# Sample Practice Question Set (Problem Solving)

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Quizzes Created: 6 | Total Attempts: 5,380
Questions: 10 | Attempts: 303  Settings  • 1.

### 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?

• A.

119/44 meter

• B.

189/44 meter

• C.

29/4 meter

• D.

31/4 meter

• E.

21/4 meter

E. 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 (< (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)

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• 2.

### 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?

• A.

1/4

• B.

1/3

• C.

1/2

• D.

2/5

• E.

3/5

B. 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 )

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• 3.

### 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?

• A.

720

• B.

570

• C.

670

• D.

560

• E.

760

E. 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.)

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• 4.

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

• A.

3

• B.

2

• C.

1

• D.

4

• E.

7

C. 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.)

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• 5.

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

• A.

(3m + 7n) / 3

• B.

Mn / 5

• C.

(m + 2) / (n + 1)

• D.

(2m + 6n) / 4

• E.

(m + 3) / (n + 7)

C. (m + 2) / (n + 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.)

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• 6.

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

• A.

15%

• B.

38%

• C.

24%

• D.

30%

• E.

18%

C. 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%.)

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• 7.

### 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?

• A.

(pm – qn)/(m + n)

• B.

X(pn – qm)/q(m + n)

• C.

Q(m + n)/(pn – qm)

• D.

Q(m + n)/(pm – qn)

• E.

(pn – qm)/(m + n)

B. 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)])

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• 8.

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

• A.

(ABC)^3

• B.

Cube root(ABC)

• C.

(ABC)^2

• D.

ABC

• E.

Sqrt (ABC)

E. 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).)

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• 9.

### 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?

• A.

1/2

• B.

1/9

• C.

1/3

• D.

2/3

• E.

2/9

E. 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.)

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• 10.

### 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?

• A.

M (xz – wy)/(z – w))

• B.

M xy/(w + z)

• C.

M xy/(xz – wy)

• D.

M (x + y)/(xz – wy)

• E.

M x (z – w)/(xz – wy)

E. 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))

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