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## Numbers -> IMPORTANT FACTS AND FORMULAE

 1. Natural Numbers : Counting numbers 1, 2, 3, 4, 5, .. are called natural numbers. II. Whole Numbers : All counting numbers together with zero form the set of whole numbers. Thus, I. 0 is the only whole number which is not a natural number. II. Every natural number is a whole number. III.Some Important Formulae : I. ( 1 + 2 + 3 + .....+ n) = n (n + 1 ) / 2 II. (1 2 + 22 + 32 + ..... + n2) = n ( n + 1 ) (2n + 1) / 6 III. (1 3 + 23 + 33 + ..... + n3) = n2 (n + 1)2 / 4

## Numbers -> SOLVED EXAMPLES

1. 217 × 217 + 183 × 183 = ? Sol. Let the speed of the motorboat in still water be x kmph. Then, Speed downstream = (x + 2) kmph; Speed upstream = (x - 2) kmph. ∴ 6/x+2 + 6/x-2 = 33/60 ⇔ 11x² - 240x - 44 = 0 ⇔ 11x² - 242x + 2x - 44 = 0 ⇔ (x - 22) (11x + 2) = 0 ⇔ x = 22. Hence, speed of motorboat in strill water = 22 kmph. A man can row 18 kmph in still water. It takes him trice as long to row up as to row down the river. Find the rate of stream. Sol. Let man’s rate upsream be x kmph. Then, his rate downstream = 3x kmph. ∴ Rate in still water = 1/2(3x + x) kmph = 2x kmph. So, 2x = 18 or x = 9. ∴ Rate upstream = 9 km/hr, Rate downstream = 27 km/hr. Hence, rate of stream = 1/2(27 - 9) km/hr = 9 km/hr. A man can row upstream at 7 kmph and downstream at 10 kmph. Find man’s rate in still water and the rate of current. Sol. Rate in still water = 1/2 (10 + 7) km/hr = 8.5 km/hr, Rate of current = 1/2 (10 - 7) km/hr = 1.5 km/hr.

## Numbers -> Exercise

31. When a certain number is multiplied by 13, the product consists entirely of fives. The smallest such number is

• A. 41625
• B. 42315
• C. 42516
• D. 42735
 Ans: D. Sol. By hit and trial, we find that a number exactly divisible by 13 and consisting entirely of fives is 555555. On dividing 555555 by 13, we get 42735 as quotient. ∴ Required number = 42735.

32. How many three digit numbers are divisible by 6 in all?

• A. 149
• B. 150
• C. 155
• D. 169
 Ans: B. Sol. Required numbers are 102, 108, 114, ....., 996. This is an A.P. with a = 102 and d = 6. Let the numbers of its terms be n. Then, a + (n - 1) d = 996 ⇔ 102 + (n - 1) x 6 = 996 ⇔ n = 150.

33. The number of times 99 is subtracted from 1111 so that the remainder is less than 99 is

• A. 10
• B. 11
• C. 12
• D. 13
 Ans: B. Sol. Let it be n times. Then, (1111 - 99n) < 99. By hit and trial, we find that n = 11.