Solution: WWW - Gradeup.co
Solution: WWW - Gradeup.co
Solution: WWW - Gradeup.co
co
Solution
1-3
1.
Ans. A
Solution
The top view of the given assembly will look like the figure above
Outermost is the sphere. Inside that there is a cube and within that there is a cone and cylinder with
same radius.
Here side of cube = a
Diameter of Sphere = body diagnol = √3 a
Radius of sphere = √3 a/2 = r1
Height of Cylinder = Height of cone = side of cube = a = h
Radius of cylinder = Radius of cone = side of cube/2 = a/2 = r2 (as shown in the figure)
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𝑉𝑠𝑝ℎ𝑒𝑟𝑒 𝜋𝑟 3
3 1
Volume of sphere/volume of cone = = 1 = 6√3:1
𝑉𝑐𝑜𝑛𝑒 𝜋𝑟 2 ℎ
3 2
2.
Ans. C
Solution
The top view of the given assembly will look like the figure above
Outermost is the sphere. Inside that there is a cube and within that there is a cone and cylinder with
same radius.
Here side of cube = a
Diameter of Sphere = body diagnol = √3 a
Radius of sphere = √3 a/2 = r1
Height of Cylinder = Height of cone = side of cube = a = h
Radius of cylinder = Radius of cone = side of cube/2 = a/2 = r2 (as shown in the figure)
𝑉𝑐𝑢𝑏𝑒 𝑎3 𝑎3
= = 2 =
𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝜋𝑟2 ℎ 𝜋(𝑎2 /4)𝑎
Put π = 22/7
= 14/11
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3.
Ans. D
Solution
The top view of the given assembly will look like the figure above
Outermost is the sphere. Inside that there is a cube and within that there is a cone and cylinder with
same radius.
Here side of cube = a
Diameter of Sphere = body diagnol = √3 a
Radius of sphere = √3 a/2 = r1
Height of Cylinder = Height of cone = side of cube = a = h
Radius of cylinder = Radius of cone = side of cube/2 = a/2 = r2 (as shown in the figure)
4-6
4.
Ans. A
Solution
1/2
Area of triangle ADC = (𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐))
Where s is the semi perimeter of triangle = (AD + DC + CA) / 2 = 15+28+41 / 2 = 42 cm
1/2
Area = (42(42 − 15)(42 − 28)(42 − 41))
= (42 ∗ 27 ∗ 14 ∗ 1)1/2
= 126 cm2
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5
Ans B
Solution
6.
Ans. C
Solution
7-8
7
Ans. D
Solution
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8
Ans. C
Solution
For equilateral triangle circumcenter and centroid are the same points
So distance from vertex = radius of circumcircle = 20√3
9-10
9
Ans. A
Solution
Let lengths, breadth and height of cuboid be l, b and h respectively
According to question
l+b+h = 22cm……(i)
and √(l2+b2+h2) = 14cm …..(ii)
Surface area of cuboid = 2(lb+bh+lh)
Squaring eq (i) gives
l2+b2+h2 + 2(lb+bh+lh) = 484
Substituting l2+b2+h2 from eq (i)
2(lb+bh+lh) = 484-196 = 288 cm2
10
Ans. C
Solution
Let lengths, breadth and height of cuboid be l, b and h respectively
According to question
l+b+h = 22cm……(i)
and √(l2+b2+h2) = 14cm …..(ii)
S = l3+b3+h3 and V = lbh
S-3V = l3+b3+h3 - 3 lbh = (l+b+h)( l2+b2+h2-[lb+bh+lh])…(iii)
As we know
Squaring eq (i) gives
l2+b2+h2 + 2(lb+bh+lh) = 484
Substituting l2+b2+h2 from eq (i)
2(lb+bh+lh) = 484-196 = 288 cm2
lb+bh+lh = 144 cm2
Putting this in eq (iii) we get
22(196-144) = 22*52 = 1144cm2
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11.
Ans. B
Solution
50 80 100
9∗60+8∗60+7.5∗ 60
Average speed = Total Distance / Total time = 50 80 100
+ +
60 60 60
= (45+64+75)/23 = 184/23
= 8 kmph
12.
Ans. C
Solution
a/(b+c) = b/(c+a) = c/(a+b)
Taking reciprocal and adding 1 to each ratio we get;
(b+c)/a + 1 = b/(c+a) + 1 = c/(a+b) + 1
Or (a+b+c)/a = (a+b+c)/b = (a+b+c)/c
So this can only be equal when a=b=c or a+b+c = 0
When a=b=c we get a/(b+c) = ½
When a+b+c = 0 we get b+c = -a
So a/(b+c) = -1
So the ratios are ½ or -1
13.
Ans. B
Solution
3521/8
As we know 32=9 will leave remainder = 1 when divided by 8
So 3521/8 = [(32)260 * 3]/8 = 1*3/8 = 3/8 Thus remainder is 3
14
Ans. D
Solution
For prime no units place cannot be occupied by even number except for 2
Thus no of digits occupying unit digit of prime numbers = 6 (1,2,3,5,7,9)
Example 2,3,5,7,11,19 in itself are prime numbers
15.
Ans. D
Solution
Let CP be Rs x
Then
1.06x – 0.94x = 6
So x = Rs 50
16.
Ans. C
Solution
12 men or 18 women can complete in 14 days
8 men and 16 women can complete in how many days
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17.
Ans. C
Solution
3x = 4y = 12z
Taking log of all 3 we get
xln3 = yln4 = zln12 = k
z = k/ln12 = k / ln(3*4) = k/ln3 + ln4 = k / (k/x +k/y) = xy / (x+y)
18.
Ans. C
Solution
(4a+7b)(4c-7d) = (4a-7b)(4c+7d)
(4a+7b)/(4a-7b) = (4c+7d)/(4c-7d)
Using componendo and dividendo
(4a+7b)+(4a-7b) / (4a+7b)-(4a-7b) = (4c+7d)+(4c-7d) / (4c+7d)-(4c-7d)
Or 8a/14b = 8c/14d
Or a/b = c/d
19.
Ans. D
Solution
Since x2 + ax + b when divided by x-1 or x+1 leaves the same remainder
So on putting x=1 and x=-1 we get the same value
1+a+b = 1-a+b
2a=0
a=0
here b can take any value as it will always get cancelled out
20
Ans. D
Solution
Let them take x hours working together
1/x = 1/10 + 1/6 = 8/30
X= 30/8 hours = 15/4 hours = 3hours 45 minutes
21.
Ans D
Solution
2 + √2 + √2 + √2 + ⋯ . = t (let)
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2 + √t = t
Or t-2 = √t
Squaring both sides
t = t2 – 4t + 4
or t2 – 5t + 4 = 0
Or t = 4,1 Now t cannot be equal to 1 as it is clear that it is always greater than 2
So t = 4
22.
Ans D
Solution
23.
Ans. C
Solution
Let his wife get a share of Rs x
Each of the 4 daughters get = Rs 2x
Each of the 5 sons get = Rs 6x
So x + 4*2x + 5*6x = 390000
So 39x = 390000
X= 10000 = wife’s share
24.
Ans. B
Solution
A = P(1 + 𝑅/100)^𝑡
3P < P(1 + 40/100)^𝑡
3 < (1.4)^t
When t = 3 ; 1.4^3 = 2.744
And when t = 4; 1.4^4 = 3.8416
T=4 is the answer
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25.
Ans. B
Solution
Let sum invested @ 5% be P1, @ 6% be P2 then @ 9% = 17200-(P1+P2)
So according to question
P1*5*2/100 = P2*6*2/100 or P1 = (6/5) P2
Also P2*6*2/100 = [17200-(P1+P2)]*9*2/100
Or 2 P2 = [17200 – (11/5)P2] * 3
Or (2 + 33/5)P2 = 17200 * 3
P2 = 17200 * 3 * 5 / 43 = 6000
So P1 = 6/5 P2 = 7200
So Sum invested @ 9% = 17200 –(6000+7200) = Rs 4000
26
Ans. A
Solution
27.
Ans. A
Solution
let n-1, n, n+1 be 3 consecutive integers
So
(n+1)2 = n2 + (n-1)2
(n+1)2- (n-1)2 = n2
4n = n2
So n = 0 or n = 4
n can’t be 0 as n-1 will be negative then
So 3,4 and 5 is the only triplet formed
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28.
Ans. C
Solution
29.
Ans. C
Solution
Initially carpet is 6×12 = 72 sq feet
Since red border is 6 inches wide from all 4 side
So area without border = 5 × 11 = 55 sq feet
Area of border = total – area without border = 72 – 55 = 17 sq feet
30.
Ans. C
Solution
Let other side and hypotenuse be 4x and 5x respectively
Shortest side2 + (4x)2 = (5x)2
Shortest side = 3x
According to question
K*3x = 12x
So k = 4
31.
Ans. B
Solution
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32.
Ans. A
Solution
4k + k + k = 6x =180 degrees
k= 30 degrees
So triangle is 30,30 and 120 degrees
Let sides of triangle be x,x and y units with y being the largest side opposite to 120 degree angle
Using cosine law
Cos 120 = - sin 30 = -1/2 = (2x2 – y2)/2x2
So 3x2 = y2 …. (i)
Given Perimeter = k (Largest side)
Or 2x+y = ky
Putting value of x from eq (i)
2y/√3 + y = ky
K = 2/√3 + 1
33.
Ans. C
Solution
Hypotenuse = 10cm
Let the other 2 perpendicular sides be a and b
Area ½ a*b = 24
So a*b = 48 cm2
Also using Pythagoras
a2 + b2 = 100
(a+b)2 = a2 + b2 + 2ab = 100 + 96 = 196
a+b = 14
Similarly
a-b = 2
So
a=8 and b=6
Now smaller side is halved and larger side is doubled
So a1 = 16 and b1 = 3
New hypotenuse = √(162+32) = √265
34.
Ans. D
Solution
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35.
Ans. C
Solution
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36.
Ans. B
AB = CD = x = Length of ladder
Let OC = y m
y2 + 3.92 = x2
(y+0.8)2 + 2.52 = x2
So y2 + 3.92 = (y+0.8)2 + 2.52
y = 5.2m
x= √(5.22+3.92)
x= 6.5m
37.
Ans. C
Solution
38
Ans. D
Solution
(1) Only one circle can be drawn through 3 non collinear points
Angle in the minor segment is always obtuse
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39
Ans. D
Solution
AC-AB<BC Or AB+BC>AC
BC-AC<AB Or AB+AC>BC
AB-BC<AC Or AC+BC>AB
Sum of 2 sides of triangle is always greater than the third side
So all three statements are true
40.
Ans. C
Solution
1. Perimeter of triangle is greater than the sum of 3 medians
Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third
side,(Theorem to be remembered)
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴ AB + BC + AC > AD + BE + CF
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD,
AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the
third side] ---- 1
In triangle ADC,
AC + DC > AD [because, the sum of any two sides of a triangle is always greater than the
third side] ---- 2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved
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41.
Ans. C
Solution
Mean = (sum of fixi )/ (sum of f) = (8*5 + 12*15 + 10*25 + P*35 + 9*45) / (8+12+10+P+9) = 25.2
(875 + 35P)/(39+P) = 25.2
P = 11
42.
Ans. C
Solution
Summation of frequencies = 6+4+5+8+9+6+4 = 42
Median = mid value = average of 21st and 22nd value
Arranging data in increasing order we get
x f
4 6
5 4
6 5
7 4
8 6
9 9
10 8
So mid value i.e 21st and 22nd value = 8
43.
Ans. B
Solution
Sum of n consecutive natural numbers = n(n+1)/2
Average of n consecutive natural numbers = (n+1)/2
For first 50 average = 51/2 = x
Last 50 average = 55/2 = x+2
44.
Ans. C
Solution
All such 2 digit numbers are 11,22,33,44……. upto 99
Forms an AP
So sum = n/2(a+l)
= 9/2(11+99)
Average = sum/9 = ½(11+99) = 55
45.
Ans. D
Solution
All three are types of data representation
Pictogram uses pictures so show different identities with different numbers
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46.
Ans. D
Solution
Primary data is information that you collect specifically for the purpose of your research project.
An advantage of primary data is that it is specifically tailored to your research needs. A
disadvantage is that it is expensive to obtain.
47.
Ans. B
Solution
15 cm corresponds to 6000 rs
Education = 480/6000 * 15 cm = 1.2cm
Miscellaneous = 1660/6000 * 15cm = 4.15 cm
48.
Ans. A
Solution
Mean of m observations is n
Mean of n-m observations is m
So total = nm + (n-m)m
Total observations = n
Mean = Total / Total observations = (2mn-m2)/n = 2m – m2/n
49.
Ans. A
Solution
An ogive (oh-jive), sometimes called a cumulative frequency polygon, is a type of frequency polygon that
shows cumulative frequencies. In other words, the cumulative percents are added on the graph from
left to right. An ogive graph plots cumulative frequency on the y-axis and class boundaries along the x-
axis. Only median can be traced using frequency polygon curve. Thus it has a graphical location on the
curve. Hence the only option correctly matched is option A.
50.
Ans. D
Solution
Area of the polygon gives sum of fixi not summation of fi
51.
Ans. C
Solution.
Let the breadth of the rectangle = x
Length of the the rectangle will be = 3 times of breadth = 3x
So the initial perimeter = 2(length + breadth) = 2(x + 3x) = 8x
New breadth after increase = x + 10x/100 = 1.1x
New length after increase = 3x + 30*3x/100 = 3.9x
New perimeter = 2(1.1x + 3.9x) = 10x
Percentage change in perimeter = ( 10x-8x )*100/8x = 25%
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52.
Ans. A
Solution
Area of triangle of = ½*a*b* sinθ = A
Where a and b are sides of the triangle and θ be the angle between them
After decreasing each side
New area = ½*(a/2)*(b/2)*sinθ = ¼ A
%decrease = [(A – ¼ A)/A ]*100 = 75%
53.
Ans. A
Solution
Let the volume of spherical balloon initially = V
New volume after increase = V + 700*V/100 = 8V
Since we know that volume of sphere is directly proportional to the radius of sphere
𝑖𝑛𝑖𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑟𝑎𝑑𝑖𝑢𝑠)3
=
𝑓𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑓𝑖𝑛𝑎𝑙 𝑟𝑎𝑑𝑖𝑢𝑠)3
𝑉 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑟𝑎𝑑𝑖𝑢𝑠)3
=
8𝑉 (𝑓𝑖𝑛𝑎𝑙 𝑟𝑎𝑑𝑖𝑢𝑠)3
Final radius = 2* initial radius
Since surface area of sphere is directly proportional to the square of the radius of sphere,()
𝑖𝑛𝑖𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑟𝑎𝑑𝑖𝑢𝑠)2
=
𝑓𝑖𝑛𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 (𝑓𝑖𝑛𝑎𝑙 𝑟𝑎𝑑𝑖𝑢𝑠)2
𝑖𝑛𝑖𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 (𝑅)2
=
𝑓𝑖𝑛𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 (2𝑅)2
Final surface area = 4*initial surface area
𝐹𝑖𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 – 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑟𝑒𝑎
% change = × 100 = 300%
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑟𝑒𝑎
54.
Ans. B
Solution.
Case – 1
When both the chords are in two different halves of the circle
Distance between chords = OM + ON = √𝑟 2 − 𝑁𝐷 2 + √𝑟 2 − 𝑀𝐵2
12 2 16 2
= √102 − ( ) + √102 − ( ) = 8𝑐𝑚 + 6𝑐𝑚 = 14𝑐𝑚
2 2
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Case – 2
When both the chords are in two different halves of the circle
Distance between chords = OM + ON = √𝑟 2 − 𝑁𝐷 2 + √𝑟 2 − 𝑀𝐵2
12 2 16 2
√ 2 √ 2
= 10 − ( ) − 10 − ( ) = 8𝑐𝑚 − 6𝑐𝑚 = 2𝑐𝑚
2 2
55.
Ans. C
Solution.
56.
Ans. A
Solution.
We know that when a+b+c = 0, then
a3 + b3 + c3 = 3abc
in the above question,
(x-y) + (y-z) + (z-x) = 0
Therefore,
(x-y)3 + (y-z)3 + (z-x)3 = 3(x-y)(y-z)(z-x)
(x − y)3 + (y − z)3 + (z − x)3
=1
3(x − y)(y − z)(z − x)
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57.
Ans. C
Solution.
ax = by = cz = k
a = k1/x
b = k1/y
c = k1/z
given b2 = ac, putting the above values of a,b,c in the equation we get
k2/y = k1/x .k1/z
2/y = 1/x + 1/z
58.
Ans. B
Solution.
In the below equation,
x2 – 15x + r = 0
sum of roots = p + q = -(-15)/1 = 15 (sum of roots for equation ax2 + bx + c is -b/a)
product of roots = pq = r/1 = r (product of roots for equation ax2 + bx + c is c/a)
given p – q = 1
also we know that p+q = 15
subtracting the squares of both
(p+q)2 + (p-q)2 = 152 – 1
p2 + q2 + 2pq – p2 – q2 +2pq = 225 -1
4pq = 224
4r = 224
r = 56
59.
Ans.D
Solution.
As we can see from the graph of the quadratic equation, that the value of the equation is greater than
zero for the values of x < 3 and x > 4
60.
Ans. C
Solution.
52n – 23n = ( 52 )n – ( 23 )n = (25)n – (8)n
We know that an – bn always have a common factor (a - b)
Therefore one of the factor is 25 – 8 = 17
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61.
Ans. B
Solution.
tan x = 1
then
x = 45o
1 1
2sin x. cos x = 2 × × =1
√2 √2
62.
Ans. C
Solution.
sin 46o. cos 44o + cos 46o. sin 44o
sin 46o. sin (90 - 44)o + cos 46o. cos (90 - 44)o
= sin2 46o + cos2 46o = 1
63.
Ans. B
Solution.
We know that,
Arithmetic mean ≥ Geometric mean
(4sin2 θ + 1)/2 ≥ √4 sin2 θ . 1
4sin2 θ + 1 ≥ 2. 2 sin θ
4sin2 θ + 1 ≥ 4sin θ
64.
Ans. B
Solution
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65.
Ans. B
Solution.
tan 1o. tan 89o = tan 1o. cot 1o = 1
similarly,
tan 2o. tan 88o = tan 2o. cot 2o = 1
tan 3o. tan 87o = tan 3o. cot 3o = 1
hence the equation will reduce to
tan 45o = 1
66.
Ans. C
Solution.
67.
Ans. A
Solution.
3tan θ = cot θ
3tan θ = 1/tan θ
tan2 θ = 1/3
tan θ = 1/√3
θ = π/6
68.
Ans.B
Solution.
sin2 25o + sin2 65o = sin2 25o + sin2 (90 – 25)o = sin2 25o + cos2 25o = 1
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69.
Ans. A
Solution.
sin6 θ + cos6 θ + 3sin2 θ .cos2 θ - 1
sin6 θ + cos6 θ + 3sin2 θ .cos2 θ. 1 - 1
sin6 θ + cos6 θ + 3sin2 θ .cos2 θ. (sin2 θ + cos2 θ ) - 1
(sin2 θ + cos2 θ )3 – 1 = 1 – 1 =0
70.
Ans. C
Solution.
Sec of any number can never be less than 1
tan can take any value from -∞ to + ∞
cosec of any number can never be less than 1
cos of any number can never be greater than 1
so option 1,3,4 are not possible
71 to 73
71.
Ans. A
Solution.
The number of people who read only I , only II and only II are
1 % + 19% + 0% = 20% of total population = 20/100 * 100000 = 20000
72.
Ans. A
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As we can see from the above venn diagram the number of people who read two or more newspapers
are 1% + 1% + 3% + 7% = 12% = 12/100 * 100000 = 12000
73.
Ans. D
Solution.
Number of people who do not read any of these newspaper = total population – number of people who
read atleast one of these newspapers.
number of people who read atleast one of these newspapers = 1% + 1% + 3% + 1% + 7% + 19% = 32% of
total population = 32000
required number of people = 100000 – 32000 = 68000
74.
Ans.C
Solution.
From the above table we can see that the power 73 is of the form 4x + 1
Therefore the unit digit according to the table = 7
75.
Ans.C
Solution.
N2 + 48 =k2
48 =k2 – N2
(k - N)(k + N) = 48
So the possible number of pairs of (k - N) and (k + N) are
(1,48),(2,24), (3,16), (4,12), (6,8)
On solving the above pairs for (k - N) and (k + N), we get the integer values of N and k as
N=1 ,k= 7
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N=4 , k=8
N=11,k=13
So the total possible values of N are three
76.
Ans. D
Solution.
4√6
x=
√2+√3
on rationalizing,
4√6 √3+√2
x= × 3− 2
√2+√3 √ √
x = 12√2 − 8√3
putting the value of x in the equation
14√2−8√3 12√2−6√3 7√2−4√3 6√2−3√3
10√2−8√3
+ 12 2−10 3 = 5 2−4 3 + 6 2−5 3
√ √ √ √ √ √
2√2 2√3
+1+1+
5√2−4√3 6√2−5√3
2√2(6√2−5√3)+2√3(5√2−4√3)
2+ (5√2−4√3)(6√2−5√3)
24−10√6+10√6−24
2+ (5√2−4√3)(6√2−5√3)
=2+ 0=2
77.
Ans. D
Solution.
x = 30% of z = 30z/100 = 3z/10
y = 40% of z = 40z/100 = 4z/10
According to the question,
(x/y)*100 = p%
3𝑧/100
p% = × 100 = 75%
4𝑧/100
78.
Ans.C
Solution.
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79.
Ans.D
Solution.
All the given statements are true. The following are the examples for all the statements
Statement 1: Both p and q may be prime numbers. E.g. 3 and 5
Statement 2 : Both p and q may be composite numbers. E.g. 4 and 9
Statement 3 : One of p and q may be prime and the other composite. E.g. 7 and 12
80.
Ans. A
Solution.
By alligation,
girls boys
24 32
30
2 : 6
1 : 3
So the number of girls will be =( 1/(1+3))*100 = 25
81.
Ans. C
Solution.
For the equation,
√(𝑎 − 𝑏)2 + √(𝑏 − 𝑎)2
Where a and b are real numbers,
The roots of number is always positive and hence it can be zero only at a=b
So the above equation is positive only when a=b
82.
Ans. C
Solution.
Let a = x then b = 6x
Also let c = y then d = 6y
𝑎 2 +𝑐 2 𝑥 2 +𝑦 2 1
𝑏2 +𝑑 2
= =
(6𝑥)2 +(6𝑦)2 36
83.
Ans. A
Solution.
. ̅53
̅̅̅ +0.53̅
= 0.5353535353….+0.5333333333…..
̅̅̅̅
= 1.068686868 = 1. 068
84.
Ans. D
Solution.
3N > N3 holds for all the natural numbers except N = 3 at which 3N = N3
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85.
Ans. D
Solution.
A number that cannot be represented in the form p/q where p and q are two integers, is known as
irrational number √59049 = 243 . Hence it is rational
231
593
is already in the form of rational number
0.4545454545……. can be represented in the form of p/q as 5/9
0.12112211122211112222……… cannot be represented in the form of p/q as there is no recurring digits
in the given number
86.
Ans. D
Solution.
The number 1729 = (18 - 1)29 when divided by 18 leaves the remainder (-1)29 = 18-1 = 17
The number 1929 = (18 + 1)29 when divided by 18 leaves the remainder (1)29 = 1
Then after adding these two the remainder will be 17+1 = 18 which is divisible by 18
Hence the remainder will be 0
87.
Ans.A
Solution.
For the number to be divisible by 10n , it must contain the same powers for 2 and 5
Power of 2 = 25+2.8+7+3.12+6+2.14 + 11 = 25+16+7+36+6+28+11 = 2109
Power of 5 = 53+6+12+14+2.15 = 565
Hence maximum possible power of 10 can be 65 only.
88.
Ans. A
Solution.
If the number is divisible by 9 the sum of all its digit is divisible by 9
4+7+9+8+6+5+A+B = 39 + A + B is divisible by 9
Possible values of B are 1,3,5,7,9 as it is given that last digit is odd
For B= 1, A=5
For B = 3 A= 3
For B= 5 , A = 1
For B = 7, A = 8
For B = 9, A= 6
89.
Ans. D
Solution.
999 x abc = def132
We can write the above equation as
(1000 – 1) x abc = def132
abc000 – abc = def000 + 132 = (def +1)x 1000 - 868
on comparing the LHS and RHS, we get
a = 8, b = 6, and c = 8 and d = a = 8 , e = b = 6 and f = c – 1 = 8 -1 = 7
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90.
Ans. A
Solution.
Distance covered by A till 6pm = 60 km
Distance covered by A till 7 pm = 120 km
Time taken by B to catch A = 60/(80-60 ) = 3 hrs
So A and B will meet at 6pm + 3 hrs = 9pm
Since we know that all three met at the same time
The time taken by C to cover 120 km difference will be = 9pm – 7pm = 2hrs
Therefore, (x – 60 )*2 = 120
x = 120 km/hr
91.
Ans. C
Solution
Let present age of Priya be p
p-4 = n3
p+4 = √k
since n is a no >1 on putting n= 2 we get p = 12
So p+4 = 16 which is square of an integral number thus consistent with given information
Now after how many years her age becomes such that age – 1 is a square and age + 1 is a cube
Using option if we add 14 years to current age , we get age = 26 years
Here 25 is a square and 27 is a cube thus making 14 the correct answer
92.
Ans. D
Solution
Option C is incorrect as 6n – 1 form can be a prime number but it is not necessarily true.
Example 35 is of form 6n-1 but is not a prime number
93.
Ans. C
Solution
For x>0 Min of x + (x+2)/2x = ?
x + (x+2)/2x = x + ½ + 1/x
So we have to find the minimum of x+1/x and add ½ to it
As AM>GM
So (x+1/x)/2 > √(x*1/x)
Or x + 1/x > 2
So min of x + (x+2)/2x = 2+1/2 = 5/2
94.
Ans. A
Solution.
1 + 𝑝𝑥 1 − 𝑞𝑥
√ =1
1 − 𝑝𝑥 1 + 𝑞𝑥
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95.
Ans. C
Solution
Let initial rent be rs 10
And initial rooms be 10
So initial collection = 10*10 = Rs 100
Now new rent = 10 + 20% of 10 = 12
New no of rooms = 10 + 20% of 10 = 12
So new collection = 12*12 = 144
% change in collection = (144-100)/100*100 = 44%
96.
Ans. C
Solution
Let the distance between be D km
Time taken by radha – Time taken by Hema = 9 mins
So D/8 – D/10 = 9/60 hrs
D = 6km
97.
Ans. B
Solution
3x+2 + 3-x = 10
Only powers of 3 that add upto 10 is
32 + 30 = 10
X+2 = 0
X= -2 solution is consistent
Or x+2 = 2
X= 0 solution is consistent
Thus x = 0, -2 are the solutions
Alternatively, we can put values from the options and check.
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98.
Ans. C
Solution
No of digits in (108)10
We have to find the log of the given number with base 10 and add one to its integral part to find the no
of digits
log (108)10 = 10 log 108 = 10 log(22 * 33) = 10[2log2 + 3log3]
= 10[2*0.301 + 3*0.477] = 20.33
Integral part = 20
No of digits = 20+1 = 21
99.
Ans. D
Solution
Let the three prime numbers be x, y, y+36
x+y+y+36 = 100
x+2y = 64
2y is an even number always
We know that
Even + even = even or odd + odd = even
So x has to be even to satisfy x+2y = 64
The only even prime no is 2
Put x=2
2y = 62
Or y = 31
So the numbers are 2, 31 , 67
Thus option D is the answer
100.
Ans. B
Solution
16 1 1 1 1 1 1
= = = = = =
23 23 7 1
1 + 16 1 1 + 2
16 1 + ( 16 ) 2 + (7)
7 1 1
1+ 1+
1 1
2 + (7) 2+( 1)
( 2 ) ( 3+
2 )
On comparing equations we get a= 1 , b = 2 and c = 3
Mean = a+b+c/3 = 6/3 = 2
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