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Test: Equations- 1 - CA Foundation MCQ


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30 Questions MCQ Test Quantitative Aptitude for CA Foundation - Test: Equations- 1

Test: Equations- 1 for CA Foundation 2024 is part of Quantitative Aptitude for CA Foundation preparation. The Test: Equations- 1 questions and answers have been prepared according to the CA Foundation exam syllabus.The Test: Equations- 1 MCQs are made for CA Foundation 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Equations- 1 below.
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Test: Equations- 1 - Question 1

Choose the most appropriate option (a) (b) (c) or (d)

The equation –7x + 1 = 5–3x will be satisfied for x equal to:

Detailed Solution for Test: Equations- 1 - Question 1

Take -7x to R.H.S. and 5 to L.H.S.,

Now on R.H.S., we have -3x+7x =4x and on L.H.S., we have -5+1= -4

So the equation becomes

-4 = 4x

=> x = -1

Test: Equations- 1 - Question 2

The Root of the equation 

Detailed Solution for Test: Equations- 1 - Question 2

LCM of 4 and 3 = 12 , then 3(x+4) +4(x-5)=11 × 12 then 140/7 = 20 Answer

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Test: Equations- 1 - Question 3

Pick up the correct value of x for X/30 =2/45

Detailed Solution for Test: Equations- 1 - Question 3

If we cross multiply; 

45x = 30*2
45x  =  60

x = 60/ 45 

x  = 4/3 

Test: Equations- 1 - Question 4

The solution of the equation 

Detailed Solution for Test: Equations- 1 - Question 4

Taking the L.C.M. on R.H.S. as 4, and then on cross-multiplication, we get

4(x+24) = 5(16+x)

=> x = 16

Test: Equations- 1 - Question 5

8 is the solution of the equation

Detailed Solution for Test: Equations- 1 - Question 5

In option 2,

Took LCM .. SO u get 9x+36+2x+20=144.. so 11x= 88.. x=8

Test: Equations- 1 - Question 6

The value of y that satisfies the equation 

Test: Equations- 1 - Question 7

The solution of the equation (p+2) (p–3) + (p+3) (p–4) = p(2p–5) is

Detailed Solution for Test: Equations- 1 - Question 7

(p+2)(p-3)+(p+3)(p-4)=p(2p-5)

p^2-3p+2p-6+p^2-4p+3p-12=2p^2-5p

2p^2-2p-18=2p^2-5p

-2p+5p=18

3p=18

p=6

 

Test: Equations- 1 - Question 8

Test: Equations- 1 - Question 9

Pick up the correct value x for which 

Detailed Solution for Test: Equations- 1 - Question 9

Put x=10
10/0.5-1/0.05+10/0.005-1/0.0005
= 20-20+2000-2000
=0

Hence, x=10 is the solution of the given equation.

Test: Equations- 1 - Question 10

The sum of two numbers is 52 and their difference is 2. The numbers are

Detailed Solution for Test: Equations- 1 - Question 10

Let the 2 nos. be x and x+2

Their sum is 52, so: 

52 = x + (x+2) 

52 = x + x + 2 

50 = x + x 

50 = 2x 

25 = x 

So your unknown numbers x and (x+2) are 25 and 27.

Test: Equations- 1 - Question 11

The diagonal of a rectangle is 5 cm and one of at sides is 4 cm. Its area is

Detailed Solution for Test: Equations- 1 - Question 11

Given :-

one side = a = 4cm

nd diagonal = h = 5cm

Let, the other side be 'b' 

Now,

By applying pythgoras theorem ;

 

find base,

b^2 = h^2 - p^2

b^2 = (5)^2 - (4)^2

b^2 = 25 - 16 

b^2 = 9

b = 3

So, the two sides are 4cm and 3cm 

Then ;

Area of rect. = length * breadth

ar. = 4 * 3

ar. = 12 cm^2

Test: Equations- 1 - Question 12

Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. The parts are.

Detailed Solution for Test: Equations- 1 - Question 12

Let the first part be x
other part is 56 - x
three times the first part exceeds one third of the second party by 48
3x = 1/3(56-x) + 48
9x = 56-x + 144
9x + x = 200
10x = 200
x = 20

First part, x =  20
Second part, 56 - x = 36

Test: Equations- 1 - Question 13

The sum of the digits of a two digit number is 10. If 18 be subtracted from it, the digits in the resulting number will be equal. The number is

Detailed Solution for Test: Equations- 1 - Question 13

x = 10's digit of a two digit number(Let)

 y = units digit(Let)

n=Resulting equal digits(Let)

Condition 1-the sum of a two digit number is 10

x + y = 10

further simplifying

y = (10-x)

 Condition 2-if 18 be subtracted from it the digits in the resulting number will be equal

10x + y - 18 = 10n + n

10x + y = 11n+ 18

Substitutituting values found above

10x + (10-x) = 11n + 18

10x - x = 11n + 18 - 10

9x = 11n+ 8

 

If we solve it we will found the values as - x = 7, y = 3, z = 5 and finally we can say the no. is 73.

 

Test: Equations- 1 - Question 14

The fourth part of a number exceeds the sixth part by 4. The number is

Detailed Solution for Test: Equations- 1 - Question 14

Let the number be 'x'.
By the problem,
(x/4)=(x/6)+4
⇒(x/4)-(x/6)=4
⇒(3x-2x)/12=4
⇒x/12=4
⇒x=4×12
⇒x=48.
Therefore, the number is 48.

Verification:
48/4=48/6+4
12=8+4
12=12
∴LHS=RHS.
So,the number is 48.
 

Test: Equations- 1 - Question 15

Ten years ago the age of a father was four times of his son. Ten years hence the age of the father will be twice that of his son. The present ages of the father and the son are.

Detailed Solution for Test: Equations- 1 - Question 15

F-10 = 4(S-10) on solving this one gets F - 4S = -30....1

F = 10 = 2(S+10) on solving this one gets F - 2S = 10....2

solve both the equation and you get F= 50 S = 20

Test: Equations- 1 - Question 16

The product of two numbers is 3200 and the quotient when the larger number is divided by the smaller is 2.The numbers are

Detailed Solution for Test: Equations- 1 - Question 16

Let two no is x and y X›y

X*y=3200

x/y=2

x=2y

2y*y=3200

2y^2=3200

y^2=1600

y=✓1600

y=40

then x=2*40

x=80

Test: Equations- 1 - Question 17

The denominator of a fraction exceeds the numerator by 2. If 5 be added to the numerator the fraction increases by unity. The fraction is.

Detailed Solution for Test: Equations- 1 - Question 17

Let numerator be: x

Let denominator be: x+2

fraction = x/(x+2)

(x+5)/(x+2) = x/(x+2) + 1

(x+5)/(x+2) - x/(x+2) = 1

(x+5 -x)/(x+2)= 1

5/(x+2) = 1

5 = x+2

5-2 = x

x = 3

The fraction is 3/5.

Test: Equations- 1 - Question 18

Three persons Mr. Roy, Mr. Paul and Mr. Singh together have Rs. 51. Mr. Paul has Rs. 4 less than Mr. Roy and Mr. Singh has got Rs. 5 less than Mr. Roy. They have the money as.

Detailed Solution for Test: Equations- 1 - Question 18

Mr. Roy has = x

Mr. Pual has = x-4

Mr. Singh has = x-5 

Together they have = 51

Mr. Roy + Mr. Pual + Mr. Singh = 51

x+(x-4)+(x-5)=51

3x-9=51

3x=51+9

3x=60

divide by 3

3x/3=60/3

x=20 

 

Mr. Roy has = x = 20

Mr. Pual has = x-4 = 20-4 = 16

Mr. Singh has = x-5 = 20-5= 15 

20+16+15=51

51=51

Test: Equations- 1 - Question 19

A number consists of two digits. The digits in the ten’s place is 3 times the digit in the unit’s place. If 54 is subtracted from the number the digits are reversed. The number is

Detailed Solution for Test: Equations- 1 - Question 19

Let the digit at Tens place be x

The digit at unit place be y

The Number is 10x + y

ACCORDING TO THE QUESTION :-

x = 3y ... eq. 01

According to the second condition :

10x + y - 54 = 10y + x

=> 9x - 9y = 54

=> 9(x - y) = 9 × 6

=> x - y = 6 ... eq.02

Now we can solve the Equation 01 and 02 by substitution Method :-

x - y = 6

=> 3y - y = 6 [ from eq.01 , x = 3y ]

=> 2y = 6

=> y = 3

Now we can substitute the value of y in eq. 01 to get value of x

x = 3y

=> x = 3 × 3

=> x = 9

Hence the number is

Hence the number is 10x + y = 90 + 3 = 93 (ANS)

Test: Equations- 1 - Question 20

One student is asked to divide a half of a number by 6 and other half by 4 and then to add the two quantities. Instead of doing so the student divides the given number by 5. If the answer is 4 short of the correct answer then the actual answer is

Detailed Solution for Test: Equations- 1 - Question 20

Test: Equations- 1 - Question 21

If a number of which the half is greater than 1/5th of the number by 15 then the number is

Detailed Solution for Test: Equations- 1 - Question 21

Let it be x

x/2-x/5=15

3x/10=15

x/10=5

x=50

Test: Equations- 1 - Question 22

The solution of the set of equations 3x + 4y = 7, 4x – y = 3 is

Detailed Solution for Test: Equations- 1 - Question 22

Here 3x+4y=7.........(1)

4x-y=3........(2)

eq(1) * 1 --- 3x +4y =7
.
eq(2) * 4--- 16x -4y =12

adding eq(1) with eq(2)

................................................................
19x=19

=>x=19/19=1 ........

from eq(1) 3x + 4y=7

=>3*1 + 4y=7

=>4y = 7-3=4

=> y= 4/4=1 .

Test: Equations- 1 - Question 23

The values of x and y satisfying the equations  are given by the pair.

Test: Equations- 1 - Question 24

 are satisfied by the values given by the pair.

Detailed Solution for Test: Equations- 1 - Question 24

According to the given question,

i ) x + y = p + q

x = p + q - y ---------( 1 )

ii ) x / p + y / q = 2

Multiply each term with pq we get

qx + py = 2pq ---------( 2 )

Put x value , from ( 1 ) in equation ( 2)

q( p + q - y) + py = 2pq

pq + q^2 - qy + py = 2pq

- qy + py = 2pq - pq - q^2

y ( - q + p) = pq - q^2

y ( p - q ) = q ( p - q )

y = q ( p - q )/ ( p - q )

y = q -----( 3 )

Put y = q in ( 1 )

x = p + q - q

x = p

Therefore ,

x = p and y = q

Test: Equations- 1 - Question 25

The solution for the pair of equations 

Detailed Solution for Test: Equations- 1 - Question 25

Put x=1/4 and y=1/3 in equation only option a is satisfied the equation LHS=RHS

Test: Equations- 1 - Question 26

 The values of x and y are given by the pair.

Detailed Solution for Test: Equations- 1 - Question 26

By option checking,

3xy = 10(y- x) so 3.2.5 =10(5-2)30=30

Test: Equations- 1 - Question 27

The pair satisfying the equations x + 5y = 36, 

Test: Equations- 1 - Question 28

Solve for x and y : x–3y =0, x+2y = 20. The values of x and y are given as

Detailed Solution for Test: Equations- 1 - Question 28

given here

x-3y = 0...(1)

x+2y = 20...(2)

subtract equations (1) and (2)

we get ,

-5y = -20

=> y = (-20)/(-5)

=> y = 4

by (1).

x-3×4= 0

=> x= 12

Thus:-

Value of X = 12

Value of Y = 4

Test: Equations- 1 - Question 29

The simultaneous equations 7x–3y = 31, 9x–5y = 41 have solutions given by

Test: Equations- 1 - Question 30

1.5x + 2.4 y = 1.8, 2.5(x+1) = 7y have solutions as

Detailed Solution for Test: Equations- 1 - Question 30

I want to show you a good method to solve these equations.  Take time to read it and follow it.  This is the method in the International standards.  It is easy when you understand it.

  1.5  x  + 2.4 y  -  1.8  =  0 

  2.5  x  -  7 y   + 2.5   =  0

For solving :  a x + b y + c  =  0     and    d x + e y + f  = 0 ,  we do as mentioned in the diagram enclosed.  This is the easier way.

   -1.8      1.5      2.4      -1.8

   2.5      2.5        -7       2.5

        y                    x

x =   (-7*-1.8  -  2.4*2.5 ) / (2.5 * 2.4  - 1.5*-7)

   = (12.6  -  6) / (6 + 10.5)  =  6.6/16.5 =  0.4  = 2/5

  y =  (2.5*1.5 -  -1.8*2.5) / (2.5*2.4 - 1.5*-7) 

   = (3.75 + 4.5) / 16.5    =  8.25 / 16.5  = 0.5 = 1/2

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