The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is = ?
(A) 91
(B) 910
(C) 1001
(D) 1911
Solution:
Required number of students = HCF of 1001 and 910 = 91
3.
Four bells begin to toll together and toll respectively at intervals of 6, 7, 8 and 9 seconds. In 1.54 hours, how many times do they toll together and in what interval (seconds)?
(A) 14, 504
(B) 14, 480
(C) 12, 504
(D) 16, 580
Solution:
Interval after which the bells will toll together = (LCM of 6, 7, 8, 9) sec = 504 sec. In 1.54 hours, they will toll together $$\eqalign{ & {\text{ = }}\left[ {\left( {\frac{{1.54 \times 60 \times 60}}{{504}}} \right) + 1} \right] \cr & {\text{ = 12 times}} \cr} $$
4.
Three girls start jogging from the same point around a circular track and each one completes one round in 24 seconds, 36 seconds and 48 seconds respectively. After how much time will they meet at one point ?
(A) 2 minutes 20 seconds
(B) 2 minutes 24 seconds
(C) 3 minutes 36 seconds
(D) 4 minutes 12 seconds
Solution:
LCM of 24, 36, 48 = 144 So, the three girls will meet at one point in 144 seconds i.e., 2 minutes 24 seconds
5.
The HCF of two numbers is 29, and the other factors of their LCM are 15 and 13. The larger of the two numbers is:
(A) 435
(B) 377
(C) 406
(D) 464
Solution:
Let the numbers are a and b HCF = 29 LCM = HCF (a × b) LCM = 29(15 × 13) Largest number = 15 × 29 = 435
6.
The HCF of two numbers , each having three digits, is 17 and their LCM is 714. The sum of the numbers will be ?
(A) 289
(B) 391
(C) 221
(D) 731
Solution:
HCF = 17 Let numbers are = 17x, 17y LCM = 17xy = 714 (given) xy = 42 Possible pairs are (1, 42), (2, 21), (3, 14), (6, 7) Possible numbers are (17, 714), (34, 357), (51, 238), (102, 119) but given that both numbers are of three digits ∴ numbers are = (102, 119) ∴ sum of numbers = 102 + 119 = 221
7.
What is the LCM of (8x3 + 80x2 + 200x) and (4x4 + 16x3 - 20x2)?
The largest 4-digit number exactly divisible by each of 12, 15, 18 and 27 is?
(A) 9690
(B) 9720
(C) 9930
(D) 9960
Solution:
LCM of (12, 15, 18, 27) ⇒ 4 × 3 × 5 × 3 × 3 = 540 largest 4-digits number = 9999 on dividing by 540 to number = $$\frac{{9999}}{{540}}$$ remainder = 279 Required number = 9999 - 279 = 9720
9.
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
(A) 1008
(B) 1015
(C) 1022
(D) 1032
Solution:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7 = 1008 + 7 = 1015
10.
The sum of two numbers is 528 and their HCF is 33. The number of pairs of numbers satisfying the above conditions is = ?
(A) 4
(B) 6
(C) 8
(D) 12
Solution:
Let the required numbers be 33a and 33b. Then 33a + 33b = 528 ⇒ a + b = 16 Now co - primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9) ∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9) The number of such pairs is 4.