ML Aggarwal Solution Class 9 Chapter 6 Problems on Simultaneous Linear Equations MCQs

 MCQs

Choose the correct answer from the given four options (1 to 8):

Question 1

Sum of digits of a two digit number is 8. If the number obtained by reversing the digits is 18 more than the original number, then the original number is

(a) 35

(b) 53

(c) 26

(d) 62

Sol :
Let unit digit of a number =x 
Then tens digit =8-x

∴Number=x+10(8-x)=x+80-10x
=80-9x

By reversing the order of digit 
Unit digit =8-x and tens digit =x

∴Number=8-x+10x
=8+9x
∴8+9x=80-9x+18
⇒9x+9x=80+18-8=90
⇒18x=90
⇒$x=\frac{90}{18}=5$

Putting value of x , we get
=8-x+10x
=8-5+10×5
=8-5+50
=3+50
=53

Ans : (b)

Question 2

The sum of two natural numbers is 25 and their difference is 7. The numbers are
(a) 17 and 8
(b) 16 and 9
(c) 18 and 7
(d) 15 and 10
Sol :
Let x and y are two natural number 
Then x+y=25 
⇒x-y=7

Adding , we get 
⇒2x=32
⇒$x=\frac{32}{2}=16$

Subtracting , 
⇒2y=18
⇒$y=\frac{18}{2}=9$

∴Numbers are 16,9
Ans : (b)

Question 3

The sum of two natural numbers is 240 and their ratio is 3 : 5. Then the greater number is
(a) 180
(b) 160
(c) 150
(d) 90
Sol :
Let x and y the natural number , then
⇒x+y=240...(i)

and ⇒$\frac{x}{y}=\frac{3}{5}$
⇒5x=3y
⇒$x=\frac{3}{5} y$

Substituting the value of x in (i)
⇒$\frac{3}{5} y+y=240$
⇒$\frac{8}{5} y=240$
⇒$y=\frac{240 \times 5}{8}=150$
∴Greater number=150

Ans : (c)

Question 4

The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed. The number
(a) 27
(b) 72
(c) 63
(d) 36
Sol :
Sum of digits of a two digit number =9
Let x be unit digit 
Then ten's digit =9-x

∴Number=x+10(9-x)
=x+90-10x
=90-9x

By reversing the digits,

Unit digit=9-x and tens digit=x
∴Number=9-x+10x=9+9x
∴90-9x+27=9+9x
⇒117-9=9x+9x
⇒18x=108
⇒$x=\frac{108}{18}=6$

∴Number=90-9x
=90-9×6
=90-54=36

Ans : (d)

Question 5

The sum of the digits of a two digit number is 12. If the number is decreased by 18, its digits get reversed. The number is
(a) 48
(b) 84
(c) 57
(d) 75
Sol :
Sum of digits of a two digits number=12
Let unit digit =x
Then tens digit =12-x
∴Number=x+10(12-x)
=x+120-10x
=120-9x

By reversing the digits,
Unit digit=12-x and tens digit=x
∴Number=12-x+10x=12+9x
∴120-9+x-18=12+9x
⇒102-12=9x-9x
⇒18x=90
⇒$x=\frac{90}{18}=5$
Number=120-9x
=120-9×5
=120-45=75
Ans : (d)

Question 6

Aruna has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of the money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively
(a) 35 and 15
(b) 35 and 20
(c) 15 and 75
(d) 25 and 25
Sol :
Total number of coins =50 and total amount of coins=₹ 75

Let coins of ₹ 1=x and coins of ₹ 2=50-x

and coins of ₹ 2=50-x

∴x×1+(50-x)×2=75
⇒x+100-2x=75
⇒-x=75-100=-25
⇒x=25
∴Number of ₹ 1 coins=25
and of ₹=50-25=25
Ans : (d)


Question 7

The age of a woman is four times the age of her daughter. Five years hence, the age of the woman will be three times the age of her daughter. The present age of the daughter is
(a) 40 years
(b) 20 years
(c) 15 years
(d) 10 years
Sol :
Let age of daughter=x years
Then age of women=4x
5 years hence,
Age of daughter=x+5
and age of woman=4x+5
∴4x+5=3(x+5)
⇒4x+5=3x+15
⇒4x-3x=15-5
⇒x=10
∴Age of daughter=10 years
Ans : (d)

Question 8

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 age 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
Sol :
Let son's age =x years 
Then age of his father =6x
4 years hence 
Age of son =x+4 and age of father =6x+4
∴6x+4=4(x+4)
⇒6x+4=4x+16
⇒6x-4x=16-4
⇒2x=12
⇒$x=\frac{12}{2}=6$
Present age of son=6 years
and age of father=6×6=36 years 
Ans : (c)

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