13-07-2025
9610WJA802282250001 JA
PART 1 : PHYSICS
SECTION-I
1) A dimensionless quantity
(A) never has a unit
(B) always has a unit
(C) may have a unit
(D) does not exit
2) An unknown quantity "α" is expressed as α =
where m = mass, a = acceleration, ℓ = length
The unit of a should be
(A) meter
(B) m/s
(C) m/s2
(D) s–1
3) Which of the following set have different dimensions?
(A) Velocity, Speed, Change in velocity
(B) Average acceleration, Acceleration, Change in acceleration
(C) Heat, Work done, Energy
(D) Force, Pressure, Density
4) The velocity of water waves may depend on their wavelength λ [L], the density of water ρ [ML–3]
and the acceleration due to gravity g [LT–2]. The method of dimensions gives the relation between
these quantities as
where k is a dimensionless constant
(A) v2 = kλ–1 g –1 ρ–1
(B) v2 = k g λ
(C) v2 = k g λ ρ
(D) v2 = k λ3 g–1 ρ–1
5) Applying the principle of homogeneity of dimensions, determine which one is correct. Where T is
time period, G is gravitational constant [M–1 L3 T–2], M is mass, r is radius of orbit.
(A)
(B)
(C)
(D)
6) If surface tension (S), Moment of inertia (I) [ML2] and Planck's constant (h) [ML2 T–1], were to be
taken as the fundamental units, the dimensional formula for linear momentum would be :-
0
(A) S3/2I1/2h
0
(B) S1/2I1/2h
(C) S1/2I1/2h–1
(D) S1/2I3/2h–1
7) If the speed v of a particle of mass m as function of time t is given by v = ωA sin .
Where A has dimension of length.
(A) Dimensional formula of ω is LT–1
(B) Dimensional formula of k is MLT–2
(C)
Dimensional formula of is T
(D) Dimentional formula of k is MT–2
8) Given where symbols have their usual meaning (F = force, t = time). The dimensions
of a and b are respectively :-
(A) [ML–1T–1] and [MLT–4]
(B) [MLT–1] and [MLT–4]
(C) [MLT–1] and [ML]
(D) [MLT–1] and [ML–1T–1]
9) Temperature can be expressed as a derived quantity in terms of which of the following.
(A) length and mass
(B) mass and time
(C) length, mass and time
(D) in terms of none of these
10) The force if given in terms of time t and displacement x by the equation
The dimensional formula of is :-
(A)
(B)
(C)
(D)
11) The density of a material in SI units is 128 kg m–3. In certain units in which the unit of length is
25 cm and the unit of mass is 50 g, the numerical value of density of the material is :
(A) 410
(B) 640
(C) 16
(D) 40
12)
Velocity of a particle depend on time t according to equation,
V= + bt + The a, b, c and d represents the following quantities in order :-
(A) Distance, distance, acceleration, time
(B) Acceleration, distance, time, distance
(C) Acceleration, distance, distance, time
(D) Distance, acceleration, distance, time
13)
Which of the following is incorrect statement?
(A) a dimensionally correct equation may be correct
(B) a dimensionally correct equation may be incorrect
(C) a dimensionally incorrect equation may be correct
(D) a dimensionally incorrect equation is incorrect
14) Density of ice is 0.9 g/ cc in the CGS system of units. The corresponding value in MKS units is
(A) 900
(B) 9
(C) 0.9
(D) 9000
15) Dimensional formula for the universal gravitational constant G is
(A) M–1L2T–2
0 0 0
(B) M L T
(C) M–1L3T–2
(D) M–1L3T–1
16) Consider three physical quantities A, B and C. Operations A + B and B–C are valid with these
physical quantities. Which of the following conclusions you can't make?
(A) The operation A ± C is also valid.
(B) If dimension of any of the three is known, dimension of other two can be predicted.
(C) If dimension of product of any two of them is known, dimension of all of them can be predicted.
(D) If dimension of quotient of any two of them is known, dimension of all of them can be predicted.
17) The volume of a gas is dependent on area and length of the container. Which of the expressions
is/are dimensionally incorrect :-
(A) V = A × ℓ
(B)
(C)
(D) V = A3ℓ–3
18) Surface tension is the property of a liquid surface. If dimension formula is S =
x y z
M L T then the value of x – z + y is :-
(A) 0
(B) 1
(C) 2
(D) 3
19) If the unit of mass becomes three times then the speed of light becomes x times, the value of x is
:
(A) 3
(B) 1
(C) 9
(D) 1/3
20) If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the
dimensional formula for energy is :
(A) [PA–1 T–2]
(B) [PA1/2T–1]
(C) [P2AT–2]
(D) [P1/2AT–1]
SECTION-II
1) If and , then angle between and is , the value of n is-
2) If the units of length and force are increased two times, by what factor is the unit of energy is
increased ?
3) Energy of particle is 5J. If unit of length and time are doubled and unit of mass is halved then
numerical value of the energy in this new system will be 5n. Value of n is.
4) If mass is expressed as vx dy az where v is velocity; d is density and a is acceleration then the value
of (x + y + z) is:
5) In a new system of units, the unit of mass is 100 g, unit of length is 4 m and unit of time is 2 s.
Find the numerical value of 2 J in this system.
PART 2 : CHEMISTRY
SECTION-I
1) Which of the following contains maximum number of atoms in 100g sample -
(A) CO2
(B) N2O
(C) NO2
(D) H2O
2) Which of the following have maximum number of atoms ?
(A) 3.2 g of O2
(B) 3.4 g of NH3
(C) 16 g of SO2
(D) 5.6 g of N2
3) How many moles of Ca are present in 10 g CaCO3 ?
(A) 1 Mol
(B) 0.1 Mol
(C) 2 Mol
(D) 10 Mol
4) If mass of electron become half, mass of proton become double, and mass of neutron become
3/4th than new mass of 8O16 will be :-
(A) 37.5 % increase
(B) Remain unchange
(C) 12.5 % increase
(D) 25 % decrease
5) Which of the following sample contains maximum number of atoms
(A) 1 mg of O2
(B) 1 mg of N2
(C) 1 mg of Na
(D) 1 mg of water
6)
A mixture of 1.65 × 1021 molecules of X and 1.85 × 1021 molecules of Y weight 0.688 g. If molecular
weight of Y is 187, what is mol.weight of X ?
(A) 46
(B) 45
(C) 42
(D) 41
7) Total number of protons present in 1 mole of H2O.
(A) 8 × NA
(B) 10 × NA
(C) 5 × NA
(D) 3 × NA
8) If the masses of Mn and O are in the ratio of 55 : 16 in MnO, what is the ratio of O that combines
with the same mass of Mn in MnO2 and Mn2O7 ?
(A) 2 : 7
(B) 4 : 7
(C) 7 : 2
(D) 7 : 4
9) If law of conservation of mass was to hold true, then 40 g NaOH on reaction with 36.5 g HCl will
produce 18 g H2O and x g NaCl. The value of x is: -
(At. mass : Na = 23, Cl = 35.5, H = 1, O = 16)
(A) 48 g
(B) 38.5 g
(C) 48.5 g
(D) 58.5 g
10) Hydrogen and oxygen combine to form H2O2 and H2O containing 5.93% and 11.2% Hydrogen
respectively. The data illustrates-
(A) Law of conservation of mass
(B) Law of constant proportions
(C) Law of gaseous volume
(D) Law of multiple proportions
11) CO, CO2, C2O3 follow
(A) Law of conservation of mass
(B) Law of multiple proportion
(C) Law of definite proportion
(D) All
12) Maximum number of atoms in 100g sample of CaCO3 is of ;
(A) Ca
(B) C
(C) O
(D) All contains equal no. of atoms
13) Which of the following contains minimum number of atoms in 1 g sample -
(A) CO2
(B) N2O
(C) NO2
(D) H2O
14) Which of the following pairs is correctly matched :
Isotopes,
(A)
(B) Isotones,
(C) Isobars,
(D) Isoelectronic,
15) Species which are isoelectronic to one another are
(a) CN⊝ (b) OH⊝ (c) (d) N2 (e) CO
Correct answer is
(A) a, b, c
(B) a, c, d
(C) a, d, e
(D) b, c, d
16) which of the following is isoneutronic
(A)
(B)
(C)
(D)
17) Which of the following is a compound :
(A) Graphite
(B) Hydrogen gas
(C) Cement
(D) marble
18) A pure substance can only be:
(A) A compound
(B) An element
(C) An element or a compound
(D) A heterogeneous mixture.
19) One part of an element A combines with two parts of another element B. Six parts of the element
C combine with four parts of the element B. If A and C combine together, the ratio of their weights
will be governed by
(A) Law of definite proportions
(B) Law of multiple proportions
(C) Law of reciprocal proportions
(D) Law of conservation of mass
20) Nitrogen forms five stable oxides with oxygen of the formula, N2O, NO, N2O3, N2O4, N2O5. The
formation of these oxides explains fully the
(A) Law of definite proportions
(B) Law of partial pressures
(C) Law of multiple proportions
(D) Law of reciprocal proportions
SECTION-II
1) The total number of protons, electrons and neutrons in 12gm of 6C12 is y × 1025. Then value of y is
(Round off to the nearest integer)
2) A certain mass of ammonium phosphate, (NH4)3 PO4 contains 18 moles of hydrogen atoms. The
number of moles of oxygen atoms in the sample is
3) A non-reacting gaseous mixture contains SO3 and SO2 in the mass ratio of 5 : 1. The ratio of the
number of moles of SO2 : SO3 is 1 : x. Find x
4) Calculate the mass of urea (NH2CONH2) containing gm-atom H.
5) If, from 10 moles NH3 and 5 moles of H2SO4, all the H-atoms are removed in order to form H2 gas,
then find the number of moles of H2 formed
PART 3 : MATHEMATICS
SECTION-I
1) If A = {1, 3, 5, 7, 11, 13, 15, 17, 19, 21}
B = {2, 4, 6, ....., 22}
U is universal set, then A' ∪ ((A ∪ B) ∩ B') is
(A) U
(B) A'
(C) A ∩ B
(D) A
2) The value of is
(A) 9
(B) 10
(C) 11
(D) 12
3) Two newspapers A and B are published in a city. It is known that 25% of the city populations
reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B
look into advertisements and 40% of those who read B but not A also look into advertisements, while
50% of those who read both A and B look into advertisements. Then the percentage of the population
who look into advertisement is :-
(A) 12.8
(B) 13.5
(C) 13.9
(D) 13
4) If A = [1, 8), B = [3, 10], C = [2, 9) then interval A ∩ B ∩ C is
(A) [3, 8)
(B) [3, 10]
(C) [2, 8)
(D) [3, 9)
5) The number 497A is divisible by 3 and the number 562B is divisible by 4. If the number of possible
ordered pairs (A, B) is λ then value of is (where A and B are single digit number)
(A) 1
(B) 2
(C) 3
(D) 4
6) Let S = {1,2}, number of elements in the set {(A,B) : A ∪ B = S}
(A) 4
(B) 6
(C) 9
(D) 10
7) If (x, y ≥ 0, x≠y), then x : y
(A) 5 : 2
(B) 2 : 5
(C) 25 : 4
(D) 4 : 25
8) Which of the following given diagram represents A ∪ B
(A)
(B)
(C)
(D)
9) A = {x : x is a multiple of 3, x < 100, x ∈ N}
B = {x : x is a multiple of 5, x < 100, x ∈ N}
then value of n(A – B) is
(A) 27
(B) 6
(C) 13
(D) 19
10) Let n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100, then n(A' ∩ B') =
(A) 400
(B) 600
(C) 300
(D) 200
11) In a study about a pandemic, data of 900 persons was collected. It was found that
190 persons had symptom of fever,
220 persons had symptom of cough,
220 persons had symptom of breathing problem,
330 persons had symptom of fever or cough or both,
350 persons had symptom of cough or breathing problem or both,
340 persons had symptom of fever or breathing problem or both,
30 persons had all three symptoms (fever, cough and breathing problem).
Number of persons having at most one symptom is _____________.
(A) 480
(B) 240
(C) 720
(D) 420
12) A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper
B. If x% of the people read both the newspapers, then a possible value of x can be :
(A) 65
(B) 37
(C) 29
(D) 55
13) If A = {x : x = 4n + 1, n ≤ 5, n ∈ N} and B {3n : n ≤ 8, n ∈ N}, then A – (A – B) is :
(A) {9, 21}
(B) {9, 12}
(C) {6, 12}
(D) {6, 21}
14) The least value of
x2 + 4y2 + 3z2 – 2x – 12y – 6z + 14, is
(A) 0
(B) 2
(C) 3
(D) 1
15) If A = {x : x2 – 5x + 6 = 0}, B = {2, 4}, C = {4, 5} then A × (B ∩ C) is-
(A) {(2, 4), (3, 4)}
(B) {(4, 2), (4, 3)}
(C) {(2, 4), (3, 4), (4, 4)}
(D) {(2, 2), (3, 3), (4, 4), (5, 5)}
16) Two finite sets have m and n elements. The number of subsets of the first set is 112 more than
that of the second set. The absolute difference of values of m and n is
(A) 2
(B) 3
(C) 4
(D) 5
17) If and , where P\Q = P ∩ QC.
If , then the value of is -
(where n(X) denotes the cardinal number of set X)
(A) 63
(B) 72
(C) 90
(D) 70
18) If x,y ∈ and xy – 3x – 2y = 54, then number of ordered pairs (x,y) is
(A) 7
(B) 8
(C) 12
(D) 24
19) is
(A) composite number
(B) even number
(C) smallest composite number
(D) smallest odd prime number
20) If a2 + b2 + c2 – ab – bc – ca ≤ 0 then is equal to (a,b,c ∈ R)
(A) 2
(B) 0
(C) 3
(D) 1
SECTION-II
1)
Let and x is prime number greater than
2} and then number of elements in is
2) If , then total number of possible ordered pair of positive integer (x, y) is
3) If a679b is a five digit number that is divisible by 72, then a + b is equal to
4) Let where p & q are coprime, then p – q is
5) In a Zoo, there are 6 Bengal white tigers and 6 Bengal royal tigers. Out of these tigers, 5 are
males and 10 are either Bengal royal tigers o males. Find the number of female Bengal white tigers
in the Zoo.
ANSWER KEYS
PART 1 : PHYSICS
SECTION-I
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. C A D B C B D B D A D D C A C D C D B B
SECTION-II
Q. 21 22 23 24 25
A. 6 4 2 4 5
PART 2 : CHEMISTRY
SECTION-I
Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. D B B A D C B B D D B C C B C A D C C C
SECTION-II
Q. 46 47 48 49 50
A. 1 6 4 5 20
PART 3 : MATHEMATICS
SECTION-I
Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. A A C A C C C B A C C D A D A B A C D D
SECTION-II
Q. 71 72 73 74 75
A. 3 6 5 7 2
SOLUTIONS
PART 1 : PHYSICS
1) Angle is dimensionless but has unit (radian or degree).
2) [α] = ..........(i)
0 0 0
=MLT
⇒ = [α] = [ℓ] = L.
3) Force = Mass × Acceleration
Pressure =
Density = .
4) [v] = [k] [λa ρb gc]
⇒ LT–1 = La Mb L–3b Lc T–2c
⇒ LT–1 = Mb La – 3b + c T–2c
⇒ a = ½, b = 0, c = ½
so, v2 = kgλ.
5) According to principle of homogeneity dimension of LHS should be equal to dimensions of
RHS so option (C) is correct.
(Dimension of G is )
6) p = k saIbhc
where k is dimensionless constant
MLT–1 = (MT–2)a(ML2)b(ML2T–1)c
a+b+c=1
2 b + 2c = 1
–2a – c = –1
a= b= c=0
1/2 1/2 0
s I h.
7) is dimensionless
[k] = MT–2
8)
9)
Conceptual
10) Dimension of
Dimension
11) n1 u1 = n2 u2
= n2
= n2
n2 = 40
12)
[b] = ⇒ acceleration
–1
[c] = LT T = L ⇒ Distance
[d] = T ⇒ Time
[a] = ⇒ Distance
13)
Conceptual
14)
15)
[G] = M–1L3T–2
16)
[A] = [B] = [C],
17)
[V] = L3
18)
0
S = M1L T–2
x = 1, y = 0, z = –2
x–z+y=3
19) Speed of light is independent of Mass.
20) Let [E] = [P]x [A]y [T]z
ML2T–2 = [MLT–1]x [L2]y [T]z
ML2T–2 = Mx Lx+2y T–x+z
→ x=1
→ x + 2y = 2
1 + 2y = 2
y=
→ –x + z = –2
–1 + z = –2
z = –1
[E] = [PA1/2 T–1]
21)
A2 + B2 + 2AB cosθ = A2
2A2 (1 + cosθ) =A2
cosθ =
θ = 120° =
22) E = Work done = F.s
23)
[E] = ML2T–2
N=5
24)
m = R vx dy az
(where k is dimensionless)
[M] = [LT–1 ] [ML–3]y [LT–2]z
Solve for x, y and z.
25)
units
PART 2 : CHEMISTRY
26)
(A) CO2 = moles =
(B) N2O = moles =
(C) NO2 = moles =
(D) H2O = moles =
H2O contains maximum number of moles.
27) 3.2 g of O2 ⇒ No. of atoms = = 0.2 NA
3.4 g of NH3 ⇒ No. of atoms =
= 0.2 × 4NA
= 0.8 NA
16 g of SO2 ⇒ No. of atoms =
= NA = 0.75 NA
5.6 g of N2 ⇒ No. of atoms = = 0.4 NA
28) Mole = = 0.1 Mol of Ca
29) % change in mass
=
16
8O = 8p + 8e + 8n
(neglect)
new mass = 8 × 2 +
30) No. of atoms ⇒
31)
32) Total proton in 1 molecule of H2O = 10
10 × NA
33) In MnO, 1 atom of Mn combines with 1 atom of O (16) . Therefore, in MnO2, 1 atom of Mn
will combine with 16 × 2 = 32 parts of O. Again 2 atoms of Mn combines with 7 atoms or 7 ×
16 = 112 parts of O in Mn2O7. Hence, 1 atom of Mn will combines with 112/2 = 56 parts of
oxygen. The ratio of O that combines with the same mass of Mn is MnO2 : Mn2O7 : : 32 : 56, i.e.,
4 : 7.
34)
According to law of conservation of mass
Mass of reactant = mass of product
40 + 36.5 = x + 18
x = 58.5 g
35) H2O2 = 5.93% H, 94.07 % O H2O = 11.2% H, 88.8% O
5.93g H reacts with = 94.07g O 11.2 g H reacts with = 88.8g O
1g H reacts with = 94.07g O 1 g H reacts with = 7.93g O
36)
CO : CO2 : C2O3 or (C1O1.5)
1 mol C = 12 g C in CO = 16 g O
=12 C in CO2 = 32 g O
=12 C in CO2 = 24 g O
Thus, ratio of O that combines with 12 g C = 16 : 32 : 24
= 2 : 4 : 3.
37) 1 mole CaCO3 = 100 g
no. of Ca atom = 1 × NA
no. of C atom = 1 × NA
no. of O atom = 3 × NA
38) (A)
(B)
(C)
(D)
NO2 contains minimum number of atoms.
39) Same number of neutrons & different number of protons.
Protons – Different
Neutron – Same (16)
40) Isoelectronic :
(a)
(b)
(c)
(d)
(e)
41) Same number of neutron
Neutron : 8, 8, 8
42) Marble = CaCO3 = compound.
(i) Graphite – Element
(ii) Cement - mixture
43) An element or a compound
44) Law of reciprocal proportions (by definition).
45) The weights of oxygen which combine with the fixed weight of nitrogen ( = 28 g) in N2O,
NO, N2O3, N2O4, N2O5 are 16, 32, 48, 64 and 80 g respectively. They are in the ratio 1 : 2 : 3 : 4
: 5. This proves the law of multiple proportions.
46) 12 gm of 6C12 = 1 mol
no. of moles of electrons, protons and neutrons in 1 mol 6C12 = 6 + 6 + 6 = 18 mol
Total number of e–, p, n = 18 × 6.022 × 1023
= 108.39 × 1023 = 1.08 × 1025
Without considering decimal places = 1 × 1025
47)
48)
49)
4 mole H → 1 mole urea
1 mole H → mole urea
mole H → mole Urea
⇒ × 60 = 5 gm
50)
Total moles of H = 10 × 3 + 5 × 2 = 40
Moles of H2 formed = 20
PART 3 : MATHEMATICS
51) (A ∪ B) ∩ B' = A
⇒ A' ∪ ((A ∪ B) ∩ B') = A' ∪ A = U
52) Rationalise Denomenators
53) Let population = 100
n(A) = 25
n(B) = 20
n(A ∩ B) = 8
5.1 + 4.8 + 4 = 13.9
54)
55) 497A is divisible by 3
∴ 20 + A is divisible by 3
⇒ A = 1, 4, 7
562B is divisible by 4,
∵ B = 0, 4, 8
Number of possible pairs (A, B) is 9
56) Combinations:
A = {1,2}
B = {1,2} (Only 1 combinations)
___________
A = {1}
B = {2} (2 combinations)
___________
A = {}
B = {1,2} (2 combinations)
___________
A = {1}
B = { 1,2} (2 combinations)
___________
A = {2}
B = {1,2} (2 combinations)
___________
({1},{2}), ({2},{1}),({1,2},{}) ({},{1,2}), ({1,2},{1,2}) ,({1},{1,2}),{{1,2},{1}}
({2},{1,2}),{{1,2},{2}}
57)
58) Do yourself
59) A = {3,6,9,.....99}
B = {5,10,15,....95}
A ∩ B = {15,30,45,60,75,90}
n(A) = 33, n(A ∩ B) = 6
n(A – B) = 27
60) n (Ac ∩ Bc) = n[({A ∪ B)c] = n(U) – n (A ∪ B)
= n(U) – [n(A) + n(B) – n (A ∩ B)]
= 700 – [200 + 300 – 100] = 300.
61)
n(U) = 900
Let A ≡ Fever, B ≡ Cough
C ≡ Breathing problem
∴ n(A) = 190, n(B) = 220, n(C) = 220
n(A ∪ B) = 330, n(B ∪ C) = 350,
n(A ∪ C) = 340, n (A ∩ B ∩ C) = 30
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ 330 = 190 + 220 – n(A ∩ B)
⇒ n(A ∩ B) = 80
Similarly,
350 = 220 + 220 – n(B ∩ C)
⇒ n(B ∩ C) = 90
and 340 = 190 + 220 – n(A ∩ C)
⇒ n(A ∩ C) = 70
∴ n(A∪B∪C) = (190 + 220 + 220) – (80 + 90 + 70) + 30
= 660 – 240 = 420
⇒ Number of person without any symptom
= n (∪) – n(A ∪ B ∪ C)
= 900 – 420 = 480
Now, number of person suffering from exactly one symptom
= (n(A) + n(B) + n(C)) – 2(n(A ∩ B) + n(B ∩ C) + n(C ∩ A)) + 3n(A ∩ B ∩ C)
= (190 + 220 + 220) – 2(80 + 90 + 70) + 3(30)
= 630 – 480 + 90 = 240
∴ Number of person suffering from atmost one symotom
= 480 + 240 = 720
62) n(B) ≤ n(A ∪ B) ≤ n(U)
⇒ 76 ≤ 76 + 63 – x ≤ 100
⇒ –63 ≤ –x ≤ –39
⇒ 63 ≥ x ≥ 39
63) A = {5, 9, 13, 17, 21} and B = {3, 6, 9, 12, 15, 18, 21, 24}
A – B = {5, 13, 17}
A – (A – B) = {9, 21}
64)
65) A = {2, 3}, B = {2, 4}, C = {4, 5}
B ∩ C = {4}
∴ A × (B ∩ C) = {(2, 4) , (3, 4)}.
66) 2m – 2n = 112
⇒ 2n(2m – n – 1) = 16 × 7
⇒ 2m – n = 8
m–n=3
67)
8n(B) = 9n(A)
k=1
68) xy – 3x – 2y = 54
⇒ xy – 3x – 2y + 6 = 54 + 6
⇒ x (y–3) – 2 (y–3) = 60
⇒ (x–2) (y–3) = 60
1 60
2 30
3 20
4 15
5 12
6 10
10 6
12 5
15 4
20 3
30 2
60 1
69)
70)
a = b = c (only)
71)
A = {1, 2, 3, 4, 5}
B = {3, 5, 7} U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C = {5, 10}
= 10 – 7 = 3
72)
4y + 5x = xy
xy – 5x – 4y = 0
(x – 4) (y – 5) = 20
Now 20 can be split in following ways
1 × 20 ⇒ (x, y) = (5, 25)
20 × 1 ⇒ (x, y) = (24, 6)
4 × 5 ⇒ (x, y) = (8, 10)
5 × 4 ⇒ (x, y) = (9, 9)
10 × 2 ⇒ (x, y) = (14, 7)
2 × 10 ⇒ (x, y) = (6, 15)
Hence, total number of possible ordered pair of positive integer (x, y) is 6.
73) Number is divisible by 9 and 8
The Number is divisible by 8 if last three digits are divisible by 8 i.e. 79b is divisible by 8 ⇒ b =
2
For divisible by 9
a + 6 + 7 + 9 + b is divisible by 9
⇒ a + 24 is divisble by 9
⇒a=3
⇒a+b=5
74) 1.23 = 1.2 +
75)
We have to find y
x+y=6 .....(1)
z+w=6 .....(2)
x+z=5 ......(3)
x + z + w = 10 .....(4)
(4) – (2) ⇒ x = 4
Now put x = 4 in (1)
⇒ 4 + y = 6 ⇒ y = 2.
