Online Aptitude Test - Aptitude Test - Random

Let the numbers be x and y.

Then, x + y = 25 and x - y = 13.

4xy = (x + y)² - (x- y)²

   = (25)² - (13)²

   = (625 - 169)

   = 456

xy = 114.

To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m.

While A covers 3 m, B covers 4 m.

While A covers 360 m, B covers		4	x 360	m	= 480 m.
		3

Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.

A wins by 20 m.

Numbers must be (14)², (13)², (12)², (11)², (10)², (9)², (8)².

So, 80 is wrong.

Each number is the proceeding number multiplied by -2.

So, the required number is -128.

Work done by the leak in 1 hour =		1	-	3		=	1	.
		2		7			14

Leak will empty the tank in 14 hrs.

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R - Q = ?.

Explanation:

R - Q = Q - T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R - Q) = ?

Here we know the value(age) of Q (25), but we don't know the age of R.

Therefore, (R-Q) cannot be determined.

Let the required number of working hours per day be x.

More pumps, Less working hours per day (Indirect Proportion)

Less days, More working hours per day (Indirect Proportion)

Pumps	4	:	3		:: 8 : x
Days	1	:	2

4 x 1 x x = 3 x 2 x 8

x =	(3 x 2 x 8)
	(4)

x = 12.

Let the required number of days be x.

Less men, More days (Indirect Proportion)

27 : 36 :: 18 : x       27 x x = 36 x 18

x =	36 x 18
	27

x = 24.

log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + 1

log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + log₁₀ 10

log₁₀ [5 (5x + 1)] = log₁₀ [10(x + 5)]

5(5x + 1) = 10(x + 5)

5x + 1 = 2x + 10

3x = 9

x = 3.

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

Their increased number is (120% of 7x) and (110% of 8x).

		120	x 7x		and		110	x 8x
		100					100

	42x	and	44x
	5		5

The required ratio =		42x	:	44x		= 21 : 22.
		5		5

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.

C.P. of 1 kg of the mixture = Rs.		100	x 68.20		= Rs. 62.
		110

By the rule of alligation, we have:

Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind.
Rs. 60	Mean Price Rs. 62	Rs. 65
3		2

Required ratio = 3 : 2.

Let the thickness of the bottom be x cm.

Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000

320 x 250 x (110 - x) = 8000 x 1000

(110 - x) =	8000 x 1000	= 100
	320 x 250

x = 10 cm = 1 dm.

From statement I:

A two digit number, greater than 9 and multiple of 51 should be 51 itself.

Because, 2 x 51 = 102 (3 digit number). Therefore, I alone sufficient to answer.

From statement II:

A two digit number, greater than 9 and sum of the digit is 6.

It can be 15, 24, 33, 42, 51. So we cannot determine the required answer from the statement II alone.

Thus, I alone give the answer while II alone not sufficient to answer.

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = ³P₃ = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = ³P₃ = 3! = 6.

Total number of ways = (6 x 6) = 36.

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways =	6!	= 360.
	(1!)(2!)(1!)(1!)(1!)

Video Explanation: https://youtu.be/2_2QukHfkYA

Rate downstream =		1	x 60	km/hr = 6 km/hr.
		10

Rate upstream = 2 km/hr.

Speed in still water =	1	(6 + 2) km/hr = 4 km/hr.
	2

Required time =		5	hrs = 1	1	hrs = 1 hr 15 min.
		4		4

Increase in 10 years = (262500 - 175000) = 87500.

Increase% =		87500	x 100	% = 50%.
		175000

Required average =		50	% = 5%.
		10

From II, let l = 4x, b = 6x and h = 5x.

Then, area of the hall = (24x²) m².

From I. Area of the hall = 24 m².

From II and I, we get 24x² = 24         x = 1.

l = 4 m, b = 6 and h = 5 m.

Thus, area of two adjacent walls = [(l x h) + (b x h)] m² can be found out and so the cost of painting two adjacent walls may be found out.

Thus, III is redundant.

Correct answer is (C).

Angle traced by hour hand in	13	hrs =		360	x	13		°	= 130°.
	3			12		3

Angle traced by min. hand in 20 min. =		360	x 20		°	= 120°.
		60

Required angle = (130 - 120)° = 10°.

  I. 2(l + b) = 110         l + b = 55.

 II. l = (b + 5)         l - b = 5.

III.	l	=	6	5l - 6b = 0.
	b		5

These are three equations in l and b. We may solve them pairwise.

Any two of the three will give the answer.

Correct answer is (B).