Method 1 to solve the given question.
Lets assume the unit's digit is y and the ten's digit is x. then, the number is 10 x + y.
According to question, after interchanging the digits, the new number is 10y + x.
Then,
New Number - 18 = Original number
10y + x = 10x + y + 18
? 9y - 9x = 18
? y - x = 2 .....................(1)
Again according to question,
Sum of digits of Original Number = 8
x + y = 8 ..............................(2)
Add the equation (1) and (2) , we will get
y - x + x + y = 2 + 8
? 2y = 10
? y = 5
Put the value of Y in equation (2) , we will get
x + 5 = 8
? x = 8 - 5 = 3
Then , the original number = 10x + y
Put the value of x and y and get the original number.
The original number = 10x + y = 10 x 3 + 5 = 30 + 5 = 35
Method 2 to solve the given question.
Given that sum of digits of original number is 8.
Let us assume that unit digit is x , then ten digit will be 8 - x. (since sum of digits is 8.)
Now Original number = 10( 8 - x) + x
After interchanging the digits , the number = 10x + (8 - x) = 9x + 8
According to question.
New number = original number + 18
? 9x + 8 = 10(8 - x) + x + 18
? 9x + 8 = 80 - 10x + x + 18
? 9x + 8 = 98 - 9x
? 9x + 9x = 98 - 8
? 18x = 90
? x = 90/18
? x = 5
The original number = 10(8 - x) + x = 10(8 - 5) + 5 = 10(3) + 5 = 30 + 5 = 35
According to the question,
Borrowed sum = [121/{1/(1 + 10/100)}] + 121/(1 + 10/100)2
= 121/(11/10) + 121/{(11/10) x (11/10)]
=110 + 100
= ? 210
Clearly, 1 is the highest common factor of 23, 32, and 15.
6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
? 36+12X2 = 1 day
60 children can do the work in 1 day
Now, 5 men = 10 children
? 10 children can do the work in 6 days.
Here, CP = ? 800 , r = 10 % and R = 12.5 %
? Marked price (MP) = CP x (100 + R)/(100 - r)
= 800 x (100 + 12.5)/(100 - 10)
= (800 x 112.5)/90 = ? 1000
Sum of the square of first n natural numbers = n(n+1)(2n+1) / 6
Given n = 35
? Required sum = ( 35 x 36 x 71 ) / 6
= 14910
Then, Sachin's age = (x - 7) years.
∴ | x - 7 | = | 7 |
x | 9 |
⟹ 9x - 63 = 7x
⟹ 2x = 63
⟹ x = 31.5
Hence, Sachin's age =(x - 7) = 24.5 years.
A : B = 200 : 160
A : C = 200 : 144
? B/C = (B/A) x (A/C) = (160/200) x (200/144)
= 160/144 = 10/9 = 200/180
Hence, B can give C 20 points
Let us assume the present age of father = F year and Son's present age = S years
According to question, 5 years ago,
Father's age = F - 5 and Son's age = S - 5.
According to the question,
The age of the father 5 years ago was 5 times the age of his son.
F - 5 = 5(S - 5)
F - 5 = 5S - 25.....................(1)
At present the father's age is 3 times that of his son.
F = 3S.................................(2)
Put the value of F from equation (2) in equation (1), we will get
? 3S - 5 = 5S - 25
? 25 - 5 = 5S - 3S
? 20 = 2S
? 10 = S
? S = 10.
Put the value of S in Equation (2). we will get,
F = 3S = 3 x 10 = 30
So the present Age of Father = 30.
Average speed = 2t1 x t2/( t1 + t2)
= (2x64x80)/(64+80)km
= (2 x 64 x 80 ) / 144 km/hr.
=71.11km/hr.
Let Sum = P, Then SI=P
As Amount A = 2 × P
where , P = Principal
Rate R = (100 × SI)/(P × T)
= (100 × P)/(P × 8) %
= 12.5 %
where , SI= Simple Interest
T= Time
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