Division of Rs. 750 into 3 parts in the ratio 4 : 5 : 6 is
The sum of the ages of 3 persons is 150 years. 10 years ago their ages were in the ratio 7 : 8 : 9. Their present ages are
The numbers 14, 16, 35, 42 are not in proportion. The fourth term for which they will be in proportion is
If x/y = z/w, implies y/x = w/z, then the process is called
If p/q = r/s = p–r/q–s, the process is called
If a/b = c/d, implies (a+b)/(a–b) = (c+d)/(c–d), the process is called
If u/v = w/p, then (u–v)/(u+v) = (w–p)/(w+p). The process is called
12, 16, *, 20 are in proportion. Then * is
4, *, 9, 13½ are in proportion. Then * is
The mean proportional between 1.4 gms and 5.6 gms is
Let 1.4 gms, x gms and 5.6 gms be in continuous proportion.
(Mean proportional terms)^2 = Product of extremes.
Therefore x^2 = 1.4 x 5.6
therefore x^2 = 7.84
therefore x= 2.8 gms.
Two numbers are in the ratio 3 : 4; if 6 be added to each terms of the ratio, then the new ratio will be 4 : 5, then the numbers are
Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
if then (bc)x + (ca)y+(ab)z is
Let X=(b+ca);y=(c+ab);z=(a+bc)
=(bc)(b+ca)+(ca)(c+ab)+(ab)(a+bc)
=b^2c^2ab+ac+c^2a^2bc+ab+a^2b^2ac+bc
=0
4x^{–1/4} is expressed as
The value of 8^{1/3} is
The value of 2 × (32) ^{1/5} is
The value of 4/(32)^{1/5} is
The value of (8/27)^{1/3} is
2^{½} .4^{¾} is equal to
has simplified value equal to
x^{a–b} × x^{b–c} × x^{c–a} is equal to
The value of is equal to
{(3^{3})^{2} × (4^{2})^{3} × (5^{3})^{2}} / {(3^{2})^{3} × (4^{3})^{2} × (5^{2})^{3}} is
Which is True ?
If x^{1/p} = y^{1/q} = z^{1/r} and xyz = 1, then the value of p+q+r is
The value of y^{a–b} × y^{b–c} × y^{c–a} × y–^{a–b} is
The True option is
The simplified value of 16x^{–3}y^{2 }× 8^{–1}x^{3}y^{–2} is
The value of (8/27)^{–1/3} × (32/243)^{–1/5} is
The value of {(x+y)^{2/3} (x–y)^{3/2}/√x+y × √ (x–y)^{3}}^{6} is
Simplified value of (125)^{2/3} × √25 × ^{3}√5^{3 }× 5^{1/2} is
[{(2)^{1/2} . (4)^{3/4} . (8)^{5/6} . (16)^{7/8} . (32)^{9/10}}4]^{3/25} is
[1–{1–(1–x^{2})^{–1}}^{–1}]^{–1/2} is equal to
ANSWER : a
Solution : [1–{1–(1–x^{2})^{–1}}^{–1}]^{–1/2}
=[1  { 1 x^{2} 1/(1x^{2}) }^{1}]^{1/2} here we used a^{n} = 1/a^{n}
= [1  {(x^2)/(1x^2)}^1]^1/2
= [ 1  { x^2 /( x^2 1) }^{1}]^{1/2}
= [ 1 (x^{2}1)/ x^{2}]^{1/2}
= [ (x^{2} x^{2}+1)/ x^{2}]^{1/2}
= [1/x^{2}]^{1/2}
= [x^{2}]^{1/2}
= x ^{2*1/2}
= x
Ajay and Raj together have Rs. 1050. On taking Rs. 150 from Ajay, Ajay will have same amount as what Raj had earlier. Find the ratio of amounts with Ajay and Raj initially.
If a^{3}–b^{3 }= (a–b) (a^{2} + ab + b^{2}), then the simplified form of
Using (a–b)^{3} = a^{3}–b^{3}–3ab(a–b) tick the correct of these when x = p^{1/3} – p^{–1/3}
On simplification, 1/(1+a^{m–n}+a^{m–p}) + 1/(1+a^{n–m}+a^{n–p}) + 1/(1+a^{p–m}+a^{p–n}) is equal to
If A:B = 2:3, B:C = 4:5 and C:D = 6:7, then find the value of A:B:C:D
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