eSaral - JEE | NEET | Class 8-10 Exam Prep App

The point of the form (a, a), where a ≠ 0, lies on

Question: The point of the form (a,a), wherea 0, lies on(a) thex-axis(b) they-axis(c) the liney=x(d) the linex+y= 0 Solution: (c) the line $y=x$ Given, a point of the form $(a, a)$, where $a \neq 0$. When $a=1$, the point is $(1,1)$ When $a=2$, the point is $(2,2)$...... and so on. Plot the points $(1,1)$ and $(2,2) \ldots$ and so on. Join the points and extend them in both the direction. You will get the equation of the line $y=x$....

Write the arithmetic progression when first term a and common difference d are as follows:

Question: Write the arithmetic progression when first termaand common differencedare as follows: (i) $a=4, d=-3$ (ii) $a=-1, d=\frac{1}{2}$ (iii) $a=-1.5, d=-0.5$ Solution: In the given problem, we are given its first term (a) and common difference (d). We need to find the A.P (i) $a=4, d=-3$ Now, as $a=4$ A.P would be represented by $a, a_{1}, a_{2}, a_{3}, a_{4}, \ldots \ldots$ So, $a_{1}=a+d$ $a_{1}=4+(-3)$ $a_{1}=1$ Similarly, $a_{2}=a_{1}+d$ $a_{2}=1+(-3)$ $a_{2}=-2$ Also, $a_{3}=a_{2}+d$ $...

A 5-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 without repetition.

Question: A 5-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is (a) 216 (b) 600 (c) 240 (d) 3125 Solution: (a) 216 A number is divisible by 3 when the sum of the digits of the number is divisible by 3. Out of the given 6 digits, there are only two groups consisting of 5 digits whose sum is divisible by 3. 1+2+3+4+5 = 15 0+1+2+4+5 = 12 Using the digits 1, 2, 3, 4 and 5, the 5 digit numbers tha...

The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive,

Question: The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive, is (a) 4! 3! (b) 4! (c) 3! 3! (d) none of these. Solution: (a) 4! 3! According to the question, 3 men have to be 'consecutive' means that they have to be considered as a single man. But, these 3 men can be arranged among themselves in 3! ways. And, the remaining 3 men, along with this group, can be arranged among themselves in 4! ways. Total number of arrangements = 4! 3!...

If in a group of n distinct objects, the number of arrangements of 4 objects is 12 times the number of arrangements of 2 objects,

Question: If in a group ofndistinct objects, the number of arrangements of 4 objects is 12 times the number of arrangements of 2 objects, then the number of objects is (a) 10 (b) 8 (c) 6 (d) none of these. Solution: (c) 6 According to the question: ${ }^{n} P_{4}=12 \times{ }^{n} P_{2}$ $\Rightarrow \frac{n !}{(n-4) !}=12 \times \frac{n !}{(n-2) !}$ $\Rightarrow \frac{(n-2) !}{(n-4) !}=12$ $\Rightarrow(n-2)(n-3)=4 \times 3$ $\Rightarrow n-2=4$ $\Rightarrow n=6$...

If in a group of n distinct objects, the number of arrangements of 4 objects is 12 times the number of arrangements of 2 objects,

Question: If in a group ofndistinct objects, the number of arrangements of 4 objects is 12 times the number of arrangements of 2 objects, then the number of objects is (a) 10 (b) 8 (c) 6 (d) none of these. Solution: (c) 6 According to the question: ${ }^{n} P_{4}=12 \times{ }^{n} P_{2}$ $\Rightarrow \frac{n !}{(n-4) !}=12 \times \frac{n !}{(n-2) !}$ $\Rightarrow \frac{(n-2) !}{(n-4) !}=12$ $\Rightarrow(n-2)(n-3)=4 \times 3$ $\Rightarrow n-2=4$ $\Rightarrow n=6$...

Which of the following functions form Z to itself are bijections?

Question: Which of the following functions form Z to itself are bijections? (a) $f(x)=x^{3}$ (b) $f(x)=x+2$ (c) $f(x)=2 x+1$ (d) $f(x)=x^{2}+x$ Solution: (a) $f$ is not onto because for $y=3 \in$ Co-domain $(Z)$, there is no value of $\mathrm{x} \in$ Domain(Z) $x^{3}=3$ $\Rightarrow x=\sqrt[3]{3} \notin Z$ $\Rightarrow f$ is not onto. So, fis not a bijection. (b) Injectivity:Letxandybe two elements of the domain (Z), such that $x+2=y+2$ $\Rightarrow x=y$ So, $f$ is one-one. Surjectivity:Letybe a...

If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary,

Question: If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is (a) 324 (b) 341 (c) 359 (d) none of these Solution: (a) 324 When arranged alphabetically, the letters of the word KRISNA are A, I, K, N, R and S.Number of words that will be formed with A as the first letter = Number of arrangements of the remaining 5 letters = 5 Number of words that will be formed with I as the first letter = Numbe...

For the following arithmetic progressions write the first term a and the common difference d:

Question: For the following arithmetic progressions write the first termaand the common differenced: (i) 5, 1, 3, 7, ... (ii) $\frac{1}{5}, \frac{3}{5}, \frac{5}{5}, \frac{7}{5}$ (iii) 0.3, 0.55, 0.80, 1.05, ...(iv) 1.1, 3.1, 5.1, 7.1, ... Solution: In the given problem, we need to write the first term (a) and the common difference (d) of the given A.P (i) 5, 1, 3, 7 Here, first term of the given A.P is (a) = 5 Now, we will find the difference between the two terms of the given A.P $a_{2}-a_{1}=...

The equation of the x-axis is

Question: The equation of thex-axis is(a)x= 0(b)y= 0(c)x=y(d)x+y= 0 Solution: Since, they-coordinate of any point onx-axis is always 0.So, the equation of thex-axis isy= 0.Hence, the correct option is (b)....

Number of all four digit numbers having different digits formed of the digits 1, 2, 3, 4 and 5 and divisible by 4 is

Question: Number of all four digit numbers having different digits formed of the digits 1, 2, 3, 4 and 5 and divisible by 4 is (a) 24 (b) 30 (c) 125 (d) 100 Solution: (a) 24 In order to make a number divisible by 4, its last two digits must be divisible by 4, which in this case can be 12, 24, 32 or 52. Since repetition of digits is not allowed, the remaining first two digits can be arranged in $3 \times 2$ ways in each case. $\therefore$ Total number of numbers that can be formed $=4 \times\{3 \...

Two students A and B contributed ₹ 100 towards the Prime Minister's Relief Fund to help the earthquake victims

Question: Two students A and B contributed ₹ 100 towards the Prime Minister's Relief Fund to help the earthquake victims. Write a linear equation to satisfy the above data and draw its graph. Solution: Let the contribution of A and B be₹xand₹y,respectively.Total contribution of A and B =₹x+₹y=₹ (x+y)It is given that the total contribution of A and Bis ₹ 100. x+y= 100This is the linear equation satisfying the the given data.x+y= 100⇒y= 100xWhenx= 10,y= 10010 = 90Whenx= 40,y= 10040 = 60Whenx=60,y=...

The number of ways to arrange the letters of the word CHEESE

Question: The number of ways to arrange the letters of the word CHEESE are (a) 120 (b) 240 (c) 720 (d) 6 Solution: (a) 120 Total number of arrangements of the letters of the word $\mathrm{CHEESE}=$ Number of arrangements of 6 things taken all at a time, of which 3 are of one kind $=\frac{6 !}{3 !}=120$...

Draw the graph for each of the equation x + y = 6 and x – y = 2

Question: Draw the graph for each of the equationx + y= 6 andx y= 2 on the same graph paper and find the coordinates of the point where the two straight lines intersect. Solution: $x+y=6$ $\Rightarrow y=-x+6$ When $x=0, y=-0+6=6$ When $x=1, y=-1+6=5$ When $x=3, y=-3+6=3$ Thus, the points on the linex + y = 6are as given in the following table: Plotting the points (0,6),(1,5) and(3,3) and drawing a line passing through these points, we obtain the graph of of the linex + y = 6. $x-y=2$ $\Rightarro...

The number of ways in which the letters of the word 'CONSTANT'

Question: The number of ways in which the letters of the word 'CONSTANT' can be arranged without changing the relative positions of the vowels and consonants is (a) 360 (b) 256 (c) 444 (d) none of these. Solution: (a) 360 The word CONSTANT consists of two vowels that are placed at the 2ndand 6thposition, and six consonants. The two vowels can be arranged at their respective places, i.e. 2ndand 6thplace, in 2! ways. The remaining 6 consonants can be arranged at their respective places in $\frac{6...

The number of arrangements of the word "DELHI" in which E precedes I is

Question: The number of arrangements of the word "DELHI" in whichEprecedesIis (a) 30 (b) 60 (c) 120 (d) 59 Solution: (b) 60 There are 4 cases where E precedes I i.e. Case 1: When E and I are together, which are possible in 4 ways whereas other 3 letters are arranged in 3!, So, the number of arrangements $=4 \times 3 !=24$ Case 2: When E and I have 1 letter in between, which are possible in 3 ways whereas other 3 letters are arranged in 3!, So,the number of arrangements $=3 \times 3 !=18$ Case 3:...

If one root of the equation 4x2 − 2x + (λ − 4) = 0

Question: If one root of the equation $4 x^{2}-2 x+(\lambda-4)=0$ be the reciprocal of the other, then $\lambda=$ (a) 8(b) 8(c) 4(d) 4 Solution: Let $\alpha$ and $\beta$ be the roots of quadratic equation $4 x^{2}-2 x+(\lambda-4)=0$ in such a way that $\alpha=\frac{1}{\beta}$ Here, $a=4, b=-2$ and,$c=(\lambda-4)$ Then, according to question sum of the roots $\alpha+\beta=\frac{-b}{a}$ $\frac{1}{\beta}+\beta=\frac{-(-2)}{4}$ $\frac{1+\beta^{2}}{\beta}=\frac{1}{2}$ $2+2 \beta^{2}=\beta$ $2+2 \beta...

If one of the equation x2 + ax + 3 = 0 is 1, then its other root is

Question: If one of the equation $x^{2}+a x+3=0$ is 1 , then its other root is (a) 3(b) 3(c) 2(d) 1 Solution: Let $\alpha$ and $\beta$ be the roots of quadratic equation $x^{2}+a x+3=0$ in such a way that $\alpha=1$ Here, $a=1, b=a$ and, $c=3$ Then, according to question sum of the roots $\alpha+\beta=\frac{-b}{a}$ $1+\beta=\frac{-a}{1}$ $\beta=-a-1$ And the product of the roots $\alpha \cdot \beta=\frac{3}{1}$ $1 \times \beta=3$ $\beta=3$ Therefore, value of other root be $\beta=3$ Thus, the co...

If one of the equation x2 + ax + 3 = 0 is 1, then its other root is

Question: If one of the equation $x^{2}+a x+3=0$ is 1 , then its other root is (a) 3(b) 3(c) 2(d) 1 Solution: Let $\alpha$ and $\beta$ be the roots of quadratic equation $x^{2}+a x+3=0$ in such a way that $\alpha=1$ Here, $a=1, b=a$ and, $c=3$ Then, according to question sum of the roots $\alpha+\beta=\frac{-b}{a}$ $1+\beta=\frac{-a}{1}$ $\beta=-a-1$ And the product of the roots $\alpha \cdot \beta=\frac{3}{1}$ $1 \times \beta=3$ $\beta=3$ Therefore, value of other root be $\beta=3$ Thus, the co...

If one root the equation 2x2 + kx + 4 = 0 is 2, then the other root is

Question: If one root the equation $2 x^{2}+k x+4=0$ is 2 , then the other root is (a) 6(b) 6(c) 1(d) 1 Solution: Let $\alpha$ and $\beta$ be the roots of quadratic equation $2 x^{2}+k x+4=0$ in such a way that $\alpha=2$ Here, $a=2, b=k$ and,$c=4$ Then , according to question sum of the roots $\alpha+\beta=\frac{-b}{a}$ $2+\beta=\frac{-k}{2}$ $\beta=\frac{-k}{2}-2$ $\beta=\frac{-k-4}{2}$ And the product of the roots $\alpha \cdot \beta=\frac{c}{a}$ $=\frac{4}{2}$ $=2$ Putting the value of $\bet...

The number of six letter words that can be formed using the letters of the word "ASSIST" in which S's alternate with other letters is

Question: The number of six letter words that can be formed using the letters of the word "ASSIST" in which S's alternate with other letters is (a) 12 (b) 24 (c) 18 (d) none of these. Solution: (a) 12 All S's can be placed either at even places or at odd places, i.e. in 2 ways. The remaining letters can be placed at the remaining places in 3!, i.e. in 6 ways. $\therefore$ Total number of ways $=6 \times 2=12$...

The number of words from the letters of the word 'BHARAT' in which B and H will never come together,

Question: The number of words from the letters of the word 'BHARAT' in which B and H will never come together, is (a) 360 (b) 240 (c) 120 (d) none of these. Solution: (b) 240 Total number of words that can be formed of the letters of the word BHARAT $=\frac{6 !}{2 !}$ = 360 Number of words in which the letters $\mathrm{B}$ and $\mathrm{H}$ are always together $=2 \times \frac{5 !}{2 !}$ = 120 $\therefore$ Number of words in which the letters $\mathrm{B}$ and $\mathrm{H}$ are never together $=360...

If one of the equation ax2 + bx + c = 0 is three times times

Question: If one of the equationax2+bx+c= 0 is three times times the other, thenb2:ac= (a) 3 : 1(b) 3 : 16(c) 16 : 3(d) 16 : 1 Solution: Let $\alpha$ and $\beta$ be the roots of quadratic equation $a x^{2}+b x+c=0$ in such a way that $\alpha=3 \beta$ Here, $a=a, b=b$ and,$c=c$ Then, according to question sum of the roots $a+\beta=\frac{-b}{a}$ $3 \beta+\beta=\frac{-b}{a}$ $4 \beta=\frac{-b}{a}$ $\beta=\frac{-b}{4 a} .$.....(1) And the product of the roots $\alpha \cdot \beta=\frac{c}{a}$ $3 \bet...

The number of different signals which can be given from 6 flags of different colours taking one or more at a time,

Question: The number of different signals which can be given from 6 flags of different colours taking one or more at a time, is (a) 1958 (b) 1956 (c) 16 (d) 64 Solution: (b) 1956 Number of permutations of six signals taking 1 at a time =6P1 Number of permutations of six signals taking 2 at a time =6P2 Number of permutations of six signals taking 3 at a time =6P3 Number of permutations of six signals taking 4 at a time =6P4 Number of permutations of six signals taking 5 at a time =6P5 Number of p...

The number of different signals which can be given from 6 flags of different colours taking one or more at a time,

Question: The number of different signals which can be given from 6 flags of different colours taking one or more at a time, is (a) 1958 (b) 1956 (c) 16 (d) 64 Solution: (b) 1956 Number of permutations of six signals taking 1 at a time =6P1 Number of permutations of six signals taking 2 at a time =6P2 Number of permutations of six signals taking 3 at a time =6P3 Number of permutations of six signals taking 4 at a time =6P4 Number of permutations of six signals taking 5 at a time =6P5 Number of p...

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