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The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively
(A) 4 and 24
(B) 5 and 30
(C) 6 and 36
(D) 3 and 24

Answer:(C) 6 and 36
Solution:

According to questions at present father’s age is six times than son’s age.
Let son age = x
Father age = y
y = 6x … (1)
After 4 years Father’s age will be four times than son’s age. That is
y + 4 = 4 (x + 4)
y + 4 = 4x + 16
4x – y + 12 = 0 …(2)
Put y = 6x in equation (2)
4x – 6x + 12 = 0
+ 2x = + 12
x = \frac{12}{2} = 6
Put x = 6 in equation (1)
y = 6 (6) = 36
y = 36
Hence present age of father is 36 and son is 6.

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Aruna has only Rs 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Rs 1 and Rs 2 coins are, respectively
(A) 35 and 15
(B) 35 and 20
(C) 15 and 35
(D) 25 and 25

Answer:(D) 25 and 25
Solution:

Let Rs. 1 coins = x
Rs. 2 coins = y
As per the question:
x + y = 50 …(1)
x + 2y = 75 … (2)
Find value of x, y using elimination method.
x + y = 50
x + 2y = 75
y = 25              (subtract equation 1 from equation 2)
put y = 25 in equation (1)
x + 25 = 50
x = 25
x = 25, y = 25
Hence, Rs. 1 coins = 25
Rs. 2 coins = 25

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If x = a, y = b is the solution of the equations x – y = 2 and x + y = 4, then the values of a and b are, respectively
(A) 3 and 5
(B) 5 and 3
(C) 3 and 1
(D) –1 and –3

Answer : (C)
Solution:

Equation are    x – y = 2 … (1)
x + y = 4 … (2)
Add equation (1) and (2)
x – y = 2
x + y = 4
2x = 6
x = 3
Put x = 3 in equation (2)
3 + y = 4
y = 4 – 3
y = 1
it is given that a = x, b = y
Hence a = 3, b = 1

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A pair of linear equations which has a unique solution x = 2, y = –3 is
(A) x + y = –1 ; 2x – 3y = –5
(B) 2x + 5y = –11 ;  4x + 10y = –22
(C) 2x – y = 1 ; 3x + 2y = 0
(D) x– 4y –14 = 0 ; 5x – y – 13 = 0

Answer: B, D
Solution:

A pair which have unique solution x = 2, y = –3 will satisfy L.H.S. = R.H.S. if we put x = 2 and y = –3
Put x = 2, y = –3 in option (A)
x + y = –1        ;           2x – 3y = –5
2 – 3 = –1        ;           2(2) – 3 (–3) = –5
–1 = –1 (True) ;           4 + 9 = –5
   13 = – 5 (False)
Put x = 2, y = –3 in option (B)
2x + 5y = –11  ;           4x + 10y = –22
2(2) + 5(–3) = –11  ;    4(2) + 10 (–3) = –22
4 – 15 = –11    ;           8 – 30 = –22
–11 = –11 (True)         –22 = –22 (True)
Put x = 2, y = – 3 in option (C)
2x – y = 1        ;           3x + 2y = 0
2(2) – (–3) = 1 ;           3(2) + 3 (–3) = 0
4 + 3 = 1          ;           6 – 6 = 0
7 = 1 (False)    ;           0 = 0 (True)
Put x = 2, y = – 3 in option (D)
x – 4y – 14 = 0            ;           5x – y – 13 = 0
2 – 4 (–3) – 14 = 0      ;           5(2) – (–3) – 13 = 0
2 + 12 – 14 = 0            ;           10 + 3 – 13 = 0
0 = 0 (True)                 ;           13 – 13 = 0
      0 = 0 (True)
By putting the value of x = 2, y = –3 in A, B, C, D; two options B, D satisfy L.H.S = R.H.S
Hence both (B, D) are correct

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One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be
(A) 10x + 14y + 4 = 0
(B) –10x – 14y + 4 = 0
(C) –10x + 14y + 4 = 0
(D) 10x – 14y = –4

Answer: D
Solution:

Given equation is
–5x + 7y – 2 = 0
a1 = –5, b1 = 7, c1 = –2
For dependent linear equation
\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}= \frac{1}{k}
\frac{-5}{a_{2}}= \frac{7}{b_{2}}= \frac{-2}{c_{2}}= \frac{1}{k}          
a2 = –5k, b2 = 7k, c2 = –2k
Put k = 2
a2 = –10, b2 = 14, c2 = –4
The revived equation is of the type a2x + b2y + c2 = 0
Put a2 = – 10, b2 = 14, c2= –4
–10x + 14y – 4 = 0
10x – 14 y = –4

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The value of c for which the pair of linear equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is
(A) 3
(B) – 3
(C) –12
(D) no value

Answer: D
Solution:

Equations are
cx – y – 2 = 0, 6x – 2y – 3 = 0
In equation
cx – y – 2 = 0
a1 = c, b1 = –1, c1 = –2
In equation
6x – 2y – 3 = 0
a2 = 6, b2 = –2, c2 = –3
Since equations have infinitely many solutions
Hence \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
\frac{c}{6}= \frac{-1}{-2}\neq \frac{-2}{-3}
Here we find that \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}} i.e., equations are inconsistent

So, we can not assign any value to c.

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If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is
(A)\frac{-5}{4}
(B)\frac{2}{5}
(C)\frac{15}{4}|
(D)\frac{3}{2}
 

Answer: C
Solution:

Equations are
3x + 2ky = 2
2x + 5y + 1 = 0
In equation
3x + 2ky – 2 = 0
a1 = 3, b1 = 2k, c1 = –2
In equation
2x + 5y + 1 = 0
a2 = 2, b2 = 5, c2 = 1
\frac{a_{1}}{a_{2}}= \frac{3}{2}
\frac{b_{1}}{b_{2}}= \frac{2k}{5}
\frac{c_{1}}{c_{2}}= \frac{-2}{1}=-2
Since lines are parallel hence
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}

\frac{3}{2}=\frac{2k}{5}\neq-2           ……(1)

\frac{3}{2}=\frac{2k}{5}         ( from equation (1))

2k=\frac{15}{2}\Rightarrow k=\frac{15}{4}

Hence the value of k is \frac{15}{4} .

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For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = –16 represent coincident lines?
(A) \frac{1}{2}
(B) –-\frac{1}{2}
(C) 2
(D) –2

Answer: C
Solution:

Equations are
3x – y + 8 = 0
6x – ky + 16 = 0
In equation
3x – y + 8 = 0
a1 = 3, b1 = –1, c1 = 8
In equation
6x – ky + 16 =0
a2 = 6, b2 = –k, c2 = 16
\frac{a_{1}}{a_{2}}= \frac{3}{6}= \frac{1}{2}
\frac{b_{1}}{b_{2}}= \frac{-1}{-k}= \frac{1}{k}
\frac{c_{1}}{c_{2}}= \frac{8}{16}= \frac{1}{2}
Since lines are coincident
Hence \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
\frac{1}{2}= \frac{1}{k}= \frac{1}{2}….(i)
1/k = 1/2     ( from equation (1))
k = 2 Hence the value of k is 2.

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The pair of equations x = a and y = b graphically represents lines which are
(A) parallel
(B) intersecting at (b, a)
(C) coincident
(D) intersecting at (a, b)

Answer: D

Solution:

The lines x = a and y = b are intersecting at (a, b)

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The pair of equations y = 0 and y = –7 has
(A) one solution
(B) two solutions
(C) infinitely many solutions
(D) no solution

(D) no solution

 

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Sushmitha JN

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