Mathematical

reasoning
1. The following table gives the percentage distribution of population of five states, P, Q, R, S and T on
the basis of poverty line and also on the basis of gender.

What will be the male population above poverty line for State P if the female population below poverty
line for State P is 2.1 million?

A 2.1 milion

B 2.3 million

C 2.7 million

D 3.3 million

Ans 3.3million

Female population below poverty line for State P = 2.1 million

Let the male population below poverty line for State P be x million.

Then, 5 : 6 = x : 21 => x =2.1 x 5/6= 1.75.

Therefore Population below poverty line for State P = (2.1 + 1.75) million = 3.85 million.

Let the population above poverty line for State P by y million.

Since, 35% of the total population of State P is below poverty line, therefore, 65% of the total population
of State P is above poverty line i.e., the ratio of population below poverty line to that above poverty line
for State P is 35 : 65.

Therefore 35 : 65 = 3.85 : y => y = 65 x 3.85/35= 7.15.

Therefore Population above poverty line for State P = 7.15 million and so, male population above poverty
line for State P=(6x 7.15)/13million= 3.3 million.
2. Which of the boxes comes next in the sequence?

Ans:C

Shaded right angle triangle rotates by 90 degrees in a clockwise direction with each alternate turn.
Unshaded diamond goes from small to big with each turn.

3. In a particular system the units of length,mass and time are chosen to be 10cm,10g and 0.1s resp.The
unit of force in this system will be equal to

A 0.1 N

B 1N

C 10N

D 100 N

Ans: 0.1 N

Force = Mass × acceleration = 10g × 10 cm / (0.1×0.1) s 2 = 10-2 kg × 10-1 m / 10-2 s2 = 10-1 kg m/s2 = 0.1 N
4.

A grain silo is built from two right circular cones and a right circular cylinder with internal measurements
represented by the figure above. Of the following, which is closest to the volume of the grain silo, in
cubic feet?

A 261.8
B 785.4
C 916.3
D 1047.2

Ans: 1047.2

The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed
(a cylinder and two cones). The silo is made up of a cylinder (with height 10 feet and base radius 5 feet)
and two cones (each with height 5 ft and base radius 5 ft). The formulas given at the beginning of the
SAT Math section:

Volume of a Cone

V= 1/3 πr2h

Volume of a Cylinder
V=πr2h

can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the
total volume, in cubic feet, of the silo is given by

Vsilo=π(52)(10)+(2)(1/3)π(52)(5)=(4/3)(250)π

which is approximately equal to 1,047.2 cubic feet.

5. if a/b=b/a=1 then what will the value of a3-b3 be?

A 4

B 8

C 27

D 0

Ans: 0

a/b+b/a=1

=(a^2+b^2)/ab=1

=a^2+b^2=ab

=a^2+b^2-ab=0

Now,a^3+b^3=(a+b)(a^2+b^2-ab)

=(a+b)*0 (Because,a^2+b^2-ab=0)

=0 Ans

6. The value of √(1+2+3+4+…..+175+174+173+….+2+1) is

A 49√5

B 175

C 175√174

D 174
Ans: 175

√(1+2+3+4+…..+175+174+173+….+2+1)

√175(174)/2+174x175/2

√175(176=174)/2

√175x175

175

7. Total number of two digit numbers increase by 18 when their digits are reversed is

A 15

B 7

C 31

D 13

Ans: 7

Let the number be 10x+y

Acc to question

10y+x-(10x+y)=18

y-x=2

so cases will be

79, 68,57,46, 35, 24, 13

So, 7 cases

8. A three digit number is being subtracted from another three digit number consisting of same digits in
reverse order gives 594. The minimum possible su of all the three digits of this number is

A 8

B 12
C 6

D Cannot be determined

Ans: 8

Let x, y, z be the hundreds, tens, units digits.

We can see

(100z+10y+x)-(100x=10y=z)=594

z-x=6

possible values of x and z are (1,7),(2,8),(3,9)

so min possible value of x+y+z=1+0+7=8

9. The number of factors in the expression 64x86x108x1210 is

A 48

B 64

C 72

D 80

Ans: 72

=(2x3)4x(23)6x(2x5)8x(22x3)10

=24x34x218x28x58x220x310

=250x314x58

=(50+14+8) prime factors

72 prime factors

10. What is the number of squares in the last section where the question mark is?
The number of boxes in the first section is 1.1.
The number of boxes in the second section is 1,1, which is the sum of all the numbers of boxes before
that.
The number of boxes in the third section is 2,2, which is the sum of all the numbers of boxes before that,
i.e.1+1=2.1+1=2.The number of boxes in the fourth section is 4,4, which is the sum of all the numbers of
boxes before that, i.e.1+1+2=4.1+1+2=4.The number of boxes in the fifth section is 8,8, which is the sum
of all the numbers of boxes before that, i.e.1+1+2+4=8.1+1+2+4=8.Thus, the number of boxes in the
sixth and last section should be 16,16, which is the sum of all the numbers of boxes before that,
i.e.1+1+2+4+8=16.1+1+2+4+8=16.Therefore, the correct answer is c

11. The following pie chart shows the amount of subscriptions generated for India Bonds from different
categories of investors.
Subscriptions Generated for India Bonds
If the total investment flows from FII's were to be doubled in the next year and the investment flows from
all other sources had remained constant at their existing levels for this year, then what would be the
proportion of FII investment in the total investment into India Bonds next year (in US $ millions) ?

A 40%

B 50%

C 60%

D 70%

Ans: 50%

FII's currently account for 33 out of 100.

If their value is doubled and all other investments are kept constant then their new value would be 66 out
of 133 = approximately equal to 50%

12. The sum of the series 1/5x9+1/9x13+1/13x17+…..+1/61x65=?

A 4/45

B 3/65

C 2/35

D 3/35

Ans: 3/65

=1/4[1/5-1/9+1/9-1/13+1/13-1/17+….+1/61-1/65}

=1/4[1/5-1/65]

=1/4[12/65]

=3/65

13. If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average
of 3m and 18, what is the average of x, y, and z in terms of m?

A  m+6
B  m+7
C  2m+14
D 3m+21
Ans: m+7

Since the average (arithmetic mean) of two numbers is equal to the sum of the two numbers divided by 2,
the equations x=(m+9)/2, y=(2m+15)/2, z=(3m+18)/2 are true. The average of x, y, and z is given
by x+y+z/3. Substituting the expressions in m for each variable (x, y, z) gives

[(m+9)/2+(2m+15)/2+(3m+18)/2]/3

This fraction can be simplified to m+7.

14. If a−b=3, and a3 –b3=117, then absolute value of |a+b| is,

A -7

B 8

C 7

D 9

Ans: 7

( a - b )³ = a³ - b³ - 3ab × (a-b)

3³ = 117 - 3ab × (3)

27 = 117 - 9ab

9ab = 117 - 27

9ab = 90

ab = 10

( a + b )² = (a - b)² + 4ab

( a + b )² = 3² + 4*10

( a + b )² = 9 + 40

( a + b )² = 49

a+b=7

15. A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months
his saving increases by Rs.40 more than the saving of immediately previous month. After how many
months his total savings from the start of service will be Rs. 11040 after:
A 18 months

B 19 months

C 20 months

D 21 months

Ans: 21 months

From the question, saving are 200, 200, 200, 240, 280,…….. upto n terms.
But 200, 240, 280,……..(n-2) terms are in AP.
Using sum of n terms of an AP, we have
400 + (n-2)/2 [2×200 + (n – 2-2)40] = 11040
=> (n-2)[200 + 20n – 60] = 10640
=> 20(n + 7)(n – 2) = 10640
=> n2 + 5n – 546 = 0
=> (n + 26)(n – 21) = 0
-ve value is not possible.
n = 21
Therefore, total time = 21 months

16.  The coefficients a,b and c of the quadratic equation, ax 2 + bx + c = 0 are obtained by throwing a dice
three times. The probability that this equation has equal roots is :

A 1/54

B 1/72

C 1/36

D 5/216

Ans: 5/216

ax2 + bx + c = 0
a,b,c ∈ {1,2,3,4,5,6}
n(s) = 6 × 6 × 6 = 216
D=0 ⇒ b2 = 4ac
ac = b2/4, If b = 2, ac = 1 ⇒ a = 1,c = 1
If b = 4, ac = 4 ⇒ a = 1, c = 4
a = 4, c = 1
a = 2, c = 2
If b = 6, ac = 9 ⇒ a = 3, c = 3
Therefore, probability = 5/216

17. Find the correct 3D shape for the given net

Ans: D
Just observe the given net and fold it up in your mind

18. The integer k, for which the inequality x2 – 2(3k – 1)x + 8k2 – 7 >0 is valid for every x in R is

A 3

B 2

C 4

D 0

Ans: 3
D<0
(2(3k-1))2 – 4(8k2 – 7) < 0
4(9k2 – 6k + 1) - 4(8k2 – 7) < 0
k2 – 6 k + 8 < 0
(k–4)(k–2)<0
2< k < 4
then k = 3

19. How many zeroes will be at the end of the expression 100 x 200 x 300 x…..x 10000
A 1000
B 224
C 483
D 292
Ans: 224
We can convert it as
100100(1x2x3x….x100)
10200(1x2x3x…..x100)
100200(100!)
So total zeroes are 200+24=224 zeroes
20. Find the correct shape for the given net.

Ans:B
Just imagine folding up the net to get option B

21. what is the highest power of 3 available is 58!-38!?

A 17
B 45
C 89
D 43
Ans: 17
We can write it as
58x57x56….40x39x38!-38!
Expression inside the bracket represents 3n-1 kind of number
Thus only the number of 3’s in 38! would matter
12+4+1=17

22. In a zoo, there are rabbits and pigeons. If heads are counted, there are 340 heads and if legs are
counted there are 1060 legs.How many pigeons are there?
A 120
B 150
C 170
D 180
Ans:150
Suppose there are all the pigeons then total no of heads are 340 and total no of legs are 680. 
 
Now, since 380 (1060-680) legs are extra, it means there will be 190 (380/2) rabbits.As we know a rabbit
has two extra legs than that of a pigeon.
 
Therefore,  number of rabbits =190
 
and number of pigeons = 340- 190 = 150

23. Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a single
cube whose diagonal is 12√3 Find the edges of the three cubes.
A 6,8,10
B 7,11,14
C 3,4,5
D 9,12,15
Ans: 6,8,10
Let the edges of three cubes (in cm) be 3x, 4x and 5x respectively.

So, the volume of the cube after melting will be

= (3x)3 + (4x)3 + (5x)3 = 9x3 + 64x3 + 125x3 = 216x3

Now, let a be the edge of the new cube so formed after melting

Then we have,

a3 = 216x3

a = 6x

We know that, Diagonal of the cube

= √(a2 + a2 + a2) = a√3 So, 12√3 = a√3 a = 12 cm x = 12/6 = 2

Thus, the edges of the three cubes are 6 cm, 8 cm and 10 cm respectively.
24. The function f(x)=x3−x2−x−11/4  is graphed in the xy-plane above. If k is a constant such that the
equation f(x)=k has three real solutions, which of the following could be the value of k?

A 4
B 5
C -6
D -3
Ans: -3
The equation f(x)=k gives the solutions to the system of equations

y=f(x)=x3−x2−x−11/4

and

y=k

A real solution of a system of two equations corresponds to a point of intersection of the graphs of the
two equations in the xy-plane.

The graph of y=k is a horizontal line that contains the point (0,k) and intersects the graph of the cubic
equation three times (since it has three real solutions). Given the graph, the only horizontal line that
would intersect the cubic equation three times is the line with the equation y=−3, or f(x)=−3.
Therefore, k is −3.
25. If equation 2x2−7x+12=02x2−7x+12=0 has two roots α and β, then the value of α/β+β/α is,

A 1/24
B 7/24
C 7/2
D 97/24
Ans: 1/24

= α/β+β/α 
= α ^2+ β2/ α. β

= (α + Bita)2 - 2. α. β / α. β

= (-b/a)^2 - 2 .c/a/c/a

=49/4-12/6

=1/4/6

=1/24

26. Which number replaces the ‘?’?

A 10

B 11

C 12

D 13

Ans: 12
The diagram contains two diamonds of 3x3 circles, arranged side by side, with the central 4 circles
overlapping.
Starting in the central left circle of each diamond, the numbers from 1 to 9 are written in lines, moving up
and to the right.

In the circles where the diamonds overlap, the values are added together to give the ones shown on the
final diagram.

27. 25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both
hockey and football. Then the number of persons playing neither hockey nor football is:

A 2
B 17
C 19
D 3
Ans: 3
the total people are 25
the people play hockey is 15
the people play foot ball is 17 
both are 10 people 
that means the people play only hockey is 15-10=5
and football is 17-10=7
sum all the individuals and combined players gives 22
that means the remaining 3 will not play either of the games

28. Operators □,◊,→ are defined by : a □ b = a−b/a+b; a ◊ b = a+b/a−b; a → b = ab. Find the value of
(66 □ 6) → (66 ◊ 6).

A -2

B 1

C 3

D 2

Ans: 1

We can see both □,◊ get cancelled if the→is used as it multiplies the reciprocals and the value becomes 1
29. The bar graph given below shows the foreign exchange reserves of a country (in million US $) from
1991 - 1992 to 1998 - 1999.
Foreign Exchange Reserves Of a Country. (in million US $)

The foreign exchange reserves in 1996-97 were approximately what percent of the average foreign
exchange reserves over the period under review?

A 95%

B 110%

C 115%

D 125%

Ans: 125%

Average foreign exchange reserves over the given period

=[(1/8)x (2640 + 3720 + 2520 + 3360 + 3120 + 4320 + 5040 + 3120)]million US $

= 3480 million US $.

Foreign exchange reserves in 1996 - 1997 = 4320 million US $.

Therefore Required percentage = (4320x 100) /3480% = 124.14% ~=125%.

30.  The total number of positive integral solutions (x, y, z) such that xyz = 24 is

A 36
B 24

C 45

D 30

Ans: 30

x.y.z = 24
x.y.z=23.31
x=2(a_1 )⋅3(b_1 )
y = 2(a_2 )⋅3(b_2 )
z = 2(a_3 )⋅3(b_3 )
a1, a2, a3 ∈ {0,1,2,3}
b1, b2, b3 ∈ {0,1}
Case 1: a1 + a2 + a3 =3
Non negative solution = (3+3-1) C(3-1) = 5C2 = 10
Case 2: b1 + b2 + b3 = 1
Non negative solution = (1+3-1) C(3-1) = 3C2 = 3
∴ Total solutions =10×3=30