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This mock test of Test: Equations- 2 for CA Foundation helps you for every CA Foundation entrance exam.
This contains 40 Multiple Choice Questions for CA Foundation Test: Equations- 2 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Equations- 2 quiz give you a good mix of easy questions and tough questions. CA Foundation
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QUESTION: 1

3x–4y+70z = 0, 2x+3y–10z = 0, x+2y+3z = 13

Solution:

QUESTION: 2

Monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenses are in the ratio 7 : 9. If each saves Rs. 50 per month find their monthly incomes.

Solution:

QUESTION: 3

Find the fraction which is equal to 1/2 when both its numerator and denominator are increased by 2. It is equal to 3/4 when both are incresed by 12.

Solution:

QUESTION: 4

The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages. Find his present age.

Solution:

Let the age of father be x and that of his son be a,b

We have x=2(a+b) and we have been given that 5 years ago we had the following situation which is given by the equation x-5=3(a+b-10)

On solving the above equations we obtain

a+b=25 thus the age of father is 50

QUESTION: 5

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed find the number.

Solution:

QUESTION: 6

The wages of 8 men and 6 boys amount to Rs. 33. If 4 men earn Rs. 4.50 more than 5 boys determine the wages of each man and boy.

Solution:

QUESTION: 7

A number consisting of two digits is four times the sum of its digits and if 27 be added to it the digits are reversed. The number is :

Solution:

QUESTION: 8

Of two numbers, 1/5th of the greater is equal to 1/3rd of the smaller and their sum is 16.The numbers are:

Solution:

QUESTION: 9

Y is older than x by 7 years 15 years back X’s age was 3/4 of Y’s age. Their present ages are:

Solution:

QUESTION: 10

The sum of the digits in a three digit number is 12. if the digits are reversedthe number is increased by 495 but reversing only of ten’s and the unit digits increases the number by 36

Solution:

QUESTION: 11

Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and 1/3 of the smaller and 1/5 of the greater number are together 21. The numbers are :

Solution:

QUESTION: 12

The demand and supply equations for a certain commodity are 4q + 7p = 17 and respectively where p is the market price and q is the quantity then the equilibrium price and quantity are:

Solution:

QUESTION: 13

If the roots of the equation 2x^{2 }+ 8x – m^{3} = 0 are equal then value of m is

Solution:

QUESTION: 14

If 2^{2x + 3} – 3^{2}. 2^{x} + 1 = 0 then values of x are

Solution:

QUESTION: 15

Solution:

QUESTION: 16

If œ ß be the roots of the equation 2x^{2} – 4x – 3 = 0 the value of α^{2} + β^{2} is

Solution:

QUESTION: 17

If the sum of the roots of the quadratic equation ax^{2} + bx + c = 0 is equal to the sum of the squares of their reciprocals then

Solution:

QUESTION: 18

The equation x^{2} –(p+4)x + 2p + 5 = 0 has equal roots the values of p will be.

Solution:

QUESTION: 19

The roots of the equation x2 + (2p–1)x + p^{2} = 0 are real if.

Solution:

QUESTION: 20

If x = m is one of the solutions of the equation 2x^{2} + 5x – m = 0 the possible values of m are

Solution:

QUESTION: 21

If p and q are the roots of x^{2 }+ x + 1 = 0 then the values of p^{3} + q^{3} becomes

Solution:

QUESTION: 22

If L + M + N = 0 and L M N are rationals the roots of the equation (M+N–L) x^{2} + (N+L–M)x + (L+M–N) = 0 are

Solution:

QUESTION: 23

If ∝ and ß are the roots of x^{2} = x+1 then value of

Solution:

QUESTION: 24

If p ≠ q and p^{2} = 5p – 3 and q^{2} = 5q – 3 the equation having roots as

Solution:

QUESTION: 25

If one rot of 5x^{2} + 13x + p = 0 be reciprocal of the other then the value of p is

Solution:

QUESTION: 26

A solution of the quadratic equation (a+b–2c)x^{2} + (2a–b–c)x + (c+a–2b) = 0 is

Solution:

QUESTION: 27

If the root of the equation x^{2}–8x+m = 0 exceeds the other by 4 then the value of m is

Solution:

QUESTION: 28

The values of x in the equation 7(x+2p)^{2} + 5p^{2} = 35xp + 117p^{2} are

Solution:

QUESTION: 29

The solutions of the equation

Solution:

QUESTION: 30

The satisfying values of x for the equation

Solution:

QUESTION: 31

Time taken by a train to slow down from 80 kmh^{-1} to 20 kmh^{-1} with a uniform deceleration of 2 ms^{-2} is

Solution:

QUESTION: 32

The values of x satisfying the equation

Solution:

QUESTION: 33

The solution of the equation 3x^{2}–17x + 24 = 0 are

Solution:

QUESTION: 34

has got the solution as

Solution:

QUESTION: 35

The equation has got two values of x to satisfy the equation given as

Solution:

QUESTION: 36

Te sum of two numbers is 8 and the sum of their squares is 34. Taking one number as x form an equation in x and hence find the numbers. The numbers are

Solution:

QUESTION: 37

The difference of two positive integers is 3 and the sum of their squares is 89. Taking the smaller integer as x form a quadratic equation and solve it to find the integers. The integers are.

Solution:

QUESTION: 38

Five times of a positive whole number is 3 less than twice the square of the number. The number is

Solution:

QUESTION: 39

The area of a rectangular field is 2000 sq.m and its perimeter is 180m. Form a quadratic equation by taking the length of the field as x and solve it to find the length and breadth of the field. The length and breadth are

Solution:

Let the length be l and the breadth be b. now area = 2000 sq.m so length * breadth = 2000; l*b=2000.

now perimeter = 180m

so 2 ( l + b ) =180;

l + b = 90;

l = 90 - b.

substituting this value of l in l * b = 2000,

( 90 - b ) * b = 2000;

90b - b^2 = 2000;

b^2 - 90b + 2000 = 0;

b^2 - 40b - 50b - 2000 = 0;

b ( b - 40 ) - 50( b - 40 ) = 0;

( b - 40 )( b - 50 ) = 0;

b = 40 or b = 50;

so l = 40 or l = 50 but the longer side is considered the length and the shorter one the breadth.

so length = 50 m and breadth = 40 m

QUESTION: 40

Two squares have sides p cm and (p + 5) cms. The sum of their squares is 625 sq. cm. The sides of the squares are

Solution:

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