21. The L.C.M. of two numbers is 20 times their H.C.F. The sum of H.C.F. and L.C.M. is 2520. If one of the numbers is 480, the other number is :
a) 400 b) 520 c) 480 d) 600
Correct Answer: (d)
Using Rule 1,
1st number × 2nd number = L.C.M. × H.C.F
Let the H.C.F. be H.
L.C.M. = 20H
Then, H + 20H = 2520
⇒ 21 H = 2520
⇒ H = 2520/20 = 120
∴ L.C.M. = 20H = 20×120= 2400
As, First number × Second number = L.C.M. × H.C.F.
⇒ 480 × Second number = 2400 × 120
⇒ Second number = (2400×120)/480 = 600
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22. The H.C.F. of two numbers, each having three digits, is 17 and their L.C.M. is 714. The sum of the numbers will be:
a) 289 b) 221 c) 391 d) 731
Correct Answer: (b)
Let the numbers be 17x and 17y
where x and y are co-prime.
LCM of 17x and 17y = 17 xy
According to the question,
17xy = 714
⇒ xy = 714/14 = 42 = 6 x 7
⇒ x = 6 and y = 7
or, x = 7 and y =6.
First number = 17x = 17 × 6 = 102
Second number = 17y =17 × 7 = 119
∴ Sum of the numbers = 102 + 119 = 221
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23. The product of the LCM and HCF of two numbers is 24. The difference of the two numbers is 2. Find the numbers?
a) 8 and 6 b) 2 and 4 c) 8 and 10 d) 6 and 4
Correct Answer: (d)
Using Rule 1,
1st number × 2nd number = L.C.M. × H.C.F
Let the numbers be x and (x + 2).
∴ Product of numbers = HCF × LCM
⇒ x (x + 2) = 24
⇒ x2 + 2x – 24 = 0
⇒ x2 + 6x – 4x – 24 = 0
⇒ x (x + 6) – 4 (x + 6) = 0
⇒ (x – 4) (x + 6) = 0
⇒ x = 4, as x ≠ 6 = 0
∴ Numbers are 4 and 6.
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24. The sum of two numbers is 45. Their difference is 1 9 of their sum. Their L.C.M. is
a) 200 b) 100 c) 250 d) 150
Correct Answer: (b)
Let the number be x and y.
According to the question,
x + y = 45 ……… (i)
Again, x – y = 1/9 ( x + y )
x – y = 1/9 x 45
or x – y = 5 ….. (ii)
By (i) + (ii) we have,
x + y = 45
x – y = 5
—–
2x
= 50
or, x = 25
y = 45 – 25 = 20.
Now, LCM of 25 and 20 = 100.
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25. A number x is divisible by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of x is:
a) 148 b) 150 c) 149 d) 147
Correct Answer: (d)
LCM of 8, 12 and 16 = 48
∴ Required number
= 48a + 3 which is divisible by 7.
x = 48a + 3 = (7 × 6a) + (6a + 3) which is divisible by 7.
i.e. 6a + 3 is divisible by 7.
When a = 3, 6a + 3 = 18 + 3 = 21
which is divisible by 7.
∴ x = 48 × 3 + 3 = 144 + 3 = 147
