When f(x) = x4 - 6x3 + 2x2 + 3x + 1 is divided by (x - 2), then remainder can be calculated by putting x - 2 = 0
i.e. x = 2
Now, put x = 2 in f(x)
f(2) = (2)4 - 6(2)3 + 2(2)2 + 3 x 2 + 1
= 16 - 48 + 8 + 6 + 1 = -17
? The remainder will be - 17.
Let the number of 25 paise coins and 50 paise cons be x and y,respectively.
According to the question,
x + y = 38
Also, 25x + 50y = 1350
On multiplying Eq.(i) by 25 and substracting from Eq.(ii), we get y =16
On substituting the value of y in Eq.(i),we get y=22
So,the number of 25 paise coins in 22.
The linear equation is y = 4x,
When, x = 1, y = 4
Since two digit number = 10p + q
According to question , ? q = 2p ? 1.. (i)
When digits are interchanged then new number = 10q + p
then original number ? [new number ? original number] = 20
? 10p + q ? [10q + p ? (10p + q)] = 20
? 10p + q ? 10q ? p + 10p + q = 20
19p ? 8q = 20
19p ? 8 (2p ? 1) = 20 (Using eq. (i))
19p ? 16p + 8 = 20
3p = 12 ? p = 4
From (i) q = 2 × 4 ? 1 ? q = 7
? original number = 10p + q = 10 × 4 + 7 = 47
Let the number be p and q.
According to question,
(p + q)2 = 4pq
? p2 + q2 + 2pq ? 4pq = 0
? (p ? q)2 = 0
? p = q
Hence , required ratio is 1 : 1 .
Since, p(x) = ax3 + 4x2 + 3x - 4 and q(x) = x3 -4x + a divided by x - 3, hence p(3) = q(3)
? a x 33 + 4 x 32 + 3 x 3 - 4 = 33 - 4 x 3 + a
? a = -1
Let the number be x. Then,
5x/2 - 2x/5 = 5208
? (25x - 4x)/10 = 5208
? 21x/10 = 5208
? 21x = 5208 x 10
? x = (5208 x 10)/21
? x = 2480
So, the original numbers was 2480.
Let the distance covered by walking be x km.
Speed while walking = 2 km/h
Time spent on walking = x/2 h
Now, distance covered by runnig = (18 - x) km
Speed while running = 6 km/h
Time spent on running = (18 - x)/6 h
?x/2 + (18 -x)/6 = 5
? (3x + 18x - x)/6 = 5
? 2x + 18 = 30
? 2x = 12
? x = 6
? The distance covered by running = 18 - 6 = 12 km
Let the number of oranges Ramesh has be x and the number of oranges Ganesh has be y.
According to the first condition,
y + 5 = 3(x - 5)
? 3x - y + 10 = 0 ...(i)
According to the second condition,
x + 5 = y - 5
? x - y + 10 = 0 ...(ii)
On subtracting Eq. (ii) from Eq. (i), we get x = 15
On substituting the value of x in Eq.(ii), we get y = 25
? The ratio of oranges to be distributed Ramesh and Ganesh = 15:25 = 3:5
Let the digit in the unit's place be y and digit in the ten's place be x.
? Number = 10x + y
Then, x = 2y + 2
? x - 2y - 2 = 0 ...(i)
Number obtained by reversing the digits = 10y + x
Then, 10y + x = 5 + 3(x + y)
? 2x - 7y + 5 = 0 ...(ii)
On multiplying Eq.(i) by 2 and then subtracting it from Eq. (ii), we get y = 3
On substituting the value of y in Eq.(i), we get x = 8
So, the two-digit number is 83.
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