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Miscellaneous Problems – Set 1

1) A window is made up of a square portion and an equilateral triangle portion above it. The base of the triangular portion coincides with the upper side of the square. If the perimeter of the window is 6 m, the area of the window in m2  is _______________

         a)  1.43                             b)   2.06

         c)  2.68                              d)  2.88

Solution:  The figure can be drawn as shown below (Figure 1):

Figure 1

From the Figure 1.

Upper side of the square ABCD is CD & it coincides with the base of equilateral triangle CDE i.e CD.

Now from this figure we can easily arrive that the sides of equilateral triangle = sides of the square.

Let the sides be a m.

So , AB = BC = CD = CE = ED = DA = a

Perimeter of window = 6 m

AB + BC + CE + ED + DA = 6 m

5a = 6 ; a = 1.2 m

Area of the window = Area of the square ABCD + Area of the equilateral triangle CDE.

=   a2    +    √3/4  a2

=  (1.2)²  +  √3 ⁄4  x  (1.2)² ≈ 2.06 m².

2) What is the next number in the series ?

12   ,    35   ,   81   ,   173   ,   357   ,   _________

Solution :

35 = 12 + 23 x 1

81 = 35 + 23 x 2

173 = 81 + 23 x 4

357 = 173 + 23 x 8

From the above pattern we can easily arrive the next term

i.e.  357 + 23 x 16 = 725

3) If log a , log b , log c are in A.P. , then a , b , c are in

a)  A.P.       b)  G.P.       c)  H.P.      d) None of these

Solution:

Since the above sequence is in Arithmetic Progression (A.P.) .Therefore , the difference between any two consecutive numbers is constant.

log b – log a = log c – log b

2 log b = log a + log c

log b² = log ac

or b² = ac 

So a , b , c are in G.P. (option b)

Note: Laws of logarithms (used in the above problem)

  • log A + log B = log AB
  • n log A = log An

4) The average age of three boys is 15 years. If their ages are in the ratio 3 : 5 : 7, the age of the youngest boy is

a)  21 years      b)  18 years      c)  15 years      d)  9 years

Solution: As the ratio is 3 : 5 : 7 , let their ages be 3X , 5X , 7X

Average age = (3X + 5X + 7X)/3

15X/3 = 15

X = 3

Therefore , age of the youngest boy = 3X = 3 x 3 = 9 years

5) Rs 2820 is to divided among A , B & C such that 3 times A’s share is equal to 4 times B’s share & 5 times C’s share. Find A’s share?

Solution:

3A = 4B = 5C

Ratio of A’s & B’s share = A : B = 4 : 3

Ratio of B’s & C’s share = B : C = 5 : 4

Therefore , their shares ratio A : B : C = 20 : 15 : 12

So A’s share = {20/(20 +15 + 12)} x 2820 = 20 x 2820 /47 = Rs 1200

6) Price of a commodity is first increased by X% and then decreased by X%. If the new price is K/100, find the original price.

a)  (X -100)/K      b)  (X² – 100²)100/K      c)  (100 – X)100/K      d)  100K/(100² – X²)

Solution:

Let the original price be Rs Y

Price increased by X% , then new price = Y + YX% = Y ( 1 + X/100)

then price decreased by X% , final new price will be =

Y (1 + X/100) (1 – X/100) = Y (1 – X²/100²)

Y (100² – X²)/100² = K/100

Y (original price) = 100K/(100² – X²) { option d}

7) Abhinav can do a work in 12 days and Abhishek in 15 days. If they work on it together for 4 days , then the fraction of the work that is left is:

(a)  3/20      (b)  3/5      (c)  2/5      (d)  2/20

Solution: 

In one day , Abhinav will complete 1/12 of the work.

whereas , Abhishek will complete 1/15 of the work.

Then, in one day , if Abhinav and Abhishek work together they will complete

1/12 + 1/15 = 3/20 of the work.

So , in 4 days together they will complete = 4 x 3/20 = 3/5 of the work.

Therefore , fraction of the work that is left = 1 – 3/5 = 2/5

8) log tan 1° + log tan 2° +………..+ log tan 89° is _________________.

(a)  1      (b)  1/√2      (c)  0      (d)  -1

Solution:

log tan 1° + log tan 2° + log tan 3° + …………+ log tan 45° + log tan 46° + …………+ log tan 89°

{ using Law of Logarithmslog A + log B = log AB }

= log ( tan 1° x tan 2° x tan 3° x ……..x tan 45° x tan 46° x tan 47° x……….x tan 88° x tan 89°)

= log { tan 1° x tan 2° x …….. x tan 45° x tan (90° – 44°) x ……. x tan (90° – 2°) x tan (90° – 1°) }

= log { tan 1° x tan 2° x ……… x tan 45° x cot 44° x cot 43° x cot 42° x ………. x cot 2° x cot 1° }

[ using formula: tan (90° – θ) = cot θ ; tan θ x cot θ = 1 ]

= log { tan 45° } = log 1 = 0

9) A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. Find the diameter of the base of the cone.

Solution: 

Radius of the Sphere , R = 12.6/2 = 6.3 cm

height of the cone , h = 25.2 cm

Let ‘r’ be the radius of cone.

In this case , Volume of the Sphere = Volume of the Cone.

4/3 ΠR³ = 1/3 Πr²h

4R³ = r²h

4(6.3)³ = r²(25.2)

4 x 6.3 (6.3)² = r² (25.2)

r (radius of the cone) = 6.3 cm

Diameter of the cone = 2r = 2 x 6.3 = 12.6 cm

10) Find the sum of all numbers divisible by 6 in between 100 to 400.

Solution: 

The numbers divisible by 6 in between 100 to 400 are 102 , 108 , 114 , 120 ……………396.

Here 1st term a = 102 , last term divisible by 6 is l = 396 & common difference d = 6

l = a + (n – 1) d

396 = 102 + (n – 1) 6

6 (n – 1) = 396 – 102

n – 1 = 294/6 = 49

n = 49 + 1 = 50

Sum of n terms of an A.P = n/2 [ 2a + ( n -1 ) d ]

= 50/2 [a + { a + ( n -1 ) d}] = 25[ a + l ]

= 25 [ 102 + 396 ] = 25 x 498 = 12450

Suffering is the essence of success.

~A.P.J Abdul Kalam

 

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About the author

Knowledge Sharing

Knowledge Sharing is purely an educational blog which is maintained by Abhinav Prasad. Abhinav is currently pursuing MBA & he predominantly writes articles related to Aptitude and General Knowledge. He is also an enthusiastic learner and loves photography.


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