Rama

(a) Given expression is 4a.
The possible factors of 4 are 1, 2, and 4.
The possible factors of a is a.

All possible factors of 4a are 1, a, 2, 2a, 4, 4a.

(b) Given expression is 10ab.
The possible factors of 10 are 1, 2, 5, and 10.
The possible factors of ab is a, b, and ab.

All possible factors of 10ab are 1, 2, 5, and 10, a, b, ab, 2a, 2b, 2ab, 5a, 5b, 5ab, 10a, 10b, 10a.

(c) Given expression is 12mn².
The possible factors of 12 are 1, 2, 3, 4, 6, and 12.
The possible factors of mn² are m, n, n², mn², mn.

All possible factors of 12mn² are 1, 2, 3, 4, 6, 12, m, n, n², mn², mn, 2m, 2n, 2n², 2mn, 2mn², 3m, 3n, 3n², 3mn, 3mn², 4m, 4n, 4n², 4mn, 4mn², 6m, 6n, 6n², 6mn, 6mn², 12m, 12n, 12n², 12mn and 12mn²

(d) Given expression is 24m²n.
The possible factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
The possible factors of m²n are m, m², n, mn, m²n.

All possible factors of 24m²n are 1, 2, 3, 4, 6, 8, 12, 24, m, m², n, mn, m²n, 2m, 2m², 2n, 2mn, 2m²n, 3m, 3m², 3n, 3mn, 3m²n, 4m, 4m², 4n, 4mn, 4m²n, 6m, 6m², 6n, 6mn, 6m²n, 8m, 8m², 8n, 8mn, 8m²n, 12m, 12m², 12n, 12mn, 12m²n, 24m, 24m², 24n, 24mn, 24m²n.

(e) Given expression is 5ab²c².
The possible factors of 5 are 1, and 5.
The possible factors of ab²c² are a, b, b², c, c², ab, ab², bc, bc², ac, ac², b²c².

All possible factors of 5ab²c² are 1, 5, a, b, b², c, c², ab, ab², bc, bc², ac, ac², b²c², 5a, 5b, 5b², 5c, 5c², 5ab, 5ab², 5bc, 5bc², 5ac, 5ac², 5b²c².

(a) Given that 4a² and 10ab
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4 and 10 is 2.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4a² and 10ab is 2a.

The final answer is 2a.

(b) Given that 21m²n and 49mn²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 21 and 49 is 7.
HCF of literal coefficients:
The lowest power of m is 1.
The lowest power of n is 1.
Therefore, the HCF of literal coefficients is a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 21m²n and 49mn² is 7mn.

The final answer is 7mn.

(c) Given that a³b² and – 5b²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1 and -5 is 1.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is b².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of a³b² and – 5b² is b².

The final answer is b².

(d) Given that 4a³, 6b² and 8c
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4, 6, and 8 is 2.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is 0.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4a³, 6b² and 8c is 2.

The final answer is 2.

(e) Given that 2a²b³, 10a³b² and 14ab
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 2, 10, and 14 is 2.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
Therefore, the HCF of literal coefficients is ab.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 2a²b³, 10a³b² and 14ab is 2ab.

The final answer is 2ab.

(f) Given that 5m³, -15m^2, and 45m
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 5, -15, and 45 is 5.
HCF of literal coefficients:
The lowest power of m is 1.
Therefore, the HCF of literal coefficients is m.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 5m³, -15m^2, and 45m is 5m.

The final answer is 5m.

(a) Given that 3a², 15ab
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3 and 15 is 3.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3a² and 15ab is 3a.

The final answer is 3a.

(b) Given that 33a²b, – 77ab²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 33 and -77 is 11.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
Therefore, the HCF of literal coefficients is ab.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 33a²b, and – 77ab² is 11ab.

The final answer is 11ab.

(c) Given that 9a²b²c, 54abc²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 9 and 54 is 9.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
The lowest power of c is 1.
Therefore, the HCF of literal coefficients is abc.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 9a²b²c and 54abc² is 9abc.

The final answer is 9abc.

(d) Given that a³b², – 5b²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1 and -5 is 1.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is b².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of a³b² and – 5b² is b².

The final answer is b².

(e) Given that 4a³, 10b³, and 6c³
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4, 10, and 6 is 2.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is 1.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4a³, 10b³, and 6c³ is 2.

The final answer is 2.

(f) Given that 2abc³, 3a²b²c, 5abc
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 2, 3, and 5 is 1.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
The lowest power of c is 1.
Therefore, the HCF of literal coefficients is abc.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 2abc³, 3a²b²c, and 5abc is abc.

The final answer is abc.

(a) Given that – 2a² – 8a³ – 20a⁴
Firstly, find the HCF of given terms.
HCF of their numerical coefficients -2, -8, and -20 is -2.
HCF of literal coefficients:
The lowest power of a is 2.
Therefore, the HCF of literal coefficients is a².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of – 2a² – 8a³ – 20a⁴ is -2a².

(b) Given that 3mn² + 9m²n² – 6m²n
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, 9, and -6 is 3.
HCF of literal coefficients:
The lowest power of m is 1.
The lowest power of n is 1.
Therefore, the HCF of literal coefficients is mn.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 3mn² + 9m²n² – 6m²n is 3mn.

(c) Given that 5a² + 25a + 50
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 5, 25, and 50 is 5.
HCF of literal coefficients:
The lowest power of a is 0.
Therefore, the HCF of literal coefficients is 1.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 5a² + 25a + 50 is 5.

(d) Given that a³b² – 8b²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1, and -8 is 1.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is b².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of a³b² – 8b² is b².

(a) The given expression is 3a⁴ – 9a⁵ – 15a³
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, -9, and -15 is 3.
HCF of literal coefficients:
The lowest power of a is 3.
Therefore, the HCF of literal coefficients is a³.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3a⁴ – 9a⁵ – 15a³ is 3a³.
Multiply and divide each term of the given expression 3a⁴ – 9a⁵ – 15a³ with 3a³.
3a³ (3a⁴/3a³ – 9a⁵/3a³ – 15a³/3a³) = 3a³ (a – 3a² – 5)

The final answer is 3a³ (a – 3a² – 5).

(b) The given expression is – 10mn³ + 30nm³ – 50m²n³
Firstly, find the HCF of given terms.
HCF of their numerical coefficients -10, 30, and -50 is -10.
HCF of literal coefficients:
The lowest power of m is 1.
The lowest power of n is 1.
Therefore, the HCF of literal coefficients is mn.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of – 10mn³ + 30nm³ – 50m²n³ is -10mn.
Multiply and divide each term of the given expression – 10mn³ + 30nm³ – 50m²n³ with -10mn.
-10mn (– 10mn³/-10mn+ 30nm³/-10mn – 50m²n³/-10mn) = -10mn (n² – 3m² + 5mn²)

The final answer is -10mn (n² – 3m² + 5mn²).

(c) The given expression is 15x² – 20y² + 25z²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 15, -20, and 25 is 5.
HCF of literal coefficients:
The lowest power of x is 0.
The lowest power of y is 0.
The lowest power of z is 0.
Therefore, the HCF of literal coefficients is 1.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 15x² – 20y² + 25z² is 5.
Multiply and divide each term of the given expression 15x² – 20y² + 25z² with 5.
5 (15x²/5 – 20y²/5 + 25z²/5) = 5 (3x² – 4y² + 5z²)

The final answer is 5 (3x² – 4y² + 5z²).

(d) The given expression is a³bc + 5ab³+ 9a³b
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 1, 5, and 9 is 1.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is ab.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of a³bc + 5ab³+ 9a³b is ab.
Multiply and divide each term of the given expression a³bc + 5ab³+ 9a³b with ab.
ab (a³bc/ab + 5ab³/ab + 9a³b/ab) = ab (a²c + 5b² + 9a²)

The final answer is ab (a²c + 5b² + 9a²).

(e) The given expression is 3a² – 9a²b – 27a³c
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, -9, and -27 is 3.
HCF of literal coefficients:
The lowest power of a is 2.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is a².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3a² – 9a²b – 27a³c is 3a².
Multiply and divide each term of the given expression 3a² – 9a²b – 27a³c with 3a².
3a² (3a²/3a² – 9a²b/3a² – 27a³c/3a²) = 3a² (1 – 3b – 9ac)

The final answer is 3a² (1 – 3b – 9ac).

(e) The given expression is 3a² – 9a²b – 27a³c
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3, -9, and -27 is 3.
HCF of literal coefficients:
The lowest power of a is 2.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is a².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3a² – 9a²b – 27a³c is 3a².
Multiply and divide each term of the given expression 3a² – 9a²b – 27a³c with 3a².
3a² (3a²/3a² – 9a²b/3a² – 27a³c/3a²) = 3a² (1 – 3b – 9ac)

The final answer is 3a² (1 – 3b – 9ac).

(f) The given expression is 7a³ + 7ab² + 7ac²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 7, 7, and 7 is 7.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
The lowest power of c is 0.
Therefore, the HCF of literal coefficients is 7a.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 7a³ + 7ab² + 7ac² is 7a.
Multiply and divide each term of the given expression 7a³ + 7ab² + 7ac² with 7a.
7a (7a³/7a + 7ab²/7a + 7ac²/7a) = 7a (a² + b² + c²)

The final answer is 7a (a² + b² + c²).

Given that M = 1/2 × a × b
To find b, rewrite the given equation by changing the subject to b.
M = 1/2 × a × b
Move 1/2 × a to the left side and divide it with M
2M/a = b
b = 2M/a.
Substitute M = 65 and a = 10 in the above equation.
b = 2(65)/10 = 130/10
b = 13

The final answer is b = 13.

Given that L = 1/5 n²r
To find r, rewrite the given equation by changing the subject to r.
L = 1/5 n²r
Move 1/5 to the left side and multiply it with L.
5L = n²r
Move n² to the left side and divide it with 5L
(5L)/n² = r
r = (5L)/n²
Substitute L = 30 and n = 5 in the above equation.
r = 5 (30)/(5)²
r = 30/5 = 6
r = 6

The final answer is r = 6.

Given that A = 7b²
To find b, rewrite the given equation by changing the subject to b.
A = 7b²
Move 7 to the left side and divide it with A
A/7 = b²
Apply square root on both sides.
√(A/7) = √(b²)
√(A/7) = b
b = √(A/7)
Substitute A = 700 in the above equation.
b = √(700/7)
b = √100
b = 10

The final answer is b = 10.

Given that R = 2 (m + n)
To find n, rewrite the given equation by changing the subject to n.
R = 2 (m + n)
Move 2 to the left side and divide it with R
R/2 = m + n
Move m to the left side and subtract it with R/2
R/2 – m = n
n = R/2 – m
Substitute m = 15 and R = 100 in the above equation.
n = R/2 – m
n = (100/2) – 15
n = 50 – 15 = 35
n = 35

The final answer is n = 35

Given that C = (S × 100)/(100 × g)
To find S, rewrite the given equation by changing the subject to S.
C = (S × 100)/(100 × g)
Move (100 × g) to the left side and multiply it with C
C(100 × g) = S × 100
Move 100 to the left side and divide it with 100
C(100 × g)/100 = S
Cg = S
S = Cg
Substitute C= 140 and G = 5 in the above equation.
S = Cg
S = 140(5)
S = 700

The final answer is S = 700

Given that P = Q + R
To find R, rewrite the given equation by changing the subject to R.
P = Q + R
Move Q to the left side and subtract it from the P.
P – Q = R
R = P – Q
Substitute P = 5700 and Q = 2640 in the above equation.
R = P – Q
R = 5700 – 2640
R = 3060

The final answer is R = 3060

Given that F = 5/2 (T – 32)
To find F, rewrite the given equation by changing the subject to F.
F = 5/2 (T – 32)
Substitute T = 64 in the above equation.
F = 5/2 (T – 32)
F = 5/2 (64 – 32)
F = 5/2 (32)
F = 5 × 16 = 80
F = 80

The final answer is F = 80.

Given that Q = a × b × c
To find a, rewrite the given equation by changing the subject to a.
Q = a × b × c
Move bc to the left side and divide it from Q
Q/bc = a
a = Q/bc
Substitute Q = 4000, b = 10, c = 10 in the above equation.
a = 4000/(10 × 10)
a = 4000/100 = 40

The final answer is a = 40

Given that G = 2k
To find k, rewrite the given equation by changing the subject to k.
G = 2k
Move 2 to the left side and divide it from G
G/2 = k
k = G/2
Substitute G = 70 in the above equation.
k = 70/2
k = 35

The final answer is k = 35.

Given that n + o + p = 190
To find o, rewrite the given equation by changing the subject to o.
n + o + p = 190
Move n + p to the right side and subtract it from 190.
o = 190 – (n + p)
Substitute n = 60 and p = 55 in the above equation.
0 = 190 – (60 + 55)
o = 190 – 115
o = 75

The final answer is o = 75

Given that a = x + yz
To find z, rewrite the given equation by changing the subject to z.
a = x + yz
Move x to the left side and subtract it from a
a – x = yz
Move y to the left side and divide it from a – x
(a – x)/y = z
z = (a – x)/y
Substitute a = 54, x = 12 and y = 3 in the above equation.
z = (54 – 12)/3
z = 42/3
z = 14

The final answer is z = 14.

Given that rs²/4π² = m
To find s, rewrite the given equation by changing the subject to s.
rs²/4π² = m
Move 4π² to the right side and multiply it with m
rs² = m(4π²)
Move r to the right side and divide it from m(4π²)
s² = m(4π²)/r
Substitute m = 49, r = 4 in the above equation. π² = 9.8
s² = 49(4 × 9.8)/4
s² = 480.2
s = √480.2
s = 21.91

The final answer is s = 21.91(approximately)

Given that Y = 2x (m + n),
To find x, rewrite the given equation by changing the subject to x.
Y = 2x (m + n)
Move 2(m + n) to the left side and divide it by Y
Y/(2(m + n)) = x
x = Y/(2(m + n))
Substitute Y = 50, m = 6, n = 4 in the above equation.
x = 50/(2(6 + 4))
x = 50/2(10)
x = 50/20
x = 5/2

The final answer is x = 5/2

Given that L = m/2 {2b + (m – 1) d}
To find d, rewrite the given equation by changing the subject to d.
L = m/2 {2b + (m – 1) d}
Move 2 to the left side and multiply it with the 2.
2L = m{2b + (m – 1) d}
Move m to the left side and divide it with the 2L.
2L/m = 2b + (m – 1) d
Move 2b to the left side and subtract it from the 2L/m.
2L/m – 2b = (m – 1) d
Move m – 1 to the left side and divide it from the 2L/m – 2b.
(2L/m – 2b)/(m – 1) = d
d = (2L/m – 2b)/(m – 1)
Substitute L = 125, m = 10, b = 2 in the above equation.
d = (2(125)/10 – 2(4))/(10 – 1)
d = (250/10 – 8)/9
d = (25 – 8)/9
d = 17/9

The final answer is d = 17/9.

Given that T = 2 πr
To find r, rewrite the given equation by changing the subject to r.
T = 2 πr
Move 2 π to the left side and divide it from the T
T/2 π = r
r = T/2 π
Substitute T = 88, π = 22/7 in the above equation.
r = 88/2(22/7)
r = 88/(44/7) = 88 × 7/44 = 2 × 7 = 14
r = 14

The final answer is r = 14.

Given that S = H × R
To find H, rewrite the given equation by changing the subject to H.
S = H × R
Move R to the left side and divide it from the S
S/R = H
H = S/R
Substitute S = 450 and R = 15 in the above equation.
H = 450/15
H = 30

The final answer is H = 30.

Given that c = (l – m)/(l + m)
To find m, rewrite the given equation by changing the subject to m.
c = (l – m)/(l + m)
Move l + m to the left side and multiply it with the c.
c(l + m) = l – m
cl + cm = l – m
Move cl to the right side and m to the left side with proper opertaions.
cm + m = l – cl
m(c + 1) = l (1 – c)
Move (1 + c) to the right side and divide it by l (1 – c)
m = l (1 – c)/(1 + c)
Substitute c = 3/7 and l = 10 in the above equation.
m = 10 (1 – 3/7)/(1 + 3/7)
m = 10 (0.4)
m = 4

The final answer is m = 4

Given that R = (A × G × E)/100
To find G, rewrite the given equation by changing the subject to G.
R = (A × G × E)/100
Move 100 to the left side and multiply it with the R.
100R = A × G × E
Move AE to the left side and divide it with the 100R.
100R/AE = G
G = 100R/AE
Substitute R = 80, A = 400, E = 10 in the above equation.
G = 100(80)/(400)(10)
G = 8000/4000
G = 2

The final answer is G = 2.

Given that A/B = 100
To find B, rewrite the given equation by changing the subject to B.
A/B = 100
Move B to the right side and multiply it by 100.
A = 100B
Move 100 to the left side and divide it with the A.
A/100 = B
B = A/100.
Substitute A = 200 in the above equation.
B = 200/100
B = 2

The final answer is B = 2.

Given that PQR/T = AB.
To find B, rewrite the given equation by changing the subject to B.
PQR/T = AB
Move A to the left side and divide it from the PQR/T.
B = APQR/T
Substitute P = 2, Q = 3, R = 4, T = 4, A = 2 in the above equation.
B = (4 × 2 × 3 × 4)/4 = 24
B = 24.

The final answer is B = 24.

(a) Given that A = l × b; make l as subject
A = l × b
Divde b on both sides
A/b = (l × b)/b
A/b = l

The final answer is l = A/b.

(b) Given that L = C.P. – S.P; make C.P. as subject
L = C.P. – S.P
Move S.P to the left side and add it to the L
C.P. = L + S.P

The final answer is C.P. = L + S.P

(c) Given that V = u + ft; make f as subject
V = u + ft
Subtract u on both sides
V – u = u – u + ft
V – u = ft
Move t to the left side and divide with (V – u)
(V – u)/t = f

The final answer is f = (V – u)/t

(d) Given that S = D/T; make T as subject
S = D/T
Move T to the left side and multiply it with S
ST = D
Now, S to the right side and divide it with D
S = D/T

The final answer is S = D/T

(e) Given that V = πrh²; make h as subject
V = πrh²
Divide πr on both sides
V/πr = πrh²/πr
V/πr = h²
h = √V/πr

The final answer is h = √V/πr

(f) Given that C = 2πt; make t as subject
C = 2πt
Move 2π to the left side and divide it with C
C/2π = t

The final answer is t = C/2π

(g) Given that P = 2 (l + b); make b as subject
P = 2 (l + b)
Move 2 to the left side and Divide it with P.
P/2 = l + b
Subtract l on both sides
P/2 – l = l – l + b
P/2 – l = b

The final answer is b = P/2 – l

(h) Given that S = n/2 (a + l); make n as subject
S = n/2 (a + l)
Move 2 to the left side and Multiply it with S.
2S = n (a + l)
Move (a + l) to the left side and divide it with 2S
n = 2S/(a + l)

The final answer is n = 2S/(a + l)

(i) Given that A = P {1 + (Rn/100)}; make R as subject
A = P {1 + (Rn/100)}
Move P to the left side and Divide it with A.
A/P = 1 + (Rn/100)
Move 1 to the left side and Subtract it with A/P.
A/P – 1 = Rn/100
Move 100 to the left side and Multiply it with A/P – 1.
100(A/P – 1) = Rn
Move n to the left side and Divide it with 100(A/P – 1).
{100(A/P – 1)}/n = R

The final answer is R = {100(A/P – 1)}/n

(j) Given that 1/x = y + z/y + 1; make z as subject
1/x = y + z/y + 1
Move 1 to the left side and Subtract it with 1/x.
1/x – 1 = y + z/y
Move y (addition) to the left side and Subtract it with 1/x – 1.
(1/x – 1) – y = z/y
Move y to the left side and Multiply it with (1/x – 1) – y.
y[(1/x – 1) – y] = z

The final answer is z = y[(1/x – 1) – y]

(a) Given that y = mx + c; make m as subject
y = mx + c
Subtract con both sides
y – c = mx + c – c
y – c = mx
Move x to the left side and divide it with y – c
(y – c)/x = m

The final answer is m = (y – c)/x

(b) Given that D = b² – 4ac; make b as subject
D = b² – 4ac
Move 4ac to the left side and add it to the D
D + 4ac = b²
Apply square root on both sides
√(D + 4ac) = √b²
√(D + 4ac) = b

The final answer is b = √(D + 4ac)

(c) Given that T = 2π √(l/g); make l as subject
T = 2π √(l/g)
Move 2π to the left side and divide it with T
T/2π = √(l/g)
Apply square on both sides
(T/2π)² = (√(l/g))²
(T/2π)² = l/g
Move g to the left side and multiply it with (T/2π)²
g(T/2π)² = l

The final answer is l = g(T/2π)²

(d) Given that S = ut + 1/2 gt²; make u as subject
S = ut + 1/2 gt²
Move 1/2 gt² to the left side and subtract it from S
S – 1/2 gt² = ut
Divide t on both sides
(S – 1/2 gt²)/t = ut/t
(S – 1/2 gt²)/t = u

The final answer is u = (S – 1/2 gt²)/t

(e) Given that C = 2πr (h + x); make r as subject
C = 2πr (h + x)
Move (h + x) to the left side and divide it with C
C/(h + x) = 2πr
Move 2π to the left side and divide it with C/(h + x)
C/2π(h + x) = r

The final answer is r = C/2π(h + x)

(f) Given that h² = p² + b²; make p as subject
h² = p² + b²
Move b² to the left side and subtract it from h²
h² – b² = p²
Apply square root on both sides
√(h² – b²) = √(p²)
√(h² – b²) = p

The final answer is p = √(h² – b²)

(g) Given that S.P. = {(100 + G%) C.P.}/100; make G% as subject
S.P. = {(100 + G%) C.P.}/100
Move 100 to the left side and multiply it with S.P.
S.P. × 100 = (100 + G%) C.P.
Move C.P. to the left side and divide it with S.P. × 100
(S.P. × 100)/C.P. = 100 + G%
Move 100 to the left side and subtract it from (S.P. × 100)/C.P.
(S.P. × 100)/C.P. – 100 = G%

The final answer is G% = (S.P. × 100)/C.P. – 100

(h) Given that HM = 2ab/(a + b); make M as subject
HM = 2ab/(a + b)
Move H to the right side and divide it with 2ab/(a + b)
M = 2ab/H(a + b)

The final answer is M = 2ab/H(a + b)

(i) Given that I = (P × R × T)/100; make P as subject
I = (P × R × T)/100
Move 100 to the left side and multiply it with I
I × 100 = PRT
Move RT to the left side and divide it with 100I
100I/RT = P

The final answer is P = 100I/RT

(j) Given that A = 1/2 (l₁ + l₂) h; make l₁ as subject
A = 1/2 (l₁ + l₂) h
Move 1/2.h to the left side and divide it with A
2A/h = l₁ + l₂
Move l₂ to the left side and subtract it from 2A/h
2A/h – l₂ = l₁

The final answer is l₁ = 2A/h – l₂

(a) Given that the area of four walls of a room(a) Given that the area of a box is equal to one-third the product of base and height.
Let the area of the box is ‘a’, Base = b, Height = h.
If the area of a box is equal to one-third the product of base and height, then a = 1/3 × bh

The final answer is a = 1/3 . bh

(b) Given that the diameter of a matchbox is twice the radius of the matchbox.
Let the diameter of a matchbox is d, the radius of the matchbox is r
If the diameter of a matchbox is twice the radius of the matchbox, then d = 2r

The final answer is d = 2r.

(c) Given that the area of a table is half the sum of the parallel sides.
Let the Area of a table = a, sum of the parallel sides = x
If the area of a table is half the sum of the parallel sides, then a = x/2.

The final answer is a = x/2.

(d) Given that the difference between the x and y is the addition of 10 and z.
x – y = 10 + z

The final answer is x – y = 10 + z

(e) The volume of a cone is equal to the product of its breadth, length, and height.
Let the volume of a cone is v, breadth is b, the length is l, and height is h.
If the volume of a cone is equal to the product of its breadth, length, and height, then v = blh.

The final answer is v = lbh.

(a) The surface area of the cube is 8 times the square of its edge.
Let the surface area of the cube is S, the square of its edge R, then S = 8R.

The final answer is S = 8R.

(b) A body of material is moved with a force is equal to the product of the mass of that material body and the acceleration produced by it.
Let the force is F, the mass of that material body is m, and the acceleration of the body is a.
If the body of material is moved with a force is equal to the product of the mass of that material body and the acceleration produced by it, then F = ma.

The final answer is F = ma.

(c) Given that the speed of a car x is equal to the distance (m) traveled by it upon the time (t) taken to cover this distance.
Speed = distance/time.
x = m/t m/sec.

The final answer is x = m/t m/sec.

(d) Given that the Multiplication of x, y, z is equal to half of m.
xyz = m/2.

The final answer is xyz = m/2.

(e) Given that the area of four walls of a room is equal to two times the product of height and perimeter of the room.
Let the area of four walls of a room is a, height is h, and perimeter of the room is p.
If the area of four walls of a room is equal to two times the product of height and perimeter of the room.
a = 2hp

The final answer is a = 2ph.

(a) Given that Six subtracted from a number gives two.
Let the unknown number is x.
x – 6 = 2.

The final answer is x – 6 = 2.

(b) Given that Five added to a number is 10.
Let the unknown number is m.
m + 5 = 10.

The final answer is m + 5 = 10.

(c) Given that Four-thirds of a number is 12.
Let the unknown number is n.
4/3 . n = 12
n = 12 . 3/4
n = 9

The final answer is n = 9.

(d) Given that One-fifth of a number is 10 more than 2.
Let the unknown number is p.
1/5 . p = 10 + 2
p/5 = 12
p = 12 . 5
p = 60

The final answer is p = 60.

(e) Given that the sum of four times a number and 12 is 64.
Let the unknown number is y.
4y + 12 = 64.
4y = 64 – 12
4y = 52
y = 52/4
y = 13

The final answer is y = 13.

(f) Given that the sum of three consecutive odd numbers is 72.
Let the unknown odd number is z.
z + (z + 2) + (z + 4) = 72
3z + 6 = 72
3z = 72 – 6
3z = 66
z = 66/3
z = 22.

The final answer is z = 22.

(g) Given that the sum of two multiples of 5 is 65.
Let the first number be b.
b + b+ 5 = 65
2b + 5 = 65
Move 5 to the right side and substract it with 65.
2b = 65 – 5
2b = 60
Move 2 to the right side and divide it with 60.
b = 60/2
b = 30.

The final answer is b = 30.

(h) Given that One number is 5 less than four times the other, if their sum is increased by 3 the result is 24.
Let the unknown number is x.
One number is 5 less than four times the other.
4x – 5.
Their sum is increased by 3 the result is 24.
x + 4x – 5 + 3 = 24
5x -2 = 24
Move 2 to the right side and add it to the 24.
5x = 24 + 2
5x = 26
Move 5 to the right side and divide it by 26.
x = 26/5

The final answer is 26/5.

(i) Given that If 2/3 is subtracted from a number and the difference is multiplied by 5, the result is 2.
Let the unknown number is y.
2/3 is subtracted from y.
y – 2/3
The above difference is multiplied by 5.
5(y – 2/3)
The result is 2.
5(y – 2/3) = 2.
Move 5 to the right side and divide it by 2.
y – 2/3 = 2/5.
Move 2/3 to the right side and add it to 2/5.
y = 2/5 + 2/3 = 16/15
y = 16/15

The final answer is y = 16/15.

(j) Given that the numerator of a fraction is 6 less than the denominator. If 1 is added to both numerator and denominator, the fraction becomes 3/4.
Let the unknown number is x.
The numerator of a fraction is 6 less than the denominator.
Numerator = x – 6, denominator = x
(x – 6)/x = 3/4
1 is added to both numerator and denominator.
(x – 6 + 1)/(x + 1) = 3/4
(x – 5)/(x + 1) = 3/4
4(x – 5) = 3(x + 1)
4x – 20 = 3x + 3
4x – 3x = 3 + 20
x = 23.

The final answer is x = 23.

(i) If the present age of a girl is y years.
(a) What will her age be after 9 years?
After 9 years, her age will be (y + 9) years.
(b) What was her age 7 years ago?
7 years ago, her age will be (y – 7) years
(c) Her father’s age is 2 more than 8 times her age. Find her father’s age.
Her father’s age is (8y + 2) years
(d) Mother is 3 years younger than her father. What is her mother’s age?
(8y + 2 – 3)  = (8y – 1) years

(ii) The length of a rectangle is 6m less than thrice its breadth. If the perimeter of the rectangle is 124m find its length and breadth. ​
Let the length is l, breadth = b, and Perimeter = p.
length = 2b – 6
Perimeter p = 2(l + b) = 2(2b – 6 + b) = 2(3b – 6) = 6b – 12
6b – 12 = 124m
6b = 124 + 12 = 136
b = 136/6
length = 2b – 6 = 2(136/6) – 6 = 136/3 – 6 = (136 – 18)/3 = 118/3

The final answer is length = 118/3, b = 136/6.

(iii). A bus covers 168 km in 4 hours. How much distance will it cover in 80 hours?
Distance d = 168km, Time t = 4 hours.
Speed s = d/t = 168/4 km/hr = 42 km/hr.
If the time t = 80 hours, distance d = 80hrs × 42 km/hr = 3360 km
Distance d = 3360 km.

The final answer is d = 3360km.

(iv) The weight of orange is 70 g and that of mango is 60 g. Find the total weight of x oranges and y mangoes?
The weight of the orange is 70 g.
For x oranges, 70xg
The weight of the mango is 60g
For y mangoes, 60yg
The total weight of x oranges and y mangoes = 70x + 60y grams.

The final answer is 70x + 60y grams.

(a) A pencil costs $x. An eraser costs $2x.
A pencil costs $x. The eraser cost is twice the pencil cost.

(b) A pen costs $y. A notebook costs $7 + y.
A pen costs $y. A notebook cost is $7 more than that pen cost.

(c) The number of girls in a class is ‘m’. The number of boys in the class is (1/3)m.
The number of girls in a class is ‘m’. The number of boys in the class is one-third of the number of girls in a class.

(d) Ram is y years old. His mother is (5y – 1) years old.
Ram is y years old. His mother’s age is one less than five times of Ram’s age.

(e) In an arrangement of flowerpots, there are n rows. Each row has 8 flower-pots.
In an arrangement of flower pots, there are 8n rows of flowerpots available. If there are n rows, how many flowerpots available for each row?

Given that Two persons are running from the same place at a speed of 6 km/hr and 3 km/hr.
The speed of a first-person is 6 km/hr.
The speed of the second person is 3 km/hr.
As the two persons moving in opposite directions, the speeds are added.
Relative Speed = 6 km/hr + 3 km/hr = 9 km/hr.
Given that the time = 3 minutes.
Convert the time from minutes to hours.
Divide minutes with 60 to convert minutes to hours.
3 minutes = 3/60 hours
3 minutes = 1/20 hours.
Let us find the distance between two persons.
Distance = Speed × Time
Distance = 9 km/hr × 1/20 hrs
Distance = 9/20 km.

The distance between two persons when they are moving each other in opposite directions is 9/20 km.

Given that Two boys are running in the same direction from the same place at the speed of 20 km/hr and 12 km/hr.
The speed of a first-person is 20 km/hr.
The speed of the second person is 12 km/hr.
As the two boys moving in the same direction, the speeds are subtracted.
Relative Speed = 20 km/hr – 12 km/hr = 8 km/hr.
Given that the time = 40 minutes.
Convert the time from minutes to hours.
Divide minutes with 60 to convert minutes to hours.
40 minutes = 40/60 hours
40 minutes = 2/3 hours.
Let us find the distance between two persons.
Distance = Speed × Time
Distance = 8 km/hr × 2/3 hours.
Distance = 16/3 km.

The distance between two boys running in the same direction from the same place is 16/3 km.

Given that a train 70 m long is running at a speed of 40 km/hr.
The man running at a speed of 4 km/hr in the same direction.
As the train and man are moving in the same direction, the speeds are subtracted.
Relative Speed = 40 km/hr – 4 km/hr = 36 km/hr.
Given that the distance = 70 m.
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
36 km/hr × 5/18 = 10 m/sec.
Distance = Speed × Time
Time = Distance/Speed
Time = 70 m/10 m/sec
Time = 7 sec

The train passes a man who is running at speed of 4 km/hr in the same direction in 7 sec.

Given that a train 75 m long is running at a speed of 33 km/hr.
The man running at a speed of 3 km/hr in the opposite direction.
As the train and man are moving in the opposite direction, the speeds are added.
Relative Speed = 33 km/hr + 3 km/hr = 36 km/hr.
Given that the distance = 75 m.
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
75 km/hr × 5/18 = 20.833 m/sec.
Distance = Speed × Time
Time = Distance/Speed
Time = 75 m/20.833 m/sec
Time = 3.6 sec

The train passes a man who is running at speed of 4 km/hr in the opposite direction in 3.6 sec.

Given that Two trains of length 120 m and 80 m respectively are running at the speed of 30 km/hr and 20 km/hr on parallel tracks in the opposite direction.
As the two trains are moving in the opposite direction, the speeds are added.
Relative Speed = 30 km/hr + 20 km/hr = 50 km/hr.
Distance = (120 m – 80 m) = 40 m.
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
50 km/hr × 5/18 = 13.888 m/sec.
Distance = Speed × Time
Time = Distance/Speed
Time = 40 m/13.888 m/sec
Time = 2.88 sec

The two trains take 2.88 sec to cross each other.

Given that Two trains of length 65 m and 125 m long are running on parallel tracks in the same direction with a speed of 35 km/hr and 40 km/hr respectively.
As the two trains are moving in the same direction, the speeds are subtracted.
Relative Speed = 40 km/hr – 35 km/hr = 5 km/hr.
Distance = (65 m + 125 m) = 190 m.
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
5 km/hr × 5/18 = 1.388 m/sec.
Distance = Speed × Time
Time = Distance/Speed
Time = 190 m/1.3888 m/sec
Time = 136.8 sec

The two trains take 136.8 sec to clear off each other from the moment they meet.

Given that Two trains are running on parallel tracks in the same direction at 20 km/hr and 15 km/hr respectively. The faster train passed a man than the slower train in 72 seconds.
As the two trains are running on parallel tracks in the same direction, the speeds are subtracted.
Relative Speed = 20 km/hr – 15 km/hr = 5 km/hr.
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
5 km/hr × 5/18 = 1.388 m/sec.
Distance = Speed × Time
Distance = 1.388 m/sec × 72 seconds = 100 sec

The length of the faster train is 100 sec.

Given that a train 160 m long crosses a bridge which is 40 m long in 20 seconds.
So, from the given data, time = 20 seconds, the length of the train is 160 m, and the length of the bridge is 40 m.
Now, find the total distance covered by the train to cross the bridge.
Distance covered by the train to cross the bridge = (160 + 40) m
Distance covered by the train to cross the bridge = 200 m
Now, find the speed of the train.
Speed of the train = distance/time
Speed of the train = 200 m/20 sec
The Speed of the train = 10 m/sec.

The Speed of the train is 10m/sec long crosses a bridge which is 40 m long in 20 seconds.

Given that a train 110 m long is running at a speed of 72 km/hr.
So, from the given data, speed = 72 km/hr, the length of the train is 110 m, and the length of the long tunnel is 55 m.
Now, find the total distance covered by the train to cross the bridge.
Distance covered by the train to cross the bridge = (110 + 55) m
Distance covered by the train to cross the bridge = 165 m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
72 km/hr × 5/18 = 20 m/sec.
Now, find the time taken by the train to cross the tunnel = distance/speed
Time taken by the train to cross the tunnel = 165 m/20 m/sec
The Time taken by the train to cross the tunnel = 8.25 sec

The time takes to cross a 55 m long tunnel is 8.25 sec.

Given that 75 m long train passes through a bridge which is 50 m long, running at a speed of 36 km/hr.
So, from the given data, speed = 36 km/hr, the length of the train is 75 m, and the length of the long tunnel is 50 m.
Now, find the total distance covered by the train to cross the bridge.
Distance covered by the train to cross the bridge = (75 + 50) m
Distance covered by the train to cross the bridge = 125 m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
36 km/hr × 5/18 = 10 m/sec.
Now, find the time taken by the train to cross the tunnel = distance/speed
Time taken by the train to cross the tunnel = 125 m/10 m/sec
The Time taken by the train to cross the tunnel = 12.5 sec

The time taken by 75 m long train passes through a bridge which is 50 m long, running at a speed of 36 km/hr is 12.5 sec.

Given that a 45 m long train is running at a speed of 27 km/hr.
So, from the given data, speed = 27 km/hr, the length of the train is 45 m, and the time to cross a platform is 30 seconds.
Now, find the total distance covered by the train to cross the platform.
Distance covered by the train to cross the platform = (45 + length of the platform) m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
27 km/hr × 5/18 = 7.5 m/sec.
The distance covered by train to cross the platform = speed × time
Distance covered by train to cross the platform = 7.5 m/sec × 30 seconds = 225 m.
(45 + length of the platform) m = 225m
Length of the platform = 225 – 45 = 180 m

The length of the platform is 180 m.

Given that a train 65 m long takes to clear the platform 45 m long if its speed is 60 km/hr.
So, from the given data, speed = 60 km/hr, the length of the train is 65 m, and the length of the platform is 45 m.
Now, find the total distance covered by the train to cross the bridge.
Distance covered by the train to cross the bridge = (65 + 45) m
Distance covered by the train to cross the bridge = 110 m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
60 km/hr × 5/18 = 50/3 m/sec.
Now, find the time taken by the train to clear the platform = distance/speed
Time taken by the train to clear the platform = 110 m/(50/3 m/sec)
The Time taken by the train to clear the platform = 6.6 sec

The time takes to cross the platform is 6.6 sec.

Given that a train 90 m long took 30 seconds to pass a 30 m long tunnel.
So, from the given data, time = 30 seconds, the length of the train is 90 m, and the length of the tunnel is 30 m.
Now, find the total distance covered by the train to pass a 30 m long tunnel.
Distance covered by the train to cross the tunnel = (90 + 30) m
Distance covered by the train to cross the tunnel = 120 m
Now, find the speed of the train.
Speed of the train = distance/time
Speed of the train = 120 m/30 sec
The Speed of the train = 4 m/sec.

The Speed of the train is 4 m/sec.

Given that a train 180 m long is running at a speed of 45 km/hr and crosses the platform in 20 seconds.
So, from the given data, speed = 45 km/hr, the length of the train is 180 m, and the time to cross a platform is 20 seconds.
Now, find the total distance covered by the train to cross the platform.
Distance covered by the train to cross the platform = (180 + length of the platform) m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
45 km/hr × 5/18 = 12.5 m/sec.
The distance covered by train to cross the platform = speed × time
Distance covered by train to cross the platform = 12.5 m/sec × 20 seconds = 250 m.
(180 + length of the platform) m = 250 m
Length of the platform = 250 – 180 = 70 m

The length of the platform is 70 m.

Given that a train moving at a speed of 120 km/hr crosses the 222 m long bridge in 30 seconds.
So, from the given data, speed = 120 km/hr, the length of the bridge is 222 m, and the time to cross a bridge is 30 seconds.
Now, find the total distance covered by the train to cross the bridge.
Distance covered by the train to cross the bridge = (222 + length of the train) m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
120 km/hr × 5/18 = 33.33 m/sec.
The distance covered by train to cross the platform = speed × time
Distance covered by train to cross the platform = 33.33 m/sec × 30 seconds = 1000 m.
(222 + length of the train) m = 1000 m
Length of the train = 1000 – 222 = 778 m

The length of the train is 778 m.

Given that a train 25 m long crosses a bridge which is 15 m long in 40 seconds.
So, from the given data, time = 40 seconds, the length of the train is 25 m, and the length of the bridge is 15 m.
Now, find the total distance covered by the train to cross the bridge.
Distance covered by the train to cross the bridge = (25 + 15) m
Distance covered by the train to cross the bridge = 40 m
Now, find the speed of the train.
Speed of the train = distance/time
Speed of the train = 40 m/40 sec
The Speed of the train = 1 m/sec.

The Speed of the train is 1 m/sec long crosses a bridge which is 15 m long in 40 seconds.

Given that a 60 m long train is running at a speed of 25 km/hr.
So, from the given data, speed = 25 km/hr, the length of the train is 60 m, and the time to cross a platform is 12 seconds.
Now, find the total distance covered by the train to cross the platform.
Distance covered by the train to cross the platform = (60 + length of the platform) m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
25 km/hr × 5/18 = 6.94 m/sec.
The distance covered by train to cross the platform = speed × time
Distance covered by train to cross the platform = 6.94 m/sec × 12 seconds = 83.33 m.
(60 + length of the platform) m = 83.33 m
Length of the platform = 83.33 – 60 = 23.333 m

The length of the platform is 23.333 m (approximately).

Given that 65 m long train to cross a tunnel which is 95 m long running at a speed of 27 km/hr.
So, from the given data, speed = 27 km/hr, the length of the train is 65 m, and the length of the long tunnel is 95 m.
Now, find the total distance covered by the train to cross the bridge.
Distance covered by the train to cross the bridge = (65 + 95) m
Distance covered by the train to cross the bridge = 160 m
Now, find the speed of the train in m/sec.
convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
27 km/hr × 5/18 = 7.5 m/sec.
Now, find the time taken by the train to cross the tunnel = distance/speed
Time taken by the train to cross the tunnel = 160 m/7.5 m/sec
The Time taken by the train to cross the tunnel = 21.33 sec (approximately)

The time taken by 65 m long train to cross a tunnel which is 95 m long running at a speed of 27 km/hr is 21.33 sec (approximately).

Given that a train 120 m long is running at a uniform speed of 36 km/hr.
So, from the given data, speed = 36 km/hr, and the length of the train is 120m.
The speed of the train = 36 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
36 km/hr × 5/18 = 10 m/sec.
Now, find the time taken by the train to cross a pole.
Time taken by the train to cross a pole = length of train/speed of the train.
Time taken by the train to cross a pole = 120m/10 m/sec = 12 sec.

The train takes 12 seconds to cross the pole.

Given that the time taken by a train 800 m long, running at a speed of 72 km/hr in crossing the pole.
So, from the given data, speed = 72 km/hr, and the length of the train is 800 m long.
The speed of the train = 72 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
72 km/hr × 5/18 = 20 m/sec.
Now, find the time taken by the train to cross a pole.
Time taken by the train to cross a pole = length of train/speed of the train.
Time taken by the train to cross a pole = 800m/20 m/sec = 4 sec.

The train takes 4 seconds to cross the pole.

Given that a train is running at a speed of 90 km/hr. It crosses a tower in 16 seconds.
So, from the given data, speed = 90 km/hr, and time is taken by a train is 16 seconds.
The speed of the train = 90 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
90 km/hr × 5/18 = 25 m/sec.
Now, find the Length of the train.
Length of the train = speed × time
Length of the train = 25 m/sec × 16 sec
The Length of the train = 400 m

The Length of the train is 400 m that crosses a tower in 16 seconds.

Given that a train is running at a speed of 63 km/hr. if it crosses a pole in just 14 seconds.
So, from the given data, speed = 63 km/hr, and the time taken by a train is 14 seconds.
The speed of the train = 63 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
63 km/hr × 5/18 = 17.5 m/sec.
Now, find the Length of the train.
Length of the train = speed × time
Length of the train = 17.5 m/sec × 14 sec
The Length of the train = 245 m

The Length of the train is 245 m that crosses a pole in 14 seconds.

Given that a train 60 m long is running at a speed of 36 km/hr.
So, from the given data, speed = 36 km/hr, and the length of the train is 60 m.
The speed of the train = 36 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
36 km/hr × 5/18 = 10 m/sec.
Now, find the time taken by the train to cross a man standing on the platform.
Time taken by the train to cross a man standing on the platform = length of train/speed of the train.
Time taken by the train to cross a man standing on the platform = 60m/10 m/sec = 6 sec.

The train takes 6 seconds to cross a man standing on the platform.

Given that a train 180 m long cross a tree in 10 seconds.
So, from the given data, the length of the train is 180 m and time = 10 seconds.
We know that Length of the train = speed × time
Speed = Length of the train/time
Speed = 180 m/10 seconds
The speed = 18 m/sec
The speed of the train = 18 m/sec.
Now, convert m/sec into km/hr.
Multiply m/sec with 18/5 to convert it into km/hr.
m/sec × 18/5 = km/hr
18 m/sec × 18/5 = 64.8 km/hr
The speed of the train is 64.8 km/hr.

The train travels at a speed of 64.8 km/hr to cross a tree in 10 seconds.

Given that a train is running at a speed of 22.5 km/hr. It crosses the pole in 20 seconds.
So, from the given data, the speed = 22.5 km/hr and time = 20 seconds.
The speed of the train = 22.5 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
22.5 km/hr × 5/18 = 6.25 m/sec.
Now, find the Length of the train.
Length of the train = speed × time
Length of the train = 6.25 m/sec × 20 sec
The Length of the train = 125 m

The Length of the train is 125 m that crosses a pole in 20 seconds.

Given that a train 80 m long is running at a uniform speed of 36 km/hr.
So, from the given data, speed = 36 km/hr, and the Length of the train is 80 m long.
The speed of the train = 36 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
36 km/hr × 5/18 = 10 m/sec.
Now, find the time taken by the train to cross a pole.
Time taken by the train to cross a pole = length of train/speed of the train.
Time taken by the train to cross a pole = 80 m/10 m/sec = 8 sec.

The train takes 8 seconds to cross the pole.

Given that a train is running at a speed of 120 km/hr. It crosses the tower in 18 seconds.
So, from the given data, speed = 120 km/hr, and the time taken by a train is 18 seconds.
The speed of the train = 120 km/hr.
Now, convert km/hr into m/sec.
Multiply km/hr with 5/18 to convert it into m/sec.
km/hr × 5/18 = m/sec
120 km/hr × 5/18 = 33.33 m/sec (approximately).
Now, find the Length of the train.
Length of the train = speed × time
Length of the train = 33.33 m/sec × 18 sec
The Length of the train = 600 m

The Length of the train is 600 m that crosses a tower in 18 seconds.

Given that a car moves at a speed of 40 km/hr.
Speed = 40 km/hr
Time = 18 minutes
1 minute = 1/60 hr
18 minutes = 18/60 hr
We know that Speed = distance/time
Distance = Speed × time
Substitute the given value in the formula.
Distance = 40 km/hr × (18/60 hr)
Distance = 12 km

The car will travel 12 km in 18 minutes.

Given that a bus covers 84 km in 2 hours.
Therefore, the distance = 84 km, the time = 2 hours.
We know that Speed = distance/time.
Speed = 84/2 km/hr = 42 km/hr.
If the time takes 40 minutes = 40/60 hr = 2/3 hr
Distance = speed × time
Distance = 42 km/hr × 2/3 hr
The distance = 28 km

A bus covers 28 km in 40 minutes

Given that a motorist moves with a speed of 15 km/hr and covers the distance from place A to place B in 2 hours 30 minutes.
Therefore, the speed = 15 km/hr, time = 2 hours 30 minutes.
Let’s convert the time from minutes into hours.
1 minute = 1/60 hr
2 hours 30 minutes = 2 hours + 30/60 hours = 2 + 1/2 hr = 2.05 hr
We know that Speed = distance/time.
Distance = speed × time
Substitute the given value in the formula.
Distance = 15 km/hr × 2.05 hr
Distance = 30.75 km

The distance between place A and place B is 30.75 km.

From the given data, Mohan drives a bike at a uniform speed of 30 km/hr
Speed = 30 km/hr
Time = 45 minutes
Let’s convert the time from minutes into hours.
1 minute = 1/60 hr
45 minutes = 45/60 hr = 9/12 hr = 3/4 hr
We know that Speed = distance/time.
Distance = speed × time
Substitute the given value in the formula.
Distance = 30 km/hr × 3/4 hr
Distance = 22.5 km

Mohan drives a bike moves 22.5 km in 45 minutes.

Given that a bus is running at a speed of 36 km/hr.
Convert km/hr into m/sec
km/hr × 5/18 = m/sec
36 km/hr × 5/18 = 10 m/sec
We know that Speed = distance/time.
Distance = speed × time
Substitute the given value in the formula.
Distance = 10 m/sec × 7 seconds = 70 m

The bus will go 70 m in 7 seconds.

Given that a bike travels at a speed of 27 km/hr.
Convert km/hr into m/sec
km/hr × 5/18 = m/sec
27 km/hr × 5/18 = 7.5 m/sec

The bike travels 7.5 meters in 1 second.

Firstly, take the speed = 50 km/hr and time = 6 hours.
Now, find the distance.
We know that Speed = distance/time.
Distance = speed × time
Substitute the given value in the formula.
Distance = 50 km/hr × 6 hours = 300 km.
Secondly, take the speed = 40 km/hr and time = 6 hours.
Now, find the distance.
We know that Speed = distance/time.
Distance = speed × time
Substitute the given value in the formula.
Distance = 40 km/hr × 6 hours = 240 km.
Difference in distance = 300 – 240 km = 60 km.

An interstate bus goes traveling 50 km/hr moves 60 km farther than an interstate bus goes traveling 40 km/hr in 6 hours.

Let us assume the distance is x km.
Now, find the time taken to cover 6 and 8 km.
Time taken to cover x km at 6 km/hr = x/6 hr
The time to cover x km at 8 km/hr = x/8 hr
Given that the distance covered by each one of them when one takes 20 minutes longer than the other.
Therefore, x/6 – x/8 = 20/60
x/6 – x/8 = 1/3
4x – 3x = 24/3
x = 8
Therefore, the distance is 8 km.

The distance covered by each one of them is 8 km when one takes 20 minutes longer than the other.

Let us assume the distance covered by Arun and Kiran is x km.
Now, find the time taken to cover 2 and 4 km.
Time taken to cover x km at 2 km/hr = x/2 hr
The time to cover x km at 4 km/hr = x/4 hr
Given that the distance covered by each one of them when one takes 10 minutes longer than the other.
Therefore, x/2 – x/4 = 10/60
x/2 – x/4 = 1/6
2x – x = 4/6
x = 0.66 km (approximately)
Therefore, the distance is 0.66 km.

The distance covered by each one of them is 0.66 km (approximately) when one takes 20 minutes longer than the other.

Given that a train travels at a speed of 126 km in one hour.
Convert km/hr into m/sec.
To convert km/hr into m/sec, multiply km/hr × 5/18 = m/sec
126 km/hr × 5/18 = 35 m/sec.
35 meters per second.

The train travels 35 meters per one second.