1).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
2).Find the ratio of speed of A,B and C if for every 11 steps taken by A,B takes 12 steps and C takes 14 steps. Also 12 steps of A are equal to 14 steps of B and 16 steps of C.
a) 154:144:147
b) 132:168:224
c) 25:24:23
d) None f these
e) 161:123:113
3).The price of a ring varies jointly, with the cube root of number of diamonds and weight of gold used in it. The price of the ring was “10,000. When it had 64 diamonds and 15 gm. of gold. Find the no. of diamonds if the price of ring was~` 10,500 and weight of gold was 21 gm.
a) 3
b) 9
c) 27
d) 81
e) 16
4).There are two containers A and B. Container A is filled with 30 liters of kerosene while container B is filled with 30 liters of water. Ten liters of kerosene is taken from container A and put in container B and then twelve liters of the mixture (of kerosene and water) is taken from container B and put in container A. Find the ratio of kerosene in A and B respectively.
a) 23:9
b) 21:5
c) 23:7
d) 1:3
e) None of these
5).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
6).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
7). 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
8).The no. of pens in three different pencil boxes in the ratio of 1:2:3. Find the ratio in which the number of pens in the first and the second boxes must be increased so that the new ratio becomes 3:2:1.
a) 1:3
b) 2:1
c) 2:3
d) 3:4
e) None of these
9).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:
1) d), 2) a), 3) c), 4) c), 5) c), 6) a), 7) b), 8) b), 9) b).
Solutions:
1).Let Pintu’s present age =x, Chintu = 2x and Mintu = x+2x =3x
Two years later [2x+2/3x+2] = (7/10)
20x+20=21x+14 = 6=x
Pintu’s present age =6
Five years ago, his age was 6-5= 1 year
Answer: d)
2.Ratio of speed is same as ratio of distance covered by them as speed and distance are directly proportional to each other. When time is constant.
A B C
(x/12)×11 (x/14)×12 (x/16)×14
Ratio is (11/12): (12/14): (14/16)
L.C.M of 12, 14, 16 is 336
=308:288:294 = 154:144:147
[What is x? Since distance covered by A in 12 steps is equal to distance covered by B in 14 steps and distance covered by C in 16 steps, hence let this distance =x. Distance covered in 1 step by A, B and C= (x/12), (x/14) and (x/16) respectively ]
Answer: a)
3. Let price =P, number of diamonds = n and weight of gold = g
When p=10000, n=64 and g=15, k=?
P= kg3√n => 10000 = k(15) (3√64)
K=(10000/15×4)=(500/3)
P=10500, g=21, k=(500/3), n=?
P=(500/3) g3√n => 10500=(500/3).213√n
3√n=10500×(3/500)×(1/21)=3=>n=27
Answer: c)
4.Container A Container B
Kerosene water Kerosene water
30 0 0 30
-10 –+10 –
20 0 10 30
When 10 litres of kerosene is taken out from A and put in B, then kerosene in A= 20 litres. Mixture in B= 40 litres in which ratio of Kerosene to water =10:30 = 1:3, when 12 litres of mixture is taken out from B, and put in A, then out of 12 litres
12×(1/4)=3 litres is kerosene and remaining 9 litres is water.
Container A Container B
Kerosene water Kerosene water
20 0 10 30
+3 +9-3 -9
23 9 -7 21
Ratio of Kerosene in A to B = 23:7
Answer: c)
5.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 (51/(95+51))×29200=10200
Answer: c)
6.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.
Answer: a)
7.(2) Let quantity of mixture taken from first be x and second be y.
Amount of kerosene oil in the resultant mixture (x+y) is
(7/12)x+(5/8)y=(3/5)(x+y)
(7/12)x-(3/5)x=(3/5)y-(5/8)y-(1/60)x=-(1/40)y=>(x/y)=(6/4)=(3/2)=3:2
Answer: b)
8.let the no. of pens in 1st, 2nd and 3rd pencil box be x, 2x and 3x respectively and let the required no. be 3y, 2y and y.
The quantity f pens in the third pencil box would remain the same, hence
3x=y or x =(y/3)
Quantity of pens in the boxes originally is x, 2x and 3x
When x=(y/3), hence quantity is x=(y/3), 2x=(2y/3), 3x=y
i.e (y/3), (2y/3), y
The required number of pens is 3y, 2y and y in
Increase in 1st box=3y-(y/3)=(8/3)y&
Increase in 2nd box = 2y-(2y/3)=(4/3)y
Ratio of increase = (8y/3)L4/3)y=2:1
Answer: b)
9.Method-1: 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
=(3/5)×5+(4/9) ×4+(3/4) ×3
=3+(16/9)+(9/4)=[(108+64+81)/36]=(253/36)
Quantity of sugar syrup is
12-(253/36)=(179/36)
Ratio of sugar syrup to water = 179:253
Answer: b)