Basics questions of ratio and proportion
Answers
Answer:
Q:
A bag contains 50 P, 25 P and 10 P coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type respectively.
A) 360, 160, 200
B) 160, 360, 200
C) 200, 360,160
D) 200,160,300
Answer: C) 200, 360,160
Explanation:
let ratio be x.
Hence no. of coins be 5x ,9x , 4x respectively
Now given total amount = Rs.206
=> (.50)(5x) + (.25)(9x) + (.10)(4x) = 206
we get x = 40
=> No. of 50p coins = 200
=> No. of 25p coins = 360
=> No. of 10p coins = 160
Q:
Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
A) 2:5
B) 3:5
C) 4:5
D) 5:4
Answer: C) 4:5
Explanation:
Let the third number be x.
=>Then, first number = 120% of x =120x/100 = 6x/5
=>Second number =150% of x = 150x/100 = 3x/2
=> Ratio of first two numbers = 6x/5 : 3x/2 => 12x : 15x = 4 : 5
Q:
Salaries of Ravi and Sumit are in the ratio 2:3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40:57. What is Sumit's salary?
A) 38000
B) 46800
C) 36700
D) 50000
Answer: A) 38000
Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,
(2x+4000) / (3x+4000) = 40 / 57
⇒ 57 × (2x + 4000) = 40 × (3x+4000)
⇒ 6x = 68,000
⇒ 3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000
Q:
A sum of Rs.312 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs. 2.40 the number of girls is
A) 35
B) 40
C) 45
D) 50
Answer: B) 40
Explanation:
Step (i): Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100 -------------- (i)
Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.60x + 2.40y = 312 -------------- (ii)
Step (iii):
Solving (i) and (ii)
3.60x + 3.60y = 360 --------- Multiply (i) by 3.60
3.60x + 2.40y = 312 --------- (ii)
1.20y = 48
y = 48 / 1.20
= 40
Answer:
In △BCA and △BAD,
∠BCA=∠BAD ....Each 90
o
∠B is common between the two triangles.
So, △BCA∼△BAD ...AA test of similarity ....(I)
Hence,
AB
BC
=
AD
AC
=
BD
AB
...C.S.S.T
And, ∠BAC=∠BDA ....C.A.S.T ....(II)
So,
AB
BC
=
BD
AB
∴AB
2
=BC×BD
Hence proved.
(ii) In △BCA and △DCA,
∠BCA=∠DCA ....Each 90
o
∠BAC=∠CDA ...From (II)
So, △BCA∼△ACD ...AA test of similarity ....(III)
Hence,
AC
BC
=
CD
AC
=
AD
AB
...C.S.S.T
So,
AC
BC
=
CD
AC
∴AC
2
=BC×DC
Hence proved.
(iii) From (I) and (III), we get
△BAD∼△ACD
Hence,
AC
AB
=
CD
AD
=
AD
BD
So, AD
2
=BD×CD
Hence proved.