Q1. Chintu, Pintu and Mintu are three friends. Chintu is twice as old as Pintu and Mintu is as old as Cintu and Pintu together. Two years later, the ratio of age of Chintu and Mintu would be 7:10 what was the age of Pintu five years ago?
A. 2 years
B. 4 years
C. 5 years
D. 1 years
E. 8 years
Answer - Option D
Explanation - Let Pintu’s present age = x, Chintu = 2x and Mintu = x + 2x = 3x
Two years later [latex][\frac{2x+2}{3x+2}][/latex] = [latex](\frac{7}{10})[/latex]
20x + 20 = 21x + 14
x = 6
Pintu’s present age = 6
Five years ago, his age was 6 - 5 = 1 year
Q2. In a school, students of class I and class II are going for a picnic to Surajkund and Badkal lake respectively. The ratio f number of students in class I and II is 5:3. Also the ratio of the contribution made by each student of class I and II is 19:17. If the total contribution made by the all the students of both the class is Rs.29200, then find the total contribution made by class II students only
A. Rs.10950
B. Rs.13789
C. Rs.10200
D. Rs.13272
E. None of these
Answer - Option C
Explanation - Let the no. of students in class II be 5x and 3x respectively and contribution made by each student of class I and class II be 19y and 17y respectively.
Hence, ratio of total contribution of class I and class II = 5x × 19y : 3x × 17y = 95 : 51
Total contribution made by students of class II ([latex]\frac{51}{(95+51)}[/latex]) × 29200 = 10200
Q3. A donkey moves at a speed of 8 kmph, when no load is put on him. Reduction in the speed of donkey varies directly to the square root of the kgs of load put on him. When only 4 kgs of load is put the speed of the donkey becomes 6 kmph. Find the minimum load that can be put on the donkey with which it cannot move.
A. 64 kg
B. 63.9 kg
C. 65.2 kg
D. Cannot be determined
E. None of these
Answer - Option A
Explanation - Speed of the donkey, Without any load = 8 kmph. With 4 kgs of load, speed becomes 6 kmph, hence speed is reduced by 2 kmph.
Reduction in speed varies directly with the square root of the load.
Hence (8 - 6) = k√4 = k ± 1
The donkey cannot move at zero speed. i.e. when his speed reduced by 8 kmph. So reduction in speed = 8 kmph
8 = 1√l => √l = 8 and l = 64, when √l = -8, l = 64
At 64 kg the donkey will stop.
Q4. A container has a mixture of kerosene and castor oil in the ratio of 7:5 and another container contains kerosene and castor oil in the ratio of 5:3. Find the proportion in which the mixtures from two containers should be mixed so that the resultant mixture has ratio of kerosene and castor oil of 3:2.
A. 2:3
B. 3:2
C. 4:1
D. 5:2
E. None of these
Answer - Option B
Explanation - Let quantity of mixture taken from first be x and second be y.
Amount of kerosene oil in the resultant mixture (x + y) is
([latex]\frac{7}{12}[/latex])x + ([latex]\frac{5}{8}[/latex])y = ([latex]\frac{3}{5}[/latex])(x + y)
([latex]\frac{7}{12}[/latex])x - ([latex]\frac{3}{5}[/latex])x = ([latex]\frac{3}{5}[/latex])y - ([latex]\frac{5}{8}[/latex])y - ([latex]\frac{1}{60}[/latex])x = -([latex]\frac{1}{40}[/latex])y => ([latex]\frac{x}{y}[/latex]) = ([latex]\frac{6}{4}[/latex]) = ([latex]\frac{3}{2}[/latex]) = 3 : 2
Q5. There are three vessels 1, 2 and 3. The ratio of the total capacity of vessels 1, 2 and 3 is 5:4:3 respectively. All the vessels are full of mixture of sugar syrup and water. In vessel 1, ratio of sugar syrup to water is 2:3. Similar ratio in case of vessel 2 and vessel 3 is 5:4 and 1:3 respectively. The mixture of all the three vessels is emptied into one bigger vessel. What is the resulting ratio of sugar syrup and water?
A. 17:25
B. 179:253
C. 253:179
D. 233:169
E. None of these
Answer - Option B
Explanation - Let the total mixture in vessel 1,2 and 3 be 5 litres and 3 litres respectively. So, quantity of water in (5 + 4 + 3) = 12 litres of mixture is
= ([latex]\frac{3}{5}[/latex])×5 + ([latex]\frac{4}{9}[/latex])×4 + ([latex]\frac{3}{4}[/latex])×3
= 3 + ([latex]\frac{16}{9}[/latex]) + ([latex]\frac{9}{4}[/latex]) = ([latex]\frac{108+64+81}{36}[/latex]) = ([latex]\frac{253}{36}[/latex])
Quantity of sugar syrup is
12 - ([latex]\frac{253}{36}[/latex]) = ([latex]\frac{179}{36}[/latex])
Ratio of sugar syrup to water = 179 : 253