write the number names in International system of numeration.
a) 9543201 b) 7000023
Ans: 9,543,201 = Nine million, five hundred forty three thousand, two hundred one.
7,000,023 = Seven million, twenty three.
6. Using divisibility test, determine which of the following numbers are divisible by 11: a) 10600601 b) 169308 c) 434434
Ans: DIVISIBILITY BY 11: First, find the difference between the sum of the
digits at odd place ( from the right ) and sum of
the digits at even place ( from the right ) of the
number. If the difference is either 0 or divisible
by 11, then the number is divisible by 11.
NUMBER SUM OF THE DIGITS
AT THE ODD PLACE
FROM THE RIGHT SUM OF THE DIGITS
AT THE EVEN PLACE
FROM THE RIGHT
DIFFERENCE
10600601 1+ 6 + 0 + 0 = 7 0 + 0 + 6 + 1 = 7 7 – 7 = 0
169308 8 + 3 + 6 = 17 0 + 9 + 1 = 10 17 – 10 = 7
434434 4 + 4 + 3 = 11 3 + 4 + 3 = 10 11 – 10 = 1
10600601 = Yes, 10600601 is divisible by 11.
169308 = No, 169308 is not divisible by 11.
434434 = No, 434434 is not divisible by 11.
7. Find the first three common multiples of 9,12 and 18
9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108
12 = 12, 24, 36, 48, 60, 72, 84, 96, 108
18 = 18, 36, 54, 72, 90, 108
The first three common multiples of 9, 12, 18 are 36, 72 and 108
8. Find the common factors of 75,100 and 60
Ans: Factors of 75 are 1, 3, 5, 15, 25 and 75
Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 30 and 60
Thus, common factors of 75, 100 and 60 are 1 and 5
9. Find the HCF of 54 and 81 by prime factorization method.
Ans: 54
2 54
3 27
3 9
3 3
1
81
3 81
3 27
3 9
3 3
1
Thus, 54 = 2 x 3 x 3 x 3
81 = 3 x 3 x 3 x 3
10. State the divisibility rules for 4 and 8 and check if the number 684572 is divisible by 4 and 8.
Ans: DIVISIBILITY BY 4: A number with 3 or more digits is divisible by 4 if
the number formed by its last two digits is
divisible by 4.
DIVISIBILITY BY 8: A number with 4 or more digits is divisible by 8, if the
number formed by the last three digits is divisible by 8.
Thus, 684572 is divisible by 4 and 8.
11. Test the divisibility of the following number by 6
a) 5634 b) 65990
Ans: DIVISIBILITY BY 6: If a number is divisible by 2 and 3 both, then it is
divisible by 6
5634: Yes, 5634 is divisible by 6. 5634 is divisible by both 2 and 3 so it is
divisible by 6 also.
65990: No, 65990 is not divisible by 6. 65990 is divisible by 2 but 65990 is not divisible by 3.
12. Test the divisibility of the following number by 9
a) 94567 b) 91827
Ans: DIVISIBILITY BY 9: If the sum of the digits is divisible by 9, then the
number itself is divisible by 9.
94567: 94567 is not divisible by 9.
91827: Yes, 91827 is divisible by 9
13. If a pencil box costs Rs 85 and a packet of sketch pens costs Rs 24, what is the total cost of 120 pencil boxes and 300 packets of sketch pens?
Ans: Cost of one pencil box
14. Fruit baskets each weighing 4kg 500g are loaded in a van. How many such baskets can be loaded in a van which can carry only 900kg?
Solution: 1Kg = 1000g
Weight of one Fruit basket = 4Kg 500g
= 4 x 1000g + 500g
= 4000g + 500g
= 4500g
Weight the van can carry = 900Kg
= 900 x 1000g
= 9, 00, 000
Number of baskets that can = 900, 000 4,500
be loaded in the van. = 200
Ans: Thus, 200 Fruit baskets can be loaded in the van.
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