ACS IBA Quantitative Reasoning Chapter 1: Numbers PDF

ACS IBA QUANTITATIVE Table of Contents Contents Odd & Even...................................................................................................................................... 1 Truth Condition (Can be/Must be)............................................................................................ 10 Sequence......................................................................................................................................... 18 Arithmetic Series......................................................................................................................... 18 Geometric Series......................................................................................................................... 19 Prime Numbers.............................................................................................................................. 24 Consecutive Numbers.................................................................................................................. 28 Value Identification...................................................................................................................... 36 Reciprocal and Number line....................................................................................................... 41 Reciprocals................................................................................................................................... 41 Number Line............................................................................................................................... 44 HCF LCM........................................................................................................................................ 47 Lowest Common Multiple (LCM)............................................................................................ 47 Highest Common Factor (HCF)................................................................................................ 48 Properties of HCF and LCM:.................................................................................................... 50 Divisibility...................................................................................................................................... 55 Some Divisibility Criteria.......................................................................................................... 55 Last Digit Problems...................................................................................................................... 63 Digits 0, 1, 5 & 6.......................................................................................................................... 63 Digits 4 & 9.................................................................................................................................. 63 Digits 2, 3, 7 & 8.......................................................................................................................... 64 Exponents & roots......................................................................................................................... 69 Exponents.................................................................................................................................... 69 General Form of Exponents...................................................................................................... 69 Rules of Exponents..................................................................................................................... 70 Roots............................................................................................................................................. 73 Properties of Square root........................................................................................................... 73 ACS IBA QUANTITATIVE Odd & Even Even-Odd: Any number divisible by 2 is called an even number, and any number not divisible by 2 is called an odd number. All even numbers end with the digits 0, 2, 4, 6, or 8, while odd numbers end with the digits 1, 3, 5, 7, or 9. For example, the numbers 432, 10, 16, 500, 34, and 58 are all even numbers. The numbers 87, 539, 555, 73, and 51 are odd numbers. All even numbers, no matter how many digits the number has, when divided by 2, will give a remainder of 0. This means, all even numbers will be divisible by 2 and there will be no remainder (ভাগশেষ). A simple hack is to just check whether the last number is divisible by 2 or not. For example, whether 5247418 is even or not, you can just test it by dividing 8 by 2 and testing whether the remainder is 0 or not. We can see, the digit “8” is divisible by 2. So, the whole number is divisible by 2 and it is an Even number. Similarly, all odd numbers, no matter how many digits, when divided by 2, will not give a remainder of 0, i.e., an even number will not be evenly divisible by 2. The remainder will be anything but a 0, there will always remain some remainders. It is important to remember the following hacks: Even + Even = Even 2+2=4 Odd + Odd = Even 3+3=6 Even + Odd = Odd 2+3=5 Odd + Even = Odd 3+2=5 Similarly, Even x Even = Even 2x2=4 Odd x Odd = Odd 3x3=9 Even x Odd = Even 2x3=6 Odd x Even = Even 3x2=6 1|Page ACS IBA QUANTITATIVE You don’t need to memorize these. Just take two small numbers (2 as even and 3 as odd), and determine what the answer is. One property of numbers is their behaviour under exponentiation. EvenEven = Even 22 = 4 OddOdd = Odd 33 = 27 EvenOdd = Even 23 = 8 OddEven = Odd 32 = 9 If you raise any odd number to any power/exponent (সূচক/ঘাত), the result will always be an odd number. Example: 32 = 9 (odd) 33 = 27 (odd) 34 = 81 (odd) No matter the exponent, the outcome remains odd. Similarly, if you raise any even number to any power, the result will always be an even number. Example: 42 = 16 (even) 43 = 64 (even) 44 = 256 (even) No matter the exponent, the outcome remains even. 2|Page ACS IBA QUANTITATIVE Past Paper Questions 1. If m is an integer, and y=(m-1) (m+2)-(m-1) (m-2), then which of the following must be true? [IBA BBA 10-11] A. y is odd only when m is odd B. y is odd only when m is even C. y is even only when m is even D. y is even only when m is odd 2. If x is an odd integer, for which of the following equations must y be an even integer? [IBA BBA 03-04] A. xy=15 B. x+2y= 15 C. 2x+y=15 D. 3x+y= 15 E. None of these 3. If X is a positive integer and y = (x-1) (x-2)-(x-3) (x-4), which of the following must be true? [IBA MBA 18-19] A. y is either odd or even B. y is even, only when x is odd C. y is even only when x is even D. y is even E. none 4. If x is a positive even integer and y is a positive odd integer, which of the following will be even? [IBA BBA 14-15] A. x3 +y3 B. 3x +y + y2 C. (x - y)(x + y) D. xy + y2 E. none of these 5. a, b, c, d, and e are five consecutive integers in increasing order of value. Which one of the following expressions must not be odd? [IBA MBA 18-19] A. a+b+c B. ab + c C. ab + d D. ac + e E. None of these 6. If p and q are positive integers and p(p + q) is even, which of the following must be true? [IBA MBA 15-16] A. if p is odd, then q is odd B. if p is odd, then q is even C. if p is even, then q is even D. if p is even, then q is odd E. None of these 7. If a, b and c are odd integers, which of the following expressions must be an even integer? [IBA MBA 13-14] A. ab + bc +ca B. a(b + c -1) C. a2 - b2 + c2 D. 3(ac-bc) E. none of these 8. If x is an even integer and y and z are odd integers, which of the following CANNOT be an integer? [IBA MBA 16-17] 𝑥2 𝑦 𝑧 𝑥𝑧 A. B. C. 𝑦 D. E. None of these 𝑦 𝑧 𝑦 9. If x is an even integer and x/12 is an odd integer, which of the following is NOT an even integer? [IBA MBA 15-16] 𝑥2 𝑥2 𝑥2 𝑥2 A. B.. 12 C. 24 D. 36 E. None of these 9 3|Page ACS IBA QUANTITATIVE 10. If x, y and z are integers and x = 2y-7+ 3z, which of the following must be odd? [IBA MBA 12-13] A. y B. z C. xy-1 D. xz-1 E. none of these 11. The positive difference between the squares of any two consecutive integers is always— [BUP 19-20] A. an even integer B. an odd number C. a prime number D. the square of an integer E. None of these Answers of past paper questions 1.E 2.D 3.D 4.B 5.D 6.D 7.D 8.E 9.E 10.E 4|Page ACS IBA QUANTITATIVE Solution to Past Paper Questions 1. If m is an integer, it can be either odd or even Let, m is odd (m -1) (m+2) – (m - 1) (m - 2) = (odd - 1) (odd + 2) - (odd - 1) (odd-2) = (even x odd) - (even x odd) = even - even = even Again, let m is even (m-1) (m+2) - (m-1) (m-2) = (even-1) (even + 2) - (even-1) (even -2) = (odd x even) - (odd x even) = even - even = even So, option (E) is the correct answer. 2. Given, x = odd 15 𝑜𝑑𝑑 (A) xy = 15 or, y = = 𝑜𝑑𝑑= odd 𝑥 (B) x+2y= 15 or, 2y = 15-x or, y = 15 - x / 2 𝑜𝑑𝑑 𝑒𝑣𝑒𝑛 odd - = = even or odd 2 2 8 10 [example 4=4= even; =5= odd] 2 (C) 2x+y=15 or, y = 15-2x = odd -2x odd = odd-even = odd (D) 3x+y= 15 or, y = 15-3x= odd - odd x odd = odd - odd = even. So, option (D) is the correct answer. 3. If x is an even number, x-1= odd, x-2= even, x-3= odd, x-4= even ∴ y= (x-1) (x-2) -(x-3) (x-4) = (odd x even) - (odd x even) = even-even = even Again, if x is odd x-1= even, x-2= odd, x-3= even, x-4 = odd ∴y= (x-1) (x-2) -(x-3) (x-4) = (even x odd) - (even x odd) = even-even = even That means, y is even in every case. So, option (D) is the correct answer. 5|Page ACS IBA QUANTITATIVE 4. (B) Given, x = positive even , y= positive odd let, x=2 and y=3 x3+y2=23+32=17 (odd) 3x+y2+y =3(2) +3+32 = 18 (even) (x-y) (x+y) =x2-y2=22-32=-5 (Odd) xy+y2=32+32= 15 (Odd) 5. Assume the values of a, b, c, d, e to be Case 1: 1, 2, 3, 4, 5 or Case2: 2, 3, 4, 5, 6. Check for all the options. A. a+b+c => taking case1, 1 + 2 + 3 = 6; which is not an odd. But if we take the values in case 2, a + b + c = 2 + 3 + 4 = 9, which is an even. The answer must not be odd. So, this option A is not an answer. Similarly, we check the rest of the options B. ab + c => taking case 1, 1 x 2 + 3 = 5; which is an odd. So, this, too is not an answer C. ab + d => taking case 1, 1 x 2 + 4 = 6; which is not odd. Taking case 2, 2 x 3 + 5 = 11, which is odd. So, C is not an option D. ac + e => taking option 1, 1 x 3 + 5 = 8, which is not odd. Taking case 2, 2 x 4 + 6 = 14; which also is not odd. Only ac + e is always even So, option (D) is the correct answer. 6. Given, p (p + q) = even If the value of p is even, then this multiplication will also be even, in that case, we cannot come to any conclusive decision. But, if p is odd, then q must be odd, otherwise this multiplication will not be even. For example, If P = 3 and q = 4, then, 3 (3 + 4) = 3 x 7 = 21 ≠ even again, If P = 3 and q = 5, 3 (3+5) = 3x8=24 = even So, option (A) is the correct answer. 6|Page ACS IBA QUANTITATIVE 7. a, b, c are odd integers. So, ‘ac' will be odd and 'bc' too will be odd. But their subtraction will be (ac-bc) = even. So, 3(ac-bc) must be even, because odd x even= even. So, option (D) is the correct answer. 𝑥 𝑒𝑣𝑒𝑛 8. (E) 𝑦= = can be an integer 𝑜𝑑𝑑 𝑦 𝑧 𝑜𝑑𝑑 or 𝑦 = 𝑜𝑑𝑑= can be an integer 𝑧 𝑥𝑧 (𝑒𝑣𝑒𝑛×𝑜𝑑𝑑) 𝑒𝑣𝑒𝑛 = = = can be an integer 𝑦 𝑜𝑑𝑑 𝑜𝑑𝑑 So, all of the above can be integers. 9. To determine which expression is not an even integer, we need to examine each option and see if it can be expressed as a product of an even integer and another integer. 𝑥2 Option A: Since x is even, x² is also even. Dividing an even number by 9 does not change 9 𝑥2 its evenness. So, is even. 9 𝑥2 Option B: 12 Since x is even, x² is also even. Dividing an even number by 12 does not change 𝑥2 its evenness. So, 12 is even. 𝑥2 Option C: 24 Since x is even, x² is also even. Dividing an even number by 24 does not change 𝑥2 its evenness. So, 24 is even. 𝑥2 Option D: 36 Since x is even, x² is also even. Dividing an even number by 36 does not change 𝑥2 its evenness. So, 36 is even. Therefore, the answer is (E). None of these. 10. x=2y−7+3z Option A: y could be either odd or even depending on x and z. Option B: z could be either odd or even. Option C: xy−1 If x is odd and y is odd, xy is odd, so xy−1 is even. If x is even and y is odd, xy is even, so xy−1 is odd. Option D: xz−1 if x is odd and z is odd, xz is odd, so xz−1 is even. If x is even and z is odd, xz is even, so xz−1 is odd. Not enough information to conclude which is odd So, option (E) is the correct answer. 7|Page ACS IBA QUANTITATIVE 11. (x+1)2 - x2 =x2+2x+1-x2 = 2x+1 2x is an even number. So, (2x+1) must be an odd number. So, option (B) is the correct answer. Practice Problems 1. If n is an odd integer, which of the following must also be odd? I. n + n II. n + n + n III. n x n x n A. I only B. II only C. III only D. II and III only E. I, II and III 2. Which of the following must be odd? I. Even × Even x Odd II. Odd × Odd x Odd III. Even + Odd +Even A. None B. I only C. II and III only D. I and III only E. I, II and III 3. If x is a positive integer, what is the units digit of (249)4x+3 x (525)3x x (423)3? A. 2 B. 5 C. 4 D. 0 E. None of these 4. If n is an even integer, which of the following must be an odd integer? 16𝑛−24 6𝑛−12 A. 7n-2 B. 5(n-2) C. D. E. none of these 8 3 𝒙 5. If x is an even integer and is an odd integer, which of the following is NOT an even 𝟏𝟐 integer? 𝑥2 𝑥2 𝑥2 𝑥2 A. B. 12 C. 24 D. E. None of these 9 3 Answers of Practice Problems 1.D 2.C 3.B 4.C 5.E 8|Page ACS IBA QUANTITATIVE Solution to Practice Problems 1. Let, n= 3, I. n + n = 6 (even) I. n + n + n = 9(odd) I. n x n x n = 27 (odd) So, option (d) is the correct answer. 2. I. Even x Even x Odd = Even x Odd = Even II. Odd x Odd x Odd = Odd x Odd = Odd III. Even + Odd + Even = Even + Odd = Odd So, option (c) is the correct answer. 3. 4x+3 will always be odd, regardless of the value of x. So, the units digit of (249)4x+3 = 9 (Last digit when 9odd) Again, units digit of (525)3x = 5 (Last digit of 5n will always be 5) And, units digit of 4233 = 7 (From the power cycle of 3) So, the required units digit = unit digit of (9 x 5 x 7), which is 5 So, option (b) is the correct answer. 4. A. 7n even; even – even= even B. n-2 is even; odd x even= even 16𝑛−24 C. = 2n-3, 2n is even, even - odd is odd. 8 6𝑛−12 D. = 2n-4, even - even = even 3 So, option (C) is the correct answer. 5. X = even Again, 𝑥 = odd integer 12 ⇒x=12x odd integer So values of x could be: 12 x 1 = 12; 12 × 3 = 36; 12 x 5=60 etc..... Putting the values of x in the answer options, you’ll see that all of the options are integers So, option (E) is the correct answer. 9|Page ACS IBA QUANTITATIVE Truth Condition (Can be/Must be) You have come across some of the questions in the previous type asked which option “must be true”? "Must be true" questions are made out of scenarios where the equation will act or result in the same manner in every possible condition. Basically, the required answer has to be applicable to all possible scenarios. For example, If given, x3 = 27 and x is a positive integer; it must be true that x = 3. No other integer fulfills the conditions. Again, say x + 3y = 10, and if the value of y is 2, then the value of x must be 4. "Can be true" questions are made from scenarios where the equation might get the accurate result under only one or more conditions, but not all. So, the equation should fulfil at least one condition. For example, when only x2 = 36 is stated, x = 6 can be true. It can also be true that x = – 6. In both cases, the value of the x can satisfy the condition. Another example, If x and y are integers, and 3x + 2y = 13, which of the following could be the value of y? a)0 b)1 c)2 d)3 e)-1 13 Back calculations here will be the easiest. Using 0 we get 3x = 13; x = which is not a whole 3 number (integer). This is how we are going to check all the options. 9 The right answer is 2, since 3x + 2(2) = 13; 3x = 13 - 4 = 9; x = = 3. 3 In some questions you will be asked “all the following must be true except?”. Then, you have to find out which of the following options must be false. Same logic goes for “all of the following could be true except?”. For example, A jar contains 12 pencils. Some sharpened and some unsharpened. Each of the following could be the ratio of sharpened to unsharpened pencils except. a) 2:1 b) 3:1 c) 4:1 d) 5:1 e) 1:1 10 | P a g e ACS IBA QUANTITATIVE Option (a) implies 2 + 1 = 3, which is a factor of 12. So, it cannot be the answer. Option (b) implies 3 + 1 = 4, which is a factor of 12. So, it cannot be the answer. Option (c) implies 4 + 1 = 5, which is not a factor of 12. So, it can be the answer. Option (d) implies 5 + 1 = 6, which is a factor of 12. So, it cannot be the answer. Option (e) implies 1 + 1 = 2, which is a factor of 12. So, it cannot be the answer. So, option (c) is the correct answer. Past Paper Questions 1. If n is an even integer, which of the following must be an odd integer? [IBA BBA 01-02] (16𝑛−24) 6𝑛−12 A. 7n – 2 B. 5(n - 2) C. D. E. none of these 8 3 2. If xy>0 and y pq B. ps> qr C. (q-p)(s-r) E. none of these 5. If the product of 6 integers is negative, at most how many of the integers can be negative? [IBA BBA 11-12] A. 2 B. 3 C. 4 D. 5 E. None of these 6. If x is an even integer and y and z are odd integers, which of the following CANNOT be an integer? [IBA MBA 16-17] 𝑥 𝑦 𝑧 𝑥𝑧 A.𝑦 B. C.𝑦 D. E. None of these 𝑧 𝑦 7. If (x+8) (y+5)= 0, which of the following must be true? [IBA BBA 18-19] A. x t>-16. Not true, consider t=-20. E. -t < -12 --> t>12. Not true. So, option (B) is the correct answer. 13 | P a g e ACS IBA QUANTITATIVE 4. Given that, p 2√x, which of the following must be true? A. x < 1 B. x < 2 C. x < 3 D. x < 4 E. none of these 2. if ab < 0, then all the following must be true EXCEPT, 𝑎 A. 𝑏 < 0 B. a² + b² > 0 𝑏 C. a3 + b3 < 0 D. 𝑎 < 0 E. None of these 3. A jar contains 12 pencils. Some sharpened and some unsharpened. Each of the following could be the ratio of sharpened to unsharpened pencils except. A. 2:1 B. 3:1 C. 4:1 D. 5:1 E. 1:1 4. If x and y are positive integers and x - y = 7, which of the following can be the value of x+y? A. 15 B. 36 C. 42 D. 64 E. none of these 5. If 5 and 11 are the lengths of two sides of a triangular region, which of the following can be the length of the third side? I. 6 II. 11 III. 16 A. ΙI only B. ΙII only C. I and ΙI only D. II and ΙII only E. I, ΙΙ, and ΙΙI Answers of Practice Problems 1.D 2.C 3.C 4.A 5.A 16 | P a g e ACS IBA QUANTITATIVE Solution to Practice Problems 1. We solve the equation: 2 + √x > 2√x => 2 > 2√x - √x => 2 > √x => 4 > x; or, x < 4 So, option (D) is the correct answer. 2. a and b both have different signs (positive and negative). So a, b, d - each must be true, but the value of a3 + b3 can be either greater than 0, or less than 0. So, option C is not true, thus the correct answer is C So, option (C) is the correct answer. 3. The total number of pencils (12) must be divisible by the sum of the parts in the ratio. This is to make sure we don't end up with fractions of pencils. Option (a) implies 2 + 1 = 3, which is a factor of 12. So, it cannot be the answer. Option (b) implies 3 + 1 = 4, which is a factor of 12. So, it cannot be the answer. Option (c) implies 4 + 1 = 5, which is not a factor of 12. So, it can be the answer. Option (d) implies 5 + 1 = 6, which is a factor of 12. So, it cannot be the answer. Option (e) implies 1 + 1 = 2, which is a factor of 12. So, it cannot be the answer. So, option (C) is the correct answer. 4. (A) if x-y = 7, a combination of values can be 11-4 = 7. In this case, x+y = 15, which is in option A So, option (A) is the correct answer. 5. The only possibility for the third side of the triangle is 11 Because, we know one property of the triangle- the sum of any 2 sides must be larger than the third side. (a+b>c) So, option (A) is the correct answer. 17 | P a g e ACS IBA QUANTITATIVE Sequence A sequence is an arrangement of any objects or a set of numbers in a particular order followed by some rule. If 𝑎1 , a ², a ³, a ⁴ ……… etc. denote the terms of a sequence, then 1,2,3, 4…...denotes the position of the term. There can be two types of series: 1. Arithmetic series 2. Geometric series For our discussion, arithmetic series is relatively a bit more important than geometric series. Arithmetic Series A sequence in which every term is created by adding or subtracting a definite number to the preceding number is an arithmetic sequence. The most important characteristic of this type of series is that the difference between two consecutive numbers is always the same, it’s constant. For example, the sequence 5, 7, 9, 11, 13, 15 is an arithmetic progression If the first term in an arithmetic progression is 'a' and the common difference between all the numbers in the sequence ‘d’. The general form of that Arithmetic Progression (AP) is a, a + d, a + 2d, a + 3d and so on. Here, the 2nd term is (a+d) The 3rd term is (a +2d) The 4th term is (a +3d) Thus, nth term of an AP series is Tn = {a + (n - 1) d} 𝒏 Sum of first n terms of an AP, S = ( ) x {2a + (n - 1) d} 𝟐 For example, 1. What is the sum of the first 50 odd numbers? Answer: We know, 𝒏 Sum of first n terms of an AP, S = ( 𝟐) x {2a + (n - 1) d} Here, n = 50 a=1 d = 3-1 = 5=3 = 2 50 So, summation = ( 2 ) x {2 x + (50 - 1) x 2} = 2550 18 | P a g e ACS IBA QUANTITATIVE 2. If 4,7,10,13,16,19,22……is a sequence, Find: a. Common difference b.21st term Answer: a. The common difference = 7 – 4 = 3 b. 21st term as: T21 = 4 + (21-1) x 3 = 4+60 = 64. Some additional formulas: (𝒂𝟏 +𝒂𝒏 ) a. Average of first n terms = ; where an is the nth term 𝟐 (𝒂𝒏 −𝒂𝟏 ) b. Number of terms up to nth term = + 𝟏; where 𝑫 - an is the highest number between a and b that is divisible by D - a1 is the lowest number between a and b that is divisible by D 𝒏(𝒏+𝟏) 2 𝒏(𝒏+𝟏)(𝟐𝒏+𝟏) c. Addition of the first n natural numbers = 1 +22+32+42+..... = 𝟐 𝟐 [𝒏(𝒏+𝟏)] d. 13+23+33+43+..... = 𝟐 Geometric Series A sequence in which every term is obtained by multiplying or dividing a definite number with the preceding number is known as a geometric sequence. In this type of series, there exists a certain ratio between any two consecutive terms this ratio is termed as common ratio, or r. For example, a series can be 2, 4, 8, 16, 32, 64,.... Here, the common ratio in this series is 4 8 32 = = = 2. Another example can be 0.5, 0.25, 0.125, 0.0625,..... Here, the common ratio 2 4 16.25 125 is =. =.5..5.25 For any geometric series, if a1 = first term r = common ratio n = number of terms Then, nth term = arn-1 𝒂𝒓𝒏−𝟏 Sum up to n terms, Sn = (𝟏−𝒓) 19 | P a g e ACS IBA QUANTITATIVE Past Paper Question 1. In an arithmetic series, the 2nd term is 27 and 5th term is 84. What is the First term? [IBA MBA 19-20] A. 7 B. 8 C. 9 D. 10 E. none 2. If the 1st number in a series of consecutive odd numbers is 8 less than the last number in the series, how many numbers are there in the series? [IBA BBA 04-05] A. 4 B. 5 C. 7 D. 8 E. none of these 3. Find the odd one out of the series: 4, 5, 7, 10, 14, 18, 25, 32. [JU IBA 14-15] A. 10 B. 25 C. 4 D. 32 E. 18 4. Find the missing number in the series 3, 6, 18, 72, 360, ___ [JU IBA 19-20] A. 720 B. 2160 C. 1440 D. 2340 E. None of these Answers of Past Paper Questions 1.A 2.B 3.E 4.B 20 | P a g e ACS IBA QUANTITATIVE Solutions to Past Paper Questions 1. The nth term of an arithmetic progression is given by: nth term a +(n-1)d Where, a first term, d = common difference According to the question, 27=a+(2-1)d or, 27=a+d-------------------(1) and, 84=a+(5-1)d Or, 84=a+4d--------------------(2) Now, on subtracting equation (1) from (2), we get: 84-27=(a+4d)-(a+d) 57=3d ∴d=19 Now, on replacing the value of d in equation (1), we get: 27=a+(19) a=27-19 ∴a=8 So, option (A) is the correct answer. 2. Let us assume that the series is 1,3,5,7,9,11,13,15...... There is a difference of 8 between the numbers 1 and 9. Hence, there are 5 numbers in the series. So, option (B) is the correct answer. 3. 2nd = (1st + 1) : 3rd = (2nd + 2); 4th = (3rd + 3); 5th = (4th +4). But 18 = 6th term not equal ≠ 5th + 5 = 14 + 5 = 19. So, option (E) is the correct answer. 4. 3 x 2 = 6 6 x 3 = 18 18 x 4 = 72 72 x 5 = 360 360 x 6 = 2160 So, option (B) is the correct answer. 21 | P a g e ACS IBA QUANTITATIVE Practice Problems 1. The sum of 3rd and 15th elements of a series in arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero? A. 1st B. 9th C.12th D. 15th E. None of These 2. (51+ 52 + 53 +......... + 100) is equal to: A. 2525 B. 2975 C. 3225 D. 3775 E. None of these 3. Find the odd number out of the following number series - 2, 4, 8, 11.5, 18.25, 28.375 A. 4 B. 28.375 C. 8 D. 11.5 E. None of these 4. What is the next number in the following number series 3, 5, 8, 15, 28, 51, ? A. 98 B. 102 C. 87 D. 75 E. Cannot be determined Answers of Practice Problems 1.C 2.D 3.C 4.A 22 | P a g e ACS IBA QUANTITATIVE Solution to Practice Problems 1. Let's consider the third term to be ‘x' and the common difference of the given arithmetic progression to be 'd' The 15th term will be (x + 12d) 6th term will be (x + 3d) 11th term will be (x + 8d) and 13th term will be (x + 10d) Thus, as per the given condition, 2x + 12d = 3x + 21d. Or x + 9d = 0 ; x + 9d will be the 12th term. So, option (C) is the correct answer. 𝑛 2. We know, Sum of first n terms of an AP: S =( 2)x[2a + (n – 1)d] Here, n = number of terms = 50 a = first term of the series = 51 d = Common difference = 1 50 So, S = ( 2 ) x [(51 X 2) + (50 – 1) X 1] Or, S = 25 X (102 + 49) So, S = 3775 So, option (D) is the correct answer. 3. Given Number series follows a pattern that, 2 2 x 1.5 + 1 = 4 4 x 1.5 + 1 = 7 (not equal to 8) 7 x 1.5 + 1 = 11.5 11.5 x 1.5 + 1 = 18.25 18.25 x 1.5 + 1 = 28.375 So, option (C) is the correct answer. 4. The given number series is = 3, 5, 8, 15, 28, 51,?Here is the pattern it follows is 3 3 + 2 = 5 (add prime number less than 3 i.e, 2) 5 + 3 = 8 (3 is a prime number less than 5) 8 + 7 = 15 (7 is a prime number less than 8) 15 + 13 = 28 (13 is a prime number less than 15) 28 + 23 = 51 (23 is a prime number less than 28) 51 + 47 = 98 (47 is a prime number less than 51) So, option (A) is the correct answer. 23 | P a g e ACS IBA QUANTITATIVE Prime Numbers A prime number is a positive integer, divisible by only 1 and itself. There is no number other than 1 and itself that divides a prime number. So a prime number has exactly 2 different positive divisors, not more, not less. Take 23 for example, 23 can only be divided by 1 and 23 itself hence, 23 is a prime number. 25 on the other hand, is not a prime number because 25 can be divided by 1, 5, and 25, i.e. 3 numbers. Some important properties to keep in mind while dealing with prime numbers are: Prime numbers are positive numbers greater than 1. For a number to be a prime number, it must be a non-zero whole number. 1 is not a prime number because it has only one positive divisor 0 is not a prime number because it has no divisor 2 is the only even prime number out there, all the other prime numbers are odd There are a total of 25 prime numbers between 1 and 100. These are the only numbers you need to know. These are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. All these numbers are divisible by only 1 and the number itself. A table on how many prime numbers are there between each 10 numbers is presented in the next page. There is an amazing sequence that you should memorize to know how many prime numbers are there between any two numbers. This sequence is 4422322321 - That means there are 4 prime numbers between 1 and 10, 4 between 11 and 20. Similarly, there is just 1 prime number between 91 and 100, and it is 97. Now, if you were asked to find out how many prime numbers are there between 47 and 81, you can easily figure it out by adding 2, 2, and 3 together to find out there are 7 prime numbers - 53, 59, 61, 67, 71, 73, and 79. Take a look at the table to correlate yourself with the sequence and what’s happening. 24 | P a g e ACS IBA QUANTITATIVE Prime Numbers between 1 and 100 Prime numbers between 1 and 10 2, 3, 5, 7 Prime numbers between 10 and 20 11, 13, 17, 19 Prime numbers between 20 and 30 23, 29 Prime numbers between 30 and 40 31, 37 Prime numbers between 40 and 50 41, 43, 47 Prime numbers between 50 and 60 53,59 Prime numbers between 60 and 70 61, 67 Prime numbers between 70 and 80 71, 73, 79 Prime numbers between 80 and 90 83,89 Prime numbers between 90 and 100 97 To determine whether a number is prime or not, follow these steps: Determine the rough approx. square root of that number. Divide the number by all the primes less than the approx. square root. If the number is not divisible by any of the primes, it is a prime. Let’s take 181 for example. The closest square numbers to 181 are 169 (132) and 196 (142). So let’s take 14 for example. All the prime numbers less than 14 are 2, 3, 5, 7, 11, 13. If we divide 181 by all of these, we see that 181 is not evenly divisible by any of these. So, 181 is a prime number. Had we taken 187 for example, we would have seen that it was evenly divisible by 11, so it cannot be a prime number. 25 | P a g e ACS IBA QUANTITATIVE Practice & Past Paper Questions 1. What is the least integer that is a sum of three different primes, each greater than 20? [BUP 19-20] A. 69 B. 73 C. 75 D. 83 E. 85 2. Fermat primes are prime numbers that can be written in the form 𝟐𝒌+𝟏 , where k is an integer and a power of 2. Which of the following is NOT a Fermat prime? A. 3 B. 5 C. 17 D. 31 E. 257 3. How many prime numbers are between 1 and 45? A. 11 B. 12 C. 13 D. 14 E. None of These 4. What is the closest prime number to 35? A. 31 B. 33 C. 37 D. 39 E. 41 5. If f and g are distinct prime numbers less than 10. Which of the following cannot be the product of f and g? [JU IBA 12-13] A. 6 B. 9 C. 10 D. 14 E. None of these Answers of Practice and past paper Questions 1.D 2.D 3.D 4.D 5.B 26 | P a g e ACS IBA QUANTITATIVE Solutions to Practice & Past Paper Questions 1. A Prime number is a positive integer with exactly two distinct natural number divisors: 1 and itself. Primes greater than 20 are 23, 29, 31, 37,... The least integer that is a sum of three different primes each greater than 20 is thus 23 + 29 + 31 = 83 So, option (D) is the correct answer. 2. Consider whether each answer choice can be written in the form 2x + 1. A 3 = 21+1; where k=1 = 20 B 5 = 22+1; where k=2=21 C 17 = 24+1; where k=4 = 22 D 31 = 30+1; 30 CANNOT be expressed as an integer power of 2 E 257=28+1 where k=8=23 So, option (D) is the correct answer. 3. There are 14 prime numbers from 1 to 45: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 We can use the 4422322321 rule as well. So, option (D) is the correct answer. 4. 37 is the closest. So, option (D) is the correct answer. 5. Prime numbers between 1 and 10 are 2, 3, 5, 7. Check for all the possible multiplications. 9 is the multiple of 3 and 3, but the question asked for distinct numbers, so the answer is 9 So, option (B) is the correct answer. 27 | P a g e ACS IBA QUANTITATIVE Consecutive Numbers Consecutive numbers are numbers that follow each other in order. They have a difference of 1 between every two numbers. In a set of consecutive numbers, the mean, and the median are equal. If n is a number, then n, n+1, and n+2 would be consecutive numbers Examples. 1, 2, 3, 4, 5, or -2, -1, 0, 1, 2, etc. However, keep in mind that consecutive even or consecutive odd numbers have a common difference of 2. For example, 1, 3, 5, 7, 9 or 2, 4, 6, 8, 10 etc. In these cases, if the first number is n, then the second number is n+2, then n+4, then n+6 and so on. Some additional things to keep in mind while dealing with consecutive numbers are: The formula to get a consecutive integer is n + 1, For odd consecutive integers, the general form of a consecutive odd integer is 2n+1, For even consecutive integers, the general form of a consecutive even integer is 2n. Past Paper Questions 1. If x is a product of 4 consecutive integers, and x is divisible by 11, which of the following is not necessarily a divisor of x? [IBA MBA 17-18] A. 12 B. 22 C. 24 D. 33 E. None of these 2. If x, y and z are consecutive positive integers and if x < y b>c, which of the following has the maximum value? [JU IBA 21] 𝑏 𝑐 A. 𝑎 + 𝑐 B. 𝑏 + 𝑎 𝑏 𝑎 𝑐 C. 𝑐 − 𝑎 D. 𝑐 + 𝑏 E. 𝑎 − 𝑏 8. Two sets of 4 consecutive positive integers have exactly one integer in common. The sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set? [BUP 12-13] A. 4 B. 7 C. 8 D. 12 E. Cannot be determined 9. Of 10 consecutive numbers, the Sum of the first five is 560. What will be the sum of the last 5 numbers? [BUP 12-13] A. 585 B. 570 C. 575 D. 565 E. none of these 10. If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is: [BUP 19-20] A. A+1 B. A+5 C. A+25 D. 2A E. A/2 Answers of Past Paper Questions 1.E 2.E 3.C 4.B 5.A 6.D 7.A 8.D 9.A 10.C 29 | P a g e ACS IBA QUANTITATIVE Solutions to Past Paper Questions 1. The cases here can be Case 1: 8 x 9 x 10 x 11 Case 2: 9 x 10 x 11 x 12 Case 3: 10 x 11 x 12 x 13 Case 4: 11 x 12 x 13 x 13 Here, we can see that in each of the 4 cases, the numbers are divided by the options. That means all of the options (12, 22, 24, 33) can divide any of the cases 1, 2, 3, or 4. So, none of them are necessarily a divisor of x So, option (E) is the correct answer. 2. x, y, z consecutive positive integer ∴ x < y X+X+2 = X+4+9 => 2X-X= 9+4-2 => X=11 => Z=11+4=15 So, option (B) is the correct answer. 30 | P a g e ACS IBA QUANTITATIVE 5. Let's take the consecutive integers as x, x+2, x+4, x+6. They all are even integers. [𝑥+ (𝑥+2)+ (𝑥+4)+ (𝑥+6)] Their average= = x+3 4 As x is even, so x+3 will always be odd. So, option (A) is the correct answer. 6. (E) Let, the largest number = x So, according to question, Sum of these 6 integers, sum = 2x + 38 Or, 2x = sum - 38 𝑠𝑢𝑚 − 38 Or, x = 2 The sum is to be an odd number. So if you back calculate from the answer choice, you’ll see that none of the numbers are odd 43 − 38 Option A: 43. = 4.5; which is not odd 2 50 − 38 Option B: 50. = 6; which is not odd 2 64 − 38 Option C: 64. = ; which is not odd 2 43 − 38 Option D: 72. = 13; which is odd 2 So, option (D) is the correct answer. Alternative way Let, the numbers are: x, x + 2, x + 4, x + 6, x + 8, x + 10. Their sum: (x + x + 2 + x + 4 + x + 6 + x + 8 + x + 10) = 6x + 30 According to question, 6x + 30 = 2(x + 10) + 38 Or, 6x + 30 = 2x + 20 + 38 Or, 4x = 28 Or, x = 7 So, sum of the 6 numbers = 6 x 7 + 30 = 72 So, option (D) is the correct answer. 7. To get the maximum value from a fraction, we would need the numerator to be as high as possible and the denominator to be as low as possible. As option A has c as the denominator, which is the lowest integer. So, option (A) is the correct answer. 31 | P a g e ACS IBA QUANTITATIVE 8. Let two sets of 4 consecutive positive integers with one number in common be: n + (n + 1) + (n + 2) + (n + 3 ) and (n + 3) + (n + 4) + (n + 5) + (n + 6) Note that each term in the second set is 3 more than the equivalent term in the first set. Difference Set 1 -Set 2= 3+3+3+3=12 (D) Since there are four terms the total of the differences will be 4 x 3 = 12 So, option (D) is the correct answer. 9. Let the numbers be n-2,n-1,n,n+1,n+2,…,n+7 Given , sum of first 5 integers 560. So we have Sum of n-2,n-1,n,n+1,n+2 as 560 i.e. 5n = 560 n=112 Sum of last 5 integers will be 5n +3+4+5+6+7 ie 5x112 + 25 = 585 So, option (A) is the correct answer. 10. Let 1st number =n The 2nd. 3rd, 4th and 5th numbers must correspondingly be:: n+1, n+2, n+3 & n+4 Thus, the sum of the five (5) consecutive numbers (A) = n+(n+1)+(n+2)+(n+3)+(n+4) Or, A = 5n + 10 ……….(1) From equation (1) 𝐴 − 10 n= ……(2) 5 Let the next five (5) consecutive numbers 6th, 7th, 8th, 9th & 10th be as follows; n+5, n+6, n+7, n+8 & n+9 Let the sum of theses 5 consecutive numbers be B Thus, B = 5n + 35..,,…(3) 𝐴−10 Substitute from (2) for n in (3) 5 𝐴−10 Hence,B = 5( ) +35 …….(4) 5 Hence: B = A + 25 So, option (C) is the correct answer. 32 | P a g e ACS IBA QUANTITATIVE Practice Problems 1. In a series of 7 consecutive even numbers, the average of the first three is 10. What is the average of the last five numbers in the series? A. 12 B. 22 C. 24 D. 16 E. None of these 2. The average of 7 consecutive numbers is 33. The largest of these numbers is: A. 36 B. 33 C. 30 D. 28 E. None of these 3. The numbers of products a store sold on 4 consecutive days were x, x + 5, x + 3 and x + 12. If the daily average of the products sold was 13, the highest number of products sold in a day is? A. 16 B. 17 C. 19 D. 20 E. None of these 4. On a road, three consecutive traffic lights change after 40, 48 and 56 seconds respectively. If the lights are first switched on at 10:00 AM sharp, at what time will they change simultaneously? A. 10:35 a.m. B. 10:28 a.m. C. 10:40 a.m. D. 10:43 a.m. E. None of these 5. If 14 is the average of 6 consecutive odd numbers, what is the smallest number? A. 9 B. 11 C. 1 D. 13 E. 17 6. From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers? A. (-10)20 B. (-10)10 C. 0 D. –(10)19 E. –(10)20 Answers of Practice Problems 1.D 2.A 3.D 4.B 5.A 6.E 33 | P a g e ACS IBA QUANTITATIVE Solutions to Practice Problems 1. The average of three consecutive even numbers is 10. So, the first three numbers in the series are 8,10 and 12 So, the series is; 8,10,12,14,16,18, 20. So, the average of the last 5 numbers will be 16. So, option (D) is the correct answer. 2. Let the seven numbers be n1, n2, n3…, n7 Let, n1=x Therefore, n2=x+1, n3=x+2, and so on up to n7, n7= n1+6 So, the series would be x, x+1, x+2, x+3, x+4, x+5, x+6 Now, [(𝑥)+(𝑥+1)+(𝑥+2)+(𝑥+3)+(𝑥+4)+(𝑥+5)+(𝑥+6)] = 33 7 [ 7𝑥 + 21 ] = 33 7 Therefore, x= 30 Now, the largest number would be n7 And we know that n7= n1 + 6 = 30+6 = 36 So, option (A) is the correct answer. 3. Here, 𝑥 + 𝑥 + 5 + 𝑥 + 3 + 𝑥 + 12 = 13 4 Or, 4x + 20 = 52 So, x = 8 So, the highest items sold in a day = (8 + 12) = 20 So, option (D) is the correct answer. 4. For the lights to change simultaneously again, the time elapsed must be a common multiple of all three intervals. Specifically, it must be the Least Common Multiple (LCM) of the intervals. The LCM is the smallest number that each interval divides evenly into. LCM of 40, 48, 56 = 1680 sec 1680 𝑠𝑒𝑐 Hence, the lights will change simultaneously after= = 28 minutes. 60 So, option (B) is the correct answer. 34 | P a g e ACS IBA QUANTITATIVE 5. Given, 14 is the average of 6 consecutive odd numbers. So, 14 is the middle number which also happens to be even, and there are 3 numbers to its right and 3 numbers to its left. As these numbers are odd, they are - 9,11,13,15,17,19. So, the smallest number is 9 So, option (A) is the correct answer. 6. If any of the 20 integers chosen is 0, the product of all integers will be 0. Since 0 is included in the range from −10 to 10, it is possible to choose 0 at least once. Therefore, one choice is to include 0 in our selections. In this case, the product is=0 Select 10 odd numbers of times, and -10 the remaining number of times, this will ensure that the product will be negative. For example, select 10 once and -10 nineteen times, then the product will be 101×(−10)19=−1020. So, option (E) is the correct answer. 35 | P a g e ACS IBA QUANTITATIVE Value Identification The problems in this type require the use of basic understanding of numbers which have been discussed in the previous sections. All you need to do is develop equations using which you can determine the value of the unknown variables. Practice & Past Paper Questions 1. If x and k are integers and (12x)(42x+1)=(2k)(32), what is the value of k? A. 5 B. 7 C. 10 D. 12 E. 14 𝒙 2. When a positive integer x is divided by a positive integer y the remainder is 9. If = 𝒚 96.12, then what is the value of y? [BUP 19] A. 96 B. 25 C. none D. 75 E. 69 3. The integers x and y are greater than 1. If (4x) (7y)=756, what is the value of x+y? [IBA MBA 21-22] A. 3 B. 9 C. 12 D. 28 E. None 4. If x and y are both positive integers and 10 x=96.12y --> x=96y+0.12y (so q above equals to 96); So, 0.12y=9 --> y=75 So, option (D) is the correct answer. 756 3. given, (4x)(7y) 756⟶28 xy = 756⟶ xy = 28 = 27 Since x and y are positive integers, there are only four possible solutions: Case 1: x=1& y=27, or Case 2: x=3&y=9, or Case 3: x=9&y=3,or Case 4: x = 27 & y=1 Since x and y are greater than 1, only cases 2 and 3 are possible. In both cases, x+y=12 So, option (B) is the correct answer. 37 | P a g e ACS IBA QUANTITATIVE 4. Given, x & y are both positive integer 10 < x 0, what is the least possible value for x+(1/x)? [ JU IBA 14 ] A. 0.5 B. 1 C. 1.5 D. 2 E. None of these 4. If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is, A. 3 B. 5 C. 7 D. 9 E. 11 5. If S is the sum of the reciprocals of the 10 consecutive integers from 21 to 30, then S is between which of the following two fractions? 1 1 1 1 1 1 1 1 1 1 A. 3 and 2 B. 4 and 3 C. 5 and 4 D. 6 and 5 E. 7 and 6 Answers of Practice & Past Paper Questions 1.a 2.A 3.D 4.C 5.A 41 | P a g e ACS IBA QUANTITATIVE Solution to Practice & Past Paper Questions 1. Let, the required proper fraction = x 1 So, the reciprocal 𝑥 1 Since x is a proper fraction, 𝑥 > x. 1 9 Now, (𝑥) – x = 20 1−𝑥 2 9 Or, = 20 𝑥 Or, 20- 20 𝑥 2 =9x Or, 20 𝑥 2 + 9x - 20= 0 Or, 20 𝑥 2 + 25x - 16x - 20 = 0 Or, 5(4x+5) - 4(4x+5) =0 Or, (4x+5) (5x-4) =0 5 4 So, x= –4 or, 5 But x is a proper fraction 4 So , x= 5 1 5 =4 𝑥 1 4 5 16+25 41 Sum of the fraction and its reciprocal, x + (𝑥) = 5 + 4 = = 20 20 So, option (A) is the correct answer. 2. Let the numbers are a and b Sum of the numbers, a + b = 12 Product of the numbers, ab = 35 1 1 𝑎+𝑏 12 Sum of the reciprocals of the numbers ( ) + ( ) = = 𝑎 𝑏 𝑎𝑏 35 [putting the values of a + b and ab] So, option (A) is the correct answer. 1 3. 𝑥 + 𝑥 x is reciprocal of 1/x & vice versa. To have x+1/x the lowest, we require to find the number whose reciprocal is the same as that of it 1 x=𝑥 x2=1 x=1 1 is the only number whose reciprocal = 1 Least addition = 1 + 1 = 2 So, option (D) is the correct answer. 42 | P a g e ACS IBA QUANTITATIVE 4. We can let the first odd integer = x and the next odd integer = x + 2; thus, the reciprocals are 1 1 and 𝑥+2. Thus, 𝑥 1 1 12 + 𝑥+2 = 35 𝑥 Multiplying by 35x(x+2), we have: 35(x + 2) + 35x = 12x(x + 2) 35x + 70 + 35x = 12𝑥 2 + 24x 12𝑥 2 - 46x - 70 = 0 6𝑥 2 - 23x - 35 = 0 (6x + 7)(x - 5) = 0 7 Thus, x = − 6 or x = 5. Since x is an integer, x must be 5 and the greater integer is x + 2 = 7. Alternate solution: We are given that the sum of the reciprocals of two consecutive odd integers is 12/35. We see that the denominator is 35. It’s not difficult to conjecture that the integers have to be 5 1 1 12 and 7 since 5 x 7 = 35. Finally, we can check the sum of and to see if they sum to : 5 7 35 1 1 7 5 12 + = + = 5 7 35 35 35 12 Since they do add up to , the larger integer is 7. 35 So, option (C) is the correct answer. 1 1 1 1 1 1 1 1 1 1 5. S=21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30. 1 1 Notice that 21 is the largest term and 30 is the smallest term. 1 1 1 If all 10 terms were equal to , then the sum would be 10× 30 = 3, but since the actual sum 30 1 is more than that, then we have that S > 3. 1 1 10 If all 10 terms were equal to 21, then the sum would be 10× 21 = 21, but since the actual sum 10 is less than that, then we have that S < 21. Therefore, 𝟏 𝟏𝟎 10 1 1 10 1 < 𝑺 < 𝟐𝟏 (notice that 21 < 2 , so 3 < S < < 2). 𝟑 21 So, option (A) is the correct answer. Number lines are the horizontal straight lines in which the integers are placed in equal intervals. All the numbers in a sequence can be represented in a number line. This line extends indefinitely at both ends. 43 | P a g e ACS IBA QUANTITATIVE Number Line -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 A number line is a picture of numbers that are lined up in a straight line. It is a list of numbers that can be used to compare and sort them. This number can stand for any real number, such as a natural number or a whole number. To review, a whole number is a collection of all the counting numbers (1, 2, 3,4,5,6...) and zero (0). A natural number, on the other hand, is a collection of all the counting numbers, such as 1, 2, 3,4,5,6... Putting numbers on a number line helps you see how they relate. We can see that the numbers on the left are less than the numbers on the right in the picture above. As an example, 0 is less than 1, -1 is less than 0, -2 is less than -1, and so on. A number line is a better way to show how to do math with numbers. First, you need to know how to find numbers on a number line. One hundred is in the middle of a number line. When you look at the number line, all positive numbers are to the right of zero, and all negative numbers are to the left of zero. The value of a number goes down as we move to the left. Let's say that 1 is greater than -2. On a number line, it's also easy to show whole numbers, parts, and decimals. The absolute value or modulus of a number is the numerical value of the number, without regard to its sign. So, for example, the absolute value of 10 is 10, and the absolute value of -10 is also 10. The absolute value of a number may be thought of as its distance from zero. Since distance can never be negative, absolute value can never be negative. Absolute value of 15 = |15| = 15 (in the number line, the distance from 0 to +15 is 15) Absolute value of -15 = |-15| = 15 (in the number line, the distance of 0 to -15 is 15) 44 | P a g e ACS IBA QUANTITATIVE Practice & Past Paper Questions 1. On a real number line, x1 = –5 and x2= 17. What is the distance between these two points? A. 4 B. -22 C. 22 D. 10 E. None of these 2. What is the absolute value of twice the difference of the roots of the equation 5y2 - 20y + 15=0? [ BUP 13 ] A. 0 B. 1 C. 2 D. 3 E. 4 3. If x is a number such that -2≤ x ≤2, which of the following has the largest possible absolute value? [IBA MBA 07-08 ] A. 3x-1 B. x²+1 C. 3-c D. x-3 E. x²-x Answers of Practice & Past Paper Questions 1.C 2.E 3.A 45 | P a g e ACS IBA QUANTITATIVE Solution to Practice & Past Paper Questions 1. The distance between two points is always positive. We calculate lx×2 – x×1l, which will give us the distance between the points. |17 – (–5)| = |17 + 5| = |22| = 22 So, option (C) is the correct answer. 2. In this equation, after doing the middle term, we can simplify it as 5 (y-1) (y-3) = 0; i.e. the roots of the equation are 1 and 3. So, the difference between the roots are (3-1) = 2. So, twice of the difference is 2 x 2 = 4; whose absolute value is also 4. So, option (E) is the correct answer. 3. Pot the range of the condition, i.e. -2 and 2 in the options 3x – 1 : At ends, value is 5 and -7 2x + 1 : At ends, value is 5 and global min value is at x=0, where it is 1 3 – x : At ends, value is 1, 5 x – 3 : At ends, value is -5,-1 2x– x : At ends, value is 2,6. Global min value is at x=0.5, where it is -0.25 So, option (A) is the correct answer. 46 | P a g e ACS IBA QUANTITATIVE HCF LCM Factors of any number are numbers that we can multiply with an integer to get that initial number. Again, multiple of a number is what we get after multiplying that number by an integer (not a fraction), Let, 'x' and 'y' are 2 integers and x = 10y So, 'y' is a factor of 'x' Again,'x' is a multiple of 'y' Any number itself is its largest factor and smallest multiple. Lowest Common Multiple (LCM) The least or smallest common multiple of any two or more given natural numbers are termed as LCM. For example, Take three numbers 5, 15, and 25 The multiples of 5 are 5, 10, 15, 20, 25,....., 70, 75, 80,..... The multiples of 15 are 15, 30, 45, 60, 75, 90,..... The multiples of 25 are 25, 50, 75, 100, …. The least of these common multiples is 75 Therefore, the LCM of 5, 15, and 25 is 75. LCM can be determined by two ways: 1. Prime factorization method To calculate the LCM of two numbers 60 and 45. Out of other ways, one way to find the LCM of given numbers is as below:List the prime factors of each number first. 60 = 2 × 2 x 3 × 5 45 = 3 × 3 × 5 Then multiply each factor the most number of times it occurs in any number.If the same multiple occurs more than once in both the given numbers, then multiply the factor by the most number of times it occurs. The occurrence of Numbers in the above example: 2: two times 3: two times 5: one time LCM = 2 × 2 x 3 × 3 × 5 = 180 47 | P a g e ACS IBA QUANTITATIVE 2. LCM by division method Let us see with the same example, which we used to find the LCM using prime factorization. Solve LCM of (60,45) by division method Therefore, LCM of 60 and 45 = 2 × 2 x 3 × 3 × 5 = 180 2 60, 45 2 < , 45 30 2 15, 45 2 5, 15 2 5, 5 1, 1, Highest Common Factor (HCF) The largest or greatest factor common to any two or more given natural numbers is termed as HCF of given numbers. For example, 4, 8, 12, 16 The factors of 4 are 1, 2, and 4 The factors of 8 are 1, 2, 4, and 8 The factors of 12 are 1, 2, 3, 4, 6, and 12 The factors of 16 are 1, 2, 4, 8, and 16 Here, the highest of these common factors is 4. So, HCF of 4, 8, 12, and 16 is 4 The important thing to keep in mind is that, if you cannot find a single common factor amongst the numbers, then the HCF of that series of numbers is 1. For example, take the number series 4, 8, 12, 16, 15. Here, the highest of their common factors is 1 only. So the HCF of this series is 1. 48 | P a g e ACS IBA QUANTITATIVE HCF can be determined by two ways 1. Prime factorization method Take an example of finding the highest common factor of 144, 104 and 160. Now let us write the prime factors of 144, 104 and 160. 144 = 2 × 2 × 2 × 2 × 3 × 3 104 = 2 × 2 × 2 × 13 160 = 2 × 2 × 2 × 2 × 2 × 5 The common factors of 144, 104 and 160 are 2 × 2 × 2 = 8 Therefore, HCF (144, 104, 160) = 8 2. HCF by division method Steps to find the HCF of any given numbers; 1. Larger number/ Smaller Number 2. The divisor of the above step / Remainder 3. The divisor of step 2 / remainder. Keep doing this step till R = 0(Zero). 4. The last step’s divisor will be HCF. The above steps can also be used to find the HCF of more than 3 numbers. Example: Find the HCF of 144 and 160 by division method. Since 160>144, so the dividend will be 160 and the divisor will be 144. By using the division method, we get: Hence, we can see here that 16 is the highest number which divides 160 and 144. Therefore, HCF (144, 160) = 16 144) 160 (1 144 16) 144 (9 XXX 49 | P a g e ACS IBA QUANTITATIVE Properties of HCF and LCM: The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers. ○ LCM of two numbers x HCF of the two numbers = Product of the two numbers. ○ Suppose A and B are two numbers, then, LCM (A & B) x HCF (A & B) = A x B H.C.F. and L.C.M. of Fractions 𝑳𝑪𝑴 𝒐𝒇 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓𝒔 ○ LCM of fractions = 𝑯𝑪𝑭 𝒐𝒇 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓𝒔 𝑯𝑪𝑭 𝒐𝒇 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓𝒔 ○ HCF of fractions = 𝑳𝑪𝑴 𝒐𝒇 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓𝒔 50 | P a g e ACS IBA QUANTITATIVE Practice & Past Paper Questions 1. A certain number is divided by 16, 18, 24 and 36, and we get 3 as remainder, every time. What is the number? A. 75 B. 69 C. 147 D. 141 E. 96 2. If a certain number of apples are distributed among 18, 24 and 36 children then 15, 21 and 33 apples remain respectively, what is the smallest possible number of apples? A. 69 B. 75 C. 144 D. 147 E. 147 3. What is the minimum number of chocolates that can be distributed among 8, 9, 12, and 16 people so that 3 chocolates remain every time? A. 75 B. 147 C. 141 D. 69 E. None of these 4. Bell A rings after every 12 minutes, bell B rings after every 15 minutes and bell C rings after every 20 minutes. The bells rang together at 10:00. When will they ring together again? A. 10:30 B. 10:40 C. 11:30 D. 11:00 E. None of these 5. What is the sum of all integers from 1 to 1000 that are divisible by both 3 and 5? A. 33165 B. 33333 C. 33466 D. 33500 E. 33766 6. Bouquets are to be made using white tulips and red tulips, and the ratio of the number of white tulips to the number of red tulips is to be the same in each bouquet. If there are 15 white tulips and 85 red tulips available for the bouquets, what is the greatest number of bouquets that can be made using all the tulips available? A. 3 B. 5 C. 8 D. 10 E. 13 7. When x is divided by 13, the answer is y with a remainder of 3. When x is divided by 7, the answer is z with a remainder of 3. If x, y, and z are all possible integers, what is the 𝐲𝐳 remainder of 𝟏𝟑? [IBA BBA 15-16] A. 0 B. 3 C. 4 D. 7 E. none of these 8. Mr. Saif has n luxury apartments, where n is an integer such that 20 0, and all the variables are non-zero numbers, which of the following must be true? [IBA BBA 16-17] A. ab > 0 B. ab < 0 C. cd > 0 D. cd < 0 E. none of these 2. The expression √(𝟐 + √(𝟐 + √(𝟐 + √𝟐 + ⋯ … … ))) extends to an infinite number of roots. Which of the following choices most closely approximates the value of this expression [IBA BBA 15-16] A. √3 B. 2 C. 1+√2 D. 1+√3 E. none of these 3. If both 𝟓𝟐 and 𝟑𝟐 are factors of x where x = n × 𝟐𝟓 × 𝟔𝟐 × 𝟕𝟑 , what is the smallest possible positive value of n? [IBA MBA 07-08] A. 25 B. 27 C. 45 D. 75 E. None of these 4. If x = (0.08)𝟐 , y = 1 / (0.08)𝟐 and z = (1 – 0.08)𝟐 – 1, which of the following is true? [IBA BBA 05-06] A. x = y = z B. y < z < x C. z < x < y D. y < x and x = z E. None of these 5. What is the greatest positive integer n such that 2n is a factor of 1210? [IBA BBA 00-01] A. 10 B. 20 C. 30 D. 40 E. 50 6. At room temperature, bacteria doubles every 30 minutes. In four hours, what will be the total number of bacteria from an initial number of 4? [IBA MBA 94-95] A. 512 B. 1024 C. 2048 D. 4096 E. None of these 7. If x > 0.7, which of the following could be the value of x? [IBA MBA 11-12] A. √0.49 B. √0.7 C.(𝟎. 𝟕) 𝟐 D. (1-√0.2) E. none of these 8. Which of the columns below has the greatest value? Column 1 Column 2 Column 3 Column 4 (−𝟔)𝟑 (−𝟔)𝟒 (−𝟔)𝟓 (−𝟔)𝟐 [IBA MBA 09-08] A. Column 1 B. Column 2 C. Column 3 D. Column 4 E. All are equal Answers of Practice & Past Paper Questions 1.C 2.B 3.A 4.C 5.B 6.B 7.B 8.D 75 | P a g e ACS IBA QUANTITATIVE Solutions to Practice & Past Paper Questions 1. Given, (ax)16 ×(by)12 ×(cxy)17 ×(dxy)11 >0 a16 x b12 x c17 x d11 x x16+17+11 x y12+17+11 > 0 a16 x b12 x c17 x d11 x x11 x y40 > 0. Now a, b, x and y have even power. Raised to even power, they would always be POSITIVE. But without power, each could be positive or negative. Therefore, we cannot comment on their PRODUCT without powers. So we cannot comment on ab. Now all other terms with a, b,x and y are positive, so c17 x d11 also should be > 0... Therefore, cd>0. So, option (C) is the correct answer. 2. Let,x= √(𝟐 + √(𝟐 + √(𝟐 + √𝟐 + ⋯ … … ))) We can write, x= √2+x [Because this is an infinite series, and we replaced √(𝟐 + √(𝟐 + √(𝟐 + √𝟐 + ⋯ … … ))) this part by x] x2=2+x [Squaring both sides] or,x2-x-2=0 or,x-2x+x-2 = 0 or,x (x-2)+ 1 (x-2)= 0 or,(x-2) (x+1) = 0 or,x=2,-1 x cannot be negative. The reason x cannot be negative is that, the original expression involves square roots, and square roots always produce non-negative numbers. Hence, x=2 So, option (B) is the correct answer. 76 | P a g e ACS IBA QUANTITATIVE 3. We need to determine the smallest possible positive value of “n” Given, x is divisible by 52 or 25 and 32 or 9. Now, 32 is present ( in 6 x 6) but 25 is not present in the other factors i.e. 25 x 62 x 73 So, 25 is the minimum value of n. So, option (A) is the correct answer. 4. X = 0.0064, [This value is less than 1] 1 y= [This value will be more than 1] 0.0064 z = (1 – 0.08)2 – 1= (.92)2 - 1 We won’t need to find the value of Z as we can see it will be a negative value. Here, x < 1, y > 1 and z < 0 So, z < x < y Option (C) is the correct answer. 5. The given number is 1210 =(2𝑥2𝑥3)10 =(220 )x(310 ) So the greatest possible value for n such that 2𝑛 can be factor of given number is 20. So, option (B) is the correct answer. 6. Since the bacteria doubles every 30 minutes, the number of 30-minute intervals in four hours 240 𝑚𝑖𝑛 is = = 8 times 30 𝑚𝑖𝑛 So, total number of bacteria = 4 x 28 = 4 x 256= 1024 So, option (B) is the correct answer. 7. we know, if th 0 < x < 1, then with the increasing power of x, the value decreases and the decreasing power of x, the value increases so, as x > 0.7, the option b which is clearly greater than 0.7 as √0.7 > 0.7, option B is the answer So, option (B) is the correct answer. 8. the values in column 1 and 3 are negative and the powers are odd. so the values will also be negative. as we are looking for the greatest value, we can omit options A and C then, the values in options B and D are negative but the powers here are even. so these values are actually positive. between them, 6 > 4, so (−6)6 > (−6)4. So, option (D) is the correct answer. 77 | P a g e

ACS IBA Quantitative Reasoning Chapter 1: Numbers PDF

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