Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Let P_{1}, P_{2}, ......, P_{15} be 15 points on a circle. The number of distinct triangles formed by points P_{i}, P_{j}, P_{k} such that i +j + k $$\ne$$ 15, is :

A

12

B

419

C

443

D

455

Total number of triangles = $${}^{15}{C_3}$$

i + j + k = 15 (Given)

Number of possible triangles using the vertices P_{i}, P_{j}, P_{k} such that i + j + k $$\ne$$ 15 is equal to $${}^{15}{C_3}$$ $$-$$ 12 = 443

Option (c)

i + j + k = 15 (Given)

Number of possible triangles using the vertices P

Option (c)

2

If $${}^n{P_r} = {}^n{P_{r + 1}}$$ and $${}^n{C_r} = {}^n{C_{r - 1}}$$, then the value of r is equal to :

A

1

B

4

C

2

D

3

$${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$$

$$ \Rightarrow (n - r) = 1$$ .....(1)

$${}^n{C_r} = {}^n{C_{r - 1}}$$

$$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$$

$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$$

$$ \Rightarrow n - r + 1 = r$$

$$ \Rightarrow n + 1 = 2r$$ ..... (2)

From (1) and (2), $$ 2r - 1 - r = 1 \Rightarrow r = 2$$

$$ \Rightarrow (n - r) = 1$$ .....(1)

$${}^n{C_r} = {}^n{C_{r - 1}}$$

$$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$$

$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$$

$$ \Rightarrow n - r + 1 = r$$

$$ \Rightarrow n + 1 = 2r$$ ..... (2)

From (1) and (2), $$ 2r - 1 - r = 1 \Rightarrow r = 2$$

3

The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :

A

26664

B

122664

C

122234

D

22264

Total possible numbers using 1, 2, 2 and 3 is

= $${{4!} \over {2!}}$$ = 12

When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are

= $${{3!} \over {2!}}$$ = 3

When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are

= 3! = 6

When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are

= $${{3!} \over {2!}}$$ = 3

$$ \therefore $$ Sum of unit places of all (3 + 6 + 3) 12 numbers is

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3)

Similarly,

When 10^{th} place is 1, the total possible numbers using remaining 2, 2 and 3 are

= $${{3!} \over {2!}}$$ = 3

When 10^{th} place is 2, the total possible numbers using remaining 1, 2 and 3 are

= 3! = 6

When 10^{th} place is 3, the total possible numbers using remaining 1, 2 and 2 are

= $${{3!} \over {2!}}$$ = 3

$$ \therefore $$ Sum of 10^{th} places of all (3 + 6 + 3) 12 numbers is

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 10

Similarly,

Sum of 100^{th} places of all (3 + 6 + 3) 12 numbers is

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 100

and Sum of 1000^{th} places of all (3 + 6 + 3) 12 numbers is

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 1000

$$ \therefore $$ Total sum = ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) + ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 10

+ ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 100 + ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 1000

= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $$ \times $$ 24 = 26664

= $${{4!} \over {2!}}$$ = 12

When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are

= $${{3!} \over {2!}}$$ = 3

When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are

= 3! = 6

When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are

= $${{3!} \over {2!}}$$ = 3

$$ \therefore $$ Sum of unit places of all (3 + 6 + 3) 12 numbers is

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3)

Similarly,

When 10

= $${{3!} \over {2!}}$$ = 3

When 10

= 3! = 6

When 10

= $${{3!} \over {2!}}$$ = 3

$$ \therefore $$ Sum of 10

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 10

Similarly,

Sum of 100

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 100

and Sum of 1000

= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 1000

$$ \therefore $$ Total sum = ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) + ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 10

+ ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 100 + ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 1000

= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $$ \times $$ 24 = 26664

4

If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :

A

240

B

360

C

333

D

364

Total number of triangles

= $${}^{14}{C_3} - {}^3{C_3} - {}^5{C_3} - {}^6{C_3}$$

= 364 – 31 = 333

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