Question
If Anil covers a certain journey at 75% of his usual
speed, he reaches his destination 53 minutes late. The usual time (in minutes) taken by him to reach the destination is:Solution
Let the usual speed and time be 'x' km/hr and 't' minutes, respectively. ATQ, x X 0.75 X {(t + 53)/60} = x X (t/60) Or, 75/100 x (t + 53) = t Or, 3t + 159 = 4t So, 't' = 159 Alternate solution To cover a certain distance, ratio of time taken is reciprocal of ratio of speeds Since, speed is decreased to 75% Therefore, ratio of usual speed to that of decreased speed = 100:75 = 4:3 Or, ratio of usual time taken to the time taken with decreased speed = 3:4 Required time taken = 53 X {3/(4 - 3)} = 159 minutes
9/5 × 18/25 ÷ 42/21 = ? - 82/75
15 × 18 + 25 × 12 + 30 × 24 = ?% of 1720
√324 * 6 – 20% of 180 + ? = 130% of 150
400 % of 20 + 65 % of 620 - 92 × 5 = ?
- Find the value of the expression:
15 + 10 – 6 × [20 + 8 – 2 × (50 – 35)] What will come in place of (?) in the given expression.
(3/4 of 64) + (1/2 of 48) = ?If 1210 ÷ 22 + 1332 ÷ 37 - y + 54 × 3 = 980 ÷ 20 × 144 ÷ 48, then the value of y is:
600 ÷ 8 + 12 % of 250 + ? * 14 = 50 * √49
What will come in the place of question mark (?) in the given expression?
23 X 35 - ? = (132 + 16) X 3 + 25
- What will come in the place of question mark (?) in the given expression?
435 + 729 - 282 x 2 = ? + 225 + 125
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