CA Foundation Maths - Test Series
Chapter 1
Question 1
Anand earns Rs 80 in 7 hours and Promode Rs 90 in 12 hours. The ratio of their earnings is
A
32:21
B
23:12
C
8:9
D
None of these
SOLUTION
32:21
Hint :
Anand earns Rs. 80 in 7 hours, so his hourly earning is 80/7.
Promode earns Rs. 90 in 12 hours, so his hourly earning is 90/12.
The ratio of their hourly earnings is 80/7 : 90/12 = 32 : 21.
Therefore, the ratio of their total earnings is 32 : 21.
So the answer is (1).
CORRECT OPTION
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A
B
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16.16 %
37.37 %
Question 2
If ap = bq = cr = ds and ab = cd then the value of
A
1/a
B
1/b
C
0
D
1
+ + + reduces to
SOLUTION
0
Hint : To answer option 3, "0," for the value of 1/p + 1/q + 1/r + 1/s, we need to find a condition or set of values that would make the expression
equal to 0. Let's analyze the given equations again:
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CA Foundation Maths - Test Series
ap = bq = cr = ds ---(1)
ab = cd ---(2)
From equation (2), we can rewrite it as:
a/b = c/d
This implies that a/b and c/d are equivalent fractions. For two fractions to be equal, their numerators and denominators must be proportional.
Therefore, we can say:
a = kc
b = kd
where k is a constant.
Now, substituting these values into equation (1):
=ap = bq = cr = ds
kc * p = kd * q = cr = ds
Dividing the first equation by kc and the second equation by kd, we get:
p = q = c/d * r/k = s
Therefore, if p, q, r, and s are all equal, the expression 1/p + 1/q + 1/r + 1/s becomes:
1/p + 1/q + 1/r + 1/s = 1/p + 1/p + 1/p + 1/p = 4/p
If p, q, r, and s are all equal, the expression simplifies to 4/p.
For the expression to equal 0, we need p to be infinite or p to be equal to 0. However, since we cannot divide by 0, p cannot be 0. Therefore, the
only possibility is p being infinite.
In conclusion, to answer option 3, "0," for the value of 1/p + 1/q + 1/r + 1/s, p must be infinite.
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C
D
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Question 3
if a2 + b2 = 7ab then the value of log
A
0
B
1
C
-1
D
7
-
-
SOLUTION
0
Hint : To find the value of the expression log((a+b)/3) - log(a/2) - log(b/2), we can simplify it using logarithm rules and the given equation.
First, let's simplify each term individually:
log((a+b)/3) = log(a+b) - log(3)
log(a/2) = log(a) - log(2)
log(b/2) = log(b) - log(2)
Substituting these values back into the expression:
log((a+b)/3) - log(a/2) - log(b/2) = (log(a+b) - log(3)) - (log(a) - log(2)) - (log(b) - log(2))
Using the property log(a) - log(b) = log(a/b), we can rewrite the expression as:
= log(a+b) - log(3) - log(a) + log(2) - log(b) + log(2)
= log(a+b) - log(a) - log(b) - log(3) + log(2) + log(2)
Now, let's substitute the given equation a^2 + b^2 = 7ab:
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CA Foundation Maths - Test Series
a^2 + b^2 = 7ab
Rearranging the terms, we get:
a^2 - 7ab + b^2 = 0
This equation can be factored as:
(a - b)(a - 7b) = 0
So, either (a - b) = 0 or (a - 7b) = 0.
Case 1: (a - b) = 0
In this case, a = b.
Substituting this back into the expression:
log(a+b) - log(a) - log(b) - log(3) + log(2) + log(2) = log(2a) - log(a) - log(a) - log(3) + log(2) + log(2)
= log(2a) - 2log(a) - log(3) + 2log(2)
= log(2a) - 2log(a) + 2log(2) - log(3)
= log(2a) - log(a^2) + log(4) - log(3)
= log(2a/a^2) + log(4/3)
= log(2/a) + log(4/3)
= log((2/a) * (4/3))
= log(8/3a)
Now, we can see that this expression is not necessarily equal to 0, 1, -1, or 7.
Case 2: (a - 7b) = 0
In this case, a = 7b.
Substituting this back into the expression:
log(a+b) - log(a) - log(b) - log(3) + log(2) + log(2) = log(8b) - log(7b) - log(b) - log(3) + log(2) + log(2)
= log(8b) - 2log(b) - log(3) + 2log(2)
= log(8b) - log(b^2) - log(3) + log(4)
= log(8b/b^2) + log(4/3)
= log(8/b) + log(4/3)
= log((8/b) * (4/3))
= log(32/3b)
Again, we can see that this expression is not necessarily equal to 0,
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15.15 %
75.25 %
Question 4
The ratio compounded of duplicate ratio of 4 : 5, triplicate ratio of 1: 3, sub-duplicate ratio of 81: 256 and sub-triplicate ratio of 125: 512 is
A
4:512
B
3:32
C
1:12
D
None of these
SOLUTION
None of these
Let us find the individual ratios first.
Duplicate ratio of 4: 5 is (4)2 : (5)2 = 16:25
Triplicate ratio of 1:3 is (1)3 : (3)3 = 1: 27
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CA Foundation Maths - Test Series
Sub-duplicate ratio of 81 : 256 is
:
Sub-triplicate ratio of 125: 512 is
= 9 : 16
:
= 5 :8
We know that,
The ratio compounded of the two ratios a : b and c:d is ac: bd.
Therefore, the required compounded ratio =
=
=
=
= 1: 120
Hence, the correct answer is none of these.
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15.66 %
43.94 %
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Question 5
4, *, 9, 13 are in proportion. Then * is
A
6
B
8
C
9
D
None of these
SOLUTION
6
Hint:Given: 4, *, 9, 1312
are in proportion
Let * is represented by x
Now, by cross product rule
Product of extremes = Product of means
i.e., 4 × 13 =9x
→4
=9×x
→ X=
→X = 6
Therefore, the value of * is 6.
Hence, the correct answer is option (1) i.e., 6.
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CA Foundation Maths - Test Series
1
0.25
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45.45 %
Question 6
If A: B = 3: 2 and B: C = 3: 5, then A : B : C is
A
9:6:10
B
6:9:10
C
10: 9: 6
D
None of these
SOLUTION
9:6:10
Hint (1)
Given, A : B = 3: 2 and B : C = 3: 5
i.e.,
and =
and
=
=
and =
= A: B=9: 6 and B: C = 6:10
= A: B: C=9:6:10
Hence, the correct answer is option (1) i.e., 9:6:10
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40.4 %
Question 7
Two numbers are in the ratio 7: 8. If 3 is added to each of them, their ratio becomes 8 : 9. The numbers are:
A
14,16
B
24,27
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CA Foundation Maths - Test Series
C
21,24
D
16,18
SOLUTION
21,24
Hint: Go by choices
(2) and (4) are not in the ratio 7 : 8 So (2) & (4) are not answer For (1) it is added then
=
(a) is not answer
(c) is answer Detail Method:
Let x is common in the ratio
∴ Numbers are 7x & 8x
Now
=
or 64x + 24 = 63x + 27
or 64x – 63x = 27-24
or x = 3
1st number = ix = 7 3 = 21
2nd number = 8x = 8 3 = 24
(3) is correct
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40.91 %
Question 8
Ratio of earnings of A and B is 4 : 7. If the earnings of A increase by 50% and those of B decrease by 25%, the new ratio of their earning becomes 8
: 7. What is A’s earning ?
A
21,000
B
26,000
C
28,000
D
Data inadequate
SOLUTION
Data inadequate
Hint: Detailed Method Let x is common in the ratio
∴ A’s and B’s present earnings are 4x and 7x respectively
From question
or
=
=
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CA Foundation Maths - Test Series
x cannot be found.
Data is inadequate
∴ (4) is Correct
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20 %
18.33 %
61.67 %
Question 9
In 40 litres mixture of glycerine and water, the ratio of glycerine and water is 3:1. The quantity of water added in the mixture in order to make this
ratio 2:1 is :
A
15 litres
B
10 litres
C
8 litres
D
5 litres
SOLUTION
5 litres
Hint: Glycerine =
Water =
= 30 litres.
x 1 = 10 litres 4
Let x litres of water is added to the mixture
Then
=
or, 2x + 20 = 30
or x = 5
∴ (4) is Correct
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108 secs
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22.73 %
22.22 %
55.05 %
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CA Foundation Maths - Test Series
Question 10
If A: B= 2: 5, then (10A + 3B) : (5A + 2B) is equal to:
A
7: 4
B
7: 3
C
6: 5
D
7: 9
SOLUTION
7: 4
Hint: It A : B = 2 : 5 Then
=
= =
= 7: 4
(1) is Correct
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56.57 %
2.02 %
41.41 %
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Question 11
Which of the numbers are not in proportion?
A
6, 8, 5, 7
B
7, 14, 6, 12
C
18, 27,12, 18
D
8, 6, 12, 9
SOLUTION
6, 8, 5, 7
Hint: (1) Go by choices
For (1) =
(1) is not in proportion
CORRECT OPTION
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A
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42.42 %
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46.97 %
Question 12
If the ratio of two numbers is 7 : 11. If 7 is added to each number then the new ratio will be 2 : 3 then the numbers are.
A
49,77
B
42,45
C
43,42
D
39,40
SOLUTION
49,77
GBC (Go by Choices) (1)
Hint: 49 7 = 7
77 11 = 7 both must be equal.
Here it is correct.
Now:
= =
Divide 56 by numerator (2) and 84 by Denominator (3) we get same value “28”
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55.42 %
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41.25 %
Question 13
is the duplicate ratio of
then find the value of x :
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CA Foundation Maths - Test Series
A
6
B
2
C
5
D
9
SOLUTION
6
Given
=
=
Hint: Go by choices
for option (a) putting x = 6 in LHS; we get
=
∴ (1) is correct.
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44.44 %
5.56 %
50 %
Question 14
If (a + b): (b + c): (c + a) = 7 : 8 : 9 and a + b + c = 18 then a : b : c = .
A
5:4:3
B
3:4:5
C
4:3:5
D
4: 5 : 3
SOLUTION
4:3:5
Hint: Go by choices.
(3) Let a : b : c = 4 : 3 : 5
It is in ratio. So, it should must satisfy given ratio (a + b): (b + c): (c + a) = 7 : 8 : 9
i.e. (4 + 3): (3 + 5): (5 + 4) = 7 : 8 : 9 (True) Avoid 2nd condition.
In detail it will take too much time.
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1
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11.11 %
73.23 %
Question 15
A bag contains 23 number of coins in the form of 1 rupee, 2 rupee and 5 rupee coins. The total sum of the coins is
and 2 rupees coins is 3 : 2. Then the number of 1 rupee coins is:
A
12
B
8
C
10
D
16
43. The ratio between 1 rupee
SOLUTION
12
Hint: Go by choices
Let option (1) is correct.
Let x is common in the ratio.
So, 1 coins = 3x = 12 ; So, x = 4
No. of 2 coins = 2 4 = 8
Hence no. of coins of 5 coins = 23 – 12 – 8= 3
Total money = 12 1 + 8 2 + 3 5 = 43
Satisfied. So (1) is correct.
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31.25 %
5.42 %
63.33 %
Question 16
x, y, z together starts a business, if x invests 3 times as much as y invests and y invests two third of what z invests, then the ratio of capitals of x, y, z
is:
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CA Foundation Maths - Test Series
A
3:9:2
B
6: 3: 2
C
3:6:2
D
6: 2: 3
SOLUTION
6: 2: 3
Hint: Go by choices
6 = 3 2 and 2 = 3 2/3
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108 secs
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33.33 %
13.13 %
53.54 %
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Question 17
(2p2 – q2) = 7pq, where p, q are positive then p : q.
A
5:6
B
5:7
C
3:5
D
3:7
SOLUTION
5:6
5:6 is correct 15(2p2 – q2) = 7pq
Hint: Go by choices
For (a) put p = 5; q = 6 we get
15[2 52 – 62] = 3 5 6
or 15 14 = 210
or 210 = 210
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CA Foundation Maths - Test Series
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80.81 %
Question 18
The first, second and third month salaries of a person are in the ratio 2:4:5. The difference between the product of the salaries of first 2 months &
last 2 months is 4,80,00,000. Find the salary of the second month
A
4,000
B
6,000
C
12,000
D
8,000
SOLUTION
8,000
Hint: Let x is common in the ratio.
1st, 2nd and 3rd month salaries of a person = 2x ; 4x ; 5x
From Qts.
4x 5x – 2x 4x = 4,80,00,000.
or, 12x2 = 4,80,00,000.
or, x2 = 4000000
x = 2000.
2nd month salary = 4x = 4 2000
= 8000
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24.24 %
8.59 %
67.17 %
Question 19
If X varies inversely as square of Y and given that Y=2 for X = 1, then the value of X for Y =6 will be:
A
3
B
9
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CA Foundation Maths - Test Series
C
1/3
D
6
SOLUTION
6
⇒ x = K.
x
x=k
, where k = proportional constant
where k = proportional constant
When x = 1 Then y = 2
⇒k=4∴x=
1=
When y = 6, Then x =
=
x=
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SUGGESTED TIME
YOUR TIME
AVERAGE TIME
90 secs
0 secs
42.7 secs
AVERAGE CORRECT
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5%
32.5 %
62.5 %
Question 20
If
A
= 0; then find the value of
=
9abc
B
C
abc
D
SOLUTION
abc
Hint:
Let a = -1; b= -1 and c =8,
Because = -1 – 1 + 2 = 0(R.H.S) ∴
= (2)3 = 8
=
= (-1).(-1).(8) = abc
∴ (3) is correct
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1
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SUGGESTED TIME
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108 secs
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25.14 secs
AVERAGE CORRECT
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11.11 %
20.2 %
68.69 %
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B
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Question 21
Simplification of
A
xm
B
x-m
C
xn
D
x-n
is :
SOLUTION
x-m
Reason
= xm+3n+4m-9n-6m+6n = x-m
Question 22
On simplification
+
+
reduces to :
A
B
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CA Foundation Maths - Test Series
C
1
D
0
SOLUTION
1
Hint:
1/1+Za b+Za c+1/1+Zb c+Zb a+1/1+Zc a+Zc a
= 1 [it is in cyclic order]
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27.78 %
11.11 %
61.11 %
Question 23
The recurring decimal 2.7777 ………. can be expressed as:
A
24/9
B
22/9
C
26/9
D
25/9
SOLUTION
25/9
Reason
Go by choices.
By calculator
= 2.666 ………………….
= 2.444 ………..
2.777
2.777
= 2.888 ……………
2.77
= 2.777
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SUGGESTED TIME
YOUR TIME
AVERAGE TIME
108 secs
0 secs
16.04 secs
AVERAGE CORRECT
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AVERAGE SKIPPED
48.48 %
2.02 %
49.49 %
CORRECT OPTION
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21.72 %
9.09 %
69.19 %
Question 24
The value of
A
1/5
B
1/6
C
1/4
D
1/9
is equal to:
SOLUTION
1/6
Hint: Put n = 0
=
=
=
Detailed method
=
Question 25
If 2x =
=
then
A
By Prof. Virendra Singh Thakur
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CA Foundation Maths - Test Series
B
C
D
None
SOLUTION
Hint:
∵ 2x2 = 3y2 = 12z2 ………..(1) (Given)
Factorize 12 in terms of 2 & 3. We get
22 31 = 121 ……….(2)
Always write as power of base of (2) Power on same base of 1 ; put “+”
Sign at the place of “ ” Sign. So;
So (c) is correct.
Let
=
So; 2=
=
; 3=
=
; 12=
Now
12 =
Or
Or
OR
CORRECT OPTION
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C
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SUGGESTED TIME
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AVERAGE TIME
120 secs
0 secs
31.27 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
4.58 %
27.5 %
67.92 %
Question 26
If u5x = v5y = w5z and u2 = vw then xy + zx – 2yz
A
0
B
1
By Prof. Virendra Singh Thakur
18
CA Foundation Maths - Test Series
C
2
D
None of these
SOLUTION
0
Reason
u5x = v5y = w5z ⇒ ux = vy = wz
Tricks : See Quicker BMLRS Chapter : Indices
u2 = vw;
∴
=
or ; xy + zx = 2yz
or; xy + zx -2yz = 0
CORRECT OPTION
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A
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1
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AVERAGE TIME
108 secs
0 secs
28.04 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
16.16 %
8.59 %
75.25 %
Question 27
If abc = 2, then the value of
A
1
B
2
C
D
SOLUTION
1
Reason
“Put a = 1, b = 2 & c = 1. So that abc = 2” in the given question. We get
CORRECT OPTION
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A
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By Prof. Virendra Singh Thakur
19
CA Foundation Maths - Test Series
1
0.25
Medium
SUGGESTED TIME
YOUR TIME
AVERAGE TIME
108 secs
0 secs
19.52 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
19.7 %
6.06 %
74.24 %
Question 28
The value of
A
y
B
-1
C
1
D
None
SOLUTION
1
We have:
Code snippet
(y^a/y^b )^ (a^2+ab+b^2 ) (y^b/y^c )^(b^2+bc+c^2 ) (y^c/y^a )^(c^2+ca+a^2 ) =
y^(a^2+ab+b^2) / y^(b^2+bc+c^2) * y^(b^2+bc+c^2) / y^(c^2+ca+a^2) * y^(c^2+ca+a^2) / y^(a^2+ab+b^2)
Use code with caution. Learn more
Since the base is the same, we can cancel the terms:
Code snippet
= y^(a^2+ab+b^2 - b^2-bc-c^2) * y^(b^2+bc+c^2 - c^2-ca-a^2) * y^(c^2+ca+a^2 - a^2-ab-b^2)
Use code with caution. Learn more
The terms in the exponents cancel out, so the expression is equal to 1.
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AVERAGE TIME
108 secs
0 secs
16.55 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
38.38 %
2.53 %
59.09 %
Question 29
By Prof. Virendra Singh Thakur
20
CA Foundation Maths - Test Series
If 3x = 5y = (75)z then
A
B
C
D
None
SOLUTION
Reason
See Short Cut Tricks Book “QUICKER BMLRS”
3x = 5y = (75)z ………(1)
31 52 = 751 ………(2)
Tricks:
Power of (2) power of (1)
and put + sign at the place of “ ”
We get
CORRECT OPTION
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A
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AVERAGE TIME
108 secs
0 secs
12.69 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
20.71 %
12.12 %
67.17 %
Question 30
A
x-(ap+bq+cr)
B
xa+b+c
By Prof. Virendra Singh Thakur
21
CA Foundation Maths - Test Series
C
x(ap+bq+cr)
D
xabc
SOLUTION
xa+b+c
Reason
…………=
Similarly doing as above; we get
= xa.xb.xc
= xa+b+c
CORRECT OPTION
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B
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AVERAGE TIME
120 secs
0 secs
17.82 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
6.57 %
9.09 %
84.34 %
Question 31
7log
A
0
B
1
C
log 2
D
log 3
+ 5log
+ 3log
is equal to
SOLUTION
log 2
Reason
7log
= log+
+ 5log
log
+ 3log
+ log
= log
Calculation Tricks
I Type i6 15 button Then push = button 6 time then push M+ button.
By Prof. Virendra Singh Thakur
22
CA Foundation Maths - Test Series
II Type 25 24 button Then push = button 4 times
III Then Push button Then MRC button 2 time Then Push M+ button
IV Type 81 80 button Then = button 2 times
V Then Push button and Then MRC button 2 times Then = button
We get; it is approx 2
So. Value = log 2
CORRECT OPTION
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C
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0.25
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AVERAGE TIME
120 secs
0 secs
10.48 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
11.11 %
12.12 %
76.77 %
CORRECT OPTION
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A
Skipped
0
MARKS
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LEVEL
1
0.25
Easy
SUGGESTED TIME
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AVERAGE TIME
90 secs
0 secs
27.47 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
23.74 %
13.64 %
62.63 %
Question 32
The value of the expression
A
t
B
abcdt
C
(a + b + c + d + t)
D
None
SOLUTION
t
Hint:
alogab.logbc.logcd.logdt
= a1logat = t1 = t
(1) is correct
By Prof. Virendra Singh Thakur
23
CA Foundation Maths - Test Series
Question 33
+
A
0
B
1
C
2
D
-1
+
is equal to:
SOLUTION
2
Hint:
1logab(abc)
+1logbc(abc)
+1logca(abc)
= logabc (abJ)c.ca) = logabc (abc)2
= 2 logabc (abc) = 2 1 = 2
∴ (3) is correct
CORRECT OPTION
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C
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AVERAGE TIME
120 secs
0 secs
14.09 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
8.08 %
30.3 %
61.62 %
Question 34
log 144 is equal to:
A
2 log 4 + 2 log 2
B
4 log 2 + 2 log 3
C
3 log 2 + 4 log 3
D
3 log 2 -4 log 3
SOLUTION
4 log 2 + 2 log 3
Reason
log 144 = log(16 9)
= log 16 + log 9
= log 24 +log 32
= 4 log 2 + 2 log 3
By Prof. Virendra Singh Thakur
24
CA Foundation Maths - Test Series
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B
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AVERAGE TIME
108 secs
0 secs
19.12 secs
AVERAGE CORRECT
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AVERAGE SKIPPED
32.83 %
7.58 %
59.6 %
CORRECT OPTION
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A
Skipped
0
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LEVEL
1
0.25
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SUGGESTED TIME
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AVERAGE TIME
108 secs
0 secs
22.42 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
22.22 %
11.62 %
66.16 %
Question 35
log (m+n) = log m+ log n, m can be expressed as :
A
m=
B
m=
C
m=
D
m=
SOLUTION
m=
Reason
log(m + n) = log m + log n
or log (m + n) = log(mn)
or m + n = mn
or m – mn = -n
or m (1 – n) = -n
or m =
By Prof. Virendra Singh Thakur
25
CA Foundation Maths - Test Series
Question 36
The value of 2log x + 2 log x2 + 2 log x3 + ………. + 2 log xn will be:
A
B
n(n + 1)log x
C
n2 log x
D
None of these
SOLUTION
n(n + 1)log x
Reason
Detail Method:
2 log x + 2 log x2 + 2logx3 + ………….. + 2 log xn
=2 log x + 2.2 logx + 2.3logx + ……….. + 2.n.logx
= 2 logx. [1 + 2 + 3 + ………. + n]
= 2 log x.
= n(n+1)logx
CORRECT OPTION
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B
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AVERAGE TIME
108 secs
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21.07 secs
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AVERAGE WRONG
AVERAGE SKIPPED
16.67 %
15.66 %
67.68 %
Question 37
(512):
A
128:61
B
2:3
C
3 :2
D
None
324 =
SOLUTION
3 :2
Reason
By Prof. Virendra Singh Thakur
26
CA Foundation Maths - Test Series
Calculator Tricks
(512) = 5 + 1
Type 2 2 button = button.
Then press x button then continue pressing = button until to get 512
Here = button has been pressed 5 times. So log value
= No. of = button pressings + 1
Similarly For
324
Type 3 2 button = button then
x = button 3 times ; we get
324 value = 3 + 1=4
So;
(512):
324
= 6: 4 = 3: 2
CORRECT OPTION
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C
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SUGGESTED TIME
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AVERAGE TIME
120 secs
0 secs
18.85 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
20.83 %
13.33 %
65.83 %
CORRECT OPTION
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YOU GOT
C
Skipped
0
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LEVEL
1
0.25
Easy
SUGGESTED TIME
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AVERAGE TIME
90 secs
0 secs
26.77 secs
Question 38
log2 log2 log216 = ?
A
0
B
3
C
1
D
2
SOLUTION
1
Reason
log2 log2 log216 = log2 log24 = log22 = 1
By Prof. Virendra Singh Thakur
27
CA Foundation Maths - Test Series
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
34.34 %
8.08 %
57.58 %
CORRECT OPTION
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YOU GOT
C
Skipped
0
MARKS
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LEVEL
1
0.25
Medium
SUGGESTED TIME
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AVERAGE TIME
108 secs
0 secs
27.85 secs
AVERAGE CORRECT
AVERAGE WRONG
AVERAGE SKIPPED
8.59 %
15.15 %
76.26 %
Question 39
If log3[log4(logxx)] = 0 then X =
A
4
B
8
C
16
D
32
SOLUTION
16
Hint: GBC
for option (3) log3[log4(log2x)]
= log3[log4(log216)]
= log3(log44) = log31 = 0
Question 40
Given log 2 = 0.3010 and log 3 = 0.4771 then the value of log 24
A
1.3081
B
1.1038
C
1.3801
By Prof. Virendra Singh Thakur
28
CA Foundation Maths - Test Series
D
1.8301
SOLUTION
1.3801
Reason
Calculator Tricks:
Type 24 then button 19 times – 1 x 227695 = button. We will get the required value of log 24.
CORRECT OPTION
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YOU GOT
C
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1
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SUGGESTED TIME
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AVERAGE TIME
108 secs
0 secs
25.28 secs
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AVERAGE WRONG
AVERAGE SKIPPED
30.81 %
7.07 %
62.12 %
By Prof. Virendra Singh Thakur
29