Ration Example 3 :
Ration based problem are very important for Competitive here is some problems which are given in exams that is in bag contain some coins like 50 p, 25 p, 10 p, in a ratio and total amounting is given and we need to find each type of coins in a bag,Now we discuss this example in ratio example 3.
Example 1:
A money bag contains 50 p, 25 p, and 10 p coins in the ratio 5 : 9 : 4, and the total amounting to Rs.206.
Find the individual number of coins of each type.
Answer :
Step 1: Let the number of 50 p ,25 p, and 10 p coins be 5x, 9x, 4x respectively.
Then, 5x / 2 + 9x / 4 + 4x / 10 = 206
= 50x + 45x + 8x = 4120
= 103x = 4120
= x = 40.
Step 2: Number of 50 p coins is ( 5 x 40 = 200 ),
Number of 25 p coins is( 9 x 40 = 360 ),
Number of 10 p coins ( 4 x 40 = 160 ),
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Ration based problem are very important for Competitive here is some problems which are given in exams that is in bag contain some coins like 50 p, 25 p, 10 p, in a ratio and total amounting is given and we need to find each type of coins in a bag,Now we discuss this example in ratio example 3.
Example 1:
A money bag contains 50 p, 25 p, and 10 p coins in the ratio 5 : 9 : 4, and the total amounting to Rs.206.
Find the individual number of coins of each type.
Answer :
Step 1: Let the number of 50 p ,25 p, and 10 p coins be 5x, 9x, 4x respectively.
Then, 5x / 2 + 9x / 4 + 4x / 10 = 206
= 50x + 45x + 8x = 4120
= 103x = 4120
= x = 40.
Step 2: Number of 50 p coins is ( 5 x 40 = 200 ),
Number of 25 p coins is( 9 x 40 = 360 ),
Number of 10 p coins ( 4 x 40 = 160 ),
If You Have any question regarding this topic then please do comment on below section. You can also send us message on facebook.
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