Quantitative Aptitude

Number System

616 Questions
All HCF and LCM Word Problems 416 Remainder Theorems 122 Number Properties and Series 21 Successive Division 20 Divisibility Rules 19 Digit Word Problems 9 Decimal and Fraction Operations 6 Unit Digit and Last Digits 2 BODMAS Simplification 1

Number System Questions

Multiple choice

Find the least number which when divided by 16, 24, 32 and 40 leaves the remainder 7, 15, 23 and 31 respectively.

  1. 489

  2. 231

  3. 471

  4. 480
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The divisors are 16, 24, 32, and 40. The differences between each divisor and its respective remainder are 16-7=9, 24-15=9, 32-23=9, and 40-31=9. The required number is the Least Common Multiple (LCM) of (16, 24, 32, 40) minus the common difference 9. The LCM is 480, and 480 - 9 = 471.

Multiple choice

Find the greatest number of four digits which when divided by 10, 11, 15 and 22 leaves 3, 4, 8 and 15 as remainders, respectively.

  1. 9907

  2. 9903

  3. 9893

  4. 9993
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The differences between divisors and remainders are (10-3)=7, (11-4)=7, (15-8)=7, (22-15)=7. Find LCM(10, 11, 15, 22) = 330. The largest 4-digit number is 9999. 9999 / 330 = 30 with remainder 99. The largest multiple is 9999 - 99 = 9900. Subtracting the common difference 7, we get 9900 - 7 = 9893.

Multiple choice

If x is a number of five-digits which when divided by 8, 12, 15 and 20 leaves respectively 5, 9, 12 and 17 as remainders, then find x such that it is the lowest such number.

  1. 10017

  2. 10057

  3. 10097

  4. 10077
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The divisors are 8, 12, 15, 20. The remainders are 5, 9, 12, 17. Note that (divisor - remainder) is constant at 3. LCM(8, 12, 15, 20) = 120. The number is 120k - 3. We need a 5-digit number, so 120k - 3 >= 10000. 120k >= 10003, k >= 83.35. For k=84, 120(84) - 3 = 10080 - 3 = 10077.

Multiple choice

Which is the largest six-digit number, which when divided by 12, 15, 20, 24 and 30 leaves the remainders 8, 11, 16, 20 and 26, respectively?

  1. 9,99,956

  2. 9,99,960

  3. 9,99,964

  4. 9,99,982
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Find LCM of (12, 15, 20, 24, 30) = 120. The difference between divisor and remainder is constant (12-8=4, 15-11=4, etc). We need the largest 6-digit number N such that N = 120k - 4. Largest 6-digit number is 999999. 999999 / 120 = 8333 with remainder 39. 999999 - 39 = 999960. Subtract 4: 999960 - 4 = 999956.

Multiple choice

Let y be the least 4 digit-number between 2000 and 2500 that when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder of 1 in each case. What is the sum of the digits of y?

  1. 9

  2. 15

  3. 10

  4. 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The number y must be 1 more than a multiple of LCM(2, 3, 4, 5, 6, 7) = 420. Multiples of 420 are 420, 840, 1260, 1680, 2100, 2520. The range 2000-2500 includes 2100. Thus, y = 2100 + 1 = 2101. The sum of the digits is 2 + 1 + 0 + 1 = 4.