Sample questions for ratio and proportion topic

This article is in continuation to the last article focusing on the ratio and proportion topic. Here are some practice questions along with solutions that will help you in your preparation for the State government recruitment tests.

1. Three cats are roaming in a zoo in such a way that when cat A takes 5 steps, B takes 6 steps and C takes 7 steps. But the 6 steps of A are equal to the 7 steps of B and 8 steps of C. What is the ratio of their speeds?

a. 140:144:147
b. 40:44:47
c. 15:21:28 d. 252:245:240

Ans: a

Solution:
Frequency of step of A: B: C = 5: 6: 7
But in terms of size of step, 6A = 7B = 8C
Therefore, ratio of speeds of A, B and C = 5/6: 6/7: 7/8 = 140: 144: 147

 

2. The ratio of the angles of a triangle is 3: 4: 5. The three angles of a quadrilateral is equal to three angles of this triangle. What is the sum of the largest angle and second smallest angle of the quadrilateral?

a. 225 degree
b. 210 degree
c. 205 degree
d. 245 degree

Ans: a

Solution:
Angle of triangle = 180 × 3/12 = 45 degree
Other angles of the triangle are = 60 degree, 75 degree
Angle of quadrilateral
= 360 – (45 + 60 + 75)
= 180 degree
Therefore, the four angles of the quadrilateral are 45, 60, 75 and 180 degrees.
Hence, the sum of largest and smallest angle of the quadrilateral is 180 45 = 225 degree.

3. In an exam, a candidate secured 504 marks out of the maximum mark of ‘M’. If the maximum mark ‘M’ is converted into 800 marks, he would have secured 384 marks. What is the value of ‘M’?

a. 910
b. 1000
c. 970
d. None of the above

Ans: d

Solution:
504/M = 384/800
(504 x 800) / 384 = M
M = 1050

4. A man spends Rs 810 in buying trousers at Rs 70 each and shirts at Rs 30 each. What will be the ratio of trouser and shirt when the maximum number of trouser is purchased?

a. 1:3 b. 2:1
c. 3:2 d. 2:3

Ans: c

Explanation:
Let us assume S as number of shirts and T as number of trousers Given that each trouser cost
= Rs 70 and that of shirt = Rs 30
Therefore, 70 T 30 S = 810
=> 7T 3S = 81…… (1)
T = (81 – 3S)/7
We need to find the least value of S which will make (81 – 3S) divisible by 7 to get maximum value of T Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7
Hence T = (81-3 × S)/7
= (81-3 × 6)/7 = 63/7 = 9
Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.

M Venkat
Director
MVK Publications
Dilsukhnagar
7671002120