Mathematical

reasoning
1. The following pie-charts show the distribution of students of graduate and post-graduate levels in seven
different institutes in a town.
Distribution of students at graduate and post-graduate levels in seven institutes:

How many students of institutes of M and S are studying at graduate level?

A
7516

B 8463

C 9127

D
9404

Ans: 8463

Students of institute M at graduate level= 17% of 27300 = 4641.

Students of institute S at graduate level = 14% of 27300 = 3822.

 Total number of students at graduate in institutes M = (4641 + 3822)

and S
 = 8463.

2. An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right
circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylinderical part is 240 cm
high and the conical part in 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8
grams. Round off your answer.

A 395366 g

B 456783 g

C 338925 g

D 378954 g

Ans: 395366 g

Height of cylinder = 240 cm

Radius of cylinder = 8 cm]

Volume of cylinder = 22/7x64x240

Volume of cylinder = 48274.28 cm^3

Height of cone =36 cm

Radius of cone = 8 cm

Volume of cone = 1/3x22/7x64x36

Volume of cone = 2413.71 cm^3

Total volume of pillar = Volume of cone + Volume of cylinder

Total volume of pillar = 2413.7142 + 48274.2857 = 50687.99

1 cubic cm = 7.8 grams

So, 50687.9999 cubic cm = 395366.322 g= 395366 g

3. A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to the
probability of getting 9 heads, then the probability of getting 2 heads is:

A 15 / 212

B 15 / 213
C 15 / 214

D 15 / 28

Ans: 15 / 213

P (x = 9) = P (x = 7)
⇒ nC9 (1 / 2)n-9 (1 / 2)9 = nC7 (1 / 2)n-7 (1 / 2)7
⇒ nC9 (1 / 2)n = (1 / 2)n x nC7
⇒ n = 9 + 7 = 16
P (x = 2) = 16C2 (1 / 2)14 (1 / 2)2
= 16C2 (1 / 2)16
= 15 / 213

4. What number replaces the ‘?’?

A 12

B 9

C 16

D 7

Ans: 16

Starting with the top left number, and working down one row at a time, alternating between left and right,
double the number each time.

Repeat this sequence, starting with the top right number.

5. A player throws a ball at a basket kept at a distance. The probability that the ball falls into the basket in
a single attempt is 0.3. The player attempts to throw the ball twice. Considering each attempt to be
independent, the probability that this player puts the ball into the basket only in the second attempt is?

A 0.1

B 0.7

C 0.07

D 0.001

Ans: 0.09
P(E)=0.1

P(e)=1-P(E)=1-0.3=0.7

Probability=P9does not fall)xP(falls)

=0.1x0.7=0.07

6. The resistance of a wire varies directly as its length and inversely as the area of cross section. The
resistance is 1 ohm, when the length is 50 mm and the cross sectional area is 0.25 mm 2. Find the
resistance, when cross-sectional are is 0.5mm 2 and the length of the wire is 200mm

A 2 ohm

B 1.2 ohm

C 0.2 ohm

D 0.12 ohm

Ans: 2 ohm

R=kL/A

K=1/200

R=L/200R=200/200x0.5=2 ohm

7. In a class of 60 students, the number of boys and girls participating in the annual sports is in the ratio
3:2 respectively. The number of girls not participating in the sports is 5 more than the number of boys not
participating in the sports. If the number of boys participating is 15, then how many girls are there in the
class?

A 20

B 30

C 25

D Inadequate data

Ans: 30

No of boys=x

Number of girls not participating is =x+5

Total students not in sports=60-15-10=35

So, x=15

Total number of girls=20+10=30

8. The total expenses of a boarding house are partly fixed ans partly varying with the number of boarders.
The avg expense per boarder is $700, when there are 25 boarders and $600, when there ae 50 boarders.
What is the avg expense per boarder when there are 100?

A $540

B $550

C $580

D $570

Ans:$550

700x25=a+25b

30000=a+50b

17500=a+25b

Solving

A=5000

B=5000

When 100 boarders

E=5000+500(100)=55000

Average expense=$550

9. The captain of a football team of 11 players is 25 yrs old and gk is 3 yrs older to him. If the ages of
these two players are excluded, the avg age of the remaining players is 1 yr less than avg age of whole
team. What is the avg age of the whole team?

A 22.5 yrs

B 23.5 yrs

C 22 yrs
D 25 yrs

Ans: 22 yrs

Let avg age be x

Sum of all ages is 11x

Avg of 9 players=11x-25-28/9

x-1=11x-53/9

2x=44

X=22

10. A sum of $5000 was lent partly at 6% and patly at 9% simple interest. If the interest received after one
year is $390, Find the ratio in which money was lent

A 2:3

B 3:4

C 1:2

D 3:2

Ans:2:3

(6xXx1/100)+(5000-x)x9x1/100=390

6x+45000-9x=39000

X=2000

Ans 5000-x=3000

Ratio 2000:3000=2;:3

11. A cistern is provided by two taps A and B. A can fill it in 20 mins and B can in25 mins, both the taps
are kept open for 5 mins and then the tap B is turned off. The cistern will be completely filled in another

A 11 mins

B 10 mins

C 15 mins
D 12 mins

Ans: 11 mins

A takes 20 mins

B takes 25 mins

In 1 min it fills 1/25 together in 5 min=(1/20+1/25)x5=9/20

Work left to be done=1-9/20=11/20

In 20 mins 11/20=11 mins

12. A man travelled 5 miles in the second hour of his trip. This was 1/4 th more than he travelled in the first
hour. In the third hour, he travelled 1/5th more than he did in the second. How far does he travel in the 3
hours?

A 6 miles

B 20 miles

C 13 4/5 miles

D 15 miles

Ans:D

Distance in 2nd hour=5 miles

Let distance in 1st hour be n miles

N+4/n=5

N=4miles

Third hour=5+5/5 miles=6miles

Total=5+4+6= 15 miles

13. Points A and B are 70 km apart on a highway , a car starts from point A and another car starts from
point B simultaneously. If they travel in the same direction, they meet in 7 hrs, but if they travel towards
each other, they meet in 1 hour.Find the speed of 2 cars ?

A A=30, B=40

B A=40, B=30
C A=20, B=50

D A=10, B=60

Ans: A=40, B=30

let the speed of A be x and B be y.

if they travel towards each other →← they meet in 1 hour.

when they are moving in same direction ,the speed is x-y.

when they are moving towards each other,the speed is x+y.

Distance = 70km

7(x-y)=70

x-y=10---›(i)

similarly

1(x+y)=70

x+y=70---›(ii)

Solving (i) & (ii)

x+y=70
x -y=10
_____
2x=80.

x=80/2=40 Km/ hr.

substitute x in x+y=70

so,40+y=70;y=30 km/hr.

The speed of Car A is 40km/hr.

The speed of Car B is 30 km/hr.

14. In the circle given below, let OA = 1 unit, OB = 13 unit and PQ perpendicular to OB. Then, the area
of the triangle PQB (in square units) is :

A 26 √3

B 24 √2

C 24 √3

D 26 √2

Ans:  24 √3

OC = 13 / 2 = 6.5
AC = CO - AO
= 6.5 - 1
= 5.5
In △PAC
PA = √(6.52 - 5.52)
PA = √12
PQ = 2PA = 2 √12
Now, area of △PQB = (1 / 2) * PQ * AB
= (1 / 2) * 2 √12 * 12
= 12 √12
= 24 √3

15. The number of integral values of ‘k’ for which the equation 3 sin x + 4 cos x = k + 1 has a solution, k
∈ R is ______.

A 12

B 34

C 15

D 11

Ans: 11

3 sin x + 4 cos x = k + 1
⇒ −5 ≤ k + 1 ≤ 5
⇒ −6 ≤ k ≤ 4
−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4 → 11 integral values

16. Let (λ, 2, 1) be a point on the plane which passes through the point (4, – 2, 2). If the plane is
perpendicular to the line joining the points (– 2, – 21, 29) and (– 1, – 16, 23), then (λ / 11) 2 - (4λ / 11) - 4
is equal to ______.

A 10

B 8

C 15

D 13

Ans: 8
AB is perpendicular to PQ.
[(4 - λ) i - 4j + k] . [i + 5j - 6k] = 0
4 - λ - 20 - 6 = 0
λ = - 22
Now, (λ / 11) = - 2
(λ / 11)2 - (4λ / 11) - 4 = 4 + 8 - 4 = 8

17. Study the following table and answer the questions.

Classification of 100 Students Based on the Marks Obtained by them in Physics and Chemistry in an
Examination.

The percentage of number of students getting at least 60% marks in Chemistry over those getting at least
40% marks in aggregate, is approximately?

A 21%

B 27%
C 29%
D 31%

Ans:C

Number of students getting at least 60% marks in Chemistry

= Number of students getting 30 and above marks in Chemistry

= 21.

Number of students getting at least 40% marks in aggregate

= Number of students getting 20 and above marks in aggregate

= 73.

Required percentage=(21x 100)/73%

= 28.77%

~ 29%.

18. What time is the last appointment?

A 1

B 0

C 8

D 5

Ans: 0

You’ll quickly notice a pattern if you add up the digits in each time; ignore the decimal points. The digits
in the first “time” add up to 11; the second, 12; the third, 13; and, well, you’ve now got a seriously simple
pattern. So you know that the digits in the fifth time must add up to 15. And when you add them…they
already do! So the question mark must be replaced with a 0, to keep the sum at 15.

19  If the first day of January is Tuesday what will be the fourteenth day of this month?
A Saturday

B Tuesday

C Friday

D Monday

Ans: Monday

After adding 14 days with Tuesday excule Tuesday itself and the 14 th day will be Monday

20. In an examination, a candidate gets 2 marks for the right answer & loose 1 marks for the wrong
answer. He got 80 marks by answering 100 questions. How many of his answer were right ?
A 30

B 35

C 40

D 60

Ans: 60
Use trial and error with the given options

21. The area bounded by the lines y = |x - 1| - 2 and X axis is ______.

A 4

B 6

C 8

D 9

Ans: 4

Required area area of △PQR

Area = (1 / 2) * 4 * 2 = 4
If the second curve is x - axis, then answer will be 4
22. Select the correct net.

Ans: D
Just unfold the figure

23. Which number replaces the ‘?’?

A 13
B 2
C 11
D 1
Ans: 1
Taking any series of 3 numbers in a straight line in the diagram, their total is always 19.
24.

What is the length of a running train?

I. The train crosses a man in 9 seconds.

 II. The train crosses a 240 metre long platform in 24 seconds.

A I alone sufficient while II alone not sufficient to answer

B II alone sufficient while I alone not sufficient to answer

C Either I or II alone sufficient to answer

D Both I and II are necessary to answer

Ans: D

Time taken by train to cross a man =Length of train/Speed of train

Speed =l/9 ....(i)

Time taken by train to cross a platform =(Length of train +Length of platform)/Speed of train

Speed =(l + 240)/24 ....(ii)

From (i) and (ii), we get

l/9=(l + 240)/24

Thus, l can be obtained. So both I and II are necessary to get the answer.

25. If the pth term of an AP be 1/q and qth term be 1/p, then the sum of its pq th terms will be

A (pq-1)/2

B (1-pq)/2

C (pq+1)/2

D -(pq+1)/2

Ans: (pq+1)/2
Let a be the first term and d be the common difference of an AP.
We know an = a+(n-1)d
Given ap =1/q
a+(p-1)d = 1/q…(i)
aq = 1/p
a+(q-1)d = 1/p….(ii)
(i) – (ii)
(p-1-q+1)d = 1/q – 1/p
(p-q)d = (p-q)/pq
So d = 1/pq
Substitute d in (i)
a+(p-1)(1/pq) = 1/q
a + (1/q) – 1/pq = 1/q
So a = 1/pq
Sum, Sn = (n/2)(2a+(n-1)d)
Spq = (pq/2)(2a+(pq-1)d)
= (pq/2)((2/pq)+(pq-1)(1/pq))
= (pq+1)/2

26. The difference between degree and order of a differential equation that represents the family of curves
given by y2 = a (x + (√a / 2)), a > 0 is ______.

A 1

B 2

C 3

D 4

Ans: 2

y2 = a (x + (√a / 2))

Differentiating w.r.t.
2yy’ = a
y2 = 2yy’ [x + (√2yy’ / 2)]
y = 2y’ [(x + √yy’ / 2)]
y - 2xy’ = √2y’ √yy’
[y - 2x (dy / dx)]2 = 2y (dy / dx)3
Degree = 3 and Order = 1
Degree - Order = 3 - 1 = 2

27. A seven-digit number is formed using the digit 3, 3, 4, 4, 4, 5, 5. The probability, that number so
formed is divisible by 2, is:

A 6/7

B 4/7

C 3/7

D 1/7

Ans: 3/7

n (S) = 7! / [2! 3! 2!]

n (E) = 6! / [2! 2! 2!]
P (E) = n (E) / n (S) = [6! / 7!] * [[2! 3! 2!] / [2! 2! 2!]] = 3 / 7

28. In a huge pile of apples and oranges, both ripe and unripe mixed together, 15% are unripe fruits. Of
the unripe fruits, 45% are apples. Of the ripe ones, 66% are oranges. If the pile contains a total of
5692000 fruits, how many of them are apples?

A 2029198

B 2467482

C 2789080

D 3577422

Ans: 2029198
Let T = total no of fruits = 5692000

R = Ripe fruits

U = Unripe fruits

A = Apple        

O = Oranges

Given U = 15% of T :  15/100×5692000= 853800

R = T – U= 4838200

A(U) = 45% of U:  45/100×853800=384210

A(R) = (100-66)% of R: 34/100×4838200=1644988

∴ A(U) + A(R) = 2029198

29. S, M, E and F are working in shifts in a team to finish a project. M works with twice the efficiency of
others but for half as many days as E worked. S and M have 6 hour shifts in a day, whereas E and F have
12 hours shifts. What is the ratio of contribution of M to contribution of E in the project?

A 1:1

B 1:2

C 1:4

D 2:1

Ans: 1:2

Let the efficiency of E be x then the efficiency of M is 2x.

Let M work for y day. Since E works for twice as long, E must work for 2y days.

Thus, amount of work done by M = 2x × y × 6

 and the amount of work done by E = x × 2y × 12

The ratio of work done by M to E is

⇒  ME=2x × y ×6/x × 2y ×12=1:2
30. Five different books (P, Q, R, S, T) are to be arranged on a shelf. The books R and S are to be
arranged first second, respectively from the right side of the shelf. The number of different orders in
which P, Q and T may be arranged is ________

A 2
B 6
C 12
D 120
Ans: 6

According to the question, the arrangement will be as shown:

__ __ __ S R
Now P, Q, T has to come in one of these 3 places

Starting with P, it can come in 3 places.

After placing P, Q can come in 2 places.

After placing P and Q, T can come in 1 place.

Total ways = 3 × 2 × 1 = 6 ways