Answer :
Total Score = Average * Number of matches
Total score of 13 matches = 13 × 42 = 546
Total score of first 5 matches = 5 × 54 = 270
Therefore, total score of last 8 matches = 546 – 270 = 276
Average = 276/8 = 34.5
Answer :
Let P be the sum
Therefore,
3584 = (P * 7 * 4) / 100 => P = 12800/-
CI = 12800(1 + (4/100))2 - 12800
=> 13844.48 - 12800 = 1044.48/-
Question 3. 6*136/8+132 / 628/16-26.25=?
Answer :
6*136/8+132 / 628/16-26.25
⇒ 234/13=18
Question 4. (1331)1/3/ 275 *52 = ?
Answer :
(1331)1/3/275 *(5)2
= (113)1/3 / 11*25 *(25)
=1
Question 5. (43)2 + 841 = (?)2 + 1465?
Answer :
Using BODMAS
2 = (43)2 + 841 – 1465
= 1849 + 841 – 1465 = 1225
Therefore, = √1225 = 35
Question 6. (216)4 ÷ (36)4 × (6)5 = (6) ?
Answer :
(6) = (216)4 ÷ (36)4 × (6)5 =
2164/364 * 65 =
612 / 68 * 65=
69 Therefore, = 9
Question 7. (1097.63 + 2197.36 - 2607.24) ÷ 3.5 = ?
Answer :
(1097.63 + 2197.36 - 2607.24) ÷ 3.5 = 687.75 ÷ 3.5
= 196.5
Answer :
Let x be the cost of each capsule and y be the no of capsules
Therefore, x = 216/y and x - 10 = 216/(y+15)
Putting the value of x from eq. 1 into eq. 2, we get
216/y - 216/(y+15) = 10
y2 +15y - 324 = 0
Solving for y, we get y = 12
Question 9. What Will Come In Place Of Question Mark (?)in The Given Question?
Answer :
Here the difference between each term is in the form of 13+1, 33+1, 53+1, 73+1 and so on
4 6 34 160 504 1234
+2 +28 +126 +344 +730
13 + 1 33 + 1 53 + 1 73 + 1 93 + 1
So = 160
Answer :
We are given that
6B2 = A2 + 540 and A : B = 3 : 2
Put the value of A from 2nd equation into the first equation, we get
6B2 = (1.5B)2 + 540
→ (6 – 2.25)B2 = 540 → 3.75B2 = 540 → B2 = 144 → B = 12
Answer :
Net part filled in 1 hour = ( 1/20 – 1/30 – 1/40) = 7 / 120
Therefore tank will be filled in = 120/7 = 17 1/7 minutes.
Answer :
⇒ 5 mangoes + 4 oranges = 7 mangoes + 1 orange
⇒4 orange – 1 orange = 7 mangoes – 5 mangoes
⇒ 3 orange = 2 mangoes
Hence Required Ratio is
⇒ Cost of mango: cost of orange = 3 : 2
Answer :
Total number of shoes=6 + 4=10;
Let S be the sample space.
Then, n(s) = Number of ways of drawing 2 shoes out of
10 =10C2 =(10 x 9)/(2 x 1) =45
Let E = Event of drawing 2 shoes which are red.
Therefore n(E) = Number of ways of drawing 2 red shoes out of
6 shoes.= 6C2 =(6 x 5)/(2 x 1) =15 P(E) = n(E)/n(S) = 1/3
Answer :
Case 1: Initial quantity of Water = 15 litres Quantity of Alcohol = 20% = (20/100) * 15 = 3 litres
Case 2: Quantity of Water after adding 5 litres of water = 15 + 5 = 20 litres There of the % of alcohol in water = (3 /20) * 100 = 15%
Answer :
Let x and y are the digits of the number with x in ten’s place and y in one’s place.
So, 10x+y is the value of the number.
Given 10x + y = 3 (x + y)
⇒7x – 2y = 0 …………….(i)
And also given that,
10x + y + 45 = 10y + x (∵ digits are reversed⇒ x in one’s and y in ten’s place).
⇒ 9x – 9y = -45
⇒ x – y = -5 ………………(ii)
Put value of y from eq. (i) to eq. (ii)
⇒ x – (7x/2) = -5
⇒ x = 2
And y = 7
So the number is 27
Answer :
We know that, formula:
Volume of any Cuboid = Length × Breadth × Height
According to the given question:
⇒ Volume of the water required in the tank = (200 × 150 × 2) m3 = 60000 m3 . . .
∴ Length of water column flown in1 min =(20 × 1000)/60 m = 1000/3 m
∴ Volume flown per minute = 1.5 × 1.25 × (1000/3) m3 = 625 m3 .
∴ Required time = (60000/625)min = 96min.
Hence, the required answer is 96 minutes.
Answer :
Rate downstream= 20 /4 = 5 kmph
Rate upstream= 18 / 6 = 3 kmph
Speed in still water = 1/2 (5 + 3) = 4 kmph.
Question 18. A Person Is 80 Years Old In 490 And Only 70 Years Old In 500 In Which Year Is He Born ?
Answer :
Since we are given that a person is elder in 490 than in 500.
Hence, he must have been born in BC.
Hence, we can say that he must have been born in BC 490+80
= BC 570
Answer :
Let quantity of grass initially = g
Let r be the rate at which grass grow in 1 day,
Let c be the quantity of grass a cow eat in 1 day
we can deduce from ‘. It takes 40 days for 40 cows and 60 days for 30 cows to eat the whole of the grass’ that
⇒ g + 40r = 40 × 40c = 1600c-----------[1]
and
⇒ g + 60r = 60 × 30c = 1800c
⇒ g = 1800c - 60r----------------2
Putting this value of g in eqn [1]
we have
⇒ 1800c – 60r + 40r = 1600c
⇒ 200c = 20r
⇒ C = 0.1 r
⇒ r = 10c
Let m be the number of days required by 20 cows to eat the entire of the field = m
Then
We have the eqn as
⇒ g + 96r = 96nc
⇒ 96nc = 1800c - 60r + 96r (as we have g=1800c-60r from eqn 2)
⇒ 96nc = 1800c + 36r
⇒ 96nc = 1800c+ 36 × 10c
⇒ 96nc = 1800c + 360c
⇒ 96nc = 2160c
Hence we have 96nc = 2160c
Dividing both the sides by 96c
We get
⇒ n = 22.5
Answer :
We have SP = 75 per litre
Profit Per cent = 37.5%
Hence
⇒ CP = SP × 100/(100 + 37.5)
⇒ CP = 7500/137.5 = 54.5454
Hence, to make the CP from 60 to 54.54,
Hence,
Required ratio is
= 54.54 : 60 - 54.54
= 54.54 : 5.454
= 10:1
Answer :
We need to mix rice varieties costing Rs 9.3 and Rs 10.8 per kg to make the mixture at Rs 10 /kg
To do so,
We need to add these varieties in the ratio
⇒10.8 - 10 : 10 - 9.3
= 0.8 : 0.7
= 8 : 7
Answer :
Let the quantity of tea a = x
Let the quantity of tea b = y
Let the quantity of tea c = z
Given : y = z ...... (i)
x + y + z = 100 ....... (ii)
Cost of Mixture = (Sum of Cost of individual items in a mixture / total quantity of mixture)
90 = (95x + 100y + 70z) / (x + y + z)
90x + 90y + 90z = 95x + 100y + 70z
5x + 10y = 20z
5x = 20y - 10y = 10y............................... (from Eq. (i), y =z )
x = 2y .................... (iii)
From Equation (i), (ii) & (iii)
x + y + z = 100
2y + y + y = 100
4y = 100
y = 25
Answer :
% of Liquid in Solution A = 20%
⇒ % of Water in Solution A = 80%
Volume of Solution A = 12 Litres
Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres
% of Liquid in Solution B = 30%
⇒ % of Water in Solution A = 70%
Volume of Solution B = 10 Litres
Volume of Water in Solution B = (70/100) × 10 = 7 Litres
If Solution A is mixed with Solution B
⇒Net Total Volume of Solution = (12 + 10) = 22 Ltrs
⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs
% of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100
% of Water in New Mixture = (16.6 / 22) × 100 = 75.45%
Answer :
Let quantity of solution sold be 1 litre
When the solution is sold for 297, the profit percent = 10%
Hence CP of of solution = SP × 100/(100+P%)
⇒CP = 297 × 100/(100+10)
⇒CP = 270
Ratio of quantities of brand A and brand B is 3:7
Let their costs be 2c and 3c.
⇒ 0.3×2c + 0.7×3c = 270
⇒ 2.7c = 270
⇒ c = 100
Hence cost of Brand B = 3c = 300
Answer :
Work done by 9 men in a day = 1/20
∴Work done by 1 man in a day for 7.5 hours = (1/20) ÷ 9 = 1/180
∴ Work done by 1 man in an hour = (1/180) ÷ (7.5) = 1/1350
Given that 2 men of latter type are equal to 3 men of former type ⇒ 1 man of latter type = 1.5 men of former type.
∴ Work done by 1 man of latter type in an hour = 1.5 × (1/1350) = 1/900
∴ Work done by 1 man in 6 hours = 6 × (1/900) = 1/150
∴ Work done by 12 men in a day working 6 hours/day = 12 × (1/150) = 2/25
Days required to finish the work = 25/2 = 12.5 days.
Answer :
Basic pay per hour of normal work = 200/40 = Rs.5
Over time is paid at 25% more = (125/100) × 5 =Rs.6.25 per hour
Man was paid Rs.200 for his regular work and Rs.100 for over time (total Rs.300).
Number of regular hours = 40
Number of overtime hours at 6.25 per hour = 100 ÷ 6.25 = 16 hours.
Total number of hours worked = 40+16 = 56 hours.
Answer :
Selling price = cost price + profit
Cost price = 225 + 15 = 240
i.e. 240 + X * 240 * 1/100 = 300 240 + 24X/ 10 = 300
(taking LCM) 2400 + 24X = 3000
24 X = 3000 – 2400
24X = 600
X = 600/ 24
X = 25
Answer :
Both truck and car covers a distance = 640 km
Time taken by truck is 10 hrs.
∴ Speed of truck = 640/10 = 64 km/hr
Time taken by car is 8 hrs.
∴ Speed of car = 640/8 = 80 km/hr
∴ Ratio between speed of truck and car =
Speed of truck : speed of car = 64 : 80 = 4 : 5
Answer :
We know that
Distance = time × speed
Suppose first train travelled the distance x kms from A.
Then second train will travel the distance (x + 120) kms from B.
Let they met each other after t time
∴ (x + 120)/60 = (x/50)
⇒ x = 600 kms
But the distances between two point is (x + x + 120)
= 2x +120
= 2× 600 + 120
= 1320 km
Answer :
Cyclist speed is 10kmph and he'll take 30 min to cover 5 kms.
If Cyclist come 25 min early he would be 5 min before crossing.
So it means train takes 5 min to reach crossing.
We have speed of train calculated as
⇒ Speed= distance/time
⇒ Speed = 5km/5 min
⇒ Speed = 1 km/min
Or 60 km/hr
Answer :
Given, distance between Salem and Koyambedu = 342 km.
Speed of bus = 57 kmph
Time = Distance/Speed
⇒ Time needed by the bus to travel from Koyambedu to Salem = 342/57 hours = 6 hours.
Time actually taken by the bus = 6 hours and 35 minutes.
Bus takes 35 minutes extra ⇒ The compulsory halt time of bus = 35 minutes
Question 32. Find The Equation Whose Roots Are 9 And 5 ?
Answer :
Finding the roots of given equations x2 – 14x + 45 = 0 x2 – 9 x – 5 x + 45 = 0 (x – 5) (x – 9) = 0
Answer :
Given expression is-
76% of 1285 = 35% of 1256 +?
⇒ (76/100) × 1285 = (35/100) × 1256 +?
⇒ 97660/100 = 43960/100 +?
⇒ ? = (97660/100) – (43960/100)
⇒ ? = 53700/100 = 537
Question 34. What Will Come In Place Of Question Mark In The Following Equation?
Answer :
(5863 - √2704) × 0.5 = ?
(5863 - √2704) × 0.5 = x
⇒ (5863 – 52) × 0.5 = x
⇒ x = 2905.5
Answer :
Let those 2 numbers be p and q .
We are given that HCF (p, q) = 13 and
LCM (p, q) = 273.
We know that product of numbers = HCF × LCM
Hence, p × q = 13 × 273
⇒ p × q = 13 × 13 × 3 × 7
P cannot be 13 × 3 or 13 × 13 or 7 × 3 or 13 or 3 or 7 as these numbers do not come between 60 and 140.
Hence we can say that p = 13 × 7 = 91
Answer :
Let’s find a number which is completely divisible by 4, 6, 8, 12 and 16.
To find such number, we need to find the LCM of all these numbers
LCM of (4,6) , 8 , 12 , 16
= 12 , 8 , 12 , 16
= LCM of (12 , 8) , LCM of (12 , 16)
= 24 , 48
= LCM of (24 , 48)
= 48
Hence 48 + 2 ie 50 will leave a remainder of 2 if divided by any of the following numbers : 4, 6, 8, 12 and 16
Answer :
We have interior angle as 40 degrees.
Hence Let n be the number of sides of that polygon,
Then formula is
⇒180 × (n-2) = 40 × n
⇒180n – 360 = 40n
⇒140n = 360
Since, no integer value of n is possible, hence such a polygon can not exist
Answer :
We are given a 5 digit number
Let first digit be 'X'
then 5th digit is '3X'
let 2nd digit be 'Y'
then 3rd digit is 'Y - 3'
and 4th digit is 'Y + 4'
then the no is '(X)(Y)(Y - 3)(Y + 4)(3X)'
from the above we can say 3X <= 9
so X<=3 and any of the digit in the number is <= 9
and also given that 3 pairs sum is 11...
Also Y–3 ≥ 0 and Y+4 ≤ 9
⇒ Y ≥ 3 and Y ≤ 5 i.e. Y = 3,4 or 5
for x = 1, all the conditions won’t be satisfied.
For x = 2,
Let Y = 5
∴ The number is 25296
Question 39. What Will Come In Place Of Question Mark In The Following Equation? 4910+311+715= ?
Answer :
4 + 9/10 + 3/11 + 7/15
= 4 + 9/10 + 3/11 + 7/15
=1617+90+154330=1861330
Zensar Technologies Aptitude Related Practice Tests |
---|
Aptitude Practice Tests |
All rights reserved © 2020 Wisdom IT Services India Pvt. Ltd
Wisdomjobs.com is one of the best job search sites in India.