{"id":17858,"date":"2020-01-04T12:33:12","date_gmt":"2020-01-04T07:03:12","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17858"},"modified":"2021-07-02T15:31:18","modified_gmt":"2021-07-02T10:01:18","slug":"kseeb-solutions-for-class-5-maths-chapter-2-part-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-5-maths-chapter-2-part-2\/","title":{"rendered":"KSEEB Solutions for Class 5 Maths Chapter 2 Division"},"content":{"rendered":"

Students can Download Maths Chapter 2 Division Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 5 Maths Chapter 2 Division<\/h2>\n

KSEEB Class 5 Maths Division Revision Exercise<\/h3>\n

I. Encircle the objects in as shown in the example and write answer in given box:<\/span><\/p>\n

\"KSEEB<\/p>\n

II. Divide the following using repeated substruction method:<\/span><\/p>\n

a. 12 \u00f7 4
\n12 – 4 = 8
\n8 – 4 = 4
\n4 – 4 = 0
\nSubtract is done 3 times
\n\\(\\frac{12}{4}\\) = 3<\/p>\n

\"KSEEB<\/p>\n

b. 25 \u00f7 5
\n25 – 5 = 20
\n20 – 5 = 15
\n15 – 5 = 10
\n10 – 5 = 5
\n5 – 5 = 0
\nSubtract is done 5 times
\n25 \u00f7 = 5<\/p>\n

c. 42 \u00f7 7
\n42 – 7 = 35
\n35 – 7 =28
\n28 – 7 = 21
\n21 – 7 = 14
\n14 – 7 = 7
\n7 – 7 = 0
\nSubtract is done 6 times<\/p>\n

d. 30 \u00f7 10
\n30 – 10 = 20
\n20 – 10 = 10
\n10 – 10 = 0
\nSubtract is done 3 times
\n30 \u00f7 10 = 3<\/p>\n

e. 75 \u00f7 15
\n75 – 10 = 60
\n60 – 15 = 45
\n45 – 15 = 30
\n30 – 15 = 15
\n15 – 15 = 0
\nSubtract is done 5 times
\n5 \u00f7 15 = 5<\/p>\n

KSEEB Class 5 Maths Division Ex 2.1<\/h3>\n

I. Find the quotient and the remainder:<\/span><\/p>\n

1. 48 \u00f7 6<\/p>\n

\"KSEEB
\nQuotient = 8
\nRemainder =0<\/p>\n

2. 36 \u00f7 3<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n36 = (3 \u00d7 12) +0
\n36 = 36 + 0
\n36 = 36
\nQuotient = 12
\nRemainder =0<\/p>\n

3. 55 \u00f7 4<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n55 = (4 \u00d7 13) + 3
\n55 = 52 + 3
\n55 = 55
\nquotient = 13
\nRemainder = 3<\/p>\n

4. 72 \u00f7 7<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n72 = (7 \u00d7 10) + 2
\n72 = 70 + 2
\n72 = 72
\nquotient = 10
\nRemainder = 02<\/p>\n

II. Find the quotient and the remainder:<\/span><\/p>\n

1. 232 \u00f7 4<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n232 = (4 \u00d7 58) + 0
\n232 = 232 + 0
\n232 = 232
\nquotient = 58
\nRemainder = 3<\/p>\n

\"KSEEB<\/p>\n

2. 474 \u00f7 6<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n474 = (6 \u00d7 79) + 0
\n474 = 474 + 0
\n474 = 474
\nquotient = 79 3
\nRemainder =0<\/p>\n

3. 255 \u00f7 11<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n255 = (11 \u00d7 23) + 2
\n255 = 253 + 2
\n255 = 255
\nquotient = 23
\nRemainder = 2<\/p>\n

4. 527 \u00f7 12<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n527 = (12 \u00d7 43) + 11
\n527 = 516 + 11
\n527 = 527
\nquotient = 43
\nRemainder = 11<\/p>\n

III. Find the quotient and the remainder:<\/span><\/p>\n

1. 1,653 \u00f7 8<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n1653 = (8 \u00d7 206) + 5
\n1653 = 1648 + 5
\n1653 = 1653
\nquotient = 206
\nRemainder = 5<\/p>\n

2. 1,325 \u00f7 2<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n1325 = (2 \u00d7 662) + 1
\n1325 = 1324 + 1
\n1325 = 1325
\nquotient = 662
\nRemainder = 1<\/p>\n

3. 1,435 \u00f7 15<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n1435 = (15 \u00d7 95) + 10
\n1435 = 1425+10
\n1435 = 1435
\nquotient = 95
\nRemainder =10<\/p>\n

\"KSEEB<\/p>\n

4. 2,647 \u00f7 13<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n2647 = (13 \u00d7 203) + 8
\n2647 = 2639 + 8
\n2647 = 2647
\nquotient = 203
\nRemainder = 8<\/p>\n

IV. Find the quotient and the remainder:<\/span><\/p>\n

1. 24,658 \u00f7 2<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n24658 = (2 \u00d7 12329) + 0
\n24658 = 24658 + 0
\n24658 = 24658
\nquotient = 12329
\nRemainder = 0<\/p>\n

2. 14,005 \u00f7 7<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n14005 = (7 \u00d7 2000) + 5
\n14005 = 14000 + 5
\n14005 = 14005
\nquotient = 2000
\nRemainder = 5<\/p>\n

3. 32,745 \u00f7 10<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n32745 = (10 \u00d7 3274) + 5
\n32745 = 32740 + 5
\n32745 = 32745
\nquotient = 3274
\nRemainder =5<\/p>\n

\"KSEEB<\/p>\n

4. 12,056 \u00f7 12<\/p>\n

\"KSEEB
\nDividend = (divisor \u00d7 quotient) + Remainder
\n12056 = (12 \u00d7 1004) +8
\n12056 = 12048 + 8
\n12056 = 12056
\nquotient = 1004
\nRemainder = 8<\/p>\n

V. Solve the following problems:<\/span><\/p>\n

Question 1.
\nA tailor has 18 metres of cloth. He stitch 9 shirts from this cloth. Find the length of cloth required to stitch one shirt.
\nAnswer:
\nA tailor has 18 meter of cloth
\nTailor has stitch 9 shirts
\nThe length of cloth required to stitch one shirt = \\(\\frac{18}{9}\\) = 2 meter<\/p>\n

Question 2.
\nA bike can cover 240 km with 5 litres of petrol. Find how many kilometres does it cover with 1 litre of petrol.
\nAnswer:
\nA bike can cover 240 Km with 5 litres of petrol
\nIn one litre of petrol, bike covers the distance = \\(\\frac{240}{5}\\) = 48 km<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\n25,004 bags of cement are loaded equally into 14 railway wagons. Find the number of bags in each wagon.
\nAnswer:
\n25,004 bags of cement are loaded equally into 14 railway wagons
\nThe number of bags
\n= \\(\\frac{25004}{14}\\) = 1,786
\n\"KSEEB<\/p>\n

Question 4.
\nMr. Sridhar purchased 11 toys of same price for Rs. 946 from a toy shop. Find the cost of each toy.
\nAnswer:
\nThe cost of a toy =?
\nMr. Sridhar purchase 11 toys of same price = Rs 946
\nThe cost of each toy
\n= \\(\\frac{946}{11}\\) = 86 Rs.<\/p>\n

Question 5.
\nThe annual income of a person is Rs.90,912. Find his monthly income.
\nAnswer:
\nThe Annual income of a peusen Rs 90,912
\nMonthly income = \\(\\frac{90912}{12}\\) = 7,576 Rs
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nA car manufacturing company manufactures 14,820 cars in 13 months. Find the number of cars manufactured in one month.
\nAnswer:
\nA car manufacturing company manufactures 14820 cars in 13 months. For one month, the number of cars = \\(\\frac{14820}{13}\\) = 1,140
\n\"KSEEB<\/p>\n

KSEEB Class 5 Maths Division\u00a0Ex 2.2<\/h3>\n

I. Solve<\/span><\/p>\n

Question 1.
\nAn orange garden has 82 orange plants. A farmer plucks 60 oranges from each plant and packs 12 oranges in each box. Calculate the number of boxes required to pack all oranges.
\nAnswer:
\nAn orange garden has 82 oranges . A farmer plunks 60 oranges from each plant
\nThe number of boxes = ?
\nThe number of oranges in each box = 12
\nThe number of oranges in an orange garden = 410
\n\"KSEEB<\/p>\n

Question 2.
\n15 school children hire a cab for a day\u2019s excursion at Rs. 9 per km. If they travel a distance of 325 km, find the amount to be shared by each one of them.
\nAnswer:
\nA cab of charges Rs 9 Km
\nThe childrens travel a distance 325Km
\n= 325 \u00d7 9
\n= 2925 Km
\nThe amount shared by each student
\n= 2925 \u00f7 15 = Rs 195
\n\"KSEEB<\/p>\n

Question 3.
\nAnita has a cow which yields 8 litres of milk in a day. The selling price of one litre of milk is Rs. 18. She wants to divide the amount earned in one month (30days) among her 4 sons. What amount does each one get?
\nAnswer:
\nOne litre of milk = Rs 18
\nA cow which yields 8 litres of milk in a day for 30 days, the number of litres of milk
\n18 \u00d7 8 = Rs. 144
\nThe amount of milk
\n144 \u00d7 30 = Rs. 4320
\nThe amount of each son 1,080
\n4320 \u00f7 4
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Division Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 5 Maths Chapter 2 Division KSEEB Class 5 Maths Division Revision Exercise I. Encircle the objects …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[86],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17858"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=17858"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17858\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17858"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17858"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17858"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}