Playing With Numbers
Exercise 2.10 (Some Application of L.C.M)
Question 1 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We know that the smallest number divisible by 24, 36 and 54 is their L.C.M. Therefore, the required number must be 5 more than their L.C.M.
L.C.M = 2x2x2x3x3x3 = 216 Hence, the required number = 216+5 = 221 |
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Question 2 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We know that the smallest number divisible by 33 and 39 is their L.C.M. Therefore, the required number must be 5 more than their L.C.M.
L.C.M = 3x11x13 = 429 Hence, the required number = 429+5 = 434
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Question 3 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We know that the least number divisible by 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 is their L.C.M.
L.C.M = 1x2x2x2x3x3x5x7 = 2520
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Question 4 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We know that the smallest number divisible by 35, 56 and 91 is their L.C.M. Therefore, the required number must be 7 more than their L.C.M.
L.C.M = 2x2x2x5x7x13 = 3640 Hence, the required number = 3640+7 = 3647
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Question 5 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Section A – 32 Students Section B – 36 Students We know that the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B is L.C.M of 32 and 36.
L.C.M = 2x2x2x2x2x3x3 = 288 Hence, the required number = 288 books |
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Question 6 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We know that the minimum distance each should walk so that he can cover the distance in complete steps is the L.C.M of 80cm, 85cm and 90cm.
L.C.M = 2x2x2x2x3x3x5x17 = 12240 cm or 122m 40 cm
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Question 7 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We first compute L.C.M of 8, 15 and 21.
L.C.M = 2x2x2x3x5x7 = 840 840 x 119 = 99960 840 x 120 = 100800 (nearest to 100000 but greater than 100000) 840 x 121 = 101640 |
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Question 8 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We first compute L.C.M of 6 and 8.
L.C.M = 2x2x2x3 = 24 Therefore, the bus stop at the 24th block of flats.
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Question 9 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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We first compute L.C.M 220m and 300m
L.C.M = 2x2x3x5x5x11 = 3300m Next heap which lies at the foot of a pole will at the distance of 3300m.
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Question 10 (Refer Book - Mathematics Class VI R.D. Sharma) |
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Solution: |
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L.C.M of 28 and 32
L.C.M = 2x2x2x2x2x7 = 224 Smallest number which leaves remainder 8 and 12 when divided by 28 and 32 = LCM –(8+12) = 224 - 20 = 204
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