Elitmus Quantitative Ability 2

Question 1

If both 11² and 3³ are factors of the number a * 4³ * 6² * 13¹¹, then what is the smallest possible value of a? 

		A.	256
		B.	589
		C.	2
		D.	363

Explanation :

11² is a factor of the given number. 
In the given expression, a * 4³ * 6² * 13¹¹ none of the other factors, viz., 4, 6 or 13 is either a power or multiple of 11. 

Hence, "a" should include 11²
The question states that 3³ is a factor of the given number. 6² is a part of the number. 

62 can be expressed as 3² * 2². 

Therefore, if 3³ has to be a factor of the given number a * 4³ * 6² * 13¹¹, then we will need another 3

as part of the number.

Therefore, "a" should be at least 11² * 3 = 363 if the given number has to have 11² and 3³ as its factors.

Hence (D) is the correct answer.

Question 2

Joe's age, Joe's sister's age and Joe’s fathers age sums up to a century. When son is as old as his father, Joe's sister will be twice as old as now. When Joe is as old as his father then his father is twice as old as when his sister was as old as her father.Find the age of Father?

		A.	45
		B.	48
		C.	50
		D.	55

Explanation :

Joe+sister+father=100

After x years lets joe age is equal to his father

Joe + x = father

Therefore, Sister + x = 2 * Sister

=>                Sister = x

Joe + Sister = Father

Therefore,

=>                  2 * Father  = 100

=>                      Father = 50

Question 3

A change-making machine contains one-rupee, two-rupee and five-rupee coins. The total number of coins is 300. The amount is Rs. 960. If the numbers of one-rupee coins and two-rupee coins are interchanged, the value comes down by Rs. 40. The total number of five-rupee coins is

		A.	100
		B.	140
		C.	60
		D.	150

Explanation :

Let the number of 5 rupee , 2 rupee and 1 rupee coins be x, y and z respectively.

x + y + z = 300.

5x + 2y + z = 960           .......(i)

5x - y + 2z = 920             ......(ii)

On subtracting (ii) from (i) we get, 

y - z = 40.

And, x + 2y = 340.

Using the options, 

if x= 140, y= 100 and z= 60, this satisfies all the given conditions.

Question 4

Shyam visited Ram during his brief vacation. In the mornings they both would go for yoga. In the evenings they would play tennis. To have more fun, they indulge only in one activity per day, i.e. either they went for yoga or played tennis each day. There were days when they were lazy and stayed home all day long. There were 24 mornings when they did nothing, 14 evenings when they stayed at home, and a total of 22 days when they did yoga or played tennis. For how many days Shyam stayed with Ram?

		A.	32
		B.	24
		C.	30
		D.	None of these

Explanation :

Let , 

P days : they play tennis.

Y days : they went to yoga.

T days : total duration for which Ram and Shyam were together.

=> P + Y = 22.

Also, (T - Y ) = 24  & (T - P) = 14.

Adding all of them,

=> 2T = 22 + 24 + 14.

=> T = 30 days.

Question 5

100 identical coins, each with probability P of showing  'heads' are tossed once. If 0<P<1 and the probability of 'heads' showing on 50 coins is equal to that of 'heads' showing on 50 coins, then find the value of P.

		A.	51/101.
		B.	51/100
		C.	32/81
		D.	None of these

Explanation :

Let X be the number of coins showing 'heads', then X follows a binomial distribution with parameters n = 100 and P.

As given, P(X=50) = P(X=51)

=> ¹⁰⁰C₅₀P⁵⁰(1-P)⁵⁰ = ¹⁰⁰C₅₁P⁵¹(1-P)⁴⁹ .

=> 51/50 = P / (1-P)

=> p = 51/101.

Hence, the value of p is 51/101.

Question 6

A tea party is arranged for 16 people along the two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and two on the other side. In how many ways can they be seated ?

		A.	8P4 x 8P2
		B.	8P4 x 8P2 x 10P10.
		C.	8P2 x 10P10.
		D.	8P4 x 8P2 / 10P10.

Explanation :

Let the persons be P1, P2,P3, P4, P5, P6, P7, P8 and Pi, Pii Piii, Piv, Pv, Pvi, Pvii, Pviii.

Here, the order of seating arrangement is as below :-

1.    4 persons ( P1, P2,P3, P4 ) wish to sit on one side in any of the 8 chairs.

2.    2 persons ( Pi, Pii) wish to sit at other side in any of the 8 chairs.

3.    Rest 10 persons can sit at any of the remaining 10 chairs.

Now, 

1.    4 persons can be arranged in 8 chairs in  N1 = ⁸P₄.

2.    2 persons can be arranged in 8 chairs on the other side of the table in N2 = ⁸P₂.

3.    Rest 10 persons can be arranged in remaining 10 chairs (after sitting arrangement of above 6 persons are complete ) in N3 = ¹⁰P₁₀.

Hence, the required number of seating arrangements is :-

=> N1 x N2 x N3.

=>  ⁸P₄ x ⁸P₂ x ¹⁰P₁₀.

Question 7

Find the number of six-digit telephone numbers in a city if at least one of their digits is repeated and 

(i) Zero (0) is allowed at the beggining of telephone number.

(ii) Zero (0) can't  initiate the number.

		A.	635241
		B.	763920
		C.	900000
		D.	None of these

Explanation :

Here, all the ten digits 0, 1, 2, ....9 have equal importance since '0' can also start the telephone number.

Now, 

No. of six-digit telephone number = 10⁶. ( when any digit may be repeated)

and, 

No. of six-digit telephone number = ¹⁰P₆. (repetition of digits not allowed)

Hence, the required no. of telephone numbers ( at least one of the digits is repeated) is :-

=> 10⁶ - ¹⁰P₆.

=> 8,48,800.

(ii) Here, 0(zero) can't come at the start of the telephone number.

Now, no. of six-digit telephone number = 9 x 10⁵. 

(since, except 0, other 9 digits (if repetition of digit allowed) are allowed at first place.

no. of six-digit telephone number = 9 x ⁹P₅.

Hence, required no. of telephone numbers (at least one of the digits is repeated )

=> 9 x 10⁵ - 9 x ⁹P₅.

=> 7,63,920.

Question 8

The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will

		A.	remain the same
		B.	decrease by 13.64%
		C.	decrease by 15%
		D.	decrease by 18.75%
		E.	decrease by 30%

Explanation :

2 walls (facing each other) have length of the room and height of the room as their components. The other 2 walls (facing each other) have breadth of the room and height of the room as their components.

Area of 1 set of walls is l * h (rectangle area), and the other set of walls is b * h.

So, the total area of the 1st 2 walls are 2*l*h, while that of the next 2 walls are 2*b*h. (as there are 2 walls each)

Total area of 4 walls now are 2 * l * h + 2 * b * h = 2 * h * (l + b).

We know, l: b: h = 3:2:1. That means, l = 3x, b = 2x, h = x.

So, total area of 4 walls = 2 * x * (5 * x) = 10x².  (Applying the formula derived = 2*h*(l+b) )

After the change,

l = 6x,  (length is doubled) b = x,  (breadth is halved) h = x/2. (height is halved)

So, total area of 4 walls = x *(7*x) = 7x².

Total decrease = 3x².

Percentage change = Change / Original * 100 = 3/10 * 100 = 30%.

Question 9

An equilateral triangle BPC is drawn inside a square ABCD. What is the value of the angle APD in degrees?

		A.	75
		B.	90
		C.	120
		D.	150

Explanation :

First, get the diagram right :-

We know that BPC is a equilateral triangle, and hence ∠BPC = ∠PCB = ∠PBC = 60^o.

Also BC = BP = PC. Since ABCD is a square, AB = BC = CD = AD.

So BP = AB (both are equal to BC). So ∠BPA = ∠BAP = x^o. (Angles opposite to equal sides are equal). Similarly, CP = CD, so ∠CPD = ∠CDP = y. (Angles opposite to equal sides are equal)

∠ABC = 90^o, since it is a square. ∠PBC = 60^o, so ∠ABP = 30^o.

Now, in ∆APB, sum of all angles is 180^o. Two angles are equal to x and the 3rd angle is 30^o.

So, 30 + 2x = 180, x = 75^o. Similarly, we get y = 75^o. (When you consider ∆CPD, where two angles are y, and 3rd angle is 30^o)

That means ∠PAB and ∠PDC is 75.

So ∠PAD and ∠PDA is 15^o. ( 90^o – 75^o)

Now in ∆APD, ∠APD = 180^o – 15^o – 15^o = 150^o.

Question 10 of 20  

A semi-circle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semi-circle at D. Given that AC = 2 cm and CD = 6 cm, the area of the semi-circle (in sq. cm.) will be:

		A.	32π
		B.	50π
		C.	40.5π
		D.	undeterminable

Explanation :

The diagram according to the question is as follows :- 

joining AD and BD we will get 3 right angle triangles where ∠ADB=∠ACD=∠BCD=90⁰.

Now,

=> CD² = AC * CB.

or, 6² = 2 * CB.

or, CB = 18. 

so AB = AC + CB = 2+ 18=20

So, radius=10 and area π*r²/2 = π*10²/2 = 50π.

Question 11

If n is such that 36 ≤ n  ≤ 72, then x = (n² + 2√n(n + 4) + 16) / (n+ 4√n+ 4) satisfies 

		A.	20 < x < 54
		B.	23 < x < 58
		C.	25 < x < 64
		D.	28 < x < 60

Explanation :

36 ≤ n  ≤ 72

x = (n² + 2√n(n + 4) + 16) / (n+ 4√n+ 4) 

Put x = 36,

x = (36² + 2√36(36 + 4) + 16) / (36+ 4√36+ 4)  

i.e which is least value for n = 28.

Question 12

The interior angle of an octagon ABCDEFGH are in AP. if the largest and second largest have an average of 153, find the average of the least two?

		A.	117
		B.	131
		C.	141
		D.	can't be determined

Explanation :

Let the angles be a , (a + d) , (a +2d) , (a + 3d) , (a + 4d) , (a + 5d) , (a + 6d) and (a + 7d).

Sum of all angles in octagon = 1080 = 8a + 28d.
Average of (a + 7d) and (a + 6d) = 153 => 2a + 13d = 306.

Solving both , a = 114 , d = 6.
=> Average of the smallest two = (2a + d)/2 = 117.

Question 13

A basketball is dropped from a height of 20 feet.it bounces back each time to a height which is one half of the height of the last bounce. How far approximately will the ball have to be traveled before it comes to rest?

		A.	30 ft
		B.	40 ft
		C.	60 ft
		D.	can't be determined

Explanation :

1st time = 20 ft
2nd time = 10ft down + 10 ft up
3rd time = 5ft up + 5ft down
and so on....

Total distance traveled = 20 + (10 + 10) + (5 + 5) + (2.5 + 2.5) + . . .
= 20 + (10 + 5 + 2.5 + ...) + (10 + 5 + 2.5 + ...)
Assuming the ball comes to rest at infinity,

Total distance traveled = 20 + [ 10/ (1 - 0.5) ] + [ 10/ ( 1 - 0.5) ]
= 20 + 20 + 20 = 60.

Question 14 of 20  You did not attempt this Question.

What is the approx. value of W, if W=(1.5)*11? Given log2=0.301, log 3=.477.

		A.	68
		B.	86
		C.	105
		D.	125

Explanation :

If W =(1.5)*11 then simply,W = 16.5
If, W=(1.5)¹¹
Take log both sides,
=> logW = 11*log(1.5) =11*log(3/2)
=> logW = 11(log3 - log2)= 11*0.176 = 1.936.
=> W = 10^(1.936) 
=> W = 86 (approx.)

Question 15

A train cross a 450 m bridge in 42 sec and another bridge 650 m length in 50 s what is approximate speed of train in km/h?

		A.	90 km/hr
		B.	45km/hr
		C.	80 km/hr
		D.	None of these

Explanation :

Let the length of the train be  L.
Then,

=> (450 + L)/42 = (650+L)/50.
=> L= 600m.
Now, speed = (450+600)/42.

or, (650+600)/50 = 25 m/s. i.e 25*(18/5) = 90Km/hr.

Question 16

A cow was standing on a bridge,  5m away from middle of the bridge. A train was coming towards the bridge from the end nearest to the cow. seeing this the cow ran towards the train and managed to escape when the train was 2m away from the bridge. if it had run in the opposite direction (i.e away from the train ) it would have been hit by the train 2m before the end of the bridge , what is the length of the bridge in meters assuming speed of train is 4 times that of the cow?

		A.	32
		B.	36
		C.	40
		D.	Can't be determined

Explanation :

Let the length of bridge be 2x and the train is coming from an end at a distance of y from one end of the bridge.
Let the speed of the cow is c , then speed of the train will be 4c.

Now the time taken by the cow to escape from bridge and time taken by train to reach at first point will be equal  i.e.,
=> (x-5)/c=(y-2)/4c.
=> 4x-y = 18.
Now again the time taken by the train to hit the cow and time taken by cow to reach at second point will be equal i.e.,
=> (5+x-2)/c = (y+x+x-2)/4c.
=> 2x-y = 15.
Solving both the equations, we will get the value of 2x as 32 that is the required answer.

Question 17

5 Women can paint a building in 30 working hours. After 16 hours work, 2 women decided to leave. How long will it take for the work to be finished ?

		A.	24
		B.	55
		C.	118/3
		D.	None of the above

Explanation :

5 women paints the wall in 30 hours. 
1 women will paint in 5*30=150 hours.
3 women will paint in 150/3=50 hours.
After 16 hours work done by 5 womens will be:16x/30=8x/15.
Remaining work is:-     x-(8x/15) = 7x/15.
Now this work has to be finished by remaining 3 womens.
Time taken by 3 womens to finish the entire work is 50 hours so 7x/15 work part will be finished by them in:
(50/x)*(7x/15)=70/3.
Hence the total time taken by them to finish the entire work is 16+(70/3)=118/3.

Question 18

A city has a population of 3,00,000 out of which 1,80,000 are males. 50% of the population is literate. If 70% of the males are literate, the number of literate females is :

		A.	20000
		B.	24000
		C.	30000
		D.	34000

Explanation :

Total No. of male=1,80,000
Total No. of female=1,20,000
Since 70% of the male are literate,
so, 1,80,000*70/100=1,26,000
Total 50% population of literate=1,50,000
So, total No. of female=1,50,000-1,26,000=24,000. 

Question 19

How many 6 digit no. can be formed using digits 0 to 5,without repetition such that number is divisible by digit at its unit place?

		A.	420
		B.	426
		C.	432
		D.	None of the above.

Explanation :

when unit place is 1 then numbers which are divisible by 1 = 96 (4*4*3*2*1*1 ways)
when unit place is 2 then numbers which are divisible by 2 = 96 (4*4*3*2*1*1 ways)
when unit place is 3 then numbers which are divisible by 3 = 96 (4*4*3*2*1*1 ways)

when unit place is 4 and ten place is 2 means (24) then numbers which are divisible by 4 = 18 (3*3*2*1*1*1 ways )
when unit place is 4 and ten place is 0 means (04)then number which are divisible by 4 also= 24 (4*3*2*1*1*1 ways )

when unit place is 3 then numbers which are divisible by 5 = 96 (4*4*3*2*1*1 ways)
Total=96+96+96+18+24+96=426.

Question 20

Shyam went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is 3/4 times the distance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was half as much again as that from Chandigarh to Shimla. If the average speed for the entire journey was 49 kmph. What was the average speed from Chandigarh to Shimla?

		A.	39.2 kmph
		B.	63 kmph
		C.	42 kmph
		D.	None of these

Explanation :

It is clear that the ratio of the distances between (Delhi-Chandigarh) : (Chandigarh-Shimla) = 3 : 4.

The ratio of the speeds between (Delhi-Chandigarh) : (Chandigarh-Shimla) = 3 : 2.

Let the distances be 3x & 4x respectively and speeds be 3y and 2y.

So the time taken will be (x/y) and (2x/y) respectively.

Since average speed is given as (Total Distance) / (Total Time) = (7x)/(x/y + 2x/y) = 7y/3 = 49.

Hence, y = 21.

So the average speed from Chandigarh to Shimla = 2y = 42 kmph.