APTITUDE TEST 2

Problem 1 :

Find the number of prime factors of

6¹⁰ X 7¹⁷ X 55²⁷  

(A) 90              (B) 91           (C) 92            (D) 93

Solution :

From 6¹⁰ × 7¹⁷ × 55²⁷, we have to write each base in terms of multiplication of its prime factors.

That is

= (2 x 3)¹⁰ × (7)¹⁷ × (5 x 11)²⁷

= 2¹⁰ X 3¹⁰ X 7¹⁷ X 5²⁷ X 11²⁷

The no. of prime factors = sum of the exponents

= 10+10+17+27+27

= 91

Hence, the number of prime factors is 91.

Problem 2 :

Two trains running at 60 kmph and 48 kmph cross each other in 15 seconds when they run in opposite direction. When they run in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. Find the length of the two trains (in meters). 

   (A) 300, 130    (B) 310, 115   (C) 320, 110    (D) 330, 120

Solution :

When the two train running in opposite direction, relative speed

= 60 + 48

= 108 kmph

= 108 x 5/18 m/sec

= 30 m/sec

Sum of the lengths of the two trains = sum of the distances covered by the two trains in the above relative speed

Sum of the lengths of the two trains

= 30 x 15

= 450 m

When the two trains running in the same direction, relative speed

= 60-48 =12 kmph

= 12X5/18

= 10/3 m/sec

When the two trains running in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. The distance he covered in 36 seconds in the relative speed is equal to the length of the slower train.

Length of the slower train = 36 x 10/3

= 120 m

Length of the faster train = 450 - 120

= 330 m

Hence, the length of the two trains are 330 m and 120 m

Problem 3 :

A & B can do a work in 15 days, B & C in 30 days and A & C in 18 days. They work together for 9 days and then A left. In how many more days, can B and C finish the remaining work ?

 (A) 9 days     (B) 8 days     (C) 7 days    (D) 6 days

Solution :

We can apply L.C.M method to solve this problem

Total work = 90 units (L.C.M of 15,30,18,9)

• (A + B) can complete 6 units/day (90/15 = 6)
• (B + C) can complete 3 units/day (90/30 = 3)
• (A + C) can complete 5 units/day (90/18 = 5)

By adding, we get 2(A+B+C)

= 14 units/day

(A+B+C) = 14/2

= 7 units/day

work done by A = (A + B + C) - (B + C) = 7 - 3 = 4 units/day

work done by B = (A + B + C) - (A + C) = 7 - 5 = 2 units/day

work done by C = (A + B + C) - (A + B) = 7 - 6 = 1 unit/day

work done by (A + B + C) in 9 days = 9 x 7 = 63 units.

Balance work = 90-63

= 27 units

This 27 units of work to be completed B & C. Because A left after 9 days of work.

No. of days taken by B & C to complete the balance 27 units of work

= 27/3

= 9 days.

(Because B&C can complete 3 units of work per day)

Problem 4 :

A man traveled from the village to the post office at the rate of 25 k mph and walked back at the rate of 4 kmph. If the entire journey had taken 5 hours 48 minutes, find the required distance of the post office from the village.

(A) 20 km       (B) 21 km       (C) 22 km     (D) 23 km

Solution :

Average speed = 2pq/(p+q),

here p = 25, q = 4

= (2 x 25 x 4)/(25 + 4)

Therefore, average speed = 200/29 km/hr

And 5 hour 48 min = 5  48/60 hrs = 29/5 hours

Distance = Speed x Time

Distance = (200/29) x (29/5) = 40 km

Distance covered in 5 hrs 48 min = 40 km

Hence, distance of the post office from the village

= 40/2

= 20 km

Problem 5 :

The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.

(A) 35 yrs      (B) 34 yrs      (C) 33 yrs    (D) 32 yrs

Solution :

Let "x" be the present age of the son.Then the present age of father is (3x+3)

3 years hence, father's age will be 10 years more than twice the age of the son

(3x+3)+3 = 2(x+3)+10

3x+6 = 2x+16

x = 10

To find the present age of the father, plug x = 10 in

(3x + 3)

Present age of the father = 3(10) + 3

= 33 years

Problem 6 :

The average of four consecutive even numbers is 27. Find the largest of these numbers. 

(A) 28        (B) 30         (C) 32             (D) 34

Solution :

If "x' be the first even number, then the four consecutive even numbers are x, x + 2, x + 4, x + 6

Average of the four consecutive even numbers = 27

(x + x + 2 + x + 4 + x + 6)/4 = 27

(4x + 12) = 108

4x = 96 ===> x = 24

Hence the largest even number = x+6

= 30

Problem 7 :

A boat takes 19 hours for traveling downstream from point A to point B and coming back to a point C midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph, what is the distance between A and B ? 

(A) 150 km        (B) 160 km        (C) 170 km     (D) 180 km

Solution :

From the given information, we have

• Speed downstream = (14+4) = 18 kmph
• Speed upstream = (14-4) = 10 kmph

Let "x" be the distance between A and B

x/18 + (x/2)/10 = 19 (Hint: Time = Distance/Speed)

x/18 + x/20 = 19

(10x + 9x)/180 = 19 (L.C.M of 18,20 is 180)

19x/180 = 19

x = 180

Hence the distance between A and B is 180 km.

Problem 8 :

John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7:6, find his new weight. 

(A) 48.6 kg       (B) 49.6 kg      (C) 50.6 kg      (D) 51.6 kg

Solution :

Original weight of John = 56.7 kg (given)

He is going to reduce his weight in the ratio 7 : 6

[Hint: If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a']

His new weight = (6 x 56.7)/7

= 6x8.1

= 48.6 kg

Problem 9 :

Find the ratio in which, water to be mixed with milk to gain 20% by selling the mixture at cost price. 

(A) 1:5         (B) 5:1        (C) 1:7        (D) 1:7

Solution :

Let the cost price of 1 ltr pure milk be $1

Now we take some quantity of milk (less than 1 ltr),add some water and make it to be 1 ltr mixture

Let "x" be the money we invest for milk in 1 ltr of milk-water mixture

Since the gain is 20%, selling price = x + 20% of x

Selling price = 120% of x

= (120/100)x

= (6/5)x

But the mixture is sold at the cost price of pure milk

So, we have (6/5) x = 1

x = 5/6

Therefore cost price of the milk in the mixture = $(5/6

Cost price of the water = $0

Rule to find the ratio for producing mixture

= (d-m):(m-c)

(d-m):(m-c) = 1-5/6:5/6-0

= 1/6:5/6

= 1:5

So, water and milk to be mixed in the ratio to gain 20% is 1:5

Problem 10 :

15% of income of A is equal to 25% of income of B and 10% of income of B is equal to 30% of income of C. If income of C is $ 1600, then total income of A, B and C is  

(A) $ 14200    (B) $ 14300   (C) $ 14400     (D) $ 14500

Solution :

Let A,B and C be the incomes of A,B and C respectively

From the given information, we have C = $1600

10% of B = 30% of C

(10/100)B = (30/100)X1600

B = $ 4800

15% of A = 25% of B

(15/100)A = (25/100)X4800

A = $8000

A+B+C = 8000 + 4800 + 1600

= 14400

Hence, the total income of A, B and C is $14400

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.