Practice questions for SI exams
By Banda Ravipal Reddy Hyderabad: This article is in continuation to the last article on preparation for the Sub-Inspector of Police recruitment exams. Here are some practice questions and answers on ratio and proportion along with explanations. 1) The price of a scooter and a television set are in the Ratio 3:2. If a scooter […]
By Banda Ravipal Reddy
Hyderabad: This article is in continuation to the last article on preparation for the Sub-Inspector of Police recruitment exams. Here are some practice questions and answers on ratio and proportion along with explanations.
1) The price of a scooter and a television set are in the Ratio 3:2. If a scooter costs Rs. 10,000 more than the television set, Then the price of a scooter is(SI 2016)
a) 30,000 b) 40, 000 c) 45,000 d) 50,000
Ans: A
Explanation:
The prices of a scooter and television set are in the ratio 3:2. If a scooter cost
Rs.10000 more than the television set.
Let the price of the scooter be Rs.3x and the price of television set Rs.2x.
Difference = 3x – 2x = x = Rs.10000
The price of the scooter = 3×10000 = Rs.30000
2) x: y = 3: 4, them (7x 3y): (7x – 3y) =? (SI 2018)
a) 5: 2 b) 4:3 c) 11:3 d) 37:19
Ans: c
Explanation:
x: y = 3:4 x/y=3/4
7x 3y/7x-3y
Divide every term with y
7(x/y) 3 / 7(x/y)-3=7(3/4) 3 / 7(3/4)-3
= = = 11 : 3
Short cut method:
x: y= 3: 4
directly substitute x=3 and y=4 in the given equation i.e., 7x 3y/7x-3y
= 7(3) 3(4)/ 7(3)-3(4) = 21 12 / 21-12
= 33/9 = 11/3 = 11 : 3
3) a: b = 1:2, b:c=3:5, c:d = 5: 4 and e:d = 5: 6then a:b:c:d:e =
(SI 2018)
a) 3: 6 :10 :8 :7 b) 9: 18:30: 24 :20 c) 15: 30:50:40: 48
d) 1: 2: 3: 4: 5
Ans: b
Explanation:
There is a simple way of doing these kinds of problems by multiplying both sides with multiples of a number make a particular variable common in both the ratio
A:B = 1: 2, B:C = 3: 5
Now multiply A:B with 3 and multiply B:C with 2
Then A:B = 3:6 and B:C = 6:10
Now combining both A:B:C = 3:6: 10
Again C:D = 5: 4
Now multiplying A:B:C with 1 and C:D with 2
Then A:B:C = 3: 6: 10
and C:D = 10: 8
Now combining the above two
A:B:C:D = 3: 6: 10: 8
Again E: D = 5:6 d:e=6: 5
Now by multiplying A:B:C:D by 3 and D:E by 4 we get,
A:B:C:D = 9:18: 30: 24
and D:E = 24: 20
So again, combining the above two
A:B:C:D:E = 9:18: 30: 24: 20
4) If x: y: z = 7: 8: 9, then x/y:y/z:z/x=(SI 2018)
a) 72 :63 :56 b) 56 :63 :72 c) 684: 448:441 d) 441: 448 :648
Ans: d
Explanation:
x: y: z = 7: 8: 9
directly substitute x=7, y=8 and z=9 in the given equation i.e.,x/y:y/z:z/x
=7/8: 8/9: 9/7
To convert the ratio which is in fraction into normal ratio multiply the ratio with l.c.m of denominators
l.c.m of (7, 8, 9) = 504
=7/8 * (504): 8/9* (504): 9/7* (504)=441: 448: 648
5) Two positive integers whose sum is 143, cannot be in the ratio(SI 2018)
a) 2:7 b) 8:3 c) 6:7 d) 4:9
Ans: A
Explanation:
In these kinds of questions if we add antecedent and consequent, we will get some value.
if that value is factor of the given number, then that ratio will possible for the given number.
At the same time if that value is not a factor of the given number, then that ratio will not possible for the given number.
Here option a)2 7=9 which not a factor of 143 which is the right answer.
6)(8^(3 )× (27)^4×6^5)/((36)^2×9^4×(18)^2 ) = 2a×3b the a:b =(SI 2018)
a) 1:8 b) 2:3 c) 8:1 d) 3:2
Ans: C
Explanation:
First of all, we need to convert all the values into power of 2 and 3.
83= (23)3= 29, 274= (33)4=312 and 65=25*35
362= (32*22)2= 34*24, 94= (32)4= 38 and 182=(2*32)2= 22*34
Substitute these values in the given fraction
(8^(3 )× (27)^4×6^5)/((36)^2×9^4×(18)^2 )= (2^9× 3^12×2^5×3^5)/(2^4× 3^4× 3^8×2^2×3^4 )
=(2^14×3^17)/(2^6×3^16 ) =2^(14-6)×3^(17-16)= 28×31
A=8 and b=1
Ratio will be 8:1
Author is the (Director at Sai Institute of General Mental Ability)
Hyderabad
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