# Cocubes Aptitude Questions for Cocubes Aptitude Test 2019-20 (Numerical Reasoning)

Cocubes aptitude questions with answers are given here. These are the most repeated questions in Cocubes aptitude test. Before you start practicing Cocubes aptitude questions, it is important to know the Cocubes registration process and Cocubes syllabus 2018.

## Skills Tested in Cocubes Aptitude Questions

The following skills are tested in Cocubes Aptitude Questions.
1. Should be able to solve business problems and think in a structured manner
2. Should be good with numbers and be able to formulate equations and mathematical relationships
3. Should be able to identify relevant information from large volumes of data and generate meaningful business insights.

 Skill Aptitude question types Solving business problem Data sufficiency questions Statement and conclusions Flowcharts Formulating equations and mathematical relations Time-work-distance Arithmetic Algebra Ratio Profit-and-loss Equations Progressions Identifying relevant information Data interpretation based problems (graphs, charts and tables)

## Cocubes Aptitude Section Pattern & Syllabus

Below is Cocubes syllabus for aptitude section i.e Numerical Reasoning section.

 Module Number of Questions Durations Syllabus Aptitude (Numerical Reasoning) 15 15 mins Concepts of Mathematics including Time & Work, Speed & Distance, Algebra, Equations, Progressions, Profit & Loss, Ratios, Averages, Geometry and Data Interpretation

## Cocubes Aptitude Questions with Answers

1) A and B are two alloys in which ratios of gold and copper are 5:3 and 5:11 respectively. If these equally amount of two alloys are melted and made alloy C. What will be the ratio of gold and copper in alloy C?
a) 20:21
b) 21:20
c) 15:17
d) 17:16

Ans:  c
Given the ratio of Gold and Copper in Alloy A = 5 : 3, Amount of Gold in Alloy A = 5/8 and Amount of Copper in A = 3/8
Ratio of Gold and Copper in Alloy B = 5 :11, Amount of Gold In Alloy B = 5/16 and Amount of Copper in B = 11/16
Amount of Gold In C = (Amount of gold in A + Amount of gold in B)
= (5/8) +(5/16) = (10+5)/16 = 15/16.
Amount of Copper in C = Amount of Copper in A + Amount of Copper in B
= (3/8) + (11/6) = 17/16.
So, Ratio of Gold and Copper in C = 15/16 : 17/16 = 15 : 16

2) A, B and C can all together do piece of work in 20 days, in which B takes twice as long as A and C together do the work and C takes twice as long as A and B together take to do the work. In how many days B can alone do the work?
a) 40
b) 35
c) 60
d) 45

Ans:  c
(A+C) in x days so B completes in 2x days then (1/x) + (1/2x) = 1/20
solve, x = 30, so B 2x =60days

3) A project manager hired 15 men to complete a project in 40 days. However, after 30 days, he realized that only 1/2 of the work is completed. How many more men does he need to hire to complete the project on time?
a) 15
b) 30
c) 20
d) 25

Ans:  a
15 Men complete a work in 40days.
15*40/1 = (40-30)10*x/(1/2)=x=30Men
Men required=30-15=15Men.

4) The cost price of item B is Rs. 200/- more than the cost price of item A. Item A was sold at a profit of 20% and item B was sold at a loss of 30%. If the respective ratio of selling prices of items A and B is 6 : 7, what is the cost price of item B?
a) Rs. 520
b) Rs. 430
c) Rs. 400
d) Rs. 360

Ans:  c
Let the CP of item A be x, CP of item B is x+200.
(120/100*x)/(x+200)*70/100 =6/7
120x/(x+200)*70=6/7
20x/10(x+200) = 1, X=Rs200.
CP of item B is 200+200 = Rs. 400.

5) In a certain store, the profit is 270% of the cost. If the cost increases by 30% but the selling price remains constant, approximately what percentage of the selling price is the profit.
a) 68%
b) 72%
c) 50%
d) 65%

Ans:  d
Let C.P.= Rs. 100. Then, Profit = Rs. 270,
S.P. = Rs. 370. New C.P. = 130% of Rs. 100 = Rs. 130
New S.P. = Rs. 370, Profit = Rs. (370 – 130) = Rs. 240
Required percentage = (240/370) * 100 = 64.86 =65%(approx)

6) A certain sum is invested for certain time. It amounts to Rs. 600 at 10% per annum. But when invested at 5% per annum, it amounts to Rs. 400. Find the time.
a) 40 years
b) 75 years
c) 50 years
d) 60 years

Ans:  a
600-P=P*10*t/100 —>1===>6000-10P=Pt
400-P=P*5*t/100—->2===>8000-20P=Pt
Equate 1 and 2, 6000-10P=8000-20P==>P=200
Substitute P in 1 then 400 = 200*5*t/100 ==> 40 years

7) A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. What part of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
a) 7/11
b) 6/7
c) 1/5
d) 2/7

Ans:  c
Suppose the vessel initially contains 8 liters of liquid.
Let x liters of this liquid be replaced with water.
Water in new mixture = (3-3x/8+x)
Syrup in new mixture = (5-5x/8)
Then (3-3x/8+x) = (5-5x/8)
5x + 24 = 40 – 5x
10x=16==>x=8/5
So part of mixture replaced is 8/5*1/8=1/5

8) Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
a) 2 : 3 : 4
b) 6 : 7 : 8
c) 6 : 8 : 9
d) 7 : 6: 9

Ans:  a
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively. Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
[140/(100 * 5x)] , [150/(100 * 7x) and [175/(100 * 8x)
7x, 21x/2 and 14x
The required ratio = 7x : 21x/2 : 14x
14x : 21x : 28x
2 : 3 : 4

9) In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
a) 50
b) 100
c) 150
d) 200

Ans:  c
Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, sum of their values = Rs. [25x/100 + 10*2x/100 + 5*3x/100] = Rs. 60x/100
60x/100 = 30 à x = 50
Hence, the number of 5p coins = (3 x 50) = 150

10) The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20m towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
a) √3 m
b) 5√3 m
c) 10√3 m
d) 20√3 m

Ans:  c

11) Efficiency of A is 25% more then B and B takes 25 days to complete a piece of work. A started a work alone and then B joined her 5 days before actual completion of the work. For how many days A worked alone?
a) 9
b) 11
c) 10
d) 15

Ans:  b
Efficiency (A : B) = 5 : 4
Number of days(A : B) = 4x : 5x = 4x : 25
Number of days required by A to finish the work alone = 4x = 4 x 5 = 20.
A and B work together for last 5 days = 5 x, à 9 = 45%
Efficiency of A = 5% and B’s efficiency = 4%
No. of days taken by A to complete 55% work = 55/5 = 11 days

12) A person borrows Rs. 3000 for 2 years at 5% p.a. simple interest. He immediately lends it to another person at 6 ¼ % p.a for 2 years. Find his gain in the transaction per year.
a) Rs. 42
b) Rs. 39.25
c) 35
d) 37.5

Ans:  d
Gain in 2 yrs = [(3000*25/4*2/100)-
(3000*2*5/100)] 375-300=75.
Gain in 1yr=75/2=37.5

13) Two spheres of their radious in the ratio 4 : 3. Find its volumes ratio?
a) 64 : 26
b) 64 : 25
c) 64 : 27
d) 64 : 23

Ans:  c
R – radius of larger sphere; r – radius of smaller sphere
Sphere volume (V) = 4/3 πR3 : 4/3 π r3
= 43 : 33
= 64 : 27

14) There are three pipes, A, B and C, attached to container. A and B can fill the container alone in 20 and 30 mins, respectively whereas C can empty the container alone in 45 mins. The three pipes are kept opened alone for one minute each in the the order A, B and C. The same order is followed subsequently. In how many minutes will the reservoir be full?
a) 25 min
b) 35 min
c) 20 min
d) 47 min

Ans:  d
LCM of 20, 30 and 45 is 180
A : 20  – 9 units (i.e., 180/20)
B : 30  – 6 units
C : 5    –  4 units
1st Minute => A is opened => fills 9 L
2nd Minute => B is opened =>fills another 6 L
3rd Minute => C is opened => empties 4L
Hence every 3 minutes => (9 + 6 – 4) = 11 litres are filled into the container. So in 45 minutes (11 × 15 =) 165 litres are filled. In the 46th minute A is opened and it fills 9 litres. In the 47th minute B is opened and it fills 6 litres. Hence the container will be full in 47 minutes.

15) A discount of 20% is given on the marked price of an article. The shopkeeper charges sales tax of 10% on the discounted price. If the selling price be Rs 1848, what is the marked price (in rupees) of the article?
a) 2500
b) 3200
c) 3600
d) 2100

Ans: d
Let the MP be x, then x* 80/100(20%discount)*110/100(10%sales) =1848
X=2100

16) A man arranges to pay off a debt of Rs 3600 by 40 annual installments which are in A.P. When 30 of the installments are paid he dies leaving one-third of the debt unpaid.
The value of the 8th installment is:
a) Rs. 35
b) Rs. 50
c) Rs. 65
d) Rs. 70

Ans: c
Let the first installment be ‘a’ and the common difference between any two consecutive installments be ‘d’, using the formula for the sum of an A.P:
S = n/2 [ 2a + (n-1)d]

3600 = 40/2 [2a + 39d]
180 = 2a + 39d
And 2400 = 30/2 [2a + 29d]
160 = 2a + 29d
On solving both the equations we get, d = 2 and a = 51.
Value of 8th installment = 51 + (8-1)2 = Rs. 65

17) A student was asked to divide a number by 6 and add 12 to the quotient. He, however first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been:
a) 122
b) 118
c) 114
d) 124

Ans:  a
Let the number be x, then operations undertook by the student: (x + 12)/6 = 112
x = 660, hence 660 /6 + 12 = 122

Directions (18-19): These questions are to be answered on the basis of the following table giving the bank rates for 100 units of various foreign currencies converted to Indian rupees.

18) A man wants to convert Rs. 10,000 into foreign currency. He wants to buy the foreign currency which gives him the maximum number of units. Which currency should he buy?
a) Pound Sterling
b) U.A.E. Dirham
c) Japanese Yen
d) French Franc

Ans:  c

19) What is the approximate ratio of the buying rate of Australian Dollar of that to the U.S. Dollar?
a) 15
b) 1.02
c) 1.09
d) 0.67

Ans: d

Required ratio = 2080/3120 = 208//312 =0.67

20) The average of 13 numbers is 60. Average of the first 7 of them is 57 and that of the last 7 is 61. Find the 8th number?
a) 46
b) 32
c) 68
d) 51

Ans:  a
Sum of all the 13 numbers = 13 * 60 = 780,
Sum of the first 7 of them = 7 * 57 = 399,
Sum of the last 7 of them = 7 * 61 = 427,
So, the 8th number = 427 + 399 – 780 = 46.