Solve: 4y = 20
Solve: 3x = 2x+ 18
Solve: x = 4/5(x+10)
The sum of three consecutive multiples of 7 is 357. Find the smallest multiple.
Given,
Sum of multiples of 7 = 357
Let the numbers be x, x + 7 , x + 14
ATP,
x + x + 7 + x + 14 = 357
3x + 21 = 357
3x = 357  21
3x = 336
x = 336/3
x = 112
Therefore,
the numbers are 112, 119, 136
In an equation the values of the expressions on the LHS and RHS are _______.
The sum of two digit number and the number formed by interchanging its digit is 110. If ten is subtracted from the first number, the new number is 4 more than 5 times of the sum of the digits in the first number. Find the first number.
Let the unit digit be y & tens digit be x.
Original number = (10x+y)
After interchanging the digits
New number = (10y+x)
(10x+y) + (10y+x) = 110
11x +11y = 110
11(x+y)= 110
x+y = 110/11
x+y= 10 (1)
x= 10y (2)
(10x+y)  10 = 4+ 5(x+y)
(10x+y)  10 = 4+ 5(10)
(10x+y) = 4+ 50+10
(10x+y) = 64
10(10y) +y = 64
10010y +y= 64
100 9y = 64
9y = 64100
9y = 36
y= 36/9= 4
y= 4
putting the value of y in eqn 2
x= 10y
x= 104
x= 6
Hence , the first number is 6 & second number is 4.
Original Number is 10x+y = 10* 6+4= 60+4= 64
Solve: x − 2 = 7
Solve: 5t – 3 = 3t – 5
5t3=3t55t3t=5+32t=2 which gives t=1
What will be the solution of these equations ax+by = ab, bxay = a+b
ax+by=ab
ax=abby
x=(abby)/a
bxay=a+b
substituting
b((abby)/a)ay=a+b
(abb²b²ya²y)/a=a+b
abb²(b²+a²)y=a²+ab
(b²+a²)y=a²+abab+b²
y=(a²+b²)/(a²+b²)
y= 1
substituting value of y
x=(abb(1))/a
x=a/a
x=1
The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.
Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solve: 7x = 21
Solve: 5x+ 9 = 5 + 3x
Solve: y + 3 = 10
y + 3 = 10
y = 10  3
y = 7
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Let the number be x.
So, the other number is (x + 15).
(x + 15) + x = 95
2x + 15 = 95
2x = 95  15
2x = 80
x = 40
So, the numbers are 40 and 55.
The value of the expression on one side of the equality sign is _____ to the value of the expression on the other side.
This says that any given equation says that it’s LHS is equal to RHS.
Solve: 5a = 30
Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.
8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish the work in 14 days. find the time taking by 1 man alone and 1 boy alone to finish the same work.
Let the time taken by man be x days
time taken by boys be y days
work done by 1 man in one day=1/x
work done by 1 boy in one day=1/y
8/x+12/y = 1/10  eq 1
6/x+ 8/y = 1/14  eq2
Let u = 1/x & v = 1/y
8u+ 12y = 1/10  eq3
6u+8v = 1/14  eq4
Multiplying eq3 by 2 & eq4 by 3
16u+ 24y = 2/10  eq5
18u+24v = 3/14  eq6
subtract eq 5 & 6
2u = 3/14+2/10
2u = (3×10 +2×14)/140
2u = (30+28)/140
2u = 2/140
u = 1/140
put this value in eq 5
16u+ 24y = 2/10
16×1/140 +24y = 1/5
4/35+24y = 1/5
24y = 1/54/35
24y = (1×74)/35
24y = (74)/35
24y = 3/35
y= 3/35×24
y= 1/280
u=1/x
1/140 = 1/x
x = 140
v = 1/y
1/280 = 1/y
y = 280
A man complete the work in 140 days
A boy can complete the work in 280 days
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