This set of Aptitude Questions and Answers (MCQs) focuses on “HCF – Set 2”.
1. Find the greatest number which on dividing 1543 and 2125 leaves remainder 6 and 5 respectively.
a) 43
b) 53
c) 47
d) 61
View Answer
Explanation: The greatest number that will divide a, b and c leaving remainders x, y and z respectively = HCF (a-x, b-y and c-z).
Therefore, the greatest number which on dividing 1543 and 2125 leaves remainder 6 and 5 respectively = HCF (1543-6, 2125-5) = HCF (1537, 2120) = 53.
2. Find the largest number which divides 397, 612 and 1300 to leave same remainder in each case.
a) 43
b) 59
c) 37
d) 53
View Answer
Explanation: The largest number which divides a, b and c to leave same remainder in each case = HCF (a-b, b-c, c-a).
Therefore, the largest number which divides 399, 751 and 924 to leave same remainder in each case = HCF (612-397, 1300-612, 1300-397) = HCF (215, 688, 903) = 43.
3. A person has completely put each of three liquids, 558 liters of petrol, 682 liters of diesel and 589 liters of kerosene in bottles of equal size without mixing with any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required?
a) 56
b) 59
c) 63
d) 66
View Answer
Explanation: HCF of 558, 682 and 589 is 31.
Therefore, the least possible number of bottles required = \(\frac{558}{31}+\frac{682}{31}+\frac{589}{31}\)=59.
4. If N is the greatest number that divides 2238, 4474 and 7828, leaving same remainder in each case, then what is the product of the digits of N?
a) 8
b) 10
c) 12
d) 15
View Answer
Explanation: The largest number which divides a, b and c to leave same remainder in each case = HCF (a-b, b-c, c-a).
The greatest number that divides 2238, 4474 and 7828, leaving same remainder in each case = HCF (4474-2238, 7828-4474, 7828-2238) = HCF (2236, 3354, 5590) = 1118.
Therefore, product of the digits of 1118 = 1*1*1*8 = 8.
5. If r is the remainder when each of 1037, 1796 and 2808 is divided by the greatest number d, then find the value of (d-r).
a) 278
b) 253
c) 228
d) 258
View Answer
Explanation: The largest number which divides a, b and c to leave same remainder in each case = HCF (a-b, b-c, c-a).
HCF (1796-1037, 2808-1796, 2808-1037) = HCF (759, 1012, 1771) = 253 = d.
r = 253*4 – 1037 = 1012 – 1037 = 25.
Therefore, d-r = 253 – 25 = 228.
6. Find the number of pairs whose sum is 88 and HCF is 8.
a) 6
b) 10
c) 9
d) 5
View Answer
Explanation: The pairs whose sum is 88 and HCF is 8 are (8, 88), (16, 72), (24, 64), (32, 56) and (40, 48).
Therefore, the number of pairs are 5.
7. Find the greatest number that divides 2069, 3121, 4409 and 5711.
a) 181
b) 63
c) 16
d) 1
View Answer
Explanation: All the given 4 numbers are prime.
Hence, HCF = 1.
8. Find the HCF of (79-1) and (49-1).
a) 7
b) 6
c) 48
d) 24
View Answer
Explanation: We know that HCF (am-1) and (an-1) is (aHCF of m and n-1).
Therefore, the HCF of (79-1) and (49-1) = HCF of (79-1) and (72-1) = 7 – 1 = 6.
9. What is the greatest possible rate at which a man can walk 1650 mts, 1089 mts and 2541 mts in exact number of minutes?
a) 59 mts/min
b) 81 mts/min
c) 71 mts/min
d) 33 mts/min
View Answer
Explanation: HCF of 1650, 1089 and 2541 is 33.
Therefore, the greatest possible rate at which a man can walk is 33 mts/min.
10. A school has 504 girl students and 1176 boy students. The school is divided into strictly boys or strictly girls’ section. All sections in the school have same number of students. Given this information, what are the minimum number of sections in the school?
a) 15
b) 10
c) 12
d) 11
View Answer
Explanation: 504 = 23*32*7 and 1176 = 23*3*72.
HCF of 504 and 1176 is 23*3*7 = 168.
Therefore, the minimum number of sections in the school = \(\frac{504}{168}+\frac{1176}{168}\)=10.
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