i request you to edit . it may help many studentsA train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is
(A) 0 (B) 1 (C) 2 (D) 3 (E) None of the above
The train leaves X at A hours, B minutes; i.e. at (60A + B) minutes.
It reaches Y at B hours, C minutes; i.e. at (60B + C) minutes.
The time interval between the two is given as C hours, A minutes; i.e. (60C + A) minutes.
Hence, we have,
60C + A = 60B + C − (60A + B)
∴ 61A = 59(B − C)
∴ A = 59(B − C)/61
Hence, the only possible value of A is 0, when B = C. Thus, there is one possible value of
A.
Hence, option 2.
Two circles of radius 1 cm touch at point P. A third circle is drawn through the points A, B and C such that PA is the diameter of the first circle, and BC - perpendicular to AP - is the diameter of the second circle. The radius of the third circle is
A. 9/5
B. 7/4
C. 5/3
D. root 10/2
(E) 2
As third circle is passing through the points A, B and C, the center (say G) of the third
circle must lie on the segment AD.
Let AG = BG = CG = x cm
∴ AG^2 = BG^2
∴ x^2 = BD^2 + GD^2
∴ x^2 = 12 + (3 – x)^2
Solving this, we get, x=5/3
Hence, option 3
Or
Let the center of third circle be O and radius be r.
OA = OB
= OC = r
Mid point of BC is D
DC = 1 cm, OD = 3-r
OC^2 = OD^2 + DC^2
r*r = 1 + (3-r)^2
r = 5/3 cm
In a triangle ABC, AB = 3, BC = 4 and CA = 5. Point D is the midpoint of AB, point E is on segment AC and point F is on segment BC. If AE = 1.5 and BF = 0.5, then ∠DEF =
(A) 30° (B) 45° (C) 60° (D) 75° (E) Cannot be determined
Let ∠DEF = y, ∠AED
= x, and ∠FEC = z
Let ∠DEF = y, ∠AED
= x, and ∠FEC = z
∠AED
+ ∠DEF + ∠FEC = x + y + z = 180° …(i)
In ΔAED, AE = AD = 1.5 and in ΔCEF, CE = CF = 3.5
So ∠AED
= ∠ADE = x and ∠CEF = ∠CFE = z
In quadrilateral BDEF, ∠DEF + ∠EFB + ∠FBD + ∠BDE
= 360°
y + (180° − z) + 90° + (180° − x) = 360°
450° + y − (x + z) = 360°
450° + y − (180° − y) = 360° …From (i)
2y = 90°
y = 45°
Hence, option 2.
or
angle BCA=x (say)
angle BAC
=(90-X)
NOW tRIANGLE cef AND ade ARE
ISOS
use this to calculate the angles,den use angles on a st line aka property and ul get angle DEF=180-(90+x)/2-(180-x)/2=45
Consider a sequence −6, −12, 48, 24, −30, −36, 42 … If sum of the first n terms of the sequence is 132, then the value of n is?
(A) 11 (B) 13 (C) 18 (D) 22 (E) 24
sum of terms taken 4 at a time = 24
Total sum = 132
=> n*24/4 = 132
=> n = 22
lternative method:
6*2/1 =12; 12*3/2 = 18; 18* 4/3 = 24; 24* 5/4= 30; 30*6/5 = 36;& so on........Also, the signs follow a pattern of −, −, +, +, −, −, +, + … and so on.
The number of possible real solution(s) of y in equation y^2 − 2ycos x + 1 = 0 is (A) 0 (B) 1 (C) 2 (D) 3 (E) None of the above
4 cos^2 x -4>=0
=>Cos
^2 x>=1
x=0,pi,2pi,...
at dis value equation is either (y-1)^2=0 or (y+1)^2=0
=>2 values
A.M (x,1/x) >= GM (x,1/x)
since x is a positive real number
(x+1/x) / 2 >= Sqrtof(x*1/x)
i.e x+1/x >= 2
therefore min. value of x+1/x is >= 2
Let X = {a, b, c} and Y = {l, m}.
Consider the following four subsets of X × Y.
F1 = {(a, l), (a, m), (b, l), (c, m)}
F2 = {(a, l), (b, l), (c, l)}
F3 = {(a, l), (b, m), (c, m)}
F4 = {(a, l), (b, m)}
Which one, amongst the choices given below, is a representation of functions from X to Y?
(A) F1, F2 and F3
(B) F2, F3 and F4
(C) F2 and F3
(D) F3 and F4
(E) None of the above
We are supposed to find the representation of functions from X to Y,
∴ X will be considered as the domain and Y will be considered as the range.
We will consider functions satisfying only many to one and one to one relationships.
In F1 since a is paired with l and m ∴ It satisfies one to many relationship and hence is
not a representation of function from X to Y.
Elements in F2 only satisfy many to one relationship and hence F2 is valid
Elements in F3 satisfy one to one and many to one relationship and hence F3 is valid.
Elements in F4 only satisfy one to one relationship. However, F4 does not contain 'c',
and a function is defined as a rule which attributes to every number present in the
domain, exactly one Real number. Hence, F4 is not a representation of function from X to
Y.
∴ Only F2 and F3 are representations of functions from X to Y.
Hence, option 3.
Question No. 53-56.
During a four-week period, each one of seven previously unadvertised products – G, H, J, K, L, M and O – will be advertised. A different pair of these products will be advertised each week. Exactly one of the products will be a member of two of these four pairs. None of the other products gets repeated in any pair. Further, the following constraints must be observed:
J is not advertised during a given week unless H is advertised during the immediately preceding week.
The product that is advertised twice is advertised during week 4 but is not advertised during week 3.
G is not advertised during a given week unless either J or O is also advertised that week.
K is advertised during one of the first two weeks.
O is one of the products advertised during week 3.
53. Which one of the following could be the schedule of the advertisements?
(A) Week 1: G, J; week 2: K, L; week 3: O, M; week 4: H, L
(B) Week 1: H, K; week 2: J, G; week 3: O, L; week 4: M, K
(C) Week 1: H, K; week 2: J, M; week 3: O, L; week 4: G, M
(D) Week 1: H, L; week 2: J, M; week 3: O, G; week 4: K, L
(E) Week 1: K, M; week 2: H, J; week 3: O, G; week 4: L, M
54. If L is the product that is advertised during two of the weeks, which one of the following is a product that MUST be advertised during one of the weeks in which L is advertised?
(A) G (B) H (C) J (D) K (E) M
55. Which one of the following is a product that could be advertised in any of the four weeks?
(A) H (B) J (C) K (D) L (E) O
56. Which one of the following is a pair of products that could be advertised during the same week?
(A) G and H (B) H and J (C) H and O (D) K and O (E) M and O
54..∵ L is the product that is advertised during two of the weeks
∴ L has to be advertised once in the 4th week and once in the first two weeks.
∴ There are two cases possible:
Case 1: L is advertised during the first and the fourth week
In this case the only schedule possible is:
Week 1: H, L
Week 2: J, K
Week 3: G, O
Week 4: L, M
Case 2: L is advertised during the second and the fourth week
In this case the only schedule possible is:
Week 1: H, K
Week 2: J, L
Week 3: G, O
Week 4: M, L
In both the cases M is definitely advertised in the same week as L.
Hence, option 5.
56. Consider option 1:
G and H cannot be advertised in the same week, since G is not advertised during a
given week unless either J or O is also advertised that week.
∴ This option is not valid.
Consider option 2:
H and J cannot be advertised together as J is not advertised during a given week unless H is advertised during the immediately preceding week and if H is advertised twice then it cannot be done in consecutive weeks.
∴ This option is also not valid.
Consider option 3:
If H and O are advertised together in the third week then G and J have to be advertised in the fourth week and then one of G or J is advertised twice, which is not possible.
∴ This option is also not valid.
Consider option 4:
K cannot be advertised in the third week with O as K has to be advertised in the first two weeks and it cannot be advertised twice since the repeated product is not advertised in the third week.
∴ This option is not valid.
Consider option 5:
M and O can be advertised together satisfying all the given conditions.
Week 1: H, K
Week 2: G, J
Week 3: O, M
Week 4: L, H or L, K
Hence, option 5.
Good Explanation. ThankYou!!
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