289794
Mark the correct alternative in the following: If the number 2345 a 60b is exactly divisible by 3 and 5, then the maximum value of a + b is:
1 12
2 13
3 14
4 15
Explanation:
13 A number is divisible by 5 if its last digit is either 0 or 5 out of which 5 is maxim \(\therefore\) b = 5 A number is divisible by 3 if the sum of its digits is divisible by 3 2 + 3 + 4 + 5 + 6 + 0 + 5 = 25 So, we can add maximum 8 to 25 which will give us 33 which is divisible by 3 \(\therefore\) a = 8 Now, a + b = 8 + 5 = 13 Hence, the correct answer is option (b).
12. PLAYING WITH NUMBERS
289795
Mark the correct alternatiue in the following: If x and y are two co-primes, then their LCM is
1 xy
2 x+y
3 \(\frac{\text{x}}{\text{y}}\)
4 1
Explanation:
xy The LCM of two co-prime numbers is equal to their product. Thus, LCM of 'x' and 'y' will be xy.
12. PLAYING WITH NUMBERS
289796
x is twice the difference between the 6th and 10th multiple of 7. Find the value of x.
1 38
2 56
3 60
4 28
Explanation:
56 6th multiple of 7 = 42 10th multiple of 7 = 70 Now, x = 2 × (70 - 42) = 2 × 28 = 56
289794
Mark the correct alternative in the following: If the number 2345 a 60b is exactly divisible by 3 and 5, then the maximum value of a + b is:
1 12
2 13
3 14
4 15
Explanation:
13 A number is divisible by 5 if its last digit is either 0 or 5 out of which 5 is maxim \(\therefore\) b = 5 A number is divisible by 3 if the sum of its digits is divisible by 3 2 + 3 + 4 + 5 + 6 + 0 + 5 = 25 So, we can add maximum 8 to 25 which will give us 33 which is divisible by 3 \(\therefore\) a = 8 Now, a + b = 8 + 5 = 13 Hence, the correct answer is option (b).
12. PLAYING WITH NUMBERS
289795
Mark the correct alternatiue in the following: If x and y are two co-primes, then their LCM is
1 xy
2 x+y
3 \(\frac{\text{x}}{\text{y}}\)
4 1
Explanation:
xy The LCM of two co-prime numbers is equal to their product. Thus, LCM of 'x' and 'y' will be xy.
12. PLAYING WITH NUMBERS
289796
x is twice the difference between the 6th and 10th multiple of 7. Find the value of x.
1 38
2 56
3 60
4 28
Explanation:
56 6th multiple of 7 = 42 10th multiple of 7 = 70 Now, x = 2 × (70 - 42) = 2 × 28 = 56
289794
Mark the correct alternative in the following: If the number 2345 a 60b is exactly divisible by 3 and 5, then the maximum value of a + b is:
1 12
2 13
3 14
4 15
Explanation:
13 A number is divisible by 5 if its last digit is either 0 or 5 out of which 5 is maxim \(\therefore\) b = 5 A number is divisible by 3 if the sum of its digits is divisible by 3 2 + 3 + 4 + 5 + 6 + 0 + 5 = 25 So, we can add maximum 8 to 25 which will give us 33 which is divisible by 3 \(\therefore\) a = 8 Now, a + b = 8 + 5 = 13 Hence, the correct answer is option (b).
12. PLAYING WITH NUMBERS
289795
Mark the correct alternatiue in the following: If x and y are two co-primes, then their LCM is
1 xy
2 x+y
3 \(\frac{\text{x}}{\text{y}}\)
4 1
Explanation:
xy The LCM of two co-prime numbers is equal to their product. Thus, LCM of 'x' and 'y' will be xy.
12. PLAYING WITH NUMBERS
289796
x is twice the difference between the 6th and 10th multiple of 7. Find the value of x.
1 38
2 56
3 60
4 28
Explanation:
56 6th multiple of 7 = 42 10th multiple of 7 = 70 Now, x = 2 × (70 - 42) = 2 × 28 = 56
289794
Mark the correct alternative in the following: If the number 2345 a 60b is exactly divisible by 3 and 5, then the maximum value of a + b is:
1 12
2 13
3 14
4 15
Explanation:
13 A number is divisible by 5 if its last digit is either 0 or 5 out of which 5 is maxim \(\therefore\) b = 5 A number is divisible by 3 if the sum of its digits is divisible by 3 2 + 3 + 4 + 5 + 6 + 0 + 5 = 25 So, we can add maximum 8 to 25 which will give us 33 which is divisible by 3 \(\therefore\) a = 8 Now, a + b = 8 + 5 = 13 Hence, the correct answer is option (b).
12. PLAYING WITH NUMBERS
289795
Mark the correct alternatiue in the following: If x and y are two co-primes, then their LCM is
1 xy
2 x+y
3 \(\frac{\text{x}}{\text{y}}\)
4 1
Explanation:
xy The LCM of two co-prime numbers is equal to their product. Thus, LCM of 'x' and 'y' will be xy.
12. PLAYING WITH NUMBERS
289796
x is twice the difference between the 6th and 10th multiple of 7. Find the value of x.
1 38
2 56
3 60
4 28
Explanation:
56 6th multiple of 7 = 42 10th multiple of 7 = 70 Now, x = 2 × (70 - 42) = 2 × 28 = 56
