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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Write the decimal expainsion of ` 73/ (((2^(4) xx 5^(3)))` |
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Answer» Correct Answer - 0.0365 ` 73/((2^(4) xx 5^(3)))= ( 73xx 5)/(2^(4) xx5^(4)) = 365/((2xx5(^(4)) = 365/((10)^(4)) = 365/10000 = 0.0365` |
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| 102. |
Express each of the following as a rational number in simplest form. (a) ` 0.overline(6)` (ii) ` 1.overline(8)` (iii) ` 0.1overline6` |
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Answer» Let ` x= 0.overline6`. Then , x = 0.666….. 10x = 6.666 On subtracting (i) from (ii), we get 9x= 6 ` Rightarrow x = 6/9 = 2/3` Hence ` 0.overline(6) = 2/3` (b) Let ` x= 1.overline(8)` then, x= 1.888 10x= 18.888 On subtracting (i) from (ii),we get ` 9x = 17 Rightarrow x = 17/9 = 1""8/9` Hence, ` 1.overline(8) = 1""8/9` (c) Let `x= 0.1overline(6)` , then x = 0.1666 10x= 1.6666 And , 100x = 16.6666 On subtracting (ii) from (iii),we get 90x = 15 ` Rightarrow x 15/90 = 1/6` ` 0.1overline6= 1/6` |
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| 103. |
Express each of the following in the simplest form : (i) `0.bar(36)` (ii) `1.bar(046)` |
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Answer» (i) Let `x = 0.bar(36)` `rArr` x = 0.363636……. `" "`…(1) `rArr` 100 x = 36.363636……. `" "` …(2) Substracting equation (1) from (2), we get 99 x = 36 `rArr` `x = (36)/(99) = (4)/(11)` `rArr` `0.bar(36) = (4)/(11)` (ii) Let `x = 1.bar(046)` `rArr` x = 1.046046046..... `" "` ...(1) `rArr` 1000x = 1046.046046046.... `" "` ...(2) Subtracting equation (1) from (2), we get 999x = 1045 `rArr` `x = (1045)/(999) ` `rArr` `1.bar(046) = (1045)/(999)` (See the shortcut method in "Gold coins") |
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| 104. |
Find the successor of the number `-10`. |
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Answer» Successor of `-10 is -10+1=-9`. The number obtained by subtracting 1 form the given integer is called predecessor of the given integer. |
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| 105. |
Show that each of the following are non-terminating repeating decimal : (i) `(5)/(12)` (ii) `(7)/(75)` |
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Answer» (i) `(5)/(12) = (5)/(2 xx 2 xx 3) = (5)/(2^(2) xx 3)` `because` the denominator `2^(2) xx 3` is not in the form of `2^(m) xx 5^(n)` `therefore` it is non-terminating repeating decimal. ` " "`Hence Proved. (ii) `(7)/(75) = (7)/(3 xx 5 xx 5) = (7)/(3 xx 5^(2))` `because` the denominator `3 xx 5^(2)` is not in the form `2^(m) xx 5^(m)` `therefore` it is non-terminating repeating decimal. `" "` Hence Proved. |
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| 106. |
Verify wheter 223 is a prime number or not . |
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Answer» Given number `n=223` now `sqrt(223)˜15` We have `m=14`. Prime number below `m={2,3,5,7,11,13}`. Clearly 223, is not divisible by any of the above prime numbers Therefore, 223 is a prime number. |
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| 107. |
Find the predecessor of `-7`. |
| Answer» Predecessor of `-7 is -7-1=-8`. | |
| 108. |
Express each of the following in the simplest form : (i) `0.bar(6)` (ii) `3.bar(3)` |
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Answer» (i) Let `x = 0.bar(6)` `rArr` x = 0.666……. `" "`…(1) `rArr` 10 x = 6.666……. `" "` …(2) Substracting equation (1) from (2) we get 9 x = 6 `rArr` `x = (6)/(9) = (2)/(3)` `rArr` `0.bar(6) = (2)/(3)` (ii) Let `x = 3.bar(3)` `rArr` x = 3.333....... `" "` ...(1) `rArr` 10x = 33.333...... `" "` ...(2) Subtracting equation (1) from (2), we get 9x = 30 `rArr` `x = (30)/(9) = (10)/(3)` `rArr` `3.bar(3) = (10)/(3)` |
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| 109. |
The decimal expansionof the rational number `(43)/(2^4xx5^3)`will terminate after how many places of decimals? |
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Answer» `(43)/(2^(4) xx 5^(3)) = (43 xx 5)/(2^(4) xx 5^(4)) = (215)/(10^(4)) = 0.0215` `therefore` It will terminate after 4 places of decimals. |
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| 110. |
Find two rational numbers between `(3)/(4)and (2)/(7)`. |
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Answer» The rational number obtained from `((3)/(4)+(2)/(7))/(2)i.e., (29)/(56)` lies between `(3)/(2) and (2)/(7)` Similarly the rational number obtained from `((3)/(4)+(29)/(56))/(2)i.e., (71)/(112)` lies between `(3)/(4) and (2)/(7)` `therefore` The required rational numbers are `(29)/(56) and (71)/(112)`. |
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| 111. |
Find the value of `(0.424242...)-(0.353535...)`. |
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Answer» (a) HCF= 1 and LCM =product of numbers, (b) `mn=4199`. |
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| 112. |
A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1,2, 3 (Fig.), which are equally likely What is the probability that the atrow outcomes Wi at (i) an odd number (ii) a number greater than 3 (iii) a number less than 9 |
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Answer» Here, total number of outcomes`(n(S)) = 8` (i) Number of possible outcomes that arrow will point at odd number ` = 4` `:.` Probability that arrow will point at odd number ` = 4/8 = 1/2` (ii)Number of possible outcomes that arrow will point at a number greater than `3 = 5` `:.` Probability that arrow will point at a number greater than `3 = 5/8` (iii)Number of possible outcomes that arrow will point at a number less than `9 = 8` `:.` Probability that arrow will point at a number less than `9 = 8/8 = 1` |
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| 113. |
Consider the numbers `4^n`, where n is a natural number. Cheek whether there is any value of n for which `4^n`ends with the digit zero. |
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Answer» `4^n = (2*2)^n = 2^n*2^n` For a number to be end with `0`, it should be divided by `2 and 5`. But, as we can see `4^n` can be divided by only `2 or 2^n`. It can not be divided by `5`. So, `4^n` can not end with digit zero for any natural number. |
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| 114. |
By what number should 1365 be divided to get 31 as quotient and 32 as remainder ? |
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Answer» Correct Answer - 43 Let ther required divisor be x. ltbr. Then, dividend = ( divisor ` xx` quotient) + remainder `Rightarrow 1365 = (x xx 31) + 32` ,find x . |
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| 115. |
Prove that `sqrtp+sqrtq` is an irrational, where `p and q` are primes. |
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Answer» Let us assume that` sqrtp + sqrtq` is rational. Then , there exist co-primes a and b such that ` sqrtp + sqrtq = a/b` ` Rightarrow sqrtp = a/b -sqrtq` ` Rightarrow (sqrtp)^(2) = (a/b -sqrtq)^(2)` ` Rightarrow p = a^(2)/b^(2) - (2a)/bsqrtq + q` ` Rightarrow (2a)/(b) sqrtq = a^(2)/b^(2) + q -p` ` Rightarrow sqrtw = (a^(2) +b^(2)q - b^(2)p) xx b/(2a) = (a^(2)b+b^(3)(q-p))/(2a)` Since, a,b,p,q are intergers so `(a^(2)b+b^(3)(q-p))/(2a)` is rational Thus, ` sqrtq` is also rational. But, q being prime `sqrtq` is irrational. Since, a contradicition arises so our assumption is incorrect. Hence, `(sqrtp +sqrtq)` is irrational. |
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| 116. |
Every recurring decimal is a ____ number. |
| Answer» Correct Answer - Rational | |
| 117. |
The absolute value of `|x-2|+|x+2|`, if `0lt xlt 2 ` isA. `2x`B. 4C. `2(x+1)`D. |
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Answer» Correct Answer - B (i) Concept of modules (ii) `lt x lt 2 then |x-2|=-(x-2) and |x+2|=x+2` |
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| 118. |
The absolute value of `25-(25+10)+25div125xx25` isA. -5B. 3C. 15D. 5 |
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Answer» Correct Answer - D Simplify using BODMAS rule and evaluate its absolute value. |
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| 119. |
Find the absolute values of `x- 2`, where `x < 2`.A. `x-1`B. `x+2`C. `2-x`D. `2x` |
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Answer» Correct Answer - C Write the positive form of the given numbers. Also use, `|x-a|=x-a if x gt a` `=a-x if x lt a`. |
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| 120. |
What is the absolute value of -10? |
| Answer» Correct Answer - `|-10|=10` | |
| 121. |
What is the least natural number? |
| Answer» Correct Answer - 1 | |
| 122. |
All odd numbers are prime numbers. Is this statement true or false? |
| Answer» Correct Answer - False | |
| 123. |
The greatest common factor of relatively prime numbers is |
| Answer» Correct Answer - 1 | |
| 124. |
Which of the following values are even ? (A)`21+18+9+2+19` (B) `34xx28xx37xx94xx12712` (C) `33xx35xx37xx39xx41xx43` (D) `11xx11xx11xx11xx11xx.....` (E) `1^10` (F)` 39-24`A. A,B,CB. D,E,FC. BD. A,B,D,E |
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Answer» Correct Answer - C Recall the properties of odd and even numbers. |
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| 125. |
What is the number which has at least three factors called ? |
| Answer» Correct Answer - Composite number | |
| 126. |
If a, b, c and d are four positive real numbers such that sum of a, b and c is even the sum of b,c and d is odd, then `a^2-d^2` is necessarilyA. oddB. EvenC. PrimeD. Either (a) or (b) |
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Answer» Correct Answer - A (i) Subtract one condition from another and proceed. (ii) `(a+b+c)-(b+c+d)=` even-odd `rArr a-d` is odd. (iii) `a+d` is also odd. |
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| 127. |
What is the sum of ten odd numbers and eleven even numbers? (even/odd) |
| Answer» Correct Answer - Even number | |
| 128. |
Two relatively prime numbers need not be prime numbers. Is this statement true? |
| Answer» Correct Answer - True | |
| 129. |
The product of two positive integers and eight negative integers is a |
| Answer» Correct Answer - Positive number | |
| 130. |
Find the value of `k` such that `x^2+2x+9k-5=0` has both roots negative. |
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Answer» For both roots to be negative, `D ge 0`. `=>b^2 - 4ac ge 0` `=>2^2 - 4(9k-5) ge 0` `=>4-36k+20 ge 0` `=>k le 24/36=>k le 2/3->(1)` Also, for roots to be negative, their product should be greater than `0`.`:. 9k-5 gt 0=> k gt 5/9` So, value of k should be greater than `5/9` and less than or equal to `2/3`. |
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| 131. |
If the sets of equations `z^x=y^(2x),2^x=2.4^x,x+y+z=16,` the integral roots in the order `x,y,z` |
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Answer» Here, `z^x = y^(2x)` `=>z^x = (y^2)^x` `=>z = y^2->(1)` `2^z = 2.4^x` `=>2^z = 2.(2^2)^x=2^(2x+1)` `=>z = 2x+1` `=>x = (z-1)/2->(2)` From (1), `x = (y^2-1)/2` `x+y+z = 16` `=>(y^2-1)/2+y+y^2 = 16` `=>y^2-1+2y+2y^2 = 32` `=>3y^2+2y-33=0` `=>3y^2+11y-9y-33 = 0` `=>(3y+11)(y-3) = 0` `y = 3 and y = -11/3` But, we have to find only integer root. So, `y = 3` `x = (9-1)/2 = 4` `z = 3^2 = 9` |
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| 132. |
In the sets of equations `z^x =y^(2x)`, `2^z = 2. 4^x`, x+y+z = 16 , the integral roots in the order x, y, z are : |
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Answer» Here, `z^x = y^(2x)` `=>z^x = (y^2)^x` `=>z = y^2->(1)` `2^z = 2.4^x` `=>2^z = 2.(2^2)^x=2^(2x+1)` `=>z = 2x+1` `=>x = (z-1)/2->(2)` From (1), `x = (y^2-1)/2` `x+y+z = 16` `=>(y^2-1)/2+y+y^2 = 16` `=>y^2-1+2y+2y^2 = 32` `=>3y^2+2y-33=0` `=>3y^2+11y-9y-33 = 0` `=>(3y+11)(y-3) = 0` `y = 3 and y = -11/3` But, we have to find only integer root. So, `y = 3` `x = (9-1)/2 = 4` `z = 3^2 = 9` |
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| 133. |
I rode one-third of a journey at 10 km an hour, one-third more at 9, and the rest at 8 km an hour; if I had ridden half the journey at 10, and the other half at 8 km per hour, I should have been half a minute longer on the way. I ride for x distance. Then x is : |
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Answer» Time taken in first case `T_1 = x/(3*10)+x/(3*9)+x/(3*8) = x/30+x/27+x/24` Time taken in second case `T_2 = x/(2*10)+x/(2*8) = x/20+x/16` Now, we are are given, difference between `T_2` and `T_1` is `1/2` minute or `1/120` hour. `:. x/20+x/16-x/30-x/27-x/24 = 1/120` `=>(x(108+135-72-80-90))/2160 = 1/120` `=>x = 2160/120 km` `x = 18km` |
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| 134. |
Express `0.2overline25` as a rational number. |
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Answer» In `0.overline225` , periodicity `=2`. So, multiply the decimal fraction by 100 an subtract the original number form the proudct. let `0.overline225 =x` . Then , `x=0.2252525......` and `100x=22.525252......` Subtracing x from `100x`, we get `100x =22.5252...` `(-)x=2252.....` `99x=22.3` `x=(22.3)/(99)=(223)/(990)`. |
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| 135. |
Express `12.0bar5` as rational number. |
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Answer» Here the whole obtained by writing digits in their order in 1205 and the whole number made by non-reacurriring decimal in oreder is 12. The number of digits after decimal point =2 and the number of digits after the decimal point that do not recur `=0`. `12.05=(125)/(10^2-10^0)=(1193)/(99)` |
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| 136. |
Verify that 3,-1,-1/3 are the zeros of the cubic polynomial `p(x)=3x^3-5x^2-11x-3`. verify the relationship between the zeros and its coefficients. |
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Answer» `F(x)=(x-3)(x+1)(x+1/3)` `=(x^2-3x+x-3)(x+1/3)` `=(x^2-2x-3)(x+1/3)` `=x^3-2x^2-3x+1/3x^2-2/3x-1` `F(x)=x^3-5/3x^2-11/3x-1` |
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| 137. |
Express the following rational numbers as decimals and identify which of them are terminating ? (A) `(1)/(2)` (B) `-(2)/(3)` (C) `(15)/(5)` (D) `(5)/(6)` (E) `(1)/(7) (F) `17(5)/(6)` (G) `3(1)/(2)+8(1)/(4)` (H) `2^-4xx3` .A. (A) and (B)B. (C)and (D)C. (A) ,(C) and (G)D. (A),(C),(G) and (H) |
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Answer» Correct Answer - D If the denominator of a fraction is a factor of only 2(or) only 5(or ) of both 2 and 5, then it is a terminating decimal otherwise nonterminating. |
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| 138. |
The number 1.732 isA. an irrational numberB. a rational numberC. an integerD. a whole number |
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Answer» Correct Answer - B 1.732 = ` 1732/1000 ` which is a rational number. |
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| 139. |
If `1 leq k leq 25`, how many prime numbers are there which are of the form `6k + 1?`A. 15B. 16C. 17D. 18 |
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Answer» Correct Answer - B (i) Write all the prime numbers, which are less than or equal to 151. (ii) Multiply in the equation with 6. i.e., `6 k le 150`. `rArr 7 le 6k+1le 151`. Find the prime numbers from 7 to 151. |
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| 140. |
From the set `{2, 3, 4, 5, 6, 7, 8, 9}`, how many pairs of co-primes can be formed? |
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Answer» Recall the defination of primers and proceed. (b) Form all the pairs of co-primers with each of the sets like `(2,3),(2,5)`, etc. |
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| 141. |
Which pair of numbers below are twin primes ?A. 8 and 9B. 2 and 3C. 3 and 7D. 41 and 43 |
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Answer» Correct Answer - D Twin primes differ by 2. |
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| 142. |
Pairs of numbers of the form `(6k-1)` and `(6k+1)` are twin primes. Is this statement true ? |
| Answer» Correct Answer - No | |
| 143. |
In figure, the the A the shows shape of a solid copper piece (made of two pieces) with dimensions is shown. The face ABCDEF is the uniform cross-section. Assume that, the angles right angles. at A, B, C, D, E and F are Calculate the volume of the piece. |
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Answer» Area of shape=Area of AEFG+Area of CDEG =AF*FG+CD*ED =8*2+8*3 =40`cm^2` Volume =Area*depth =40*22=880`cm^3`. |
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| 144. |
If `23457 a68` divisible by 9, then the lest value of a is _____ |
| Answer» Correct Answer - 1 | |
| 145. |
Prime numbers differeing by 2 are called _____ |
| Answer» Correct Answer - Twin primes | |
| 146. |
If x is divisible by both 3 and 5 ,then x is divisible by 15. is this statement true ? |
| Answer» Correct Answer - Yes | |
| 147. |
Show that one and only one out of n,( n+2) and ( n +4) is divisble by 3, where n is any positive interger. |
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Answer» When , n is dividied by 3 , let q and r be the quotient and remainder respectively . `therefore " "` n = 3q + r where , `0 le r lt 3 ` i.e., r = 0 or r =1 or r =2 (i) When , r = 0 then n = 3q , which is divisible by 3 . n + 1 = 3q +1 which leaves a remainder 1 when divided by 3. i.e., (n +1) is not divisible by 3 . n +2 = 3q +2 which leaves a remainder 2 when divided by 3 . i.e., (n +2) is not divisible by 3 . So , only n = 3q is divisible by 3 when r = 0 . `" " .... (1)` (ii) When r = 1 , then n = 3q +1 which is not divisible by 3 . `" " (because` it leaves a remainder 1) n + 1 = 3q +2 which is not divisible by 3. `" " (because` remainder= 2) n + 2 = 3q + 3 = 3(q+1) which is divisible by 3 . So , only n +2 is divisible by 3 when r= 1 . `" "....(2)` (iii) When , r=2 then n = 3q +2 which is not divisible by 3 . `" " (because ` remainder =2) `implies " " n + 1 = 3q + 3 = 3(q +1)` which is divisible by 3. n +2 = 3q + 4 = 3q + 3 + 1 = 3(q+1) + 1 which is not divisible by 3 `" " (because` remainder =1) So , only n+1 is divisible by 3 when r =2 . `" " .... (3)` Hence , from equation (1) , (2) and (3) , we can say that one and only one out of n , (n +1) and (n +2) is divisible by 3. |
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| 148. |
`(2- 1/3)(2 - 3/5)(2-5/7).....(2-997/999)` is equal to |
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Answer» `(2-1/3)(2-3/5)(2-5/7)....(2-997/999)` `5/3*7/5*9/7*11/9...1001/999` `1001/3`. |
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| 149. |
In a seminar. the number of participants in Hindi, English and Mathematics are 60,84 and 108 respectively. Find the minimum number of rooms required if, in each room the same number of participants are to be seated and all of them being in the same subject. |
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Answer» Correct Answer - 21 maximum number of partipants in each room = HCF ( 60,84,108) =12 . Minimum number of rooms required = `( 60/12 + 84/12 + 108/12) = 21` |
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| 150. |
Find the remainder when `1^2013+2^2013+3^2013+...+2012^2013` is divisible by 2013 |
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Answer» `1^2013+2^2013+3^2013+...+2012^2013` `=(1^2013+2012^2013)+(2^2013+2011^2013)+...(1006^2013+1007^2013)` So, these terms are of the form `a^n+b^n` and they will be divided by `a+b` as `n = 2013` and it is odd. `:.` Remainder of the given expression when divided by `2013` is `0`. |
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