
Need Help with this question. Thanks in advance.
Need Help with this question. Thanks in advance.
Just putting the solution in branch diagram.
Step 1: Branch out a diagram and write down information:
Total lamps 155
Table 100% ( ) Standing 100% ( )
40% =2/5 ( ) 60%=3/5 ( ) 80%= 4/5 ( ) 20%=1/5 ( )
KIV (Keep In View): 60% of table lamps is 30 more than 80% of standing lamps
Step 2: Compare what we need to compare
Total lamps 155
Table Lamps 100% ( ) Standing Lamps 100% ( )
40% =2/5 ( ) 60%=3/5 ( ) 80%= 4/5 ( ) 20%=1/5 ( )
Step 3: “Quick and dirty” way is LCM of 3units and 4parts is 12u. “convenient” no. of u
See sequential plucking in of units.
Total lamps 155 (#7. 35u+50)
Table Lamps 100% (#6. 20u+50) Standing Lamps 100% (#4. 15u)
40% =2/5 (#5. 8u+20) 60%=3/5 (#2. 12u+30) 80%= 4/5 (#1. 12u) 20%=1/5 (#3. 3u)
[note for #5: working quick with %. 60% = 12u+30. Divide by 3 is 20% = 4u+10. Multiply by 2 is 40%=8u+20]
Step 4: Solve u and find answers
35u+50 = 155 –> u=3
Write down total.
Total lamps 155 (35u+50)
Table 100% (20u+50)=110 Standing 100% (15u)=45
40% =2/5 (8u+20) 60%=3/5 (12u+30) 80%= 4/5 (12u) 20%=1/5 (3u)
44 66 36 9
Solve: 40% of table lamps is 44 –> 10% is 11
Remainder after 30% removed is 70% left = 77 table lamps left#
If there are 10u of table lamp
80% of standing lamp = 6u -30
Total standling lamp = (6u-30)/0.8 = 7.5u – 37.5
10u + 7.5u -37.5 = 155
17.5u = 192.5
u = 11
Number of table lamp = 10u = 110
If 30% of the table lamp were removed, there will be 70% left
0.7 X 110 = 77 left ###