Sign in
Please select an account to continue using cracku.in
↓ →
Student Raju got 30 marks more than the student Raman and the marks of Raju were equal to the 60 percent of the sum of their marks. What are the marks obtained by Raju and Raman respectively?
If (5t + 3u) : (5t - 3u) = 5 : 3 then what is the value of t : u?
If p percent of p is 729, then what is the value of p?
If A percent of A is 9604, then what is the value of A?
The price of 3 notebooks and 6 pen is Rs. 3000. With the same money one can buy a notebooks and 12 pen, Raju wants to buy 15 pen, how much will he have to pay?
If 60 is subtracted from a number, then number becomes 85 percent of itself. What is the number?
Rewa has some hens and some goats. If the total number of animal heads is 100 and the total number of animal feet is 348, then what is the total number of goats with Rewa?
For what value of m will the system of equations 17x+my+102=0 and 23x+299y+138=0 have infinite number of solutions?
For what value of m will the system of equations 18x-72y+13=0 and 7x-my-17=0 have no solution?
If 20% of (A + B) = 30% of (A − B), then what percentage of B is equal to A?
If two LED TVs and one mobile phone cost ₹31,000, while two mobile phones and one LED TV cost ₹35,000, then the value of one mobile phone is:
If a - b = 5 and $$a^2 + b^2 = 45$$, then the value of ab is:
Given that a-b = 5 and $$a^2 +b^2 = 45 $$ then ab = ?
we know that $$ (a-b)^2 = a^2 +b^2 -2ab $$
putting the value of a-b and $$ a^2 +b^2 $$
so $$(5)^2 = 45 -2ab$$
$$\Rightarrow 25 = 45 -2ab $$
$$\Rightarrow 2ab = 20 $$
$$\Rightarrow ab =10 Ans $$
A and B are positive integers. If $$A + B + AB = 65$$, then what is the difference between $$A$$ and $$B (A, B \leq 15)?$$
$$A + B + AB = 65$$
$$A(1+B)+B=65$$
$$A=(65-B)/(1+B)$$
As A and B < 15
let B=10 then A=55/11
=5
Therefore A=5 and B=10 will satisfy and difference =10-5=5
If $$a^3 - b^3 = 1603$$ and $$(a - b) = 7$$, then $$(a + b)^2 - ab$$ is equal to:
If $$x + y + z = 22$$ and $$xy + yz + zx = 35$$, then what is the value of $$(x - y)^2 + (y - z)^2 + (z - x)^2$$?
If $$a^3 — b^3 = 208$$ and $$a - b = 4$$, then $$(a + b)^2 — ab$$ is equal to:
If $$\alpha$$ and $$\beta$$ are the roots of the equation $$x^2 + x - 1 = 0$$, then what is the equation whose roots are $$\alpha^5$$ and $$\beta^5$$?
If $$\frac{(x + y)}{z} = 2$$, then what is the value of $$\left[\frac{y}{(y - z)}\right] + \left[\frac{x}{(x - z)}\right]$$?
If $$x + \frac{1}{x} = 7$$, then $$x^3 + \frac{1}{x^3}$$ is equal to:
We have $$x+\frac{1}{x}=7$$
Taking cubes on both sides we get :
$$x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=343$$
we get $$x^3+\frac{1}{x^3}=343-21=322$$
If $$\alpha$$ and $$\beta$$ are the roots of equation $$x^2 - 2x + 4 = 0$$, then what is the equation whose roots are $$\frac{\alpha^3}{\beta^2}$$ and $$\frac{\beta^3}{\alpha^2}$$?
If $$ab + bc + ca = 8$$ and $$a+b +c = 12$$ then $$(a^2 + b^2 + c^2)$$ is equal to:
If one root of the equation $$Ax^2 +Bx + C = 0$$ is two and a half times the others, then which of the following is TRUE?
If $$x + (\frac{1}{x}) = \frac{(\surd3 + 1)}{2}$$, then what is the value of $$x^4 + (\frac{1}{x^4})$$?
If $$a + a^2 + a^3 - 1 = 0$$, then what is the value of $$a^3 + (\frac{1}{a})$$?
If $$x^2 - 12x + 33 = 0$$, then what is the value of $$(x - 4)^2 + \left[\frac{1}{(x - 4)^2}\right]$$?
If $$a - \left(\frac{1}{a}\right) = b, b - \left(\frac{1}{b}\right) = c$$ and $$c - \left(\frac{1}{c}\right) = a$$, then what is the value of $$\left(\frac{1}{ab}\right) + \left(\frac{1}{bc}\right) + \left(\frac{1}{ca}\right)$$?
If $$a^4 + 1 = \left[\frac{a^2}{b^2}\right] (4b^2 - b^4 - 1)$$, then what is the value of $$a^4 + b^4$$?
If $$3\sqrt{\frac{1-a}{a}} + 9 =19 - 3\sqrt{\frac{a}{1-a}}$$, then what is the value of $$a$$?
If the roots of the equation $$a(b - c)x^2 +b(c - a)x + c(a - b) = 0$$ are equal, then which of the following is true?
If $$a + b =10$$ and $$\sqrt{\frac{a}{b}} - 13 = -\sqrt{\frac{b}{a}} -11$$, then what is the value of $$3ab + 4a^2 + 5b^2$$
If $$[\surd(a^2 + b^2 + ab)] + [\surd(a^2 + b^2 - ab)] = 1$$, then what is the value of $$(1 - a^2)(1 - b^2)$$?
If $$3x + 4y - 11 = 18$$ and $$8x - 6y + 12 = 6$$, then what is the value of $$5x - 3y - 9$$?
we have:
3x+4y =29 (1)
8x-6y =-6 (2)
Multiplying (1) by 3 and (2) by 2 and adding we get
9x+12y+16x-12y =75
we get 25x=75
x=3
Now substituting x=3
we get y=5
Now 5x-3y-9 =15-15-9 =-9
If $$3x + 4y - 2z + 9 = 17, 7x + 2y + 11z + 8 = 23$$ and $$5x + 9y + 6z - 4 = 18$$, then what is the value of $$x + y + z - 34$$?
If $$3x + 5y + 7z = 49$$ and $$9x + 8y + 21z = 126,$$ then what is the value of y?
$$3x + 5y + 7z = 49$$
multiplying with 3 on both sides
$$9x + 15y + 21z = 147$$
$$9x + 8y + 21z = 126$$
Subtracting we get
7y=21
y=3
Cost of 4 pens, 6 note books and 9 files is Rs 305. Cost of 3 pens, 4 notebooks and 2 files is Rs 145. What is the cost (in Rs) of 5 pens, 8 notebooks and 16 files?
let the cost of pen be 'p',note book be 'n' and of file be 'f'
given 4p+6n+9f=305
8p+12n+18f=610
3p+4n+2f=145
Subtracting we get
5p+8n+16f=465
If $$2x + 3y - 5z = 18, 3x + 2y + z = 29$$ and $$x + y + 3z = 17$$, then what is the value of $$xy + yz + zx$$?
x+y+3z=17..........(1)
2x+3y-5z=18.......(2)
3x+2y+z=29........(3)
Now, (1)×2-(2) :
11z-y=6......(4)
And, (3)×2-(2)×3 :
17z-5y=4............(5)
Now,4×(4)-(5) :
z=2 .
Now, put this value in (4) ,we get :
y=6.
From equation (1) we get : x=5.
So,xy+yz+xz=10+30+12=52.
B is correct choice.
If $$3x + 6y + 9z = \frac{20}{3}, 6x + 9y + 3z = \frac{17}{3}$$ and $$18x + 27y - z = \frac{113}{9}$$, then what is the value of $$75x + 113y$$?
$$3x+6y+9z=\frac{20}{3}.$$
or, $$x+2y+3z=\frac{20}{9}...............\left(1\right)$$
$$6x+9y+3z=\frac{17}{3}.$$
or, $$2x+3y+z=\frac{17}{9}................\left(2\right)$$
$$18x+27y-z=\frac{113}{9}.$$
or, $$72x+108y-4z=\frac{452}{9}.......................\left(3\right)$$
By adding (1),(2) & (3) :
$$75x+113y=\frac{452}{9}+\frac{20}{9}+\frac{17}{9}=\frac{489}{9}=\frac{163}{3}.$$
A is correct choice.
If $$a + b + c = \frac{7}{12}, 3a - 4b + 5c = \frac{3}{4}$$ and $$7a - 11b - 13c = \frac{-7}{12}$$, then what is the value of $$a + c$$?
If $$x + 3y - \frac{2z}{4} = 6, x + \frac{2}{3}(2y + 3z) = 33$$ and $$\frac{1}{7}(x + y + z) + 2z = 9$$, then what is the value of $$46x +131y$$
If $$x + y + z = 1, x^2 + y^2 + z^2 = 2$$ and $$x^3 + y^3 + z^3 = 3$$,then what is the value of $$xyz$$?
$$\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\ .$$
Or, $$\left(1\right)^2=2+2\left(xy+yz+xz\right).$$
Or, $$\left(xy+yz+xz\right)=-\frac{1}{2}.$$
$$x^3+y^3+z^3=(x+y+z)[x^2+y^2+z^2−xy−yz−xz]+3xyz\ .$$
Or, $$3=(1)[2+\frac{1}{2}]+3xyz\ .$$
Or, $$3xyz=3-\frac{5}{2}=\frac{1}{2}\ .$$
Or, $$xyz=\frac{1}{6}\ .$$
B is correct choice.
If x + y = 1 and xy(xy - 2) = 12, then the value of $$x^4 + y^4$$ is:
Given that,
$$x + y = 1$$
$$\Rightarrow xy(xy-2)=12$$
$$\Rightarrow x^2 y^2-2xy=12$$
$$\Rightarrow x + y = 1$$
Now, squaring both side,
$$\Rightarrow (x + y)^2 = 1$$
$$\Rightarrow x^2+y^2+2xy = 1$$
$$\Rightarrow x^2+y^2 = 1-2xy$$
Again squaring both side,
$$\Rightarrow (x^2+y^2)^2 = (1-2xy)^2$$
$$\Rightarrow x^4+y^4+2x^2y^2 = 1+4x^2y^2-4xy$$
$$\Rightarrow x^4+y^4 = 1+4x^2y^2-4xy-2x^2y^2$$
$$\Rightarrow x^4+y^4 = 1+2x^2y^2-4xy$$
$$\Rightarrow x^4+y^4 = 1+2(x^2y^2-2xy)=1+2\times 12=25$$
If x+y =12 and xy = 27, x > y, then the value of $$(x^3 - y^3)$$ is:
Given that,
x+y =12 and xy = 27
We know that $$(x-y)=\sqrt{(x+y)^2-4xy}$$
$$\Rightarrow (x-y)=\sqrt{12^2-4\times 27}=\sqrt{144-108}$$
$$\Rightarrow (x-y)=\sqrt{36}=6$$
Now, We know that $$(x-y)^3=x^3-y^3-3xy(x-y)$$
$$\Rightarrow 6^3=x^3-y^3-3\times 27\times(6)$$
$$\Rightarrow 216=x^3-y^3-486$$
$$\Rightarrow x^3-y^3=486+216=702$$
The value of k for which the graph of (k -1) x + y - 2 = 0 and (2 - k) x - 3y + 1) = 0 are parallel is
The value of x which satisfies the equation $$\frac{x + a^2 + 2c^2}{b + c} + \frac{x + b^2 + 2a^2}{c + a} + \frac{x + c + 2b^2}{a + b} = 0$$ is
If $$N=(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}})$$, then what is the value of $$\frac{1}{N}$$ ?
Given : $$N=\frac{\sqrt7-\sqrt5}{\sqrt7+\sqrt5}$$
=> $$\frac{1}{N}=\frac{\sqrt7+\sqrt5}{\sqrt7-\sqrt5}$$
Rationalizing the denominator, we get :
= $$\frac{\sqrt7+\sqrt5}{\sqrt7-\sqrt5}\times\frac{\sqrt7+\sqrt5}{\sqrt7+\sqrt5}$$
= $$\frac{(\sqrt7+\sqrt5)^2}{(\sqrt7-\sqrt5)(\sqrt7+\sqrt5)}$$
= $$\frac{7+5+2(\sqrt7)(\sqrt5)}{7-5}$$
= $$\frac{12+2\sqrt{35}}{2}=6+\sqrt{35}$$
=> Ans - (B)
If $$\frac{1}{N}=\frac{(\sqrt{6}+\sqrt{5})}{\sqrt{6}-\sqrt{5}}$$, then what is the value of N ?
Given : $$\frac{1}{N}=\frac{\sqrt6+\sqrt5}{\sqrt6-\sqrt5}$$
=> $$N=\frac{\sqrt6-\sqrt5}{\sqrt6+\sqrt5}$$
Rationalizing the denominator, we get :
=> $$N=\frac{\sqrt6-\sqrt5}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}$$
=> $$N=\frac{(\sqrt6-\sqrt5)^2}{(\sqrt6+\sqrt5)(\sqrt6-\sqrt5)}$$
=> $$N=\frac{6+5-2(\sqrt6)(\sqrt5)}{6-5}$$
=> $$N=11-2\sqrt{30}$$
=> Ans - (C)
If N = (√6 - √5)/(√6 + √5), then what is the value of N + (1/N)?
Given N = (√6 - √5)/(√6 + √5)
Now, N + (1/N) = $$\frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$$ + $$\frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}}$$
$$\Rightarrow 6 + 5 - 2\sqrt{30} + 6 + 5 + 2\sqrt{30}$$ = 11 + 11 = 22
Hence, option D is the correct answer.
If P = ($$\sqrt{7}-\sqrt{6})/(\sqrt{7}+\sqrt{6})$$, then what is the value of P + (1/P)?
Given : $$P=\frac{\sqrt7-\sqrt6}{\sqrt7+\sqrt6}$$ ---------------(i)
=> $$\frac{1}{P}=\frac{\sqrt7+\sqrt6}{\sqrt7-\sqrt6}$$ ----------(ii)
Adding equations (i) and (ii),
=> $$P+\frac{1}{P}=(\frac{\sqrt7-\sqrt6}{\sqrt7+\sqrt6})+(\frac{\sqrt7+\sqrt6}{\sqrt7-\sqrt6})$$
= $$\frac{(\sqrt7-\sqrt6)^2+(\sqrt7+\sqrt6)^2}{(\sqrt7+\sqrt6)(\sqrt7-\sqrt6)}$$
= $$\frac{(13-2\sqrt{42})+(13+2\sqrt{42})}{7-6}$$
= $$13+13=26$$
=> Ans - (D)
Which value among $$\sqrt[4]{7}, \sqrt[3]{11}\ and\ \sqrt[12]{1257}\ $$is the largest ?
Terms : $$\sqrt[4]{7}, \sqrt[3]{11}\ and\ \sqrt[12]{1257}\ $$
L.C.M. of exponents (4,3,12) = 12
Multiplying the exponents by 12, we get :
$$\equiv(7)^{\frac{12}{4}},$$ $$(11)^{\frac{12}{3}}$$ and $$(1257)^{\frac{12}{12}}$$
$$\equiv(7)^3,(11)^4,(1257)^1$$
$$\equiv343,14641,1257$$
Thus, largest number = $$14641\equiv\sqrt[3]{11}$$
=> Ans - (A)
What is the simplified value of $$(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)$$?
Given $$(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)$$
Multiply and divide by (3 - 1)
$$\frac{(3 - 1)(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)}{3 - 1}$$
$$\frac{(3^{2} -1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)}{2}$$
$$\frac{(3^{8} - 1)(3^{8} + 1)(3^{16} + 1)}{2}$$
$$\frac{(3^{16} - 1)(3^{16} + 1)}{2}$$
$$\frac{3^{32} - 1}{2}$$
Hence, option A is the correct answer.
What is the value of $$\ \frac{(1.1)^{3}+(0.7)^{3}}{(1.1)^{2}-1.1\times0.7+(0.7)^{2}}$$?
Expression = $$\ \frac{(1.1)^{3}+(0.7)^{3}}{(1.1)^{2}-1.1\times0.7+(0.7)^{2}}$$
Let $$x=1.1$$ and $$y=0.7$$
= $$\frac{x^3+y^3}{x^2-xy+y^2}$$
= $$\frac{(x+y)(x^2-xy+y^2)}{x^2-xy+y^2}$$
= $$x+y=1.1+0.7=1.8$$
=> Ans - (D)
What is the value of positive square root of $$69+28\sqrt{5}$$ ?
Let $$x=69+28\sqrt5$$
=> $$x=69+2(2)(7)(\sqrt{5})$$
=> $$x=(49)+(20)+2(7)(2\sqrt5)$$
=> $$x=(7)^2+(2\sqrt5)^2+2(7)(2\sqrt5)$$
=> $$x=(7+2\sqrt5)^2$$
=> Positive square root of $$x=7+2\sqrt5$$
=> Ans - (A)
What is the value of $$\ \sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\sqrt{1+\frac{1}{4^{2}}+\frac{1}{5^{2}}}\ $$?
Expression = $$\ \sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\sqrt{1+\frac{1}{4^{2}}+\frac{1}{5^{2}}}\ $$
= $$\ \sqrt{1+\frac{1}{4}+\frac{1}{9}}+\sqrt{1+\frac{1}{9}+\frac{1}{16}}+\sqrt{1+\frac{1}{16}+\frac{1}{25}}\ $$
= $$\sqrt{\frac{36+9+4}{36}}+\sqrt{\frac{144+16+9}{144}}+\sqrt{\frac{400+25+16}{400}}$$
= $$\sqrt{\frac{49}{36}}+\sqrt{\frac{169}{144}}+\sqrt{\frac{441}{400}}$$
= $$\frac{7}{6}+\frac{13}{12}+\frac{21}{20}$$
= $$\frac{70+65+63}{60}=\frac{198}{60}=\frac{33}{10}$$
=> Ans - (D)
For what value of N, 270N will be a perfect square, where 270N is a 4 digit number??
Number = 270N
We know that $$(50)^2=2500$$ and $$(51)^2=2601$$
Thus, $$(52)^2=2704$$
=> $$N=4$$
=> Ans - (C)
What is the value of $$\ (203+107)^{2}-(203 - 107)^{2}\ $$?
Expression = $$\ (203+107)^{2}-(203 - 107)^{2}\ $$
Let $$x=(203+107)$$ and $$y=(203-107)$$
=> $$x^2-y^2=(x+y)(x-y)$$
= $$[(203+107)+(203-107)][(203+107)-(203-107)]$$
= $$406\times214=86884$$
=> Ans - (B)
What is the value of $$3^{2}+7^{2}+13^{2}+17^{2}-1^{2}-5^{2}-9^{2}-11^{2}-15^{2}$$?
Expression = $$3^{2}+7^{2}+13^{2}+17^{2}-1^{2}-5^{2}-9^{2}-11^{2}-15^{2}$$
= $$(9+49+169+289)-(1+25+81+121+225)$$
= $$516-453=63$$
=> Ans - (D)
Which value among √11 + √5, √14 + √2, √8 + √8 is the largest?
Given √11 + √5, √14 + √2, √8 + √8
If we add the numbers of each element we get same value i.e 16 (11 + 5 = 14 + 2 = 8 + 8 = 16)
Now multiply the numbers of each element,
11 x 5 = 55, 14 x 2 = 28, 8 x 8 = 64
Element with highest value will be the highest number
√8 + √8 is the highest value
Hence, option C is the correct answer.
If 46N is divisible by 18, then what is the value of N?
For a number to be divisible by 18, it should be even and sum of its digits should be divisible by 9.
Sum of digits of 46N = $$4+6+N=(N+10)$$
=> $$N+10=18$$
=> $$N=18-10=8$$
=> Ans - (D)
What is the value of $$\frac{(0.5^{3})-(0.1)^{3}}{(0.5)^{2}+0.5 \times 0.1+(0.1)^{2}}$$?
Given $$\frac{(0.5^{3})-(0.1)^{3}}{(0.5)^{2}+0.5 \times 0.1+(0.1)^{2}}$$......(1)
we know that $$a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$$......(2)
Substitute (2) in (1)
$$\frac{(0.5-0.1)(0.5)^{2} + 0.5 \times 0.1 + (0.1)^{2}}{(0.5)^{2}+0.5 \times 0.1+(0.1)^{2}}$$
$$0.5 - 0.1 = 0.4$$
Hence, option B is the correct answer.
What is the value of $$\frac{(0.7)^{3}-(0.4)^{3}}{(0.7)^{2}+\ 0.7\times0.4+(0.4)^{2}}\ $$?
Expression = $$\frac{(0.7)^{3}-(0.4)^{3}}{(0.7)^{2}+\ 0.7\times0.4+(0.4)^{2}}\ $$
Let $$x=0.7$$ and $$y=0.4$$
= $$\frac{x^3-y^3}{x^2+xy+y^2}$$
= $$\frac{(x-y)(x^2+xy+y^2)}{x^2+xy+y^2}$$
= $$x-y=0.7-0.4=0.3$$
=> Ans - (A)
What is the value of $$999\frac{1}{3}+999\frac{1}{6}+999\frac{1}{12}+999\frac{1}{20}+999\frac{1}{30}$$?
Expression : $$999\frac{1}{3}+999\frac{1}{6}+999\frac{1}{12}+999\frac{1}{20}+999\frac{1}{30}$$
= $$(999+999+999+999+999)+(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30})$$
= $$(4995)+(\frac{20+10+5+3+2}{60})$$
= $$4995+\frac{40}{60}$$
= $$4995\frac{4}{6}$$
=> Ans - (D)
If 2A = 3B, then what is the value of (A + B)/A?
Given, 2A = 3B i.e
$$\frac{A}{B} = \frac{3}{2}$$ (or) $$\frac{B}{A} = \frac{2}{3}$$
Now, $$\frac{A+B}{A} \Rightarrow (\frac{A}{A} + \frac{B}{A}$$) $$\Rightarrow (1 + \frac{2}{3}) = \frac{5}{3}$$
Hence, option D is the correct answer.
If 3A = 5B, then what is the value of (A + B)/B?
Given : $$3A=5B$$
=> $$\frac{A}{B}=\frac{5}{3}$$
Let $$A=5$$ and $$B=3$$
To find : $$\frac{(A+B)}{B}$$
= $$\frac{(5+3)}{3}=\frac{8}{3}$$
=> Ans - (A)
If a - b : b - c : c - d = 1 : 2 : 3, then what is the value of (a + d) : c?
Given = $$(a-b):(b-c):(c-d)=1:2:3$$
=> $$\frac{a-b}{b-c}=\frac{1}{2}$$
=> $$2a-2b=b-c$$
=> $$2a+c=3b$$ -----------(i)
Similarly, $$\frac{b-c}{c-d}=\frac{2}{3}$$
=> $$3b-3c=2c-2d$$
=> $$3b=5c-2d$$ ----------(ii)
Substituting above value in equation (i), we get :
=> $$2a+c=5c-2d$$
=> $$2a+2d=5c-c$$
=> $$2(a+d)=4c$$
=> $$\frac{a+d}{c}=\frac{4}{2}=2$$
$$\therefore$$ $$(a+d):c=2:1$$
=> Ans - (B)
A trader sells two articles for Rs 14784 each. If he gains 12% on one and losses 12% on other, then what is the value (in Rs) of the loss?
Selling price of each article = Rs. 14,784
Profit % on one article = 12%
=> Cost price of first article = $$\frac{14784}{(100+12)}\times100$$
= $$\frac{14784}{1.12}=Rs.$$ $$13,200$$
Similarly, cost price of second article = $$\frac{14784}{(100-12)}\times100$$
= $$\frac{14784}{0.88}=Rs.$$ $$16,800$$
Thus, total cost price = $$13,200+16,800=Rs.$$ $$30,000$$
Total selling price = $$14,784+14,784=Rs.$$ $$29,568$$
$$\therefore$$ Total loss = $$30,000-29,568=Rs.$$ $$432$$
=> Ans - (C)
P is 20% more than Q and 40% less than R. If value of Q is Rs 150, then what is the value of R (in Rs)?
Given P is 20% more than Q,
P = 120% of Q (or) P = (6/5) x 150 (Q = 150)
P = 180
P is 40% less than R,
P = 60% of R (or) 180 = (3/5) x R
R = 300
Hence, option A is the correct answer.
X is 30% more than Y and 25% less than Z. If value of Y is Rs 300, then what is the value (in Rs) of Z ?
Given : $$Y=300$$
Also, X is 30% more than Y
=> $$X=300+(\frac{30}{100}\times300)$$
= $$300+90=390$$
And, X is 25% less than Z
=> $$Z=\frac{390}{(100-25)}\times100$$
= $$\frac{390}{3}\times4=520$$
=> Ans - (D)
The price of motor cycle depreciates every year by 10%. If the value of the motor cycle after 3 years will be Rs 36450, then what is the present value (in Rs) of the motor cycle?
Let the present value of motor cycle = Rs. $$x$$
Rate of depreciation = 10%
=> Value of the motor cycle after 3 years = $$x\times(1-\frac{10}{100})\times(1-\frac{10}{100})(1-\frac{10}{100})$$
=> $$x\times(\frac{90}{100})^3=36450$$
=> $$x=36450\times\frac{100^3}{90^3}$$
=> $$x=0.05\times1000000=50000$$
$$\therefore$$ The present value (in Rs) of the motor cycle = Rs. 50,000
=> Ans - (B)
The price of motorcycle depreciates every year by 8%. If the value of the motorcycle after 2 years will be Rs 84640, then what is the present value (in Rs) of the motorcycle?
Given that price of motorcycle depreciates every year by 8% then after 2 years,
92% of 92% of $$x = 84,640$$
$$\frac{92}{100}$$ x $$\frac{92}{100}$$ x $$x = 84,640$$
$$x = 100000$$
Hence, option D is the correct answer.
A certain sum becomes Rs 1020 in 5 years and Rs 1200 in 8 years at simple interest. What is the value of principal?
Simple interest for 3 years = 1200 - 1020 = 180 (or 60 for 1 year)
Then simple interest for 5 years = 300
Sum becomes 1020 in 5 years, hence, principal amount = 1020 - 300 = 720.
Hence, option D is the correct answer.
If the angles of a triangle are (2x - 8)$$^{o}$$, (2x + 18)$$^{o}$$and 6x$$^{o}$$. What is the value of 3x (in degrees)?
Sum of angles of a triangle = $$180^\circ$$
=> $$(2x-8)^\circ+(2x+18)^\circ+(6x)^\circ=180^\circ$$
=> $$10x+10=180$$
=> $$10x=180-10=170$$
=> $$x=\frac{170}{10}=17$$
$$\therefore$$ $$3x=3\times17=51$$
=> Ans - (C)
If $$(1/x) + (1/y) + (1/z) = 0$$ and $$x + y + z = 11$$, then what is the value of $$x^{3}+y^{3}+z^{3}-3xyz$$ ?
Given : $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$
=> $$\frac{yz+zx+xy}{xyz}=0$$
=> $$xy+yz+zx=0$$ -------------(i)
Also, $$x+y+z=11$$ ------------(ii)
Squaring both sides, we get :
=> $$(x+y+z)^2=(11)^2$$
=> $$(x^2+y^2+z^2)+2(xy+yz+zx)=121$$
Substituting value from equation (i),
=> $$x^2+y^2+z^2=121$$ ------------(iii)
To find : $$x^{3}+y^{3}+z^{3}-3xyz$$
= $$(x+y+z)[(x^2+y^2+z^2)-(xy+yz+zx)]$$
Substituting values from equations (i), (ii) and (iii),
= $$(11)(121-0)$$
= $$11\times121=1331$$
=> Ans - (A)
If $$x^{2} -7x + 1 = 0$$, then what is the value of $$x + \frac{1}{x}$$?
Given : $$x^2-7x+1=0$$
Dividing both sides by $$'x'$$
=> $$x-7+\frac{1}{x}=0$$
=> $$x+\frac{1}{x}=7$$
=> Ans - (A)
If $$\ (x/y)^{5a-3}\ $$ =$$\ (y/x)^{17-3a}$$, then what is the value of a?
Given : $$\ (x/y)^{5a-3}\ $$ =$$\ (y/x)^{17-3a}$$
=> $$\ (x/y)^{5a-3}\ =\ (x/y)^{3a-17}$$
=> $$5a-3=3a-17$$
=> $$5a-3a=-17+3$$
=> $$2a=-14$$
=> $$a=\frac{-14}{2}=-7$$
=> Ans - (A)
What is the value of $$[\frac{1}{1-x^{(p-q)}}+\frac{1}{1-x^{(q-p)}}]\ $$?
Expression = $$[\frac{1}{1-x^{(p-q)}}+\frac{1}{1-x^{(q-p)}}]\ $$
= $$(\frac{1}{1-\frac{x^p}{x^q}})+(\frac{1}{1-\frac{x^q}{x^p}})$$
= $$(\frac{x^q}{x^q-x^p})+(\frac{x^p}{x^p-x^q})$$
= $$(\frac{x^q}{x^q-x^p})-(\frac{x^p}{x^q-x^p})$$
= $$\frac{x^q-x^p}{x^q-x^p}=1$$
=> Ans - (B)
When a = 61, b = 63 and c = 65, then what is the value of a$$^{3}$$ + b$$^{3}$$ + c$$^{3}$$ - 3abc?
We know that $$(a + b + c)^{3} - 3abc= \frac{1}{2}(a + b + c)((a - b)^{2} + (b - c)^{2} + (c - a)^{2})$$
$$\Rightarrow \frac{1}{2} (61 + 63 + 65)((61 - 63)^{2}$$ + $$(63 - 65)^{2}$$ + $$(65 - 61)^{2})$$
$$\Rightarrow \frac{1}{2} (189)(2^{2} + 2^{2} + 4^{2})$$ = $$ \frac{1}{2} (189)(24) = 2268$$
Hence, option B is the correct answer.
If $$x^{2}- 2\sqrt{10}x+ 1 = 0$$, then what is the value of $$x - \frac{1}{x}$$?
Given : $$x^2-2\sqrt{10}x+1=0$$
Dividing both sides by $$'x'$$
=> $$x+\frac{1}{x}=2\sqrt{10}$$
Squaring both sides, we get :
=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=40$$
=> $$x^2+\frac{1}{x^2}=40-2=38$$
=> $$(x-\frac{1}{x})^2+2(x)(\frac{1}{x})=38$$
=> $$(x-\frac{1}{x})^2=38-2=36$$
=> $$x-\frac{1}{x}=\sqrt{36}=6$$
=> Ans - (B)
If $$\ x^{2}\ - 3x + 1 = 0$$, then what is the value of $$x + \frac{1}{x} $$?
Given : $$x^2-3x+1=0$$
Dividing both sides by $$'x'$$
=> $$x-3+\frac{1}{x}=0$$
=> $$x+\frac{1}{x}=3$$
=> Ans - (A)
What is the simplified value of $$\ [\frac{(1+x^{3})}{(x^{2}-1)}\div\frac{(x^{2}+1-x)}{(x+1)}] \times (x-1)$$?
Expression = $$\ [\frac{(1+x^{3})}{(x^{2}-1)}\div\frac{(x^{2}+1-x)}{(x+1)}]\times(x-1)$$
= $$[\frac{1+x^3}{(x-1)(x+1)}\times\frac{x+1}{x^2+1-x}](x-1)$$
= $$[\frac{(x+1)(x^2+1-x)}{(x-1)}\times\frac{1}{(x^2+1-x)}](x-1)$$
= $$\frac{(x+1)}{(x-1)}\times(x-1)=(x+1)$$
=> Ans - (C)
What is the value of $$\ (\frac{x^{2}-x-6}{x^{2}+x-12})\div(\frac{x^{2}+5x+6}{x^{2}+7x+12})?$$
Expression = $$\ (\frac{x^{2}-x-6}{x^{2}+x-12})\div(\frac{x^{2}+5x+6}{x^{2}+7x+12})$$
= $$\ (\frac{x^{2}-x-6}{x^{2}+x-12})\times(\frac{x^{2}+7x+12}{x^{2}+5x+6})$$
= $$\ (\frac{(x-3)(x+2)}{(x+4)(x-3)})\times(\frac{(x+4)(x+3)}{(x+3)(x+2)})$$
= $$1$$
=> Ans - (A)
If $$a + b + c = - 11$$, then what is the value of $$\ (a + 4)^{3}+(b+ 5)^{3}+(c + 2)^{3}-3(a + 4)(b + 5)(c + 2)\ $$?
Given : $$a+b+c=-11$$
=> $$a+b+c+11=0$$
=> $$(a+4)+(b+5)+(c+2)=0$$
Let $$(a+4)=x$$, $$(b+5)=y$$ and $$(c+2)=z$$
=> $$x+y+z=0$$ -----------(i)
To find : $$\ (a + 4)^{3}+(b+ 5)^{3}+(c + 2)^{3}-3(a + 4)(b + 5)(c + 2)\ $$
= $$x^3+y^3+z^3-3xyz$$
= $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
Substituting value from equation (i), we get :
= $$(0)(x^2+y^2+z^2-xy-yz-zx)=0$$
=> Ans - (C)
If $$\ \frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}$$= 0, then what is the value of $$\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ $$?
Given : $$\ \frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}=0$$
=> $$(3-\frac{1}{x})+(5-\frac{1}{y})+(7-\frac{1}{z})=0$$
=> $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3+5+7$$
=> $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=15$$
=> Ans - (C)
If $$\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}$$= 62, then what is the value of $$x$$ ($$x < 0$$) ?
Expression : $$\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}=62$$
=> $$\frac{(x+\sqrt{x^2-1})^2+(x-\sqrt{x^2-1})^2}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}=62$$
=> $$\frac{(x^2+x^2-1+2x\sqrt{x^2-1})+(x^2+x^2-1-2x\sqrt{x^2-1})}{(x^2)-(x^2-1)}=62$$
=> $$\frac{4x^2-2}{1}=62$$
=> $$4x^2-2=62$$
=> $$4x^2=62+2=64$$
=> $$x^2=\frac{64}{4}=16$$
=> $$x=\sqrt{16}=\pm4$$
$$\because x<0$$, => $$x=-4$$
=> Ans - (A)
If$$\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}\ $$for x > 0, then what is the value of $$\ x^{3}+\frac{1}{x^{3}}\ $$
Given : $$\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}\ $$
=> $$(x+\frac{1}{x})^2-2(x)(\frac{1}{x})=\frac{7}{4}$$
=> $$(x+\frac{1}{x})^2=\frac{7}{4}+2=\frac{15}{4}$$
=> $$x+\frac{1}{x}=\frac{\sqrt{15}}{2}$$
Cubing both sides, we get :
=> $$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=(\frac{\sqrt{15}}{2})^3$$
=> $$x^3+\frac{1}{x^3}+3(\frac{\sqrt{15}}{2})=\frac{15\sqrt{15}}{8}$$
=> $$x^3+\frac{1}{x^3}=\frac{15\sqrt{15}}{8}-\frac{3\sqrt{15}}{2}$$
=> $$x^3+\frac{1}{x^3}=\frac{3\sqrt{15}}{8}$$
=> Ans - (C)
If$$\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$$, then what is the value of $$\frac{x}{x+1}+\frac{4}{y+2}+\frac{z}{z+2662}?$$
Given $$\frac{1}{x+2}=\frac{1}{3}$$
We get x = 1 .....(1)
$$\frac{3}{y+3}=\frac{1}{3}$$
we get y = 6 .....(2)
$$\frac{1331}{z+1331}=\frac{1}{3}$$
we get z = 2662 ....(3)
We need to find $$\frac{x}{x+1}+\frac{3}{y+3}+\frac{z}{z+2662}$$
Substitute equations (1), (2) and (3) in the above equation
= $$\frac{1}{1+1}+\frac{3}{6+3}+\frac{2662}{2662+2662}$$
= $$\frac{1}{2}+\frac{3}{9}+\frac{1}{2}$$
= $$\frac{3}{2}$$
Hence, option C is the correct answer.
If $$\ \sqrt{7x+12}+\sqrt{7x-12}=3+\sqrt{33}\ $$, then what is the value of x?
Expression : $$\ \sqrt{7x+12}+\sqrt{7x-12}=3+\sqrt{33}\ $$
$$(7x-12)$$ must be greater than equal to 0.
=> $$7x-12\geq0$$
=> $$x\geq\frac{12}{7}$$
Thus, first two options are eliminated. Putting $$x=3$$ in above equation,
=> $$\sqrt{7(3)+12}+\sqrt{7(3)-12}$$
=> $$\sqrt{33}+\sqrt9$$
= $$3+\sqrt{33}=$$ R.H.S.
=> Ans - (C)
If $$\ x^{2}\ - 3x + 1 = 0$$, then what is the value of $$\ x^{4}\ $$+ $$\ \frac{1}{x^{4}}$$?
Given : $$x^2-3x+1=0$$
Dividing both sides by $$'x'$$
=> $$x+\frac{1}{x}=3$$
Squaring both sides, we get :
=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=9$$
=> $$x^2+\frac{1}{x^2}=9-2=7$$
Again squaring both sides,
=> $$x^4+\frac{1}{x^4}+2(x^2)(\frac{1}{x^2})=49$$
=> $$x^4+\frac{1}{x^4}=49-2=47$$
=> Ans - (C)
If $$x^{2} - 8x + 1 = 0$$, then what is the value of $$x^{2}+\frac{1}{x^{2}}$$?
Given : $$x^2-8x+1=0$$
Dividing both sides by $$'x'$$
=> $$x+\frac{1}{x}=8$$
Squaring both sides, we get :
=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=64$$
=> $$x^2+\frac{1}{x^2}=64-2=62$$
=> Ans - (D)
If$$\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}$$ for x > 0, then what is the value of x + $$\frac{1}{x}$$?
Let,
$$(x + \frac{1}{x})^{2}$$ = $$\ x^{2}+\frac{1}{x^{2}} + 2(x)(\frac{1}{x})$$
$$(x + \frac{1}{x})^{2}$$ = $$\frac{7}{4} + 2$$
$$(x + \frac{1}{x})^{2}$$ = $$\frac{15}{4}$$
$$(x + \frac{1}{x})$$ = $$\frac{\sqrt{15}}{2}$$
Hence, option B is the correct answer.
If$$x^{2} - 3x + 1 = 0$$, then what is the value of $$x^{2}+ \frac{1}{x^{2}}$$?
Given : $$x^2-3x+1=0$$
Dividing both sides by $$'x'$$
=> $$x+\frac{1}{x}=3$$
Squaring both sides, we get :
=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=9$$
=> $$x^2+\frac{1}{x^2}=9-2=7$$
=> Ans - (B)
If $$x + \frac{1}{x} = 3\sqrt{2}$$, then what is the value of $$x^{5}+\frac{1}{x^5}$$
Given : $$x+\frac{1}{x}=3\sqrt2=k$$
Now, $$x^5+\frac{1}{x^5}=[(x^3+\frac{1}{x^3})\times(x^2+\frac{1}{x^2})]-(x+\frac{1}{x})$$
= $$[(x+\frac{1}{x})^3-3(x+\frac{1}{x})\times(x+\frac{1}{x})^2-2(x)(\frac{1}{x})]-(x+\frac{1}{x})$$
= $$[(k^3-3k)\times(k^2-2)]-(k)$$
= $$[(54\sqrt2-9\sqrt2)\times(18-2)]-(3\sqrt2)$$
= $$(45\sqrt2\times16)-3\sqrt2$$
= $$720\sqrt2-3\sqrt2=717\sqrt2$$
=> Ans - (D)
If x + y + z = 0, then what is the value of $$\frac{x^{2}}{yz}+\frac{y^{2}}{xz}+\frac{z^{2}}{xy}$$?
Given x + y + z = 0. So, x + y = -z, y + z = -x, z + x = -y -----------(1)
$$\frac{x^{2}}{yz}+\frac{y^{2}}{xz}+\frac{z^{2}}{xy}$$
$$\frac{x^{3} + y^{3} + z^{3}}{xyz}$$ -----------(2)
We know that, $$a^{3} + b^{3} + c^{3} = (a+b+c)^{3} - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 6abc$$
Hence, equation (2) can be written as,
$$\frac{(x+y+z)^{3} - 3xy(x+y) - 3yz(y+z) - 3zx(x+z) - 6xyz}{xyz}$$
Now substitute equation (1) in the above equation,
$$\frac{(0)^{3} + 3xy(z) + 3yz(x) + 3zx(y) - 6xyz}{xyz}$$
$$\frac{3xyz}{xyz}$$ = 3
Hence, option D is the correct answer.
If $$\ x^{4}+\frac{1}{x^{4}}\ $$= 98 and $$x > 1$$, then what is the value of $$x-\frac{1}{x}\ ?$$
Given : $$\ x^{4}+\frac{1}{x^{4}}\ =98$$
=> $$(x^2+\frac{1}{x^2})^2-2(x^2)(\frac{1}{x^2})=98$$
=> $$(x^2+\frac{1}{x^2})^2=98+2=100$$
=> $$x^2+\frac{1}{x^2}=\sqrt{100}=10$$
=> $$(x-\frac{1}{x})^2+2(x)(\frac{1}{x})=10$$
=> $$(x-\frac{1}{x})^2=10-2=8$$
=> $$x-\frac{1}{x}=\sqrt8=2\sqrt2$$
=> Ans - (B)
What is the simplified value of $$\ (x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x-\frac{1}{x})(x^{16}+\frac{1}{x^{16}})(x+\frac{1}{x})(x^{4}+\frac{1}{x^{4}})$$?
Expression = $$\ (x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x-\frac{1}{x})(x^{16}+\frac{1}{x^{16}})(x+\frac{1}{x})(x^{4}+\frac{1}{x^{4}})$$
= $$\ (x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^2-\frac{1}{x^2})(x^{16}+\frac{1}{x^{16}})(x^{4}+\frac{1}{x^{4}})$$
Multiply and divide by $$(x^2+\frac{1}{x^2})$$, we get :
= $$\frac{1}{x^2+\frac{1}{x^2}}\times\ (x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^2-\frac{1}{x^2})(x^2+\frac{1}{x^2})(x^{16}+\frac{1}{x^{16}})(x^{4}+\frac{1}{x^{4}})$$
= $$\frac{1}{x^2+\frac{1}{x^2}}\times\ (x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^{16}+\frac{1}{x^{16}})(x^{4}+\frac{1}{x^{4}})(x^4-\frac{1}{x^4})$$
= $$\frac{1}{x^2+\frac{1}{x^2}}\times\ (x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x^{16}+\frac{1}{x^{16}})(x^{8}-\frac{1}{x^{8}})$$
= $$\frac{1}{x^2+\frac{1}{x^2}}\times\ (x^{32}+\frac{1}{x^{32}})(x^{16}+\frac{1}{x^{16}})(x^{16}-\frac{1}{x^{16}})$$
= $$\frac{1}{x^2+\frac{1}{x^2}}\times\ (x^{32}+\frac{1}{x^{32}})(x^{32}-\frac{1}{x^{32}})$$
= $$\frac{1}{x^2+\frac{1}{x^2}}\times\ (x^{64}-\frac{1}{x^{64}})$$
=> Ans - (B)
What is the value of $$\ \frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$$?
Expression : $$\ \frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$$
= $$\ \frac{(a-b)[(a^{2}+b^{2})-(a-b)]}{(ab)(a-b)}$$
= $$\frac{(a^2+b^2)-(a-b)}{ab}$$
= $$\frac{(a-b)^2+2ab-(a-b)}{ab}$$
If P is the circum-center in ΔABC, ∠BPC = 130°, then what is the value (in degrees) of ∠BAC?
Given : P is the circum-center in ΔABC and ∠BPC = 30°
To find : $$\angle$$ BAC = ?
Solution : In a circle, angle subtended by an arc at the centre is double the angle subtended by the same arc on any other point on the circle
=> $$\angle BPC = 2 \times \angle BAC$$
=> $$\angle$$ BAC = $$\frac{130^\circ}{2}=65^\circ$$
=> Ans - (C)
In the given figure, O is the center of the circle, $$\ \angle$$PQR = 100$$^\circ\ $$and $$\angle$$STR = 105$$^\circ$$. What is the value (in degrees) of $$\ \angle$$OSP?
Given : $$\ \angle$$PQR = 100$$^\circ\ $$and $$\angle$$STR = 105$$^\circ$$
To find : $$\ \angle$$OSP = ?
Solution : Quadrilateral PQRS is cyclic quadrilateral, hence opposite angles are supplementary.
=> $$\angle$$ PQR + $$\angle$$ PSR = $$180^\circ$$
=> $$\angle$$ PSR = $$180^\circ-100^\circ=80^\circ$$ --------------(i)
Also, angle at the centre is double the angle at any point on the circumference of the circle in the same segment.
=> reflex ($$\angle$$ SOR) = $$2$$ $$\times$$ $$\angle$$ STR
=> reflex ($$\angle$$ SOR) = $$2\times105^\circ=210^\circ$$
Thus, $$\angle$$ SOR = $$360^\circ-210^\circ=150^\circ$$
Now, in $$\triangle$$ SOR, OS = OR = radius
=> $$\angle$$ OSR = $$\angle$$ ORS = $$15^\circ$$ ----------(ii)
Subtracting equation (ii) from (i), we get :
$$\therefore$$ $$\angle$$ OSP = $$80^\circ-15^\circ=65^\circ$$
=> Ans - (D)
In ΔPQR, ∠P : ∠Q : ∠R = 1 : 3 : 5. What is the value (in degrees) of ∠R - ∠P?
Let, ∠P = 1x, ∠Q = 3x, ∠R = 5x
Sum of interior angles of a triangle = 180 degrees i.e
∠P + ∠Q + ∠R = 180 degrees
$$\Rightarrow$$ 9x = 180 (or) x = 20
∠R - ∠P = 5x - 1x
∠R - ∠P = 4x (or) 4 x 20
∠R - ∠P = 80
Hence, option B is the correct answer.
In the given figure, EF = CE = CA, What is the value (in degrees) of $$\angle$$EAC?
Given : EF = CE = CA
=> $$\angle$$ CAE = $$\angle$$ CEA = $$x$$ and $$\angle$$ ECF = $$\angle$$ EFC = $$y$$
To find : $$\angle$$ EAC = $$x=?$$
Solution : Using exterior angle property, => $$\angle$$ CAE + $$\angle$$ CFE = $$\angle$$ ACD
=> $$x+y=96^\circ$$ -----------------(i)
Also, $$\angle$$ CEF = $$(180^\circ-2y)=180^\circ-x$$
=> $$x=2y$$ ------------(ii)
Substituting it in equation (i), => $$2y+y=3y=96^\circ$$
=> $$y=\frac{96}{3}=32^\circ$$
$$\therefore$$ $$x=2\times32=64^\circ$$
=> Ans - (B)
If ΔPQR is right angled at Q, PQ = 12 and ∠PRQ = 30°, then what is the value of QR?
Given : ΔPQR is right angled at Q, PQ = 12 and ∠PRQ = 30°
To find : QR = ?
Solution : $$tan(\angle R)=\frac{PQ}{QR}$$
=> $$tan(30^\circ)=\frac{12}{QR}$$
=> $$\frac{1}{\sqrt3}=\frac{12}{QR}$$
=> $$QR=12\sqrt3$$ cm
=> Ans - (A)
In ΔABC, ∠C = 54°, the perpendicular bisector of AB at D meets BC at E. If ∠EAC = 42°, then what is the value (in degrees) of ∠ABC?
Given : ED is the perpendicular bisectors of AB, $$\angle C=54^\circ$$ and $$\angle EAC=y=42^\circ$$
To find : $$\angle B=x=?$$
Solution : In $$\triangle$$ EAC, using exterior angle property,
=> $$\angle$$ AEB = $$\angle$$ C + $$y$$
=> $$\angle$$ AEB = $$54^\circ+42^\circ=96^\circ$$
Thus, in $$\triangle$$ AEB, => $$x+z=180^\circ-96^\circ=84^\circ$$ ------------(i)
Also, in $$\triangle$$ EAD and $$\triangle$$ BDE
AD = DB (DE bisects AB)
$$\angle$$ EAD = $$\angle$$ EDB = $$90^\circ$$
DE = DE (Common)
Thus, $$\triangle$$ EAD $$\cong$$ $$\triangle$$ BDE (By SAS criterion)
=> $$x=z$$ (By CPCT)
Substituting above value in equation (i), we get :
=> $$x+x=2x=84^\circ$$
=> $$x=\frac{84}{2}=42^\circ$$
=> Ans - (B)
In the given figure, O is the centre of the circle, $$\angle$$PQO=30 $$^\circ$$and$$\ \angle$$\ QRO = $$\ 45^\circ\$$. What is the value (in degrees) of $$\angle$$POR?
In an isosceles triangle DEF, ∠D = 130°. If I is the incentre of the triangle, then what is the value (in degrees) of ∠EIF?
Given : I is the incentre of $$\triangle$$ DEF and $$\angle$$ D = 130°
To find : $$\angle$$ EIF = $$\theta$$ = ?
Incentre of a triangle = $$90^\circ+\frac{\angle D}{2}$$
=> $$\theta=90^\circ+\frac{130^\circ}{2}$$
=> $$\theta=90^\circ+65^\circ$$
=> $$\theta=155^\circ$$
=> Ans - (C)
In an isosceles triangle PQR, ∠P = 130$$^\circ$$. If I is the in-centre of the triangle, then what is the value (in degrees) of ∠QIR?
Given : I is the incentre of $$\triangle$$ PQR and $$\angle$$ BAC = 130°
To find : $$\angle$$ QIR = $$\theta$$ = ?
Incentre of a triangle = $$90^\circ+\frac{\angle P}{2}$$
=> $$\theta=90^\circ+\frac{130^\circ}{2}$$
=> $$\theta=90^\circ+65^\circ$$
=> $$\theta=155^\circ$$
=> Ans - (C)
In the given figure, O is the centre of the circle, OQ is perpendicular to RS and $$\angle$$SRT = $$30^\circ\ $$. If RS=10$$\surd2$$, then what is the value of PR$$^{2}$$?
$$2\frac{1}{5}x^{2}\ $$= 2750, find the value of x ?
$$2\frac{1}{5}x^{2}=2750$$
==> $$\frac{11}{5}x^{2}=2750$$
==> $$x^{2}= \frac{2750\times5}{11}$$
==> $$x^{2}= 1250$$
==> $$x = \sqrt{1250}=\sqrt{625\times2}$$
==> $$x = 25\sqrt{2}$$
If ΔABC is right angled at B, AB = 30 and ∠ACB = 60°, then what is the value of AC?
It is given that $$\triangle$$ ABC is a right angled at B, AB = 30 cm and $$\angle C=60^\circ$$
Now, in right $$\triangle$$ ABC,
=> $$sin(\angle C)=\frac{AB}{AC}$$
=> $$sin(60^\circ)=\frac{30}{AC}$$
=> $$\frac{\sqrt3}{2}=\frac{30}{AC}$$
=> $$AC=\frac{2}{\sqrt3}\times30=20\sqrt3$$ cm
=> Ans - (B)
In the following figure, O is the centre of the circle and $$\ \angle$$PRQ = $$\ 50^\circ$$. What is the value (in degrees) of $$\ \angle$$PTQ?
Given : $$\ \angle$$PRQ = $$\ 50^\circ$$
To find : $$\ \angle$$PTQ = ?
Solution : Quadrilateral PRQT is a cyclic quadrilateral, in which opposite angles are supplementary.
=> $$\angle$$ PRQ + $$\angle$$ PTQ = $$180^\circ$$
=> $$\angle$$ PTQ = $$180^\circ-50^\circ=130^\circ$$
=> Ans - (C)
If 2cos$$^{2}$$ θ - 1 = 0 and θ is acute, then what is the value of (cot$$^{2}$$ θ - tan$$^{2}$$ θ)?
Given : $$2cos^2\theta-1=0$$
=> $$cos^2\theta=\frac{1}{2}$$
=> $$cos\theta=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt2}$$
=> $$\theta=cos^{-1}(\frac{1}{\sqrt2})=45^\circ$$
$$\therefore$$ $$cot^2\theta-tan^2\theta$$
= $$cot^2(45^\circ)-tan^2(45^\circ)$$
= $$1-1=0$$
=> Ans - (A)
If 4sin$$^{2}$$ 2θ - 3 = 0 and θ is acute, then what is the value of (cot$$\ ^{2}\ $$ θ + tan $$^{2}$$ θ)?
Given : $$4sin^22\theta-3=0$$
=> $$sin^22\theta=\frac{3}{4}$$
=> $$sin2\theta=\sqrt{\frac{3}{4}}=\frac{\sqrt3}{2}$$
=> $$sin2\theta=sin(60^\circ)$$
=> $$2\theta=60$$
=> $$\theta=\frac{60}{2}=30^\circ$$
$$\therefore$$ $$cot^2\theta+tan^2\theta$$
= $$cot^2(30^\circ)+tan^2(30^\circ)$$
= $$(\sqrt3)^2+(\frac{1}{\sqrt3})^2$$
= $$3+\frac{1}{3}=\frac{10}{3}$$
=> Ans - (C)
If sec$$\theta=\frac{13}{12}\ and\ \theta\ $$is acute, then what is the value of $$(\sqrt{cot\theta+tan \theta})$$?
Given,
sec$$\theta = \frac{13}{12}$$
Then, cot$$\ \theta = \frac{12}{5}$$ and tan$$\ \theta = \frac{5}{12}$$
$$(\sqrt{cot\theta+tan \theta})$$ = $$(\sqrt{\frac{12}{5}+\frac{5}{12}})$$ = $$(\sqrt{\frac{169}{60}})$$ = $$\frac{13}{2\sqrt{15}}$$
Hence, option A is the correct answer.
If $$\sin\theta+\cos\theta=\sqrt{3}\cos(90-\theta)$$, then what is the value of $$\tan\theta$$ ?
Given : $$sin\theta+cos\theta=\sqrt3cos(90^\circ-\theta)$$
=> $$sin\theta+cos\theta=\sqrt3sin\theta$$
=> $$cos\theta=sin\theta(\sqrt3-1)$$
=> $$\frac{sin\theta}{cos\theta}=\frac{1}{\sqrt3-1}$$
=> $$tan\theta=\frac{1}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1}$$
=> $$tan\theta=\frac{\sqrt3+1}{3-1}$$
=> $$tan\theta=\frac{\sqrt3+1}{2}$$
=> Ans - (C)
If tan $$\ \theta\ $$= $$\ \frac{2}{3}\ $$, then what is the value of $$\ \frac{15sin^{2} \theta-3cos^{2} \theta}{5sin^{2} \theta+3cos^{2} \theta}$$?
Given : $$tan\ \theta=\frac{2}{3}$$
=> $$\frac{sin\ \theta}{cos\ \theta}=\frac{2}{3}$$
Let $$sin\ \theta=2$$ and $$cos\ \theta=3$$
To find : $$\ \frac{15sin^{2} \theta-3cos^{2} \theta}{5sin^{2} \theta+3cos^{2} \theta}$$
= $$\frac{15(2)^2-3(3)^2}{5(2)^2+3(3)^2}$$
= $$\frac{60-27}{20+27}$$
= $$\frac{33}{47}$$
=> Ans - (C)
If $$(1+ tan^{2} \theta) = \frac{625}{49}\ and\ \theta\ $$is acute, then what is the value of $$\ (\sqrt{sin \theta+cos\ \theta})\ $$?
Given,
$$(1+ tan^{2}\theta)$$ = $$\frac{625}{49}$$ (or) $$sec^{2}\theta$$ = $$\frac{625}{49}$$
$$sec\ \theta$$ = $$\frac{25}{7}$$
$$Cos\ \theta = \frac{7}{25}$$ and $$sin\ \theta = \frac{24}{25}$$
$$\ \sqrt{sin \theta+cos\ \theta}\ = \sqrt{\frac{7 + 24}{25}}$$ = $$\sqrt{\frac{31}{25}}$$ = $$\frac{\sqrt{31}}{5}$$
Hence, option C is the correct answer.
If √5 tan θ = 5 sin θ, then what is the value of (sin$$^{2}$$ θ - cos$$^{2}$$θ)?
Given : $$\sqrt5tan\ \theta=5sin\ \theta$$
=> $$\frac{sin\ \theta}{cos\ \theta}=\sqrt5sin\ \theta$$
=> $$cos\ \theta=\frac{1}{\sqrt5}$$
=> $$cos^2\ \theta=\frac{1}{5}$$ ---------------(i)
Now, $$sin^2\ \theta=1-cos^2\ \theta$$
=> $$sin^2\ \theta=1-\frac{1}{5}=\frac{4}{5}$$ --------------(ii)
Subtracting equation (i) from (ii), we get :
$$\therefore$$ $$(sin^2\ \theta-cos^2\ \theta)=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}$$
=> Ans - (A)
If $$\ \frac{1}{cos\theta+sec\theta}=\frac{1}{2}\ $$, then what is the value of $$\ cos^{100}\ \theta+sec^{100}\ \theta\ $$?
Given : $$\ \frac{1}{cos\theta+sec\theta}=\frac{1}{2}\ $$
=> $$\ \frac{1}{cos\theta+\frac{1}{cos\theta}}=\frac{1}{2}\ $$
=> $$\ \frac{cos\theta}{cos^2\theta+1}=\frac{1}{2}\ $$
=> $$cos^2\theta+1-2cos\theta=0$$
=> $$(cos\theta-1)^2=0$$
=> $$cos\theta=1$$
Also, $$sec\theta=\frac{1}{cos\theta}=1$$
$$\therefore$$ $$\ cos^{100}\ \theta+sec^{100}\ \theta\ $$
= $$(1)^{100}+(1)^{100}=1+1=2$$
=> Ans - (C)
If $$\ \frac{cos\theta}{1+sin\theta}+\frac{cos\theta}{1-sin\theta}\ $$= 4 and $$\ \theta\ $$is acute, then what is the value of (in degrees) of $$\ \theta$$?
Expression : $$\ \frac{cos\theta}{1+sin\theta}+\frac{cos\theta}{1-sin\theta}\ =4$$
=> $$cos\theta( \frac{1}{1+sin\theta}+\frac{1}{1-sin\theta}) =4$$
=> $$cos\theta(\frac{(1-sin\theta)+(1+sin\theta)}{(1+sin\theta)(1-sin\theta)})=4$$
=> $$cos\theta\times\frac{2}{1-sin^2\theta}=4$$
=> $$cos\theta\times\frac{1}{cos^2\theta}=\frac{4}{2}$$
=> $$\frac{1}{cos\theta}=2$$
=> $$cos\theta=\frac{1}{2}$$
=> $$\theta=cos^{-1}(\frac{1}{2})=60^\circ$$
=> Ans - (C)
If tan A = $$\frac{1}{3}\ $$and tan B = $$\frac{2}{5}$$, then what is the value of tan (2A + B) ?
Given : $$tanA=\frac{1}{3}$$ and $$tanB=\frac{2}{5}$$
=> $$tan2A=\frac{2tanA}{1-tan^2A}$$
=> $$tan2A=\frac{2\times\frac{1}{3}}{1-(\frac{1}{3})^2}$$
=> $$tan2A=\frac{\frac{2}{3}}{\frac{8}{9}}$$
=> $$tan2A=\frac{2}{3}\times\frac{9}{8}=\frac{3}{4}$$
To find : $$tan(2A+B)$$
= $$\frac{tan(2A)+tan(B)}{1-tan(2A)tan(B)}$$
= $$\frac{\frac{3}{4}+\frac{2}{5}}{1-(\frac{3}{4})(\frac{2}{5})}$$
= $$\frac{\frac{(15+8)}{20}}{1-\frac{3}{10}}$$
= $$\frac{23}{20}\times\frac{10}{7}$$
= $$\frac{23}{14}$$
=> Ans - (D)
if$$\ \frac{1}{sin \theta+cosec \theta}=\frac{1}{2}\ $$, then what is the value of sin$$^{100}\ \theta\ +\ cosec^{100}\ \theta$$?
Given : $$\ \frac{1}{sin \theta+cosec \theta}=\frac{1}{2}\ $$
=> $$\ \frac{1}{sin\theta+\frac{1}{sin\theta}}=\frac{1}{2}\ $$
=> $$\ \frac{sin\theta}{sin^2\theta+1}=\frac{1}{2}\ $$
=> $$sin^2\theta+1-2sin\theta=0$$
=> $$(sin\theta-1)^2=0$$
=> $$sin\theta=1$$
Also, $$cosec\theta=\frac{1}{sin\theta}=1$$
$$\therefore$$ $$\ sin^{100}\ \theta+cosec^{100}\ \theta\ $$
= $$(1)^{100}+(1)^{100}=1+1=2$$
=> Ans - (D)
If sin θ + $$sin^{2}$$ θ = 1, then what is the value of $$(cos^{12}\ \theta\ + 3\ cos^{10}\ \theta+3\ cos^{8}\ \theta\ + cos^{6}\ \theta -1)$$
Given $$sin \theta + sin^{2} \theta = 1$$
$$\Rightarrow sin$$ θ = 1 - $$sin^{2}$$ θ $$\Rightarrow$$ $$sin$$ θ = $$cos^{2}$$ θ .....(1)
Now, $$(cos^{12}\ \theta\ + 3\ cos^{10}\ \theta+3\ cos^{8}\ \theta\ + cos^{6}\ \theta -1)$$
$$(cos^{4} \theta$$ + $$cos^{2} \theta)^{3}$$ - 1
Substitute equation (1) in the above equation
$$(sin^{2} \theta$$ + $$cos^{2} \theta)^{2}$$ - 1
1 - 1 = 0
Hence, option B is the correct answer.
What is the simplified value of 1 + cot A cot $$(\frac{A}{2})$$?
Now We have :
$$1+\cot A\ \cot\ \frac{A}{2}$$
Now we get 1+$$\frac{\cos A}{\sin A}\times\ \frac{\cos\ \frac{A}{2}}{\sin\ \frac{A}{2}}=\ \frac{\cos A}{2\sin\ \frac{A}{2}\cos\ \frac{A}{2}}\times\ \frac{\cos\ \frac{A}{2}}{\sin\ \frac{A}{2}}$$
we get $$1+\frac{\cos A}{2\sin^2\ \frac{A}{2}}$$
Now cos A = $$2\sin^2\ \frac{A}{2}-1$$
substituting we get $$\frac{1}{2\sin^2\ \frac{A}{2}}=\frac{1}{2}\operatorname{cosec}^2\ \frac{A}{2}$$
What is the simplified value of [(1 + sec 2θ) $$tan^{2}$$ θ] + 1?
What is the simplified value of $$\ \frac{7}{sec^{2} \theta}+ \frac{3}{1+cot^{2} \theta}+ 4\ sin^{2} \theta$$?
Expression : $$\ \frac{7}{sec^{2} \theta}+ \frac{3}{1+cot^{2} \theta}+ 4\ sin^{2} \theta$$
= $$7cos^2\ \theta+\frac{3}{cosec^2\ \theta}+4sin^2\ \theta$$
= $$7cos^2\ \theta+3sin^2\ \theta+4sin^2\ \theta$$
= $$7cos^2\ \theta+7sin^2\ \theta$$
= $$7(cos^2\ \theta+sin^2\ \theta)$$
= $$7\times1=7$$
=> Ans - (D)
If cot $$\theta=\sqrt{11}\ and\ \theta\ $$is acute, then what is the value of $$(\frac{cosec^{2}\ \theta\ +\ sec^{2}\ \theta}{cosec^{2}\ \theta\ -\ sec^{2}\ \theta})$$?
Given, cot $$\theta=\sqrt{11}$$
$$(\frac{cosec^{2}\ \theta\ +\ sec^{2}\ \theta}{cosec^{2}\ \theta\ -\ sec^{2}\ \theta})$$ = $$(\frac{1 + cot^{2}\ \theta\ +\ 1 + tan^{2}\ \theta}{1 + cot^{2}\ \theta\ -\ 1 + tan^{2}\ \theta})$$
$$\Rightarrow (\frac{1 + 11\ +\ 1 + \frac{1}{11}}{1 + 11\ -\ 1 + \frac{1}{11}})$$ = $$(\frac{\frac{144}{11}}{\frac{120}{11}})$$ = $$\frac{144}{20}$$
$$\Rightarrow \frac{6}{5}$$
Hence, option B is the correct answer.
If tan$$\ (\frac{\theta}{2})tan(\frac{2\theta}{5})\ $$= 1, then what is the value (in degrees) of $$\theta$$?
Given : $$tan(\frac{\theta}{2})tan(\frac{2\theta}{5})=1$$ --------------(i)
Now, we know that $$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$$
=> $$tan(\frac{\theta}{2}+\frac{2\theta}{5})=[tan(\frac{\theta}{2})+tan(\frac{2\theta}{5})]\div[1-tan(\frac{\theta}{2})tan(\frac{2\theta}{5})]$$
Substituting value from equation (i), we get :
=> $$tan(\frac{\theta}{2}+\frac{2\theta}{5})=\frac{tan(\frac{\theta}{2})+tan(\frac{2\theta}{5})}{0}$$
=> $$tan(\frac{\theta}{2}+\frac{2\theta}{5})=tan(90^\circ)$$
=> $$\frac{\theta}{2}+\frac{2\theta}{5}=90$$
=> $$\frac{5\theta+4\theta}{10}=90$$
=> $$\theta=90\times\frac{10}{9}$$
=> $$\theta=100^\circ$$
=> Ans - (C)
What is the simplified value of [1+ $$cos^{2}$$ θ $$cosec^{2}$$ θ] - $$cot^{2}$$ θ?
Expression : $$[1+cos^2\theta cosec^2\theta]-cot^2\theta$$
= $$[1+\frac{cos^2\theta}{sin^2\theta}]-cot^2\theta$$
= $$1+cot^2\theta-cot^2\theta=1$$
=> Ans - (C)
What is the value of $$\ \frac{tan^{2}25^\circ}{cosec^{2}65^\circ}+\frac{cot^{2}25^\circ}{sec^{2}65^\circ}\ $$+ 2tan$$\ 20^\circ\ tan\ 45^\circ\ tan\ 70^\circ\ $$?
Expression = $$\ \frac{tan^{2}25^\circ}{cosec^{2}65^\circ}+\frac{cot^{2}25^\circ}{sec^{2}65^\circ}\ + 2tan\ 20^\circ\ tan\ 45^\circ\ tan\ 70^\circ\ $$
= $$\ \frac{tan^{2}25^\circ}{cosec^{2}(90^\circ-25^\circ)}+\frac{cot^{2}25^\circ}{sec^{2}(90^\circ-25^\circ)}\ + 2tan\ 20^\circ\ tan\ 45^\circ\ tan\ (90^\circ-20^\circ\ )$$
Using, $$cosec(90^\circ-\theta)=sec\theta$$
= $$\ \frac{tan^{2}25^\circ}{sec^{2}25^\circ}+\frac{cot^{2}25^\circ}{cosec^{2}25^\circ}\ + 2tan\ 20^\circ\ tan\ 45^\circ\ cot\ 20^\circ\ $$
= $$(\frac{sin^225^\circ}{cos^225^\circ}\times cos^225^\circ)+(\frac{cos^225^\circ}{sin^225^\circ}\times sin^225^\circ)+(2tan20^\circ.cot20^\circ.tan45^\circ)$$
$$\because tan\ \theta. cot\ \theta=1$$
= $$(sin^225^\circ+cos^225^\circ)+(2\times1)$$
= $$1+2=3$$
=> Ans - (C)
If $$(\cos \theta + 31^\circ)=\sin47^\circ$$, then what is the value of $$\sin 5\theta $$?
Given : $$cos(\theta+31^\circ)=sin(47^\circ)$$
=> $$cos(\theta+31^\circ)=sin(90^\circ-43^\circ)$$
Using, $$sin(90^\circ-x)=cos (x)$$
=> $$cos(\theta+31^\circ)=cos(43^\circ)$$
=> $$\theta+31^\circ=43^\circ$$
=> $$\theta=43^\circ-31^\circ$$
=> $$\theta=12^\circ$$
$$\therefore$$ $$sin(5\theta)=sin(5\times12^\circ)$$
= $$sin(60^\circ)=\frac{\sqrt3}{2}$$
=> Ans - (C)
If cosec θ + 3 sec θ = 5 cosec θ, then what is the value of cot θ?
Given : $$cosec\theta+3sec\theta=5cosec\theta$$
=> $$3sec\theta=5cosec\theta-cosec\theta$$
=> $$3sec\theta=4cosec\theta$$
=> $$\frac{3}{cos\theta}=\frac{4}{sin\theta}$$
=> $$\frac{cos\theta}{sin\theta}=\frac{3}{4}$$
=> $$cot\theta=\frac{3}{4}$$
=> Ans - (B)
If cosec θ + $$cosec^{2}$$ θ = 1, then what is the value of ($$cot^{12}\ θ - 3\ cot^{10}\ θ + 3 cot^{8}\ θ - cot^{6}\ θ)$$?
$$If\ \frac{cos\ \theta}{1+sin\ \theta}+ \frac{cos\ \theta}{1-sin\ \theta}=2\sqrt{2}\ and\ \theta\ $$is acute, then what is the value (in degrees) of $$\theta$$?
Given,
$$ \frac{cos\ \theta}{1+sin\ \theta}+ \frac{cos\ \theta}{1-sin\ \theta}=2\sqrt{2}\ $$
$$\cos\theta(\frac{1}{1+sin\ \theta}+ \frac{1}{1-sin\ \theta})=2\sqrt{2}\ $$
$$\cos\theta (\frac{1 - \sin\theta + 1 + \sin\theta}{1^{2} - \sin^{2}\theta}) = 2\sqrt{2}$$
$$\cos\theta (\frac{2}{\cos^{2}\theta}) = 2\sqrt{2}$$
$$\cos\theta = \frac{1}{\sqrt{2}}$$
$$\theta = 45^{\circ}$$
Hence, option B is the correct answer.
What is the simplified value of$$\ \frac{3}{cosec^{2}\theta}+\frac{5}{1+tan^{2}\theta}-2cos^{2}\theta$$?
Expression = $$\ \frac{3}{cosec^{2}\theta}+\frac{5}{1+tan^{2}\theta}-2cos^{2}\theta$$
= $$3sin^{2}\theta+\frac{5}{sec^{2}\theta}-2cos^{2}\theta$$
= $$3sin^{2}\theta+5cos^{2}\theta-2cos^{2}\theta$$
= $$3sin^{2}\theta+3cos^{2}\theta$$
= $$3(sin^{2}\theta+cos^{2}\theta)$$
= $$3\times1=3$$
=> Ans - (A)
The value of $$\sqrt{2\sqrt[3]{4}\sqrt{2\sqrt[3]{4}}\sqrt[4]{2\sqrt[3]{4}}.....}$$ is
To find : $$y=\sqrt{2\sqrt[3]{4}\sqrt{2\sqrt[3]{4}}\sqrt[4]{2\sqrt[3]{4}}.....}$$
Let $$2\sqrt[3]4=x$$
=> $$y=\sqrt{(x)\times(\sqrt{x})\times(\sqrt[4]{x})\times.......}$$
=> $$y^2=(x)^{[1+\frac{1}{2}+\frac{1}{4}+......+\infty]}$$
Now, sum of infinite G.P. = $$\frac{a}{(1-r)}$$, where first term = $$a=1$$ and common ratio = $$r=\frac{1}{2}$$
=> $$y^2=(x)^{\frac{1}{1-\frac{1}{2}}}$$
=> $$y^2=(x)^2$$
=> $$y=x$$
$$\therefore$$ $$\sqrt{2\sqrt[3]{4}\sqrt{2\sqrt[3]{4}}\sqrt[4]{2\sqrt[3]{4}}.....}=2\sqrt[3]4$$
=> Ans - (A)
The value of$$\ \frac{3\sqrt{2}}{(\sqrt{3}+\sqrt{6})}-\frac{4\sqrt{3}}{(\sqrt{6}+\sqrt{2})}+\frac{\sqrt{6}}{(\sqrt{2}+\sqrt{3})}\ $$is
$$\frac{3\sqrt{2}}{(\sqrt{3}+\sqrt{6})}-\frac{4\sqrt{3}}{(\sqrt{6}+\sqrt{2})}+\frac{\sqrt{6}}{(\sqrt{2}+\sqrt{3})}$$
= $$\frac{6\sqrt{6}+18+6\sqrt{2}+6\sqrt{3}-(12\sqrt{2}+24+12\sqrt{3}+12\sqrt{6})+6\sqrt{3}+6+6\sqrt{6}+6\sqrt{2}}{(\sqrt{3}+\sqrt{6})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})}$$
= $$\frac{6\sqrt{6}+18+6\sqrt{2}+6\sqrt{3}-12\sqrt{2}-24-12\sqrt{3}-12\sqrt{6}+6\sqrt{3}+6+6\sqrt{6}+6\sqrt{2}}{(\sqrt{3}+\sqrt{6})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})}$$
=0
If$$\ a^{2}+b^{2}+c^{2}$$= 2(a-b-c)-3, then the value of 4a - 3b + 5c is
$$\ a^{2}+b^{2}+c^{2}$$= 2(a-b-c)-3
We can write the above equation as
$$ a^{2}-2a+1+b^{2}+2b+1+c^{2}+2c+1$$=0
$$\Rightarrow (a-1)^{2}+(b+1)^{2}+(c+1)^{2}$$=0
$$(a-1)^{2}$$=0$$\Rightarrow$$a=1
$$(b+1)^{2}$$=0$$\Rightarrow$$ b=-1
$$(c+1)^{2}$$=0$$\Rightarrow$$ c=-1
4a-3b+5c= 4(1)-3(-1)+5(-1)=4+3-5=2
If 2x + $$\ \frac{2}{x}\ $$= 3 then the value of $$\ x^{3}+\frac{1}{x^{3}}$$+ 2 is
Given 2x+$$\frac{1}{x} =$$ 3
2(x+$$\frac{1}{x}) =$$ 3
$$\Rightarrow$$ x+$$\frac{1}{x} = \frac{3}{2}$$
Cubing on both sides
(x+$$\frac{1}{x})^{3} = \frac{27}{8}$$
x$$^{3}$$+$$\frac{1}{x^{3}}$$+3$$\times$$x$$\times$$ $$\frac{1}{x}$$(x+$$\frac{1}{x}$$) $$= \frac{27}{8}$$
$$\Rightarrow x^{3}+\frac{1}{x^{3}}+3(\frac{3}{2}) = \frac{27}{8}$$
$$\Rightarrow x^{3}+\frac{1}{x^{3}} = \frac{27}{8}-\frac{9}{2} = \frac{-9}{8}$$
$$x^{3}+\frac{1}{x^{3}}+2 = \frac{-9}{8}+2 = \frac{7}{8}$$
If X = $$\ \sqrt[3]{5}\ $$+ 2, then the value of $$\ x^{3}-6x^{2}\ $$+ 12x - 13 is
Given x=\sqrt[3]{5}+2
$$\ x^{3}-6x^{2}\ $$+ 12x - 13
= $$(\sqrt[3]{5}+2)^{3}-6(\sqrt[3]{5}+2)^{2}+12(\sqrt[3]{5}+2)-13$$
= $$(5+8+6\times5^{\frac{2}{3}}+12\times5^{\frac{1}{3}})-6[5^{\frac{1}{3}}+4+4\times5^{\frac{1}{3}}]+12(5^{\frac{1}{3}}+2)-13$$
= $$13+6\times5^{\frac{2}{3}}+12\times5^{\frac{1}{3}}-6\times5^{\frac{2}{3}}-24-24\times5^{\frac{1}{3}}+12\times5^{\frac{1}{3}}+24-13$$
= 0
Find the value of $$\ \sqrt{30+\sqrt{30}+...}$$
Let X=$$ \sqrt{30+\sqrt{30}+...}$$
Above equation can be written as
X=$$\Rightarrow\sqrt{30+X}$$
Squaring on both sides
$$X^{2}$$=30+X
$$X^{2}$$-X-30=0
$$X^{2}$$-6X+5X-30=0
X(X-6)+5(X-6)=0
(X-6)(X+5)=0
X=-5,6
Taking positive value
X=6
Sin A + Sin$$^{2}\ $$A = 1, then the value of cos$$^{2}\ $$A + cos$$^{4}\ $$A is
Given sin A+$$sin^{2}$$ A=1
==> sin A = 1-$$sin^{2}$$ A
==> sin A = $$cos^{2}$$ A ($$\because cos^{2}A+sin^{2}A$$=1)
$$cos^{2}$$ A=sin A ==> $$cos^{4}A$$=$$sin^{2}A $$
$$\therefore cos^{2}A+cos^{4}A=1 ( \because sin A+sin^{2}A=1)$$
The value of $$\frac{(75.8)^{2}-(35.8)^{2}}{40}$$ is
$$\frac{75.8^{2}-35.8^{2}}{40}$$=$$\frac{(75.8+35.8)(75.8-35.8)}{40}$$
=$$\frac{111.6\times40}{40}$$=111.6
If $$x + y = 4$$, then what is the value of $$x^{3} +y^{3} +12xy$$?
Given : $$x+y=4$$ -----------(i)
Cubing both sides, we get :
=> $$(x+y)^3=(4)^3$$
=> $$x^3+y^3+3xy(x+y)=64$$
=> $$x^3+y^3+3xy(4)=64$$
=> $$x^3+y^3+12xy=64$$
=> Ans - (C)
If $$x^{4}+\frac{1}{x^{4}}= 198$$ and $$x>0$$, then what is the value of $$x^{2}-\frac{1}{x^{2}}$$?
Given : $$\ x^{4}+\frac{1}{x^{4}}\ =198$$
=> $$(x^2-\frac{1}{x^2})^2+2(x^2)(\frac{1}{x^2})=198$$
=> $$(x^2-\frac{1}{x^2})^2=198-2=196$$
=> $$x^2-\frac{1}{x^2}=\sqrt{196}=14$$
=> Ans - (A)
If $$3x - \frac{1}{3x} = 9$$, then what is the value of $$x^{2} + \frac{x^2}{81}$$?
Given : $$3x-\frac{1}{3x}=9$$
Dividing both sides by 3, => $$x-\frac{1}{9x}=3$$
Squaring both sides, we get :
=> $$(x-\frac{1}{9x})^2=(3)^2$$
=> $$x^2+\frac{1}{81x^2}-2(x)(\frac{1}{9x})=9$$
=> $$(x^2+\frac{1}{81x^2})-\frac{2}{9}=9$$
=> $$(x^2+\frac{1}{81x^2})=9+\frac{2}{9}$$
=> $$(x^2+\frac{1}{81x^2})=\frac{83}{9}$$
=> Ans - (B)
If $$x^{3} - y^{3} = 112$$ and $$x - y = 4$$, then what is the value of $$x^{2} + y^{2}$$?
Given : $$x^3-y^3=112$$ --------------(i)
Also, $$x-y=4$$ -------------(ii)
Cubing both sides, we get :
=> $$(x-y)3=(4)^3$$
=> $$(x^3-y^3)-3(x)(y)(x-y)=64$$
Substituting values from equations (i) and (ii),
=> $$112-3xy(4)=64$$
=> $$12xy=112-64=48$$
=> $$xy=\frac{48}{12}=4$$ -----------(iii)
Now, squaring equation (ii), we get :
=> $$x^2+y^2-2xy=16$$
=> $$x^2+y^2=16+8=24$$
=> Ans - (C)
If $$x = 5 - \frac{1}{x} $$, then what is the value of $$x^{5} + \frac{1}{x^{5}}$$?
Given : $$x=5-\frac{1}{x}$$
=> $$x+\frac{1}{x}=5=k$$
Now, $$x^5+\frac{1}{x^5}=[(x^3+\frac{1}{x^3})\times(x^2+\frac{1}{x^2})]-(x+\frac{1}{x})$$
= $$[(x+\frac{1}{x})^3-3(x+\frac{1}{x})\times(x+\frac{1}{x})^2-2(x)(\frac{1}{x})]-(x+\frac{1}{x})$$
= $$[(k^3-3k)\times(k^2-2)]-(k)$$
= $$[(125-15)\times(25-2)]-(5)$$
= $$(110\times23)-5$$
= $$2530-5=2525$$
=> Ans - (C)
In ΔABC, ∠A : ∠B : ∠C = 3 : 3 : 4. A line parallel to BC is drawn which touches AB and AC at P and Q respectively. What is the value of ∠AQP - ∠APQ?
Given : ∠A : ∠B : ∠C = 3 : 3 : 4 and PQ is parallel to BC
To find : ∠AQP - ∠APQ = ?
Solution : Let $$\angle A=3x$$, $$\angle B=3x$$ and $$\angle C=4x$$
Thus, in $$\triangle$$ ABC,
=> $$\angle A+\angle B+\angle C=180^\circ$$
=> $$3x+3x+4x=180^\circ$$
=> $$x=\frac{180^\circ}{10}=18^\circ$$
$$\because$$ PQ $$\parallel$$ BC, => $$\angle$$ APQ = $$\angle$$ B and $$\angle$$ AQP = $$\angle$$ C (Corresponding angles)
$$\therefore$$ $$\angle$$ AQP - $$\angle$$ APQ = $$4x-3x=x=18^\circ$$
=> Ans - (B)
In the given figure, O is the center of the circle, $$\angle$$CAO = 35$$^\circ$$. What is the value (in degrees) of $$\ \angle$$AOB?
The side QR of ΔPQR is produced to S. If ∠PRS = 105° and ∠Q = (1/2)∠P, then what is the value of ∠P?
Let $$\angle Q=x$$, => $$\angle P=2x$$
Using exterior angle property in $$\triangle$$ PQR,
=> $$\angle$$ P + $$\angle$$ Q = $$\angle$$ PRS
=> $$2x+x=105^\circ$$
=> $$x=\frac{105^\circ}{3}=35^\circ$$
$$\therefore$$ $$\angle P=2\times35^\circ=70^\circ$$
=> Ans - (C)
What is the simplified value of$$\ \sqrt{\frac{sec^{2} \theta+cosec^{2} \theta}{4}}$$?
Expression = $$\ \sqrt{\frac{sec^{2} \theta+cosec^{2} \theta}{4}}$$
= $$\ \sqrt{\frac{(\frac{1}{cos^2\ \theta})+(\frac{1}{sin^2\ \theta})}{4}}$$
= $$\ \sqrt{\frac{(\frac{sin^2\ \theta+cos^2\ \theta)}{sin^2\ \theta.cos^2\ \theta}}{4}}$$
= $$\sqrt{\frac{1}{4sin^2\ \theta cos^2\ \theta}}$$
= $$\sqrt{(\frac{1}{2sin\ \theta cos\ \theta})^2}$$
= $$\frac{1}{sin2\theta}=cosec2\theta$$
=> Ans - (A)
If$$\ \frac{x-xtan^{2}15^\circ}{1+tan^{2}15^\circ}$$= sin $$60^\circ\ $$+ cos 30$$^\circ\ $$, then what is then what is the value of x?
$$tan15^\circ=\frac{\sqrt3-1}{\sqrt3+1}$$
Expression = $$\ \frac{x-xtan^{2}15^\circ}{1+tan^{2}15^\circ}$$= sin $$60^\circ\ $$+ cos 30$$^\circ\ $$
=> $$\ \frac{x(1-tan^{2}15^\circ)}{1+tan^{2}15^\circ}= \frac{\sqrt3}{2}+\frac{\sqrt3}{2}$$
=> $$x\times\frac{1-(\frac{\sqrt3-1}{\sqrt3+1})^2}{1+(\frac{\sqrt3-1}{\sqrt3+1})^2}=\sqrt3$$
=> $$x\times\frac{(\sqrt3+1)^2-(\sqrt3-1)^2}{(\sqrt3+1)^2+(\sqrt3-1)^2}=\sqrt3$$
=> $$x\times\frac{(3+1+2\sqrt3)-(3+1-2\sqrt3)}{(3+1+2\sqrt3)+(3+1-2\sqrt3)}=\sqrt3$$
=> $$x\times\frac{4\sqrt3}{8}=\sqrt3$$
=> $$x=\frac{8}{4}=2$$
=> Ans - (A)
What is the simplified value of $$\ \frac{2sin^{3}\ \theta\ - sin\theta}{cos\theta\ -\ 2\ cos^{3}\ \theta}$$?
Expression : $$\ \frac{2sin^{3}\ \theta\ - sin\theta}{cos\theta\ -\ 2\ cos^{3}\ \theta}$$
= $$\ \frac{sin\ \theta(2sin^{2}\ \theta\ - 1)}{cos\ \theta(2 -\ 2\ cos^{2}\ \theta)}$$
= $$\frac{sin\ \theta(cos2\ \theta)}{cos\ \theta(cos2\ \theta)}$$
= $$\frac{sin\ \theta}{cos\ \theta}=tan\ \theta$$
=> Ans - (A)
If $$tan(\theta)tan(5\theta)=1$$, then what is the value of $$sin 2\theta$$ ?
Given : $$tan(\theta)tan(5\theta)=1$$
Using, $$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$$
$$tan(\theta+5\theta)=\frac{tan(\theta)+tan(5\theta)}{1-tan(\theta)tan(5\theta)}$$
=> $$tan(6\theta)=\frac{tan(\theta)+tan(5\theta)}{1-1}$$
=> $$tan(6\theta)=\infty$$
=> $$tan(6\theta)=tan(90^\circ)$$
=> $$6\theta=90^\circ$$
=> $$\theta=\frac{90^\circ}{6}=15^\circ$$
$$\therefore$$ $$sin(2\theta)=sin(2\times15^\circ)$$
= $$sin(30^\circ)=\frac{1}{2}$$
=> Ans - (B)
The value of 0.65 x 0.65 + 0.35 x 0.35+0.70 x 0.65 is
Expression : $$(0.65\times0.65)+(0.35\times0.35)+(0.70\times0.65)$$
= $$(0.65)^2+(0.35)^2+2(0.35)(0.65)$$
Comparing with : $$(x)^2+(y)^2+2(x)(y)=(x+y)^2$$
= $$(0.65+0.35)^2=(1)^2=1$$
=> Ans - (B)
If$$\ x^{2}$$- 3x + 1 = 0, then the value of $$\ x^{5}+\frac{1}{x^{5}}\ $$is equal to
$$x^{2}-3x+1$$=0
Taking 'x' common
x(x-3+$$\frac{1}{x})$$=0
$$\Rightarrow x+\frac{1}{x}$$=$$3\rightarrow(1)$$
Squaring on both sides
$$x^{2}+\frac{1}{x^{2}}+2\times x\times\frac{1}{x}$$=9
$$\Rightarrow x^{2}+\frac{1}{x^{2}}$$=$$7\rightarrow(2)$$
Cubing equation(1) on both sides
$$x^{3}+\frac{1}{x^{3}}+3\times x\times\frac{1}{x}(x+\frac{1}{x})$$=27
$$x^{3}+\frac{1}{x^{3}}$$+$$3\times 1\times3$$=27($$\because x+\frac{1}{x}$$=3)
$$x^{3}+\frac{1}{x^{3}}$$=27-9=$$18\rightarrow(3)$$
Squaring equation(2) on both sides
$$x^{4}+\frac{1}{x^{4}}+2\times x^{2}\times\frac{1}{x^{2}}$$=49
$$x^{4}+\frac{1}{x^{4}}$$=$$47\rightarrow(4)$$
Multiplying equation(1) and equation(4)
$$(x^{4}+\frac{1}{x^{4}})(x+\frac{1}{x}$$)=$$47\times3$$
$$x^{5}+\frac{1}{x^{5}}+x^{3}+\frac{1}{x^{3}}$$=$$47\times3$$=141
$$x^{5}+\frac{1}{x^{5}}+18$$=141($$\because x^{3}+\frac{1}{x^{3}}$$)
$$\therefore x^{5}+\frac{1}{x^{5}}$$=123
The value of cosec$$^{2}\ 18^\circ\ - \frac{1}{cot^{2}72^\circ}\ $$is
cosec$$^{2}\ 18^\circ\ - \frac{1}{cot^{2}72^\circ}\ $$
= cosec$$^{2} 18^\circ - tan^{2} 72^\circ$$ ($$\because \frac{1}{cot^{2} \ominus}$$=$$tan^{2}\ominus$$)
= cosec$$^{2} 18^\circ$$ - $$tan^{2} (90-72)^\circ$$
= cosec$$^{2} 18^\circ$$ - $$sec^{2} 18^\circ$$ ($$\because sec^{2}\ominus= tan^{2}(90^\circ-\ominus)$$)
cosec$$^{2} 18^\circ$$ - $$sec^{2} 18^\circ$$=1($$\because cosec^{2} \ominus$$ - $$sec^{2} \ominus$$=1)
The bar graph given below represents the revenue of a firm for 8 years. All the revenue figures have been shown in terms of Rs crores.
What is the total value of revenue of the firm (in crores Rs) in years 2010, 2011 and 2012?
Total value of revenue of the firm in years 2010, 2011 and 2012 = 260 + 350 + 320 = 930
Hence, option B is the correct answer.
If $$\dfrac{x^{2}}{yz}+\dfrac{y^{2}}{zx}+\dfrac{z^{2}}{xy}=3$$, then what is the value of $$(x+y+z)^{3}$$ ?
Given : $$\dfrac{x^{2}}{yz}+\dfrac{y^{2}}{zx}+\dfrac{z^{2}}{xy}=3$$
=> $$\dfrac{x^3+y^3+z^3}{xyz}=3$$
=> $$x^3+y^3+z^3=3xyz$$
=> $$x^3+y^3+z^3-3xyz=0$$
=> $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$$
=> $$x+y+z=0$$
Cubing both sides, we get :
=> $$(x+y+z)^3=0$$
=> Ans - (A)
If x + (1/x) = 2, then what is the value of $$x^{64}+x^{121}$$ ?
x + (1/x) = 2
we can substitute x = 1
$$x^{64}+x^{121}$$=$$1^{64}+1^{121}$$=$$1+1$$=$$2$$
so the answer is option C.
If a+b+c=27, then what is the value of $$(a-7)^{3}+(b - 9)^{3}+(c - 11)^{3}-3(a - 7)(b - 9)(c - 11)$$ ?
If x+y+z = 0, $$(x)^{3}+(y)^{3}+(z)^{3}-3(x)(y)(z)=0$$
puty x = a-7, y = b-9, z = c-11,
then x+y+z = (a+b+c)-(7+11+9) = 27 - 27 = 0,
so $$(x)^{3}+(y)^{3}+(z)^{3}-3(x)(y)(z)=0$$
$$(a-7)^{3}+(b - 9)^{3}+(c - 11)^{3}-3(a - 7)(b - 9)(c - 11)=0$$
so the answer is option A.
If $$\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b} - \sqrt{a-2b}}=\frac{\sqrt{3}}{1}$$, find the value of $$\frac{a}{b}$$
here in this question $$\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b-}\sqrt{a-2b}}=\frac{\sqrt{3}}{1}$$
using componendo and dividendo, we will get
$$\frac{\sqrt(a+2b)}{\sqrt(a-2b)} = \frac{\sqrt3 + 1 }{\sqrt3 - 1}$$
now on squaring both side and solving, we will get
16 b = 4a$$\surd3$$
$$\frac{a}{b}$$ = $$\frac{4}{\surd3}$$
A point in the 4th quadrant is 6 unit away from x-axis and 7 unit away from y-axis. The point is at
A point in the 4th quadrant will be in the form of $$(x,-y)$$
Since, the point is 6 units away from x axis, => y coordinate = 6
and the point is 7 units away from y axis, => x coordinate = 7
=> Point = (7,-6)
If $$\frac{3}{4}$$ of a number is 7 more then $$\frac{1}{6}$$ of the number, then $$\frac{5}{3}$$ of the number is
Let the number be $$x$$
Acc to ques :
=> $$\frac{3x}{4} = \frac{x}{6} + 7$$
=> $$\frac{14x}{24} = 7$$
=> $$x = 12$$
=> $$\frac{5}{3}$$ of the number = $$\frac{5}{3}$$ * 12 = 20
The area of the triangle formed by the graphs of the equations x= 0, 2x+ 3y= 6 and x+ y= 3 is :
AC represents $$x+y=3$$
BC represents $$2x+3y=6$$
AB represents $$x=0$$
=> ABC is the required triangle.
Base AB = 1 unit and height OC = 3 units
=> Area of $$\triangle$$ABC = $$\frac{1}{2}$$ * AB * OC
= $$\frac{1}{2}$$ * 1 * 3 = 1$$\frac{1}{2}$$ sq. unit
A number exceeds its two fifth by 75. The number is
Let the number be $$x$$
Acc to ques :
=> $$x - \frac{2}{5}x = 75$$
=> $$\frac{3x}{5} = 75$$
=> $$x = 125$$
Given that $$x^{3} + y^{3} = 72$$ and $$xy = 8$$ with $$x > y$$. Then the value of $$(x - y)$$ is
Given : $$x^{3} + y^{3} = 72$$ and $$xy = 8$$
Solution : $$(x+y)^3 = x^3 + y^3 + 3xy(x+y)$$
=> $$(x+y)^3 = 72 + 3.8(x+y)$$
=> $$(x+y)^3 - 24(x+y) - 72 = 0$$
This is a cubic equation in terms of $$(x+y)$$ which has one real root = 6
=> $$x+y = 6$$
Now, $$(x-y)^2 = (x+y)^2 - 4xy$$
=> $$(x-y) = \sqrt{6^2 - 4*8} = \sqrt{4}$$
=> $$(x-y) = 2$$
If the sum of two numbers, one of which is $$\frac{2}{5}$$ times the other, is 50, then the numbers are
Let the number be $$x$$
=> Other number will be $$\frac{2x}{5}$$
Acc to ques :
=> $$x$$ + $$\frac{2x}{5}$$ = 50
=> $$7x$$ = 250
=> $$x$$ = $$\frac{250}{7}$$
and second number = $$\frac{2}{5} * \frac{250}{7} = \frac{100}{7}$$
The sum of four numbers is 48. When 5 and 1 are added to the first two; and 3 and 7 are subtracted from the 3rd and 4th, all the four numbers will be equal. The numbers are
Let the numbers be $$a,b,c,d$$
=> $$a+b+c+d$$ = 48 --------Eqn(1)
When 5 & 1 are added to first two, => $$(a+5) and (b+1)$$
and when 3 & 7 are subtracted from last two, => $$(c-3) and (d-7)$$
According to question :
=> $$a+5 = b+1 = c-3 = d-7 = k$$ (let)
Now, in eqn(1)
$$(a+5) + (b+1) + (c-3) + (d-7)$$ = 48 + (5+1-3-7)
=> $$k+k+k+k$$ = 48-4
=> $$k$$ = 11
=> Numbers are : $$a = k-5 = 11-5 = 6$$
Similarly, $$b$$ = 10
$$c$$ = 14
$$d$$ = 18
The length of the portion of the straight line 3x + 4y = 12 intercepted between the axes is
Equation : $$3x + 4y = 12$$
To find $$x$$-intercept, put $$y$$=0
=> $$3x + 0 = 12$$
=> $$x$$ = 4
Similarly to find $$y$$-intercept, we need to put $$x$$ = 0
=> $$0 + 4y = 12$$
=> $$y$$ = 3
Thus, the line passes through (4,0) in x-axis and (0,3) in y-axis
Using, $$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
=> $$x_1 = 0 , x_2 = 4 , y_1 = 0 , y_2 = 3$$
=> $$d = \sqrt{4^2 + 3^2}$$
=> $$d = \sqrt{25} = 5$$
If x = 332, y = 332, z = 332, then the value of $$x^3 + y^3 + z^3 - 3xyz$$ is
when x = y = z= 332 , then $$(x^2 + y^2 + z^2 - xy - yz - xz)$$ = 0
and hence $$x^3 + y^3 + z^3 - 3xyz$$ = 0 as $$x^3 + y^3 + z^3 - 3xyz$$ = (x+y+z) $$(x^2 + y^2 + z^2 - xy - yz - xz)$$
and hence the answer for this question is = 0
2x- ky + 7 = 0 and 6x- 12y+ 15 = 0 has no solution for
NOTE : - For the pair of equations : $$a_1x + b_1y + c_1 = 0$$ and $$a_2x + b_2y + c_2 = 0$$
The equations have no solution only if : $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$
Equations : $$2x- ky + 7 = 0$$ and $$6x- 12y+ 15 = 0$$
Comparing with above formula, we get :
=> $$\frac{2}{6} = \frac{-k}{-12}$$
=> $$\frac{k}{12} = \frac{1}{3}$$
=> $$k = 4$$
If $$2x+3y=\frac{11}{2}$$ and $$xy=\frac{5}{6}$$ then the value of $$8x^{3}+27y^{3}$$ is
$$2x+3y=\frac{11}{2}$$
cubing on both sides
$$(2x+3y)^{3}=(\frac{11}{2})^{3}$$
$$8x^{3}+27y^{3}+3(2x)(8y)(2x+3y)=\frac{1331}{8}$$
$$8x^{3}+27y^{3}+3(16xy)(2x+3y)=\frac{1331}{8}$$
$$8x^{3}+27y^{3}+3(16)(\frac{5}{6})(\frac{11}{2})=\frac{1331}{8}$$
$$8x^{3}+27y^{3}+220 =\frac{1331}{8}$$
$$8x^{3}+27y^{3} =\frac{1331}{8} - 220$$
$$8x^{3}+27y^{3} =\frac{1331}{8} - \frac{1760}{8} = - \frac{429}{8}$$
If $$x=3t, y=\frac{1}{2} (t+1)$$ then the value of t for which $$x=2y$$ is
Given that $$x=3t, y=\frac{1}{2} (t+1)$$
if $$x=2y$$
3t = 2y
2y =2$$\frac{1}{2}\times(t+1)$$
3t = t+1
t = $$\frac{1}{2}$$
The graphs of x = a and y = b intersect at
Clearly, the lines x = a and y = b meets at a point (a,b)
The mean of 20 items is 55. If two items 45 and 30 are removed, the new mean of the remaining items is
The mean of 20 items is 55
=> Sum of 20 items = 55*20 = 1100
If 45 and 30 are removed
=> New sum = 1100-45-30 = 1025
and number of items remained = 20-2 = 18
=> New mean = $$\frac{1025}{18}$$ = 56.9
If $$x = k^{3} - 3k^{2}$$ and $$y= 1 - 3k,$$ then for what value of $$k$$. will be $$x= y$$?
If $$x = k^{3} - 3k^{2}$$ and $$y= 1 - 3k$$
then for $$x = y$$
=> $$k^3 - 3k^2 = 1 - 3k$$
=> $$k^3 - 1 - 3k^2 + 3k = 0$$
=> $$(k-1)^3 = 0$$
=> $$k = 1$$
The area in sq. unit. of the triangle formed by the graphs ofx= 4, y= 3 and 3x+ 4y= 12 is
Clearly, x=4 is a vertical line passing through (4,0) and y=3 is a horizontal line passing through the point (0,3)
The line 3x+ 4y= 12 also passes through (4,0) and (0,3) .
Hence, it is a right angled triangle with base=4 units and height=3 units.
The area of the triangle is = $$\frac{1}{2}bh$$
= $$\frac{1}{2}\times4\times3$$
=6 units
The equations 3x+ 4y = 10 and -x+ 2y = 0 have the solution (a, b). The value of a + b is
3x + 4y = 10 ---> (1)
x - 2y = 0
2x - 4y = 0 ---> (2)
5x = 10
x = 2
y = 1
x + y = 3
If $$x^{3}+\frac{3}{x}=4(a^{3}+b^{3})$$ and $$3x+\frac{1}{x^3}=4(a^{3}-b^{3})$$, then $$a^{2}-b^{2}$$ is equal to
Given : $$x^{3}+\frac{3}{x}=4(a^{3}+b^{3})$$ and $$3x+\frac{1}{x^3}=4(a^{3}-b^{3})$$
Adding the above equations, we get :
=> $$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = 8a^3$$
=> $$(x + \frac{1}{x})^3 = (2a)^3$$
=> $$x + \frac{1}{x} = 2a$$ ------------------Eqn(1)
Similarly, subtracting the above equations, we get :
=> $$x - \frac{1}{x} = 2b$$ ---------------Eqn(2)
Now, adding eqns(1) & (2)
=> $$2(a + b) = 2x$$
=> $$(a + b) = x$$
SUbtracting eqn(2) from (1)
=> $$2(a - b) = \frac{2}{x}$$
=> $$(a - b) = \frac{1}{x}$$
To find : $$a^{2}-b^{2}$$
= $$(a + b) (a - b)$$
= $$(x) (\frac{1}{x}) = 1$$
If x= 6 + 1 , then the value of $$x^{4}+\frac{1}{x^4}$$
If $$x,y,z \neq 0$$ and $$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}$$ = $$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$$ then the relation among x, y, z is
Expression : $$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}$$ = $$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$$
Taking L.C.M on both sides
= $$\frac{x^2 + y^2 + z^2}{x^2 y^2 z^2} = \frac{xy + yz + zx}{x^2 y^2 z^2}$$
= $$x^2 + y^2 + z^2 - xy - yz - zx = 0$$
= $$\frac{1}{2} [(x-y)^2 + (y-z)^2 + (z-x)^2] = 0$$
=> $$(x = y)$$ and $$(y = z)$$ and $$(z = x)$$
=> $$x = y = z$$
If $$a^{2} + b^{2} + c^{2} + 3 = 2 (a- b - c)$$, then the value of $$2 a - b + c$$ is :
Expression : $$a^{2} + b^{2} + c^{2} + 3 = 2 (a- b - c)$$
=> $$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$$
=> $$(a-1)^2 + (b+1)^2 + (c+1)^2 = 0$$
$$\therefore$$$$ a-1 = 0 => a = 1$$
and $$b+1 = 0 => b = -1$$
and $$c+1 = 0 => c = -1$$
To find : $$2a - b + c$$
= $$2*1 - (-1) + (-1)$$
= $$2 + 1 - 1 = 2$$
Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is
Let no. of correct questions be $$x$$ and incorrect be $$y$$
=> $$x + y = 30$$
Also, $$3x - 2y = 40$$
Solving above equations, we get :
=> $$x = 20$$ and $$y = 10$$
=> No. of correct sums = 20
A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6 . What is the fraction ?
Let the fraction be $$\frac{x}{y}$$
=> $$\frac{x+2}{y+2} = \frac{9}{11}$$
=> $$11x + 22 = 9y + 18$$
=> $$11x - 9y + 4 = 0$$
Also, $$\frac{x+3}{y+3} = \frac{5}{6}$$
=> $$6x + 18 = 5y + 15$$
=> $$6x - 5y + 3 = 0$$
Solving above equations, we get $$x = 7$$ and $$y = 9$$
=> Required fraction = $$\frac{7}{9}$$
If a* b= a+ b+ a/b, then the value of 12 * 4 is :
If $$a*b = a + b + \frac{a}{b}$$
=> $$12*4 = 12 + 4 + \frac{12}{4}$$
= $$16 + 3 = 19$$
If $$x^{2}+y^{2}+z^{2}=2(x-y-z)-3$$, then the value of $$2x-3y+4z$$ is [Assume that x, y, z are all real numbers) :
Expression : $$x^{2}+y^{2}+z^{2}=2(x-y-z)-3$$
=> $$x^2 - 2x + y^2 + 2y + z^2 + 2z + 3 = 0$$
=> $$(x^2 - 2x + 1) + (y^2 + 2y + 1) + (z^2 + 2z + 1) = 0$$
=> $$(x-1)^2 + (y+1)^2 + (z+1)^2 = 0$$
=> $$x = 1 , y = -1 , z = -1$$
To find : $$2x-3y+4z$$
= $$(2*1) - (3* -1) + (4* -1)$$
= $$2+3-4 = 1$$
If (a + b + c) = 0, then $$(\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab})$$ is
given (a + b + c) = 0
if (a + b + c) = 0,
$$a^{3}+b^{3}+c^{3}=3abc$$
$$\frac{a^{3}+b^{3}+c^{3}}{abc}=3$$
$$\frac{a^{3}}{abc}+\frac{b^{3}}{abc}+\frac{c^{3}}{abc}=3$$
$$\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}=3$$
so the answer is option A.
If x and y are positive real numbers and xy = 8, then the minimum value of 2x + y is
As $$x$$ & $$y$$ are positive real numbers, we can use AM $$\geq$$ GM
Let two numbers be $$2x$$ and $$y$$
=> $$\frac{2x + y}{2} \geq \sqrt{2xy}$$
=> $$2x + y \geq 2\sqrt{16}$$
=> $$2x + y \geq 8$$
=> Minimum value of $$(2x + y)$$ is 8
The graphs of 2x + 1 = 0 and 3y- 9 = 0 intersect at the point
Equations : $$2x + 1 = 0$$ and $$3y - 9 = 0$$
Solving above equations, we get :
$$x = \frac{-1}{2}$$ and $$y = 3$$
=> The graphs of above equations will intersect at $$(\frac{-1}{2} , 3)$$
The total cost of 8 buckets and 5 mugs is 92 and the total cost of 5 buckets and 8 mugs is 77. Find the cost of 2 mugs and 3 buckets.
Let cost of 1 bucket be $$x$$ and one mug be $$y$$
=> $$8x + 5y = 92$$
and $$5x + 8y = 77$$
Solving above equations, we get $$x = 9$$ and $$y = 4$$
=> Cost of 2 mugs and 3 buckets = $$3x + 2y$$
= 3*9 + 2*4 = 35
The total cost of 8 buckets and 5 mugs is 92 and the total cost of 5 buckets and 8 mugs is 77. Find the cost of 2 mugs and 3 buckets.
Let cost of 1 bucket be $$x$$ and one mug be $$y$$
=> $$8x + 5y = 92$$
and $$5x + 8y = 77$$
Solving above equations, we get $$x = 9$$ and $$y = 4$$
=> Cost of 2 mugs and 3 buckets = $$3x + 2y$$
= 3*9 + 2*4 = 35
Find the maximum number of trees which can be planted, 20 metres apart, on the two sides of a straight road 1760 metres long
To understand this, let's focus on one side to begin with.
If the straight road was 20 meters, you could have planted two trees.
If it was 40 meters, you could have planted three trees
If it was 'n' meters, you could have planted $$(\frac{n}{20}+1)$$ trees
Number of trees that can be planted on one side = $$\frac{1760}{20}+1$$ = 88 + 1 = 89
Total trees = 89*2 = 178
If 4x/3 + 2P = 12, for what value of P, x = 6?
Expression : $$\frac{4x}{3} + 2P = 12$$
Putting x = 6, we get :
=> $$8 + 2P = 12$$
=> $$P = \frac{4}{2} = 2$$
If $$\frac{4x}{3}+2p=12$$ for what value of $$p, x=6$$ ?
Expression : $$\frac{4x}{3}+2p=12$$
Putting x = 6
=> $$8 + 2p = 12$$
=> $$p = \frac{4}{2} = 2$$
If $$x^{2} - y^{2} = 80$$ and x- y= 8, then the average of x and y is
$$x^{2} - y^{2} = 80$$ and $$x - y= 8$$
=> $$x + y = \frac{x^2 - y^2}{x - y} = \frac{80}{8}$$
= 10
$$\therefore$$ Required average = 10/2 = 5
The straight line 2x + 3y = 12 passes through :
The line above represents the equation $$2x + 3y = 12$$
Clearly, it passes through 1st, 2nd and 4th quad.
If a + b+c= 0, then the value $$(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$$ $$(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$$ is:
Given : $$a+b+c = 0$$
Let $$a = 1 , b = 1$$ and $$c = -2$$ [We can take any values that satisfy above equation]
To find : $$(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$$ $$(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$$
= $$(\frac{2}{-2} + \frac{-1}{1} + \frac{-1}{1}) (\frac{1}{-1} + \frac{1}{-1} + \frac{-2}{2})$$
= $$(-3) (-3) = 9$$
If a, b, c are non-zero, $$a+\frac{1}{b}=1$$ and $$b+\frac{1}{c}=1$$ then the value of abc is :
It is given that : $$b + \frac{1}{c} = 1$$
=> $$b = (1 - \frac{1}{c})$$ -------------Eqn(1)
Also, $$a + \frac{1}{b} = 1$$
=> $$a = 1 - \frac{1}{b}$$
=> $$a = 1 - \frac{1}{1 - \frac{1}{c}}$$ [Using Eqn(1)]
=> $$a = (1 - \frac{c}{c-1})$$ ---------------Eqn(2)
To find : $$abc$$
Using eqn(1) and (2)
= $$(1 - \frac{c}{c-1}) (1 - \frac{1}{c}) (c)$$
= $$(\frac{-1}{c-1}) (\frac{c-1}{c}) (c)$$
= $$-1$$
The value of a machine depreciates every year at the rate of 10% on its value at the beginning of that year. If the current value of the machine is r 729, its worth 3years ago was:
Let the value of the machine 3 years ago be $$100x$$
Since, every year value decreases by 10%
=> Current value = $$(0.9)^3 * 100x = 729$$
=> $$100x = \frac{729}{0.729} = 1000$$
=> $$x = 10$$
=> Value 3 years ago = Rs. 1,000
A three digit number 4a3 is added to another three digit number 984 to give the four digit number 13b7 which is divisible by 11. Then the value of (a+b) is:
it is given that three digit number 4a3 is added to another three digit number 984 to give the four digit number 13b7 which is divisible by 11
400 + 10a + 3 + 984 = 1300 + 10b + 7
10a - 10 b = -80
b-a=8.....(1)
13b7 is divisible by 11 and hence using its divisibility rule we can say that 9-b will be of the form 11k and hence b can only take 9 value
and so a = 1
a+b = 1+9 = 10
If 50% of (x - y) = 30% of (x + y), then what per cent of x is y?
Expression : 50% of (x - y) = 30% of (x + y)
=> $$\frac{50}{100} (x - y) = \frac{30}{100} (x + y)$$
=> $$5x - 5y = 3x + 3y$$
=> $$2x = 8y$$
=> $$\frac{y}{x} = \frac{1}{4}$$
$$\therefore \frac{y}{x}\% = \frac{1}{4} * 100 = 25\%$$
When 7 is subtracted from thrice a number, the result is 14. What is the number ?
Let the number be $$x$$
Acc to ques :
=> $$3x - 7 = 14$$
=> $$x = \frac{21}{3} = 7$$
A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. Litres of liquid A contained by the can initially was
$$\frac{QNR}{total} = \frac{QNR(initial)}{total} (1- \frac{\textrm{Replaced Quantity}}{total})$$
QNR is the quantity which has not entered again
so here QNR is A
$$\frac{A}{12y} = \frac{7y}{12y} (1- \frac{9}{12y})$$
A = 7/4 (4y-3)
B = 12y - 7y + 5.25 = 5y + 5.25
$$\frac{7/4 (4y-3)}{5y + 5.25} = \frac{7}{9}$$
y = 7/4
so volume = 12y = 21 ltr
The mean of 50 numbers is 30. Later it was discoverd that two entries were wrongly entered as 82 and 13 instead of 28 and 31. Find the correct mean.
average = $$\frac{Sum}{NumberofElements}$$
it is given that mean of 50 numbers = 30
Sum of 50 numbers = 50 x 30 = 1500
Corrected Sum = 1500 - (82 + 13) + (28 + 31) = 1464
New and corrected average = $$\frac{1464}{50}$$ = 29.28
Terms of Service
Educational materials for CAT preparation
Day-wise Structured & Planned Preparation Guide
By proceeding you agree to create your account
Free CAT Schedule PDF will be sent to your email address soon !!!
