10th Maths One Mark Questions with Answers

10th Maths One Mark Questions with answers: every one-mark question from the Samacheer Kalvi textbook, chapter by chapter.

Chapter 1: Relations and Functions

  1. If n(A×B)=6n(A \times B) = 6 and A={1,3}A = \{1,3\} then n(B)n(B) is

    1. (1) 1
    2. (2) 2
    3. (3) 3
    4. (4) 6
    Show answer

    Answer: (3) 3

    1. n(A×B)=n(A)×n(B)n(A\times B)=n(A)\times n(B).

    2. 6=2×n(B)6=2\times n(B).

    3. n(B)=3n(B)=3.

    Question 1 of 15
  2. A={a,b,p}A = \{a,b,p\}, B={2,3}B = \{2,3\}, C={p,q,r,s}C = \{p,q,r,s\} then n[(A∪C)×B]n[(A \cup C) \times B] is

    1. (1) 8
    2. (2) 20
    3. (3) 12
    4. (4) 16
    Show answer

    Answer: (3) 12

    1. A∪C={a,b,p,q,r,s}A\cup C=\{a,b,p,q,r,s\}, so n(A∪C)=6n(A\cup C)=6.

    2. n(B)=2n(B)=2.

    3. n[(A∪C)×B]=6×2=12n[(A\cup C)\times B]=6\times 2=12.

    Question 2 of 15
  3. If A={1,2}A = \{1,2\}, B={1,2,3,4}B = \{1,2,3,4\}, C={5,6}C = \{5,6\} and D={5,6,7,8}D = \{5, 6, 7, 8\} then state which of the following statement is true.

    1. (1) (A×C)⊂(B×D)(A \times C) \subset (B \times D)
    2. (2) (B×D)⊂(A×C)(B \times D) \subset (A \times C)
    3. (3) (A×B)⊂(A×D)(A \times B) \subset (A \times D)
    4. (4) (D×A)⊂(B×A)(D \times A) \subset (B \times A)
    Show answer

    Answer: (1) (A×C)⊂(B×D)(A \times C) \subset (B \times D)

    1. A={1,2}⊂B={1,2,3,4}A=\{1,2\}\subset B=\{1,2,3,4\} and C={5,6}⊂D={5,6,7,8}C=\{5,6\}\subset D=\{5,6,7,8\}.

    2. So every pair of A×CA\times C is also a pair of B×DB\times D.

    3. Hence (A×C)⊂(B×D)(A\times C)\subset (B\times D).

    Question 3 of 15
  4. If there are 1024 relations from a set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\} to a set BB, then the number of elements in BB is

    1. (1) 3
    2. (2) 2
    3. (3) 4
    4. (4) 8
    Show answer

    Answer: (2) 2

    1. Number of relations from AA to BB is 2n(A)⋅n(B)2^{n(A)\cdot n(B)}.

    2. 25 n(B)=1024=2102^{5\,n(B)}=1024=2^{10}.

    3. 5 n(B)=10⇒n(B)=25\,n(B)=10\Rightarrow n(B)=2.

    Question 4 of 15
  5. The range of the relation R={(x,x2)∣xR = \{(x, x^2) \mid x is a prime number less than 13}\} is

    1. (1) {2,3,5,7}\{2,3,5,7\}
    2. (2) {2,3,5,7,11}\{2,3,5,7,11\}
    3. (3) {4,9,25,49,121}\{4,9,25,49,121\}
    4. (4) {1,4,9,25,49,121}\{1,4,9,25,49,121\}
    Show answer

    Answer: (3) {4,9,25,49,121}\{4,9,25,49,121\}

    1. Primes less than 13 are 2,3,5,7,112,3,5,7,11.

    2. Range consists of the squares x2x^{2}.

    3. Range ={4,9,25,49,121}=\{4,9,25,49,121\}.

    Question 5 of 15
  6. If the ordered pairs (a+2,4)(a + 2, 4) and (5,2a+b)(5, 2a + b) are equal then (a,b)(a,b) is

    1. (1) (2,−2)(2, -2)
    2. (2) (5,1)(5,1)
    3. (3) (2,3)(2,3)
    4. (4) (3,−2)(3, -2)
    Show answer

    Answer: (4) (3,−2)(3, -2)

    1. Equal ordered pairs: a+2=5a+2=5 and 2a+b=42a+b=4.

    2. a=3a=3.

    3. 2(3)+b=4⇒b=−22(3)+b=4\Rightarrow b=-2.

    Question 6 of 15
  7. Let n(A)=mn(A) = m and n(B)=nn(B) = n then the total number of non-empty relations that can be defined from AA to BB is

    1. (1) mnm^n
    2. (2) nmn^m
    3. (3) 2mn−12^{mn} - 1
    4. (4) 2mn2^{mn}
    Show answer

    Answer: (3) 2mn−12^{mn} - 1

    1. Total number of relations from AA to BB is 2mn2^{mn}.

    2. Remove the empty relation.

    3. Non-empty relations =2mn−1=2^{mn}-1.

    Question 7 of 15
  8. If {(a,8),(6,b)}\{(a,8),(6,b)\} represents an identity function, then the value of aa and bb are respectively

    1. (1) (8,6)(8,6)
    2. (2) (8,8)(8,8)
    3. (3) (6,8)(6,8)
    4. (4) (6,6)(6,6)
    Show answer

    Answer: (1) (8,6)(8,6)

    1. For an identity function f(x)=xf(x)=x, each pair is of the form (x,x)(x,x).

    2. (a,8)⇒a=8(a,8)\Rightarrow a=8.

    3. (6,b)⇒b=6(6,b)\Rightarrow b=6.

    Question 8 of 15
  9. Let A={1,2,3,4}A = \{1,2,3,4\} and B={4,8,9,10}B = \{4,8,9,10\}. A function f:A→Bf : A \to B given by f={(1,4),(2,8),(3,9),(4,10)}f = \{(1,4),(2,8),(3,9),(4,10)\} is a

    1. (1) Many-one function
    2. (2) Identity function
    3. (3) One-to-one function
    4. (4) Into function
    Show answer

    Answer: (3) One-to-one function

    1. Images are 4,8,9,104,8,9,10 — all different.

    2. Different elements of AA have different images.

    3. So ff is one-to-one.

    Question 9 of 15
  10. If f(x)=2x2f(x) = 2x^2 and g(x)=13xg(x) = \dfrac{1}{3x}, then f∘gf \circ g is

    1. (1) 32x2\dfrac{3}{2x^2}
    2. (2) 23x2\dfrac{2}{3x^2}
    3. (3) 29x2\dfrac{2}{9x^2}
    4. (4) 16x2\dfrac{1}{6x^2}
    Show answer

    Answer: (3) 29x2\dfrac{2}{9x^2}

    1. (f∘g)(x)=f(g(x))=f(13x)(f\circ g)(x)=f\big(g(x)\big)=f\left(\dfrac{1}{3x}\right).

    2. =2(13x)2=2\left(\dfrac{1}{3x}\right)^{2}.

    3. =29x2=\dfrac{2}{9x^{2}}.

    Question 10 of 15
  11. If f:A→Bf : A \to B is a bijective function and if n(B)=7n(B) = 7, then n(A)n(A) is equal to

    1. (1) 7
    2. (2) 49
    3. (3) 1
    4. (4) 14
    Show answer

    Answer: (1) 7

    1. A bijection pairs each element of AA with exactly one element of BB.

    2. So n(A)=n(B)n(A)=n(B).

    3. n(A)=7n(A)=7.

    Question 11 of 15
  12. Let ff and gg be two functions given by f={(0,1),(2,0),(3,−4),(4,2),(5,7)}f = \{(0,1),(2,0),(3,-4),(4,2),(5,7)\}, g={(0,2),(1,0),(2,4),(−4,2),(7,0)}g = \{(0,2),(1,0),(2,4),(-4,2),(7,0)\} then the range of f∘gf \circ g is

    1. (1) {0,2,3,4,5}\{0,2,3,4,5\}
    2. (2) {−4,1,0,2,7}\{-4,1,0,2,7\}
    3. (3) {1,2,3,4,5}\{1,2,3,4,5\}
    4. (4) {0,1,2}\{0,1,2\}
    Show answer

    Answer: (4) {0,1,2}\{0,1,2\}

    1. (f∘g)(0)=f(2)=0(f\circ g)(0)=f(2)=0, (f∘g)(1)=f(0)=1(f\circ g)(1)=f(0)=1.

    2. (f∘g)(2)=f(4)=2(f\circ g)(2)=f(4)=2, (f∘g)(−4)=f(2)=0(f\circ g)(-4)=f(2)=0, (f∘g)(7)=f(0)=1(f\circ g)(7)=f(0)=1.

    3. Range ={0,1,2}=\{0,1,2\}.

    Question 12 of 15
  13. Let f(x)=1+x2f(x) = \sqrt{1 + x^2} then

    1. (1) f(xy)=f(x).f(y)f(xy) = f(x).f(y)
    2. (2) f(xy)≥f(x).f(y)f(xy) \geq f(x).f(y)
    3. (3) f(xy)≤f(x).f(y)f(xy) \leq f(x).f(y)
    4. (4) None of these
    Show answer

    Answer: (3) f(xy)≤f(x).f(y)f(xy) \leq f(x).f(y)

    1. f(xy)=1+x2y2f(xy)=\sqrt{1+x^{2}y^{2}}.

    2. f(x)⋅f(y)=(1+x2)(1+y2)=1+x2+y2+x2y2f(x)\cdot f(y)=\sqrt{(1+x^{2})(1+y^{2})}=\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}.

    3. Since x2+y2≥0x^{2}+y^{2}\ge 0,   1+x2y2≤1+x2+y2+x2y2\;1+x^{2}y^{2}\le 1+x^{2}+y^{2}+x^{2}y^{2}.

    4. Hence f(xy)≤f(x)⋅f(y)f(xy)\le f(x)\cdot f(y).

    Question 13 of 15
  14. If g={(1,1),(2,3),(3,5),(4,7)}g = \{(1,1),(2,3),(3,5),(4,7)\} is a function given by g(x)=αx+βg(x) = \alpha x + \beta then the values of α\alpha and β\beta are

    1. (1) (−1,2)(-1,2)
    2. (2) (2,−1)(2, -1)
    3. (3) (−1,−2)(-1, -2)
    4. (4) (1,2)(1,2)
    Show answer

    Answer: (2) (2,−1)(2, -1)

    1. g(1)=1⇒α+β=1g(1)=1\Rightarrow \alpha+\beta=1.

    2. g(2)=3⇒2α+β=3g(2)=3\Rightarrow 2\alpha+\beta=3.

    3. Subtracting, α=2\alpha=2, then β=−1\beta=-1.

    Question 14 of 15
  15. f(x)=(x+1)3−(x−1)3f(x) = (x + 1)^3 - (x - 1)^3 represents a function which is

    1. (1) linear
    2. (2) cubic
    3. (3) reciprocal
    4. (4) quadratic
    Show answer

    Answer: (4) quadratic

    1. (x+1)3=x3+3x2+3x+1(x+1)^{3}=x^{3}+3x^{2}+3x+1, (x−1)3=x3−3x2+3x−1(x-1)^{3}=x^{3}-3x^{2}+3x-1.

    2. Subtracting, f(x)=6x2+2f(x)=6x^{2}+2.

    3. Highest power is 22, so it is quadratic.

    Question 15 of 15

Chapter 2: Numbers and Sequences

  1. Euclid's division lemma states that for positive integers aa and bb, there exist unique integers qq and rr such that a=bq+ra = bq + r, where rr must satisfy.

    1. (1) 1<r<b1 < r < b
    2. (2) 0<r<b0 < r < b
    3. (3) 0≤r<b0 \leq r < b
    4. (4) 0<r≤b0 < r \leq b
    Show answer

    Answer: (3) 0≤r<b0 \leq r < b

    1. In a=bq+ra = bq + r, the remainder rr can be zero (when bb divides aa) but must be less than the divisor bb.

    2. So 0≤r<b0 \le r < b.

    Question 1 of 15
  2. Using Euclid's division lemma, if the cube of any positive integer is divided by 99 then the possible remainders are

    1. (1) 0,1,80, 1, 8
    2. (2) 1,4,81, 4, 8
    3. (3) 0,1,30, 1, 3
    4. (4) 1,3,51, 3, 5
    Show answer

    Answer: (1) 0,1,80, 1, 8

    1. Any positive integer is of the form 3q3q, 3q+13q+1 or 3q+23q+2.

    2. (3q)3=27q3=9(3q3)⇒(3q)^3 = 27q^3 = 9(3q^3) \Rightarrow remainder 00.

    3. (3q+1)3=27q3+27q2+9q+1=9(3q3+3q2+q)+1⇒(3q+1)^3 = 27q^3+27q^2+9q+1 = 9(3q^3+3q^2+q)+1 \Rightarrow remainder 11.

    4. (3q+2)3=27q3+54q2+36q+8=9(3q3+6q2+4q)+8⇒(3q+2)^3 = 27q^3+54q^2+36q+8 = 9(3q^3+6q^2+4q)+8 \Rightarrow remainder 88.

    Question 2 of 15
  3. If the HCF of 6565 and 117117 is expressible in the form of 65m−11765m - 117, then the value of mm is

    1. (1) 44
    2. (2) 22
    3. (3) 11
    4. (4) 33
    Show answer

    Answer: (2) 22

    1. 117=65×1+52117 = 65 \times 1 + 52, 65=52×1+1365 = 52 \times 1 + 13, 52=13×4+052 = 13 \times 4 + 0.

    2. So HCF(65,117)=13(65,117) = 13.

    3. 65m−117=13⇒65m=130⇒m=265m - 117 = 13 \Rightarrow 65m = 130 \Rightarrow m = 2.

    Question 3 of 15
  4. The sum of the exponents of the prime factors in the prime factorization of 17291729 is

    1. (1) 11
    2. (2) 22
    3. (3) 33
    4. (4) 44
    Show answer

    Answer: (3) 33

    1. 1729=7×13×19=71×131×1911729 = 7 \times 13 \times 19 = 7^1 \times 13^1 \times 19^1.

    2. Sum of exponents =1+1+1=3= 1 + 1 + 1 = 3.

    Question 4 of 15
  5. The least number that is divisible by all the numbers from 11 to 1010 (both inclusive) is

    1. (1) 20252025
    2. (2) 52205220
    3. (3) 50255025
    4. (4) 25202520
    Show answer

    Answer: (4) 25202520

    1. The least such number is the LCM of 1,2,…,101,2,\ldots,10.

    2. LCM=23×32×5×7=2520\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 2520.

    Question 5 of 15
  6. 74k≡7^{4k} \equiv _____ (mod 100)(\text{mod } 100)

    1. (1) 11
    2. (2) 22
    3. (3) 33
    4. (4) 44
    Show answer

    Answer: (1) 11

    1. 72=497^2 = 49, so 74=492=24017^4 = 49^2 = 2401.

    2. 2401=24×100+1⇒74≡1(mod100)2401 = 24 \times 100 + 1 \Rightarrow 7^4 \equiv 1 \pmod{100}.

    3. Hence 74k=(74)k≡1k≡1(mod100)7^{4k} = (7^4)^k \equiv 1^k \equiv 1 \pmod{100}.

    Question 6 of 15
  7. Given F1=1F_1 = 1, F2=3F_2 = 3 and Fn=Fn−1+Fn−2F_n = F_{n-1} + F_{n-2} then F5F_5 is

    1. (1) 33
    2. (2) 55
    3. (3) 88
    4. (4) 1111
    Show answer

    Answer: (4) 1111

    1. F3=F2+F1=3+1=4F_3 = F_2 + F_1 = 3 + 1 = 4.

    2. F4=F3+F2=4+3=7F_4 = F_3 + F_2 = 4 + 3 = 7.

    3. F5=F4+F3=7+4=11F_5 = F_4 + F_3 = 7 + 4 = 11.

    Question 7 of 15
  8. The first term of an arithmetic progression is unity and the common difference is 44. Which of the following will be a term of this A.P.

    1. (1) 45514551
    2. (2) 1009110091
    3. (3) 78817881
    4. (4) 1353113531
    Show answer

    Answer: (3) 78817881

    1. a=1a = 1, d=4d = 4, so tn=1+(n−1)4=4n−3t_n = 1 + (n-1)4 = 4n - 3.

    2. Every term leaves remainder 11 when divided by 44.

    3. 7881=4×1971−37881 = 4 \times 1971 - 3, so 78817881 is the 1971st1971^{\text{st}} term.

    Question 8 of 15
  9. If 66 times of 6th6^{\text{th}} term of an A.P. is equal to 77 times the 7th7^{\text{th}} term, then the 13th13^{\text{th}} term of the A.P. is

    1. (1) 00
    2. (2) 66
    3. (3) 77
    4. (4) 1313
    Show answer

    Answer: (1) 00

    1. 6t6=7t7⇒6(a+5d)=7(a+6d)6t_6 = 7t_7 \Rightarrow 6(a+5d) = 7(a+6d).

    2. 6a+30d=7a+42d⇒a+12d=06a + 30d = 7a + 42d \Rightarrow a + 12d = 0.

    3. t13=a+12d=0t_{13} = a + 12d = 0.

    Question 9 of 15
  10. An A.P. consists of 3131 terms. If its 16th16^{\text{th}} term is mm, then the sum of all the terms of this A.P. is

    1. (1) 16 m16\,m
    2. (2) 62 m62\,m
    3. (3) 31 m31\,m
    4. (4) 312 m\dfrac{31}{2}\,m
    Show answer

    Answer: (3) 31 m31\,m

    1. t16=a+15d=mt_{16} = a + 15d = m.

    2. S31=312 [2a+30d]=312×2(a+15d)S_{31} = \dfrac{31}{2}\,[2a + 30d] = \dfrac{31}{2} \times 2(a + 15d).

    3. =31(a+15d)=31m= 31(a+15d) = 31m.

    Question 10 of 15
  11. In an A.P., the first term is 11 and the common difference is 44. How many terms of the A.P. must be taken for their sum to be equal to 120120?

    1. (1) 66
    2. (2) 77
    3. (3) 88
    4. (4) 99
    Show answer

    Answer: (3) 88

    1. Sn=n2[2(1)+(n−1)4]=n2(4n−2)=n(2n−1)S_n = \dfrac{n}{2}[2(1) + (n-1)4] = \dfrac{n}{2}(4n - 2) = n(2n-1).

    2. n(2n−1)=120⇒2n2−n−120=0n(2n-1) = 120 \Rightarrow 2n^2 - n - 120 = 0.

    3. (n−8)(2n+15)=0⇒n=8(n-8)(2n+15) = 0 \Rightarrow n = 8.

    Question 11 of 15
  12. If A=265A = 2^{65} and B=264+263+262+⋯+20B = 2^{64} + 2^{63} + 2^{62} + \cdots + 2^{0} which of the following is true?

    1. (1) BB is 2642^{64} more than AA
    2. (2) AA and BB are equal
    3. (3) BB is larger than AA by 11
    4. (4) AA is larger than BB by 11
    Show answer

    Answer: (4) AA is larger than BB by 11

    1. B=20+21+⋯+264B = 2^0 + 2^1 + \cdots + 2^{64} is a G.P. with a=1a = 1, r=2r = 2, n=65n = 65.

    2. B=1(265−1)2−1=265−1B = \dfrac{1(2^{65} - 1)}{2 - 1} = 2^{65} - 1.

    3. A−B=265−(265−1)=1A - B = 2^{65} - (2^{65} - 1) = 1.

    Question 12 of 15
  13. The next term of the sequence 316,18,112,118,⋯\dfrac{3}{16}, \dfrac{1}{8}, \dfrac{1}{12}, \dfrac{1}{18}, \cdots is

    1. (1) 124\dfrac{1}{24}
    2. (2) 127\dfrac{1}{27}
    3. (3) 23\dfrac{2}{3}
    4. (4) 181\dfrac{1}{81}
    Show answer

    Answer: (2) 127\dfrac{1}{27}

    1. 1/83/16=1624=23\dfrac{1/8}{3/16} = \dfrac{16}{24} = \dfrac{2}{3}, 1/121/8=23\dfrac{1/12}{1/8} = \dfrac{2}{3}, 1/181/12=23\dfrac{1/18}{1/12} = \dfrac{2}{3}.

    2. It is a G.P. with r=23r = \dfrac{2}{3}.

    3. Next term =118×23=127= \dfrac{1}{18} \times \dfrac{2}{3} = \dfrac{1}{27}.

    Question 13 of 15
  14. If the sequence t1,t2,t3,…t_1, t_2, t_3, \ldots are in A.P. then the sequence t6,t12,t18,…t_6, t_{12}, t_{18}, \ldots is

    1. (1) a Geometric Progression
    2. (2) an Arithmetic Progression
    3. (3) neither an Arithmetic Progression nor a Geometric Progression
    4. (4) a constant sequence
    Show answer

    Answer: (2) an Arithmetic Progression

    1. Let the A.P. have first term aa and common difference dd.

    2. t6=a+5dt_6 = a+5d, t12=a+11dt_{12} = a+11d, t18=a+17dt_{18} = a+17d.

    3. Difference =6d= 6d, a constant, so the new sequence is an A.P.

    Question 14 of 15
  15. The value of (13+23+33+⋯+153)−(1+2+3+⋯+15)(1^3 + 2^3 + 3^3 + \cdots + 15^3) - (1 + 2 + 3 + \cdots + 15) is

    1. (1) 1440014400
    2. (2) 1420014200
    3. (3) 1428014280
    4. (4) 1452014520
    Show answer

    Answer: (3) 1428014280

    1. 13+23+⋯+153=[15×162]2=1202=144001^3+2^3+\cdots+15^3 = \left[\dfrac{15 \times 16}{2}\right]^2 = 120^2 = 14400.

    2. 1+2+⋯+15=15×162=1201+2+\cdots+15 = \dfrac{15 \times 16}{2} = 120.

    3. 14400−120=1428014400 - 120 = 14280.

    Question 15 of 15

Chapter 3: Algebra

  1. A system of three linear equations in three variables is inconsistent if their planes

    1. (1) intersect only at a point
    2. (2) intersect in a line
    3. (3) coincides with each other
    4. (4) do not intersect
    Show answer

    Answer: (4) do not intersect

    1. Inconsistent means the system has no solution.

    2. No solution happens only when the three planes have no common point.

    3. So the planes do not intersect.

    Question 1 of 20
  2. The solution of the system x+y−3z=−6x + y - 3z = -6, −7y+7z=7-7y + 7z = 7, 3z=93z = 9 is

    1. (1) x=1, y=2, z=3x = 1,\ y = 2,\ z = 3
    2. (2) x=−1, y=2, z=3x = -1,\ y = 2,\ z = 3
    3. (3) x=−1, y=−2, z=3x = -1,\ y = -2,\ z = 3
    4. (4) x=1, y=−2, z=3x = 1,\ y = -2,\ z = 3
    Show answer

    Answer: (1) x=1, y=2, z=3x = 1,\ y = 2,\ z = 3

    1. From 3z=93z = 9, z=3z = 3.

    2. Put z=3z=3 in −7y+7z=7-7y + 7z = 7: −7y+21=7⇒y=2-7y + 21 = 7 \Rightarrow y = 2.

    3. Put y=2, z=3y=2,\ z=3 in x+y−3z=−6x + y - 3z = -6: x+2−9=−6⇒x=1x + 2 - 9 = -6 \Rightarrow x = 1.

    Question 2 of 20
  3. If (x−6)(x - 6) is the HCF of x2−2x−24x^2 - 2x - 24 and x2−kx−6x^2 - kx - 6 then the value of kk is

    1. (1) 33
    2. (2) 55
    3. (3) 66
    4. (4) 88
    Show answer

    Answer: (2) 55

    1. x2−2x−24=(x−6)(x+4)x^2 - 2x - 24 = (x-6)(x+4), so (x−6)(x-6) is a factor.

    2. Since (x−6)(x-6) is the HCF, it must divide x2−kx−6x^2 - kx - 6.

    3. So x=6x = 6 makes it zero: 36−6k−6=036 - 6k - 6 = 0.

    4. 6k=30⇒k=56k = 30 \Rightarrow k = 5.

    Question 3 of 20
  4. 3y−3y÷7y−73y2\dfrac{3y - 3}{y} \div \dfrac{7y - 7}{3y^2} is

    1. (1) 9y7\dfrac{9y}{7}
    2. (2) 9y3(21y−21)\dfrac{9y^3}{(21y - 21)}
    3. (3) 21y2−42y+213y3\dfrac{21y^2 - 42y + 21}{3y^3}
    4. (4) 7(y2−2y+1)y2\dfrac{7(y^2 - 2y + 1)}{y^2}
    Show answer

    Answer: (1) 9y7\dfrac{9y}{7}

    1. 3y−3y÷7y−73y2=3(y−1)y×3y27(y−1)\dfrac{3y-3}{y} \div \dfrac{7y-7}{3y^2} = \dfrac{3(y-1)}{y} \times \dfrac{3y^2}{7(y-1)}

    2. Cancel (y−1)(y-1) and one yy.

    3. =9y7= \dfrac{9y}{7}

    Question 4 of 20
  5. y2+1y2y^2 + \dfrac{1}{y^2} is not equal to

    1. (1) y4+1y2\dfrac{y^4 + 1}{y^2}
    2. (2) (y+1y)2\left(y + \dfrac{1}{y}\right)^2
    3. (3) (y−1y)2+2\left(y - \dfrac{1}{y}\right)^2 + 2
    4. (4) (y+1y)2−2\left(y + \dfrac{1}{y}\right)^2 - 2
    Show answer

    Answer: (2) (y+1y)2\left(y + \dfrac{1}{y}\right)^2

    1. (y+1y)2=y2+2+1y2\left(y + \dfrac{1}{y}\right)^2 = y^2 + 2 + \dfrac{1}{y^2}, which is 22 more than y2+1y2y^2 + \dfrac{1}{y^2}.

    2. Options (A), (C), (D) all simplify to y2+1y2y^2 + \dfrac{1}{y^2}.

    3. So (B) is the one that is not equal.

    Question 5 of 20
  6. xx2−25−8x2+6x+5\dfrac{x}{x^2 - 25} - \dfrac{8}{x^2 + 6x + 5} gives

    1. (1) x2−7x+40(x−5)(x+5)\dfrac{x^2 - 7x + 40}{(x - 5)(x + 5)}
    2. (2) x2+7x+40(x−5)(x+5)(x+1)\dfrac{x^2 + 7x + 40}{(x - 5)(x + 5)(x + 1)}
    3. (3) x2−7x+40(x2−25)(x+1)\dfrac{x^2 - 7x + 40}{(x^2 - 25)(x + 1)}
    4. (4) x2+10(x2−25)(x+1)\dfrac{x^2 + 10}{(x^2 - 25)(x + 1)}
    Show answer

    Answer: (3) x2−7x+40(x2−25)(x+1)\dfrac{x^2 - 7x + 40}{(x^2 - 25)(x + 1)}

    1. x2−25=(x+5)(x−5)x^2 - 25 = (x+5)(x-5) and x2+6x+5=(x+5)(x+1)x^2 + 6x + 5 = (x+5)(x+1).

    2. LCM =(x+5)(x−5)(x+1)= (x+5)(x-5)(x+1).

    3. x(x+1)−8(x−5)(x+5)(x−5)(x+1)=x2+x−8x+40(x2−25)(x+1)\dfrac{x(x+1) - 8(x-5)}{(x+5)(x-5)(x+1)} = \dfrac{x^2 + x - 8x + 40}{(x^2-25)(x+1)}

    4. =x2−7x+40(x2−25)(x+1)= \dfrac{x^2 - 7x + 40}{(x^2-25)(x+1)}

    Question 6 of 20
  7. The square root of 256x8y4z1025x6y6z6\dfrac{256 x^8 y^4 z^{10}}{25 x^6 y^6 z^6} is equal to

    1. (1) 165∣x2z4y2∣\dfrac{16}{5}\left|\dfrac{x^2 z^4}{y^2}\right|
    2. (2) 16∣y2x2z4∣16\left|\dfrac{y^2}{x^2 z^4}\right|
    3. (3) 165∣yxz2∣\dfrac{16}{5}\left|\dfrac{y}{xz^2}\right|
    4. (4) 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|
    Show answer

    Answer: (4) 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|

    1. 256x8y4z1025x6y6z6=256 x2z425 y2\dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2}

    2. Taking square root: 165∣xz2y∣\dfrac{16}{5}\left|\dfrac{x z^2}{y}\right|

    Question 7 of 20
  8. Which of the following should be added to make x4+64x^4 + 64 a perfect square

    1. (1) 4x24x^2
    2. (2) 16x216x^2
    3. (3) 8x28x^2
    4. (4) −8x2-8x^2
    Show answer

    Answer: (2) 16x216x^2

    1. For (x2+8)2=x4+16x2+64(x^2 + 8)^2 = x^4 + 16x^2 + 64.

    2. Given expression has x4+64x^4 + 64, so the missing middle term is 16x216x^2.

    Question 8 of 20
  9. The solution of (2x−1)2=9\left(2x - 1\right)^2 = 9 is equal to

    1. (1) −1-1
    2. (2) 22
    3. (3) −1, 2-1,\ 2
    4. (4) None of these
    Show answer

    Answer: (3) −1, 2-1,\ 2

    1. (2x−1)2=9⇒2x−1=±3(2x-1)^2 = 9 \Rightarrow 2x - 1 = \pm 3.

    2. 2x−1=3⇒x=22x - 1 = 3 \Rightarrow x = 2.

    3. 2x−1=−3⇒x=−12x - 1 = -3 \Rightarrow x = -1.

    Question 9 of 20
  10. The values of aa and bb if 4x4−24x3+76x2+ax+b4x^4 - 24x^3 + 76x^2 + ax + b is a perfect square are

    1. (1) 100,120100,120
    2. (2) 10,1210,12
    3. (3) −120,100-120,100
    4. (4) 12,1012,10
    Show answer

    Answer: (3) −120,100-120,100

    1. Let 4x4−24x3+76x2+ax+b=(2x2−6x+c)24x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2.

    2. Expanding: 4x4−24x3+(36+4c)x2−12cx+c24x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2.

    3. Compare x2x^2 terms: 36+4c=76⇒c=1036 + 4c = 76 \Rightarrow c = 10.

    4. So a=−12(10)=−120a = -12(10) = -120 and b=102=100b = 10^2 = 100.

    Question 10 of 20
  11. If the roots of the equation q2x2+p2x+r2=0q^2x^2 + p^2x + r^2 = 0 are the squares of the roots of the equation qx2+px+r=0qx^2 + px + r = 0, then qq, pp, rr are in _______

    1. (1) A.PA.P
    2. (2) G.PG.P
    3. (3) Both A.PA.P and G.PG.P
    4. (4) none of these
    Show answer

    Answer: (2) G.PG.P

    1. Let α,β\alpha,\beta be roots of qx2+px+r=0qx^2 + px + r = 0: α+β=−pq\alpha+\beta = -\dfrac{p}{q}, αβ=rq\alpha\beta = \dfrac{r}{q}.

    2. For q2x2+p2x+r2=0q^2x^2 + p^2x + r^2 = 0: α2+β2=−p2q2\alpha^2 + \beta^2 = -\dfrac{p^2}{q^2}, α2β2=r2q2\alpha^2\beta^2 = \dfrac{r^2}{q^2}.

    3. Now α2+β2=(α+β)2−2αβ=p2q2−2rq\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \dfrac{p^2}{q^2} - \dfrac{2r}{q}.

    4. So p2q2−2rq=−p2q2⇒2p2q2=2rq⇒p2=qr\dfrac{p^2}{q^2} - \dfrac{2r}{q} = -\dfrac{p^2}{q^2} \Rightarrow \dfrac{2p^2}{q^2} = \dfrac{2r}{q} \Rightarrow p^2 = qr.

    5. Hence q,p,rq, p, r are in G.P.

    Question 11 of 20
  12. Graph of a linear equation is a _______

    1. (1) straight line
    2. (2) circle
    3. (3) parabola
    4. (4) hyperbola
    Show answer

    Answer: (1) straight line

    1. A linear equation in two variables has degree one, and its graph is a straight line.

    Question 12 of 20
  13. The number of points of intersection of the quadratic polynomial x2+4x+4x^2 + 4x + 4 with the XX axis is

    1. (1) 00
    2. (2) 11
    3. (3) 00 or 11
    4. (4) 22
    Show answer

    Answer: (2) 11

    1. x2+4x+4=(x+2)2x^2 + 4x + 4 = (x+2)^2.

    2. It has equal roots x=−2,−2x = -2, -2.

    3. So the curve touches the XX axis at exactly one point.

    Question 13 of 20
  14. For the given matrix A=(135724689111315)A = \begin{pmatrix} 1 & 3 & 5 & 7 \\ 2 & 4 & 6 & 8 \\ 9 & 11 & 13 & 15 \end{pmatrix} the order of the matrix ATA^T is

    1. (1) 2×32 \times 3
    2. (2) 3×23 \times 2
    3. (3) 3×43 \times 4
    4. (4) 4×34 \times 3
    Show answer

    Answer: (4) 4×34 \times 3

    1. AA has 33 rows and 44 columns, so order of AA is 3×43 \times 4.

    2. Transpose interchanges rows and columns, so ATA^T is 4×34 \times 3.

    Question 14 of 20
  15. If AA is a 2×32 \times 3 matrix and BB is a 3×43 \times 4 matrix, how many columns does ABAB have

    1. (1) 33
    2. (2) 44
    3. (3) 22
    4. (4) 55
    Show answer

    Answer: (2) 44

    1. A2×3B3×4A_{2\times 3} B_{3\times 4} gives order 2×42 \times 4.

    2. So ABAB has 44 columns.

    Question 15 of 20
  16. If number of columns and rows are not equal in a matrix then it is said to be a

    1. (1) diagonal matrix
    2. (2) rectangular matrix
    3. (3) square matrix
    4. (4) identity matrix
    Show answer

    Answer: (2) rectangular matrix

    1. A matrix with number of rows ≠\ne number of columns is a rectangular matrix.

    Question 16 of 20
  17. Transpose of a column matrix is

    1. (1) unit matrix
    2. (2) diagonal matrix
    3. (3) column matrix
    4. (4) row matrix
    Show answer

    Answer: (4) row matrix

    1. A column matrix has order m×1m \times 1.

    2. Its transpose has order 1×m1 \times m, which is a row matrix.

    Question 17 of 20
  18. Find the matrix XX if 2X+(1357)=(5795)2X + \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix}

    1. (1) (−2−22−1)\begin{pmatrix} -2 & -2 \\ 2 & -1 \end{pmatrix}
    2. (2) (222−1)\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}
    3. (3) (1222)\begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix}
    4. (4) (2122)\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}
    Show answer

    Answer: (2) (222−1)\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}

    1. 2X=(5795)−(1357)=(444−2)2X = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix}

    2. X=12(444−2)=(222−1)X = \dfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}

    Question 18 of 20
  19. Which of the following can be calculated from the given matrices A=(123456)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}, B=(123456789)B = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, (i) A2A^2 (ii) B2B^2 (iii) ABAB (iv) BABA

    1. (1) (i) and (ii) only
    2. (2) (ii) and (iii) only
    3. (3) (ii) and (iv) only
    4. (4) all of these
    Show answer

    Answer: (3) (ii) and (iv) only

    1. Order of AA is 3×23 \times 2 and order of BB is 3×33 \times 3.

    2. A2=A AA^2 = A\,A: (3×2)(3×2)(3\times2)(3\times2) — columns of first ≠\ne rows of second, not defined.

    3. B2B^2: (3×3)(3×3)(3\times3)(3\times3) — defined.

    4. ABAB: (3×2)(3×3)(3\times2)(3\times3) — not defined.

    5. BABA: (3×3)(3×2)(3\times3)(3\times2) — defined.

    Question 19 of 20
  20. If A=(123321)A = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}, B=(102−102)B = \begin{pmatrix} 1 & 0 \\ 2 & -1 \\ 0 & 2 \end{pmatrix} and C=(01−25)C = \begin{pmatrix} 0 & 1 \\ -2 & 5 \end{pmatrix}. Which of the following statements are correct? (i) AB+C=(5555)AB + C = \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix} (ii) BC=(012−3−410)BC = \begin{pmatrix} 0 & 1 \\ 2 & -3 \\ -4 & 10 \end{pmatrix} (iii) BA+C=(2530)BA + C = \begin{pmatrix} 2 & 5 \\ 3 & 0 \end{pmatrix} (iv) (AB)C=(−820−813)(AB)C = \begin{pmatrix} -8 & 20 \\ -8 & 13 \end{pmatrix}

    1. (1) (i) and (ii) only
    2. (2) (ii) and (iii) only
    3. (3) (iii) and (iv) only
    4. (4) all of these
    Show answer

    Answer: (1) (i) and (ii) only

    1. AB=(5470)AB = \begin{pmatrix} 5 & 4 \\ 7 & 0 \end{pmatrix}, so AB+C=(5555)AB + C = \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix} — (i) is correct.

    2. BC=(012−3−410)BC = \begin{pmatrix} 0 & 1 \\ 2 & -3 \\ -4 & 10 \end{pmatrix} — (ii) is correct.

    3. BABA is of order 3×33\times3 and CC is 2×22\times2, so BA+CBA + C is not defined — (iii) is wrong.

    4. (AB)C=(5470)(01−25)=(−82507)(AB)C = \begin{pmatrix} 5 & 4 \\ 7 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} -8 & 25 \\ 0 & 7 \end{pmatrix} — (iv) is wrong.

    Question 20 of 20

Chapter 4: Geometry

  1. If in triangles ABCABC and EDFEDF, ABDE=BCFD\dfrac{AB}{DE} = \dfrac{BC}{FD} then they will be similar, when

    1. (1) ∠B=∠E\angle B = \angle E
    2. (2) ∠A=∠D\angle A = \angle D
    3. (3) ∠B=∠D\angle B = \angle D
    4. (4) ∠A=∠F\angle A = \angle F
    Show answer

    Answer: (3) ∠B=∠D\angle B = \angle D

    1. ABDE=BCFD\dfrac{AB}{DE}=\dfrac{BC}{FD}

    2. In ΔABC\Delta ABC the sides ABAB, BCBC include ∠B\angle B.

    3. In ΔEDF\Delta EDF the sides DEDE, FDFD include ∠D\angle D.

    4. By SAS similarity, the included angles must be equal.

    5. So ∠B=∠D\angle B=\angle D.

    Question 1 of 15
  2. In ΔLMN\Delta LMN, ∠L=60∘\angle L = 60^\circ, ∠M=50∘\angle M = 50^\circ. If ΔLMN∼ΔPQR\Delta LMN \sim \Delta PQR then the value of ∠R\angle R is

    1. (1) 40∘40^\circ
    2. (2) 70∘70^\circ
    3. (3) 30∘30^\circ
    4. (4) 110∘110^\circ
    Show answer

    Answer: (2) 70∘70^\circ

    1. In ΔLMN\Delta LMN, ∠N=180∘−(60∘+50∘)=70∘\angle N = 180^\circ-(60^\circ+50^\circ)=70^\circ.

    2. ΔLMN∼ΔPQR\Delta LMN \sim \Delta PQR gives ∠R=∠N\angle R = \angle N.

    3. ∠R=70∘\angle R = 70^\circ.

    Question 2 of 15
  3. If ΔABC\Delta ABC is an isosceles triangle with ∠C=90∘\angle C = 90^\circ and AC=5AC = 5 cm, then ABAB is

    1. (1) 2.52.5 cm
    2. (2) 55 cm
    3. (3) 1010 cm
    4. (4) 525\sqrt{2} cm
    Show answer

    Answer: (4) 525\sqrt{2} cm

    1. ∠C=90∘\angle C=90^\circ, so ABAB is the hypotenuse and the equal sides are AC=BC=5AC=BC=5 cm.

    2. AB2=AC2+BC2=52+52=50AB^2=AC^2+BC^2=5^2+5^2=50

    3. AB=50=52AB=\sqrt{50}=5\sqrt{2} cm.

    Question 3 of 15
  4. In a given figure ST∥QRST \parallel QR, PS=2PS = 2 cm and SQ=3SQ = 3 cm. Then the ratio of the area of ΔPQR\Delta PQR to the area of ΔPST\Delta PST is

    1. (1) 25:425 : 4
    2. (2) 25:725 : 7
    3. (3) 25:1125 : 11
    4. (4) 25:1325 : 13
    Show answer

    Answer: (1) 25:425 : 4

    1. PQ=PS+SQ=2+3=5PQ=PS+SQ=2+3=5 cm.

    2. ST∥QR⇒ΔPST∼ΔPQRST\parallel QR \Rightarrow \Delta PST \sim \Delta PQR.

    3. [ΔPQR][ΔPST]=PQ2PS2=5222=254\dfrac{[\Delta PQR]}{[\Delta PST]}=\dfrac{PQ^2}{PS^2}=\dfrac{5^2}{2^2}=\dfrac{25}{4}.

    Question 4 of 15
  5. The perimeters of two similar triangles ΔABC\Delta ABC and ΔPQR\Delta PQR are 3636 cm and 2424 cm respectively. If PQ=10PQ = 10 cm, then the length of ABAB is

    1. (1) 6236\dfrac{2}{3} cm
    2. (2) 1063\dfrac{10\sqrt{6}}{3} cm
    3. (3) 662366\dfrac{2}{3} cm
    4. (4) 1515 cm
    Show answer

    Answer: (4) 1515 cm

    1. In similar triangles, ratio of perimeters = ratio of corresponding sides.

    2. ABPQ=3624\dfrac{AB}{PQ}=\dfrac{36}{24}

    3. AB=10×3624=15AB=10\times\dfrac{36}{24}=15 cm.

    Question 5 of 15
  6. If in ΔABC\Delta ABC, DE∥BCDE \parallel BC. AB=3.6AB = 3.6 cm, AC=2.4AC = 2.4 cm and AD=2.1AD = 2.1 cm then the length of AEAE is

    1. (1) 1.41.4 cm
    2. (2) 1.81.8 cm
    3. (3) 1.21.2 cm
    4. (4) 1.051.05 cm
    Show answer

    Answer: (1) 1.41.4 cm

    1. DE∥BCDE\parallel BC, so by Thales theorem ADAB=AEAC\dfrac{AD}{AB}=\dfrac{AE}{AC}.

    2. 2.13.6=AE2.4\dfrac{2.1}{3.6}=\dfrac{AE}{2.4}

    3. AE=2.1×2.43.6=1.4AE=\dfrac{2.1\times 2.4}{3.6}=1.4 cm.

    Question 6 of 15
  7. In a ΔABC\Delta ABC, ADAD is the bisector of ∠BAC\angle BAC. If AB=8AB = 8 cm, BD=6BD = 6 cm and DC=3DC = 3 cm. The length of the side ACAC is

    1. (1) 66 cm
    2. (2) 44 cm
    3. (3) 33 cm
    4. (4) 88 cm
    Show answer

    Answer: (2) 44 cm

    1. ADAD bisects ∠BAC\angle BAC, so ABAC=BDDC\dfrac{AB}{AC}=\dfrac{BD}{DC}.

    2. 8AC=63=2\dfrac{8}{AC}=\dfrac{6}{3}=2

    3. AC=82=4AC=\dfrac{8}{2}=4 cm.

    Question 7 of 15
  8. In the adjacent figure ∠BAC=90∘\angle BAC = 90^\circ and AD⊥BCAD \perp BC then

    1. (1) BD⋅CD=BC2BD \cdot CD = BC^2
    2. (2) AB.AC=BC2AB.AC = BC^2
    3. (3) BD⋅CD=AD2BD \cdot CD = AD^2
    4. (4) AB⋅AC=AD2AB \cdot AC = AD^2
    Show answer

    Answer: (3) BD⋅CD=AD2BD \cdot CD = AD^2

    1. ∠BAC=90∘\angle BAC=90^\circ and AD⊥BCAD\perp BC, so ADAD is the altitude to the hypotenuse.

    2. Then ΔABD∼ΔCAD\Delta ABD \sim \Delta CAD.

    3. BDAD=ADCD\dfrac{BD}{AD}=\dfrac{AD}{CD}

    4. BD⋅CD=AD2BD\cdot CD = AD^2.

    Question 8 of 15
  9. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?

    1. (1) 1313 m
    2. (2) 1414 m
    3. (3) 1515 m
    4. (4) 12.812.8 m
    Show answer

    Answer: (1) 1313 m

    1. Difference of heights =11−6=5=11-6=5 m.

    2. Distance between feet =12=12 m.

    3. Distance between tops =122+52=144+25=169=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}

    4. =13=13 m.

    Question 9 of 15
  10. In the given figure, PR=26PR = 26 cm, QR=24QR = 24 cm, ∠PAQ=90∘\angle PAQ = 90^\circ, PA=6PA = 6 cm and QA=8QA = 8 cm. Find ∠PQR\angle PQR

    1. (1) 80∘80^\circ
    2. (2) 85∘85^\circ
    3. (3) 75∘75^\circ
    4. (4) 90∘90^\circ
    Show answer

    Answer: (4) 90∘90^\circ

    1. In right ΔPAQ\Delta PAQ, PQ2=PA2+QA2=62+82=100PQ^2=PA^2+QA^2=6^2+8^2=100, so PQ=10PQ=10 cm.

    2. Now PQ2+QR2=102+242=100+576=676PQ^2+QR^2=10^2+24^2=100+576=676.

    3. PR2=262=676PR^2=26^2=676.

    4. Since PQ2+QR2=PR2PQ^2+QR^2=PR^2, by converse of Pythagoras theorem ∠PQR=90∘\angle PQR=90^\circ.

    Question 10 of 15
  11. A tangent is perpendicular to the radius at the

    1. (1) centre
    2. (2) point of contact
    3. (3) infinity
    4. (4) chord
    Show answer

    Answer: (2) point of contact

    1. The radius drawn to the point where the tangent touches the circle is perpendicular to the tangent.

    2. So it is perpendicular at the point of contact.

    Question 11 of 15
  12. How many tangents can be drawn to the circle from an exterior point?

    1. (1) one
    2. (2) two
    3. (3) infinite
    4. (4) zero
    Show answer

    Answer: (2) two

    1. From a point outside a circle exactly two tangent lines can be drawn.

    2. (From a point on the circle only one, from an interior point none.)

    Question 12 of 15
  13. The two tangents from an external points PP to a circle with centre at OO are PAPA and PBPB. If ∠APB=70∘\angle APB = 70^\circ then the value of ∠AOB\angle AOB is

    1. (1) 100∘100^\circ
    2. (2) 110∘110^\circ
    3. (3) 120∘120^\circ
    4. (4) 130∘130^\circ
    Show answer

    Answer: (2) 110∘110^\circ

    1. OA⊥PAOA\perp PA and OB⊥PBOB\perp PB, so ∠OAP=∠OBP=90∘\angle OAP=\angle OBP=90^\circ.

    2. In quadrilateral OAPBOAPB, ∠AOB+∠APB+90∘+90∘=360∘\angle AOB+\angle APB+90^\circ+90^\circ=360^\circ.

    3. ∠AOB=360∘−180∘−70∘=110∘\angle AOB=360^\circ-180^\circ-70^\circ=110^\circ.

    Question 13 of 15
  14. In figure CPCP and CQCQ are tangents to a circle with centre at OO. ARBARB is another tangent touching the circle at RR. If CP=11CP = 11 cm and BC=7BC = 7 cm, then the length of BRBR is

    1. (1) 66 cm
    2. (2) 55 cm
    3. (3) 88 cm
    4. (4) 44 cm
    Show answer

    Answer: (4) 44 cm

    1. Tangents from CC: CQ=CP=11CQ=CP=11 cm.

    2. BQ=CQ−BC=11−7=4BQ=CQ-BC=11-7=4 cm.

    3. Tangents from BB: BR=BQBR=BQ.

    4. BR=4BR=4 cm.

    Question 14 of 15
  15. In figure if PRPR is tangent to the circle at PP and OO is the centre of the circle, then ∠POQ\angle POQ is

    1. (1) 120∘120^\circ
    2. (2) 100∘100^\circ
    3. (3) 110∘110^\circ
    4. (4) 90∘90^\circ
    Show answer

    Answer: (1) 120∘120^\circ

    1. OP⊥PROP\perp PR (radius ⊥\perp tangent), so ∠OPR=90∘\angle OPR=90^\circ.

    2. ∠OPQ=90∘−60∘=30∘\angle OPQ=90^\circ-60^\circ=30^\circ.

    3. OP=OQOP=OQ (radii), so ∠OQP=∠OPQ=30∘\angle OQP=\angle OPQ=30^\circ.

    4. ∠POQ=180∘−(30∘+30∘)=120∘\angle POQ=180^\circ-(30^\circ+30^\circ)=120^\circ.

    Question 15 of 15

Chapter 5: Coordinate Geometry

  1. The area of triangle formed by the points (−5,0)(-5,0), (0,−5)(0,-5) and (5,0)(5,0) is

    1. (1) 00 sq.units
    2. (2) 2525 sq.units
    3. (3) 55 sq.units
    4. (4) none of these
    Show answer

    Answer: (2) 2525 sq.units

    1. The points (−5,0)(-5,0) and (5,0)(5,0) lie on the X axis, so the base =10=10 units.

    2. The third point (0,−5)(0,-5) is at a distance 55 from the X axis, so height =5=5 units.

    3. Area =12×10×5=25=\dfrac{1}{2}\times 10\times 5=25 sq.units.

    Question 1 of 15
  2. A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the YY axis. The path travelled by the man is

    1. (1) x=10x = 10
    2. (2) y=10y = 10
    3. (3) x=0x = 0
    4. (4) y=0y = 0
    Show answer

    Answer: (1) x=10x = 10

    1. The wall is the Y axis, and his distance from it is always 1010 units.

    2. So his xx coordinate is always 1010.

    3. Path is the line x=10x=10.

    Question 2 of 15
  3. The straight line given by the equation x=11x = 11 is

    1. (1) parallel to XX axis
    2. (2) parallel to YY axis
    3. (3) passing through the origin
    4. (4) passing through the point (0,11)(0,11)
    Show answer

    Answer: (2) parallel to YY axis

    1. x=11x=11 means every point on it has xx coordinate 1111, whatever be yy.

    2. Such a line is vertical, hence parallel to the Y axis.

    Question 3 of 15
  4. If (5,7)(5,7), (3,p)(3,p) and (6,6)(6,6) are collinear, then the value of pp is

    1. (1) 33
    2. (2) 66
    3. (3) 99
    4. (4) 1212
    Show answer

    Answer: (3) 99

    1. Slope of the line joining (5,7)(5,7) and (6,6)(6,6) is 6−76−5=−1\dfrac{6-7}{6-5}=-1.

    2. For collinearity, slope of (5,7)(5,7) and (3,p)(3,p) is also −1-1.

    3. p−73−5=−1⇒p−7=2\dfrac{p-7}{3-5}=-1\Rightarrow p-7=2

    4. p=9p=9

    Question 4 of 15
  5. The point of intersection of 3x−y=43x - y = 4 and x+y=8x + y = 8 is

    1. (1) (5,3)(5,3)
    2. (2) (2,4)(2,4)
    3. (3) (3,5)(3,5)
    4. (4) (4,4)(4,4)
    Show answer

    Answer: (3) (3,5)(3,5)

    1. 3x−y=43x-y=4 and x+y=8x+y=8.

    2. Adding: 4x=12⇒x=34x=12\Rightarrow x=3.

    3. Then 3+y=8⇒y=53+y=8\Rightarrow y=5.

    4. Point of intersection is (3,5)(3,5).

    Question 5 of 15
  6. The slope of the line joining (12,3)(12,3), (4,a)(4,a) is 18\dfrac{1}{8}. The value of 'aa' is

    1. (1) 11
    2. (2) 44
    3. (3) −5-5
    4. (4) 22
    Show answer

    Answer: (4) 22

    1. Slope =a−34−12=a−3−8=\dfrac{a-3}{4-12}=\dfrac{a-3}{-8}.

    2. a−3−8=18⇒a−3=−1\dfrac{a-3}{-8}=\dfrac{1}{8}\Rightarrow a-3=-1

    3. a=2a=2

    Question 6 of 15
  7. The slope of the line which is perpendicular to a line joining the points (0,0)(0,0) and (−8,8)(-8,8) is

    1. (1) −1-1
    2. (2) 11
    3. (3) 13\dfrac{1}{3}
    4. (4) −8-8
    Show answer

    Answer: (2) 11

    1. Slope of the line joining (0,0)(0,0) and (−8,8)(-8,8) is 8−0−8−0=−1\dfrac{8-0}{-8-0}=-1.

    2. Product of slopes of perpendicular lines =−1=-1.

    3. Required slope =−1−1=1=\dfrac{-1}{-1}=1.

    Question 7 of 15
  8. If slope of the line PQPQ is 13\dfrac{1}{\sqrt{3}} then slope of the perpendicular bisector of PQPQ is

    1. (1) 3\sqrt{3}
    2. (2) −3-\sqrt{3}
    3. (3) 13\dfrac{1}{\sqrt{3}}
    4. (4) 00
    Show answer

    Answer: (2) −3-\sqrt{3}

    1. The perpendicular bisector of PQPQ is perpendicular to PQPQ.

    2. Slope of PQ=13PQ=\dfrac{1}{\sqrt{3}}.

    3. Required slope =−1(1/3)=−3=-\dfrac{1}{(1/\sqrt{3})}=-\sqrt{3}.

    Question 8 of 15
  9. If AA is a point on the YY axis whose ordinate is 8 and BB is a point on the XX axis whose abscissae is 5 then the equation of the line ABAB is

    1. (1) 8x+5y=408x + 5y = 40
    2. (2) 8x−5y=408x - 5y = 40
    3. (3) x=8x = 8
    4. (4) y=5y = 5
    Show answer

    Answer: (1) 8x+5y=408x + 5y = 40

    1. AA is on the Y axis with ordinate 88, so A(0,8)A(0,8); BB is on the X axis with abscissa 55, so B(5,0)B(5,0).

    2. Intercept form: x5+y8=1\dfrac{x}{5}+\dfrac{y}{8}=1.

    3. Multiplying by 4040: 8x+5y=408x+5y=40.

    Question 9 of 15
  10. The equation of a line passing through the origin and perpendicular to the line 7x−3y+4=07x - 3y + 4 = 0 is

    1. (1) 7x−3y+4=07x - 3y + 4 = 0
    2. (2) 3x−7y+4=03x - 7y + 4 = 0
    3. (3) 3x+7y=03x + 7y = 0
    4. (4) 7x−3y=07x - 3y = 0
    Show answer

    Answer: (3) 3x+7y=03x + 7y = 0

    1. Slope of 7x−3y+4=07x-3y+4=0 is −7−3=73\dfrac{-7}{-3}=\dfrac{7}{3}.

    2. Slope of the perpendicular line =−37=-\dfrac{3}{7}.

    3. Through the origin: y=−37x⇒3x+7y=0y=-\dfrac{3}{7}x\Rightarrow 3x+7y=0.

    Question 10 of 15
  11. Consider four straight lines (i) l1; 3y=4x+5l_1;\ 3y = 4x + 5 (ii) l2; 4y=3x−1l_2;\ 4y = 3x - 1 (iii) l3; 4y+3x=7l_3;\ 4y + 3x = 7 (iv) l4; 4x+3y=2l_4;\ 4x + 3y = 2 Which of the following statement is true ?

    1. (1) l1l_1 and l2l_2 are perpendicular
    2. (2) l1l_1 and l4l_4 are parallel
    3. (3) l2l_2 and l4l_4 are perpendicular
    4. (4) l2l_2 and l3l_3 are parallel
    Show answer

    Answer: (3) l2l_2 and l4l_4 are perpendicular

    1. Slopes: l1=43l_1=\dfrac{4}{3}, l2=34l_2=\dfrac{3}{4}, l3=−34l_3=-\dfrac{3}{4}, l4=−43l_4=-\dfrac{4}{3}.

    2. m2×m4=34×(−43)=−1m_2\times m_4=\dfrac{3}{4}\times\left(-\dfrac{4}{3}\right)=-1.

    3. So l2l_2 and l4l_4 are perpendicular.

    Question 11 of 15
  12. A straight line has equation 8y=4x+218y = 4x + 21. Which of the following is true

    1. (1) The slope is 0.5 and the yy intercept is 2.6
    2. (2) The slope is 5 and the yy intercept is 1.6
    3. (3) The slope is 0.5 and the yy intercept is 1.6
    4. (4) The slope is 5 and the yy intercept is 2.6
    Show answer

    Answer: (1) The slope is 0.5 and the yy intercept is 2.6

    1. 8y=4x+21⇒y=48x+2188y=4x+21\Rightarrow y=\dfrac{4}{8}x+\dfrac{21}{8}.

    2. y=0.5x+2.625y=0.5x+2.625.

    3. Slope =0.5=0.5, yy intercept =2.625≈2.6=2.625\approx 2.6.

    Question 12 of 15
  13. When proving that a quadrilateral is a trapezium, it is necessary to show

    1. (1) Two sides are parallel.
    2. (2) Two parallel and two non-parallel sides.
    3. (3) Opposite sides are parallel.
    4. (4) All sides are of equal length.
    Show answer

    Answer: (2) Two parallel and two non-parallel sides.

    1. A trapezium has exactly one pair of parallel sides.

    2. So we must show two sides are parallel and the other two are not parallel.

    Question 13 of 15
  14. When proving that a quadrilateral is a parallelogram by using slopes you must find

    1. (1) The slopes of two sides
    2. (2) The slopes of two pair of opposite sides
    3. (3) The lengths of all sides
    4. (4) Both the lengths and slopes of two sides
    Show answer

    Answer: (2) The slopes of two pair of opposite sides

    1. In a parallelogram both pairs of opposite sides are parallel.

    2. Using slopes, we find the slopes of the two pairs of opposite sides and show each pair is equal.

    Question 14 of 15
  15. (2, 1)(2,\ 1) is the point of intersection of two lines.

    1. (1) x−y−3=0; 3x−y−7=0x - y - 3 = 0;\ 3x - y - 7 = 0
    2. (2) x+y=3; 3x+y=7x + y = 3;\ 3x + y = 7
    3. (3) 3x+y=3; x+y=73x + y = 3;\ x + y = 7
    4. (4) x+3y−3=0; x−y−7=0x + 3y - 3 = 0;\ x - y - 7 = 0
    Show answer

    Answer: (2) x+y=3; 3x+y=7x + y = 3;\ 3x + y = 7

    1. Substitute x=2, y=1x=2,\ y=1 in option (B).

    2. x+y=2+1=3x+y=2+1=3 ✓

    3. 3x+y=6+1=73x+y=6+1=7 ✓

    4. Both are satisfied, so (2,1)(2,1) is their point of intersection.

    Question 15 of 15

Chapter 6: Trigonometry

  1. The value of sin⁡2θ+11+tan⁡2θ\sin^2\theta + \dfrac{1}{1+\tan^2\theta} is equal to

    1. (1) tan⁡2θ\tan^2\theta
    2. (2) 11
    3. (3) cot⁡2θ\cot^2\theta
    4. (4) 00
    Show answer

    Answer: (2) 11

    1. 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta, so 11+tan⁡2θ=cos⁡2θ\dfrac{1}{1+\tan^2\theta}=\cos^2\theta.

    2. sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1.

    Question 1 of 15
  2. tan⁡θ cosec2θ−tan⁡θ\tan\theta\,\mathrm{cosec}^2\theta - \tan\theta is equal to

    1. (1) sec⁡θ\sec\theta
    2. (2) cot⁡2θ\cot^2\theta
    3. (3) sin⁡θ\sin\theta
    4. (4) cot⁡θ\cot\theta
    Show answer

    Answer: (4) cot⁡θ\cot\theta

    1. tan⁡θ cosec2θ−tan⁡θ=tan⁡θ(cosec2θ−1)\tan\theta\,\mathrm{cosec}^2\theta-\tan\theta=\tan\theta(\mathrm{cosec}^2\theta-1).

    2. cosec2θ−1=cot⁡2θ\mathrm{cosec}^2\theta-1=\cot^2\theta.

    3. =tan⁡θ⋅cot⁡2θ=cot⁡θ=\tan\theta\cdot\cot^2\theta=\cot\theta.

    Question 2 of 15
  3. If (sin⁡α+cosec α)2+(cos⁡α+sec⁡α)2=k+tan⁡2α+cot⁡2α(\sin\alpha + \mathrm{cosec}\,\alpha)^2 + (\cos\alpha + \sec\alpha)^2 = k + \tan^2\alpha + \cot^2\alpha, then the value of kk is equal to

    1. (1) 99
    2. (2) 77
    3. (3) 55
    4. (4) 33
    Show answer

    Answer: (2) 77

    1. (sin⁡α+cosec α)2=sin⁡2α+2+cosec2α(\sin\alpha+\mathrm{cosec}\,\alpha)^2=\sin^2\alpha+2+\mathrm{cosec}^2\alpha.

    2. (cos⁡α+sec⁡α)2=cos⁡2α+2+sec⁡2α(\cos\alpha+\sec\alpha)^2=\cos^2\alpha+2+\sec^2\alpha.

    3. Sum =(sin⁡2α+cos⁡2α)+4+(1+cot⁡2α)+(1+tan⁡2α)=(\sin^2\alpha+\cos^2\alpha)+4+(1+\cot^2\alpha)+(1+\tan^2\alpha).

    4. =7+tan⁡2α+cot⁡2α=7+\tan^2\alpha+\cot^2\alpha, so k=7k=7.

    Question 3 of 15
  4. If sin⁡θ+cos⁡θ=a\sin\theta + \cos\theta = a and sec⁡θ+cosec θ=b\sec\theta + \mathrm{cosec}\,\theta = b, then the value of b(a2−1)b(a^2-1) is equal to

    1. (1) 2a2a
    2. (2) 3a3a
    3. (3) 00
    4. (4) 2ab2ab
    Show answer

    Answer: (1) 2a2a

    1. a2−1=(sin⁡θ+cos⁡θ)2−1=2sin⁡θcos⁡θa^2-1=(\sin\theta+\cos\theta)^2-1=2\sin\theta\cos\theta.

    2. b=sec⁡θ+cosec θ=sin⁡θ+cos⁡θsin⁡θcos⁡θ=asin⁡θcos⁡θb=\sec\theta+\mathrm{cosec}\,\theta=\dfrac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\dfrac{a}{\sin\theta\cos\theta}.

    3. b(a2−1)=asin⁡θcos⁡θ×2sin⁡θcos⁡θ=2ab(a^2-1)=\dfrac{a}{\sin\theta\cos\theta}\times 2\sin\theta\cos\theta=2a.

    Question 4 of 15
  5. If 5x=sec⁡θ5x = \sec\theta and 5y=tan⁡θ\dfrac{5}{y} = \tan\theta, then x2−1y2x^2 - \dfrac{1}{y^2} is equal to

    1. (1) 2525
    2. (2) 125\dfrac{1}{25}
    3. (3) 55
    4. (4) 11
    Show answer

    Answer: (2) 125\dfrac{1}{25}

    1. x=sec⁡θ5x=\dfrac{\sec\theta}{5} and 1y=tan⁡θ5\dfrac{1}{y}=\dfrac{\tan\theta}{5}.

    2. x2−1y2=sec⁡2θ−tan⁡2θ25=125x^2-\dfrac{1}{y^2}=\dfrac{\sec^2\theta-\tan^2\theta}{25}=\dfrac{1}{25}.

    Question 5 of 15
  6. If sin⁡θ=cos⁡θ\sin\theta = \cos\theta, then 2tan⁡2θ+sin⁡2θ−12\tan^2\theta + \sin^2\theta - 1 is equal to

    1. (1) −32\dfrac{-3}{2}
    2. (2) 32\dfrac{3}{2}
    3. (3) 23\dfrac{2}{3}
    4. (4) −23\dfrac{-2}{3}
    Show answer

    Answer: (2) 32\dfrac{3}{2}

    1. sin⁡θ=cos⁡θ⇒tan⁡θ=1⇒θ=45∘\sin\theta=\cos\theta\Rightarrow\tan\theta=1\Rightarrow\theta=45^\circ.

    2. 2tan⁡245∘+sin⁡245∘−1=2(1)+12−1=322\tan^2 45^\circ+\sin^2 45^\circ-1=2(1)+\dfrac{1}{2}-1=\dfrac{3}{2}.

    Question 6 of 15
  7. If x=atan⁡θx = a\tan\theta and y=bsec⁡θy = b\sec\theta then

    1. (1) y2b2−x2a2=1\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1
    2. (2) x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1
    3. (3) x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1
    4. (4) x2a2−y2b2=0\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0
    Show answer

    Answer: (1) y2b2−x2a2=1\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1

    1. xa=tan⁡θ\dfrac{x}{a}=\tan\theta, yb=sec⁡θ\dfrac{y}{b}=\sec\theta.

    2. sec⁡2θ−tan⁡2θ=1⇒y2b2−x2a2=1\sec^2\theta-\tan^2\theta=1\Rightarrow\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1.

    Question 7 of 15
  8. (1+tan⁡θ+sec⁡θ)(1+cot⁡θ−cosec θ)(1+\tan\theta+\sec\theta)(1+\cot\theta-\mathrm{cosec}\,\theta) is equal to

    1. (1) 00
    2. (2) 11
    3. (3) 22
    4. (4) −1-1
    Show answer

    Answer: (3) 22

    1. Write the second bracket as 1+cos⁡θsin⁡θ−1sin⁡θ=sin⁡θ+cos⁡θ−1sin⁡θ1+\dfrac{\cos\theta}{\sin\theta}-\dfrac{1}{\sin\theta}=\dfrac{\sin\theta+\cos\theta-1}{\sin\theta}.

    2. First bracket =cos⁡θ+sin⁡θ+1cos⁡θ=\dfrac{\cos\theta+\sin\theta+1}{\cos\theta}.

    3. Product =(sin⁡θ+cos⁡θ)2−1sin⁡θcos⁡θ=2sin⁡θcos⁡θsin⁡θcos⁡θ=2=\dfrac{(\sin\theta+\cos\theta)^2-1}{\sin\theta\cos\theta}=\dfrac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2.

    Question 8 of 15
  9. acot⁡θ+b cosec θ=pa\cot\theta + b\,\mathrm{cosec}\,\theta = p and bcot⁡θ+a cosec θ=qb\cot\theta + a\,\mathrm{cosec}\,\theta = q then p2−q2p^2 - q^2 is equal to

    1. (1) a2−b2a^2 - b^2
    2. (2) b2−a2b^2 - a^2
    3. (3) a2+b2a^2 + b^2
    4. (4) b−ab - a
    Show answer

    Answer: (2) b2−a2b^2 - a^2

    1. p2−q2=(acot⁡θ+b cosec θ)2−(bcot⁡θ+a cosec θ)2p^2-q^2=(a\cot\theta+b\,\mathrm{cosec}\,\theta)^2-(b\cot\theta+a\,\mathrm{cosec}\,\theta)^2.

    2. =(a2−b2)cot⁡2θ−(a2−b2)cosec2θ=(a^2-b^2)\cot^2\theta-(a^2-b^2)\mathrm{cosec}^2\theta (the 2ab2ab terms cancel).

    3. =(a2−b2)(cot⁡2θ−cosec2θ)=(a2−b2)(−1)=b2−a2=(a^2-b^2)(\cot^2\theta-\mathrm{cosec}^2\theta)=(a^2-b^2)(-1)=b^2-a^2.

    Question 9 of 15
  10. If the ratio of the height of a tower and the length of its shadow is 3:1\sqrt{3}:1, then the angle of elevation of the sun has measure

    1. (1) 45∘45^\circ
    2. (2) 30∘30^\circ
    3. (3) 90∘90^\circ
    4. (4) 60∘60^\circ
    Show answer

    Answer: (4) 60∘60^\circ

    1. tan⁡θ=heightshadow=31=3\tan\theta=\dfrac{\text{height}}{\text{shadow}}=\dfrac{\sqrt3}{1}=\sqrt3.

    2. θ=60∘\theta=60^\circ.

    Question 10 of 15
  11. The electric pole subtends an angle of 30∘30^\circ at a point on the same level as its foot. At a second point 'bb' metres above the first, the depression of the foot of the pole is 60∘60^\circ. The height of the pole (in metres) is equal to

    1. (1) 3 b\sqrt{3}\,b
    2. (2) b3\dfrac{b}{3}
    3. (3) b2\dfrac{b}{2}
    4. (4) b3\dfrac{b}{\sqrt{3}}
    Show answer

    Answer: (2) b3\dfrac{b}{3}

    1. Let dd be the horizontal distance from the pole and hh its height.

    2. From the point bb m high, depression of the foot is 60∘60^\circ: tan⁡60∘=bd⇒d=b3\tan 60^\circ=\dfrac{b}{d}\Rightarrow d=\dfrac{b}{\sqrt3}.

    3. From the first point, elevation of the top is 30∘30^\circ: tan⁡30∘=hd\tan 30^\circ=\dfrac{h}{d}.

    4. h=13×b3=b3h=\dfrac{1}{\sqrt3}\times\dfrac{b}{\sqrt3}=\dfrac{b}{3}.

    Question 11 of 15
  12. A tower is 60 m heigh. Its shadow reduces by xx metres when the angle of elevation of the sun increases from 30∘30^\circ to 45∘45^\circ then xx is equal to

    1. (1) 41.92 m
    2. (2) 43.92 m
    3. (3) 43 m
    4. (4) 45.6 m
    Show answer

    Answer: (2) 43.92 m

    1. Shadow at 30∘30^\circ: 60tan⁡30∘=603\dfrac{60}{\tan 30^\circ}=60\sqrt3 m.

    2. Shadow at 45∘45^\circ: 60tan⁡45∘=60\dfrac{60}{\tan 45^\circ}=60 m.

    3. x=603−60=60(1.732−1)=43.92x=60\sqrt3-60=60(1.732-1)=43.92 m.

    Question 12 of 15
  13. The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30∘30^\circ and 60∘60^\circ respectively. The height of the multistoried building and the distance between two buildings (in metres) is

    1. (1) 20, 10320,\ 10\sqrt{3}
    2. (2) 30, 5330,\ 5\sqrt{3}
    3. (3) 20, 1020,\ 10
    4. (4) 30, 10330,\ 10\sqrt{3}
    Show answer

    Answer: (4) 30, 10330,\ 10\sqrt{3}

    1. Let hh be the height of the multistoried building and dd the distance between them.

    2. tan⁡60∘=hd⇒h=d3\tan 60^\circ=\dfrac{h}{d}\Rightarrow h=d\sqrt3.

    3. tan⁡30∘=h−20d⇒h−20=d3\tan 30^\circ=\dfrac{h-20}{d}\Rightarrow h-20=\dfrac{d}{\sqrt3}.

    4. d3−d3=20⇒2d3=20⇒d=103d\sqrt3-\dfrac{d}{\sqrt3}=20\Rightarrow \dfrac{2d}{\sqrt3}=20\Rightarrow d=10\sqrt3.

    5. h=103×3=30h=10\sqrt3\times\sqrt3=30 m.

    Question 13 of 15
  14. Two persons are standing 'xx' metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is

    1. (1) 2 x\sqrt{2}\,x
    2. (2) x22\dfrac{x}{2\sqrt{2}}
    3. (3) x2\dfrac{x}{\sqrt{2}}
    4. (4) 2x2x
    Show answer

    Answer: (2) x22\dfrac{x}{2\sqrt{2}}

    1. Let the shorter height be hh; the other is 2h2h. The observer is at the midpoint, x2\dfrac{x}{2} from each.

    2. tan⁡θ=2hx/2=4hx\tan\theta=\dfrac{2h}{x/2}=\dfrac{4h}{x} and tan⁡(90∘−θ)=cot⁡θ=hx/2=2hx\tan(90^\circ-\theta)=\cot\theta=\dfrac{h}{x/2}=\dfrac{2h}{x}.

    3. tan⁡θ⋅cot⁡θ=1⇒4hx×2hx=1\tan\theta\cdot\cot\theta=1\Rightarrow\dfrac{4h}{x}\times\dfrac{2h}{x}=1.

    4. 8h2=x2⇒h=x228h^2=x^2\Rightarrow h=\dfrac{x}{2\sqrt2}.

    Question 14 of 15
  15. The angle of elevation of a cloud from a point hh metres above a lake is β\beta. The angle of depression of its reflection in the lake is 45∘45^\circ. The height of location of the cloud from the lake is

    1. (1) h(1+tan⁡β)1−tan⁡β\dfrac{h(1+\tan\beta)}{1-\tan\beta}
    2. (2) h(1−tan⁡β)1+tan⁡β\dfrac{h(1-\tan\beta)}{1+\tan\beta}
    3. (3) htan⁡(45∘−β)h\tan(45^\circ-\beta)
    4. (4) none of these
    Show answer

    Answer: (1) h(1+tan⁡β)1−tan⁡β\dfrac{h(1+\tan\beta)}{1-\tan\beta}

    1. Let the cloud be at height HH above the lake and dd the horizontal distance.

    2. Elevation: tan⁡β=H−hd\tan\beta=\dfrac{H-h}{d}; depression of reflection: tan⁡45∘=H+hd\tan 45^\circ=\dfrac{H+h}{d}.

    3. So d=H+hd=H+h and (H−h)=(H+h)tan⁡β(H-h)=(H+h)\tan\beta.

    4. H(1−tan⁡β)=h(1+tan⁡β)⇒H=h(1+tan⁡β)1−tan⁡βH(1-\tan\beta)=h(1+\tan\beta)\Rightarrow H=\dfrac{h(1+\tan\beta)}{1-\tan\beta}.

    Question 15 of 15

Chapter 7: Mensuration

  1. The curved surface area of a right circular cone of height 1515 cm and base diameter 1616 cm is

    1. (1) 60π60\pi cm2^2
    2. (2) 68π68\pi cm2^2
    3. (3) 120π120\pi cm2^2
    4. (4) 136π136\pi cm2^2
    Show answer

    Answer: (4) 136π136\pi cm2^2

    1. r=162=8r=\dfrac{16}{2}=8 cm, h=15h=15 cm.

    2. l=r2+h2=64+225=289=17l=\sqrt{r^2+h^2}=\sqrt{64+225}=\sqrt{289}=17 cm.

    3. CSA =πrl=π×8×17=136π=\pi r l=\pi\times 8\times 17=136\pi cm2^2.

    Question 1 of 15
  2. If two solid hemispheres of same base radius rr units are joined together along their bases, then curved surface area of this new solid is

    1. (1) 4πr24\pi r^2 sq. units
    2. (2) 6πr26\pi r^2 sq. units
    3. (3) 3πr23\pi r^2 sq. units
    4. (4) 8πr28\pi r^2 sq. units
    Show answer

    Answer: (1) 4πr24\pi r^2 sq. units

    1. Two hemispheres joined along their bases form a sphere of radius rr.

    2. CSA of sphere =4πr2=4\pi r^2 sq. units.

    Question 2 of 15
  3. The height of a right circular cone whose radius is 55 cm and slant height is 1313 cm will be

    1. (1) 1212 cm
    2. (2) 1010 cm
    3. (3) 1313 cm
    4. (4) 55 cm
    Show answer

    Answer: (1) 1212 cm

    1. l2=r2+h2l^2=r^2+h^2

    2. 132=52+h2⇒h2=169−25=14413^2=5^2+h^2\Rightarrow h^2=169-25=144

    3. h=12h=12 cm.

    Question 3 of 15
  4. If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

    1. (1) 1:21:2
    2. (2) 1:41:4
    3. (3) 1:61:6
    4. (4) 1:81:8
    Show answer

    Answer: (2) 1:41:4

    1. New volume =π(r2)2h=πr2h4=\pi\left(\dfrac{r}{2}\right)^2h=\dfrac{\pi r^2h}{4}.

    2. Ratio =πr2h/4πr2h=14=\dfrac{\pi r^2h/4}{\pi r^2h}=\dfrac{1}{4}.

    Question 4 of 15
  5. The total surface area of a cylinder whose radius is 13\dfrac{1}{3} of its height is

    1. (1) 9πh28\dfrac{9\pi h^2}{8} sq.units
    2. (2) 24πh224\pi h^2 sq.units
    3. (3) 8πh29\dfrac{8\pi h^2}{9} sq.units
    4. (4) 56πh29\dfrac{56\pi h^2}{9} sq.units
    Show answer

    Answer: (3) 8πh29\dfrac{8\pi h^2}{9} sq.units

    1. Given r=h3r=\dfrac{h}{3}.

    2. TSA =2πr(r+h)=2πh3(h3+h)=2\pi r(r+h)=2\pi\dfrac{h}{3}\left(\dfrac{h}{3}+h\right)

    3. =2πh3×4h3=8πh29=2\pi\dfrac{h}{3}\times\dfrac{4h}{3}=\dfrac{8\pi h^2}{9}.

    Question 5 of 15
  6. In a hollow cylinder, the sum of the external and internal radii is 1414 cm and the width is 44 cm. If its height is 2020 cm, the volume of the material in it is

    1. (1) 5600π5600\pi cm3^3
    2. (2) 1120π1120\pi cm3^3
    3. (3) 56π56\pi cm3^3
    4. (4) 3600π3600\pi cm3^3
    Show answer

    Answer: (2) 1120π1120\pi cm3^3

    1. R+r=14R+r=14, width R−r=4R-r=4, h=20h=20.

    2. Volume of material =π(R2−r2)h=π(R+r)(R−r)h=\pi(R^2-r^2)h=\pi(R+r)(R-r)h

    3. =π×14×4×20=1120π=\pi\times 14\times 4\times 20=1120\pi cm3^3.

    Question 6 of 15
  7. If the radius of the base of a cone is tripled and the height is doubled then the volume is

    1. (1) made 66 times
    2. (2) made 1818 times
    3. (3) made 1212 times
    4. (4) unchanged
    Show answer

    Answer: (2) made 1818 times

    1. New volume =13π(3r)2(2h)=13π⋅9r2⋅2h=\dfrac{1}{3}\pi(3r)^2(2h)=\dfrac{1}{3}\pi\cdot 9r^2\cdot 2h

    2. =18×13πr2h=18\times\dfrac{1}{3}\pi r^2h.

    3. So the volume is made 1818 times.

    Question 7 of 15
  8. The total surface area of a hemi-sphere is how much times the square of its radius.

    1. (1) π\pi
    2. (2) 4π4\pi
    3. (3) 3π3\pi
    4. (4) 2π2\pi
    Show answer

    Answer: (3) 3π3\pi

    1. TSA of hemisphere =3πr2=3\pi r^2.

    2. So it is 3π3\pi times the square of the radius.

    Question 8 of 15
  9. A solid sphere of radius xx cm is melted and cast into a shape of a solid cone of same radius. The height of the cone is

    1. (1) 3x3x cm
    2. (2) xx cm
    3. (3) 4x4x cm
    4. (4) 2x2x cm
    Show answer

    Answer: (3) 4x4x cm

    1. Volume of sphere == volume of cone

    2. 43πx3=13πx2h\dfrac{4}{3}\pi x^3=\dfrac{1}{3}\pi x^2h

    3. h=4xh=4x cm.

    Question 9 of 15
  10. A frustum of a right circular cone is of height 1616cm with radii of its ends as 88cm and 2020cm. Then, the volume of the frustum is

    1. (1) 3328π3328\pi cm3^3
    2. (2) 3228π3228\pi cm3^3
    3. (3) 3240π3240\pi cm3^3
    4. (4) 3340π3340\pi cm3^3
    Show answer

    Answer: (1) 3328π3328\pi cm3^3

    1. V=13πh(R2+Rr+r2)V=\dfrac{1}{3}\pi h(R^2+Rr+r^2)

    2. =13π(16)(400+160+64)=\dfrac{1}{3}\pi(16)(400+160+64)

    3. =163π(624)=3328π=\dfrac{16}{3}\pi(624)=3328\pi cm3^3.

    Question 10 of 15
  11. A shuttle cock used for playing badminton has the shape of the combination of

    1. (1) a cylinder and a sphere
    2. (2) a hemisphere and a cone
    3. (3) a sphere and a cone
    4. (4) frustum of a cone and a hemisphere
    Show answer

    Answer: (4) frustum of a cone and a hemisphere

    1. The cork base is a hemisphere and the feather part widens like a frustum of a cone.

    2. So it is frustum of a cone and a hemisphere.

    Question 11 of 15
  12. A spherical ball of radius r1r_1 units is melted to make 88 new identical balls each of radius r2r_2 units. Then r1:r2r_1 : r_2 is

    1. (1) 2:12:1
    2. (2) 1:21:2
    3. (3) 4:14:1
    4. (4) 1:41:4
    Show answer

    Answer: (1) 2:12:1

    1. Volume is unchanged: 43πr13=8×43πr23\dfrac{4}{3}\pi r_1^3=8\times\dfrac{4}{3}\pi r_2^3

    2. r13=8r23⇒r1=2r2r_1^3=8r_2^3\Rightarrow r_1=2r_2

    3. r1:r2=2:1r_1:r_2=2:1.

    Question 12 of 15
  13. The volume (in cm3^3) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 11 cm and height 55 cm is

    1. (1) 43π\dfrac{4}{3}\pi
    2. (2) 103π\dfrac{10}{3}\pi
    3. (3) 5π5\pi
    4. (4) 203π\dfrac{20}{3}\pi
    Show answer

    Answer: (1) 43π\dfrac{4}{3}\pi

    1. The greatest sphere has radius equal to the base radius =1=1 cm.

    2. V=43π(1)3=43πV=\dfrac{4}{3}\pi(1)^3=\dfrac{4}{3}\pi cm3^3.

    Question 13 of 15
  14. The height and radius of the cone of which the frustum is a part are h1h_1 units and r1r_1 units respectively. Height of the frustum is h2h_2 units and radius of the smaller base is r2r_2 units. If h2:h1=1:2h_2 : h_1 = 1 : 2 then r2:r1r_2 : r_1 is

    1. (1) 1:31 : 3
    2. (2) 1:21 : 2
    3. (3) 2:12 : 1
    4. (4) 3:13 : 1
    Show answer

    Answer: (2) 1:21 : 2

    1. The small cone cut off has height h1−h2h_1-h_2 and radius r2r_2.

    2. By similar triangles r2r1=h1−h2h1\dfrac{r_2}{r_1}=\dfrac{h_1-h_2}{h_1}.

    3. Given h2=h12h_2=\dfrac{h_1}{2}, so r2r1=h1−h12h1=12\dfrac{r_2}{r_1}=\dfrac{h_1-\frac{h_1}{2}}{h_1}=\dfrac{1}{2}.

    Question 14 of 15
  15. The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is

    1. (1) 1:2:31:2:3
    2. (2) 2:1:32:1:3
    3. (3) 1:3:21:3:2
    4. (4) 3:1:23:1:2
    Show answer

    Answer: (4) 3:1:23:1:2

    1. Same diameter and same height ⇒h=2r\Rightarrow h=2r.

    2. Cylinder =πr2(2r)=2πr3=\pi r^2(2r)=2\pi r^3

    3. Cone =13πr2(2r)=23πr3=\dfrac{1}{3}\pi r^2(2r)=\dfrac{2}{3}\pi r^3, Sphere =43πr3=\dfrac{4}{3}\pi r^3

    4. Ratio =2:23:43=6:2:4=3:1:2=2:\dfrac{2}{3}:\dfrac{4}{3}=6:2:4=3:1:2.

    Question 15 of 15

Chapter 8: Statistics and Probability

  1. Which of the following is not a measure of dispersion?

    1. (1) Range
    2. (2) Standard deviation
    3. (3) Arithmetic mean
    4. (4) Variance
    Show answer

    Answer: (3) Arithmetic mean

    1. Dispersion measures how spread out the data is: range, standard deviation, variance.

    2. Arithmetic mean is a measure of central tendency, not dispersion.

    Question 1 of 15
  2. The range of the data 8,8,8,8,8...88, 8, 8, 8, 8. . . 8 is

    1. (1) 00
    2. (2) 11
    3. (3) 88
    4. (4) 33
    Show answer

    Answer: (1) 00

    1. Largest value =8= 8, smallest value =8= 8.

    2. Range =L−S=8−8=0= L - S = 8 - 8 = 0.

    Question 2 of 15
  3. The sum of all deviations of the data from its mean is

    1. (1) Always positive
    2. (2) always negative
    3. (3) zero
    4. (4) non-zero integer
    Show answer

    Answer: (3) zero

    1. ∑(xi−xˉ)=∑xi−nxˉ\sum (x_i-\bar{x}) = \sum x_i - n\bar{x}.

    2. Since xˉ=∑xin\bar{x}=\dfrac{\sum x_i}{n}, we get nxˉ=∑xin\bar{x}=\sum x_i.

    3. So the sum of deviations =0=0.

    Question 3 of 15
  4. The mean of 100100 observations is 4040 and their standard deviation is 33. The sum of squares of all observations is

    1. (1) 4000040000
    2. (2) 160900160900
    3. (3) 160000160000
    4. (4) 3000030000
    Show answer

    Answer: (2) 160900160900

    1. σ2=∑x2n−(xˉ)2\sigma^2=\dfrac{\sum x^2}{n}-\left(\bar{x}\right)^2

    2. 9=∑x2100−16009=\dfrac{\sum x^2}{100}-1600

    3. ∑x2100=1609\dfrac{\sum x^2}{100}=1609

    4. ∑x2=160900\sum x^2=160900

    Question 4 of 15
  5. Variance of first 2020 natural numbers is

    1. (1) 32.2532.25
    2. (2) 44.2544.25
    3. (3) 33.2533.25
    4. (4) 3030
    Show answer

    Answer: (3) 33.2533.25

    1. Variance of first nn natural numbers =n2−112=\dfrac{n^2-1}{12}.

    2. =202−112=39912=33.25=\dfrac{20^2-1}{12}=\dfrac{399}{12}=33.25

    Question 5 of 15
  6. The standard deviation of a data is 33. If each value is multiplied by 55 then the new variance is

    1. (1) 33
    2. (2) 1515
    3. (3) 55
    4. (4) 225225
    Show answer

    Answer: (4) 225225

    1. Multiplying each value by 55 multiplies the S.D. by 55.

    2. New S.D. =5×3=15=5\times 3=15.

    3. New variance =152=225=15^2=225.

    Question 6 of 15
  7. If the standard deviation of x, y, zx,\ y,\ z is pp then the standard deviation of 3x+53x+5, 3y+53y+5, 3z+53z+5 is

    1. (1) 3p+53p+5
    2. (2) 3p3p
    3. (3) p+5p+5
    4. (4) 9p+159p+15
    Show answer

    Answer: (2) 3p3p

    1. Adding a constant does not change the S.D.; multiplying by 33 multiplies the S.D. by 33.

    2. New standard deviation =3p=3p.

    Question 7 of 15
  8. If the mean and coefficient of variation of a data are 44 and 87.5%87.5\% then the standard deviation is

    1. (1) 3.53.5
    2. (2) 33
    3. (3) 4.54.5
    4. (4) 2.52.5
    Show answer

    Answer: (1) 3.53.5

    1. C.V. =σxˉ×100=\dfrac{\sigma}{\bar{x}}\times 100

    2. 87.5=σ4×10087.5=\dfrac{\sigma}{4}\times 100

    3. σ=87.5×4100=3.5\sigma=\dfrac{87.5\times 4}{100}=3.5

    Question 8 of 15
  9. Which of the following is incorrect?

    1. (1) P(A)>1P(A) > 1
    2. (2) 0≤P(A)≤10 \leq P(A) \leq 1
    3. (3) P(ϕ)=0P(\phi) = 0
    4. (4) P(A)+P(Aˉ)=1P(A) + P(\bar{A}) = 1
    Show answer

    Answer: (1) P(A)>1P(A) > 1

    1. Probability of any event always lies between 00 and 11.

    2. So P(A)>1P(A)>1 is impossible.

    Question 9 of 15
  10. The probability a red marble selected at random from a jar containing pp red, qq blue and r green marbles is

    1. (1) qp+q+r\dfrac{q}{p+q+r}
    2. (2) pp+q+r\dfrac{p}{p+q+r}
    3. (3) p+qp+q+r\dfrac{p+q}{p+q+r}
    4. (4) p+rp+q+r\dfrac{p+r}{p+q+r}
    Show answer

    Answer: (2) pp+q+r\dfrac{p}{p+q+r}

    1. Total marbles =p+q+r=p+q+r; red marbles =p=p.

    2. P(red)=pp+q+rP(\text{red})=\dfrac{p}{p+q+r}

    Question 10 of 15
  11. A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 77 is

    1. (1) 310\dfrac{3}{10}
    2. (2) 710\dfrac{7}{10}
    3. (3) 39\dfrac{3}{9}
    4. (4) 79\dfrac{7}{9}
    Show answer

    Answer: (2) 710\dfrac{7}{10}

    1. Units digit can be 0,1,2,…,90,1,2,\ldots,9, so n(S)=10n(S)=10.

    2. Digits less than 77: 0,1,2,3,4,5,60,1,2,3,4,5,6, so n(A)=7n(A)=7.

    3. P(A)=710P(A)=\dfrac{7}{10}

    Question 11 of 15
  12. The probability of getting a job for a person is x3\dfrac{x}{3}. If the probability of not getting the job is 23\dfrac{2}{3} then the value of xx is

    1. (1) 22
    2. (2) 11
    3. (3) 33
    4. (4) 1.51.5
    Show answer

    Answer: (2) 11

    1. P(A)+P(Aˉ)=1P(A)+P(\bar{A})=1

    2. x3+23=1\dfrac{x}{3}+\dfrac{2}{3}=1

    3. x3=13⇒x=1\dfrac{x}{3}=\dfrac{1}{3}\Rightarrow x=1

    Question 12 of 15
  13. Kamalam went to play a lucky draw contest. 135135 tickets of the lucky draw were sold. If the probability of Kamalam winning is 19\dfrac{1}{9}, then the number of tickets bought by Kamalam is

    1. (1) 55
    2. (2) 1010
    3. (3) 1515
    4. (4) 2020
    Show answer

    Answer: (3) 1515

    1. Let the number of tickets bought be nn.

    2. n135=19\dfrac{n}{135}=\dfrac{1}{9}

    3. n=1359=15n=\dfrac{135}{9}=15

    Question 13 of 15
  14. If a letter is chosen at random from the English alphabets {a,b,…,z}\{a,b,\ldots,z\}, then the probability that the letter chosen precedes xx

    1. (1) 1213\dfrac{12}{13}
    2. (2) 113\dfrac{1}{13}
    3. (3) 2326\dfrac{23}{26}
    4. (4) 326\dfrac{3}{26}
    Show answer

    Answer: (3) 2326\dfrac{23}{26}

    1. Total letters n(S)=26n(S)=26.

    2. Letters before xx are aa to ww, that is 2323 letters.

    3. P=2326P=\dfrac{23}{26}

    Question 14 of 15
  15. A purse contains 10 notes of ₹20002000, 1515 notes of ₹500500, and 2525 notes of ₹200200. One note is drawn at random. What is the probability that the note is either a ₹500500 note or ₹200200 note?

    1. (1) 15\dfrac{1}{5}
    2. (2) 310\dfrac{3}{10}
    3. (3) 23\dfrac{2}{3}
    4. (4) 45\dfrac{4}{5}
    Show answer

    Answer: (4) 45\dfrac{4}{5}

    1. Total notes =10+15+25=50=10+15+25=50.

    2. Favourable notes =15+25=40=15+25=40.

    3. P=4050=45P=\dfrac{40}{50}=\dfrac{4}{5}

    Question 15 of 15

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