Sample Practice Question Set (Problem Solving)

Explanation
ANS: (E) (Since largest circle is to be found , then the triangle should also be the largest possible. 2 sides are given. We know that any side of triangle is always less than the sum of other two sides. So 3rd side will be 16 meter (< (6 + 11)).
Now the circle is formed from the triangle only, so the perimeter of triangle and circumference of circle would be same.
Perimeter of triangle = 6 + 11 + 16 = 33 meter.
Circumference of circle = 2 Pi r = 33
Or 2 x (22/7) r = 33 or r = 21/4 meter)

Explanation
ANS: (B) (Before overhauling, the service time required per one and half month = (4/5) x 1.5 = 6/5 hour
Reduction in service time after overhauling = 6/5 – 4/5 = 2/5 hour
Fraction of service time saved = (2/5) / (6/5) = 1/3 )

Explanation
ANS: (E) (Let A be the total no. of seats which were occupied at first station.
At 2nd station, no. of passengers in the train = A – (1/4) A + 150 = (3/4)A +150
At 3rd station, no. of passengers in the train = [(3/4) A + 150] – (1/3)[ [(3/4)A + 150] + 48 Or = (2/3){ [(3/4) A + 150] + 48
Or = (1/2) A + 100 + 48 = (1/2)A + 148
Now (1/2) A + 148 = 528, Or A = 760.)

Explanation
ANS: (C) (When the last digit of a number is 6 or 5, then when the number is multiplied by itself any number of time, then the last digit remains same (6x6 = 36, 6x6x6 = 216, 5x5 =25, 5x5x5 =125). And difference of 6 & 5 is 1.)

Explanation
ANS: (C) (3m + 7n is even and even number can be divisible by 3, so (A) may be integer. mn is odd and it can be divisible by 5, so (B) may be integer. m + 2 is odd while n + 1 is even and an odd number can never be divisible by an even, so (C) is not an integer. 2m + 6n is even and can be divided by 4, so (D) may be integer. m + 3 is even and n + 7 is also even, an even number may be divisible by an even number, so (E) may be an integer.)

Explanation
Ans: (c) (Let the lined portion be a. So unlined portion of ABC is 2a.
The lined and unlined portion of PQR are in the ratio 45 : 55 or 9 : 11.
So unlined portion of PQR is (11/9)a
Now whole portion = 2a + a + (11/9)a = (38/9)a
So (lined portion/whole figure) = a / (38/9)a = 9/38
Or [(9 x 100) / 38 ]% = 23.68 %, approximately 24%.)

Explanation
Ans: (b) (Ratio of A and B = m : n
Quantity of A = mx/(m + n), where x is the quantity of mixture.
Quantity of B = nx/(m + n)
Let y liter of A is added so that the new ratio of A and B becomes p : q.
New quantity of A = [mx/(m + n)] + y
So [mx/(m + n)] + y : nx/(m + n) = p : q
Or [mx/(m + n)] + y = (p/q)[ nx/(m + n)]
Or y = (p/q)[ nx/(m + n)] - [mx/(m + n)]
Or y = [x(pn – qm)/q(m + n)])

Explanation
ANS: (e) (Volume of a cuboid = Length (L)x Breadth(B)x Height (H)
Areas of adjacent faces are LxB, LxH and BxH.
So AxBxC = (LxBxC)^2
Or ABC = (Volume)^2
So Volume = Sqrt(ABC).)

Explanation
ANS: (e) (Total no. of ways in which 3 men can enter 3 gates = 3x3x3 = 27. (1st can enter in any of the 3 gates, 2nd can do same & 3rd can also do same.)
No. of possible combinations in which 3 men can enter 3 different gates = 3P3
(ABC, ACB, BCA, BAC, CAB, CBA)
Or = 3!/1! = 6
So the probability that all 3 men will enter from different entrances = 6/27 = 2/9.)

Explanation
ANS: (e) (Let total income of both is $I.
So A’s income = I x /(x + y) and B’s income = I y/(x + y)
Also expenditure of A = income – saving
= [I x /(x + y)] – M
And expenditure of B = [I y /(x + y)] – M
The ratio of expenditures is w : z
So {[I x /(x + y)] – M} : {[I y /(x + y)] – M} = w : z
Or [I x – M (x + y)] : [I y – M (x + y)] = w : z
Or z [I x – M (x + y)] = w [I y – M (x + y)]
Or I (xz – wy) = M (x + y) (z – w)
Or I = M (x + y) (z – w) / (xz – wy)
So A’s income = I x /(x + y)
Or = M x (z – w)/(xz – wy))