Maharashtra Board Class 5 Maths Chapter 4 Multiplication and Division Set 14 Solutions

Std 5 Maths Chapter 4 Multiplication And Division

Question 1. Multiply the following:
(1) 327 × 92
Solution:
   327
 X  92
------
  654
+29430
------
30084
Answer: 30084
In simple words: To multiply 327 by 92, first multiply 327 by the unit digit 2, then multiply 327 by the tens digit 9 (placing a zero in the units place), and finally add the two results to get the total product.

🎯 Exam Tip: Accurate alignment of partial products is crucial for correct addition in multi-digit multiplication problems.

Question 1. Multiply the following:
(2) 807 × 126
Solution:
   807
 X 126
------
 4842
+16140
+80700
-------
101682
Answer: 101682
In simple words: For 807 multiplied by 126, first multiply 807 by 6, then by 20, then by 100, and sum up these three partial products to find the final answer.

🎯 Exam Tip: When multiplying by a three-digit number, remember to correctly place zeros in the partial products for the tens and hundreds digits.

Question 1. Multiply the following:
(3) 567 × 890
Solution:
  567
X 890
-----
  000
+51030
+453600
-------
504630
Answer: 504630
In simple words: To calculate 567 times 890, multiply 567 by each digit of 890 (0, 9, and 8, considering their place values), then add the resulting partial products.

🎯 Exam Tip: When a multiplier ends in zero, you can multiply by the non-zero digits and then append the appropriate number of zeros to the product for efficiency.

Question 1. Multiply the following:
(4) 4317 × 824
Solution:
   4317
 X  824
-------
 17268
+ 86340
+3453600
--------
3557208
Answer: 3557208
In simple words: To multiply 4317 by 824, compute the partial products by multiplying 4317 by 4, then by 20, and finally by 800; then, add these partial products together.

🎯 Exam Tip: Double-check each partial product calculation before summing them up, as a small error can lead to a large incorrect final answer.

Question 1. Multiply the following:
(5) 6092 × 203
Solution:
   6092
 X  203
-------
 18276
+ 00000
+1218400
--------
1236676
Answer: 1236676
In simple words: For the product of 6092 and 203, multiply 6092 by 3, then by 0 (tens place), and then by 200, making sure to align the digits correctly before adding the results.

🎯 Exam Tip: When a zero appears in the middle of the multiplier, ensure you correctly account for its place value by writing zeros in the partial product line.

Question 1. Multiply the following:
(6) 1177 × 99
Solution:
  1177
 X  99
-------
 10593
+105930
--------
116523
Answer: 116523
In simple words: To find the product of 1177 and 99, multiply 1177 by 9, then by 90, and add the two resulting values.

🎯 Exam Tip: Multiplication by numbers like 99 can sometimes be simplified by multiplying by 100 and then subtracting the original number once (e.g., 1177 × 100 - 1177).

Question 1. Multiply the following:
(7) 456 × 187
Solution:
  456
X 187
-----
 3192
+36480
+45600
------
85272
Answer: 85272
In simple words: To solve 456 multiplied by 187, calculate partial products by multiplying 456 by 7, by 80, and by 100, then sum them up.

🎯 Exam Tip: Maintaining neatness and clear separation of digits in your written work helps prevent errors, especially when adding multiple partial products.

Question 1. Multiply the following:
(8) 6543 × 79
Solution:
   6543
 X  79
--------
 58887
+458010
--------
516897
Answer: 516897
In simple words: To multiply 6543 by 79, perform two multiplications: 6543 by 9, and 6543 by 70, then add these results.

🎯 Exam Tip: Pay close attention to carrying over numbers in each step of multiplication to ensure accuracy, especially with larger numbers.

Question 1. Multiply the following:
(9) 2306 × 832
Solution:
   2306
 X  832
-------
  4612
+ 69180
+1844800
--------
1918592
Answer: 1918592
In simple words: To calculate 2306 multiplied by 832, multiply 2306 by each digit (2, 30, and 800), and then combine these intermediate products.

🎯 Exam Tip: When multiplying by a digit in the tens or hundreds place, always shift the partial product to the left by one or two places, respectively, before adding.

Question 1. Multiply the following:
(10) 6429 × 509
Solution:
   6429
 X  509
-------
 57861
+ 00000
+3214500
--------
3272361
Answer: 3272361
In simple words: For 6429 times 509, multiply 6429 by 9, then by 00 (hundreds place), and then by 500, and finally add the three resulting numbers.

🎯 Exam Tip: Recognizing a zero in the multiplier's tens place simplifies the partial product for that digit to all zeros, saving computation time while requiring correct place value alignment.

Question 1. Multiply the following:
(11) 4,321 × 678
Solution:
   4321
 X  678
-------
 34568
+302470
+2592600
--------
2929638
Answer: 2929638
In simple words: To find the product of 4321 and 678, multiply 4321 by 8, then by 70, then by 600, and add the three partial products together.

🎯 Exam Tip: Proper mental arithmetic or quick calculations for smaller multiplications reduce the chance of errors in complex problems.

Question 1. Multiply the following:
(12) 20,304 × 87
Solution:
   20304
 X  87
--------
 142128
+1624320
--------
1766448
Answer: 1766448
In simple words: To multiply 20304 by 87, first multiply 20304 by 7, then by 80, and sum these two partial results to get the final answer.

🎯 Exam Tip: Pay extra attention when a number contains zeros in the middle, ensuring place values are maintained during multiplication to avoid mistakes.

Question 2. As part of the 'Avoid Plastic' campaign, each of 745 students made 25 paper bags. What was the total number of paper bags made ?
Solution:
  745 Number of students
X  25 bags made by each
---------------------
 3725
+14900
--------
18625
Answer: 18,625 bags made
In simple words: To find the total number of bags, multiply the number of students by the number of bags each student made.

🎯 Exam Tip: For word problems, identify the key numbers and the operation needed (multiplication in this case) to solve for the unknown quantity.

Question 3. In a plantation, saplings of 215 medicinal trees have been planted in each of the 132 rows of trees. How many saplings are there in the plantation altogether ?
Solution:
  215 Saplings in each now
X 132 Number of rows
--------------------
  430
+ 6450
+21500
-------
28380
Answer: Altogether there is 28,380 saplings.
In simple words: To calculate the total number of saplings, multiply the number of saplings in one row by the total number of rows.

🎯 Exam Tip: Clearly label your steps and final answer in word problems to show a complete understanding of the solution.

Question 4. One computer costs 27,540 rupees. How much will 18 such computers cost?
Solution:
  27540 Cost of 1 computer
X  18 No. of computers
--------------------
 220320
+275400
--------
495720
Answer: Rs. 4,95,720 cost of 18 computers.
In simple words: To find the total cost, multiply the cost of one computer by the number of computers.

🎯 Exam Tip: Pay attention to the currency symbol (Rs.) and place value when writing out large numbers in the answer.

Question 5. Under the 'Inspire Awards' scheme, 5000 rupees per student were granted for the purchase of science project materials. If 154 students in a certain taluka were covered under the scheme, find the total amount granted to that taluka.
Solution:
  Rs. 5000 Granted per student
X  154 Number of students
------------------------
 20000
+250000
+500000
---------
Rs. 770000
Answer: Rs. 7,70,000 granted totally
In simple words: The total amount granted is found by multiplying the grant per student by the total number of students.

🎯 Exam Tip: When multiplying numbers ending in multiple zeros, you can multiply the non-zero parts and then append the total count of zeros to the result.

Question 6. If a certain two-wheeler costs 53,670 rupees, how much will 35 such two-wheelers cost?
Solution:
  53760 Cost of 1 two-wheeler
X  35 No. of two-wheelers
---------------------
 268800
+1612800
---------
1881600
Answer: Rs. 18,81,600 is the total cost of 35- two-wheelers
In simple words: To determine the total expense for multiple two-wheelers, multiply the price of one two-wheeler by the quantity purchased.

🎯 Exam Tip: When dealing with monetary values, ensure your final answer includes the correct currency unit (Rs.) and is clearly formatted.

Question 7. One hour has 3,600 seconds. How many seconds do 365 hours have ?
Solution:
  3600 Seconds of 1 hour
X  365 No. of hours
-------------------
 18000
+216000
+1080000
---------
1314000
Answer: 13,14,000 seconds for 365 hours.
In simple words: To find the total seconds, multiply the number of seconds in one hour by the total number of hours.

🎯 Exam Tip: Convert units consistently; if the question asks for seconds, ensure all calculations are done in seconds or converted at the end.

Question 8. Frame a multiplication word problem with the numbers 5473 and 627 and solve it.
Solution:
Cost of one mobile is 5,473. What is the cost of such 627 mobiles?
  5473 Cost of 1 mobile
X  627 Number of mobiles
--------------------
 38311
+109460
+3283800
---------
3431571
Answer: 34,31,571 cost for 627 mobiles.
In simple words: We created a problem about finding the total cost of multiple items, then multiplied the cost per item by the quantity of items to solve it.

🎯 Exam Tip: A good word problem should clearly state the given values and pose a question that requires a specific mathematical operation to solve.

Question 9. Find the product of the biggest three-digit number and the biggest four-digit number.
Solution:
9999 Biggest four-digit no.
X  999 Biggest three-digit no.
-----------------------
  89991
+ 899910
+8999100
----------
9989001
Answer: 99,89,001
In simple words: The biggest three-digit number is 999, and the biggest four-digit number is 9999; their product is found by multiplying these two values.

🎯 Exam Tip: Remember that the 'biggest' N-digit number is simply N nines (e.g., biggest three-digit is 999), which simplifies setting up the multiplication.

Question 10. One traveller incurs a cost of 7,650 rupees for a certain journey. What will be the cost for 26 such travellers?
Solution:
  7650 Cost of one traveller
X  26 No. of travellers
--------------------
 45900
+153000
--------
198900
Answer: Rs. 1,98,900 cost of 26 travellers.
In simple words: To find the total cost for 26 travellers, multiply the cost for one traveller by 26.

🎯 Exam Tip: Always include the correct unit (Rs.) with your final answer when solving word problems involving money.

Pairing Off Objects From Two Groups In Different Ways

(1) Ajay wants to travel light. So he took with him three shirts - one red, one green and one blue and two pairs of trousers - one white and one black. How many different ways does he have of pairing off a shirt with trousers?
Writing 'S' for shirt and 'T' for trousers, the possible different pairs are :
(Red S, Black T), (Green S, Black T), (Blue S, Black T)
(Red S, White T), (Green S, White T), (Blue S, White T)
Altogether 6 different pairs.
Answer: Ajay has 6 different ways of pairing a shirt with trousers.
In simple words: To find the total number of pairings, you multiply the number of shirts by the number of trousers. Ajay has 3 shirts and 2 trousers, so 3 multiplied by 2 gives 6 possible pairings.

🎯 Exam Tip: When forming pairs from two distinct groups, the total number of possible pairs is the product of the number of items in each group.

(2) Suresh has three balls of different colours marked A, B and C and three bats marked P, Q and R. He wishes to take only one bat and one ball to the playground. In how many ways can he pair off a ball and a bat to take with him?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सुरेश के पास उपलब्ध गेंदों और बल्लों के विभिन्न संयोजनों को दर्शाता है। इसमें तीन गेंदें (A, B, C) और तीन बल्ले (P, R) दिखाए गए हैं, जहाँ प्रत्येक गेंद को एक बल्ले के साथ जोड़ा गया है। उदाहरण के लिए, गेंद A को बल्ले P और R के साथ, गेंद B को P और R के साथ, और गेंद C को P और R के साथ जोड़ा गया है, जो कुल संभावित संयोजनों को दर्शाता है।
How many different pairs have been shown here?
Answer: There are 9 different pairs shown here. (A-P, A-Q, A-R, B-P, B-Q, B-R, C-P, C-Q, C-R. The image only shows P and R, but the question mentions P, Q, R so assuming all three bats are considered for each ball.)
In simple words: If you have 3 balls and 3 bats, you multiply the number of balls by the number of bats to find all the unique combinations Suresh can make, which is 3 times 3, equaling 9 pairs.

🎯 Exam Tip: For combination problems, visually listing out all possible pairs can help verify the product obtained through multiplication, especially for smaller sets.

(3) The three friends, Sanju, John and Ali went to the fair. A shop there, had five different types of hats. Each of the boys had photos taken of himself, wearing every type of hat, in turn. Find how many photographs were taken in all.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तीन दोस्तों - संजू, जॉन और अली - को दिखाता है जो पांच अलग-अलग प्रकार की टोपियाँ पहनकर तस्वीरें खिंचवा रहे हैं। प्रत्येक पंक्ति एक दोस्त को दर्शाती है, और प्रत्येक दोस्त के बगल में विभिन्न टोपियों के साथ उसकी कई तस्वीरें दिखाई गई हैं, जो यह दर्शाती हैं कि हर दोस्त ने हर टोपी पहनकर एक तस्वीर ली है।
How many different pairs were formed ? That is, how many photos were taken ?
Answer: 15 photos were taken in all.
In simple words: To find the total photos, multiply the number of friends by the number of different hats available.

🎯 Exam Tip: Recognize that each person trying on each hat is a multiplication problem: (number of people) × (number of hats).

Take two collections, each containing the given number of objects. Make as many different pairs as possible, taking one object from each collection every time. Thus, complete the table below.

Objects in one collection	Objects in the other collection	How many ways to form pairs?
3	2	6
5	2	10
2	4	8
3	4	12

What does this table tell us ?
The number of different pairs formed by pairing off objects from two groups is equal to the product of the number of objects in the two groups.
Answer: The table demonstrates that the total number of unique pairs formed by selecting one item from each of two distinct collections is always the product of the number of items in each collection.
In simple words: The table shows that if you want to pair items from two groups, you simply multiply the number of items in the first group by the number of items in the second group to find all possible pairs.

🎯 Exam Tip: Understanding the concept of multiplication for combinations is fundamental; it directly relates to probability and basic counting principles.

Division

Teacher: You have learnt some things about division. For example, we know that division means making equal parts of a given number, or, subtracting a number repeatedly from a given number. What else do you know ?

Shubha : We know that we get two divisions from one multiplication. From 9 × 4 = 36, we get the divisions 36 ÷ 4 = 9 and 36 ÷ 9 = 4.

Teacher: Very good! Right now, there's nothing new to learn about division. Only the number of digits in the dividend and the divisor will grow. Tell me what is 354 ÷ 6?

Sarang: 354 = 300 + 54. 300 divided by six is 50. And 54 ÷ 6 = 9. Hence the quotient is 50 + 9 = 59.

Teacher: Right! Now let's learn, step by step, how to divide a four-digit number by a one-digit number. So now, divide 4925 by 7 and tell me the quotient and the remainder.

Shubha : We cannot divide 4 thousands by 7 into whole thousands. Now, 4 Th = 40 H. So let us instead take the 40 hundreds together with 9 hundreds and divide 49 hundreds. 49 ÷ 7 = 7. So, everyone gets 7 hundreds. Now, we cannot divide 2T equally among 7 people. So we must write 0 in the tens place in the quotient. Then on dividing 25 by seven, we get quotient 3 and the remainder is 4. Thus, the answer is quotient 703, remainder 4.

	0	8	2	6
9 )	7	4	3	9
-	0
	7	4
-	7	2
		2	3
-		1	8
			5	9
-			5	4
				5

Teacher: Very good! Now divide 7439 by 9.

Sarang: It's difficult to do this mentally. I'll write it down on paper. The quotient is 826 and the remainder, 5.

Teacher : We use the same method to divide a four-digit number by a two-digit number. If necessary, we can prepare the table of the divisor before we start.

Study the solved examples shown below.

Example (1): Divide 4254 by 25Solution:

	0	1	7	0
25 )	4	2	5	4
-	0
	4	2
-	2	5
	1	7	5
-	1	7	5
	0	0	0	4
-	0	0	0	0
				4

25 x 1 = 25
25 x 2 = 50
25 x 3 = 75
25 x 4 = 100
25 x 5 = 125
25 x 6 = 150
25 x 7 = 175
Answer: Quotient 170, Remainder 4
In simple words: This example shows long division of a four-digit number by a two-digit number, step by step, resulting in a quotient and a remainder.

🎯 Exam Tip: Always write down the multiplication table of the divisor on the side for accuracy, especially with larger numbers, to easily find suitable multiples.

Example (2): Divide 9783 by 32Solution:

	0	3	0	5
32 )	9	7	8	3
-	0
	9	7
-	9	6
		1	8
-		0	0
		1	8	3
-		1	6	0
			2	3

32 × 1 = 32
32 × 2 = 64
32 × 3 = 96
32 × 4 = 128
32 × 5 = 160
32 × 6 = 192
Answer: Quotient 305, Remainder 23
In simple words: This example illustrates how to perform long division when a digit in the dividend requires placing a zero in the quotient, specifically when the partial dividend is smaller than the divisor.

🎯 Exam Tip: Remember to place a zero in the quotient if a partial dividend is too small to be divided by the divisor; this ensures correct place value in the final answer.

Example (3): Divide. 9842 ÷ 45Solution:

	0	2	1	8
45 )	9	8	4	2
-	9	0
	0	8	4
-		4	5
		3	9	2
-		3	6	0
		0	3	2

We can prepare the 45 times table to do this division.
But when the divisor is a big number, we can solve the example by first guessing what the quotient will be. Let us see how to do that.
We have 0 in the thousands place in the quotient.
Now, to guess the quotient when dividing 98 by 45, look at the first digits in both - the dividend and the divisor. These are 9 and 4, respectively.
Dividing 9 by 4, we will get 2 in the quotient. Let us see if 2 times 45 can be subtracted from 98. 45 × 2 = 90. 90 < 98. So, we write 2 in the hundreds place in the quotient.
Next, dividing 84 by 45 we can easily see that as 90 > 84, we have to write 1 in the tens place in the quotient.
Now, we have to divide 392 by 45. As 3 < 4, let us look at 39, the number formed by the first 2 digits, to guess the next digit in the quotient.
4 x 9 = 36 and 36 < 39. Let us check if the next digit can be 9. 45 × 9 = 405 and 405 > 392. Therefore, 9 cannot be the next digit in the quotient.
Let us check for 8. 45 × 8 = 360. 360 < 392. So, we write 8 in the units place of the quotient.
We subtract 8 × 45 from 392 and complete the division.
Answer: The quotient is 218 and the remainder, 32.
In simple words: This example demonstrates how to perform long division with a two-digit divisor by using an estimation strategy based on the leading digits of the dividend and divisor.

🎯 Exam Tip: When dividing by larger numbers, estimate the quotient digit by looking at the leading digits, then verify by multiplying and adjusting if the product is too large or too small.

Example (4): If 35 kilograms of wheat cost 910 rupees, what is the rate of wheat per kg?Solution:

		2	6
35 )	9	1	0
-	7	0
	2	1	0
-	2	1	0
	0	0	0

Weight of wheat in kg × rate of wheat per kg = cost of wheat
\( \implies \) 35 × rate per kg = 910
Therefore, when we divide 910 by 35, we will get the per kg rate of wheat.
Answer: The rate per kilogram of wheat is 26 rupees.
In simple words: To find the cost per kilogram of wheat, divide the total cost by the total weight.

🎯 Exam Tip: For word problems involving rates, remember the formula: Total Quantity = Rate × Number of Units. Use division to find the rate if the total quantity and number of units are given.

Multiplication And Division Problem Set 14 Additional Important Questions And Answers

Multiply the following:

Question (1). 2132 x 231Solution:

	2	1	3	2
x		2	3	1
	2	1	3	2
+	6	3	9	6	0
+	4	2	6	4	0	0
	4	9	2	4	9	2

Answer: 492492
In simple words: To multiply 2132 by 231, multiply 2132 by each digit of 231 (units, tens, hundreds) and add the partial products.

🎯 Exam Tip: When multiplying by a three-digit number, ensure you align the partial products correctly by adding zeros for the tens and hundreds place multiplications.

Question (2). 1863 x 432Solution:

	1	8	6	3
x		4	3	2
	3	7	2	6
+	5	5	8	9	0
+	7	4	5	2	0	0
	8	0	4	8	1	6

Answer: 804816
In simple words: This is a multiplication problem where a four-digit number is multiplied by a three-digit number, requiring the sum of three partial products for the final answer.

🎯 Exam Tip: Pay careful attention to carrying over numbers during multiplication and maintaining correct place values when adding partial products to avoid errors.

Solve the following word problems:

Question (1). A factory manufactures 34,796 pairs of socks in one hour. How many pairs will the factory manufacture in one day?Solution:
34796 Pairs of socks per hour
x 24 No. of hours in a day

	3	4	7	9	6
X				2	4
	1	3	9	1	8	4
+	6	9	5	9	2	0
	8	3	5	1	0	4

Answer: 8,35,104 pairs of socks manufactured in one day
In simple words: To find the total number of socks manufactured in a day, multiply the number of socks made per hour by the total number of hours in a day (24).

🎯 Exam Tip: Carefully read word problems to identify the "rate" and "time period" to determine if multiplication or division is needed. Remember that "one day" usually refers to 24 hours.

Question (2). There are 375 toffees in a box. How many toffees will be there in 632 such boxes?Solution:
375 No. of toffees in each box
x 632 No. of boxes

		3	7	5
X		6	3	2
			7	5	0
+		1	1	2	5	0
+	2	2	5	0	0	0
	2	3	7	0	0	0

Answer: There will be 2,37,000 toffees.
In simple words: To find the total number of toffees, multiply the number of toffees in one box by the total number of boxes.

🎯 Exam Tip: When solving problems involving 'total in multiple units', use multiplication. Ensure all partial products are correctly calculated and summed.

Question (3). There are 144 articles in a gross. How many articles are there in 2174 gross?Solution:
2174 No. of gross
x 144 Articles in 1 gross

	2	1	7	4
X		1	4	4
	8	6	9	6
+	8	6	9	6	0
+	2	1	7	4	0	0
	3	1	3	0	5	6

Answer: There are 3,13,056 articles in 2174 gross.
In simple words: To determine the total articles, multiply the number of articles in one gross by the given number of gross.

🎯 Exam Tip: Understand common units of measurement (like "gross" for 144 units) to correctly set up the multiplication for such conversion problems.