It’s not as difficult as it sounds.
The long statement, short procedures. Here’s how you can find the numbers:
Let, the smaller number be X & larger number is Y.
4X=3Y-5
3Y-4X=5 .. equation 1
X+Y=6(Y-X)+6
5Y-7X=-6 .. equation 2
Solving equation 1 & 2
X=43
Y=59
You can if it’s right or not by replacing X and Y in any of equations 1 or 2, with the values you found out.
In this article, I’ll be talking about all the questions similar to these and would brainstorm. It would be a great journey.
Let’s get to it.
Other Ways To Solve This Question
Let S be the smaller number; and, L is the larger number.
Then, as per the question,
3S – 4L = 5 (Eq.1)
Also,
S + L = 6 * (L – S) + 6
Or, S + L = 6y – 6x + 6
Or, 6S – L -6 S – S = – 6
Or, 5L – 7S = -6 (Eq. 2)
By multiplying both sides of (Eq. 1) by 5 and (Eq. 2) by 3, we arrive at the following result:
15 L – 20 S = 25 (Eq 3)
15 L – 21 S = – 18 (Eq. 4)
Subtracting (Eq. 4) from (Eq. 3), we have :
S = 43
When we replace this value for S in (Eq. 1), we get:
3 L – (4 * 43) = 5
Or, 3L – 172 = 5
Or, 3L = 5 + 172 = 177
Or, L = 177 / 3 = 59.
So, 43 is the smaller number while 59 is the larger number.
As you can see, the answer is the same while the method is different.
How Can We Check If The Answer Is Right Or Wrong?
You know what, we can check the answer by using the same equations we used to find out the values. Isn’t it fantastic?
Have a look at this:
(3 * 59) – (4 * 43) = 177 – 172 = 5 (It’s Correct)
(Sum of Numbers)- 6 multiplied by (Diff of numbers)
= (43 + 59) – 6 * ( 59 – 43) = 102 – (6 * 16) = 102 – 96 = 6
Wow, this is correct too!
How Can We Solve The Query Given Earlier By Any Alternative Way?
Let, smaller number be U and bigger number be V
As per questions, 4V – 3U = 5………(1)
(U + V) – 6 (U – V) = 6
Or, – 5U + 7V = 6…..……..(2)
By solving 1 and 2, we get U = – 11 and V = -7.
Hence the numbers are -11 and -7
What Are The Two Numbers That Are So Close To One Another That 105 Is Equal To 6 Times The Smaller + 3 Times The Larger?
Here, we can solve it by designating the smaller consecutive number as S and the larger one as S+1.
According to the condition given we get:
6S+ 3(S + 1) = 105— Equation 1
9S = 102
3S = 34
S= 34/3
S + 1 = 37/3
As I’ve already stated we can check the result by substituting the values of S and S+1 in the equation given above. If the right-hand side becomes equal to the left-hand side, the answer we found, is correct.
| Arithmetic Sequence | Simultaneous Equations |
| 5, 8, 11, 14, … | 2x + 4y = 14 |
| 80, 75, 70, 65, 60, … | 4x − 4y = 4 |
| π/2, π, 3π/2, 2π, | 3h + 2i = 8 |
An Alternative Way
Let the numbers are C and C+1
Then 3C +6(C+1)=105—-Equation Number 1
3C+6C+6=105
9C=99
C=11 and C+1=12
Thus the two consecutive numbers would be 11 and 12 respectively.
Proof
3(11)+6(11+1)=? (It should be 105)
33+66+6=105
so it is proved that the value of C is right!
How To Solve The Question Framed Above?
X and (X + 1):
6 * X + 3 * (X + 1) = 105
6X + 3X + 3 = 105
9X = 102
X = 102/9
X= 34/3
The numbers are 34/3 and 37/3.
To Solve The Question Regarding Integers, What Other Methods Can We Use?
There isn’t another method specifically, unless you mean two odd consecutive integers. It would also work if the question was phrased as 6 times the larger + 3 times the smaller, but it’s obvious that the question is not phrased that way.
Because of this, the question cannot be answered as it is.
The two numbers should be Z and Z+1.
6Z+3(Z+1)=105
6Z+3Z+3=105
9Z+3 -3 =105 -3
9Z=102.
Since 102 is not divisible by 9, no answer answers the question.
The two would be Y and Y+2 if you had asked for two odd consecutive numbers instead. In that circumstance:
6Y+3(Y+2)=105
6Y+3Y+6=105
9Y+6 -6 =105 -6
9Y=99
Y=11, Y+2=13
Therefore, if the question asked for two consecutive odd numbers, 11, and 13, would meet that requirement. If the question had instead read “6 times the larger + 3 times the smaller,” then:
6(y+1)+3y=105
6y+6+3y=105
9y+6 -6 =105 -6
9y=99 , y=11, y+1=12
Although these are consecutive integers, the way the question is worded precludes any workable answers.
Two Numbers Added Together Equals 28. Two Times The Smaller Number Added To One Is Equal To Two More Than Three Times The Smaller Number Multiplied By The Larger Number; What Are The Numbers?
Possible answers are E. 13 and 15; F. 12 and 16; G. 10 and 18; and H. 11 and 17. The right response is H. (17,11) for the inputs x and y. The issue in the question stem is read wrongly, as should be noted.
Scenario:
x+y=28 and 2x+1=3y+2
Assumptions
Suppose x and y are two numbers.
Let y<x
Calculations
y+x=28 and y+1+2x=3y+2
An equation with two simultaneous unknowns
In terms oy y, x=28-y
In terms of x, y=28-x
(x+y=28)-(2x-3y=1)
(2y+2x)-(2y-3x)=56-1
2y-(-3y)=56-1
2y+3y=55\5 v=55
5y/5=55/5
y=11
and,
Since x=28-y,=28-11
x=17 thus,
x, y={17, 11}
Proof
The simultaneous equation returns if x, y=17, 11
x+y+2x-3y=28+1=28+1
17+11+2(17)-3(11)=28+1
3(17)-2(11)=28+1
51-22=28+1 and
29=29 establishes the roots (zeros) of the simultaneity of x, y=17, 11.
L.H.S= R.H.S
Two Integers Added Together Equal 24; Seven Times The Smaller Number Is Equivalent To Five Times The Larger Amount. Which Figure Is Smaller?
Let X be the lesser number.
Thus 24-X is a greater number.
Being that. 7X=5(24-X) =>7X=120–5
=>7X+5X=120
=>12X=120
=>X =120/12=10
Consequently, 10 is the smaller number.
The Smaller Of Two Integers That Is Twice As Large Is Three Times Smaller Than The Opposite. Four Times The Larger Plus Three Times The Smaller Add Out To 71. What Numbers Are There?
x>y
2x=3+5y
so 2x-5y=3
4x-10y=6
4x +3y=71
3y-(-10y)=3y +10y =13y
=71-6=65
y=5, x=14
The Difference Between Two Numbers Is 5.8 Less Than 3 Times The Smaller Number Is Equal To Twice The Larger Amount. What Numbers Are There?
As per the conditions:
2 * A= 3B – 8; B = 5 + y
Change B into the second:
2*(5+B) = 3B – 8
10 + 2B = 3B – 8
10 + 8 = 3B – 2B
18 = B
A = 5 + 18 = 23
A Number Less Than Five Times A Smaller Number Equals A Larger Number. These Two Numbers Add Up To A Total Of 17. Which Two Numbers Are They?
Let’s omit all the long, fatiguing procedures and try to solve them in our heads. It’ll just take a few minutes.
If we multiply by 1, we get five times the smaller amount, or
17 plus 1 equals 18 (lower number * 1 plus 5).
18/6 = 3
I spend more time writing it, rather than performing it.
Two Integers Added Together Equal 32. One Number Is 4 Less Than The Other, Which Is 5 Times The First. How Do You Discover The Two Figures?
It’s a typical algebraic problem.
You’re aware of two things regarding the two numbers X and Y.
You are aware that x = 5x-4 and x + y = 32. Therefore, change the first fact into the second one as follows: 5y-4 + y = 32.
You get 6y = 36 by adding four to each side and combining like phrases. To determine y = 6, divide both sides by 6.
If you replace either of the previous facts with this new fact, you will either get x + 6 = 32 or x = 5*6–4.
If you choose the first option, you can take 6 off of either side of the equals sign to get x = 26, and if you choose the second, you can perform the math yourself to get x = 26.
Two Numbers Add Up To 56. One Amount Is Three Times Bigger Than The Other. What Figures Are There?
This is a straightforward arithmetic problem. Calculating the answer doesn’t require any specialized knowledge.
Although there are many aspects of this question that I appreciate, I find the constant barrage of questions to be confusing and make me wonder if any of them are indeed asking for anything.
Many could do this in their heads because it is so simple.
Since A + B = 56 and 3A = B,
4A = 56, 56/4 = 14,
14 * 3 = 42, and
A = 14, B = 42.
Conclusion
To conclude, I’d like to say:
- Many scenario-based questions lead to several algebraic calculations.
- I’ve already discussed some of the solutions to the questions, including integers, consecutive numbers, etc.
- All we need to do is to imagine the numbers as X, Y, or any other alphabet.
- Then we make equations according to the conditions we’re provided with.
- After that, we solve the equations simultaneously and discover our values for the supposed answers.
- Lastly, we put the numbers into the equations we’ve figured out by the conditions provided in the questions.
- These are some of the basic steps needed to solve any scenario-based queries, with a slight difference for integers and consecutive numbers.
You can pen down these solutions and memorize them so that you don’t have to start from scratch if you encounter any such mathematical questions.
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