TS 6th Class Maths Solutions Chapter 2 Whole Numbers Exercise 2.2<\/h2>\n
Question 1.
\nGive the results without actually performing the operations using the given information.
\n(i) 28 \u00d7 19 = 532 then 19 \u00d7 28 =
\n(ii) 1 \u00d7 47 = 47 then 47 \u00d7 1 =
\n(iii) a \u00d7 b = c then b \u00d7 a =
\n(iv) 58 + 42 = 100 then 42 + 58 =
\n(v) 85 + 0 = 85 then 0 + 85 =
\n(vi) a + b = d then b + a =
\nAnswer:
\n(i) 28 \u00d7 19 = 532 then 19 \u00d7 28 = 532
\n(ii) 1 \u00d7 47 = 47 then 47 \u00d7 1 = 47;
\n(iii) a \u00d7 b = c then b \u00d7 a = c
\n(iv) 58 + 42 = 100 then 42 + 58 = 100
\n(v) 85 + 0 = 85 then 0 + 85 = 85
\n(vi) a + b = d then b + a = d<\/p>\n
Question 2.
\nFind the sum by suitable rearrangement:
\n(i) 238 + 695 + 162
\nAnswer:
\n238 + 695 + 162 = (238 + 162) + 695
\n= 400 + 695 = 1095<\/p>\n
(ii) 154 + 197 + 46 + 203
\nAnswer:
\n154 + 197 + 46 + 203
\n= (154 + 46) + (197 + 203) = 200 + 400 = 600<\/p>\n
![\"TS](\"https:\/\/tsboardsolutions.in\/wp-content\/uploads\/2022\/12\/TS-Board-Solutions.png\")
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Question 3.
\nFind the product by suitable rearrangement.
\n(i) 25 \u00d7 1963 \u00d7 4
\nAnswer:
\n25 \u00d7 1963 \u00d7 4 = (25 \u00d7 4) \u00d7 1963
\n= 100 \u00d7 1963
\n= 196300<\/p>\n
(ii) 20 \u00d7 255 \u00d7 50 \u00d7 6
\nAnswer:
\n20 \u00d7 255 \u00d7 50 \u00d7 6 = (20 \u00d7 50) \u00d7 (255 \u00d7 6)
\n= 1000 \u00d7 1530
\n= 1530000<\/p>\n
Question 4.
\nFind the value of the following:
\n(i) 368 \u00d7 12 + 18 \u00d7 368
\nAnswer:
\n(368 \u00d7 12) + (18 \u00d7 368)
\n= 368 \u00d7 [12 + 18]
\n= 368 \u00d7 30
\n= 11040<\/p>\n
(ii) 79 \u00d7 4319 + 4319 \u00d7 11
\nAnswer:
\n(79 \u00d7 4319) + (4319 \u00d7 11)
\n= 4319 \u00d7 [79 + 11]
\n= 4319 \u00d7 90
\n= 388710<\/p>\n
Question 5.
\nFind the product using suitable properties:
\n(i) 205 \u00d7 1989
\nAnswer:
\n1989 \u00d7 205 = 1989 \u00d7 (200 + 5) = (1989 \u00d7 200) + (1989 \u00d7 5)
\n= 397800 + 9945 = 407745<\/p>\n
(ii) 1991 \u00d7 1005
\nAnswer:
\n1991 \u00d7 1005 = 1991 \u00d7 [1000 + 5] = (1991 \u00d7 1000) + (1991 \u00d7 5)
\n= 1991000 + 9955 = 2000955<\/p>\n
Question 6.
\nA milk vendor supplies 56 liters of milk in the morning and 44 liters of milk in the evening to a hostel. If the milk costs \u20b9 30 per liter, how much money he gets per day?
\nAnswer:
\nQuantity of milk supplied to the hostel in the morning = 56 litres
\nQuantity of milk supplied to the hostel in the evening = 44 litres
\nQuantity of milk supplied to the hostel per day = (56 + 44) litres = 100 litres
\nCost of 1 litre of milk = \u20b9 30
\nCost of 100 litres of milk = \u20b9 30 \u00d7 100 = \u20b9 3000<\/p>\n
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Question 7.
\nChandana and Venu purchased 12 note books and10 note books respectively. The cost of each note book is \u20b9 15,then how much amount should they pay to the shop keeper?
\nAnswer:
\nCost of one note book = \u20b9 15
\nNumber of note books that Chandana bought =12
\nAmount that Chandana has to pay the shopkeeper = \u20b9 15 \u00d7 12
\nNumber of notebooks that Venu bought = 10
\nAmount that Venu has to pay to the shopkeeper = \u20b9 15 \u00d7 10
\nTotal amount that Chandana and Venu have to pay the shopkeeper =
\n\u20b9 15 \u00d7 12 + \u20b9 15 \u00d7 10 = \u20b9 15 [12 + 10] = \u20b9 15 \u00d7 22 = \u20b9 330<\/p>\n
Question 8.
\nMatch the following:<\/p>\n