If the equation x2 − bx + 1 = 0 does not possess real roots, then
Question: If the equation $x^{2}-b x+1=0$ does not possess real roots, then (a) $-3b3$ (b) $-2b2$ (c) $b2$ (d) $b-2$ Solution: The given quadric equation is $x^{2}-b x+1=0$, and does not have real roots. Then find the value ofb. Here, $a=1, b=-b$ and,$c=1$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=-b$ and, $c=1$ $=(-b)^{2}-4 \times 1 \times 1$ $=b^{2}-4$ The given equation does not have real roots, if $D0$ $b^{2}-40$ $b^{2}4$ $b\sqrt{4}$ $b\pm 2$ Therefore, the value of $-2b2$...
Read More →In how many ways can the letters of the word 'ARRANGE' be arranged so that the two R's are never together?
Question: In how many ways can the letters of the word 'ARRANGE' be arranged so that the two R's are never together? Solution: The word ARRANGE consists of 7 letters including two Rs and two As, which can be arranged in $\frac{7 !}{2 ! 2 !}$ ways. Total number of words that can be formed using the letters of the word ARRANGE = 1260 Number of words in which the two Rs are always together = Considering both Rs as a single entity = Arrangements of 6 things of which two are same (two As) $=\frac{6 !...
Read More →In how many ways can the letters of the word 'ARRANGE' be arranged so that the two R's are never together?
Question: In how many ways can the letters of the word 'ARRANGE' be arranged so that the two R's are never together? Solution: The word ARRANGE consists of 7 letters including two Rs and two As, which can be arranged in $\frac{7 !}{2 ! 2 !}$ ways. Total number of words that can be formed using the letters of the word ARRANGE = 1260 Number of words in which the two Rs are always together = Considering both Rs as a single entity = Arrangements of 6 things of which two are same (two As) $=\frac{6 !...
Read More →If x = 3 and y = 4 is a solution of the equation
Question: Ifx= 3 andy= 4 is a solution of the equation 5x 3y=k, find the value ofk. Solution: Given:5x 3y=kSincex= 3 andy= 4 is a solution of the given equation so, it should satisfy the equation. $5(3)-3(4)=k$ $\Rightarrow 15-12=k$ $\Rightarrow 3=k$...
Read More →If the roots of the equations (a2+b2)x2−2b(a+c)x+(b2+c2)=0 are equal, then
Question: If the roots of the equations $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$ are equal, then (a) $2 b=a+c$ (b) $b^{2}=a c$ (c) $b=\frac{2 a c}{a+c}$ (d) $b=a c$ Solution: The given quadric equation is $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+b^{2}+c^{2}=0$, and roots are equal. Here, $a=\left(a^{2}+b^{2}\right), b=-2 b(a+c)$ and, $c=b^{2}+c^{2}$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=\left(a^{2}+b^{2}\right), b=-2 b(a+c)$ and, $c=b^{2}+c^{2}$ ...
Read More →How many number of four digits can be formed with the digits 1, 3, 3, 0?
Question: How many number of four digits can be formed with the digits 1, 3, 3, 0? Solution: The given digits are 1, 3, 3, 0. Total numbers that can be formed with these digits $=\frac{4 !}{2 !}$ Now, these numbers also include the numbers in which the thousand's place is 0. But, to form a four digit number, this is not possible. $\therefore$ Numbers in which the thousand's place is fixed as zero $=$ Ways of arranging the remaining digits $(1,3$ and 3$)$ in three places $=\frac{3 !}{2 !}$ $\ther...
Read More →Find five different solutions of each of the following equations:
Question: Find five different solutions of each of the following equations: (i) $2 x-3 y=6$ (ii) $\frac{2 x}{5}+\frac{3 y}{10}=3$ (iii) $3 y=4 x$ Solution: (i) 2x 3y = 6 (ii) $\frac{2 x}{5}+\frac{3 y}{10}=3$ (iii) 3y= 4x...
Read More →How many different signals can be made from 4 red,
Question: How many different signals can be made from 4 red, 2 white and 3 green flags by arranging all of them vertically on a flagstaff? Solution: We have to arrange 9 flags, out of which 4 are of one kind (red), 2 are of another kind (white) and 3 are of the third kind (green). $\therefore$ Total number of signals that can be generated with these flags $=\frac{9 !}{4 ! 2 ! 3 !}=1260$...
Read More →If the equations (a2+b2)x2−2(ac+bd)x+c2+d2=0 has equal roots, then
Question: If the equations $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+c^{2}+d^{2}=0$ has equal roots, then (a) $a b=c d$ (b) $a d=b c$ (c) $a d=\sqrt{b c}$ (d) $a b=\sqrt{c d}$ Solution: The given quadric equation is $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+c^{2}+d^{2}=0$, and roots are equal. Here, $a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=c^{2}+d^{2}$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=c^{2}+d^{2}$ $=\{-2(a c...
Read More →How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?
Question: How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places? Solution: There are 4 odd digits $(1,3,3$ and 1$)$ that are to be arranged in 4 odd places in $\frac{4 !}{2 ! 2 !}$ ways. The remaining 3 even digits 2,2 and 4 can be arranged in 3 even places in $\frac{3 !}{2 !}$ ways. By fundamental principle of counting: Required number of arrangements $=\frac{4 !}{2 ! 2 !} \times \frac{3 !}{2 !}=18$...
Read More →How many words can be formed by arranging the letters of the word 'MUMBAI' so that all M's come together?
Question: How many words can be formed by arranging the letters of the word 'MUMBAI' so that all M's come together? Solution: The word MUMBAI consists of 6 letters taht include two Ms. When we consider both the Ms as a single entity, we are left with 5 entities that can be arranged in 5! ways. Total number of words that can be formed with all the Ms together = 5! = 120...
Read More →How many words can be formed with the letters of the word 'PARALLEL'
Question: How many words can be formed with the letters of the word 'PARALLEL' so that all L's do not come together? Solution: The word PARALLEL consists of 8 letters that include two As and three Ls. Total number of words that can be formed using the letters of the word PARALLEL $=\frac{8 !}{2 ! 3 !}=3360$ Number of words in which all the Ls come together is equal to the condition if all three Ls are considered as a single entity. So, we are left with total 6 letters that can be arranged in $\f...
Read More →A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more,
Question: A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train. Solution: Let us assume that the speed of the train be x km/hr. We are also given that the distance covered during the journey is 360 km. Now, time taken during the journey $=\left(\frac{360}{x}\right) \mathrm{hr}$ Time taken for the new journey $=\left(\frac{360}{x+5}\right) \mathrm{hr}$ Accordi...
Read More →Check which of the following are the solutions of the equation 5x – 4y = 20.
Question: Check which of the following are the solutions of the equation 5x 4y= 20. (i) $(4,0)$ (ii) $(0,5)$ (iii) $\left(-2, \frac{5}{2}\right)$ (iv) $(0,-5)$ (v) $\left(2, \frac{-5}{2}\right)$ Solution: The equation given is5x 4y= 20.(i) (4, 0)Putting the value in the given equation we have LHS : $5(4)-4(0)=20$ RHS : 20 LHS = RHS Thus, (4, 0) is a solution of the given equation.(ii) (0, 5)Putting the value in the given equation we have LHS : $5(0)-4(5)=0-20=-20$ RHS : 20 $\mathrm{LHS} \neq \ma...
Read More →How many words can be formed with the letters of the word 'PARALLEL'
Question: How many words can be formed with the letters of the word 'PARALLEL' so that all L's do not come together? Solution: The word PARALLEL consists of 8 letters that include two As and three Ls. Total number of words that can be formed using the letters of the word PARALLEL $=\frac{8 !}{2 ! 3 !}=3360$ Number of words in which all the Ls come together is equal to the condition if all three Ls are considered as a single entity. So, we are left with total 6 letters that can be arranged in $\f...
Read More →An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Banglore.
Question: An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Banglore. If the average speed of the express train is 11 km/hr more than that of the passenger train, from the quadratic equation to find the average speed of express train. Solution: Now let us assume that the speed of the express train be x km/hr. Therefore according to the question speed of the passenger train will be x11 km/hr. Now we know that the total distance travelled by both the tra...
Read More →Find the total number of arrangements of the letters in the expression
Question: Find the total number of arrangements of the letters in the expressiona3b2c4when written at full length. Solution: When expanded,a3b2c4would result in total 9 letters. This is same as permuting 9 things, of which 3 are similar to the first kind, 2 are similar to the second kind and four are similar to the third kind, i.e. threeas , twobs and fourcs. Required number of arrangements $=\frac{9 !}{3 ! 2 ! 4 !}=1260$...
Read More →How many words can be formed with the letters of the word 'UNIVERSITY',
Question: How many words can be formed with the letters of the word 'UNIVERSITY', the vowels remaining together? Solution: The word UNIVERSITY consists of 10 letters that include four vowels of which two are same. Thus, the vowels can be arranged amongst themselves in $\frac{4 !}{2 !}$ ways. Keeping the vowels as a single entity, we are left with 7 letters, which can be arranged in 7! ways. By fundamental principle of counting, we get, Number of words $=7 ! \times \frac{4 !}{2 !}=60480$...
Read More →The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm,
Question: The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle. Solution: Now, since we have to find out base, let us assume the base to be x cm.Therefore the height of the triangle becomes x7.It is also given that the hypotenuse is 13 cm.By Pythagoras Theorem, $x^{2}+(x-7)^{2}=(13)^{2}$ $x^{2}+x^{2}+49-14 x=169$ $2 x^{2}-14 x-120=0$ $x^{2}-7 x-60=0$ Hence, this is our required quadratic equation....
Read More →Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
Question: Express each of the following equations in the formax + by + c= 0 and indicate the values ofa, b, cin each case. (i)x= 6(ii) 3xy=x 1(iii) 2x+ 9 = 0(iv) 4y= 7(v)x + y= 4 (vi) $\frac{x}{2}-\frac{y}{3}=\frac{1}{6}+y$ Solution: (i)x = 6 In the form of $a x+b y+c=0$ we have $x+0 y+(-6)=0$ where $a=1, b=0$ and $c=-6$ (ii) $3 x-y=x-1$ In the form of $a x+b y+c=0$ we have $2 x+(-1 y)+1=0$ where $a=2, b=-1$ and $c=1$ (iii) $2 x+9=0$ In the form of $a x+b y+c=0$ we have $2 x+(-1 y)+1=0$ where $a...
Read More →In how many ways can the letters of the word 'ALGEBRA'
Question: In how many ways can the letters of the word 'ALGEBRA' be arranged without changing the relative order of the vowels and consonants? Solution: The relative positions of all the vowels and consonants is fixed. The first letter is a vowel. It can be selected out of the 3 three vowels, of which two are same. So, the vowels can be arranged in selecting 3 things, of which two are of the same kind $\Rightarrow \frac{3 !}{2 !}$ The second, third, fifth and sixth letters are consonants that ca...
Read More →A cottage industry produces a certain number of toys in a day.
Question: A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. Ifxdenotes the number of toys produced that day, from the quadratic equation of findx. Solution: Now we know that x denotes the total number of toys produced in that day. But, the cost of production of a single toy is 55 minus the number of toys prod...
Read More →John and Jivanti together have 45 marbles. Both of them lost 5 marbles each,
Question: John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John andxmarbles. Solution: It is given that John had x marbles.We are also given that both John and Javanti had 45 marbles together.So, Javanti should have 45 x marbles with her.Now, it is given that both of them lose 5 marbles each.So in the new situation, John will h...
Read More →Find the number of words formed by permuting all the letters of the following words:
Question: Find the number of words formed by permuting all the letters of the following words: (i) INDEPENDENCE (ii) INTERMEDIATE (iii) ARRANGE (iv) INDIA (v) PAKISTAN (vi) RUSSIA (vii) SERIES (viii) EXERCISES (ix) CONSTANTINOPLE Solution: (i) This word consists of 12 letters that include three Ns, two Ds and four Es. The total number of words is the number of arrangements of 12 things, of which 3 are similar to one kind, 2 are similar to the second kind and 4 are similar to the third kind. $\Ri...
Read More →The product of two consecutive positive integers is 306.
Question: The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, ifxdenotes the smaller integer. Solution: Since it is given in the question that the numbers we have to find are consecutive positive integer numbers, therefore the difference between the two numbers should be equal to 1.For e.g. 7 and 8 or 26 and 27 are consecutive numbers. Let us assume the first number to be x. So our next consecutive number should be x+ 1. Now the question als...
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