Hi,
The two numbers should be 12 and 30. 12=2x2x3 while 30=2x3x5.
Their HCF is 2x3=6 and their LCM is 2x3x2x5. Because of their HFC, we know that they are both multiple of 6. Also, the question says they both are GREATER than 6, so they can’t be 6 but are 6 times “something”. Thanks to the LCM, we know that “something” is equal to 2 for the first number and to 5 for the second one, the numbers hence being 12 and 30.
I hope this helps. If I was not clear enough or if you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.
I know the 1st and last one but i’m not sure on the rest sorry :(
Answer:
70
Step-by-step explanation:
- for the value of difference to be largest, the minuend should be maximum(most possibly) and the subtrahend should be minimum
[in A-B=X, A is minuend and B is subtrahend ]
- so, $a.b should be maximum. as there is a condition that 4 digits should be distinct, the product will be maximum if we choose 2 maximum valued numbers from the given numbers. so, one of them should be 9 and the other should be 8.
therefore, $a.b=9*8=72
- as mentioned above, c.d$ should be minimum. this will be possible only when we choose 2 minimum valued numbers from the given numbers. so, one of them should be 1 and the other should be 2.
therefore, c.d$ = 1*2 = 2
- hence, the difference = 72-2 = 70
- thus, the largest possible value of the difference $a.b - c.d$ = 70
No they are not.
48/2 = 24, 60/3=20.
24/=20. (24 does not equal 20)
Answer:
The price of uniform U= $145
price of each pair of cleats C= $16
Step-by-step explanation:
Let:
The Price of Each Uniform = U
The Price of Each Pair of Cleats = C
Rigo spent $451, before taxes, and purchased three uniforms and one pair of cleats.
→ Equation A
Ian spent $757, before taxes, and purchased five uniforms and two pair of cleats.
→ Equation B
Let's calculate → 2(Equation A) - (Equation B)
2(3U+C)-(5U+2C)= 2(451) -757
6U+2C-5U-2C= 145
U=$ 145
3U+C= 451
3(145)+C= 451
C= 451-435
C= $16
