Sample Practice Question Set (Problem Solving)

1) Find the last digit of the expression (4526)⁴ – (4525)³ ?

3

2

1

4

7

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) (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.)

2) A train starts a journey from a station with all seats occupied. At the next station, 1/4 of the passengers get down while 150 new passengers are added. At third station, 1/3 are of the remaining passengers are dropped and new 48 passengers are added. At fourth and final station, 528 passengers get down and train becomes empty. What is the total number of seats in the train?

720

570

670

560

760

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: (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.)

3) If m & n are odd integers, which of the following can not be an integer?

(3m + 7n) / 3

Mn / 5

(m + 2) / (n + 1)

(2m + 6n) / 4

(m + 3) / (n + 7)

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) (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.)

4) If in x liters of mixture of A & B, the ratio of A and B is m : n. What quantity of A should be added to the mixture so that the ratio of A and B becomes p : q?

(pm – qn)/(m + n)

X(pn – qm)/q(m + n)

Q(m + n)/(pn – qm)

Q(m + n)/(pm – qn)

(pn – qm)/(m + n)

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: (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)])

5) If a triangle is made of wire with 6 meter and 11 meter as its two sides. If the wire is rearranged into a circle, what will be the radius of the largest circle which can be formed from the triangle?

119/44 meter

189/44 meter

29/4 meter

31/4 meter

21/4 meter

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 ( 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: (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 ( 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)

6) A car requires 4/5 hour of service time every month. But after overhauling, the service time becomes 4/5 hour every one and half month. What fraction of service time is saved after overhauling?

1/4

1/3

1/2

2/5

3/5

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: (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 )

7)

In the figure, the lined portion is 1/3^rd of triangle ABC and is 45% of triangle PQR. Near to which of the following percent is the lined portion of the whole figure?

15%

38%

24%

30%

18%

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: (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%.)

8) The areas of three adjacent faces of a cuboid are A, B &C. What is the volume of the cuboid?

(ABC)^3

Cube root(ABC)

(ABC)^2

ABC

Sqrt (ABC)

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) (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).)

9) Incomes of A and B are in the ratio x : y and their expenditure are in the ratio of w : z. If both save equal amount of $M each, what will be the income of A?

M (xz – wy)/(z – w))

M xy/(w + z)

M xy/(xz – wy)

M (x + y)/(xz – wy)

M x (z – w)/(xz – wy)

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))

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))

10) An auditorium has 3 entrances. 3 men have to get into the auditorium. What is the probability that all 3 men will enter from different entrances?

1/2

1/9

1/3

2/3

2/9

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) (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.)