HCF and LCM MCQ Questions

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Practice HCF and LCM MCQ Questions with Details Solution

 

Q21. The least number which when divided by 12, 16 and 18 leaves 5 as remainder in each case. Find the number.

(a) 139

(b) 144

(c) 149

(d) 154

(e) None of the above

Answer: (c) 149

Solution: According to the formula,

Required number = (LCM of x, y and z) + r, where r = remainder

= (LCM of 12, 16, 18) + 5 = 144 + 5 = 149

Rules: The least number which when divided by x, y and z leaves the same remainder r in each case = L.C.M of (x, y and z) + r

Q22. Find the largest number which divides 1305, 4665 and 6905 leaving same remainder in each case. Also, find the common remainder.

(a) 1210, 158

(b) 1120, 158

(c) 1120, 185

(d) 1210, 185

(e) None of the above

Answer: (c) 1120, 185

Solution: Required number = HCF of (4665 – 1305), (6905 – 4665) and (6905 – 1305) = 1120

On dividing 1305 by 1120, remainder is 185. On dividing 4665 by 1120, remainder is 185. On dividing 6905 by 1120, remainder is 185.

.’. Common remainder =185

Rules: The greatest number that will divide x, y and z leaving the same remainder in each case = H.C.F of (x – y), (y – z) and (z – x).

Q23. What is the greatest four digit number which when divided by 10, 15, 21 and 28 leaves remainders 4, 9, 15 and 22, respectively?

(a) 9654

(b) 9666

(c) 9664

(d) 9864

(e) None of the above

Answer: (a) 9654

Solution: LCM of 10, 15, 21 and 28 = 420 and Largest number of 4-diqits = 9999

Now remainder when 9999 divided by 420 = 339

Four-digit number divisible by 10, 15, 21 and 28

= 9999 – 339 = 9660

Also, k= 10 – 4 = 15 – 9 = 21 – 15 = 28 – 22 = 6

∴ Required number = (9660 – k) = (9660 – 6) = 9654

Q24. How many numbers are there between 4000 and 6000 which are exactly divisible by 32, 40, 48 and 60?

(a) 2

(b) 3

(c) 4

(d) 5

Answer: (c) 4

Solution: LCM of 32, 40, 48 and 60 = 480

The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760.

Hence, required number of numbers are 4.

Q25. In a fire range, 4 shooters are firing at their respective targets. The first, the second, the third and the fourth shooter hit the target once in every 5 s, 6 s, 7 s and 8 s, respectively. If all of them hit their targets at 9:00 am, when will they hit their targets together again?

(a) 9 : 04 am

(b) 9: 08 am

(c) 9 : 14 am

(d) None of the above

Answer: (c) 9: 14 am

Solution: Time after which they will hit the target again together = LCM (5, 6, 7 and 8) = 840s = 14 min.

They will hit together again at 9: 14 am.

Q26. There are four prime numbers written in the ascending order of magnitude. The product of first three numbers and that of the last three number is 385 and 1001 respectively. The fourth prime number is:

(a) 11

(b) 13

(c) 17

(d) 19

Answer: (b) 13

Solution: Let p, q, r and s be the four prime numbers,

so that p × q × r = 385 and q × r × s =1001

⇒ q × r = H.C.F. of 385 and 1001 = 77.

∴ s = 1001 ÷ 77 = 13

Q27. The HCF of (x3 – x2 – 2x) and (x3 + x2) is

(a) x3 – x2 – 2x

(b) x2 + x

(c) x4 -x3 -2x2

(d) x-2

Answer: (b) x2 + x

Solution: f(x) = x3 – x2 – 2x = x (x2 – x – 2) = x {x2 – 2x + x – 2) = x {x (x – 2) + 1 (x – 2)} = x (x + 1) (x – 2)

g(x) =(x3 + x2) = x2 (x + 1)

HCF of f(x) and g(x) =   x (x + 1) = x2 + x

Q28. Monica, Veronica and Sonia begin to jog around a circular stadium. They complete their revolutions in 42 s, 56 s and 63 s, respectively. After how many seconds will they be together at the starting point?

(a) 366

(b) 252

(c) 504

(d) Couldn’t be determined

(e) None of the above

Answer: (c) 504

Solution: Required time = (LCM of 42, 56 and 63) Sec = 504 Sec

Q29. Number of students who have opted the subjects A, B, C are 60, 84, and 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions?

(a) 28

(b) 60

(c) 12

(d) 21

Answer: (d) 21

Solution: H.C.F of 60, 84 and 108 is 12 so each room contain 12 students at minimum

So that each room contains students of only 1 subject

∴ Number of rooms = $\frac{60}{12}+\frac{108}{12}+\frac{84}{12}=21$

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