Problems on Numbers

1. 24 is divided into two parts such that 7 times the first part added to 5 times the second part gives 146. The first part is 
a. 11
b. 13
c. 16
d. 17
Correct Option: B
Explanation:
Let the first and second parts be a and 24  a, then
7a+5(24−a)=146
⇒7a+120−5a=146
⇒2a=26 or a = 13

2. The product of two numbers is 120.  The sum of their squares is 289.  The sum of the two numbers is :
a. 20
b. 23
c. 169
d. None
Correct Option: B
Explanation:
Let the number be x and y . Then
(x+y)2=(x2+y2)+2xy=289+2x120
=289+240=529⇒x+y=529−−−√=23

3. The sum of squares of two numbers is 68 and the square of their difference is 36.  The product of the two numbers is 
a. 16
b. 32
c. 58
d. 104
Correct Option:a
Explanation:
Let the numbers be x and y. Then
x2+y2=68&(x−y)2=36
But (x−y)2=36⇒x2+y2−2xy=36
⇒68−2xy=36⇒2xy=32
⇒xy=16

4. The sum of seven numbers is 235.  The average of the first three is 23 and that of the last three is 42.  The fourth number is 
a. 40
b. 126
c. 69
d. 195
Correct Option: A
Explanation:
(23×3+a+42×3)=235⇒a=40

5. Two numbers are such that the ratio between them is 3:5 but if each is increased by 10, the ratio between them becomes 5 : 7, the numbers are
a. 3, 5
b. 7, 9
c. 13, 22
d. 15, 25
Correct Option: D
Explanation:
Let the numbers be 3a and 5a
Then 3a+105a+10=57
⇒7(3a+10)=5(5a+10)⇒a=5
The numbers are 15 & 25

LCM or Least common factor and Highest common factor (HCF)or Greatest common divisor (GCD)

LCM is defined as the least number which is divisible by all the given divisors.  Take 4,6 as two divisors which divide 12, 24, 36... perfectly with no remainder.  So 12, 24, 36 are called common multiples of 4 and 6.  In other words, 4 and 6 are factors of all these number.  Of all these common multiples, 12 is the least number.  So we can say 12 is Least common multiple of all the given numbers or LCM of 4, 6.

Finding LCM: 

There are two ways to find LCM.  First one is division method, second one is Factorization method.  

1. Division Method: LCM of 15, 18, 27

In division method we have to continue the division until the numbers in the last row become co - primes with each other.  So LCM = 3 x 3 x 5 x 2 x 3 =270

2. Factorization Method: 

Here we can write all the given numbers in their prime factorization format.

15 = 3 x 5

18 = 2×32

27 = 33

Now take all primes number the given numbers and write their maximum powers. So LCM of 15, 18, 27 = 2×33×5= 270

Formula 1: If r is the remainder in each case when N is divided by x, y, z then the general format of the number is N= K x [LCM (x, y, z)] + r here K is a natural number

Example: A teacher when distributed certain number of chocolates to 4 children, 5 children, 7 children, left with 1 chocolate.  Find the least number of chocolates the teacher brought to the class

Ans:  N = K (LCM (4, 5, 7) + 1 = 140K + 1.  Where K = natural number.  When we substitute K = 1, we get the least number satisfies the condition. So minimum chocolates = 141

 

Formula 2: If x1,y1,z1 are the remainders when N is divided by x, y, z and x−x1=y−y1=z−z1=a then the general format of the number is given by N= K x [LCM (x, y, z)] - a

Example: When certain number of marbles are divided into groups of 4, one marble remained.  When the same number of marbles are divided into groups of 7 and 12 then 4, 9 marbles remained. If the total marbles are less than 10,000 then find the maximum possible number of marbles.

Ans:  In this case the difference between the remainders and divisors is constant.  i.e., 3. so  N = K (LCM (4, 7, 12) -  3 = 84K - 3.  Where K = natural number.  

But we know that 84K - 3 < 10,000 ⇒ 84 x 119  - 3 < 10,000 ⇒ 9996 - 3 = 9993