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Wednesday, 11 October 2017

MATHS (SOLVED IN STEPS) HCF & LCM - (Page -2)

8. Which of the following integers has the most number of divisors?
A. 101 B. 99 C. 182 D. 176
Answer : Option D
Explanation :
 99 = 1 × 3 × 3 × 11
=> Divisors of 99 are 1, 3, 11, 9, 33 and
99
101 = 1 × 101
=> Divisors of 101 are 1 and 101
182 = 1 × 2 × 7 × 13
=> Divisors of 182 are 1, 2, 7, 13, 14,
26, 91 and 182
176 = 1 × 2 × 2 × 2 × 2 × 11
=> Divisors of 176 are 1, 2, 11, 4, 22, 8, 44, 16, 88, 176
Hence 176 has most number of divisors


9. The least number which should be added to 28523 so that the sum is exactly divisible by 3, 5, 7 and 8 is
A. 41 B. 42 C. 32 D. 37
Answer : Option D
Explanation :
LCM of 3, 5, 7 and 8 = 840
28523 ÷ 840 = 33 remainder = 803
Hence the least number which should be added = 840 - 803 = 37

10. What is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ?
A. 1286 B. 1436 C. 1216 D. 1386
Answer : Option D
Explanation :
LCM of 12, 14, 18 and 22 = 2772
Hence the least number which will be exactly divisible by 12, 14, 18 and 22 =
2772
2772 ÷ 2 = 1386
=> 1386 is the number which when doubled, we get 2772
Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ?


11. What is the greatest possible length which can be used to measure exactly
the lengths 8 m, 4 m 20 cm and 12 m 20 cm?
A. 10 cm B. 30 cm C. 25 cm D. 20 cm
Answer : Option D
Explanation :
Required length = HCF of 800 cm, 420
cm, 1220 cm = 20 cm

12. The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers ?
A. 26, 78 B. 39, 52 C. 13, 156 D. 36, 68
Answer : Option B
Explanation :
Let the numbers be 13x and 13y (? HCF of the numbers = 13)
13x × 13y = 2028
=> xy = 12
co-primes with product 12 are (1, 12) and (3, 4) (? we need to take only co-primes with product 12. If we take two numbers with product 12, but not coprime, the HCF will not remain as 13)
Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)
Given that the numbers are 2 digit numbers
Hence numbers are 39 and 52

13. N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N?
A. 4 B. 3 C. 6 D. 5
Answer : Option A
Explanation :
If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number
6905 - 1305 = 5600
6905 - 4665 = 2240
4665 - 1305 = 3360
Hence, the greatest number which divides
1305, 4665 and 6905 and gives the same remainder, N
= HCF of 5600, 2240, 3360
= 1120
Sum of digits in N
= Sum of digits in 1120
= 1 + 1 + 2 + 0
= 4


14. A boy divided the numbers 7654, 8506 and 9997 by a certain largest number and he gets same remainder in each case. What is the common remainder?
A. 156 B. 199 C. 211 D. 231
Answer : Option B
Explanation : If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required largest number
9997 - 7654 = 2343
9997 - 8506 = 1491
8506 - 7654 = 852
Hence, the greatest number which divides
7654, 8506 and 9997 and leaves same remainder
= HCF of 2343, 1491, 852
= 213
Now we need to find out the common remainder.
Take any of the given numbers from 7654,
8506 and 9997, say 7654
7654 ÷ 213 = 35, remainder = 199
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