How many 3-digit numbers are there, with distinct digits,
Question: How many 3-digit numbers are there, with distinct digits, with each digit odd? Solution: The hundred's place can be filled by {1, 3, 5, 7, 9), i.e. 5 digits. The ten's place can now be filled by 4 digits (as one digit is already used in the hundred's place and repetition is not allowed ) Similarly, the unit's place can be filled by 3 digits. Total number of 3 -digit numbers $=5 \times 4 \times 3=60$...
Read More →Two squares have sides x cm and (x + 4) cm.
Question: Two squares have sides $x \mathrm{~cm}$ and $(x+4) \mathrm{cm}$. The sum of their areas is $656 \mathrm{~cm}^{2}$. Find the sides of the squares. Solution: Given that sides of the squares are $=x \mathrm{~cm}$ and $=(x+4) \mathrm{cm}$. Then According to question, Sum of the areas of square $=656 \mathrm{~cm}^{2}$ So, $x^{2}+(x+4)^{2}=656$ $x^{2}+x^{2}+8 x+16=656$ $2 x^{2}+8 x+16-656=0$ $2 x^{2}+8 x-640=0$ $2\left(x^{2}+4 x-320\right)=0$ $x^{2}+4 x-320=0$ $x^{2}-16 x+20 x-320=0$ $x(x-16...
Read More →How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7
Question: How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed? Solution: Since the number is less than 1000, it could be a three-digit, two-digit or single-digit number. Case I: Three-digit number: Now, the hundred's place cannot be zero. Thus, it can be filled withthreedigits, i.e. 3, 5 and 7. Also, the unit's place cannot be zero. This is because it is an odd number and one digit has already been used to fill the hundred's p...
Read More →The length of a hall is 5 m more than its breadth.
Question: The length of a hall is $5 \mathrm{~m}$ more than its breadth. If the area of the floor of the hall is $84 \mathrm{~m}^{2}$, what are the length and breadth of the hall? Solution: Let the breadth of the rectangular hall be $=x$ metres and the length $=(x+5)$ metres Then And area of the rectangle length $\times$ breadth $=84$ $(x+5) x=84$ $x^{2}+5 x-84=0$ $x^{2}-7 x+12 x-84=0$ $x(x-7)+12(x-7)=0$ $(x-7)(x+12)=0$ $(x-7)=0$ $x=7$ or $(x+12)=0$ $x=-12$ Sides of the rectangular hall never ar...
Read More →How many 9-digit numbers of different digits can be formed?
Question: How many 9-digit numbers of different digits can be formed? Solution: Since the first digit cannot be zero, number of ways of filling the first digit = 9 Number of ways of filling the second digit = 9 (as repetition is not allowed or the digits are distinct) Number of ways of filling the third digit = 8 Number of ways of filling the fourth digit = 7 Number of ways of filling the fifth digit = 6 Number of ways of filling the sixth digit = 5 Number of ways of filling the seventh digit = ...
Read More →Factorize:
Question: Factorize: $27 a^{3}-b^{3}+8 c^{3}+18 a b c$ Solution: $27 a^{3}-b^{3}+8 c^{3}+18 a b c=(3 a)^{3}+(-b)^{3}+(2 c)^{3}-3 \times(3 a) \times(-b) \times(2 c)$ $=[3 a+(-b)+2 c]\left[(3 a)^{2}+(-b)^{2}+(2 c)^{2}-3 a(-b)-(-b) 2 c-3 a \times 2 c\right]$ $=(3 a-b+2 c)\left(9 a^{2}+b^{2}+4 c^{2}+3 a b+2 b c-6 a c\right)$...
Read More →In how many ways can six persons be seated in a row?
Question: In how many ways can six persons be seated in a row? Solution: Number of seats available to the first person = 6 Number of seats available to the second person = 5 Number of seats available to the third person = 4 Number of seats available to the fourth person = 3 Number of seats available to the fifth person = 2 Number of seats available to the sixth person = 1 Total number of ways of making the seating arrangement $=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$...
Read More →The perimeter of a rectangular field is 82 m and its area
Question: The perimeter of a rectangular field is $82 \mathrm{~m}$ and its area is $400 \mathrm{~m}^{2}$. Find the breadth of the rectangle. Solution: Let the breadth of the rectangle be $=x$ metres . Then Perimeter $=82$ metres 2 (length+breadth ) $=82$ (length $+x$ ) $=41$ length $=41-x$ And area of the rectangle length $\times$ breadth $=400$ $(41-x) x=400$ $41 x-x^{2}=400$ $x^{2}-41 x+400=0$ $x^{2}-25 x-16 x+400=0$ $x(x-25)-16(x-25)=0$ $(x-25)(x-16)=0$ $(x-25)=0$ $x=25$ or $(x-16)=0$ $x=16$ ...
Read More →How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000,
Question: How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed? Solution: Since the number has to be greater than 8000, the thousand's place can be filled by only two digits, i.e. 8 and 9. Now, the hundred's place can be filled with the remaining 4 digits as the repetition of the digits is not allowed. The ten's place can be filled with the remaining 3 digits. The unit's place can be filled with the remaining ...
Read More →Factorize:
Question: Factorize: $216+27 b^{3}+8 c^{3}-108 a b c$ Solution: $216+27 b^{3}+8 c^{3}-108 a b c=(6)^{3}+(3 b)^{3}+(2 c)^{3}-3 \times 6 \times 3 b \times 2 c$ $=(6+3 b+2 c)\left[6^{2}+(3 b)^{2}+(2 c)^{2}-6 \times 3 b-3 b \times 2 c-2 c \times 6\right]$ $=(6+3 b+2 c)\left(36+9 b^{2}+4 c^{2}-18 b-6 b c-12 c\right)$...
Read More →How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 7000,
Question: How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 7000, if repetition of digits is not allowed? Solution: Since the number has to be greater than 7000, the thousand's place can only be filled by three digits, i.e. 7, 8 and 9. Now, the hundred's place can be filled with the remaining 4 digits as the repetition of the digits is not allowed. The ten's place can be filled with the remaining 3 digits. The unit's place can be filled with the remai...
Read More →Factorize:
Question: Factorize: $1+b^{3}+8 c^{3}-6 b c$ Solution: $1+b^{3}+8 c^{3}-6 b c=(1)^{3}+(b)^{3}+(2 c)^{3}-3 \times 1 \times b \times 2 c$ $=(1+b+2 c)\left[1^{2}+b^{2}+(2 c)^{2}-1 \times b-b \times 2 c-1 \times 2 c\right]$ $=(1+b+2 c)\left(1+b^{2}+4 c^{2}-b-2 b c-2 c\right)$...
Read More →The diagonal of a rectangular field is 60 meters more than the shorter side.
Question: The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field. Solution: Let the length of smaller side of rectangle be $=x$ metres then larger side be $=x+30$ metres and their diagonal be $=x+60$ metres Then, as we know that Pythagoras theorem $x^{2}+(x+30)^{2}=(x+60)^{2}$ $x^{2}+(x+30)^{2}=(x+60)^{2}$ $x^{2}+x^{2}+60 x+900=x^{2}+120 x+3600$ $2 x^{2}+60 x+900-x^{2}-120 x-3600=0$ $x^{...
Read More →How many different five-digit number licence plates can be made if
Question: How many different five-digit number licence plates can be made if (i) first digit cannot be zero and the repetition of digits is not allowed, (ii) the first-digit cannot be zero, but the repetition of digits is allowed? Solution: (i) Since the first digit cannot be zero, the number of ways of filling the first digit = 9 Number of ways of filling the second digit = 9 (Since repetition is not allowed) Number of ways of filling the third digit = 8 Number of ways of filling the fourth dig...
Read More →How many three-digit odd numbers are there?
Question: How many three-digit odd numbers are there? Solution: Available digits for filling any place = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0} Since the thousand's place cannot be zero, the number of ways of filling the thousand's place is 9. Number of ways of filling the ten's place = 10 Number of ways of filling the unit's place = 5 {1, 3, 5, 7, 9} Total 3-digit odd numbers $=9 \times 10 \times 5=450$...
Read More →How many three-digit numbers are there?
Question: How many three-digit numbers are there? Solution: Available digits for filling any place = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0} Since the thousand's place cannot be zero, available digits to fill the thousand's place = 9 Number of ways of filling the ten's digit = 10 Similarly, number of ways of filling the unit's digit = 10 $\therefore$ Total number of three digit numbers $=9 \times 10 \times 10=900$...
Read More →How many three-digit numbers are there with no digit repeated?
Question: How many three-digit numbers are there with no digit repeated? Solution: The thousand's place cannot be zero. Number of ways of selecting the thousand's digit = 9 Number of ways of selecting the ten's digit = 9 ( as repetition of digits is not allowed and one digit has already been used in the thousand's place) Similarly, number of ways of selecting the unit's digit = 8 (as two digits have been used for the thousand's and the ten's places) $\therefore$ Total three digit number that can...
Read More →From among the 36 teachers in a college, one principal,
Question: From among the 36 teachers in a college, one principal, one vice-principal and the teacher-incharge are to be appointed. In how many ways can this be done? Solution: Total number of teachers in the college = 36 Number of ways of selecting a principal = 36 Number of ways of selecting a vice-principal = 35 (as one of the teacher is already being selected for the post of principal) Similarly, number of ways of selecting the teacher-incharge = 34 $\therefore$ Total number of ways of select...
Read More →How many A.P.'s with 10 terms are there whose first term is in the set {1, 2, 3}
Question: How many A.P.'s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}? Solution: Number of ways of selecting the first term from the set {1, 2, 3} = 3 Corresponding to each of the selected first terms, the number of ways of selecting the common difference from the set {1, 2, 3, 4, 5} = 5 $\therefore$ Total number of AP's that can be formed $=3 \times 5=15$...
Read More →Twelve students complete in a race. In how many ways first three prizes be given?
Question: Twelve students complete in a race. In how many ways first three prizes be given? Solution: Number of competitors in the race = 12 Number of competitors who can come first in the race = 12 Number of competitors who can come second in the race = 11 (as one competitor has already come first in the race) Number of competitors who can come third in the race = 10 $\therefore$ Total number of ways of awarding the first three prizes $=12 \times 11 \times 10=1320$...
Read More →A team consists of 6 boys and 4 girls and other has 5 boys and 3 girls.
Question: A team consists of 6 boys and 4 girls and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl plays against a girl? Solution: A boy can be selected from the first team in 6 ways and from the second team in 5 ways. $\therefore$ Number of ways of arranging a match between the boys of the two teams $=6 \times 5=30$ Similarly, A girl can be selected from the first team in 4 ways and from the second team in 3 ...
Read More →Given 7 flags of different colours, how many different signals can be generated if a signal requires the use of two flags,
Question: Given 7 flags of different colours, how many different signals can be generated if a signal requires the use of two flags, one below the other? Solution: Number of flags = 7 Number of ways of selecting one flag = 7 Number of ways of selecting the other flag = 6 (as only 6 colours are available for use) A signal requires use of two flags $\therefore$ Total number of signal that can be generated $=7 \times 6=42$...
Read More →There are 5 books on Mathematics and 6 books on Physics in a book shop. In how many ways can a students buy :
Question: There are 5 books on Mathematics and 6 books on Physics in a book shop. In how many ways can a students buy : (i) a Mathematics book and a Physics book (ii) either a Mathematics book or a Physics book? Solution: Number of books on mathematics = 5 Number of books on physics = 6 Number of ways of buying a mathematics book = 5 Similarly, number of ways of buying a physics book = 6 (i) By using fundamental principle of multiplication: Number of ways of buying a mathematics and a physics bo...
Read More →Factorize:
Question: Factorize: $a^{3}+8 b^{3}+64 c^{3}-24 a b c$ Solution: $a^{3}+8 b^{3}+64 c^{3}-24 a b c=a^{3}+(2 b)^{3}+(4 c)^{3}-3 \times a \times 2 b \times 4 c$ $=(a+2 b+4 c)\left[a^{2}+(2 b)^{2}+(4 c)^{2}-a \times 2 b-2 b \times 4 c-4 c \times a\right]$ $=(a+2 b+4 c)\left(a^{2}+4 b^{2}+16 c^{2}-2 a b-8 b c-4 c a\right)$...
Read More →There are 6 multiple choice questions in an examination.
Question: There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next three have 2 each? Solution: Number of ways of answering the first three questions = 4 each Number of ways of answering the remaining three questions = 2 each $\therefore$ Total number of ways of answering all the questions $=4 \times 4 \times 4 \times 2 \times 2 \times 2=512$...
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