Previous Year LCM and HCF Questions for SSC CGL Exams-2017

Dear Students, SSC has released the notification for CGL 2017. This time again the competition is going to be very stiff. Questions from Quant are asked in Tier-1 and Tier-II as well. Hence, you need to focus on this subject more. The only trick to master Quant is “practice‘. So, Practice daily. We are providing topic-wise quant quizzes, solve, learn, succeed.

Q1. Two jars of capacity 50 l and 80 l are filled with oil. What must be the capacity of a mug that can completely measure the oil of the two jars?
(a) 10 l
(b) 50 l
(c) 80 l
(d) 40 l

S1. Ans.(a)
Sol. Factors of 50 = 52 × 2
Factors of 80 = 51 × 24
H.C.F. of 50 & 80 = 51×2=10 l
The capacity of the mug must be 10 l

Q2. Find the largest number which divides 34, 90, 104 leaving the same remainder in each case?
(a) 14
(b) 16
(c) 28
(d) 42

S2. Ans.(a)
Sol. Difference between number = 90 – 34 = 56 and 104 – 90 = 14
HCF of 56 and 14 = 14
So, 14 is the largest number which divides the above-given numbers leaving the same remainder in each case.

Q3. The smallest multiple of 7 which when divided by 6, 9, 15 and 18 respectively, leaves 4 as the remainder in each case?
(a) 364
(b) 360
(c) 356
(d) none of these

Q4. The length of a circular path is 20 kms. Three runners start running from a point in the same direction with speed of 4 km/h, 5 km/h and 8 km/h respectively. After how many hours will they be together at the starting point again?
(a) 20
(b) 60
(c) 15
(d) None of these

Q5. Two numbers, both greater than 29, have HCF 29 and LCM 4147. Find the sum of the numbers?
(a) 696
(b) 144
(c) 169
(d) 225

S5. Ans.(a)
Sol. H.C.F. = 29
L.C.M. = 4147
We have,
L.C.M. × H.C.F. = 1st no. × 2nd no.
4147 × 29 = 29x × 29y
29xy = 4147
xy = 143
(x, y) = (11 × 13), (1, 143).
Given,
The numbers are greater than 29.
So, the possible pair must be  (x, y) = (11, 13)
Thus, numbers are = 29 × 11 and 29 × 13
Sum = 696

Q6. The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then find the difference between both the numbers.
(a) 5
(b) 10
(c) 15
(d) 20

S6. Ans.(b)
Sol. L.C.M = 495
H.C.F = 5
Let the numbers be 5x and 5y
Now,
495 × 5 = 5x × 5y
5xy = 495
xy = 99
(x, y) = (9, 11), (1, 99)
So, the numbers are  (45, 55) and (5, 495)
But the sum = 100
So, possible pair must be = 45, 55
Difference = 10

Q7. The LCM of two numbers is 45 times their HCF. If one of the numbers is 125 and sum of HCF and LCM is 1150. The other number is–
(a) 225
(b) 235
(c) 240
(d) 250

Q8. Find the least number which when divided by 6, 7, 8, 9 and 12 leaves the remainder 1 in each case.
(a) 404
(b) 303
(c) 505
(d) 202

S8. Ans.(c)
Sol. L.C.M. of 6, 7, 8, 9, 12
LCM = 504
Required number (LCM) + 1 = 504 + 1
= 505

Q9. Find the least number which when divided by 5, 6, 7, 8 leaves remainder 3 but when divided by 9, leaves no remainder.
(a) 843
(b) 2523
(c) 1683
(d) 3363

S9. Ans.(c)
Sol. LCM of 5, 6, 7, 8 = 840
According to the question,
Given number will be of the form 840 K + 3
Since the number is completely divisible by 9, Using hit & trial
Put K = 2.
840 × 2 + 3 = 1683, which is divisible by 9.
Required number = 1683

Q10. Find the least number which when divided by 20, 25, 35, 40 leaves remainder 14, 19, 29, 34 respectively.
(a) 1220
(b) 1394
(c) 1365
(d) 1470

S10. Ans.(b)
Sol. 20 – 14 = 25 – 19 = 35 – 29 = 40 – 34 = 6
Required number (LCM of 20, 25, 35, 40) – 6
= 1400 – 6 = 1394

Q11. The LCM of two prime number x and y is 161 (where x > y). Find out the value of 3y – x.
(a) –2
(b) 2
(c) –4
(d) 4

Q12. Find the largest 5 digits number which is exactly divisible by 12, 15, 18, 27.
(a) 90000
(b) 99999
(c) 99010
(d) 99900


Q13. 3 different containers contain 496 L, 403 L and 713 L mixture of milk and water. What biggest measure of a container can measure all the 3 quantities exactly?
(a) 31
(b) 41
(c) 51
(d) 52

S13. Ans.(a)
Sol. To find the biggest measure, we have to find the HCF of 496, 403 and 713.
HCF of 496, 403 and 713 = 31

Q14. Three numbers are in ratio 2: 3: 4 and their HCF is 12. The LCM of the number is
(a) 144
(b) 192
(c) 96
(d) 72

S14. Ans.(a)
Sol. Let the numbers be 2x, 3x and 4x respectively.
∴ HCF = x = 12
∴ Numbers are
2x = 2 × 12 = 24, 3x = 3 × 12 = 36 and 4x = 4 × 12 = 48
LCM of 24, 36, 48 = 144

Q15. If the students of a class can be grouped exactly into 6 or 8 or 10, then the minimum number of students in the class must be
(a) 60
(b) 120
(c) 180
(d) 240

S15. Ans.(b)
Sol. The required number of students will be the LCM of 6, 8 and 10, i.e. 120.

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