RRB Mathematics - HCF and LCM
Exam Duration: 45 Mins Total Questions : 30
\(1095\over1168\) when expressed in simplest form is:
- (a)
\(13\over16\)
- (b)
\(15\over16\)
- (c)
\(17\over26\)
- (d)
\(25\over26\)
The H.C.F of \(2^{4}\times3^{2}\times5^{3}\times7\),\(2^{3}\times3^{3}\times5^{2}\times7^{2} \) and \(3\times5\times7\times11\) is :
- (a)
105
- (b)
1155
- (c)
2310
- (d)
27720
H.C.F = Product of lowest powers of common factors=\(3\times5\times7=105\)
Which of the following is a pair of co-primes?
- (a)
(16,62)
- (b)
(18,25)
- (c)
(21,35)
- (d)
(23,92)
H.C.F of 18 and 25 is 1.So, they are co-primes.
Find the lowest common multiple of 24,36 and 40.
- (a)
120
- (b)
240
- (c)
360
- (d)
480
The L.C.M of 22,54,108,135 and 198 is:
- (a)
330
- (b)
1980
- (c)
5940
- (d)
11880
The H.C.F of \({9\over10},{12\over25},{18\over35}\) and \(\frac { 21 }{ 40 } \) is
- (a)
\(3\over5\)
- (b)
\(252\over5\)
- (c)
\(3\over2800\)
- (d)
\(63\over700\)
Required L.C.M.=\(\frac { H.C.F.of9,12,18,21 }{ L.C.M.of10,25,35,40 } =\frac { 3 }{ 2800 } \)
The G.C.D of 1.08,0.36 and 0.9 is :
- (a)
0.03
- (b)
0.9
- (c)
0.18
- (d)
0.108
Given numbers are 1.08,0.36 and 0.90.H.C.F of 108,36 and 90 is 18
\(\therefore\) H.C.F of given numbers=0.18
The L.C.M of 3,2.7 and 0.09 is :
- (a)
2.7
- (b)
0.27
- (c)
0.027
- (d)
27
Given numbers are 3.00,2.70 and 0.09. L.C.M of 300,270 and 9 is 2700.
\(\therefore\) L.C.M of given numbers = 27.00=27
The H.C.F. of two numbers is 12 and their difference is 12.The numbers are :
- (a)
66,78
- (b)
70,82
- (c)
94,106
- (d)
84,96
Out of the given numbers,two with H.C.F. 12 and difference 12 are 84 and 96.
Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073.The sum of the three numbers is:
- (a)
75
- (b)
81
- (c)
85
- (d)
89
Since the numbers are co-prime,they contain only 1 as the common factor. Also,the given two products have the middle number im common.
So,middle number =H.C.F of 551 and 1073 =29
First number =\(({551\over29})=19\); Third number=\(({1073\over29})=37\)
Required sum =(19+29+37)=85
The H.C.F of two numbers is 11 and their L.C.M is 7700.If one of the numbers is 275,then the other is :
- (a)
279
- (b)
283
- (c)
308
- (d)
318
Other number =\(({11\times7700\over275})=308\)
The sum of two numbers is 2000 and their L.C.M. is 21879.The two numbers are:
- (a)
1993,7
- (b)
1991,9
- (c)
1989,11
- (d)
1987,13
Let the numbers be x and (2000-x).Then, their L.C.M. = x(2000-x).
So,x(2000-x)=21879 \(\Leftrightarrow \)x2-2000x+21879 = 0
\(\Leftrightarrow \) (x-1989) (x-11) = 0 \(\Leftrightarrow \)x = 1989 or x = 11
Hence, the numbers are 1989 and 11
The L.C.M. of two numbers is 495 and their H.C.F. is 5 .If the sum of the numbers is 10,then their difference is:
- (a)
10
- (b)
46
- (c)
70
- (d)
90
Let the numbers be x and (100-x).
Then, x(100-x)=\(5\times495\)\(\Leftrightarrow \) x2-100x + 2475 = 0
\(\Leftrightarrow \)(x-55) (x-45) = 0 \(\Leftrightarrow \)x = 55 or x = 45
Hence, the numbers are 45 and 55
Required difference = (55-45) = 10.
The HCF and LCM of two numbers are 50 and 250 respectively.If the first number is divided by 2,the quotient is 50 .The second number is :
- (a)
50
- (b)
100
- (c)
125
- (d)
250
First number =\((50\times2)=100\) Second number = \(({50\times250\over100})=125\)
The LCM of three different numbers is 120.Which of the following cannot be their HCF?
- (a)
8
- (b)
12
- (c)
24
- (d)
35
Since H.C.F. is always a factor of L.C.M. we cannot have three numbers with H.C.F. 35 and L.C.M. 120
The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14.The larger of the two numbers is:
- (a)
276
- (b)
299
- (c)
322
- (d)
345
Clearly, the numbers are \((23\times13)\) and \((23\times14)\)
\(\therefore\) Larger number =\((23\times14)\) =322
The HCF and LCM of two numbers are 11 and 385 respectively .If one number lies between 75 and 125,then that number is:
- (a)
77
- (b)
88
- (c)
99
- (d)
110
Product of numbers=\(11\times 385=4235\).
Let the numbers be 11a and 11b. Then. 11a \(\times \)11b = 4235 \(\Rightarrow \)ab=35.
Now, co-primes with product 35 are (1,35) and (5,7).
So, the numbers are \((11\times 1,\quad 11\times 35)\)and \((11\times 5,11\times 7).\)
Since one number lies between 75 and 125, the suitable pair is (55,77).
Hence, required numbers= 77
LCM of two prime numbers x and y (x>y) is 161.The value of 3y-x is :
- (a)
-2
- (b)
-1
- (c)
1
- (d)
2
H.C.F. of two prime numbers is 1. Product of numbers=(1 ×161)=161.
Let the numbers be a and b. Then, ab =161.
Now, co-primes with product 161 are (1,161) and (7,23).
Since x and y are prime numbers and x>y, we have x=23 and y=7.
\(\therefore \) 3y-x=(3 \(\times \) 7) - 23= -2.
Three different containers contain 496 litres,403 litres and 713 litres of mixtures of milk and water respectively.What biggest measure can measure all the different quantities exactly?
- (a)
1 litre
- (b)
7 litres
- (c)
31 litres
- (d)
41 litres
Required measurement = (H.C.F. of 496,403,713) litres = 31 litres.
The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pencil is:
- (a)
91
- (b)
910
- (c)
1001
- (d)
1911
Required number of students = H.C.F. of 1001 and 910 =91.
Let N be the greatest number that will divide 1305,4665 and 6905,leaving the same remainder in each case.Then sum of the digits in N is:
- (a)
4
- (b)
5
- (c)
6
- (d)
8
N= H.C.F. of (4665-1305), (6905-4665) and (6905-1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = (1+1+2+0)=4.
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is :
- (a)
123
- (b)
127
- (c)
235
- (d)
305
Required number = H.C.F. of (1657-6) and (2037-5)
= H.C.F. of 1651 and 2032 = 127.
The least number of five digits which is exactly divisible by 12,15 and 18 is :
- (a)
10010
- (b)
10015
- (c)
10020
- (d)
10080
Least number of 5 digits is 10000. L.C.M. of 12,15, and 18 is 180.
On dividing 10000 by 180, the remainder is 100.
\(\therefore \) Required number=10000+(180-100)=10080.
The greatest number of four digits which is divisible by 15,25,40 and 75 is
- (a)
9000
- (b)
9400
- (c)
9600
- (d)
9800
Greatest number of 4 digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600.
On diviging 9999 by 600, the remainder is 399.
\(\therefore \)Required number =(9999-399) = 9600.
The least number which should be added to 2497 so that the sum is exactly divisible by 5,6,4 and 3 is :
- (a)
3
- (b)
13
- (c)
23
- (d)
33
LCM of 5,6,4 and 3 =60.On dividing 2497 by 60 ,the remainder is 37.
\(\therefore \) Number to be added =(60-37)=23
The smallest number which when perfect square and is diminished by 12,16,18,21 and 45 is :
- (a)
1008
- (b)
1015
- (c)
1022
- (d)
1032
Required number =(L.C.M. of 12,16,18,21,28)+7 = 1008+7 = 1015
The least number increased by 5 is divisible each one of 24,32,36 and 54 is:
- (a)
427
- (b)
859
- (c)
869
- (d)
4320
Required number =(L.C.M. of 24,32,36,54)-5 = 864-5 = 859
Find the least multiple of 23,which divided by 18,21 and 24 leaves remainders 7,10 and 13 respectively.
- (a)
3002
- (b)
3013
- (c)
3024
- (d)
3036
Find the least number which when divided by 16,18,20 and 25 leaves 4 as remainder in each case,but when divided by 7 leaves no remainder
- (a)
17004
- (b)
18000
- (c)
18002
- (d)
18004
Four different electronic devices make a beep after every 30 minutes, 1 hour ,\(1{1\over2}\) hour and 1 hour 45 minutes respectively.All the devices beeped together at 12 noon.They will again beep together at:
- (a)
12 midnight
- (b)
3 a.m
- (c)
6 a.m
- (d)
9 a.m
Interval after which the devices will beep together = (L.C.M. of 30, 60,90,105) min. = 1260 min. = 21 hrs.