Ratio and Proportion Class 10 OP Malhotra Exe-6C ICSE Maths Solutions of Ch-6 questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Ratio and Proportion Class 10 OP Malhotra Exe-6C ICSE Maths Solutions
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 10th |
| Chapter-6 | Ratio and Proportion |
| Writer | OP Malhotra |
| Exe-6C | Solving proportional by k method |
| Edition | 2024-2025 |
Solving proportional by k method
The direct proportion formula says if the quantity y is in direct proportion to quantity x, then we can say y = kx, for a constant k. y = kx is also the general form of the direct proportion equation. where, k is the constant of proportionality
Exercise- 6C
Ratio and Proportion Class 10 OP Malhotra Exe-6C ICSE Maths Solutions of Ch-6
Que-1: If x/a = y/b = z/c, show that
(i) x3/a3 – y3/b3 + z3/c3 = xyz/abc
(ii) ((a2x2 + b2y2 + c2z2) / (a3x + b3y + c3z))3/2 = √(xyz / abc)
(iii) (xz + ac) / (xz – ac) = (yz + bc) / (yz – dc)
Sol: (i) Let x/a = y/b = z/c = k
=> x = ak, y = bk, z = ck
L.H.S = (x³/a³) + (y³/b³) + (z³/c³)
= (ak)³/a³ + (bk)³/b³ + (ck)³c³
= [(a³k³)/a³] + (b³k³)/b³] + [(c³k³)/c³]
= k³+k³+K³
= 3k³
R.H.S = 3xyz/abc
= [3(ak)(bk)(ck)]/abc
= 3k³
= L.H.S
=> L.H.S = R.H.S
(ii) Let x/a = y/b = z/c = k
=> x = ak, y = bk, z = ck
L.H.S. = [{a²x²+b²y²+c²z²}/{a²x+b³y+c³z}]³
= [{a².a²k²+b².b²l²+c².c²k²}/{a³.a.k+b³.bk+c³.ck}]³
= [{a⁴k²+b⁴k²+c⁴k²}/{a⁴k+b⁴k+c⁴k}]³
= [{k²(a⁴+b⁴+c⁴)}/{k(a⁴+b⁴+c⁴)}]³ = k3
R.H.S. = xyz/abc
= (ak.bk.ck)/abc = k3
∴ L.H.S. = R.H.S.
(iii) Let x/a = y/b = z/c = k
=> x = ak, y = bk, z = ck
LHS
= (xz+ac)/(xz-ac)
= (ak.ck + ac) / (ak.ck – ac)
= ac(k² + 1) /ac(k² – 1)
= (k² + 1) /(k² – 1)
RHS
=(yz+bc)/(yz-bc0
= (bk.ck + bc) / (bk.ck – bc)
= bc(k² + 1) /bc(k² – 1)
= (k² + 1) /(k² – 1)
(k² + 1) /(k² – 1)= (k² + 1) /(k² – 1)
=> LHS = RHS
Que-2: If a/b = c/d = e/f, prove the following
(i) (pa3 + qc3 + re3) / (pb3 + qd3 + rf3) = ace/bdf
(ii) √((a4 + c4)/(b4 + d4)) = (pa2 + qc2)/(pb2 + qd2)
Sol: (i) a/b = c/d = e/f = K (say)
a = kb, c = dk, e = fk
L.H.S. = (pa3 + qc3 + re3)/(pb3 + qd3 + rf3)
{p(kb)³+q(kd)³+r(kf)³}/{pb³+qd³+rf³}
{pk³b³+qk³d³+rk³f³}/{pb³+qd³+rf³}
[k³{pb³+qd³+rf³}]/{pb³+qd³+rf³}
= k³
R.H.S. = ace/bdf
{kb.dk.fk}/(bdf)
{k³bdf}/bdf
= k³
L.H.S. = R.H.S.
Hence Proved
(ii) a/b = c/d = e/f = K (say)
a = kb, c = dk, e = fk
L.H.S. = √((a⁴ + c⁴)/(b⁴ + d⁴))
√[{(kb)⁴+(kd⁴)}/(b⁴+d⁴)]
√[{k⁴b⁴+k⁴d⁴}/(b⁴+d⁴)]
√[{k⁴(b⁴+d⁴)}/(b⁴+d⁴)]
√k⁴ = k²
R.H.S. = (pa² + qc²)/(pb² + qd²)
{p(kb)²+q(kd)²}/(pb²+qd²)
{pk²b²+qk²d²}/{pb²+qd²}
{k²(pb²+qd²)}/(pb²+qd²)
k².
L.H.S. = R.H.S.
Hence Proved.
Que-3: If a, b, c are in continued proportion, then prove that
(i) (a + b + c)(a – b + c) = a2 + b2 + c2
(ii) (a2 + b2) / (b2 + c2) = a / c
(iii) (a3 + b3 + c3) / a2b2c2 = 1/a3 + 1/b3 + 1/c3
Sol: (i) As a, b, c, are in continued proportion
Let a/b = b/c = k
L.H.S. = (a + b + c) (a – b + c)
= (ck2 + ck + c)(ck2 – ck + c)
= c(k2 + k + 1)c(k2 – k + 1)
= c2(k2 + k + 1)(k2 – k + 1)
= c2(k4 + k2 + 1)
R.H.S. = a2 + b2 + c2
= (ck2)2 + (ck)2 + (c)2
= c2k4 + c2k2 + c2
= c2(k4 + k2 + 1)
∴ L.H.S. = R.H.S.
(ii) a, b and c are the continued proportion
a: b = b: c
⇒ a/b = b/c
⇒ b2 = ac
Now a/c = (a²+b²)/(b²+c²)
= a(b2 + c2) = c(a2 + b2)
L.H.S.
⇒ a(b2 + c2)
⇒ a(ac + c2)
⇒ ac(a + c)
R.H.S.
⇒ c(a2 + b2)
⇒ c(a2 + ac)
⇒ ac(a + c)
L.H.S. = R.H.S.
Hence proved.
(iii) a/b = b/c = c/d = K (say)
b² = ac
R.H.S. = 1/a3 + 1/b3 + 1/c³
(b³c³+a³c³+a³b³)/(a³b³c³)
L.H.S. = (a3 + b3 + c3) / a²b²c²
L.H.S. = (a3 + b3 + c3) / a²b²c² = R.H.S.
Que-4: If a, b, c, d are in continued proportion, then prove that
(i) (b – c)2 + (c – a)2 + (d – b)2 = (a – d)2
(ii) √[(a5 + b2c2 + a3c2) / (b4c + d4 + b2cd2) = a / d
(iii) √ab – √bc + √cd = √((a – b + c)(b – c + d))
(iv) (3a + 5d) / (5a + 7d) = (3a3 + 5d3) / (5a3 + 7d3)
Sol: (i) Since a, b, c, d are in continued proportion, we have
a/b = b/c = c/d = K (say)
∴ c = dK, b = cK = dK2 and a = bK = dK3.
L.H.S.
= (b – c)2 + (c -a)2 + (d – b)2
= (dK2 – dK)2 + (dK – dK3)2 + (d – dK2)2
= d2K2(K – 1)2 + d2K2 (1 – K2)2 + d2 (1 – K2)2
= d2 [K2(K – 1)2 + K2(K2 – 1)2 + d2(k2 – 1)2]
= d2 [K2 (K – 1)2 + K2 (K – 1)2 (K + 1)2 + (K – 1)2 (K + 1)2]
= d2 (K – 1)2 [K2 + K2 (K+ 1)2 + (K + 1)2]
= d2 (K -1)2 [K2 + K2 (K2 + 2K + 1) + K2 + 2K + 1]
= d2 (k – 1)2 [K4 + 2K3 + 3K2 + 2K + 1]
= d2(K -1)2 (K2 + K + 1)2
= d2[(K – 1) (K2 + K + 1)2]
= d2 (K3 – 1)2 = (dK3 – d)2 = (a – d)2 = (d – a)2 = R.H.S.
Hence, (b – c)2 + (c -a)2 + (d – b)2 = (d – a)2.
(ii) a/b = b/c = c/d = K (say)
∴ b = ka, c = k²a and d = k³a
L.H.S. = √[(a5 + b2c2 + a3c2)/(b4c + d4 + b2cd2)
b²c² = (ka)²(k²a)² = k²a².⁴a² = k^6 a⁴
a³c² = a³(k²a)² = a³⋅k⁴a² = k⁴a^5
b⁴c = (ka)⁴(k²a) = k⁴a⁴⋅k²a = k^6 a^5
d⁴ = (k³a)⁴ = k^12 a⁴
b²cd² = (ka)²(k²a)(k³a)² = k²a²⋅k²a⋅k^6 a² = k^10 a^5
LHS = √(a^5+k^6 a⁴+k⁴a^5)/(k^6 a^5+k^12 a⁴+k^10 a^5)
LHS = [{a^5(1+k⁴+k^6)}/{a^5(k^6+k^10+k^12)}]
LHS = (1+k⁴+k^6)/(k^6+k^10+k^12)
RHS = a/k³a = 1/k³
√[(1+k^4+k^6)/(k^6+k^10+k^12)] = 1/k³
(1+k^4+k^6)/(k^6+k^10+k^12) = 1/k^6
1+k^4+k^6 = k^6+k^10+k^12
This simplifies to an identity, as both sides are equal. Therefore, the given equation is proved.
(iii) Since a, b, c, d are in continued proportion then
a/b = b/c = c/d = k
⇒ a = bk, b = ck, c = dk
⇒ a = ck2
⇒ a = dk3, b = dk2 and c = dk
L.H.S.
= √ab – √bc + √cd
= √(dk³·dk²) – √(dk²·dk) + √(dk·d)
= d·k²√k – dk√k + d√k
= (k²-k+1)d√k.
R.H.S.
= √(a-b+c)(b-c+d)
= √{(dk³-dk²+dk)(dk²-dk+d)}
= √{d×d×k(k²-k+1)(k²-k+1)}
= (k²-k+1)d√k
L.H.S. = R.H.S.
Hence proved.
(iv) a/b = b/c = c/d = K (say)
∴ b = ka, c = k²a and d = k³a
L.H.S. = (3a + 5d)/(5a + 7d)
(3a+5k³a)/(5a+7k³a)
{a(3+5k³)}/{a(5+7k³)}
(3+5k³)/(5+7k³)
R.H.S. = (3a³ + 5d³) / (5a³ + 7d³)
(3a³+5(k³a)³)/(5a³+7(k³a)³) = (3a³+5k^9 a³)/(5a³+7k^9 a³)
{a³(3+5k^9)}/{a³(5+7k^9)}
(3+5k^9)/(5+7k^9)
When k=1, we have:
a=b=c=d
Thus, both the LHS and RHS reduce to:
(3+5)/(5+7)
= 8/12 = 2/3.
Que-5: If x/a = y/b = z/c, prove that [(ax – by)/(a + b)(x – y)] + [(by – cz)/ (b + c)(y – z)] + [(cz – ax) / (c + a)(z – x)] = 3
Sol: Let x/a = y/b = z/c = k
=> x = ak, y = bk, z = ck
L.H.S. = [(ax – by)/{(a+b)(x-y)}] + [(by – cz)/{(b+c)(y-z)}] + [(cz – ax)/{(c+a)(z-x)}]
= [(a.ak – b.bk)/{(a+b)(ak-bk)}] + [(b.bk – c.ck)/{(b+c)(bk-ck)}] + [(c.ck – a.ak)/{(c+a)(ck-ak)}]
= [(a²k-b²k)/{(a+b)k(a-b)}] + [(b²k²)/{(b+c)k(b-c)}] + [(c²k-a²k)/{(c+a)(c-a)}]
= [{k(a²-b²)}/{k(a²-b²)}] + [{k(b²-c²)}/{k(b²-c²)}] + [{k(c²-a²)}/{k(c²-a²)}]
= 1 + 1 + 1
= 3
= R.H.S.
Que-6: If x/(b + c – a) = y/(c + a – b) = z/(a + b – c), then show that each ratio is equal to (x + y + z) / (a + b + c).
Sol: x/(b + c – a) = y/(c + a – b) = z/(a + b – c) = k
x = k (b + c- a)
Y = k (c+a-b)
z = k (a+b-c)
Now ,
(x + y+z)/(a + b + c)
= [k(b + c – a)+k(c + a – b)+k(a + b – c)]/(a + b + c)
= [k(b + c – a + c + a – b + a + b – c)]/(a + b + c)
= [k(a + b + c)]/(a + b + c) = k
Hence ,
x/(b + c – a) = y/(c + a – b) = z/(a + b – c) = (x + y+z)/(a + b + c)
Proved.
Que-7: If a/(b + c) = b/(c + a) = c/(a + b), then prove that a(b – c) + b(c – a) + c(a – b) = 0
Sol: a/(b + c) = b/(c + a) = c/(a + b) = k
a = k(b + c)
b = k(c+a)
c= k(a+b)
a (b – c)+ b (c – a)+ c (a – b) = 0
LHS
a (b – c)+ b (c – a)+ c (a – b)
= k (b + c )(b – c) + k ( c + a)( c – a) + k (a + b )(a – b)
= k(b2 – C2) + k( c2 – a2) + k( a2 – b2)
= kb2 – kc2 + kc2 – ka2 + ka2 – kb2
= 0 = RHS
LHS = RHS. Hence, proved.
Que-8: If ax = by = cz, then prove that x2/yz + y2/zx + z2/xy = bc/a2 + ca/b2 + ab/c2.
Sol: Let ax = by = cz = k, then and x = k/a, y = k/b and z = k/c
L.H.S. = (x²/yz) + (y²/zx) + (z²/xy)
= [k²/{a²×(k/b)×(k/c)}] + [k²/{b²×(k/c)×(k/a)}] + [k²/{c²×(k/a)×(k/b)}]
= (bc/a²) + (ca/b²) + (ab/c²)
= R.H.S.
Hence proved.
— : Ratio and Proportion Class 10 OP Malhotra Exe-6C ICSE Maths Solutions :–
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