CPT Mock Test – 1st
Test Booklet No. – 110011
(1)
Duration : 1 Hours
Total Marks : 50
Date: - 17.08.2015
Ans. b
Explanation:
The amt to be paid without any discount will be : 7,000
(-) paid on 30/06/2006
Discount = 200
10 

1,800  
90 

Total amt. = 1800  200   2,000 
(-) paid on 30/09/2006
Discount = 150
(2850 x 5/95 )
Total amt = 2850 + 150 = 3,000
Amount to be paid in final settlement = 7000 – 2000 – 3000 = Rs. 2,000
(2)
Ans. c
Explanation:
Profit before charging managerial remuneration = Rs. 44,000
Managerial remuneration on profit after charging commission = 44000 x 10 / 110 =
Rs.4000
(3)
Ans. a
Explanation:
Profit = (Sales – Sales return) – (purchase - purchase return)
= (40000 – 5000) – (30000 – 5000) = Rs. 10000
(4)
Ans. b
Explanation:
Amt. of accrual interest
 10, 000 12% 
(5)
(6)
2
 200
12
Ans. d
Explanation:
Purchase cost =
(+) Shipping and forecasting charges
(+) Import duty =
(+)Carriage Inwards =
(+) Repair Charges =
(+) Installation Charges =
(+) Brokerage =
(+) Iron Paid =
=
Rs. 50,000
2,000
1,000
1,000
500
200
400
100
55,200
Ans. c
Explanation:
1, 00, 000 12%  12, 000 To be recorded in P&L Account
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(7)
Ans. d
Explanation:
To Balance b/d
To P&L (b/f)
Total
Provision for discount on creditors
1600
By Discount received
500
By Balance c/d
2100
Total
(8)
Ans. a
Explanation:
Because carriage inward is a direct expenses.
(9)
Ans. a
Explanation:
TRADING
To Opening stock
20,000
To Purchases
1,00,000
To Carriages
2,000
To Gross Profit (B/F)
53,000
Total
1,75,000
Selling expenses are indirect expenses
ACCOUNT
By Sales
By Closing Stock
1000
1100
2100
1,50,000
25,000
Total
175000
and it will be included in P&L A/c.
(10)
Ans. c
Explanation:
Cost of goods sold = sales – gross profit
1,00,000 = 1,25,000 - G.P.
G.P. = Rs. 25,000
(11)
Ans. d
Explanation:
Joint venture profit /loss =
Sales
Add : Value of Stock
Taken
Less: Cost of purchase
Expenses
Profit
Value of stock taken = 5 radio set @ 1000 = 5000
Proportional exp.
120,000
5250
125250
100000
5000
20250
5000
 5 = 250
100
5250
(12)
Ans. d
Explanation:
Sales values of goods
Value of Stock Taken over
90000
2700
92700
Less:
Purchase cost
Expenses
Profit
MITTAL COMMERCE CLASSES
75000
10950
6750
2|Page
(13)
Ans. a
Explanation:
Sales values of goods
Less: Cost of goods supplied by elder
Exp. Incurred by elder
Cost of goods supplied by large
Exp. Incurred by large
Commission to large@ 5% of 6000
Profit
6000
2500
200
2000
150
300
850
(14)
Ans. a
Explanation:
Profit = Sales Value of price of land - purchase price of piece of land = 6000030000= Rs. 30000
(15)
Ans. b
Explanation:
Sales
Add: Closing Stock taken by A
Less: material purchased
Wages paid
Administrative exp.
Selling Expenses
Expenses paid by A
Profit
112000
6200
65000
6000
3000
6170
1630
36400
(16)
Ans. a
Explanation:
When no spearate set of books is maintained and each co ventures records only his
own transactions then to find out profit or loss on joint venture a memorandum joint
venture account is opened which does not follows double entry system.
(17)
Ans. b
Explanation:
Capital accounts of the co- ventures represents amounts due to or due from coventures therefore these are of personal in nature.
(18)
Ans. c
Explanation:
% profit on cost, on goods sold
=
240000  160000
 100  50%
160000
Cost of remaining stock =
Total cost of purchase – Cost of goods sold
240000-160000 = 80000
Sales value of goods taken over
= 80000 + (50% of 80,000) = Rs. 120000
(19)
Ans. b
Explanation:
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Joint venture A/c is credited with the agreed value of stock taken over by coventurer i.e. Rs. 15000
(20)
Ans. a
Explanation:
When each co-ventures record only his own transactions and no separate set of
books of A/c is maintained, to find out profit or loss memorandum, joint ventures A/c
is prepared.
(21)
Ans. c
Explanation: Since after the point of inflexion the rate of increase of TP is diminishing.
(22)
Ans. b
Explanation: Since there is inverse relationship between production and cost hence
when the returns are increasing the cost would be diminishing per unit.
(23)
Ans. d
Explanation: Production function measures the relationship between inputs and
output.
(24)
Ans. d
Explanation: Since in short run at least one factor is fixed and others are variable
because there is lack of time period.
(25)
Ans. c
Explanation: Since production is having inverse relationship with cost hence at
diminishing marginal returns there would be increasing marginal cost.
(26)
Ans. a
Explanation: Since in short run when fix factors are kept fixed and units of variable
factors are increased this law is applicable.
(27)
Ans. c
Explanation: Since the supply of land is perfectly inelastic from the point of view of
the economy.
(28)
Ans. a
Explanation: Since it operates with one fixed factor and other variable factors.
(29)
Ans. c
Explanation : AFC =
TFC
Q
Q  0

TFC  240


Given
Hence it means
TFC = 240Rs.
So AFC =
(30)
240
= 120
2
Ans. c
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Explanation : The sacrificing amount is known as Opportunity cost.
(31)
Ans. b
Explanation : AFC can never be “U”shaped because TFC is fixed and when units of
quantity are increased, AFC reduces but never increases.
(32)
Ans. b
Explanation : TFC = 1000Rs.
TVC = 500Rs.
Q = 100 units
AC = ?
TC = TFC + TVC
TC = 1000+500
TC = 1500
TC
Q
1500
 15 Ans.
Hence AC =
100
We know AC =
(33)
Ans. c
Explanation :
Output
0
1
2
3
4
5
6
Diminishing marginal returns is there
between 4 to 5 units.
TC
MC
 TC 


 Q 
240
330
90
410
80
480
70
540
60
610
70
690
80
where the marginal cost starts rising and it is
(34)
Ans. a
Explanation : LAC is called planning curve as in the long run all the factors of
production can be planned.
(35)
Ans. d
Explanation : Q = 7
ATC = 150
TFC = 350
AVC = ?
ATC = AFC + AVC
AFC =
TFC 350

 50
Q
7
AVC = ATC- AFC
AVC = 150-50 (AVC = 100)
(36)
Ans. a
Explanation :
MITTAL COMMERCE CLASSES
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ACCOUNTANT
AOUA
CCNTNT
7P 7
4


222 2
7 6  5 4  321 4  32

222
2
 42  20  9

 7560 Ans.
(37)
Ans. a
Explanation :
Physics Chemistry
5
3
P1 P2 P3 P4 P5
C1 C2 C3
M1 M2
Maths
2
All books can arranged by 3P3
And Physics books can be arranged by 5
Similarly for chemistry and Maths 3 and 2
So total ways = 3  5  3  2  8640
(38)
Ans. a
Explanation :
5 × 5 ×
(39)
4
Ans. b
Explanation :
DIRECTOR
Arrangements of all letters
8

2
Arrangements with 3 vowels together
6P
 63
2
 So. No. of arrangements when three vowels never come together
MITTAL COMMERCE CLASSES
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8 6P6
 3
2 2
 18000

(40)
Ans. d
Explanation :
Let first term and Common difference are A, D respectively.
Then b = A + (a - 1)D _____
(1)
(2)
a = A + (b -1)D _____
Solve the equations and find D = -1
A = b+a-1
So Ta+b = A + (a+b -1)D
or b + a-1+(a+b -1)× -1
= 0
(41)
Ans. b
Explanation :
If a, b, c are in A.P. then
2b = a + c
So 2x = 2x +1+ x -1
x=0
(42)
Ans. b
Explanation :
Given Question is based of sum of A.P.
a = 5000
n
d = 500 Sn = 2a + (n -1)d
2
20
n = 20 =
2 × 5000 + (20 -1) × 500 = 195000
2
(43)
Ans. d
Explanation :
a = 1 r = −2/3
Sum of infinite terms of G.P.
a
1 r
1
3

=
1  (2 / 3) 5
S 
(44)
Ans. B
Explanation:
MITTAL COMMERCE CLASSES
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Given n1  k , n 2  k
x1  16 x 2  10
Combined mean
x
=
n1x1  n 2 x 2
n1  n 2
k 16  k 10 16k  20k 36k


kk
3k
3k
= 12
(45)
Ans. C
Explanation:
In this question, we will use weighted harmonic mean.
 HM =
Wi
W
 i
Xi
w1  2, w 2  3, w 3  5 , x1  2, x 2  3, x 3  5


 235 
= 

 235
2 3 5
10
=
km / hr
111
10
=
km/hr
3
(46)
Ans. B
Explanation:


 n n 
23
2 
Combined H.M. =  1

2
3
 n1  n 2 
 H H   2  1
2 
 1
   
 5 5
5
=
5  15
5 1
=

20 4
(47)
Ans. B
Explanation:
By Hit & Trial Method Checking option (b)  2 and 4
For two numbers a and b
a b 24

3
2
2
2ab 2  2  4
and HM 

ab
2 4
AM 
=
16 8

6 3
MITTAL COMMERCE CLASSES
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(48)
Ans. C
Explanation:
The algebraic Sum of deviations of a set of observations from their AM is
Example
Xi
X  X
10
20
30
−10
0
10
0
(49)
Xi
n
10  20  30
=
3
X
i
therefore
zero.

= 20

 Xi  X  0
Ans. C
Explanation:
5X  6Y  9  0 and X  6
5  6  6Y  9  0
30  6Y  9  0
6Y  39  0
6Y  39
39
6
13
Y
2
Y
= − 6.50
(50)
Ans. B
Explanation:
Wages of 8 workers in ascending order
38, 42, 45, 47, 49, 54, 55, 56
n=8
th
th
n
n 
  value    1 value
2
2 
2
median =
th
th
8
8 
  value    1 value
2
2 
2
=
th
(4) value  (5) th value
2
=
47  49
2
=
= 48
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