Alligation or Mixture

Sponsor Area

Question
SSCCGLENQA12042691

20 litres of a mixture contains 20% alcohol and the rest water. If 4 litres of water be mixed in it, the percentage of alcohol in the new mixture will be

  • 33 1 third percent sign
  • 16 2 over 3 percent sign
  • 25%

  • 12 1 half percent sign

Solution

B.

16 2 over 3 percent sign

Quantity of alcohol in 20 litres of mixture = 20% of 20 = 4 litres
Quantity of water in 20 litres of mixture = (20 - 4) = 16 litres
On adding 4 litres of water in mixture, 
New quantity of water  = (16 + 4)  = 20 litres
New quanity of mixture = 24 litres
∴  Required per cent
equals space 4 over 24 cross times space space 100 space equals space 50 over 3 space equals space 16 2 over 3 percent sign

Sponsor Area

Question
SSCCGLENQA12043160

300 grams of sugar solution has 40% of sugar in it. How much sugar should be added to make it 50% in the solution?

  • 40 gram

  • 10 gram

  • 60 gram

  • 80 gram

Solution

C.

60 gram

In 300 gm of solution,
Percentage of sugar = 40%
∴  Sugar = fraction numerator 300 space cross times space 40 over denominator 100 end fraction space equals space 120 space gm.
Let x gm of sugar be added.
Now, according to the question
  fraction numerator 120 plus straight x over denominator 300 plus straight x end fraction space equals space 1 half
rightwards double arrow space space space 240 space plus space 2 straight x space equals space 300 space plus space straight x
rightwards double arrow space space 2 straight x space minus space straight x space space equals 300 space minus space 240
rightwards double arrow space space straight x space equals space 60 space gm.

Question
SSCCGLENQA12043274

49 kg of blended tea contains Assam and Darjeeling tea in the ratio 5 : 2. Then the quantity of Darjeeling tea to be added to the mixture to make the ratio of Assam to Darjeeling tea 2 : 1 is

  • 4.5 kg

  • 3.5 kg

  • 5 kg

  • 6 kg

Solution

B.

3.5 kg

In 49 kg of mixture,
 Tea of Assam 
           equals space open parentheses 5 over 7 space cross times space 49 close parentheses space kg
equals space 35 space kg
Tea of Darjeeling = (49 - 35)kg  = 14 kg
Let x kg of Darjeeling tea be added
    therefore space space space fraction numerator 35 over denominator 14 plus straight x end fraction equals space space 2 over 1
rightwards double arrow space space space 28 space plus space 2 straight x space equals space 35
rightwards double arrow space space space 2 straight x space equals space 35 space minus space 28 space equals space 7
rightwards double arrow space space space straight x space equals space 7 over 2 space equals space 3.5 space kg

Question
SSCCGLENQA12043175

60 kg of an alloy A is mixed with 100 kg of alloy B. If alloy A has lead and tin in the ratio 3 : 2 and alloy B has tin and copper in the ratio 1 : 4, the amount of tin in the new alloy is

  • 53 kg

  • 44 kg

  • 80 kg

  • 24 kg

Solution

B.

44 kg

In 60 kg of alloy A, 
Lead = 3 over 5 space cross times space 60 space equals space 36 space kg
Tin = 2 over 5 cross times 60 space equals space 24 space kg
In 100 kg of alloy B,
Tin = 1 fifth cross times 100 space equals space 20 space kg
In 160 kg of new alloy,
Tin = 24 + 20 = 44 kg

Question
SSCCGLENQA12042860

A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C wil be

  • 5 : 7

  • 5 : 9

  • 7 : 5

  • 9 : 5

Solution

C.

7 : 5

As the equal quantities of the alloys are melted to form a third alloy C.
∴   Required ratio would be
         equals space open parentheses 7 over 9 plus 7 over 18 close parentheses space colon space open parentheses 2 over 9 plus 11 over 8 close parentheses
equals space 21 over 18 space colon space 15 over 18 space equals space 21 space colon space 15 space equals space 7 space colon space 5