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Question 1 of 5
1. Question
If p% of Rs. x is equal to t times q% of Rs. y, then what is the ratio of x to y?
Correct
Sol. (c)
By given condition, p% of x = t (q % of y)
⇒xp/100 = (yq × t)/100
⇒x/y = qt/p
Hence, x : y = =qt : p
Incorrect
Sol. (c)
By given condition, p% of x = t (q % of y)
⇒xp/100 = (yq × t)/100
⇒x/y = qt/p
Hence, x : y = =qt : p

Question 2 of 5
2. Question
A bag contains Rs. 112 in the form of Rs. 1, 50 paise and 10 paise coins in the ratio 3 : 8 : 10. What is the number of 50 paise coins?
Correct
Sol. (a)
Convert ratios in paise
100 paise:50 paise:10paise
Divide by 10
We ll get10 : 5 : 1 and new as per given ratio we get new ratio as
10 × 3 : 5 × 8 : 1 × 8 = 30 : 40 : 10 = 3 : 4 : 1
⇒3x + 4x + x = 112
⇒x = 112/8 = 14
Hence, Number of 50 paise coins = 14 x 8 = 112.
Incorrect
Sol. (a)
Convert ratios in paise
100 paise:50 paise:10paise
Divide by 10
We ll get10 : 5 : 1 and new as per given ratio we get new ratio as
10 × 3 : 5 × 8 : 1 × 8 = 30 : 40 : 10 = 3 : 4 : 1
⇒3x + 4x + x = 112
⇒x = 112/8 = 14
Hence, Number of 50 paise coins = 14 x 8 = 112.

Question 3 of 5
3. Question
The speeds of three cars are in the ratio 4 : 3 : 2. What is the ratio between the time taken by the cars to cover the same distance?
Correct
Sol. (b)
Speed ∞ 1/Time
Hence, Required ratio = 1/4 : 1/3 : 1/2 = 3 : 4 : 6
Incorrect
Sol. (b)
Speed ∞ 1/Time
Hence, Required ratio = 1/4 : 1/3 : 1/2 = 3 : 4 : 6

Question 4 of 5
4. Question
What is the ratio between times taken by a train 240 m long to cross an electric pole and a bridge of 80 m length?
Correct
Sol. (b)
Since, the speed of train is constant, then
D1/T_{1} = D2/T_{2 }
240/T_{1} = (240 + 80)/T_{2}
T_{1}/T_{2} = 240/320
⇒T1 : T2 = 3 : 4
Incorrect
Sol. (b)
Since, the speed of train is constant, then
D1/T_{1} = D2/T_{2 }
240/T_{1} = (240 + 80)/T_{2}
T_{1}/T_{2} = 240/320
⇒T1 : T2 = 3 : 4

Question 5 of 5
5. Question
‘X’ is twice as old as ‘Y’ 3 yr ago, when ‘X’ was as old as ‘Y’ today. If the difference between their ages at present is 3 yr, how old is ‘X’ at present?
Correct
Sol. (c)
Let present age of X = x yr
and present age of Y = (x – 3) yr
3 yr ago, age of X = (x – 3) yr
and age of Y = (x – 6) yr
According to the question, x – 3 = 2(x – 6)
⇒x – 3 = 2x – 12
⇒12 – 3 = 2x – x
⇒x = 9yr
Incorrect
Sol. (c)
Let present age of X = x yr
and present age of Y = (x – 3) yr
3 yr ago, age of X = (x – 3) yr
and age of Y = (x – 6) yr
According to the question, x – 3 = 2(x – 6)
⇒x – 3 = 2x – 12
⇒12 – 3 = 2x – x
⇒x = 9yr
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