Home / LCM HCF Quiz For Upcoming Exams Solution

LCM HCF Quiz For Upcoming Exams Solution

150+ RRB NTPC Previous Solved Papers PDF  Free Download Now
RRB NTPC Free Study Material PDF Free Download Now
RRB NTPC Free Mock Test Attempt It Now
Quant Booster eBook – A Complete Maths Shortcut eBook   GET Book Now
SSC CGL,CPO,GD 2018 Study Material & Book Free PDF Free Download Now

 

Print Friendly, PDF & Email

Answers :-

1. (c) LCM × HCF = 1st. Number × 2nd. number or
Product of numbers = HCF × LCM
→ LCM = 864
HCF = 144
One number x = 288
Thus Let other no. be y
Thus,, x y = LCM × HCF
→ 288 × y = 864 × 144
y =
= 432
Thus, Other no. will be = 432

2. (c) LCM = 225
HCF = 5
One number = 25
Thus, Let other number be y
Thus,. 25 × y = 225 × 5
y =
Thus, Another no. is 45

3. (c) LCM = 30
HCF = 5 (given)
One number = 10
Let another number = y
Thus, 10y = 30 × 5
y = 15

4. (b) HCF = 13
LCM = 455
Thus, Let number be 13x & 13y
Thus, LC M = 13 × y
Thus, LCM = HCF × Product of other factor
13 x y = 455
xy =
→ x y = 35
Possible co-prime Factors of x,y
→ (35, 1) (5, 7)
Thus, Numbers may be
→ 35 × 13, 1 × 13 = (455, 13)
or
→ 5 × 13, 7 × 13 = (65, 91)
→ But it is given that one number lies between
(75 & 125) so.
→ Numbers are (65, 91) and number between 75
& 125 is 91.
LCM of (65, 91) and number between 75 & 125
is 91.

5.
(c) LCM of (4, 6 , 8 , 12, 16)
→ 16 × 3 = 48
Thus, The number when divided by (4, 6 , 8 , 12,
16) leaves reminder 2 is = 48 + 2 = 50

6. (d) LCM of (12, 15, 20, 54)
→ 4 × 3 × 5 × 9 = 540
Thus, the required number is
540 + 4 = 544
→ Because when divided by LCM each is divided
completely, By adding 4 in LCM leaves remainder
4.

7. (a) 1001 pens, 910 pencils (given)
HCF of 1001, 910 is = 91
Thus, maximum no. of students are = 91

8. LCM of 4 , 6 , 8, 14 = 168 seconds
2 4 , 6, 8 , 14
2 2, 3, 4, 7
1, 3. 2 , 7
LCM = 3 × 2 × 7 × 2 × 2 = 168 seconds
= 2 minute 48 seconds
Thus, 1st they start ringing at 12.00 o’clock
→ Again they ring all together after 2 minutes 48
seconds at 12 hrs. 2 min. 48 seconds.

9. (a) LCM × HCF = 24
Thus, Product of numbers = 24
Let no. be = x, y
xy = 24
and x – y = 2 (given)
Factors of x y = 24 are (4, 6) (12, 2)
(8, 3) (24, 1)
→ Now difference between numbers be = (x- y)
= 2
So, Factor is (4, 6)

10.
(a) LCM = 495
HCF = 5 (given)
Thus, Let numbers are = 5x & 5y
Thus, LCM = 5 x y
5 x y = 495
x y = 99
thus, Possible co-prime factors are
1 , 99
9 , 11
Thus, Possible numbers are
5x , 5y = 45, 55
5, 495
Now, given that sum of numbers = 100
so, required numbers are = (45, 55)
Thus, Difference of numbers = 55 – 45 = 10

GovernmentAdda Recommends  Online Tyari Mock Test
RRB Group D Speed Test Attempt Now Attempt Now
Railway RRB Pshyco Complete Package  Get It Now
SSC CHSL, CGL Mock Test (1 free+30 Paid) Attempt Now

 

RRB JE Free Study Material PDF Books Free Download Now
FCI Free Study Material PDF Books Free Download Now
SSC JE Free Study Material PDF Books Free Download Now
Download Quant Power Bank (1000+ Pages) Free Download Now
Reasoning & GI Booster -Power Bank Book (1200+ Pages) Free Download Now

 

error: