Q.1 The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? A) 40 & 24 B) 42 & 24 C) 42 & 20 D) None of the above Ans. B Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 10 x + y in the expanded form (for example, 56 = 10(5) + 6). When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5). According to the given condition. (10x + y) + (10y + x) = 66 i.e., 11(x + y) = 66 i.e., x + y = 6 (1) We are also given that the digits differ by 2, therefore, either x – y = 2 (2) or y – x = 2 (3) If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. In this case, we get the number 42. If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. In this case, we get the number 24. Thus, there are two such numbers 42 and 24. Q.2 The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2.Find the rational number ? A) 13/ 21 B) 13/20 C) 11/21 D) 11/ 20 Ans. A Acc. to the question, 2(x + 17) = 3(x + 7) ? 2x + 34 = 3x + 21 ? 34 - 21 = 3x - 2x ?13 = x Numerator of the rational number = x = 13 Denominator of the rational number = x + 8 = 13 + 8 = 21 Q.3 Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number? A) 2 B) 3 C) 4 D) 5 Ans. C Let the number be x. According to the given question, 8x - 20 = 3x Transposing 3x to L.H.S and -20 to R.H.S, we obtain 8x - 3x = 20 5x = 20 Dividing both sides by 5, we obtain x = 4 Hence, the number is 4. Q.4 What is the number of terms in the series 117,120,123,126…333 ? A) 79 B) 76 C) 72 D) 73 Ans. D a=117, d=3 nth term = 333 Since, tn=a+(n-1)d Or 333=117+(n-1)3 333=117+3n-3 3n=333-114= 219 Or n=219/3 = 73 Q.5 If the numbers from 501 to 700 are written what is the total number of times does the digit 6 appear ? A) 141 B) 140 C) 139 D) 138 Ans. B No. of 6 between 501 to 999 at tens place = 10 No. of 6 between 501 to 599 at unit place = 10 No. of 6 between 600 to 699 at hundred place = 100 No. of 6 between 600 to 699 at tens place = 10 No. of 6 between 600 to 699 at unit place = 10 Total no. of times 6 appears =10+10+100+10+10=140

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